Datasets:

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minimario

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math-openwebmath-retrievals

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[ "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "+0 # Some old geometry problem for you to solve 0 70 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "+0 # Some old geometry problem for you to solve 0 133 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$,", "= Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"\\angle C obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"\\angle B obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: 1. The region in which $\\angle B$ is obtuse is determined by construction. 2. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"A\", A, 1.5*S, linewidth(4.5)); dot(\"B\", B, 1.5*S, linewidth(4.5)); dot(\"D\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"C_2\", C2, 1.5*N, linewidth(4.5)); dot(\"C_1\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: • If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)" ]

[ "\\mathrm{~S}$$ At node 2 , $$\\frac{\\mathbf{V}_{o}-0}{4}+2 \\mathbf{I}_{1}+\\mathbf{I}_{2}=0$$ or $$-\\mathbf{I}_{2}=0.25 \\mathbf{V}_{o}-1.5 \\mathbf{V}_{o}=-1.25 \\mathbf{V}_{o}$$ Hence, $$\\mathbf{y}_{21}=\\frac{\\mathbf{I}_{2}}{\\mathbf{V}_{1}}=\\frac{1.25 \\mathbf{V}_{o}}{-5 \\mathbf{V}_{o}}=-0.25 \\mathrm{~S}$$ Similarly, we get $\\mathbf{y}_{12}$ and $\\mathbf{y}_{22}$ using Fig. 6(b). At node 1, $$\\frac{0-\\mathbf{V}_{o}}{8}=2 \\mathbf{I}_{1}+\\frac{\\mathbf{V}_{o}}{2}+\\frac{\\mathbf{V}_{o}-\\mathbf{V}_{2}}{4}$$ But $\\mathbf{I}_{1}=\\frac{0-\\mathbf{V}_{o}}{8}$; therefore, $$0=-\\frac{\\mathbf{V}_{o}}{8}+\\frac{\\mathbf{V}_{o}}{2}+\\frac{\\mathbf{V}_{o}-\\mathbf{V}_{2}}{4}$$ or $$0=-\\mathbf{V}_{o}+4 \\mathbf{V}_{o}+2 \\mathbf{V}_{o}-2 \\mathbf{V}_{2} \\quad \\Longrightarrow \\quad \\mathbf{V}_{2}=2.5 \\mathbf{V}_{o}$$ Hence, $$\\mathbf{y}_{12}=\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{2}}=\\frac{-\\mathbf{V}_{o} / 8}{2.5 \\mathbf{V}_{o}}=-0.05 \\mathrm{~S}$$ At node 2 , $$\\frac{\\mathbf{V}_{o}-\\mathbf{V}_{2}}{4}+2 \\mathbf{I}_{1}+\\mathbf{I}_{2}=0$$ or $$-\\mathbf{I}_{2}=0.25 \\mathbf{V}_{o}-\\frac{1}{4}\\left(2.5 \\mathbf{V}_{o}\\right)-\\frac{2 \\mathbf{V}_{o}}{8}=-0.625 \\mathbf{V}_{o}$$ Thus, $$\\mathbf{y}_{22}=\\frac{\\mathbf{I}_{2}}{\\mathbf{V}_{2}}=\\frac{0.625 \\mathbf{V}_{o}}{2.5 \\mathbf{V}_{o}}=0.25 \\mathrm{~S}$$ Notice that $\\mathbf{y}_{12} \\neq \\mathbf{y}_{21}$ in this case, since the network is not reciprocal.", "0&=c_1+c_2\\\\ 1&=c_1 \\left(\\frac{5+ \\sqrt{13}}{2} \\right )+c_2 \\left(\\frac{5- \\sqrt{13}}{2} \\right ). \\end{align*} and this gives $c_1=\\frac{\\sqrt{13}}{13}, c_2=-\\frac{\\sqrt{13}}{13}.$ Thus we have $\\mathbf{u}_2=\\frac{\\sqrt{13}}{13} \\mathbf{v}_1-\\frac{\\sqrt{13}}{13}\\mathbf{v}_2. \\tag{**}$ Using (*), (**), we have \\begin{align*} &(a_i)_{i=1}^{\\infty} = a_1 \\mathbf{u}_1+ a_2\\mathbf{u}_2\\\\ &=a_1\\left(\\frac{13-5\\sqrt{13}}{26} \\mathbf{v}_1+\\frac{13+5\\sqrt{13}}{26}\\mathbf{v}_2 \\right)+a_2\\left(\\frac{\\sqrt{13}}{13} \\mathbf{v}_1-\\frac{\\sqrt{13}}{13}\\mathbf{v}_2 \\right)\\\\ &=\\left(\\frac{13-5\\sqrt{13}}{26}a_1+\\frac{\\sqrt{13}}{13}a_2\\right) \\mathbf{v}_1 +\\left( \\frac{13+5\\sqrt{13}}{26}a_1-\\frac{\\sqrt{13}}{13}a_2 \\right) \\mathbf{v}_2. \\end{align*} The coefficients in the last linear combination are the $b_1, b_2$ that we wanted to find. Therefore, the general formula for $(a_i)_{i=1}^{\\infty} \\in U$ using $a_1, a_2$ is given by \\begin{align*} &(a_i)_{i=1}^{\\infty} = \\\\ &=\\left(\\frac{13-5\\sqrt{13}}{26}a_1+\\frac{\\sqrt{13}}{13}a_2\\right) \\left( \\left(\\frac{5+ \\sqrt{13}}{2} \\right )^{i-1} \\right)_{i=1}^{\\infty}\\\\ &+\\left( \\frac{13+5\\sqrt{13}}{26}a_1-\\frac{\\sqrt{13}}{13}a_2 \\right) \\left( \\left(\\frac{5- \\sqrt{13}}{2} \\right )^{i-1} \\right)_{i=1}^{\\infty}. \\end{align*} #### Remark 1. For example, if the initial condition is $a_1=1, a_2=1$, then this formula gives the same formula we obtained in part (b). #### Remark 2. Note that we could have solved as follows. $(a_i)_{i=1}^{\\infty} = a_1 \\mathbf{u}_1+ a_2\\mathbf{u}_2 = b_1 \\mathbf{v}_1+ b_2\\mathbf{v}_2$ and wanted to express $b_1, b_2$ in terms of $a_1, a_2$. If $P$ is the change of coordinates matrix from $B’$ to $B$, we have $\\begin{bmatrix} a_1 \\\\ a_2 \\end{bmatrix}_B=P\\begin{bmatrix} b_1 \\\\ b_2 \\end{bmatrix}_{B’},$ and $P=\\begin{bmatrix} 1 & 1\\\\[6pt] \\frac{5+ \\sqrt{13}}{2} & \\frac{5- \\sqrt{13}}{2} \\end{bmatrix}.$ The inverse matix is $P^{-1}=\\begin{bmatrix} \\frac{13-5 \\sqrt{13}}{26} & \\frac{\\sqrt{13}}{13}\\\\[6pt] \\frac{13+5\\sqrt{13}}{26} & -\\frac{\\sqrt{13}}{13} \\end{bmatrix}$ and thus $\\begin{bmatrix} b_1 \\\\ b_2 \\end{bmatrix}=P^{-1}\\begin{bmatrix} a_1 \\\\ a_2 \\end{bmatrix},$ and we obtain the same $b_1, b_2$ as before.", "15. (a) $\\overrightarrow{A B}=\\vec{b}-\\vec{a}$ (b) Show (c) Show (d) $\\lambda=\\frac{2}{3}$ 16. (a)(i) Prove (ii) Show (b) $\\frac{4}{5}$ 17. (a)(i) $\\frac{2}{3} \\vec{b}-\\vec{a}$ (ii) $\\frac{3}{5} \\vec{a}+\\frac{4}{15} \\vec{b}$ (iii) $\\frac{3}{5} \\vec{a}-\\frac{11}{15} \\vec{b}$ (b) $\\lambda=\\frac{9}{11}$ (c) $\\frac{110}{27}$ 18. $(a) \\overrightarrow{C D}=\\frac{2}{3} \\vec{b}-\\frac{1}{2} \\vec{a}$ (b) $\\overrightarrow{\\mathrm{CE}}=2 \\vec{b}-\\frac{3}{2} \\vec{a}$ (c) Show 19. \\begin{aligned}&\\overrightarrow{A B}=-\\mathbf{a}+\\mathbf{b} \\\\&\\overrightarrow{P Q}=\\frac{\\mathbf{a}}{4}+\\frac{1}{2}(-\\mathbf{a}+\\mathbf{b}),=\\frac{1}{4}(2 \\mathbf{b}-\\mathbf{a}) \\text { oe } \\\\&\\overrightarrow{Q R}=\\frac{\\mu}{4}(2 \\mathbf{b}-\\mathbf{a}) \\\\&\\overrightarrow{Q R}=\\frac{1}{2}(\\mathbf{b}-\\mathbf{a})+\\lambda \\mathbf{b} \\\\&\\frac{2 \\mu}{4} \\mathbf{b}-\\frac{\\mu}{4} \\mathbf{a}=\\frac{1}{2} \\mathbf{b}-\\frac{1}{2} \\mathbf{a}+\\lambda \\mathbf{b} \\Rightarrow-\\frac{\\mu}{4}=-\\frac{1}{2} \\Rightarrow \\mu=2 \\\\&\\frac{2 \\mu}{4}=\\frac{1}{2}+\\lambda \\Rightarrow \\lambda=\\frac{1}{2} \\end{aligned}", "$dy = a\\,dx$ $\\frac{dy}{dx} = a$ $dy = \\mathbf{a}\\,d\\mathbf{x}$ $\\frac{dy}{d\\mathbf{x}} = \\mathbf{a}$ $dy = {\\rm tr}(\\mathbf{A}\\,d\\mathbf{X})$ $\\frac{dy}{d\\mathbf{X}} = \\mathbf{A}$ $d\\mathbf{y} = \\mathbf{a}\\,dx$ $\\frac{d\\mathbf{y}}{dx} = \\mathbf{a}$ $d\\mathbf{y} = \\mathbf{A}\\,d\\mathbf{x}$ $\\frac{d\\mathbf{y}}{d\\mathbf{x}} = \\mathbf{A}$ $d\\mathbf{Y} = \\mathbf{A}\\,dx$ $\\frac{d\\mathbf{Y}}{dx} = \\mathbf{A}$ 4. ^ a b c Here, $\\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 5. ^ a b Here, $\\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 6. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\\frac{\\partial \\mathbf{Y}}{\\partial x},$ by X in $\\frac{\\partial y}{\\partial \\mathbf{X}}.$ $\\frac{d\\,\\ln au}{dx} = \\frac{1}{au}\\frac{d(au)}{dx} = \\frac{1}{au}a\\frac{du}{dx} = \\frac{1}{u}\\frac{du}{dx} = \\frac{d\\,\\ln u}{dx}.$", "& = & \\frac{1}{2} \\, \\mathbf{a}_{1} + \\frac{1}{4} \\, \\mathbf{a}_{2} & = & \\frac{1}{2}a \\, \\mathbf{\\hat{x}} + \\frac{1}{4}a \\, \\mathbf{\\hat{y}} & \\left(6d\\right) & \\mbox{Si} \\\\ \\mathbf{B}_{10} & = & \\frac{1}{2} \\, \\mathbf{a}_{1} + \\frac{3}{4} \\, \\mathbf{a}_{2} & = & \\frac{1}{2}a \\, \\mathbf{\\hat{x}} + \\frac{3}{4}a \\, \\mathbf{\\hat{y}} & \\left(6d\\right) & \\mbox{Si} \\\\ \\mathbf{B}_{11} & = & \\frac{1}{2} \\, \\mathbf{a}_{2} + \\frac{1}{4} \\, \\mathbf{a}_{3} & = & \\frac{1}{2}a \\, \\mathbf{\\hat{y}} + \\frac{1}{4}a \\, \\mathbf{\\hat{z}} & \\left(6d\\right) & \\mbox{Si} \\\\ \\mathbf{B}_{12} & = & \\frac{1}{2} \\, \\mathbf{a}_{2} + \\frac{3}{4} \\, \\mathbf{a}_{3} & = & \\frac{1}{2}a \\, \\mathbf{\\hat{y}} + \\frac{3}{4}a \\, \\mathbf{\\hat{z}} & \\left(6d\\right) & \\mbox{Si} \\\\ \\mathbf{B}_{13} & = & x_{3} \\, \\mathbf{a}_{1} + x_{3} \\, \\mathbf{a}_{2} + x_{3} \\, \\mathbf{a}_{3} & = & x_{3}a \\, \\mathbf{\\hat{x}} + x_{3}a \\, \\mathbf{\\hat{y}} + x_{3}a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{Ca} \\\\ \\mathbf{B}_{14} & = & -x_{3} \\, \\mathbf{a}_{1}-x_{3} \\, \\mathbf{a}_{2} + x_{3} \\, \\mathbf{a}_{3} & = & -x_{3}a \\, \\mathbf{\\hat{x}}-x_{3}a \\, \\mathbf{\\hat{y}} + x_{3}a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{Ca} \\\\ \\mathbf{B}_{15} & = & -x_{3} \\, \\mathbf{a}_{1} + x_{3} \\, \\mathbf{a}_{2}-x_{3} \\, \\mathbf{a}_{3} & = & -x_{3}a \\, \\mathbf{\\hat{x}} + x_{3}a \\, \\mathbf{\\hat{y}}-x_{3}a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{Ca} \\\\ \\mathbf{B}_{16} & = & x_{3} \\, \\mathbf{a}_{1}-x_{3} \\, \\mathbf{a}_{2}-x_{3} \\, \\mathbf{a}_{3} & = & x_{3}a \\, \\mathbf{\\hat{x}}-x_{3}a \\, \\mathbf{\\hat{y}}-x_{3}a \\, \\mathbf{\\hat{z}}", "= \\frac13(\\mathbf{a}+\\mathbf{b}+ \\mathbf{c})$. Thus $\\frac13(AA'+BB'+CC') = \\frac13\\bigl((\\mathbf{a} - \\mathbf{p})\\mathbf{.}\\mathbf{n} + (\\mathbf{b} - \\mathbf{p})\\mathbf{.}\\mathbf{n} + (\\mathbf{c} - \\mathbf{p})\\mathbf{.}\\mathbf{n}\\bigr)$, which simplifies to $(\\mathbf{z} - \\mathbf{p})\\mathbf{.}\\mathbf{n} = ZZ'$. 3. Thanks Opalg, but actually I was looking for a solution only using Euclidean Geometry. Can you help me?", "= \\frac{\\big \\|\\mathbf T'(t) \\big \\|}{\\big \\| \\mathbf r'(t)\\big \\|}$. Since $\\mathbf r(t) = (at, bt^2, ct^3)$, $\\mathbf r'(t)= (a, 2bt, 2ct^2)$ , $\\|\\mathbf r'(t) \\| = \\sqrt{a^2+4b^2t^2+9c^2t^4}$, so $\\mathbf T(t)= \\frac{(a, 2bt, 2ct^2)}{\\sqrt{a^2+4b^2t^2+9c^2t^4}}$ and $\\mathbf T'(t)= \\frac{(0, 2b, 4ct)}{\\sqrt{a^2+4b^2t^2+9c^2t^4}}$. and $\\| \\mathbf T'(t) \\|= \\frac{\\sqrt{4b^2+16c^2t^2}}{\\sqrt{a^2+4b^2t^2+9c^2t^4}}$ Now $\\displaystyle \\kappa = \\frac{\\big \\|\\mathbf T'(t) \\big \\|}{\\big \\| \\mathbf r'(t)\\big \\|} = \\frac{\\sqrt{4b^2+16c^2t^2}}{a^2+4b^2t^2+9c^2t^4}$ For torsion do a similar thing but use the formula: $\\tau = \\frac{\\big | \\mathbf r'(t) \\ \\mathbf r''(t) \\ \\mathbf r'''(t) \\big| }{\\kappa^2}$ where the numerator is the scalar triple product.", "On comparing the ratios $\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}, \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ and $\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{\\mathbf{z}}}$ Question. On comparing the ratios $\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}, \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ and $\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$, find out whether the following pairs of linear equations are consistent, or inconsistent. (i) $3 x+2 y=5 ; 2 x-3 y=7$ (ii) $2 x-3 y=8 ; 4 x-6 y=9$ (iii) $\\frac{\\mathbf{3}}{\\mathbf{z}} \\mathrm{x}+\\frac{\\mathbf{5}}{\\mathbf{3}} \\mathrm{y}=7 ; 9 \\mathrm{x}-10 \\mathrm{y}=14$ (iv) $5 x-3 y=11 ;-10 x+6 y=-22$ (v) $\\frac{\\mathbf{4}}{\\mathbf{3}} x+2 y=8 ; 2 x+3 y=12$ Solution: (i) $3 x+2 y-5=0$ …(i) $2 x-3 y-7=0$ …(ii) $\\frac{a_{1}}{a_{2}}=\\frac{3}{2} ; \\frac{h_{1}}{b_{2}}=\\frac{2}{-3}=-\\frac{2}{3}$ $\\Rightarrow \\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}$ The equations have a unique solution. Hence, consistent. (ii) $2 x-3 y=8$ …(i) $4 x-6 y=9$ …(ii) $\\frac{a_{1}}{a_{2}}=\\frac{2}{4}, \\frac{b_{1}}{b_{2}}=\\frac{-3}{-6}, \\frac{c_{1}}{c_{2}}=\\frac{8}{9}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}=\\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}} \\neq \\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$  The equations have no solution. ​Hence inconsistent. (iii) $\\frac{\\mathbf{3}}{\\mathbf{2}} \\mathbf{x}+\\frac{\\mathbf{5}}{\\mathbf{3}} \\mathbf{y}=7$ …..(i) $9 x-10 y=14$ …..(ii) $\\frac{a_{1}}{a_{2}}=\\frac{3 / 2}{9}=\\frac{1}{6}, \\frac{b_{1}}{b_{2}}=\\frac{5 / 3}{-10}=\\frac{-1}{6}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}} \\neq \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ The equations have infinite solutions. Hence, consistent. (iv) $5 x-3 y=11$ ….(i) $-10 x+6 y=-22$ ….(ii) $\\frac{a_{1}}{a_{2}}=\\frac{5}{-10}=\\frac{-1}{2}, \\frac{b_{1}}{b_{2}}=\\frac{-3}{6}=\\frac{-1}{2}$ $\\frac{c_{1}}{c_{2}}=\\frac{11}{-22}=\\frac{-1}{2}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}=\\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}=\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$ The equations have infinite solutions. Hence, consistent. (v) $\\frac{4}{3} \\mathbf{x}+\\mathbf{2} \\mathbf{y}=8$ …..(i) $2 x+3 y=12$ ….(ii) $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$ The equations have infinite solutions. Hence, consistent. Editor", "14 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{728}$. This slightly improves upon the lower bound of 1/2124 coming from Corollary 10. Proof Combine Corollary 4 and Lemma 13. $\\Box$ Lemma 15 One has ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{2}{3}$ and $2{\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq 1$. Proof As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length $\\sqrt{3}$ and three of length $2$. If we let a denote the number of monochromatic $\\sqrt{3}$ edges and b the number of monochromatic $2$-edges, we see from inspection of all four colorings that $(a,b)$ is either $(6, 0), (4,0), (2, 1)$, or $(0,3)$. In particular, one always has $\\frac{a}{6} + \\frac{b}{3} \\geq \\frac{2}{3}$ and $2\\frac{a}{6} + \\frac{b}{3} \\geq 1$. Taking expectations, we obtain the claim. $\\Box$ Corollary 16 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{1}{6}$, ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3} ) ) \\geq \\frac{1}{4}$ and ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{28}$. Proof Combine Lemma 2, Lemma 15, and Lemma 13. $\\Box$ Lemma 17 If $a,b,c \\gt 0$ are the lengths", "## Elementary Linear Algebra 7th Edition $\\mathbf{z}=(\\frac{7}{2},3,\\frac{5}{2})$ $\\mathbf{u}=(1,2,3)\\quad\\mathbf{v}=(2,2-1)\\quad\\mathbf{w}=(4,0,-4)$ Find $\\mathbf{z}$ if $2\\mathbf{z}-3\\mathbf{u}=\\mathbf{w}$ We can solve for $\\mathbf{z}$: $2\\mathbf{z}=\\mathbf{w}+3\\mathbf{u}$ $\\mathbf{z}=\\frac{1}{2}\\mathbf{w}+\\frac{3}{2}\\mathbf{u}$. All we need to do now is plug in the vectors $\\mathbf{w}$ and $\\mathbf{u}$: $\\mathbf{z}=\\frac{1}{2}(4,0,-4)+\\frac{3}{2}(1,2,3)$ $\\mathbf{z}=(2,0,-2)+(\\frac{3}{2},3,\\frac{9}{2})=(\\frac{7}{2},3,\\frac{5}{2})$", "ago, # ^ | +11 A more intuitive way (for me) is to just write the dp as and move to the left hand side. • » » » » » » 3 years ago, # ^ | +10 Why extra \"1\" is added?? • » » » » » » » 3 years ago, # ^ | ← Rev. 4 → +10 let $k = a + b + c$$\\mathbf{E}(a,b,c) := Expected no. of trials required to reach (0,0,0) from (a,b,c)$$\\mathbf{E}(a,b,c) = \\frac{n-k}{n} * (1 + \\mathbf{E}(a,b,c)) + \\frac{a}{n} * (1 + \\mathbf{E}(a-1,b,c)) + \\frac{b}{n} * (1 + \\mathbf{E}(a+1,b-1,c)) + \\frac{c}{n} * (1 + \\mathbf{E}(a,b+1,c-1))$ or $\\mathbf{E}(a,b,c) = 1 + \\frac{n-k}{n} * \\mathbf{E}(a,b,c) + \\frac{a}{n} * \\mathbf{E}(a-1,b,c) + \\frac{b}{n} * \\mathbf{E}(a+1,b-1,c) + \\frac{c}{n} * \\mathbf{E}(a,b+1,c-1)$hence, $\\mathbf{E}(a,b,c) = \\frac{n}{k} + \\frac{a}{k} * \\mathbf{E}(a-1,b,c) + \\frac{b}{k} * \\mathbf{E}(a+1,b-1,c) + \\frac{c}{k} * \\mathbf{E}(a,b+1,c-1)$ • » » » » » » » 2 years ago, # ^ | 0 My understanding for the reason why we have to add 1: To reach state $(cnt_1, cnt_2, cnt_3)$ we can go from 4 states $(cnt_1, cnt_2, cnt_3)$, $(cnt_1-1, cnt_2, cnt_3)$, $(cnt_1+1, cnt_2-1, cnt_3)$ or $(cnt_1, cnt_2+1, cnt_3-1)$ with probability as above. For each \"from\" state, we must take the last fixed turn to reach the required state, for example if we go from", "\\mathrm{b}=\\frac{\\mathbf{1}}{\\mathbf{3}}$ Now, $\\frac{1}{\\sqrt{x}}=a=\\frac{1}{2}$ and $\\frac{1}{\\sqrt{y}}=b=\\frac{1}{3}$ $\\Rightarrow \\sqrt{\\mathbf{x}}=2, \\sqrt{\\mathbf{y}}=3$ Squaring, we get $x=4, y=9$ (iii) $\\frac{4}{x}+3 y=14$ ..(i) $\\frac{3}{x}-4 y=23$ ..(ii) Take, $\\frac{\\mathbf{1}}{\\mathbf{x}}=\\mathrm{a}$ $4 a+3 y=14$ …(iii) $3 a-4 y=23$ …(iv) Multiplying (iii) by 4 and (iv) by 3 $16 a+12 y=56$ $9 a-12 y=69$ $25 a=125$ $\\Rightarrow a=5$ Substituting a in (iii), we get $20+3 y=14$ $\\Rightarrow 3 y=-6$ $\\Rightarrow y=-2$ As, $\\frac{1}{x}=a=5$ $\\Rightarrow x=\\frac{1}{5}$ Hence, $x=\\frac{1}{5}$ and $y=-2$ (iv) $\\frac{5}{(x-1)}+\\frac{1}{(y-2)}=2$ …(i) \\frac{6}{(x-1)}-\\frac{3}{(y-2)}=1 …(ii) Take, $\\frac{\\mathbf{1}}{(\\mathbf{x}-\\mathbf{1})}=\\mathrm{a}$ and $\\frac{\\mathbf{1}}{(\\mathbf{y}-\\boldsymbol{z})}=\\mathrm{b}$ $5 a+b=2$ ..(iii) $6 a-3 b=1$ …(iv) Multiplying (iii) by 3 , we get $15 a+3 b=6$ $6 a-3 b=1$ $21 a=7$ $a=1 / 3$ So, by solving, $b=1 / 3$ As, $a=\\frac{1}{3}=\\frac{1}{x-1} \\Rightarrow x-1=3 \\Rightarrow x=4$ and $\\mathrm{b}=\\frac{\\mathbf{1}}{\\mathbf{3}}=\\frac{\\mathbf{1}}{\\mathbf{v}-\\mathbf{2}} \\Rightarrow \\mathrm{y}-2=3$ $\\Rightarrow y=5$ (v) $\\frac{7 x-2 y}{x y}=5, \\frac{8 x+7 y}{x y}=15$ By solving, we get $\\frac{7}{y}-\\frac{2}{x}=5$ (i) $\\frac{8}{y}+\\frac{7}{x}=15$ …(ii) Taking $\\frac{\\mathbf{1}}{\\mathbf{y}}=\\mathrm{u}, \\frac{\\mathbf{1}}{\\mathbf{x}}=\\mathrm{v}$ $7 u-2 v=5$ …(iii) $8 u+7 v=15$ …(iv) Multiplying (iii) by 7 and (iv) by 2 , we get $49 u-14 v=35$ $16 u+14 v=30$ $65 u=65$ $u=1$ By solving, we get $\\mathrm{v}=1$ As, $u=1=\\frac{\\mathbf{1}}{\\mathbf{y}} \\Rightarrow y=1$ and $\\mathrm{v}=1=\\frac{\\mathbf{1}}{\\mathbf{x}} \\Rightarrow \\mathrm{x}=1$ (vi) $6 x+3 y=6 x y$ $2 x+4 y=5 x y$ By solving, we get $\\frac{6}{v}+\\frac{3}{z}=6$ …(i) $\\frac{2}{y}+\\frac{4}{x}=5$ ..(ii) Take, $\\frac{1}{\\mathbf{y}}=\\mathrm{u}, \\frac{\\mathbf{1}}{\\mathbf{x}}=\\mathrm{v}$, we get $6 u+3 v=6$ ..(iii) $2 u+4 v=5$ ..(iv) Multiply (iv) by 3", "( ext{Proj}_vecs{v}vecs u = frac{32}{13}mathbf{hat i} + frac{48}{13}mathbf{hat j} ) and (vecs u_ ext{perp} = frac{33}{13}mathbf{hat i} - frac{22}{13}mathbf{hat j} ); d. ( ext{comp}_vecs{v}vecs u=frac{16}{sqrt{13}}=frac{16sqrt{13}}{13}) 49) (quadvecs u=3mathbf{hat i} +2mathbf{hat k} , quad vecs v=2mathbf{hat j} +4mathbf{hat k} ) a. ( ext{Proj}_vecs{u}vecs v =⟨frac{24}{13},0,frac{16}{13}⟩) and (vecs v_ ext{perp} = <-frac{24}{13}, frac{26}{13}, frac{36}{13}>); b. ( ext{comp}_vecs{u}vecs v=frac{8}{sqrt{13}}=frac{8sqrt{13}}{13}) c. ( ext{Proj}_vecs{v}vecs u =frac{4}{5}mathbf{hat j} + frac{8}{5}mathbf{hat k} ) and (vecs u_ ext{perp} = 3mathbf{hat i} - frac{4}{5}mathbf{hat j} + frac{2}{5}mathbf{hat k} ); d. ( ext{comp}_vecs{v}vecs u=frac{sqrt{80}}{5} = frac{4sqrt{5}}{5};) 51) Consider the vectors (vecs u=4mathbf{hat i} −3mathbf{hat j} ) and (vecs v=3mathbf{hat i} +2mathbf{hat j} .) a. Find the component form of vector ( ext{Proj}_vecs{u}vecs v) that represents the projection of (vecs v) onto (vecs u). b. Write the decomposition (vecs v=vecs w+vecs q) of vector (vecs v) into the orthogonal components (vecs w) and (vecs q), where (vecs w) is the projection of (vecs v) onto (vecs u) and (vecs q) is the vector component of (vecs v) orthogonal to the direction of (vecs u). That is, ( vecs q = vecs v_ ext{perp}). a. ( ext{Proj}_vecs{u}vecs v=⟨frac{24}{25},−frac{18}{25}⟩); b. (vecs q = vecs v_ ext{perp} =⟨frac{51}{25},frac{68}{25}⟩), (vecs v =vecs w+vecs q= ext{Proj}_vecs{u}vecs v + vecs v_ ext{perp} =⟨frac{24}{25},−frac{18}{25}⟩+⟨frac{51}{25},frac{68}{25}⟩) So we have that, (vecs v = ⟨frac{24}{25},−frac{18}{25}⟩+⟨frac{51}{25},frac{68}{25}⟩). 52) Consider vectors (vecs", "used (higher values give bad results): $$\\mathbf{p}=\\frac{1}{1000}$$: $$n\\approx\\sqrt{2\\times 2^{128} \\frac{1}{1000}}=8.24963\\times 10^{17}$$ $$\\mathbf{p}=\\frac{1}{100}$$: $$n\\approx\\sqrt{2\\times 2^{128} \\frac{1}{100}}=2.60876\\times 10^{18}$$ $$\\mathbf{p}=\\frac{1}{10}$$: $$n\\approx\\sqrt{2\\times 2^{128} \\frac{1}{10}}=8.24963\\times 10^{18}$$ These results are consistent with the previous results. ## Summary We have obtained some formulas that relate $$\\mathbf{M}$$, $$\\mathbf{n}$$ and $$\\mathbf{p}$$: $(1) p=1-\\prod_{k=0}^{\\mathbf{n}-1}(1-\\frac{k}{\\mathbf{M}}) \\\\ (2) p\\approx 1-e^{-\\frac{\\mathbf{n}(\\mathbf{n}-1)}{2\\mathbf{M}}} \\\\ (3) p\\approx \\frac{\\mathbf{n}^2}{2\\mathbf{M}} (1\\ll \\mathbf{n}^2\\ll 2\\mathbf{M}) \\\\ (4) n\\approx \\frac{1}{2}+\\sqrt{\\frac{1}{4}+2\\mathbf{M}\\log \\frac{1}{1-\\mathbf{p}}} \\\\ (5) n\\approx\\sqrt{2\\mathbf{M}\\log \\frac{1}{1-\\mathbf{p}}} (\\mathbf{M}\\mathbf{p}\\gg 1) \\\\ (6) n\\approx \\sqrt{2\\mathbf{M}\\mathbf{p}} (\\mathbf{Mp}\\gg 1, \\mathbf{p}\\ll 1)$ Tagged with: ### 2 Responses to Hash Collision Probability 1. SonOfLilit says: I felt bad with using a 1st order taylor approximation of e^x. Graphing with WA: http://www.wolframalpha.com/input/?i=e^x+-+%281+%2B+x%29 What is the relative error? http://www.wolframalpha.com/input/?i=e^x%2F%281+%2B+x%29 But since we’re using -k/M as x, and k <= n, and n < M, we're probably near 0 from the negative side. Lets ask WA \"when is the error acceptable?\": http://www.wolframalpha.com/input/?i=0.5+%3C+e^x%2F%281+%2B+x%29+%3C+1.5 It answers \"between ~-0.6 and ~1.1\". So we seem to be safe. All in all: Well played! Interesting read. • Thank you for the comment. Also remember that the relative is the biggest on the last term of our product: 1-n/M. Usually n is on the order of Sqrt(M) or lower, so the approximation is for |x| When M is big enough this number is tiny and the approximation is wonderful.", "+ (n-1)^{\\mathbf{3}} & < \\frac{n^{\\mathbf{4}}}{{\\mathbf{4}}} < 1^{\\mathbf{3}} + 2^{\\mathbf{3}} + \\cdots + n^{\\mathbf{3}} \\\\ a\\frac{b^{\\mathbf{4}}}{n^{\\mathbf{4}}} \\sum_{k=1}^{n-1} k^{\\mathbf{3}} + cb & < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb < a\\frac{b^{\\mathbf{4}}}{n^{\\mathbf{4}}} \\sum_{k=1}^{n} k^{\\mathbf{3}} + cb \\\\ a\\frac{b^{\\mathbf{4}}}{n^{\\mathbf{4}}} \\sum_{k=1}^{n-1} k^{\\mathbf{3}} + c\\frac{b(n-1)}{n} & < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb < a\\frac{b^{\\mathbf{4}}}{n^{\\mathbf{4}}} \\sum_{k=1}^{n} k^{\\mathbf{3}} + cb \\\\ s_n & < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb < S_n \\end{split}$ Sixth we assume that there is some $$A$$ such that $$s_n < A < S_n$$ and prove that $$A = a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ using the following inequalities: $$s_n < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ and $$a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb < S_n$$ If $$A > a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ and $$s_n < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ then $$S_n > a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ and $$0 < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb - s_n$$. Therefore because $$S_n - s_n > a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb - s_n$$ this implies that $$\\frac{ab^{\\mathbf{4}} + cb}{a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb - s_n} > n$$ but $$n \\geq 1$$ so there are some values of $$n$$ where $$\\frac{ab^{\\mathbf{4}} + cb}{a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb - s_n} > n$$ is false. If $$A < a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb$$ and $$a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb < S_n$$ then $$a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} + cb > s_n$$ and $$0 < S_n - a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} - cb$$. Therefore because $$S_n - a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} - cb < S_n - s_n$$ this implies that $$n < \\frac{ab^4 + cb}{S_n - a\\frac{b^{\\mathbf{4}}}{{\\mathbf{4}}} - cb}$$ but $$n", "\\mathbf{\\hat{x}} + \\frac{3}{4}a \\, \\mathbf{\\hat{y}} + \\frac{1}{4}a \\, \\mathbf{\\hat{z}} & \\left(6d\\right) & \\mbox{H_{2}O II} \\\\ \\mathbf{B}_{11} & = & \\frac{3}{4} \\, \\mathbf{a}_{1} + \\frac{1}{4} \\, \\mathbf{a}_{2} + \\frac{1}{4} \\, \\mathbf{a}_{3} & = & \\frac{3}{4}a \\, \\mathbf{\\hat{x}} + \\frac{1}{4}a \\, \\mathbf{\\hat{y}} + \\frac{1}{4}a \\, \\mathbf{\\hat{z}} & \\left(6d\\right) & \\mbox{H_{2}O II} \\\\ \\mathbf{B}_{12} & = & \\frac{1}{4} \\, \\mathbf{a}_{1} + \\frac{1}{4} \\, \\mathbf{a}_{2} + \\frac{3}{4} \\, \\mathbf{a}_{3} & = & \\frac{1}{4}a \\, \\mathbf{\\hat{x}} + \\frac{1}{4}a \\, \\mathbf{\\hat{y}} + \\frac{3}{4}a \\, \\mathbf{\\hat{z}} & \\left(6d\\right) & \\mbox{H_{2}O II} \\\\ \\mathbf{B}_{13} & = & x_{4} \\, \\mathbf{a}_{1} + x_{4} \\, \\mathbf{a}_{2} + x_{4} \\, \\mathbf{a}_{3} & = & x_{4}a \\, \\mathbf{\\hat{x}} + x_{4}a \\, \\mathbf{\\hat{y}} + x_{4}a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{O I} \\\\ \\mathbf{B}_{14} & = & \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{1} + \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{2} + x_{4} \\, \\mathbf{a}_{3} & = & \\left(\\frac{1}{2}-x_{4}\\right)a \\, \\mathbf{\\hat{x}} + \\left(\\frac{1}{2}-x_{4}\\right)a \\, \\mathbf{\\hat{y}} + x_{4}a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{O I} \\\\ \\mathbf{B}_{15} & = & \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{1} + x_{4} \\, \\mathbf{a}_{2} + \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{3} & = & \\left(\\frac{1}{2}-x_{4}\\right)a \\, \\mathbf{\\hat{x}} + x_{4}a \\, \\mathbf{\\hat{y}} + \\left(\\frac{1}{2}-x_{4}\\right)a \\, \\mathbf{\\hat{z}} & \\left(8e\\right) & \\mbox{O I} \\\\ \\mathbf{B}_{16} & = & x_{4} \\, \\mathbf{a}_{1} + \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{2} + \\left(\\frac{1}{2} - x_{4}\\right) \\, \\mathbf{a}_{3} &", "and simplify: $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} – \\mathbf{V}_{IN} (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})}{\\mathbf{R}_{1}} \\tag{3}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{4}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{5}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 – \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}]}{\\mathbf{R}_{1}} \\tag{6}\\\\$$ $$\\mathbf{I}_{1} = \\frac{-\\mathbf{V}_{IN} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{7}\\\\$$ We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation: $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\frac{- \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{8}\\\\$$ $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\mathbf{-} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{1}\\mathbf{R}_{3}} \\tag{9}\\\\$$ Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at: $$\\mathbf{R}_{IN} = \\frac{\\mathbf{V}_{IN}}{\\mathbf{I}_{1}} = \\mathbf{-} \\frac{\\mathbf{R}_{1}\\mathbf{R}_{3}}{\\mathbf{R}_{2}} \\tag{10}\\\\$$ This is the input resistance seen looking into the input of the NIC circuit. If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1. If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3.", "and simplify: $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} – \\mathbf{V}_{IN} (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})}{\\mathbf{R}_{1}} \\tag{3}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{4}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{5}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 – \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}]}{\\mathbf{R}_{1}} \\tag{6}\\\\$$ $$\\mathbf{I}_{1} = \\frac{-\\mathbf{V}_{IN} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{7}\\\\$$ We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation: $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\frac{- \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{8}\\\\$$ $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\mathbf{-} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{1}\\mathbf{R}_{3}} \\tag{9}\\\\$$ Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at: $$\\mathbf{R}_{IN} = \\frac{\\mathbf{V}_{IN}}{\\mathbf{I}_{1}} = \\mathbf{-} \\frac{\\mathbf{R}_{1}\\mathbf{R}_{3}}{\\mathbf{R}_{2}} \\tag{10}\\\\$$ This is the input resistance seen looking into the input of the NIC circuit. If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1. If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3. Description Reference Negative Impedance Converters Negative Impedance Converter, Wikipedia Use of NIC as an active load Negative Resistor Cancels Op Amp Load, Maxim Application Note 1868 Chua chaotic oscillator Improved Implementation of Chua’s Chaotic Oscillator Using Current Feedback Op Amp, A.S. Elwakil & M.P. Kennedy The use of impedance converters in active filters The Filter Wizard issue 18: Gee, I see! The Ins", "and simplify: $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} – \\mathbf{V}_{IN} (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})}{\\mathbf{R}_{1}} \\tag{3}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{4}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{5}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 – \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}]}{\\mathbf{R}_{1}} \\tag{6}\\\\$$ $$\\mathbf{I}_{1} = \\frac{-\\mathbf{V}_{IN} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{7}\\\\$$ We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation: $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\frac{- \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{8}\\\\$$ $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\mathbf{-} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{1}\\mathbf{R}_{3}} \\tag{9}\\\\$$ Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at: $$\\mathbf{R}_{IN} = \\frac{\\mathbf{V}_{IN}}{\\mathbf{I}_{1}} = \\mathbf{-} \\frac{\\mathbf{R}_{1}\\mathbf{R}_{3}}{\\mathbf{R}_{2}} \\tag{10}\\\\$$ This is the input resistance seen looking into the input of the NIC circuit. If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1. If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3. Description Reference Negative Impedance Converters Negative Impedance Converter, Wikipedia Use of NIC as an active load Negative Resistor Cancels Op Amp Load, Maxim Application Note 1868 Chua chaotic oscillator Improved Implementation of Chua’s Chaotic Oscillator Using Current Feedback Op Amp, A.S. Elwakil & M.P. Kennedy The use of impedance converters in active filters The Filter Wizard issue 18: Gee, I see! The Ins", "and simplify: $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} – \\mathbf{V}_{IN} (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})}{\\mathbf{R}_{1}} \\tag{3}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{4}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{5}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 – \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}]}{\\mathbf{R}_{1}} \\tag{6}\\\\$$ $$\\mathbf{I}_{1} = \\frac{-\\mathbf{V}_{IN} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{7}\\\\$$ We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation: $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\frac{- \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{8}\\\\$$ $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\mathbf{-} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{1}\\mathbf{R}_{3}} \\tag{9}\\\\$$ Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at: $$\\mathbf{R}_{IN} = \\frac{\\mathbf{V}_{IN}}{\\mathbf{I}_{1}} = \\mathbf{-} \\frac{\\mathbf{R}_{1}\\mathbf{R}_{3}}{\\mathbf{R}_{2}} \\tag{10}\\\\$$ This is the input resistance seen looking into the input of the NIC circuit. If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1. If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3. #### References and Additional Reading Description Reference Negative Impedance Converters Negative Impedance Converter, Wikipedia Use of NIC as an active load Negative Resistor Cancels Op Amp Load, Maxim Application Note 1868 Chua chaotic oscillator Improved Implementation of Chua’s Chaotic Oscillator Using Current Feedback Op Amp, A.S. Elwakil & M.P. Kennedy The use of impedance converters in active filters The Filter Wizard issue 18:" ]

[ "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "circle at $G$. Draw line $BG$. $A[\\Delta ABD] = A[\\Delta ABG] = 120$ (Since $\\square ABGD$ is cyclic) But $A[\\Delta ABE]$ is common in both with an area of 60. So, $A[\\Delta AED] = A[\\Delta BEG]$. \\therefore $A[\\Delta AED] \\cong A[\\Delta BEG]$ (SAS Congruency Theorem). In $\\Delta AED$, let $DI$ be the median of $\\Delta AED$. Which means $A[\\Delta AID] = 30 = A[\\Delta EID]$ Rotate $\\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\\frac{AE}{EF} = 2:1$. $AE$ is a radius and $EF$ is half of it implies $EF$ = $\\frac{radius}{2}$. Which means $A[\\Delta BEF] \\cong A[\\Delta DEI]$ Thus $A[\\Delta BEF] = \\boxed{\\textbf{(B) }30}$ ~phoenixfire & flamewavelight ## Solution 8 $[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(\"M\",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\\overline{AD}$ and $\\overline{CD}$, we find the area of $\\triangle ADB$ is $120$ and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$" ]

[ "The distance between two vectors is the magnitude of their difference. Find the value of $t$ for which the vector ${v} = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t$ is closest to ${a} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix}.$ I'm kinda confused on the magnitude part. I got $\\sqrt {901}$ for that but that doesn't sound right. I looked at the other links but didn't really get it. • $\\sqrt{9}01$ ? You mean $\\sqrt{901}$, right ? – A---B Nov 1 '17 at 0:14 • yea sorry, i''m not good at mathjax – ninjagirl Nov 1 '17 at 11:20 • You need to use curly braces around the number so that it appears wholly inside the square root symbol. – A---B Nov 1 '17 at 11:43 Consider the line $\\ell$ in $\\mathbb{R}^3$ parametrized by $$v(t) = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t,$$ and consider the point $${a} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix}.$$ Find the value of $t$ for which the point $v(t)$ on the line is closest to the point $a$. $\\textbf{Method 1}$. Let $p,q$ be two points in $\\mathbb{R}^3$ and let $d(p,q) = || p-q||$ be the distance function between $p$ and $q$, where", "? Free Version Moderate # Vector Distance: Between Normalized Vectors LINALG-S3LXJ6 If $\\boldsymbol{u} = \\begin{bmatrix} 5\\\\\\ 4\\\\\\ 3\\end{bmatrix}$ and $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ 0\\\\\\ -1\\end{bmatrix}$, then compute the distance between $\\dfrac{1}{\\Vert\\boldsymbol{u}\\Vert}\\boldsymbol{u}$ and $\\dfrac{1}{\\Vert\\boldsymbol{v}\\Vert}\\boldsymbol{v}$. A $4\\sqrt{3}$ B $\\dfrac{5\\sqrt{10}}{2}$ C $2\\sqrt{154}$ D $\\sqrt{2}$ E $\\dfrac{2\\sqrt{10}}{5}$ F $5\\sqrt{154}$", "vector components and Pythagoras’ theorem to find the magnitude of the vector. The length of a vector is its absolute value and we use the modulus symbol. Here is the formula: \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2} E.g. \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2}=\\sqrt{5^2+2^2}=\\sqrt{29} If the magnitude is equal to 1, then the vector is known as a unit vector. If the magnitude is equal to 0 , then the vector is known as a zero vector. Step-by-step guide: Magnitude of a vector ## Vector arithmetic Here is vector \\textbf{a} and vector \\textbf{b} . \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} \\textbf{b}= \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} \\textbf{a}+\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} + \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 8 \\;\\\\ \\; 1 \\; \\end{pmatrix} The vector \\textbf{a} + \\textbf{b} is known as the resultant vector. ### 5. Vector subtraction We can subtract vectors When subtracting vectors, the order is important. \\textbf{a}-\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} - \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 2 \\;\\\\ \\; 3 \\; \\end{pmatrix} Step-by-step guide: Vector subtraction ### 6. Vector multiplication We can multiply vectors by a scalar E.g. \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} 2\\textbf{a}= 2\\begin{pmatrix} \\;", "+0 # shortest distance between two lines via vector equation 0 255 4 +844 For a i got r= (-2,3,4) + λ(3,2,-7) for b so far i wanted to identify a point A on l1 and a point b on l2 i assumed the two lines are parallel as they have the same direction vector $$A=\\begin{pmatrix} -2+3λ\\\\ 3+2λ\\\\ 4-7λ \\end{pmatrix}$$ $$B= \\begin{pmatrix} 3\\\\ 1+2λ\\\\ -2-7λ \\end{pmatrix}$$ used this to get $$AB= \\begin{pmatrix} 5-3λ\\\\ -2\\\\ -6 \\end{pmatrix}$$ used this and my direction vector to find λ=53/9 via dot product substituted and calculated the magnitude which gave me 14.26 but my answer should be 3.57 please identify what went wrong, thanks Feb 10, 2019 #1 +2 The equation of the first line can be written as r1 = (-2, 3, 4) + s(3, 2, -7), (as you calculated), and the equation of the second as r2 = (0, 1, -2) + t(3, 2, -7), s and t arbitrary parameters. A vector d connecting two points, one on each line, would be given by d = r2 - r1 = (2, -2, -6) + (t - s)(3, 2 ,-7) = (2, -2, -6) + w(3, 2, -7), w for convenience. The distance between the two points will be the magnitude of this vector, so what you are looking for is the", "Free Version Moderate # Vector Distance: Equation Involving Vector Distance LINALG-JGI2ZX Find $x$ such that the distance between $\\begin{bmatrix} x\\\\\\ 0\\\\\\ -x\\end{bmatrix}$ and $\\begin{bmatrix} 1\\\\\\ 2\\\\\\ 3\\end{bmatrix}$ is $2\\sqrt{3}$. Select ALL that apply. A $-1$ B $1$ C $\\dfrac{1}{2}$ D $-\\dfrac{1}{2}$ E $2$ F $-2$", "the following vectors, $$v_1 = \\begin{pmatrix} 1 \\\\ 1 \\\\1 \\\\1 \\\\0 \\\\0 \\\\0 \\\\0\\end{pmatrix}, v_2 = \\begin{pmatrix} 0 \\\\ 0 \\\\0 \\\\0 \\\\1 \\\\1 \\\\1 \\\\1\\end{pmatrix} v_3 = \\begin{pmatrix} 1 \\\\ 1 \\\\1 \\\\1 \\\\1 \\\\1 \\\\1 \\\\1\\end{pmatrix}$$ And similarly for $\\beta$.", "Homework Help: Distance between vectors 1. Nov 13, 2005 cscott I have two vectors in the form $$\\vec{v} = v_x \\^{i} + v_y \\^{j} + v_z \\^{k}$$. What's the easiest way to find the distance between them? 2. Nov 13, 2005 marlon Err, goto yuor other post, the answer is there. The distance of two vectors (expressed in their components $$(v_x,v_y,v_z)$$) is just the magnitude of the vector you get after subtracting them. You are doing this in your other post. marlon 3. Nov 13, 2005 cscott yeah I just realized this but you beat me to it 4. Nov 13, 2005 marlon no problemo amigo marlon", "To find the distance between the vectors, we use the formula , where one vector is and the other is . Using the vectors we were given, we get Example Question #8 : Distance Between Vectors Find the distance between the vectors and . Explanation: To find the distance between the vectors, we use the formula , where one vector is and the other is . Using the vectors we were given, we get Example Question #1 : Distance Between Vectors The distance between (1, 3, -10) and (2, 5, 4) is None of the Above Explanation: We use the distance formula: Example Question #61 : Vectors And Vector Operations Given the points P:(2,1,5) andQ:(2,0,3). The magnitude of the vector from P to is: None of the Above", "# Solution to #4-1 Paper 1 Difficulty:Medium Let $\\vec{v}=\\begin{pmatrix} 1\\\\ 2\\\\ 3 \\end{pmatrix}$, and $\\vec{w}=\\begin{pmatrix} -3\\\\ 1\\\\ 2 \\end{pmatrix}$. If $\\vec{u}=\\vec{v}-2\\vec{w}$, find the magnitude of vector $\\vec{u}$. [Maximum mark:5]", "Math Distance calculations Skew lines # Distance between skew lines Unlike the distance between parallel lines, the distance between skew lines can be traced back to the Hesse normal form. Given are two skew lines: $\\text{g: } \\vec{x} = \\vec{p} + r \\cdot \\vec{m_1}$ $\\text{h: } \\vec{x} = \\vec{q} + s \\cdot \\vec{m_2}$ We calculate an orthogonal normal vector from the two direction vectors. $\\vec{n}=\\vec{m_1}\\times\\vec{m_2}$ This is normalized (unit normal vector) and inserted into the Hesse normal form together with the two support vectors $\\vec{p}$ and $\\vec{q}$. The distance $d$ is then obtained. $d = |(\\vec{p} - \\vec{a}) \\cdot \\vec{n_0}|$ i ### Method 1. Calculate normal vector 2. Form unit normal vector 3. Insert points in Hesse normal form ### Example Two skew lines are given $\\text{g: } \\vec{x} = \\begin{pmatrix} 1 \\\\ 1 \\\\4 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 0 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 1. #### Calculate normal vector ##### Variant 1 Since both direction vectors are perpendicular to the normal vector $\\vec{n}=\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$, the scalar product must result in zero. 1. $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\cdot\\color{blue}{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} = 0$ 2. $\\begin{pmatrix} x", "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "vectors from the origin and the displacement vector going from the head of J to the tail of Q. To find the distance, as mentioned above, you need to get the length of that vector. This procedure turns out to be equivalent to the distance formula. 5. May 23, 2010 ### bignum Thank you! I was really having qualms about this. Share this great discussion with others via Reddit, Google+, Twitter, or Facebook", "# Distance involving 3D lines and vectors. In this problem, a = \\begin{pmatrix} 5 \\\\ -3 \\\\ -4 \\end{pmatrix} and b = \\begin{pmatrix} -11 \\\\ 1 \\\\ 28 \\end{pmatrix} Vectors p and d exist such that the line containing a and b can be expressed in form v = p + d$t$. Additionally, for a specific value of d, it is the case that for all points v lying on the same side of a that b lies on, the distance between v and a is $t$. What is the value of d? Another problem I have no clue on as to how to solve, let alone begin. What should my \"first step\" be? Clarification on the last paragraph of the problem and hints are appreciated. • How are you defining points \"on the same side of a\"? Consider drawing a figure because it's hard to understand your description of the problem. – dpmcmlxxvi Jul 18 '15 at 4:32 • @dpmcmlxxvi; a point bisects a line – JMP Jul 18 '15 at 4:43 • @EmilianoSorbello: I'm not sure I quite understand what you mean. Can you further clarify (if possible)? – Grace Jul 18 '15 at 4:57 • i was answering a deleted comment so, mine probably makes no sense at all now – Dleep Jul 18 '15", "the distance: Finally, the distance just puts into the function defined earlier! We can utilize our Vector model methods to help make this line succinct. class Vector { /* ... */ } class RubberBand { distanceTo(p: Vector): number { /* a, b, s(t), th, ts */ return s(ts).minus(p).magnitude() } } Put it all together, and we get the following code: class Vector { x: number y: number new Vector(this.x + other.x, this.y + other.y) minus = (other: Vector) => new Vector(this.x - other.x, this.y - other.y) scaleBy = (scalar: number) => new Vector(this.x * scalar, this.y * scalar) dot = (other: Vector) => this.x * other.x + this.y * other.y magnitude = () => Math.sqrt(this.x * this.x + this.y * this.y) } class RubberBand { from: Peg to: Peg distanceTo(p: Vector): number { const a = this.from.position const b = this.to.position const s = (t: number) => a.add(b.minus(a).scaleBy(t)) const m = b.minus(a).magnitude() const th = (p.minus(a).dot(b.minus(a)) / (m * m) const ts = Math.min(Math.max(th, 0), 1) return s(ts).minus(p).magnitude() } } If you're wondering how this formula works, I recommend reading the original math exchange post, as it's summarized well, albeit rather concisely. I may decide to write something short on this topic specifically in the future. This algorithm turns out to work much better! Using our same interactive,", "Math Vectors Vector length (or magnitude) # Vector length (or magnitude) For determining the vector length (or magnitude) of a vector $|\\vec{v}|$ we use the Pythagorean theorem. The vector length (or magnitude) of $\\vec{v}=\\begin{pmatrix}v_1\\\\v_2\\end{pmatrix}$ is: $|\\vec{v}|=\\left|\\begin{pmatrix}v_1\\\\v_2\\end{pmatrix}\\right|=\\sqrt{v_1^2+v_2^2}$ In three-dimensional space: $|\\vec{v}|=\\left|\\begin{pmatrix}v_1\\\\v_2\\\\v_3\\end{pmatrix}\\right|=\\sqrt{v_1^2+v_2^2+v_3^2}$ ! ### Remember The vector length (or magnitude) is the length of its arrow and corresponds to the distance between initial point and terminal point. For determining the length of the arrow (and thus the magnitude of the vector), think of the following triangle. Using the Pythagorean theorem you will find the length of the arrow. ### Examples Determine the vector length $\\vec{a}=\\begin{pmatrix}3\\\\4\\end{pmatrix}$ $|\\vec{a}|=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ Determine the vector length $\\vec{a}=\\begin{pmatrix}2\\\\3\\\\6\\end{pmatrix}$ $|\\vec{a}|=\\sqrt{2^2+3^2+6^2}=\\sqrt{49}$ $=7$", "the distance of our projected vector from vA T = Dot(vVector2, vVector1); //If our projected distance from vA, \"t\", is greater than or equal to 0, it must be closest to the end point //vA. So we return this end point. if (T<=0) return vA; //If our projected distance from vA, \"t\", is greater than or equal to the magnitude or distance of the line //segment, it must be closest to the end point vB, so we return vB. if (T>=0) return vB; //Here we create a vector that is of length T and in the direction of vVector2 vVector3.x = vVector2.x * T; vVector3.y = vVector2.y * T; vVector3.z = vVector2.z * T; //To find the closest point on the line, we just add vVector3 to the original end point vA vClosestPoint.x = vA.x + vVector3.x; vClosestPoint.y = vA.y + vVector3.y; vClosestPoint.z = vA.z + vVector3.z; return vClosestPoint; } then just count collisions Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. PARTNERS", "# Homework Help: Distance between vectors 1. Nov 13, 2005 ### cscott I have two vectors in the form $$\\vec{v} = v_x \\^{i} + v_y \\^{j} + v_z \\^{k}$$. What's the easiest way to find the distance between them? 2. Nov 13, 2005 ### marlon Err, goto yuor other post, the answer is there. The distance of two vectors (expressed in their components $$(v_x,v_y,v_z)$$) is just the magnitude of the vector you get after subtracting them. You are doing this in your other post. marlon 3. Nov 13, 2005 ### cscott yeah I just realized this but you beat me to it 4. Nov 13, 2005 ### marlon no problemo amigo marlon", "two objects, as well as their difference, $\\vec{r}_2-\\vec{r}_1\\text{.}$ The magnitude of this latter vector is the distance between the two objects. The magnitude property of the dot product says that the magnitude of any vector is the square root of the dot product of the vector with itself. Therefore, the distance between the two objects, in coordinate independent form is: $$\\hbox{distance} = \\vert \\vec{r}_2-\\vec{r}_1\\vert = \\sqrt{(\\vec{r}_2-\\vec{r}_1)\\cdot (\\vec{r}_2-\\vec{r}_1)} .\\label{eqn-distance2}\\tag{1.15.2}$$", "it from? But, for this question, I got the same answer as Cullen_Im. aleksanderjanczewski: June 1, 2015, 11:25 p.m. I got different vectors than you, I guess you shoud try to derive identity matrix for change of basis and then multiply vector <a,b> and <c,d> by this matrix. This will give you these vectors with respect to your friend's basis. TLindberg: June 2, 2015, 4:39 p.m. A more concise way to state:$\\langle \\begin{pmatrix} a \\\\ b \\end{pmatrix} , \\begin{pmatrix} c \\\\ d \\end{pmatrix}\\rangle$is 2ac + 4bc + 4ad + 10bd.$\\\\$You can do this by computing the dot product of the the two vectors$\\begin{pmatrix} a+ 3b \\\\ a + b \\end{pmatrix} \\bullet \\begin{pmatrix} c + 3d \\\\ c + d \\end{pmatrix}\\\\\\$ I'm not sure if this is better, but this might be a safer answer on the test, because it shows that you more thoroughly understand the concept. Cullen_Im: June 3, 2015, 12:27 a.m. @tedallen: I don't believe there is a an absolute value around (v dot product w) because you're trying to find the angle between vectors and not between lines. The absolute value makes the angle be less than 90 degrees, but the angle between vectors can be greater than 90 degrees so the absolute value is not needed.", "# Magnitude of a Vector - How to Calculate It ## (also called Modulus of a Vector) Every vector has a magnitude, also called modulus. In this section we learn how to calculate the magnitude of a vector. ### Magnitude (Modulus) of a Vector Given a vector $$\\vec{v} = \\begin{pmatrix}x \\\\ y \\end{pmatrix}$$, its magnitude, also called modulus can be calculated with the formula: $\\begin{vmatrix}\\vec{v} \\end{vmatrix} = \\sqrt{x^2 + y^2}$ ### Tutorial: Magnitude of a Vector In this tutorial we learn how to calculate the magnitude of a vector. ### Example 1 Given the two vector $$\\vec{a} = \\begin{pmatrix}1 \\\\ 3 \\end{pmatrix}$$ we can calculate its magnitude as follows: \\begin{aligned} \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{1^2 + 3^2} \\\\ & = \\sqrt{1 + 9} \\\\ \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{10} \\quad \\begin{pmatrix} \\approx 3.16 \\end{pmatrix} \\end{aligned} ## Exercise 1 Find and write the exact value of the magnitude for each of the vectors, shown below. When applicable: also use your calculator to round your answers to two decimal places (2 d.p). 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$. 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$. 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$. 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$. 1. For vector $$\\vec{e} = \\begin{pmatrix}" ]

[ "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}=t\\cdot\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row $-3=t\\cdot(-3)$ $5=t\\cdot5$ $2=t\\cdot2$ If $t$ equals the same in each row (here: $t=1$) then the vectors are collinear. 2. #### Insert the vector of $h$ into $g$ Now the position vector of one line (here: $h$) is inserted into the other. $\\begin{pmatrix} -3 \\\\ 14 \\\\ 10 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Now we set up the equation system and solve it. If $r$ equals the same in every equation, then the terminal point of the position vector is on both lines and thus they are coincident. 1. $-3=3-3r$ 2. $14=4+5r$ 3. $10=6+2r$ 1. $r=2$ 2. $r=2$ 3. $r=2$ => The lines are coincident.", "0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •", "or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $$x_1$$ and $$x_2$$. The set of all linear combinations of", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "2 = ( 4 6 ) {\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}} and we wish to add these. So, v 1 + v 2 = ( 6 9 ) {\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}} . === Magnitude === The magnitude of a vector is its length in R+ == The dot product == The dot product of two vectors is defined as the sum of the products of the components. Symbolically we write ( a 1 a 2 ) ⋅ ( b 1 b 2 ) = a 1 b 1 + a 2 b 2 {\\displaystyle {\\begin{pmatrix}a_{1}\\\\a_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}b_{1}\\\\b_{2}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}} For example, ( 3 5 ) ⋅ ( 1 − 2 ) = 3 − 10 = − 7 {\\displaystyle {\\begin{pmatrix}3\\\\5\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\-2\\end{pmatrix}}=3-10=-7} The dot product of two vectors has an alternate form: a ⋅ b = | a | | b | cos ⁡ θ {\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =|\\mathbf {a} ||\\mathbf {b} |\\cos {\\theta }} The angle θ then is important, as it shows that the dot product of two vectors is related to the angle between them. More specifically, we can calculate the dot product of two vectors - if the dot product is zero we can then say that the two vectors are perpendicular. For example, consider simply ( 1 1 ) ⋅", "vector 6i+aj+bk has magnitude 21 and perpendicular to 3i+2j+2k .Find the possible values of a and b ,showing all necessary working. (a)Either: $$\\vec{PR}=2\\vec{PQ}=2(q-p)$$ $$\\vec{OR}=p+2q-2p=2q-p$$ $$\\vec{OR}=\\vec{PQ}=q-p$$ $$\\vec{OR}=\\vec{OQ}+\\vec{QR}=q+q-p=2q-p$$ (b)$$6^{2}+a^{2}+b^{2}=21^{2}$$ 18+2a+2b=0 $$a^{2}+\\left ( -a-9 \\right )^{2}=405$$ $$\\left ( 2 \\right )\\left ( a^{2}+9a-162 \\right )\\left (= 0 \\right )$$ a=9 or -18 b=-18 0r 9 Question The position vectors of points A and B relative to an origin O are a and b respectively. The position vectors of points C and D relative to O are 3a and 2b respectively. It is given that $$a=\\begin{pmatrix}2\\\\ 1\\\\ 2\\end{pmatrix}$$ and $$b=\\begin{pmatrix}4\\\\ 0\\\\ 6\\end{pmatrix}$$ (i) Find the unit vector in the direction of $$\\vec{CD}$$ (ii) The point E is the mid-point of CD. Find angle EOD. (i)$$\\vec{CD}=-3\\begin{pmatrix}2\\\\ 1\\\\ 2\\end{pmatrix}+2\\begin{pmatrix}4\\\\ 0\\\\ 6\\end{pmatrix}=\\begin{pmatrix}2\\\\ -3\\\\ 6\\end{pmatrix}$$ $$Unit vector =\\frac{1}{7}\\begin{pmatrix}2\\\\ -3\\\\ 6\\end{pmatrix}$$ (ii)$$\\vec{OE}=\\begin{pmatrix}6\\\\ 3\\\\ 6\\end{pmatrix}+\\begin{pmatrix}1\\\\-1\\tfrac{1}{2}\\\\ 3\\end{pmatrix}=\\begin{pmatrix}7\\\\ 1\\tfrac{1}{2}\\\\ 9\\end{pmatrix}$$ $$\\vec{OE}.\\vec{OD}=56+0+108=164$$ $$\\left | \\vec{OE} \\right |=\\sqrt{132.25}=11.5$$ $$\\left | \\vec{OD} \\right |=\\sqrt{208}$$ $$164=\\sqrt{132.25}\\times \\sqrt{208}\\times \\cos \\Theta$$ $$\\Theta =8.6^{\\circ}$$ Question The diagram shows a pyramid OABCD in which the vertical edge OD is 3 units in length. The point E is the centre of the horizontal rectangular base OABC. The sides OA and AB have lengths of 6 units and 4 units respectively. The unit vectors i, j and k are parallel t0 $$\\vec{OA}$$,$$\\vec{OC}$$ and $$\\vec{OD}$$ respectively. (i) Express each of the", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "{\\begin{pmatrix}1\\\\-2\\end{pmatrix}}=3-10=-7}$ The dot product of two vectors has an alternate form: ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =|\\mathbf {a} ||\\mathbf {b} |\\cos {\\theta }}$ The angle θ then is important, as it shows that the dot product of two vectors is related to the angle between them. More specifically, we can calculate the dot product of two vectors - if the dot product is zero we can then say that the two vectors are perpendicular. For example, consider simply ${\\displaystyle {\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\-1\\end{pmatrix}}=1-1=0}$ Plot these vectors on the plane and verify for yourself that these vectors are perpendicular. Vectors of 3D-lines Cartesian equation: ${\\displaystyle {\\tfrac {x-a}{l}}={\\tfrac {y-b}{m}}={\\tfrac {z-c}{n}}}$ . where a, b, and c are the coordinates on the vector line. Vector equation: r = ${\\displaystyle {\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}}$", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "END NEXT ``` Edited: 3 Feb 2013, 5:22 p.m. after one or more responses were posted Re: A quickie: improve the HP 50g IABCUV functionMessage #7 Posted by Thomas Klemm on 3 Feb 2013, 4:36 p.m.,in response to message #6 by C.Ret You may have noticed that the calculation could still be simplified: It's not necessary to compute the point of intersection M. We are only interested in the distance PM. And that's the projection of the vector u onto the blue line: $PM=u_v=\\vec{u}\\cdot\\frac{\\vec{v}}{v}$ Now we divide that by v and round the result to the closest integer k : $k=rnd(\\frac{u_v}{v})=rnd(\\frac{\\vec{u}\\cdot\\vec{v}}{v^{2}})=rnd(\\frac{\\vec{u}\\cdot\\vec{v}}{\\vec{v}\\cdot\\vec{v}})$ Finally we subtract k times the vector v from the vector u to get as close as possible to M: $\\vec{u}-k\\cdot\\vec{v}$ That's what these lines in the program do: ```\\<< u v DUP2 DOT v DUP DOT / 0 RND * - \\>> ``` Thanks for your interest Thomas The equation of the line in the sketch is: $3x+5y=44$ The vector n is perpendicular to the blue line: $\\vec{n}=\\begin{pmatrix} 3\\\\ 5 \\end{pmatrix}$ Whereas the vector v is parallel to the blue line: $\\vec{v}=\\begin{pmatrix} -5\\\\ 3 \\end{pmatrix}$ The vector u is one of the integer solutions. Here it's not the result of IABCUV because that would not fit on the drawing: $\\vec{u}=\\begin{pmatrix} -7\\\\ 13 \\end{pmatrix}$ Plug all the", "Since the distance can only be calculated if there is parallelism, you have to check whether the planes are parallel. We look to see whether the normal vectors are parallel. In contrast to the line, there is no check for perpendicularity. $\\vec{n_1}=r\\cdot\\vec{n_2}$ $\\begin{pmatrix} -4 \\\\ 4 \\\\ -8 \\end{pmatrix}=r\\cdot\\begin{pmatrix}2 \\\\ -2 \\\\ 4 \\end{pmatrix}$ $\\Rightarrow r=-2$ There is an $r$: The vectors are multiples of one another and therefore parallel. 2. #### Select point You can take any point on the plane. However, since you can easily read off the support point, this is a good option. $P(1|2|1)$ 3. #### Set up Hesse normal form $|\\vec{n}|=\\sqrt{2^2+(-2)^2+4^2}$ $=\\sqrt{24}$ $\\vec{n_0}= \\frac{\\vec{n}}{|\\vec{n}|}$ $=\\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}$ $\\text{E: } \\left(\\vec{x} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}=0$ 4. #### Insert point $\\vec{p}=\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $d=$ $\\left|\\left(\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=\\left|\\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=|-\\frac4{\\sqrt{24}}|$ $\\approx0.82$", "\\\\\\ 0 \\\\\\ 3 \\end{pmatrix} \\\\\\ \\end{pmatrix} = \\begin{pmatrix} 4 & 7 \\\\\\ 11 & 3 \\\\\\ 5 & 5 \\end{pmatrix}$$ • Algebraic properties of matrix-matrix multiplication. • $$A \\cdot B \\neq B \\cdot A$$ • $$(AB)C = A(BC)$$ • $$(AB)^T = B^T A^T$$ • $$(AB)^T \\neq A^T B^T$$ ### Null Space, Solution Space for Matrix Vector Equation • Null space of a matrix $$A$$ is $$\\{ \\boldsymbol{u} : A \\times \\boldsymbol{u} = 0 \\}$$, i.e. a vector that returns zero vectors when multiplied with matrix $$A$$. • E.g. $$\\begin{pmatrix} 1 & 2 & 4 \\\\\\ 2 & 3 & 9 \\end{pmatrix} \\times \\begin{pmatrix} 6 \\\\\\ -1 \\\\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\\\ 0 \\end{pmatrix}$$, where the null space includes $$\\begin{pmatrix} 6 \\\\\\ -1 \\\\\\ -1 \\end{pmatrix}$$. • Null space of a matrix $$\\begin{pmatrix} \\boldsymbol{a}_1 \\\\\\ ... \\\\\\ \\boldsymbol{a}_m \\end{pmatrix}$$ equals the solution set of the homogeneous linear system $$\\begin{pmatrix} \\boldsymbol{a}_1 \\cdot x = 0 \\\\\\ ... \\\\\\ \\boldsymbol{a}_m \\cdot x = 0 \\end{pmatrix}$$. • If $$\\boldsymbol{u}_1$$ is a solution to $$A \\boldsymbol{x} = \\boldsymbol{b}$$, then solution set is $$\\boldsymbol{u}_1 + V$$ where $$V = Null A$$. • E.g. given null space of $$\\begin{pmatrix} 1 & 2 & 4 \\\\\\ 2 & 3 & 9 \\end{pmatrix}$$ is $$Span \\{ \\begin{pmatrix} 6 \\\\\\ -1 \\\\\\ -1 \\end{pmatrix} \\}$$,", "the sense that the angle between $W$ and this vector is minimized) will have the largest dot product with $W$. So, we compute the three dot products: $$\\begin{array}{rcl} W \\cdot P_1 &=& (0.35)(0.35)+(0.3)(0.05)+(0.1)(0.1)+(0.05)(0.2)+(0.2)(0.3) = 0.2175\\\\ W \\cdot P_2 &=& (0.35)(0.3)+(0.3)(0.35)+(0.1)(0.05)+(0.05)(0.1)+(0.2)(0.2) = 0.2555\\\\ W \\cdot P_3 &=& (0.35)(0.2)+(0.3)(0.3)+(0.1)(0.35)+(0.05)(0.05)+(0.2)(0.1) = 0.2175 \\end{array}$$ Thus $P_2$ most closely matches $W$. 15. Which angle is smaller: $\\theta_1$, the angle between vectors $u = (.5,.3,.2)$ and $v_1=(.7,.2,.1)$; or $\\theta_2$, the angle between vectors $u$ and $v_2 = (.2,.1.,.7)$? 16. Suppose it is known that 7247 is prime. Reduce $7245!$ to its least postive residue$\\pmod{7247}$ (i.e., find $x$ where $0 \\lt x \\lt 7247$ and $7245! \\equiv x\\pmod{7247}$). 17. Suppose $M$ is a linear transformation that both operates on, and produces 2-dimensional vectors. If the following are true $$M\\begin{pmatrix}2\\\\0\\end{pmatrix} = \\begin{pmatrix}1\\\\3\\end{pmatrix} \\quad \\quad \\textrm{and} \\quad \\quad M\\begin{pmatrix}4\\\\6\\end{pmatrix} = \\begin{pmatrix}1\\\\-3\\end{pmatrix}$$ Find $\\displaystyle{M\\begin{pmatrix}2\\\\5\\end{pmatrix}}$ One way to proceed would be to realize that $M$ will have some matrix form $$M = \\left[ \\begin{array}{cc} a & b \\\\ c & d \\end{array} \\right]$$ where $$\\left[ \\begin{array}{cc} a & b \\\\ c & d \\end{array} \\right] \\left(\\begin{array}{c} 2 \\\\ 0 \\end{array} \\right) = \\left(\\begin{array}{c} 1 \\\\ 3 \\end{array} \\right) \\quad \\textrm{ and } \\quad \\left[ \\begin{array}{cc} a & b \\\\ c & d \\end{array} \\right] \\left(\\begin{array}{c} 4 \\\\ 6 \\end{array}", "# How to complete a set of independent vectors to a basis of a subspace? $\\newcommand{\\Span}{\\operatorname{Span}}$Let $V$ be a vector such that $\\dim V = n$, and let $v_1,\\ldots,v_k \\in V$ be independent vectors such that $1<k\\leq n$. Now Let $w_1,\\ldots,w_r\\in$ $\\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ be independent vectors such that $1\\leq r <k$. My question is this: How do I find the missing vectors $w_{r+1},\\ldots,w_k\\in \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ so that $\\Span\\left\\{w_1,\\ldots,w_r,w_{r+1},\\ldots,w_k\\right\\} = \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ ? Now I know they exist, I just don't know how I actually find them. For example: Lets look at $\\mathbb{R}^5$ and $$U=\\Span\\left\\{ \\begin{pmatrix} 5\\\\2\\\\3\\\\7\\\\3 \\end{pmatrix},\\begin{pmatrix} 2\\\\4\\\\4\\\\8\\\\1 \\end{pmatrix} ,\\begin{pmatrix} 3\\\\4\\\\7\\\\6\\\\1 \\end{pmatrix},\\begin{pmatrix} 5\\\\8\\\\6\\\\4\\\\8 \\end{pmatrix} \\right\\}$$ Those are all independent vectors. Now lets take $2$ independent vectors that are linear combination of those. lets say, $$w_1 = \\begin{pmatrix} 6\\\\2\\\\11\\\\15\\\\-3 \\end{pmatrix} \\; w_2=\\begin{pmatrix} 0\\\\4\\\\-2\\\\12\\\\1 \\end{pmatrix}$$ How can we complete those two vectors to form a basis of $U$ ? I would really like to understand the general idea of this. This is actually a general question of something I need it for, which is the process of finding a Jordan basis for matrices\\transformations. Thanks for any help Here's one method for the case of $V = \\Bbb R^n$: row-reduce the matrix $$\\pmatrix{w_1 & \\cdots & w_r & v_1 & \\cdots & v_k}$$ The columns of this matrix", "Math Relative position of lines Angle between two lines # Angle between two lines The angle between two lines is calculated similarly as the angle between two vectors. ! ### Remember In contrast to vectors, the angle between two lines always discribes the angle less than 90°. Two lines are given: $\\text{g: } \\vec{x} = \\vec{u} + r \\cdot \\vec{a}$ $\\text{h: } \\vec{x} = \\vec{v} + s \\cdot \\vec{b}$ For determining the angle the direction vectors of the lines are used: $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\gamma = \\cos^{-1}\\left(\\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}\\right)$ i ### Method 1. Calculate the scalar product of the direction vectors 2. Calculate the magnitude of the direction vectors 3. Insert results into the formula ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Calculate scalar product $\\vec{a}\\cdot\\vec{b}$ $=\\begin{pmatrix}9 \\\\ 8 \\\\ 7 \\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ 1 \\\\ 2\\end{pmatrix}$ $=9\\cdot1+8\\cdot1+7\\cdot2$ $=31$ 2. #### Calculate magnitude of the direction vectors $|\\vec{a}|=\\sqrt{9^2+8^2+7^2}$ $=\\sqrt{194}$ $|\\vec{b}|=\\sqrt{1^2+1^2+2^2}$ $=\\sqrt{6}$ 3. #### Insert results into the formula $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\cos(\\gamma) = \\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}$ $\\gamma = \\cos^{-1}\\left(\\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}\\right)$ $\\approx24.68°$" ]

[ "The distance between two vectors is the magnitude of their difference. Find the value of $t$ for which the vector ${v} = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t$ is closest to ${a} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix}.$ I'm kinda confused on the magnitude part. I got $\\sqrt {901}$ for that but that doesn't sound right. I looked at the other links but didn't really get it. • $\\sqrt{9}01$ ? You mean $\\sqrt{901}$, right ? – A---B Nov 1 '17 at 0:14 • yea sorry, i''m not good at mathjax – ninjagirl Nov 1 '17 at 11:20 • You need to use curly braces around the number so that it appears wholly inside the square root symbol. – A---B Nov 1 '17 at 11:43 Consider the line $\\ell$ in $\\mathbb{R}^3$ parametrized by $$v(t) = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t,$$ and consider the point $${a} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix}.$$ Find the value of $t$ for which the point $v(t)$ on the line is closest to the point $a$. $\\textbf{Method 1}$. Let $p,q$ be two points in $\\mathbb{R}^3$ and let $d(p,q) = || p-q||$ be the distance function between $p$ and $q$, where", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "shortest distance between the two. Let the line segment connecting point $A$ and vector $\\vec b$ be defined as $\\vec{v}=\\begin{pmatrix} 3s+2 \\\\ 4s-3 \\\\s+2 \\end{pmatrix}$. Since the shortest distance between the point and the line is when $\\vec v$ is orthogonal to $\\vec b$, $\\vec v \\cdot \\vec b=0$. Since the direction vector of $\\vec b$ is simply $\\vec{b}=\\begin{pmatrix} 3 \\\\ 4 \\\\1 \\end{pmatrix}$, $$\\begin{pmatrix} 3s+2 \\\\ 4s-3 \\\\s+2 \\end{pmatrix}\\cdot \\begin{pmatrix} 3 \\\\ 4 \\\\1 \\end{pmatrix}=0$$ Therefore $s={2\\over 13}$. Now by plugging in $s$ into $\\vec v$ we get $\\vec{v}=\\begin{pmatrix} {32\\over 13} \\\\ {-31\\over 13} \\\\{28\\over 13} \\end{pmatrix}$ whereas $|\\vec v |=\\sqrt {213\\over 13}={\\sqrt{2769}\\over13}$ which is the answer. Can you explain why $v$ has $(2, 3, 2)$? – Imray Nov 10 '14 at 12:47 since the shortest distance between the two can be realized at any point on one of the lines, we fix a point on one of the lines and minimize, say $$f(t)=|a(t)-b(0)|^2=|(6t-2,8t+3,2t-2)|^2$$ $$=104t^2+16t+17$$ this has a minimum where $f'(t)=208t+16=0$, so $t=-1/13$ and $f(-1/13)=213/13$. this gives a minimum distance of $\\sqrt{213/13}$. $f(-1/13)$ should be $213/13$. – David Mitra Jan 3 '12 at 0:30", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "+0 # shortest distance between two lines via vector equation 0 255 4 +844 For a i got r= (-2,3,4) + λ(3,2,-7) for b so far i wanted to identify a point A on l1 and a point b on l2 i assumed the two lines are parallel as they have the same direction vector $$A=\\begin{pmatrix} -2+3λ\\\\ 3+2λ\\\\ 4-7λ \\end{pmatrix}$$ $$B= \\begin{pmatrix} 3\\\\ 1+2λ\\\\ -2-7λ \\end{pmatrix}$$ used this to get $$AB= \\begin{pmatrix} 5-3λ\\\\ -2\\\\ -6 \\end{pmatrix}$$ used this and my direction vector to find λ=53/9 via dot product substituted and calculated the magnitude which gave me 14.26 but my answer should be 3.57 please identify what went wrong, thanks Feb 10, 2019 #1 +2 The equation of the first line can be written as r1 = (-2, 3, 4) + s(3, 2, -7), (as you calculated), and the equation of the second as r2 = (0, 1, -2) + t(3, 2, -7), s and t arbitrary parameters. A vector d connecting two points, one on each line, would be given by d = r2 - r1 = (2, -2, -6) + (t - s)(3, 2 ,-7) = (2, -2, -6) + w(3, 2, -7), w for convenience. The distance between the two points will be the magnitude of this vector, so what you are looking for is the", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "Since the distance can only be calculated if there is parallelism, you have to check whether the planes are parallel. We look to see whether the normal vectors are parallel. In contrast to the line, there is no check for perpendicularity. $\\vec{n_1}=r\\cdot\\vec{n_2}$ $\\begin{pmatrix} -4 \\\\ 4 \\\\ -8 \\end{pmatrix}=r\\cdot\\begin{pmatrix}2 \\\\ -2 \\\\ 4 \\end{pmatrix}$ $\\Rightarrow r=-2$ There is an $r$: The vectors are multiples of one another and therefore parallel. 2. #### Select point You can take any point on the plane. However, since you can easily read off the support point, this is a good option. $P(1|2|1)$ 3. #### Set up Hesse normal form $|\\vec{n}|=\\sqrt{2^2+(-2)^2+4^2}$ $=\\sqrt{24}$ $\\vec{n_0}= \\frac{\\vec{n}}{|\\vec{n}|}$ $=\\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}$ $\\text{E: } \\left(\\vec{x} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}=0$ 4. #### Insert point $\\vec{p}=\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $d=$ $\\left|\\left(\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=\\left|\\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=|-\\frac4{\\sqrt{24}}|$ $\\approx0.82$", "? Free Version Moderate # Vector Distance: Between Normalized Vectors LINALG-S3LXJ6 If $\\boldsymbol{u} = \\begin{bmatrix} 5\\\\\\ 4\\\\\\ 3\\end{bmatrix}$ and $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ 0\\\\\\ -1\\end{bmatrix}$, then compute the distance between $\\dfrac{1}{\\Vert\\boldsymbol{u}\\Vert}\\boldsymbol{u}$ and $\\dfrac{1}{\\Vert\\boldsymbol{v}\\Vert}\\boldsymbol{v}$. A $4\\sqrt{3}$ B $\\dfrac{5\\sqrt{10}}{2}$ C $2\\sqrt{154}$ D $\\sqrt{2}$ E $\\dfrac{2\\sqrt{10}}{5}$ F $5\\sqrt{154}$", "+0 # help 0 107 2 Compute the distance between the parallel lines given by $$\\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ and $$\\begin{pmatrix} -5 \\\\ 6 \\end{pmatrix} + s \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ Feb 23, 2020 #1 0 The distance is 3/2. Feb 23, 2020 #2 +25275 +1 Compute the distance between the parallel lines given by $$\\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ and $$\\begin{pmatrix} -5 \\\\ 6 \\end{pmatrix} + s \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$. $$\\text{The orthogonal vector of the lines and normalized is }\\\\ \\binom{-3}{4}*\\frac{1}{\\sqrt{(-3)^2+4^2}}~\\text{ or }~\\binom{4}{-3}*\\frac{1}{\\sqrt{3^2+(-4)^2}} \\\\ \\text{because } \\binom{4}{3}\\binom{-3}{4} = 0 ~\\text{ respectively }~\\binom{4}{3}\\binom{4}{-3} = 0\\\\$$ $$\\text{Let \\hat{n} = \\frac{1}{5}\\binom{-3}{4}}$$ The distance between the parallel lines is: $$\\begin{array}{|rcll|} \\hline d &=&\\dfrac{1}{5}\\dbinom{-3}{4}\\dbinom{-5}{6}-\\frac{1}{5}\\dbinom{-3}{4}\\dbinom{1}{4} \\\\\\\\ d &=&\\dfrac{1}{5}\\Big(15+24-( -3+16 ) \\Big) \\\\\\\\ d &=&\\dfrac{26}{5} \\\\\\\\ \\mathbf{d} &=& \\mathbf{5.2} \\\\ \\hline \\end{array}$$ Feb 24, 2020", "Free Version Moderate # Vector Distance: Between Scalar Multiples Involving Dot Products LINALG-ZERJN4 If $\\boldsymbol{u} = \\begin{bmatrix} 3\\\\\\ -1\\\\\\ -2\\end{bmatrix}$, $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 3\\end{bmatrix}$, $\\boldsymbol{w} = \\begin{bmatrix} -2\\\\\\ 1\\\\\\ -1\\end{bmatrix}$, and $\\boldsymbol{x} = \\begin{bmatrix} 0\\\\\\ 1\\\\\\ 0\\end{bmatrix}$, then compute the distance between $(\\boldsymbol{u} \\cdot \\boldsymbol{v})\\boldsymbol{w}$ and $2\\boldsymbol{x}$. A $2\\sqrt{5}$ B $\\sqrt{15}$ C $\\sqrt{14}$ D $5$ E $\\sqrt{10}$ F $\\sqrt{23}$", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "(squared) vertical distances? Thanks. Actually, there isn't any difference. The problem is set up so that minimizing the (sum of the) squared vertical distance is the same as minimizing the (Euclidean, what I guess you call \"orthogonal\") distance between two vectors. Let's say we are given $$m$$ points $$(x_1,y_1), \\dots, (x_m,y_m)$$ and we want to find $$C,D$$ such that $$y = Cx + D$$ is the \"best fit\" for those points in terms of squared vertical distance. That is, we want to find $$C,D$$ such that the expression $$\\sum_{i=1}^m \\left( Cx_i + D - y_i \\right)^2$$ is minimal. If you set up a matrix and column vectors $$A = \\begin{pmatrix} x_1 & 1 \\\\ \\vdots & \\vdots \\\\ x_m & 1 \\end{pmatrix}, \\,\\,\\, b = \\begin{pmatrix} y_1 \\\\ \\vdots \\\\ y_m \\end{pmatrix},\\,\\,X = \\begin{pmatrix} C \\\\ D \\end{pmatrix}$$ then $$AX - b = \\begin{pmatrix} Cx_1 + D \\\\ \\vdots \\\\ Cx_m + D \\end{pmatrix} - \\begin{pmatrix} y_1 \\\\ \\vdots \\\\ y_m \\end{pmatrix} = \\begin{pmatrix} Cx_1 + D - y_1 \\\\ \\vdots \\\\ Cx_m + D - y_m \\end{pmatrix}.$$ In terms of $$A,X,b$$, you want to find $$X$$ that minimizes $$\\| AX - b \\|^2$$. In other words, the vector $$AX$$ should minimize the distance to $$b$$ among all vectors of the form $$Ax$$. This implies that $$AX$$ is", "Free Version Moderate # Vector Distance: Involving Dot Products and Linear Combinations LINALG-JEUE3N If $\\boldsymbol{u} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 1\\end{bmatrix}$, $\\boldsymbol{v} = \\begin{bmatrix} 0\\\\\\ -5\\\\\\ 3\\end{bmatrix}$, $\\boldsymbol{w} = \\begin{bmatrix} -1\\\\\\ -1\\\\\\ 3\\end{bmatrix}$, $\\boldsymbol{x} = \\begin{bmatrix} 3\\\\\\ -1\\\\\\ -2\\end{bmatrix}$, and $\\boldsymbol{y} = \\begin{bmatrix} -1\\\\\\ 0\\\\\\ -3\\end{bmatrix}$ ...then compute the distance between $(\\boldsymbol{u} \\cdot \\boldsymbol{v})\\boldsymbol{w}$ and $2\\boldsymbol{x} - 3\\boldsymbol{y}$ A $4\\sqrt{110}$ B $\\sqrt{1761}$ C $\\sqrt{1762}$ D $\\sqrt{1763}$ E $42$ F $\\sqrt{1765}$", "or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $$x_1$$ and $$x_2$$. The set of all linear combinations of", "are talking about bound vector, then two bound vectors are equal if they have the same base point and end point. For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (displacement) vector. Scalar multiplication For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and Roman letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. Example Say we have ${\\displaystyle {\\begin{pmatrix}2\\\\3\\end{pmatrix}}}$ and we wish to double the magnitude. So, ${\\displaystyle 2{\\begin{pmatrix}2\\\\3\\end{pmatrix}}={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ . Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. Example Say we have ${\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ and we wish to add these. So, ${\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}}$ . Magnitude The magnitude of a vector is its length in R+ The dot product The dot product of two vectors is defined as the sum of the products of the components. Symbolically we write ${\\displaystyle {\\begin{pmatrix}a_{1}\\\\a_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}b_{1}\\\\b_{2}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}}$ For example, ${\\displaystyle {\\begin{pmatrix}3\\\\5\\end{pmatrix}}\\cdot", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "? Free Version Easy # Vector Distance: When the Third Coordinate Agrees LINALG-S83SC1 If $\\boldsymbol{u} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 3\\end{bmatrix}$ and $\\boldsymbol{v} = \\begin{bmatrix} -1\\\\\\ -1\\\\\\ 3\\end{bmatrix}$, then compute the distance between $\\boldsymbol{u}$ and $\\boldsymbol{v}$. A $\\sqrt{6}$ B $\\sqrt{7}$ C $2\\sqrt{2}$ D $3$ E $\\sqrt{5}$ F $2$", "## VECTORS In three dimensions, the base vectors are $$i=\\begin{pmatrix} 1\\\\0 \\\\0 \\end{pmatrix}$$, $$j=\\begin{pmatrix} 0\\\\1 \\\\0 \\end{pmatrix}$$, $$k=\\begin{pmatrix} 0\\\\0 \\\\1 \\end{pmatrix}$$ that are along the $$x, y$$ and $$z$$ coordinate directions, respectively, as shown in the figure. For example, $$v=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ can be written as $$-4i+3j+2k$$. The magnitude of $$v$$: $$|v|=\\sqrt{((-4)^2+3^2+2^2)}=\\sqrt{29}$$ Given that if $$v=xi+yj+zk$$ then $$|v|=\\sqrt{x^2+y^2+z^2}$$ ## UNIT VECTORS If $$\\overrightarrow{AB}$$ is a non-unit vector, then a unit vector in the direction of $$\\overrightarrow{AB}$$ is $${1\\over |\\overrightarrow{AB}|}\\overrightarrow{AB}$$ LESSON 1 Determine the unit vector in the direction of the vector $$\\overrightarrow{OA}=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$. SOLUTION • Determine the length of the vector $$|\\overrightarrow{OA}|=\\sqrt{(-4)^2+(3)^2+(2)^2}=\\sqrt{29}$$ • Use the form for a unit vector Unit Vector $${1\\over \\sqrt{29}} \\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ TRY THIS Complete Question 1 from worksheet.", "is $\\frac{\\mu\\vec a + \\lambda\\vec b}{\\lambda + \\mu}$ I assume that the units you're using are metres, so by 1 m away from the first point, you mean 1 unit. The distance between the points is $\\sqrt{51}$. So we need the point that divides the line joining $\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}$ to $\\begin{pmatrix}6\\\\5\\\\0\\end{pmatrix}$ in the ratio $1:\\sqrt{51}-1$, which is: $\\dfrac{(\\sqrt{51}-1)\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}+1\\begin{pmat rix}6\\\\5\\\\0\\end{pmatrix}}{\\sqrt{51}}$ $=\\dfrac{1}{\\sqrt{51}}\\begin{pmatrix}5\\sqrt{51}+1\\\\ 5\\\\5\\sqrt{51}-5\\end{pmatrix}$ $=\\begin{pmatrix}5+\\frac{1}{\\sqrt{51}}\\\\ \\frac{5}{\\sqrt{51}}\\\\5-\\frac{5}{\\sqrt{51}}\\end{pmatrix}$ 4. Thank you folks. I'm a bit scared by all the formulae, but I think the general idea is that the vector between my two points specifies the linear change in x y z as you go along that vector. So for 1 meters worth of change I want 1/Distance in each of x, y and z. If change in x between my two points is +1, and the 3D distance between them is root 51, then the new x position 1 m down that line is 5 + (1/ root51) = 1.140 Calculate likewise for y and z. I think this is basically what you said, but in simple speak. IF you know the changes in x y z for the whole vector, you can 'scale' those changes for each unit of the length of the vector. That's what I did. Thanks again for your help" ]

[ "\\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}$", "\\tan \\left(\\frac{\\pi x}{2} \\right) [\\ln(2-x)] = \\frac{2}{\\pi}$ Therefore, we have $\\displaystyle \\lim_{x \\to 1} (2 -x)^{\\tan \\left( \\pi x/2 \\right)} = \\lim_{x \\to 1} e^{\\ln y} = e^{2/\\pi}$", "\\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(6\\right) + \\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right)\\right)$", "( \\frac{\\pi}{6} \\right ) + sin \\left ( \\frac{\\pi}{6} \\right ) cos \\left ( \\frac{3\\pi}{4} \\right )$ I would use $tan \\left ( \\frac{\\pi}{12} \\right ) = tan \\left ( \\frac{1}{2} \\cdot \\frac{\\pi}{6} \\right )$", "^ 3\\right) \\tan \\left(\\frac{1}{x} ^ 3\\right)$", "# Find the exact value of each expression. (tan)(17pi)/12 Find the exact value of each expression. $$\\displaystyle{\\left({\\tan}\\right)}\\frac{{{17}\\pi}}{{12}}$$ • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Clara Reese 1. Split up the argument of the trigonometric function and then use the Sum formula for the Tangent function, $$\\displaystyle{\\tan{{\\left({u}+{v}\\right)}}}=\\frac{{{\\tan{{\\left({u}\\right)}}}+{\\tan{{\\left({v}\\right)}}}}}{{{1}-{\\tan{{\\left({u}\\right)}}}{\\tan{{\\left({v}\\right)}}}}}$$ , to find the value of given expression. 2. $$\\displaystyle{\\tan{{\\left(\\frac{{{17}\\pi}}{{12}}\\right)}}}={\\tan{{\\left(\\frac{{{9}\\pi}}{{12}}+\\frac{{{8}\\pi}}{{12}}\\right)}}}$$ $$\\displaystyle={\\tan{{\\left(\\frac{{{3}\\pi}}{{4}}+\\frac{{{2}\\pi}}{{3}}\\right)}}}$$ $$\\displaystyle=\\frac{{{\\tan{{\\left(\\frac{{{3}\\pi}}{{4}}\\right)}}}+{\\tan{{\\left(\\frac{{{2}\\pi}}{{3}}\\right)}}}}}{{{1}-{\\left({\\tan{{\\left(\\frac{{{3}\\pi}}{{4}}\\right)}}}{\\tan{{\\left(\\frac{{{2}\\pi}}{{3}}\\right)}}}\\right.}}}$$ $$\\displaystyle=\\frac{{-{1}+{\\left(-\\sqrt{{3}}\\right)}}}{{{1}-{\\left(-{1}\\right)}\\cdot{\\left(-\\sqrt{{3}}\\right)}}}$$ $$\\displaystyle=\\frac{{-{1}-\\sqrt{{3}}}}{{{1}-\\sqrt{{3}}}}$$ $$\\displaystyle=\\frac{{{\\left(-{1}-\\sqrt{{3}}\\right)}{\\left({1}+\\sqrt{{3}}\\right)}}}{{{\\left({1}-\\sqrt{{3}}\\right)}{\\left({1}+\\sqrt{{3}}\\right)}}}$$ $$\\displaystyle=\\frac{{-{1}-{2}\\sqrt{{3}}-{3}}}{{{1}-{3}}}$$ $$\\displaystyle=\\frac{{{4}-{2}\\sqrt{{3}}}}{{-{{2}}}}$$ $$\\displaystyle={2}+\\sqrt{{3}}$$ Result : $$\\displaystyle{\\tan{{\\left(\\frac{{{17}\\pi}}{{12}}\\right)}}}={2}+\\sqrt{{3}}$$ ###### Have a similar question? • Questions are typically answered in as fast as 30 minutes", "\\left(x\\right)$ $\\frac{2 \\tan \\left(x\\right) + 2 {\\sin}^{3} \\frac{x}{\\cos} \\left(x\\right)}{\\cos} ^ 4 \\left(x\\right) = \\frac{2 \\tan \\left(x\\right) + 2 {\\sin}^{2} \\left(x\\right) \\tan \\left(x\\right)}{\\cos} ^ 4 \\left(x\\right)$", "\\left(\\frac{1}{7}\\right)\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(\\frac{2}{5}\\right) - {\\tan}^{-} 1 \\left(\\frac{1}{7}\\right)\\right)\\right)$", "# How do you divide ( 2i -4) / ( 7 i -2 ) in trigonometric form? $\\frac{2 i - 4}{7 i - 2} = \\frac{2 \\sqrt{265}}{53} \\left[\\cos {47.48}^{\\circ} + i \\cdot \\sin {47.48}^{\\circ}\\right]$ #### Explanation: The solution: $2 i - 4 =$ $\\sqrt{4 + 16} \\left[\\cos \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right) + i \\cdot \\sin \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right)\\right]$ $\\sqrt{20} \\left[\\cos \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right) + i \\cdot \\sin \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right)\\right]$ $7 i - 2 =$ $\\sqrt{4 + 49} \\left[\\cos \\left({\\tan}^{-} 1 \\left(- \\frac{7}{2}\\right)\\right) + i \\cdot \\sin \\left({\\tan}^{-} 1 \\left(- \\frac{7}{2}\\right)\\right)\\right]$ $\\frac{2 i - 4}{7 i - 2} =$ $\\frac{\\sqrt{20}}{\\sqrt{53}} \\left[\\cos \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right) - {\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right) + i \\cdot \\sin \\left({\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right) - {\\tan}^{-} 1 \\left(- \\frac{1}{2}\\right)\\right)\\right]$ $\\frac{2 i - 4}{7 i - 2} = \\frac{2 \\sqrt{265}}{53} \\left[\\cos {47.48}^{\\circ} + i \\cdot \\sin {47.48}^{\\circ}\\right]$ God bless.....I hope the explanation is useful.", "$\\tag{12} {µ_0}=\\frac{\\sin{\\left( \\alpha \\right) }}{\\cos{\\left( \\alpha \\right) }}$ or converted to $\\tag{13} {µ_0}=\\tan{\\left( \\alpha \\right) }$", "\\frac{{11\\pi }}{6}} \\right)$$ Solution 7. $$\\displaystyle \\cos \\left( {\\frac{{8\\pi }}{3}} \\right)$$ Solution 8. $$\\displaystyle \\tan \\left( { - \\frac{\\pi }{3}} \\right)$$ Solution 9. $$\\displaystyle \\tan \\left( {\\frac{{15\\pi }}{4}} \\right)$$ Solution 10. $$\\displaystyle \\sin \\left( { - \\frac{{11\\pi }}{3}} \\right)$$ Solution 11. $$\\displaystyle \\sec \\left( {\\frac{{29\\pi }}{4}} \\right)$$ Solution", "Thus, we consider $f \\left(x\\right) = {\\tan}^{2} \\left(x\\right) = \\tan \\left(x\\right) \\tan \\left(x\\right)$ so that we can deal with $\\tan \\left(x\\right)$, for which we know the derivative. Utilizing the derivative of $\\tan \\left(x\\right)$, namely $\\frac{d}{\\mathrm{dx}} \\tan \\left(x\\right) = {\\sec}^{2} \\left(x\\right)$, and the Chain Rule $\\frac{\\mathrm{df}}{\\mathrm{dx}} = g ' \\left(x\\right) h \\left(x\\right) + g \\left(x\\right) h ' \\left(x\\right)$, we obtain: $f ' \\left(x\\right) = \\left[\\frac{d}{\\mathrm{dx}} \\left(\\tan \\left(x\\right)\\right)\\right] \\tan \\left(x\\right) + \\tan \\left(x\\right) \\left[\\frac{d}{\\mathrm{dx}} \\left(\\tan \\left(x\\right)\\right)\\right]$ $\\frac{d}{\\mathrm{dx}} \\tan \\left(x\\right) = {\\sec}^{2} \\left(x\\right)$, so... $f ' \\left(x\\right) = {\\sec}^{2} \\left(x\\right) \\tan \\left(x\\right) + \\tan \\left(x\\right) {\\sec}^{2} \\left(x\\right) = 2 \\tan \\left(x\\right) {\\sec}^{2} \\left(x\\right)$", "\\frac{\\tan 2 \\theta}{2 \\theta} \\times 2 = 2 \\left(1\\right) \\times 2 = 4$", "1 \\left(\\frac{72.0881}{-} 0.3893\\right)$ ${\\theta}_{a e} = \\pi - {\\tan}^{-} 1 \\left(\\frac{72.0881}{0.3893}\\right)$", "# Prove that $\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$ [duplicate] Prove that $$\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$$", "} } \\right) } \\right) } } =\\tan ^{ -1 }{ \\left( \\frac { x }{ \\sqrt { { x }^{ 2 }+2 } } \\right) }$$Hence$$\\displaystyle \\int _{ 0 }^{ 1 }{ \\frac { 1 }{ \\left( { x }^{ 2 }+1 \\right) \\sqrt { { x }^{ 2 }+2 } } dx } ={ \\left[ \\tan ^{ -1 }{ \\left( \\frac { x }{ \\sqrt { { x }^{ 2 }+2 } } \\right) } \\right] }_{ 0 }^{ 1 }$$$$\\displaystyle =\\frac { \\pi }{ 6 } -0=\\frac { \\pi }{ 6 }$$$$\\Rightarrow k=6$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "on the final step, but your evaluation is correct.$$-\\tan\\left(-\\frac\\pi3\\right)=\\sqrt 3$$", "is equal to in terms of $z$: $z = \\sqrt{u - 1}$ $u - 1 = {z}^{2}$ $u = {z}^{2} + 1$ This gives: $\\int \\setminus \\frac{1}{u} \\setminus \\mathrm{dz} = \\int \\setminus \\frac{1}{1 + {z}^{2}} \\setminus \\mathrm{dz}$ This is the common derivative of ${\\tan}^{-} 1 \\left(z\\right)$, so we get: $\\int \\setminus \\frac{1}{1 + {z}^{2}} \\setminus \\mathrm{dz} = {\\tan}^{-} 1 \\left(z\\right) + C$ Undoing all the substitutions, we get: ${\\tan}^{-} 1 \\left(z\\right) + C = {\\tan}^{-} 1 \\left(\\sqrt{u - 1}\\right) + C =$ $= {\\tan}^{-} 1 \\left(\\sqrt{1 + {e}^{2 x} - 1}\\right) + C = {\\tan}^{-} 1 \\left({\\left({e}^{2 x}\\right)}^{\\frac{1}{2}}\\right) + C =$ $= {\\tan}^{-} 1 \\left({e}^{x}\\right) + C$", "can say that, $\\frac{{dy}}{{dx}} = \\; - \\tan \\theta$ ⇒ ${\\left[ {\\frac{{dy}}{{dx}}} \\right]_{\\left( {\\theta = \\frac{\\pi }{4}} \\right)}} = \\; - \\tan \\frac{\\pi }{4} = \\; - 1$ Hence, option A is the correct answer.", "Show that Question: $\\int_{0}^{\\frac{\\pi}{4}} 2 \\tan ^{3} x d x=1-\\log 2$ Solution: Let $I=\\int_{0}^{\\frac{\\pi}{4}} 2 \\tan ^{3} x d x$ $I=2 \\int_{0}^{\\frac{\\pi}{4}} \\tan ^{2} x \\tan x d x=2 \\int_{0}^{\\frac{\\pi}{4}}\\left(\\sec ^{2} x-1\\right) \\tan x d x$ $=2 \\int_{0}^{\\frac{\\pi}{4}} \\sec ^{2} x \\tan x d x-2 \\int_{0}^{\\frac{\\pi}{4}} \\tan x d x$ $=2\\left[\\frac{\\tan ^{2} x}{2}\\right]_{0}^{\\frac{\\pi}{4}}+2[\\log \\cos x]_{0}^{\\frac{\\pi}{4}}$ $=1+2\\left[\\log \\cos \\frac{\\pi}{4}-\\log \\cos 0\\right]$ $=1+2\\left[\\log \\frac{1}{\\sqrt{2}}-\\log 1\\right]$ $=1-\\log 2-\\log 1=1-\\log 2$ Hence, the given result is proved." ]

[ "+0 # Really want this one solved +1 49 4 +174 Let $$\\theta$$ be an acute angle such that $$\\sin 5 \\theta = \\sin^5 \\theta$$ Compute $$\\tan 2 \\theta$$ this isn't my homework or anything, really curious how to do it Jun 6, 2021 #1 +32236 +4 Here's a helping hand: See if you can take it from here. Jun 6, 2021 #4 +113569 0 That is where I got lost. Melody Jun 7, 2021 #2 +174 +2 because of DeMoivre's Theorem: \\begin{align*} \\operatorname{cis} n \\theta &= (\\operatorname{cis} \\theta)^n \\\\ &= (\\cos \\theta + i \\sin \\theta)^n \\\\ &= \\cos^n \\theta + \\binom{n}{1} i \\cos^{n - 1} \\theta \\sin \\theta - \\binom{n}{2} \\cos^{n - 2} \\theta \\sin^2 \\theta - \\binom{n}{3} i \\cos^{n - 3} \\theta \\sin^3 \\theta + \\dotsb. \\end{align*} \\begin{align*} \\cos n \\theta &= \\cos^n \\theta - \\binom{n}{2} \\cos^{n - 2} \\theta \\sin^2 \\theta + \\binom{n}{4} \\cos^{n - 4} \\theta \\sin^4 \\theta - \\dotsb, \\\\ \\sin n \\theta &= \\binom{n}{1} \\cos^{n - 1} \\theta \\sin \\theta - \\binom{n}{3} \\cos^{n - 3} \\theta \\sin^3 \\theta + \\binom{n}{5} \\cos^{n - 5} \\theta \\sin^5 \\theta - \\dotsb. \\end{align*} \\begin{align*} \\sin 5 \\theta &= \\binom{5}{1} \\cos^4 \\theta \\sin \\theta - \\binom{5}{3} \\cos^2 \\theta \\sin^3 \\theta + \\binom{5}{5} \\sin^5 \\theta \\\\ &= 5 \\cos^4 \\theta \\sin \\theta - 10 \\cos^2 \\theta \\sin^3", "$$\\operatorname{E}\\left(\\hat{\\theta}\\right)$$ and $$\\operatorname{V}\\left(\\hat{\\theta}\\right)$$ > theta <- pi / 4 > n <- 10 > e_theta_hat <- 0 > e_theta_hat_square <- 0 > for (m1 in 0:(n-1)) + { + for (m2 in 0:(n-m1-1)) + { + e_theta_hat <- e_theta_hat + atan(sqrt((m1+m2)/(n-m1-m2)))*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + e_theta_hat_square <- e_theta_hat_square + (atan(sqrt((m1+m2)/(n-m1-m2))))^2*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + } + } > e_theta_hat [1] 0.7662692 > theta [1] 0.7853982 > > e_theta_hat_square - (e_theta_hat)^2 [1] 0.04779768 > theta <- pi / 4 > n <- 1000 > e_theta_hat <- 0 > e_theta_hat_square <- 0 > for (m1 in 0:(n-1)) + { + for (m2 in 0:(n-m1-1)) + { + e_theta_hat <- e_theta_hat + atan(sqrt((m1+m2)/(n-m1-m2)))*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + e_theta_hat_square <- e_theta_hat_square + (atan(sqrt((m1+m2)/(n-m1-m2))))^2*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + } + } > e_theta_hat [1] 0.7853983 > theta [1] 0.7853982 > > e_theta_hat_square - (e_theta_hat)^2 [1] 0.0003755647 >", "= 1$$, when the value of $$s_{n}$$ is $$n$$, it follows that the series $$1 + z + z^{2} + \\dots$$ is convergent if and only if $$r = |z| < 1$$. And its sum when convergent is $$1/(1 – z)$$. Thus if $$z = r(\\cos\\theta + i\\sin\\theta) = r\\operatorname{Cis}\\theta$$, and $$r < 1$$, we have \\begin{aligned} 1 + z + z^{2} + \\dots &= 1/(1 – r\\operatorname{Cis}\\theta), \\text{or} 1 + r \\operatorname{Cis}\\theta + r^{2} \\operatorname{Cis} 2\\theta + \\dots &= 1/(1 – r\\operatorname{Cis}\\theta)\\\\ &= (1 – r\\cos\\theta + ir\\sin\\theta)/(1 – 2r\\cos\\theta + r^{2}).\\end{aligned} Separating the real and imaginary parts, we obtain \\begin{aligned} 1 + r\\cos\\theta + r^{2}\\cos 2\\theta + \\dots &= (1 – r\\cos\\theta)/(1 – 2r\\cos\\theta + r^{2}),\\\\ r\\sin\\theta + r^{2}\\sin 2\\theta + \\dots &= r\\sin\\theta/(1 – 2r\\cos\\theta + r^{2}),\\end{aligned} provided $$r < 1$$. If we change $$\\theta$$ into $$\\theta + \\pi$$, we see that these results hold also for negative values of $$r$$ numerically less than $$1$$. Thus they hold when $$-1 < r < 1$$. Example XXXIII 1. Prove directly that $$\\phi(n) = r^{n} \\cos n\\theta$$ converges to $$0$$ when $$r < 1$$ and to $$1$$ when $$r = 1$$ and $$\\theta$$ is a multiple of $$2\\pi$$. Prove further that if $$r = 1$$ and $$\\theta$$ is not a multiple of $$2\\pi$$, then $$\\phi(n)$$ oscillates finitely;", "coefficients. \\begin{align*} (\\cos\\theta + i\\sin\\theta)^8 = \\sum_{k=0}^8 i^k \\binom{8}{k} \\sin^k\\theta \\cos^{8-k}\\theta \\\\ \\text{therefore} \\ i\\sin 8\\theta = \\binom{8}{1}i\\cos^7\\theta\\sin\\theta + \\binom{8}{3}i^3\\sin^3\\theta\\cos^5\\theta + \\binom{8}{5}i^5\\sin^5\\theta\\cos^3\\theta + \\binom{8}{7}i^7\\sin^7\\theta\\cos\\theta \\\\ = i\\left(\\binom{8}{1}\\cos^7\\theta\\sin\\theta – \\binom{8}{3}\\sin^3\\theta\\cos^5\\theta + \\binom{8}{5}\\sin^5\\theta\\cos^3\\theta – \\binom{8}{7}\\sin^7\\theta\\cos\\theta \\right) \\\\ \\text{therefore} \\sin8\\theta = \\binom{8}{1}\\cos^7\\theta\\sin\\theta – \\binom{8}{3}\\sin^3\\theta\\cos^5\\theta + \\binom{8}{5}\\sin^5\\theta\\cos^3\\theta – \\binom{8}{7}\\sin^7\\theta\\cos\\theta \\quad \\text{ as required.} \\end{align*} 15. (b) ii. \\begin{align*} \\frac{\\sin8\\theta}{\\sin2\\theta} = \\frac{\\binom{8}{1}\\cos^7\\theta\\sin\\theta – \\binom{8}{3}\\sin^3\\theta\\cos^5\\theta + \\binom{8}{5}\\sin^5\\theta\\cos^3\\theta – \\binom{8}{7}\\sin^7\\theta\\cos\\theta}{2\\sin\\theta\\cos\\theta} \\\\ = \\frac{\\binom{8}{1}}{2}\\cos^6\\theta – \\frac{\\binom{8}{3}}{2}\\cos^4\\theta\\sin^2\\theta + \\frac{\\binom{8}{5}}{2}\\cos^2\\theta\\sin^4\\theta – \\frac{\\binom{8}{7}}{2}\\sin^6\\theta \\\\ = 4(1-\\sin^2\\theta)^3 – 28\\sin^2\\theta(1-\\sin^2\\theta)^2 + 28(1- \\sin^2\\theta)\\sin^4\\theta \\\\ = 4(1 – 10\\sin^2\\theta + 24\\sin^4\\theta – 16\\sin^6\\theta) \\end{align*} 15. (c) i. \\begin{align*} LHS = x^n – 1 – n(x – 1) \\\\ = \\left(\\frac{x^n-1}{x-1}\\right)(x-1) – n(x-1) \\\\ = (x-1)\\left(\\frac{x^n-1}{x-1} – n\\right) \\\\ = (x-1)\\left(1 + x + \\cdots + x^{n-1} -n\\right) \\quad \\text{by GP Formula} \\end{align*} 15. (c) ii. Case 1, when $$x = 1$$, $$LHS = 1^n = 1 \\geq 0 = n(1-1) = 0 = RHS$$ Case 2, when $$0 < x < 1$$ \\begin{gather*} \\text{therefore} \\ x -1 < 0 \\text{ and } x^k < 1 \\text{ for } k \\in \\mathbb{Z}^+ \\\\ \\text{therefore} \\ 1 + x + \\cdots + x^{n-1} – n < 0 \\\\ \\text{therefore} \\ (x -1)(1 + x + \\cdots + x^{n-1} – n) \\geq 0 \\end{gather*} Case 3, when $$x > 1$$ \\begin{gather*} \\text{therefore} \\ x -1", "\\{A_1 e^{i\\theta_1}e^{i\\omega t}\\} + \\operatorname{Re} \\{A_2 e^{i\\theta_2}e^{i\\omega t}\\} \\ &= \\operatorname{Re} \\{A_1 e^{i\\theta_1}e^{i\\omega t} + A_2 e^{i\\theta_2}e^{i\\omega t}\\} \\ &= \\operatorname{Re} \\{(A_1 e^{i\\theta_1} + A_2 e^{i\\theta_2})e^{i\\omega t}\\} \\ &= \\operatorname{Re} \\{(A_3 e^{i\\theta_3})e^{i\\omega t}\\} \\ &= A_3 \\cos(\\omega t + \\theta_3), \\end{align} where: $A_3^2 = (A_1 \\cos{\\theta_1}+A_2 \\cos{\\theta_2})^2 + (A_1 \\sin{\\theta_1}+A_2 \\sin{\\theta_2})^2,$ $\\theta_3 = \\arctan{\\left(\\frac{A_1 \\sin{\\theta_1} + A_2 \\sin{\\theta_2}}{A_1 \\cos{\\theta_1} + A_2 \\cos{\\theta_2}}\\right)},$ or, via the law of cosines on the complex plane (or the trigonometric identity for angle differences): $A_3^2 = A_1^2 + A_2^2 - 2 A_1 A_2 \\cos(180^\\circ - \\Delta\\theta), = A_1^2 + A_2^2 + 2 A_1 A_2 \\cos(\\Delta\\theta),$ where Δθ = θ1 − θ2. A key point is that A3 and θ3 do not depend on ω or t, which is what makes phasor notation possible. The time and frequency dependence can be suppressed and re-inserted into the outcome as long as the only operations used in between are ones that produce another phasor. In angle notation, the operation shown above is written: $A_1 \\angle \\theta_1 + A_2 \\angle \\theta_2 = A_3 \\angle \\theta_3.$ Another way to view addition is that two vectors with coordinates [A1 cos(ωt+θ1), A1 sin(ωt+θ1)] and [A2 cos(ωt+θ2), A2 sin(ωt+θ2)] are added vectorially to produce a resultant vector with coordinates [A3 cos(ωt+θ3), A3 sin(ωt+θ3)]. (see animation) Phasor diagram of three waves in", "&=re^{i\\theta}\\end{align} Beauty. Theorem: Euler's formula $e^{i\\theta}=\\cos\\theta+i\\sin\\theta$ When taken as a function, the expression is sometimes denoted $$\\mathrm{cis}~\\theta$$, from \"cosine plus i sine\" Definition. The sets $$\\mathbb{E}$$ and $$\\mathbb{O}$$ denote the even and odd naturals, respectively: \\begin{align} \\mathbb{E} &= \\{n\\in\\mathbb{N}~|~(\\exists k)[k\\in\\mathbb{N}\\land n = 2k]\\}\\\\ \\mathbb{O} &= \\mathbb{N}\\backslash\\mathbb{E} \\end{align} Proof. Using power series: \\boxed{\\begin{align}\\mathbf{Recall.}\\\\\\\\ e^x &= \\sum_{n=0}^{\\infty}\\frac{x^n}{n!} \\\\\\\\ \\cos x &= \\sum_{n=0}^{\\infty}\\frac{(-1)^nx^{2n}}{2n!} \\\\\\\\ \\sin x &= \\sum_{n=0}^{\\infty}(-1)^n\\frac{x^{2n+1}}{(2n+1)!}~~~~~\\\\& \\end{align}} \\begin{align} e^{i\\theta}&=\\sum_{n=0}^{\\infty}\\frac{(i\\theta)^n}{n!} = \\sum_{n=0}^{\\infty}\\frac{i^n\\theta^n}{n!}\\\\ &= \\sum_{n\\in\\mathbb{E}}\\frac{i^n\\theta^n}{n!}+\\sum_{n\\in\\mathbb{O}}\\frac{i^n\\theta^n}{n!}\\\\ &= \\sum_{m=0}^\\infty\\frac{i^{2m}\\theta^{2m}}{2m!}+\\sum_{m=0}^\\infty\\frac{i^{2m+1}\\theta^{2m+1}}{(2m+1)!}\\\\ &= \\sum_{m=0}^\\infty\\frac{(-1)^m\\theta^{2m}}{2m!} + \\sum_{m=0}^\\infty\\frac{i(-1)^m\\theta^{2m+1}}{(2m+1)!}\\\\ &= \\sum_{m=0}^\\infty\\frac{(-1)^m\\theta^{2m}}{2m!} + i\\sum_{m=0}^\\infty\\frac{(-1)^m\\theta^{2m+1}}{(2m+1)!}\\\\ &= \\cos\\theta + i\\sin\\theta~~\\square \\end{align} ### Complex multiplication in polar form Let us take two complex numbers $$z_1$$ and $$z_2$$ of the form $$z_k = r_ke^{i\\theta_k}$$. Their product takes the form $z_1z_2=r_1r_2e^{i(\\theta_1+\\theta_2)}$ Two important observations can be made from this result. First, when multiplying complex numbers, the modulus of the product is equal to the product of the moduli of the multiplicands: $\\left|\\prod_kz_k\\right|=\\prod_k|z_k|$ Second, the argument of the product is equal to the sum of the arguments of the multiplicands: $\\arg \\prod_kz_k = \\sum_k\\arg z_k$ These properties make the multiplication of many complex numbers incredibly simple. .", "\\theta \\tan \\theta}{\\sin \\theta - \\cos \\theta} + \\frac{\\cos\\theta \\cot \\theta}{\\cos \\theta - \\sin \\theta} \\\\ &= \\frac{\\sin\\theta\\tan\\theta - \\cos\\theta \\cot \\theta}{\\sin \\theta - \\cos \\theta} \\\\&= \\frac{\\sin^3 \\theta - \\cos^3 \\theta}{\\sin \\theta \\cos \\theta (\\sin\\theta - \\cos\\theta)} \\\\&= \\frac{\\sin^2 \\theta + \\sin \\theta \\cos \\theta + \\cos^2\\theta }{\\sin \\theta \\cos \\theta}\\\\&= 1 + \\tan \\theta + \\cot \\theta\\\\&=\\boxed{\\text{D}} \\end{align*} - add comment \\begin{align} \\frac{\\tan(\\theta)}{1-\\cot(\\theta)}+\\frac{\\cot(\\theta)}{1-\\tan(\\theta)} &=\\frac{\\cot(\\theta)-\\tan^2{\\theta}}{1-\\tan(\\theta)}\\\\ &=\\cot(\\theta)\\frac{1-\\tan^3(\\theta)}{1-\\tan(\\theta)}\\\\ &=\\cot(\\theta)+1+\\tan(\\theta) \\end{align} - add comment", "that we assumed all functions are regular enough to perform change of integration and differentiation. Then, we'll have, \\begin{align}\\label{InterM} \\nabla_\\theta \\log p_\\theta(y_{1:n}) = \\int \\frac{\\nabla_\\theta p_\\theta(x_{1:n},y_{1:n})}{p_\\theta(y_{1:n})} \\mbox{d}x_{1:n} \\end{align} Now the trick is to write the following, \\begin{align}\\label{logtrick} \\nabla_\\theta p_\\theta(x_{1:n},y_{1:n}) = \\nabla_\\theta \\log p_\\theta(x_{1:n},y_{1:n}) p_\\theta(x_{1:n},y_{1:n}) \\end{align} If we put \\eqref{logtrick} into \\eqref{InterM}, we obtain, \\begin{align}\\label{InterM2} \\nabla_\\theta \\log p_\\theta(y_{1:n}) = \\int \\nabla_\\theta \\log p_\\theta(x_{1:n},y_{1:n}) \\frac{p_\\theta(x_{1:n},y_{1:n})}{p_\\theta(y_{1:n})}\\mbox{d}x_{1:n} \\end{align} Note the Bayes rule for the last term in the integral, \\begin{align}\\label{Bayes} p(x_{1:n}|y_{1:n}) = \\frac{p(x_{1:n},y_{1:n})}{p(y_{1:n})} \\end{align} then we have, \\begin{align} \\nabla_\\theta \\log p_\\theta(y_{1:n}) = \\int \\nabla_\\theta \\log p_\\theta(x_{1:n},y_{1:n}) p_\\theta(x_{1:n}|y_{1:n})\\mbox{d}x_{1:n} \\end{align} which is known as the Fisher's identity. $\\blacksquare$ Anonymous said... thanks for the notes, this helps! Deniz said... You are welcome! Unknown said... thanks you very much. Anonymous said... your post would be better if you were to mention all 'regularity conditions' in order to be able to reverse the sequence of differentiation/integration (used to move from in-between eq. 2 and 3, to 3 above) Deniz said... Thanks for the feedback. I will add them whenever I find some time.", "\\cos(\\pi - \\theta) &= -\\cos \\theta \\\\ \\tan(\\pi - \\theta) &= -\\tan \\theta \\\\ \\csc(\\pi - \\theta) &= +\\csc \\theta \\\\ \\sec(\\pi - \\theta) &= -\\sec \\theta \\\\ \\cot(\\pi - \\theta) &= -\\cot \\theta \\\\ \\end{align} Periodicity and Shifts Shift by π/2 Shift by π Period for tan and cot Shift by 2π Period for sin, cos, csc and sec \\begin{align} \\sin(\\theta + \\tfrac{\\pi}{2}) &= +\\cos \\theta \\\\ \\cos(\\theta + \\tfrac{\\pi}{2}) &= -\\sin \\theta \\\\ \\tan(\\theta + \\tfrac{\\pi}{2}) &= -\\cot \\theta \\\\ \\csc(\\theta + \\tfrac{\\pi}{2}) &= +\\sec \\theta \\\\ \\sec(\\theta + \\tfrac{\\pi}{2}) &= -\\csc \\theta \\\\ \\cot(\\theta + \\tfrac{\\pi}{2}) &= -\\tan \\theta \\end{align} \\begin{align} \\sin(\\theta + \\pi) &= -\\sin \\theta \\\\ \\cos(\\theta + \\pi) &= -\\cos \\theta \\\\ \\tan(\\theta + \\pi) &= +\\tan \\theta \\\\ \\csc(\\theta + \\pi) &= -\\csc \\theta \\\\ \\sec(\\theta + \\pi) &= -\\sec \\theta \\\\ \\cot(\\theta + \\pi) &= +\\cot \\theta \\\\ \\end{align} \\begin{align} \\sin(\\theta + 2\\pi) &= +\\sin \\theta \\\\ \\cos(\\theta + 2\\pi) &= +\\cos \\theta \\\\ \\tan(\\theta + 2\\pi) &= +\\tan \\theta \\\\ \\csc(\\theta + 2\\pi) &= +\\csc \\theta \\\\ \\sec(\\theta + 2\\pi) &= +\\sec \\theta \\\\ \\cot(\\theta + 2\\pi) &= +\\cot \\theta \\end{align} # Resources ## Active participants The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this division, you can list", "( c - b )\\operatorname{cis}(\\pm B) \\;\\frac{\\sin C}{\\sin A}\\\\[4pt] \\to\\qquad ( a - b )\\sin A &= ( c - b ) \\operatorname{cis}(\\pm B) \\sin C \\\\[4pt] \\to\\quad a \\sin A &= b \\left(\\; \\sin A - \\operatorname{cis}(\\pm B) \\sin C \\;\\right) + c\\operatorname{cis}(\\pm B) \\sin C \\end{align} We need only show that that the multiplier on $b$ has the desired form. \\begin{align} \\sin A - \\operatorname{cis}(\\pm B)\\sin C &= \\sin A - (\\cos B \\pm i \\sin B)\\sin C \\\\ &= (\\sin A - \\cos B \\sin C ) \\mp i \\sin B \\sin C \\\\ &= (\\sin(180^\\circ - B-C) - \\cos B \\sin C ) \\mp i \\sin B \\sin C \\\\ &= (\\sin(B+C) - \\cos B \\sin C ) \\mp i \\sin B \\sin C \\\\ &= \\sin B \\cos C \\mp i \\sin B \\sin C \\\\ &= \\sin B (\\cos C \\mp i \\sin C ) \\\\ &= \\sin B \\operatorname{cis}(\\mp C) \\end{align}", "r^{2}) + i\\arctan\\left(\\frac{r\\sin\\theta}{1 + r\\cos\\theta}\\right),$ we obtain \\begin{aligned} \\sum_{0}^{\\infty} \\binom{m}{n} z^{n} &= \\exp\\{\\tfrac{1}{2}m \\log(1 + 2r\\cos\\theta + r^{2})\\} \\operatorname{Cis} \\left\\{m\\arctan\\left(\\frac{r\\sin\\theta}{1 + r\\cos\\theta}\\right)\\right\\} \\\\ &= (1 + 2r\\cos\\theta + r^{2})^{\\frac{1}{2}m} \\operatorname{Cis} \\left\\{m\\arctan\\left(\\frac{r\\sin\\theta}{1 + r\\cos\\theta}\\right)\\right\\},\\end{aligned} all the inverse tangents lying between $$-\\frac{1}{2}\\pi$$ and $$\\frac{1}{2}\\pi$$. In particular, if we suppose $$\\theta = \\frac{1}{2}\\pi$$, $$z = ir$$, and equate the real and imaginary parts, we obtain \\begin{aligned} 1 – \\binom{m}{2} r^{2} + \\binom{m}{4} r^{4} – \\dots &= (1 + r^{2})^{\\frac{1}{2}m} \\cos(m\\arctan r), \\\\ \\binom{m}{1} r – \\binom{m}{3} r^{3} + \\binom{m}{5} r^{5} – \\dots &= (1 + r^{2})^{\\frac{1}{2}m} \\sin(m\\arctan r).\\end{aligned} 2. Verify the formulae of Ex. 1 when $$m = 1$$, $$2$$, $$3$$. [Of course when $$m$$ is a positive integer the series is finite.] 3. Prove that if $$0 \\leq r < 1$$ then \\begin{aligned} 1 – \\frac{1\\cdot3}{2\\cdot4} r^{2} + \\frac{1\\cdot3\\cdot5\\cdot7}{2\\cdot4\\cdot6\\cdot8} r^{4} – \\dots &= \\sqrt{\\frac{\\sqrt{1 + r^{2}} + 1}{2(1 + r^{2})}}, \\\\ \\frac{1}{2} r – \\frac{1\\cdot3\\cdot5}{2\\cdot4\\cdot6} r^{3} + \\frac{1\\cdot3\\cdot5\\cdot7\\cdot9}{2\\cdot4\\cdot6\\cdot8\\cdot10} r^{5} – \\dots &= \\sqrt{\\frac{\\sqrt{1 + r^{2}} – 1}{2(1 + r^{2})}}.\\end{aligned} [Take $$m = -\\frac{1}{2}$$ in the last two formulae of Ex. 1.] 4. Prove that if $$-\\frac{1}{4}\\pi < \\theta < \\frac{1}{4}\\pi$$ then \\begin{aligned} \\cos m\\theta &= \\cos^{m} \\theta \\left\\{1 – \\binom{m}{2} \\tan^{2} \\theta + \\binom{m}{4} \\tan^{4} \\theta – \\dots\\right\\}, \\\\ \\sin m\\theta &= \\cos^{m} \\theta \\left\\{\\binom{m}{1} \\tan\\theta – \\binom{m}{3} \\tan^{3} \\theta", "\\[Pi]/4]] As can be seen in this simple Manipulate: Manipulate[ ParametricPlot[ 1/Sqrt[2] Sec[ 1/2 ArcTan[Cot[2 (\\[Theta] - \\[Theta]0)]]] {Cos[\\[Theta]], Sin[\\[Theta]]}, {\\[Theta], 0, 2 \\[Pi]}, PlotStyle -> Directive[AbsoluteThickness[4], Orange], PlotRange -> 1.25, ImageSize -> Small, TicksStyle -> 6, Prolog -> {{Thick, Dotted, Circle[]}, Cyan, AbsoluteThickness[4], Line[{{-5, fittedradius4[\\[Theta]0]}, {5, fittedradius4[\\[Theta]0]}}], Line[{{-5, -fittedradius4[\\[Theta]0]}, {5, \\ -fittedradius4[\\[Theta]0]}}], AbsolutePointSize[5], Black, Point[{0, 0}]}], {{\\[Theta]0, 0.5, \"axial rotation\\nsquare cylinder\"}, 0, 10 \\[Pi], ImageSize -> Tiny}, TrackedSymbols :> Manipulate] Based on [1], this is the function for a polygonal ringset: polyRingsetCF = Compile[{{\\[Theta], _Real}, {r, _Real}, {\\[Theta]0, _Real}, {d, \\ _Real}, {n, _Integer}, {\\[Alpha], _Real}}, Module[{t}, t = Sec[2 ArcTan[Cot[1/2 n (\\[Theta] - \\[Theta]0)]]/ n]; {(*part1*){Cos[\\[Pi]/n] Cos[\\[Theta]] t, Cos[\\[Pi]/n] t Sin[\\[Theta]], Sec[\\[Alpha]] Sqrt[-d^2 + r^2 + 2 d Cos[\\[Pi]/n] t Sin[\\[Theta]] - Cos[\\[Pi]/n]^2 t^2 Sin[\\[Theta]]^2] - Cos[\\[Pi]/n] Cos[\\[Theta]] t Tan[\\[Alpha]]}, {(*part 2*)Cos[\\[Pi]/n] Cos[\\[Theta]] t, Cos[\\[Pi]/ n] t Sin[\\[Theta]], -Sec[\\[Alpha]] Sqrt[-d^2 + r^2 + 2 d Cos[\\[Pi]/n] t Sin[\\[Theta]] - Cos[\\[Pi]/n]^2 t^2 Sin[\\[Theta]]^2] - Cos[\\[Pi]/n] Cos[\\[Theta]] t Tan[\\[Alpha]]}}]]; ParametricPlot3D[ polyRingsetCF[t, 1., 0.5, 0., 4, 0.], {t, -\\[Pi], \\[Pi]}, PlotStyle -> {{Red, Tube[.035]}}, PlotPoints -> 500, PerformanceGoal -> \"Quality\", SphericalRegion -> True, PlotRange -> 1.1, Background -> Lighter[Gray, 0.5], ImageSize -> 300, PlotTheme -> \"Marketing\"] If we now use fittedRadius[[Theta]0] as the circumradius for the square cylinder, we can create closed ringsets as a function the axial rotation", "cos } ( A - B ) = \\operatorname { cos } ( A ) \\operatorname { cos } ( B ) - \\operatorname { cos } ( A ) \\operatorname { sin } ( B ) } \\\\ { \\operatorname { tan } ( A - B ) = \\frac { \\operatorname { tan } ( A ) - \\operatorname { tan } ( B ) } { 1 + \\operatorname { tan } ( A ) \\operatorname { tan } ( B ) } } \\end{array} \\right.$ Half-Angel Identities for Sine and Cosine $\\operatorname { sin } ( \\frac { \\theta } { 2 } ) = \\pm \\sqrt { \\frac { 1 - \\operatorname { cos } ( \\theta ) } { 2 } } \\quad \\operatorname { cos } ( \\frac { \\theta } { 2 } ) = \\pm \\sqrt { \\frac { 1 + \\operatorname { cos } \\theta } { 2 } }$ 1. $\\sin( \\frac { 7 \\pi } { 8 } )$ $\\frac{7\\pi}{8} = \\frac{7\\pi/4}{2}$ 1. $\\cos(105º)$ $105º = 210º/2$ or $105º = 45º + 60º$", "and in general terms of powers of sin θ or cos θ the following is true, and can be deduced using De Moivre's formula, Euler's formula and binomial theorem. CosineSine $\\mbox{if }n\\mbox{ is odd}$$\\cos^n\\theta = \\frac{2}{2^n} \\sum_{k=0}^{\\frac{n-1}{2}} \\binom{n}{k} \\cos{((n-2k)\\theta)}$$\\sin^n\\theta = \\frac{2}{2^n} \\sum_{k=0}^{\\frac{n-1}{2}} (-1)^{(\\frac{n-1}{2}-k)} \\binom{n}{k} \\sin{((n-2k)\\theta)}$ $\\mbox{if }n\\mbox{ is even}$$\\cos^n\\theta = \\frac{1}{2^n} \\binom{n}{\\frac{n}{2}} + \\frac{2}{2^n} \\sum_{k=0}^{\\frac{n}{2}-1} \\binom{n}{k} \\cos{((n-2k)\\theta)}$$\\sin^n\\theta = \\frac{1}{2^n} \\binom{n}{\\frac{n}{2}} + \\frac{2}{2^n} \\sum_{k=0}^{\\frac{n}{2}-1} (-1)^{(\\frac{n}{2}-k)} \\binom{n}{k} \\cos{((n-2k)\\theta)}$ ## Product-to-sum and sum-to-product identities The product-to-sum identities or Prosthaphaeresis formulas can be proven by expanding their right-hand sides using the angle addition theorems. See beat (acoustics) for an application of the sum-to-product formulæ. Product-to-sum[22] $\\cos \\theta \\cos \\varphi = {\\cos(\\theta - \\varphi) + \\cos(\\theta + \\varphi) \\over 2}$ $\\sin \\theta \\sin \\varphi = {\\cos(\\theta - \\varphi) - \\cos(\\theta + \\varphi) \\over 2}$ $\\sin \\theta \\cos \\varphi = {\\sin(\\theta + \\varphi) + \\sin(\\theta - \\varphi) \\over 2}$ $\\cos \\theta \\sin \\varphi = {\\sin(\\theta + \\varphi) - \\sin(\\theta - \\varphi) \\over 2}$ Sum-to-product[23] $\\sin \\theta \\pm \\sin \\varphi = 2 \\sin\\left( \\frac{\\theta \\pm \\varphi}{2} \\right) \\cos\\left( \\frac{\\theta \\mp \\varphi}{2} \\right)$ $\\cos \\theta + \\cos \\varphi = 2 \\cos\\left( \\frac{\\theta + \\varphi} {2} \\right) \\cos\\left( \\frac{\\theta - \\varphi}{2} \\right)$ $\\cos \\theta - \\cos \\varphi = -2\\sin\\left( {\\theta + \\varphi \\over 2}\\right) \\sin\\left({\\theta - \\varphi \\over 2}\\right)$ ### Other related identities If x, y, and", "or the multiple-angle formulae. SineCosineTangentCotangent Double-angle formulae[28][29] {\\displaystyle {\\begin{aligned}\\sin(2\\theta )&=2\\sin \\theta \\cos \\theta \\ \\\\&={\\frac {2\\tan \\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[23][30] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )\\!&=\\!-\\sin ^{3}\\theta \\!+\\!3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )\\!&=\\!\\cos ^{3}\\theta \\!-\\!3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )\\!=\\!{\\frac {3\\cot \\theta \\!-\\!\\cot ^{3}\\theta }{1\\!-\\!3\\cot ^{2}\\theta }}}$ Half-angle formulae[26][27] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1\\!-\\!\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or}}\\,\\,\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left(\\mathrm {or} \\,\\,\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[8pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[8pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[10pt]\\tan {\\frac {\\eta +\\theta }{2}}\\!&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[8pt]\\tan \\left(\\!{\\frac {\\theta }{2}}\\!+\\!{\\frac {\\pi }{4}}\\!\\right)\\!&=\\!\\sec \\theta \\!+\\!\\tan \\theta \\\\[8pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {|1-\\tan {\\frac {\\theta }{2}}|}{|1+\\tan {\\frac {\\theta }{2}}|}}\\\\[8pt]\\tan {\\frac {\\theta }{2}}\\!&=\\!{\\frac {\\tan", "e_2,(T_\\theta)^*e_1\\rangle,$$ we have $(T_\\theta)^*e_1=\\cos\\theta e_1+\\sin\\theta e_2$. Similarly, note that $$\\sin\\theta=\\langle T_\\theta e_1,e_2\\rangle=\\langle e_1,(T_\\theta)^*e_2\\rangle,$$ $$-\\cos\\theta=\\langle T_\\theta e_2,e_2\\rangle=\\langle e_2,(T_\\theta)^*e_2\\rangle,$$ we have $(T_\\theta)^*e_2=\\sin\\theta e_1-\\cos\\theta e_2$. Hence $T_\\theta=(T_\\theta)^*$, which implies that $T_\\theta$ is self-adjoint. Also note that \\begin{align*}& (T_\\theta)^2e_1=T_\\theta(\\cos\\theta e_1+\\sin\\theta e_2)\\\\ =&\\cos\\theta(\\cos\\theta e_1+\\sin\\theta e_2)+\\sin\\theta(\\sin \\theta e_1-\\cos\\theta e_2)\\\\=&(\\cos^2\\theta+\\sin^2\\theta)e_1=e_1,\\end{align*} \\begin{align*}& (T_\\theta)^2e_2=T_\\theta(\\sin\\theta e_1-\\cos\\theta e_2)\\\\ =&\\sin\\theta(\\cos\\theta e_1+\\sin\\theta e_2)-\\cos\\theta(\\sin \\theta e_1-\\cos\\theta e_2)\\\\=&(\\cos^2\\theta+\\sin^2\\theta)e_2=e_2,\\end{align*}we have $(T_\\theta)^2=\\mathrm{id}$. Therefore we have infinitely many self-adjoint operators as the square root of $\\mathrm{id}$. The construction comes from the following idea. Let $T$ be self-adjoint such that $T^2=\\mathrm{id}$. Let $\\begin{pmatrix}a & b\\\\ c & d\\end{pmatrix}$ be the matrix with respect to an orthonormal basis of $\\mb F^2$. Since $T$ is self-adjoint, we have $$\\begin{pmatrix}a & b\\\\ c & d\\end{pmatrix}=\\begin{pmatrix}\\bar a & \\bar c\\\\ \\bar b & \\bar d\\end{pmatrix}.$$ Hence $a,d\\in\\mb R$ and $b=\\bar c$. If $T^2=\\mathrm{id}$, then we have $a^2+bc=1$, $ab+bd=0$, and $d^2+bc=1$. Hence we can take $a=-d$ and $b=c$. Then $a^2+b^2=1$. Take $a=\\cos\\theta$ and $b=\\sin\\theta$, we get the construction. 10. Solution: “(a)$\\Longrightarrow$(b)” If $S$ is an isometry, so is $S^*$ by 7.42 (g)$\\iff$(a). It follows from 7.42 (a)$\\iff$(b) that $$\\langle S^*u,S^*v\\rangle =\\langle u,v\\rangle$$ for all $u,v\\in V$. “(b)$\\Longrightarrow$(c)” If $e_1,\\cdots,e_m$ is an orthonormal basis of $V$, we have $\\langle e_i,e_j\\rangle =\\delta_{ij}$. Since $$\\langle S^*u,S^*v\\rangle =\\langle u,v\\rangle$$ for all $u,v\\in V$, we have$\\langle S^*e_i,S^*e_j\\rangle=\\langle e_i,e_j\\rangle =\\delta_{ij}.$Hence $S^*e_1,\\cdots,S^*e_m$ is an orthornormal basis of $V$. “(c)$\\Longrightarrow$(d)”", "Therefore, with an eye toward a telescoping sum, I started with$$ \\begin{align} ... 12 Hint: use the fact that $\\cot{x}-2\\cot{2 x}=\\tan{x}$ 12 $$\\tan rx=\\frac{\\binom r1\\tan x-\\binom r3\\tan^3x+\\cdots}{1-\\binom r2\\tan^2x+\\cdots}\\text{ (Proof below) }$$ Now, if $\\tan rx=\\tan r\\theta\\implies rx=n\\pi+r\\theta$ where $n$ is any integer $\\implies x=n\\frac{\\pi}r+\\theta$ If $r$ is even, $=2m$(say) $$\\tan2m\\theta=\\tan2mx=\\frac{\\binom {2m}1\\tan x-\\binom {2m}3\\tan^3x+\\cdots+\\binom ... 12 Since:$$\\arctan\\frac{3n^2}{2n^4-1}=\\arctan\\frac{1}{n^2}+\\arctan\\frac{1}{2n^2}$$then:$$\\sum_{n=1}^{+\\infty}\\arctan\\frac{3n^2}{2n^4-1} = \\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{n^2}\\right)+\\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{2n^2}\\right).\\tag{1} $$Since, by the Weierstrass product for the \\sinh function,$$\\frac{\\sinh(\\pi z)}{\\pi ... 11 Let $$f(x)=\\prod_{n=0}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}\\tag{1}$$ and $$g(x)=\\prod_{n=0}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}\\tag{2}$$ Then \\begin{align} f(x)\\,g(x) &=\\prod_{n=0}^\\infty\\left(1-x^{2^{n+1}}\\right)^{1/2^n}\\\\ &=\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{2/2^n}\\\\ &=\\left(\\frac{f(x)}{1-x}\\right)^2\\tag{3} \\end{align} from ... 11 Viète's formula gives: $$\\prod_{k\\geq 1} \\cos\\left(\\frac{x}{2^k}\\right)=\\frac{\\sin x}{x}\\Rightarrow \\prod_{k\\geq 1} \\sec\\left(\\frac{x}{2^k}\\right)=\\frac{x}{\\sin x}\\Rightarrow\\sum_{k\\geq 1} \\ln \\sec\\left(\\frac{x}{2^k}\\right)=\\ln x-\\ln \\sin x$$ Differentiate twice and you have your result. 11 It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function $\\operatorname{dn}(z,k)$ has Fourier series $$\\operatorname{dn}(z,k)=\\frac{\\pi}{2K}\\left[1+2\\sum_{n=1}^{\\infty}\\frac{\\cos n\\pi\\frac{z}{K}}{\\cosh n \\pi ... 10 Actually, this sum converges for every \\alpha>0. Step I. For every x\\in\\mathbb R, the sequence s_n=\\sum_{k=1}^n\\sin kx is bounded. Indeed, if x=m\\pi, then s_n=0. If x\\ne m\\pi, then \\sin(x/2)\\ne 0, and$$ s_n=\\sum_{k=1}^n\\sin kx=\\mathrm{Im}\\left(\\mathrm{e}^{xi}+\\mathrm{e}^{2xi}+\\cdots\\mathrm{e}^{nxi}\\right)= ... 9 Use: $$\\sin \\frac{\\theta}{2} \\cdot \\cos(k \\theta) = \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k+\\frac{1}{2}\\right)\\theta\\right)}_{f_{k+1}} - \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k-\\frac{1}{2}\\right)\\theta\\right)}_{f_{k}}$$ Thus $$\\begin{eqnarray} \\sin \\frac{\\theta}{2} \\cdot \\sum_{k=1}^n \\cos(k \\theta) &=& \\sum_{k=1}^n \\left(f_{k+1} - ... 9 Recall", "\\\\ C(\\theta,\\phi) &= -\\sin\\theta \\cos\\theta (D \\cos\\phi + E \\sin\\phi) + F \\sin^2\\theta, \\end{align} where $$D,E,F$$ are constants. In summary, we have \\begin{align} K_t &= A \\left( 1 - \\frac{R_{\\text{S}}}{r} \\right), \\\\ K_r &= 0, \\\\ K_\\theta &= (-D \\sin\\phi + E \\cos\\phi) \\, r^2, \\\\ K_\\phi &= [-\\sin\\theta \\cos\\theta (D \\cos\\phi + E \\sin\\phi) + F \\sin^2\\theta] \\, r^2. \\end{align} The general solution of the Killing equations of the Schwarzschild metric will be $$$$\\begin{split} K &= g^{\\mu\\nu} K_\\mu \\partial_\\nu \\\\ &= g^{tt} K_t \\frac{\\partial}{\\partial t} + g^{\\theta\\theta} K_\\theta \\frac{\\partial}{\\partial \\theta} + g^{\\phi\\phi} K_\\phi \\frac{\\partial}{\\partial \\phi} \\\\ &= - A \\frac{\\partial}{\\partial t} +(-D \\sin\\phi + E \\cos\\phi) \\frac{\\partial}{\\partial \\theta} + [ - \\cot\\theta (D \\cos\\phi + E \\sin\\phi) + F ] \\frac{\\partial}{\\partial \\phi} \\\\ &= - A L_{(0)} + D L_{(1)} + E L_{(2)} + F L_{(3)}, \\end{split}$$$$ where \\begin{align} L_{(0)} &= \\frac{\\partial}{\\partial t}, \\\\ L_{(1)} &= -\\sin\\phi \\frac{\\partial}{\\partial \\theta} - \\cot\\theta \\cos\\phi \\frac{\\partial}{\\partial \\phi}, \\\\ L_{(2)} &= \\cos\\phi \\frac{\\partial}{\\partial \\theta} - \\cot\\theta \\sin\\phi \\frac{\\partial}{\\partial \\phi}, \\\\ L_{(3)} &= \\frac{\\partial}{\\partial \\phi} \\end{align} form a basis of the Lie algebra of the Killing fields of Schwarzschild metric. $$L_{(0)}$$ is a timelike Killing vector field that is orthogonal to a foliation of spacelike hypersurfaces, representing a static spacetime. $$L_{(1)}, L_{(2)}, L_{(3)}$$ are Killing fields of a 2-sphere, representing a spacetime", "}{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[16][24] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )&=-\\sin ^{3}\\theta +3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )&=\\cos ^{3}\\theta -3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )={\\frac {3\\cot \\theta -\\cot ^{3}\\theta }{1-3\\cot ^{2}\\theta }}}$ Half-angle formulae[20][21] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1-\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[3pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[3pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[5pt]\\tan {\\frac {\\eta +\\theta }{2}}&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[5pt]\\tan \\left({\\frac {\\theta }{2}}+{\\frac {\\pi }{4}}\\right)&=\\sec \\theta +\\tan \\theta \\\\[5pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {\\left|1-\\tan {\\frac {\\theta }{2}}\\right|}{\\left|1+\\tan {\\frac {\\theta }{2}}\\right|}}\\\\[5pt]\\tan {\\frac {\\theta }{2}}&={\\frac {\\tan \\theta }{1+{\\sqrt {1+\\tan ^{2}\\theta }}}}\\\\&{\\text{for }}\\theta \\in \\left(-{\\tfrac {\\pi }{2}},{\\tfrac {\\pi }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cot {\\frac", "\\cos \\theta \\csc \\theta}{-\\cot^2 \\theta} \\\\[5ex] \\csc \\theta = \\dfrac{1}{\\sin \\theta} ... \\:Reciprocal\\: Identity \\\\[5ex] \\tan \\theta = \\dfrac{1}{\\cot \\theta} ... \\:From \\:Reciprocal\\: Identity \\\\[5ex] \\cot \\theta = \\dfrac{\\cos \\theta}{\\sin \\theta} ... \\:Quotient\\: Identity \\\\[5ex] = \\dfrac{\\cos \\theta - \\cos \\theta * \\dfrac{1}{\\sin \\theta}}{-\\cot^2 \\theta} \\\\[5ex] = \\dfrac{\\cos \\theta - {\\dfrac{\\cos \\theta}{\\sin \\theta}}}{-\\cot^2 \\theta} \\\\[5ex] = \\dfrac{\\cos \\theta - \\cot \\theta}{-\\cot^2 \\theta} \\\\[5ex] = \\dfrac{\\cos \\theta}{-\\cot^2 \\theta} - \\dfrac{\\cot \\theta}{-\\cot^2 \\theta} \\\\[5ex] = \\dfrac{-\\cos \\theta}{\\cot^2 \\theta} - \\dfrac{\\cot \\theta}{-\\cot^2 \\theta} \\\\[5ex] = (-\\cos \\theta \\div \\cot^2 \\theta) + \\dfrac{1}{\\cot \\theta} \\\\[5ex] = (-\\cos \\theta \\div \\dfrac{\\cos^2 \\theta}{\\sin^2 \\theta}) + \\tan \\theta \\\\[5ex] = (-\\cos \\theta * \\dfrac{\\sin^2 \\theta}{\\cos^2 \\theta}) + \\tan \\theta \\\\[5ex] = \\dfrac{-\\sin^2 \\theta}{\\cos \\theta} + \\tan \\theta \\\\[5ex] = -\\sin \\theta * \\dfrac{\\sin \\theta}{\\cos \\theta} + \\tan \\theta \\\\[5ex] = -\\sin \\theta \\tan \\theta + \\tan \\theta \\\\[3ex] = \\tan \\theta - \\sin \\theta \\tan \\theta \\\\[3ex] = \\tan \\theta(1 - \\sin \\theta)$ (8.) ACT For all real values of $x$, which of the following equations is true? $F.\\:\\: \\sin(7x) + \\cos(7x) = 7 \\\\[3ex] G.\\:\\: \\sin(7x) + \\cos(7x) = 1 \\\\[3ex] H.\\:\\: 7\\sin(7x) + 7\\cos(7x) = 14 \\\\[3ex] J.\\:\\: \\sin^2(7x) + \\cos^2(7x) = 7 \\\\[3ex] K.\\:\\: \\sin^2(7x) + \\cos^2(7x) = 1$ $F.\\:\\: -\\sin(-x^\\circ) = -1 * \\sin(-x) \\\\[3ex] \\sin(-x) = -\\sin x ...Odd\\:\\: Identity" ]

[ "+0 # Really want this one solved +1 49 4 +174 Let $$\\theta$$ be an acute angle such that $$\\sin 5 \\theta = \\sin^5 \\theta$$ Compute $$\\tan 2 \\theta$$ this isn't my homework or anything, really curious how to do it Jun 6, 2021 #1 +32236 +4 Here's a helping hand: See if you can take it from here. Jun 6, 2021 #4 +113569 0 That is where I got lost. Melody Jun 7, 2021 #2 +174 +2 because of DeMoivre's Theorem: \\begin{align*} \\operatorname{cis} n \\theta &= (\\operatorname{cis} \\theta)^n \\\\ &= (\\cos \\theta + i \\sin \\theta)^n \\\\ &= \\cos^n \\theta + \\binom{n}{1} i \\cos^{n - 1} \\theta \\sin \\theta - \\binom{n}{2} \\cos^{n - 2} \\theta \\sin^2 \\theta - \\binom{n}{3} i \\cos^{n - 3} \\theta \\sin^3 \\theta + \\dotsb. \\end{align*} \\begin{align*} \\cos n \\theta &= \\cos^n \\theta - \\binom{n}{2} \\cos^{n - 2} \\theta \\sin^2 \\theta + \\binom{n}{4} \\cos^{n - 4} \\theta \\sin^4 \\theta - \\dotsb, \\\\ \\sin n \\theta &= \\binom{n}{1} \\cos^{n - 1} \\theta \\sin \\theta - \\binom{n}{3} \\cos^{n - 3} \\theta \\sin^3 \\theta + \\binom{n}{5} \\cos^{n - 5} \\theta \\sin^5 \\theta - \\dotsb. \\end{align*} \\begin{align*} \\sin 5 \\theta &= \\binom{5}{1} \\cos^4 \\theta \\sin \\theta - \\binom{5}{3} \\cos^2 \\theta \\sin^3 \\theta + \\binom{5}{5} \\sin^5 \\theta \\\\ &= 5 \\cos^4 \\theta \\sin \\theta - 10 \\cos^2 \\theta \\sin^3", "Indeed, if x=m\\pi, then s_n=0. If x\\ne m\\pi, then \\sin(x/2)\\ne 0, and$$ s_n=\\sum_{k=1}^n\\sin kx=\\mathrm{Im}\\left(\\mathrm{e}^{xi}+\\mathrm{e}^{2xi}+\\cdots\\mathrm{e}^{nxi}\\right)= ... 10 The main building block of our solution will be the formula \\begin{align*}\\prod_{k=0}^{N-1}\\left(x-e^{\\frac{2k i\\pi}{N}}\\right)=x^N-1.\\tag{0} \\end{align*} It will be convenient to rewrite (0) for odd $N=2n+1$ in the form \\begin{align*} \\prod_{k=1}^{n}\\left[x^2+1-2x\\cos\\frac{\\pi k}{2n+1}\\right]=\\frac{x^{2n+1}-1}{x-1}. \\tag{1} \\end{align*} Replacing therein ... 9 $$\\tan rx=\\frac{\\binom r1\\tan x-\\binom r3\\tan^3x+\\cdots}{1-\\binom r2\\tan^2x+\\cdots}\\text{ (Proof below) }$$ Now, if $\\tan rx=\\tan r\\theta\\implies rx=n\\pi+r\\theta$ where $n$ is any integer $\\implies x=n\\frac{\\pi}r+\\theta$ If $r$ is even, $=2m$(say) $$\\tan2m\\theta=\\tan2mx=\\frac{\\binom {2m}1\\tan x-\\binom {2m}3\\tan^3x+\\cdots+\\binom ... 9 Hint: consider$$x_{n}=a^n+b^n+c^nx_{n+3}=(a+b+c)x_{n+2}-(ab+bc+ac)x_{n+1}+abcx_{n}$$let a=\\tan^2{20}=\\tan^2{\\dfrac{\\pi}{9}},b=\\tan^2{40}=\\tan^2{\\dfrac{2\\pi}{9}},c=\\tan^2{80}=\\tan^2{\\dfrac{4\\pi}{9}} and you can add d=\\tan^2{\\dfrac{3\\pi}{9}}=3 It is easy to find$$a+b+c=30,ab+bc+ac=27, abc=3$$because you can see here lab bhattacharjee full ... 9 Recall that \\cos(x)=(e^{ix}+e^{-ix})/2 and \\sin(x)=(e^{ix}-e^{-ix})/2i, so your sum can be rewritten as$$\\sum_{j=1}^k\\biggl[\\frac{e^{j\\pi i/k}+e^{-j\\pi i/k}}2\\biggr]^n\\,\\biggl[\\frac{e^{nj\\pi i/k}-e^{-nj\\pi i/k}}{2i}\\biggr]\\,.$$Applying binomial theorem on the first factor (of each summand) your sum becomes$$\\begin{align*} ... 8 We use the Poisson summation formula. Define $f(x) \\equiv \\sin(\\pi x) / (\\pi x)$. Then the sum we are trying to solve is $$g(y) = \\sum_{n=-\\infty}^\\infty f(n-y) \\, .$$ The Poisson summation formula converts the sum over values of $f$ to a sum over values of the Fourier transform of $f$. Poisson summation Note that $g(y)$ is periodic with period $1$. The ... 8 Heuristics: Say $a_n$ is near zero but not zero, then $\\sin(x)=x-\\frac16x^3+o(x^3)$ when $x\\to0$ hence $$a_{n+1}=a_n-\\frac16a_n^3+o(a_n^3).$$ If", "being exploited. Therefore, with an eye toward a telescoping sum, I started with$$ \\begin{align} ... 12 Since: $$\\arctan\\frac{3n^2}{2n^4-1}=\\arctan\\frac{1}{n^2}+\\arctan\\frac{1}{2n^2}$$ then: $$\\sum_{n=1}^{+\\infty}\\arctan\\frac{3n^2}{2n^4-1} = \\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{n^2}\\right)+\\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{2n^2}\\right).\\tag{1}$$ Since, by the Weierstrass product for the $\\sinh$ function, $$\\frac{\\sinh(\\pi z)}{\\pi ... 12 Hint: use the fact that \\cot{x}-2\\cot{2 x}=\\tan{x} 12$$\\tan rx=\\frac{\\binom r1\\tan x-\\binom r3\\tan^3x+\\cdots}{1-\\binom r2\\tan^2x+\\cdots}\\text{ (Proof below) }$$Now, if \\tan rx=\\tan r\\theta\\implies rx=n\\pi+r\\theta where n is any integer \\implies x=n\\frac{\\pi}r+\\theta If r is even, =2m(say)$$\\tan2m\\theta=\\tan2mx=\\frac{\\binom {2m}1\\tan x-\\binom {2m}3\\tan^3x+\\cdots+\\binom ... 11 It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function $\\operatorname{dn}(z,k)$ has Fourier series $$\\operatorname{dn}(z,k)=\\frac{\\pi}{2K}\\left[1+2\\sum_{n=1}^{\\infty}\\frac{\\cos n\\pi\\frac{z}{K}}{\\cosh n \\pi ... 11 Viète's formula gives:$$\\prod_{k\\geq 1} \\cos\\left(\\frac{x}{2^k}\\right)=\\frac{\\sin x}{x}\\Rightarrow \\prod_{k\\geq 1} \\sec\\left(\\frac{x}{2^k}\\right)=\\frac{x}{\\sin x}\\Rightarrow\\sum_{k\\geq 1} \\ln \\sec\\left(\\frac{x}{2^k}\\right)=\\ln x-\\ln \\sin x$$Differentiate twice and you have your result. 11 Let$$ f(x)=\\prod_{n=0}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}\\tag{1} $$and$$ g(x)=\\prod_{n=0}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}\\tag{2} $$Then$$ \\begin{align} f(x)\\,g(x) &=\\prod_{n=0}^\\infty\\left(1-x^{2^{n+1}}\\right)^{1/2^n}\\\\ &=\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{2/2^n}\\\\ &=\\left(\\frac{f(x)}{1-x}\\right)^2\\tag{3} \\end{align} $$from ... 10 Actually, this sum converges for every \\alpha>0. Step I. For every x\\in\\mathbb R, the sequence s_n=\\sum_{k=1}^n\\sin kx is bounded. Indeed, if x=m\\pi, then s_n=0. If x\\ne m\\pi, then \\sin(x/2)\\ne 0, and$$ s_n=\\sum_{k=1}^n\\sin kx=\\mathrm{Im}\\left(\\mathrm{e}^{xi}+\\mathrm{e}^{2xi}+\\cdots\\mathrm{e}^{nxi}\\right)= ... 9 Recall that $\\cos(x)=(e^{ix}+e^{-ix})/2$ and $\\sin(x)=(e^{ix}-e^{-ix})/2i$, so your sum can be rewritten as $$\\sum_{j=1}^k\\biggl[\\frac{e^{j\\pi i/k}+e^{-j\\pi i/k}}2\\biggr]^n\\,\\biggl[\\frac{e^{nj\\pi i/k}-e^{-nj\\pi i/k}}{2i}\\biggr]\\,.$$ Applying binomial theorem on the first factor (of each summand) your", "}{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[16][24] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )&=-\\sin ^{3}\\theta +3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )&=\\cos ^{3}\\theta -3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )={\\frac {3\\cot \\theta -\\cot ^{3}\\theta }{1-3\\cot ^{2}\\theta }}}$ Half-angle formulae[20][21] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1-\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[3pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[3pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[5pt]\\tan {\\frac {\\eta +\\theta }{2}}&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[5pt]\\tan \\left({\\frac {\\theta }{2}}+{\\frac {\\pi }{4}}\\right)&=\\sec \\theta +\\tan \\theta \\\\[5pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {\\left|1-\\tan {\\frac {\\theta }{2}}\\right|}{\\left|1+\\tan {\\frac {\\theta }{2}}\\right|}}\\\\[5pt]\\tan {\\frac {\\theta }{2}}&={\\frac {\\tan \\theta }{1+{\\sqrt {1+\\tan ^{2}\\theta }}}}\\\\&{\\text{for }}\\theta \\in \\left(-{\\tfrac {\\pi }{2}},{\\tfrac {\\pi }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cot {\\frac", "Therefore, with an eye toward a telescoping sum, I started with$$ \\begin{align} ... 12 Hint: use the fact that $\\cot{x}-2\\cot{2 x}=\\tan{x}$ 12 $$\\tan rx=\\frac{\\binom r1\\tan x-\\binom r3\\tan^3x+\\cdots}{1-\\binom r2\\tan^2x+\\cdots}\\text{ (Proof below) }$$ Now, if $\\tan rx=\\tan r\\theta\\implies rx=n\\pi+r\\theta$ where $n$ is any integer $\\implies x=n\\frac{\\pi}r+\\theta$ If $r$ is even, $=2m$(say) $$\\tan2m\\theta=\\tan2mx=\\frac{\\binom {2m}1\\tan x-\\binom {2m}3\\tan^3x+\\cdots+\\binom ... 12 Since:$$\\arctan\\frac{3n^2}{2n^4-1}=\\arctan\\frac{1}{n^2}+\\arctan\\frac{1}{2n^2}$$then:$$\\sum_{n=1}^{+\\infty}\\arctan\\frac{3n^2}{2n^4-1} = \\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{n^2}\\right)+\\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{2n^2}\\right).\\tag{1} $$Since, by the Weierstrass product for the \\sinh function,$$\\frac{\\sinh(\\pi z)}{\\pi ... 11 Let $$f(x)=\\prod_{n=0}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}\\tag{1}$$ and $$g(x)=\\prod_{n=0}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}\\tag{2}$$ Then \\begin{align} f(x)\\,g(x) &=\\prod_{n=0}^\\infty\\left(1-x^{2^{n+1}}\\right)^{1/2^n}\\\\ &=\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{2/2^n}\\\\ &=\\left(\\frac{f(x)}{1-x}\\right)^2\\tag{3} \\end{align} from ... 11 Viète's formula gives: $$\\prod_{k\\geq 1} \\cos\\left(\\frac{x}{2^k}\\right)=\\frac{\\sin x}{x}\\Rightarrow \\prod_{k\\geq 1} \\sec\\left(\\frac{x}{2^k}\\right)=\\frac{x}{\\sin x}\\Rightarrow\\sum_{k\\geq 1} \\ln \\sec\\left(\\frac{x}{2^k}\\right)=\\ln x-\\ln \\sin x$$ Differentiate twice and you have your result. 11 It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function $\\operatorname{dn}(z,k)$ has Fourier series $$\\operatorname{dn}(z,k)=\\frac{\\pi}{2K}\\left[1+2\\sum_{n=1}^{\\infty}\\frac{\\cos n\\pi\\frac{z}{K}}{\\cosh n \\pi ... 10 Actually, this sum converges for every \\alpha>0. Step I. For every x\\in\\mathbb R, the sequence s_n=\\sum_{k=1}^n\\sin kx is bounded. Indeed, if x=m\\pi, then s_n=0. If x\\ne m\\pi, then \\sin(x/2)\\ne 0, and$$ s_n=\\sum_{k=1}^n\\sin kx=\\mathrm{Im}\\left(\\mathrm{e}^{xi}+\\mathrm{e}^{2xi}+\\cdots\\mathrm{e}^{nxi}\\right)= ... 9 Use: $$\\sin \\frac{\\theta}{2} \\cdot \\cos(k \\theta) = \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k+\\frac{1}{2}\\right)\\theta\\right)}_{f_{k+1}} - \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k-\\frac{1}{2}\\right)\\theta\\right)}_{f_{k}}$$ Thus $$\\begin{eqnarray} \\sin \\frac{\\theta}{2} \\cdot \\sum_{k=1}^n \\cos(k \\theta) &=& \\sum_{k=1}^n \\left(f_{k+1} - ... 9 Recall", "$$\\operatorname{E}\\left(\\hat{\\theta}\\right)$$ and $$\\operatorname{V}\\left(\\hat{\\theta}\\right)$$ > theta <- pi / 4 > n <- 10 > e_theta_hat <- 0 > e_theta_hat_square <- 0 > for (m1 in 0:(n-1)) + { + for (m2 in 0:(n-m1-1)) + { + e_theta_hat <- e_theta_hat + atan(sqrt((m1+m2)/(n-m1-m2)))*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + e_theta_hat_square <- e_theta_hat_square + (atan(sqrt((m1+m2)/(n-m1-m2))))^2*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + } + } > e_theta_hat [1] 0.7662692 > theta [1] 0.7853982 > > e_theta_hat_square - (e_theta_hat)^2 [1] 0.04779768 > theta <- pi / 4 > n <- 1000 > e_theta_hat <- 0 > e_theta_hat_square <- 0 > for (m1 in 0:(n-1)) + { + for (m2 in 0:(n-m1-1)) + { + e_theta_hat <- e_theta_hat + atan(sqrt((m1+m2)/(n-m1-m2)))*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + e_theta_hat_square <- e_theta_hat_square + (atan(sqrt((m1+m2)/(n-m1-m2))))^2*dbinom(m1, n, 0.5*(sin(theta))^2)*dbinom(m2, n, 0.5*(sin(theta))^2) + } + } > e_theta_hat [1] 0.7853983 > theta [1] 0.7853982 > > e_theta_hat_square - (e_theta_hat)^2 [1] 0.0003755647 >", "# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 74 $-2-2\\sqrt{3}i$ #### Work Step by Step Conversion of a complex number from rectangular form to polar form: For converting the complex number in rectangular form $z=a+ib$, the polar form of the complex number is given by $z=r\\left( \\cos \\theta +i\\sin \\theta \\right)$ Where \\begin{align} & r=\\sqrt{{{a}^{2}}+{{b}^{2}}} \\\\ & \\theta ={{\\tan }^{-1}}\\left( \\frac{b}{a} \\right) \\end{align} DeMoivre’s Theorem: For any complex number $z=r\\left( \\cos \\theta +i\\sin \\theta \\right)$, if n is a positive integer $\\left( z>0 \\right)$ then, \\begin{align} & {{z}^{n}}={{\\left[ r\\left( \\cos \\theta +i\\sin \\theta \\right) \\right]}^{n}} \\\\ & ={{r}^{n}}\\left( \\cos n\\theta +i\\sin n\\theta \\right) \\end{align} So, at first we will convert $1-i\\sqrt{3}$ in the polar form. Here, \\begin{align} & r=\\sqrt{{{1}^{2}}+{{\\left( -\\sqrt{3} \\right)}^{2}}} \\\\ & =2 \\end{align} And, \\begin{align} & \\theta ={{\\tan }^{-1}}\\left( -\\frac{\\sqrt{3}}{1} \\right) \\\\ & \\theta ={{\\tan }^{-1}}\\left( -\\sqrt{3} \\right) \\\\ & \\theta =300{}^\\circ \\end{align} So, $1-i\\sqrt{3}=2\\left( \\cos 300{}^\\circ +i\\sin 300{}^\\circ \\right)$. Therefore, ${{\\left( 1-i\\sqrt{3} \\right)}^{2}}={{\\left[ 2\\left( \\cos 300{}^\\circ +i\\sin 300{}^\\circ \\right) \\right]}^{2}}$. So, \\begin{align} & {{\\left[ 2\\left( \\cos 300{}^\\circ +i\\sin 300{}^\\circ \\right) \\right]}^{2}}={{2}^{2}}\\left[ \\cos 2\\left( 300{}^\\circ \\right)+i\\sin 2\\left( 300{}^\\circ \\right) \\right] \\\\ & =4\\left( \\cos 600{}^\\circ +i\\sin 600{}^\\circ \\right) \\\\ & =4\\left\\{ \\cos \\left( 720-120 \\right){}^\\circ +i\\sin \\left( 720-120 \\right){}^\\circ \\right\\} \\\\ & =4\\left( -\\frac{1}{2}-i\\frac{\\sqrt{3}}{2} \\right) \\end{align} That is,", "\\cos \\theta \\ \\\\&={\\frac {2\\tan \\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[24][31] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )&=-\\sin ^{3}\\theta +3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )&=\\cos ^{3}\\theta -3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )={\\frac {3\\cot \\theta -\\cot ^{3}\\theta }{1-3\\cot ^{2}\\theta }}}$ Half-angle formulae[27][28] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1-\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where }}B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or }}\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[3pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[3pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[5pt]\\tan {\\frac {\\eta +\\theta }{2}}&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[5pt]\\tan \\left({\\frac {\\theta }{2}}+{\\frac {\\pi }{4}}\\right)&=\\sec \\theta +\\tan \\theta \\\\[5pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {\\left|1-\\tan {\\frac {\\theta }{2}}\\right|}{\\left|1+\\tan {\\frac {\\theta }{2}}\\right|}}\\\\[5pt]\\tan {\\frac {\\theta }{2}}&={\\frac {\\tan \\theta }{1+{\\sqrt {1+\\tan ^{2}\\theta }}}}\\\\&{\\text{for }}\\theta \\in \\left(-{\\tfrac {\\pi", "due to the phase shifts and two-bounce reflections between the spheres. #### Bugfix in ART# During the initial comparison, we discovered a subtle bug in ART’s implementation of the Fresnel equations. In their system, these are implemented based on an alternative formulation [WW12], originally from [WK90]: \\begin{split}\\begin{align} F_{\\bot} &= \\frac{a^{2} + b^{2} - 2 a \\cos\\theta + \\cos^{2}\\theta}{a^{2} + b^{2} + 2 a \\cos\\theta + \\cos^{2}\\theta} \\\\ F_{\\parallel} &= \\frac{a^{2} + b^{2} - 2 a \\sin\\theta \\tan\\theta + \\sin^{2}\\theta \\tan^{2}\\theta}{a^{2} + b^{2} + 2 a \\sin\\theta \\tan\\theta + \\sin^{2}\\theta \\tan^{2}\\theta} F_{\\bot} \\\\ \\tan\\delta_{\\bot} &= \\frac{2 b \\cos\\theta}{\\cos^{2}\\theta - a^{2} - b^{2}} \\label{eq:art_fresenl_tan_s} \\\\ \\tan\\delta_{\\parallel} &= \\frac{2 \\cos\\theta \\left[ (n^{2} - k^{2})b - 2 n k a \\right]}{(n^{2} + k^{2})^{2} \\cos^{2}\\theta - a^{2} - b^{2}} \\label{eq:art_fresenl_tan_p} \\end{align}\\end{split} with \\begin{split}\\begin{align} \\eta = n + i k \\\\ 2 a^{2} &= \\sqrt{(n^{2} - k^{2} - \\sin^{2}\\theta)^{2} + 4 n^{2}k^{2}} + n^{2} - k^{2} - \\sin^{2}\\theta \\\\ 2 b^{2} &= \\sqrt{(n^{2} - k^{2} - \\sin^{2}\\theta)^{2} + 4 n^{2}k^{2}} - n^{2} + k^{2} + \\sin^{2}\\theta \\end{align}\\end{split} The Mueller matrix then follows this shape $\\begin{split}$$\\begin{bmatrix} A & B & 0 & 0 \\\\ B & A & 0 & 0 \\\\ 0 & 0 & C & S \\\\ 0 & 0 & -S & C \\end{bmatrix}$$\\end{split}$ using \\begin{split}\\begin{align} A &= \\frac{F_{\\bot} + F_{\\parallel}}{2}", "or the multiple-angle formulae. SineCosineTangentCotangent Double-angle formulae[28][29] {\\displaystyle {\\begin{aligned}\\sin(2\\theta )&=2\\sin \\theta \\cos \\theta \\ \\\\&={\\frac {2\\tan \\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[23][30] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )\\!&=\\!-\\sin ^{3}\\theta \\!+\\!3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )\\!&=\\!\\cos ^{3}\\theta \\!-\\!3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )\\!=\\!{\\frac {3\\cot \\theta \\!-\\!\\cot ^{3}\\theta }{1\\!-\\!3\\cot ^{2}\\theta }}}$ Half-angle formulae[26][27] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1\\!-\\!\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or}}\\,\\,\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left(\\mathrm {or} \\,\\,\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[8pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[8pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[10pt]\\tan {\\frac {\\eta +\\theta }{2}}\\!&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[8pt]\\tan \\left(\\!{\\frac {\\theta }{2}}\\!+\\!{\\frac {\\pi }{4}}\\!\\right)\\!&=\\!\\sec \\theta \\!+\\!\\tan \\theta \\\\[8pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {|1-\\tan {\\frac {\\theta }{2}}|}{|1+\\tan {\\frac {\\theta }{2}}|}}\\\\[8pt]\\tan {\\frac {\\theta }{2}}\\!&=\\!{\\frac {\\tan", "# 2009 AMC 12A Problems/Problem 25 ## Problem The first two terms of a sequence are $a_1 = 1$ and $a_2 = \\frac {1}{\\sqrt3}$. For $n\\ge1$, $$a_{n + 2} = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$? $\\textbf{(A)}\\ 0\\qquad \\textbf{(B)}\\ 2 - \\sqrt3\\qquad \\textbf{(C)}\\ \\frac {1}{\\sqrt3}\\qquad \\textbf{(D)}\\ 1\\qquad \\textbf{(E)}\\ 2 + \\sqrt3$ ## Solution Consider another sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ such that $a_n = \\tan{\\theta_n}$, and $0 \\leq \\theta_n < 180$. The given recurrence becomes \\begin{align*} a_{n + 2} & = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\\\ \\tan{\\theta_{n + 2}} & = \\frac {\\tan{\\theta_n} + \\tan{\\theta_{n + 1}}}{1 - \\tan{\\theta_n}\\tan{\\theta_{n + 1}}} \\\\ \\tan{\\theta_{n + 2}} & = \\tan(\\theta_{n + 1} + \\theta_n) \\end{align*} It follows that $\\theta_{n + 2} \\equiv \\theta_{n + 1} + \\theta_{n} \\pmod{180}$. Since $\\theta_1 = 45, \\theta_2 = 30$, all terms in the sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ will be a multiple of $15$. Now consider another sequence $\\{b_1, b_2, b_3...\\}$ such that $b_n = \\theta_n/15$, and $0 \\leq b_n < 12$. The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \\equiv b_{n + 1} + b_n \\pmod{12}$. As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the", "must always start with one side (either side, but usually take the more complicated side) and manipulate the side until you end up with the other side. (Some teachers will let you go down both sides until the two sides are equal). The best way to solve these is to turn everything into sin and cos. Note how we work on one side only and pull down the other side when it matches. It doesn’t matter which side we start on, but typically, it’s the most complicated. Sometimes we have to find common denominators, like in the last example. We didn’t need to turn it into sin and cos, since we only had tan and cot in the identity (although it still would have worked). And note that there may be more than one way to do these! Several answers may be “right”. Reciprocal and Quotient Trig Identities $$\\displaystyle \\sin x\\cot x\\,=\\,\\cos x$$ $$\\require{cancel} \\displaystyle \\cancel{{\\sin x}}\\left( {\\frac{{\\cos x}}{{\\cancel{{\\sin x}}}}} \\right)=\\cos x\\,\\surd$$ $$\\sin \\theta (\\cot \\theta +\\sec \\theta )=\\cos \\theta +\\tan \\theta$$ \\begin{align}\\sin \\theta \\cot \\theta +\\sin \\theta \\sec \\theta &=\\\\\\cancel{{\\sin \\theta }}\\frac{{\\cos \\theta }}{{\\cancel{{\\sin \\theta }}}}+\\sin \\theta \\frac{1}{{\\cos \\theta }}&=\\\\\\cos \\theta +\\frac{{\\sin \\theta }}{{\\cos \\theta }}&=\\cos \\theta +\\tan \\theta \\,\\surd \\end{align} $$\\displaystyle \\frac{{1+\\cot \\theta }}{{1+\\tan \\theta }}=\\,\\cot \\theta$$ \\displaystyle \\begin{align}\\frac{{1+\\frac{1}{{\\tan \\theta }}}}{{1+\\tan \\theta }}&=\\\\\\frac{{\\frac{{\\tan \\theta }}{{\\tan \\theta", "\\cos(\\pi - \\theta) &= -\\cos \\theta \\\\ \\tan(\\pi - \\theta) &= -\\tan \\theta \\\\ \\csc(\\pi - \\theta) &= +\\csc \\theta \\\\ \\sec(\\pi - \\theta) &= -\\sec \\theta \\\\ \\cot(\\pi - \\theta) &= -\\cot \\theta \\\\ \\end{align} Periodicity and Shifts Shift by π/2 Shift by π Period for tan and cot Shift by 2π Period for sin, cos, csc and sec \\begin{align} \\sin(\\theta + \\tfrac{\\pi}{2}) &= +\\cos \\theta \\\\ \\cos(\\theta + \\tfrac{\\pi}{2}) &= -\\sin \\theta \\\\ \\tan(\\theta + \\tfrac{\\pi}{2}) &= -\\cot \\theta \\\\ \\csc(\\theta + \\tfrac{\\pi}{2}) &= +\\sec \\theta \\\\ \\sec(\\theta + \\tfrac{\\pi}{2}) &= -\\csc \\theta \\\\ \\cot(\\theta + \\tfrac{\\pi}{2}) &= -\\tan \\theta \\end{align} \\begin{align} \\sin(\\theta + \\pi) &= -\\sin \\theta \\\\ \\cos(\\theta + \\pi) &= -\\cos \\theta \\\\ \\tan(\\theta + \\pi) &= +\\tan \\theta \\\\ \\csc(\\theta + \\pi) &= -\\csc \\theta \\\\ \\sec(\\theta + \\pi) &= -\\sec \\theta \\\\ \\cot(\\theta + \\pi) &= +\\cot \\theta \\\\ \\end{align} \\begin{align} \\sin(\\theta + 2\\pi) &= +\\sin \\theta \\\\ \\cos(\\theta + 2\\pi) &= +\\cos \\theta \\\\ \\tan(\\theta + 2\\pi) &= +\\tan \\theta \\\\ \\csc(\\theta + 2\\pi) &= +\\csc \\theta \\\\ \\sec(\\theta + 2\\pi) &= +\\sec \\theta \\\\ \\cot(\\theta + 2\\pi) &= +\\cot \\theta \\end{align} # Resources ## Active participants The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this division, you can list", "and $$z''$$ in terms of $$x, y, z$$ and $$x', y', z'.$$ 7. Prove the equality $(ax - by -cz -dt)^2 + (bx + ay -dz + ct)^2 + (cx + dy + az -bt)^2 + (dx - cy + bz + at)^2 =$ $(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + t^2).$ 8. Prove the following equalities: \\begin{align}\\begin{aligned}\\frac{cosn\\theta}{cos^n\\theta} = 1 - {^nC_2}tan^2\\theta + {^nC_4}tan^4\\theta - ... + A \\text{ where }\\\\A = (-1)^\\frac{n}{2}~tan^n\\theta \\text{ if } n \\text{ is even,}\\\\A = (-1)^\\frac{n - 1}{2}~{^nC_{n - 1}}tan^n\\theta \\text{ if } n \\text{ is odd;}\\\\\\frac{sinn\\theta}{cos^n\\theta} = {^nC_1}tan\\theta - {^nC_3}tan^3\\theta + {^nC_5}tan^5\\theta - ... + A \\text{ where }\\\\A = (-1)^\\frac{n - 2}{2}~{^nC_{n - 1}}tan^{n - 1}\\theta \\text{ if } n \\text{ is odd, }\\\\A = (-1)^\\frac{n}{2}~tan^n\\theta \\text{ if } n \\text{ is odd.}\\end{aligned}\\end{align} 9. Prove the following equality: $2^{2m}cos^{2m}x = \\sum_{k = 0}^{k = m - 1} 2 {2m \\choose k} cos2(m - k)x + {2m \\choose m}$ 10. Prove the following equality: $2^{2m}sin^{2m}x = \\sum_{k = 0}^{k = m - 1} (-1)^{m + k} 2 {2m \\choose k} cos2(m - k)x + {2m \\choose m}$ 11. Prove the following equality: $2^{2m}cos^{2m + 1}x = \\sum_{k = 0}^{k = m} 2 {{2m + 1} \\choose k} cos(2m - 2k + 1)x$ 12. Prove the following", "\\right)} - 1\\right) \\sin{\\left(\\beta + \\theta \\right)}}{\\tan^{2}{\\left(\\alpha \\right)}}\\\\t \\sin{\\left(\\beta + \\theta \\right)} & \\frac{d \\left(\\tan^{2}{\\left(\\alpha \\right)} + 1\\right) \\sin{\\left(\\beta + \\theta \\right)}}{\\tan{\\left(\\alpha \\right)}} + \\frac{w \\left(- \\tan^{2}{\\left(\\alpha \\right)} - 1\\right) \\cos{\\left(\\theta \\right)}}{\\tan^{2}{\\left(\\alpha \\right)}} - \\frac{w \\left(- \\tan^{2}{\\left(\\alpha \\right)} - 1\\right) \\cos{\\left(\\beta + \\theta \\right)}}{\\tan^{2}{\\left(\\alpha \\right)}}\\\\\\frac{t}{R} & \\frac{d \\left(\\tan^{2}{\\left(\\alpha \\right)} + 1\\right)}{w}\\end{matrix}\\right]$ This should give you an appreciation of how quickly the EKF become mathematically intractable. This gives us the final form of our prediction equations: \\begin{aligned} \\mathbf{\\bar x} &= \\mathbf x + \\begin{bmatrix}- R\\sin(\\theta) + R\\sin(\\theta + \\beta) \\\\ R\\cos(\\theta) - R\\cos(\\theta + \\beta) \\\\ \\beta\\end{bmatrix}\\\\ \\mathbf{\\bar P} &=\\mathbf{FPF}^{\\mathsf T} + \\mathbf{VMV}^{\\mathsf T} \\end{aligned} This form of linearization is not the only way to predict $\\mathbf x$. For example, we could use a numerical integration technique such as Runge Kutta to compute the movement of the robot. This will be required if the time step is relatively large. Things are not as cut and dried with the EKF as for the Kalman filter. For a real problem you have to carefully model your system with differential equations and then determine the most appropriate way to solve that system. The correct approach depends on the accuracy you require, how nonlinear the equations are, your processor budget, and numerical stability concerns. ### Design the Measurement Model¶ The robot's sensor provides", "\\tan(\\theta) + \\ln\\abs{\\sec(\\theta) +\\tan(\\theta) }\\Big)+C\\text{.} \\end{equation*} Thus \\begin{align*} \\amp \\int \\sqrt{4x^2-1}\\, dx =\\frac12\\int \\big(\\sec^3(\\theta) - \\sec(\\theta) \\big)\\, d\\theta\\\\ \\amp = \\frac12\\left(\\frac12\\Big(\\sec(\\theta) \\tan(\\theta) + \\ln\\abs{\\sec(\\theta) +\\tan(\\theta) }\\Big) -\\ln\\abs{\\sec(\\theta) + \\tan(\\theta) }\\right) + C\\\\ \\amp = \\frac14\\left(\\sec(\\theta) \\tan(\\theta) -\\ln\\abs{\\sec(\\theta) +\\tan(\\theta) }\\right)+C\\text{.} \\end{align*} We are not yet done. Our original integral is given in terms of $$x\\text{,}$$ whereas our final answer, as given, is in terms of $$\\theta\\text{.}$$ We need to rewrite our answer in terms of $$x\\text{.}$$ With $$a=1/2\\text{,}$$ and $$x=\\frac12\\sec(\\theta)\\text{,}$$ the reference triangle in Key Idea 6.4.3(c) shows that \\begin{equation*} \\tan(\\theta) = \\sqrt{x^2-1/4}\\Big/(1/2) = 2\\sqrt{x^2-1/4} \\text{ and } \\sec(\\theta) = 2x\\text{.} \\end{equation*} Thus \\begin{align*} \\amp \\frac14\\Big(\\sec(\\theta) \\tan(\\theta) -\\ln\\abs{\\sec(\\theta) +\\tan(\\theta) }\\Big)+C\\\\ \\amp\\quad = \\frac14\\Big(2x\\cdot 2\\sqrt{x^2-1/4} - \\ln\\abs{2x + 2\\sqrt{x^2-1/4}}\\Big)+C\\\\ \\amp\\quad = \\frac14\\Big(4x\\sqrt{x^2-1/4} - \\ln\\abs{2x + 2\\sqrt{x^2-1/4}}\\Big)+C\\text{.} \\end{align*} The final answer is given in the last line above, repeated here: \\begin{equation*} \\int \\sqrt{4x^2-1}\\, dx = \\frac14\\Big(4x\\sqrt{x^2-1/4} - \\ln\\abs{2x + 2\\sqrt{x^2-1/4}}\\Big)+C\\text{.} \\end{equation*} ###### Example6.4.9.Using Trigonometric Substitution. Evaluate $$\\ds \\int \\frac{\\sqrt{4-x^2}}{x^2}\\, dx\\text{.}$$ Solution We use Key Idea 6.4.3(a) with $$a=2\\text{,}$$ $$x=2\\sin(\\theta)\\text{,}$$ $$dx = 2\\cos(\\theta)$$ and hence $$\\sqrt{4-x^2} = 2\\cos(\\theta)\\text{.}$$ This gives \\begin{align*} \\int \\frac{\\sqrt{4-x^2}}{x^2}\\, dx \\amp = \\int \\frac{2\\cos(\\theta) }{4\\sin^2(\\theta) }(2\\cos(\\theta) )\\, d\\theta\\\\ \\amp = \\int \\cot^2(\\theta) \\, d\\theta\\\\ \\amp = \\int (\\csc^2(\\theta) -1)\\, d\\theta\\\\ \\amp = -\\cot(\\theta) -\\theta + C\\text{.} \\end{align*} We need to rewrite our answer in terms of $$x\\text{.}$$ Using", "with the sum formula that was introduced previously: $\\sin(\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$. If we let $\\alpha = \\beta = \\theta$, then we have: \\displaystyle{ \\begin{align} \\sin(\\theta + \\theta) &= \\sin \\theta \\cos \\theta + \\cos \\theta \\sin \\theta \\\\ \\sin(2\\theta) &= 2\\sin \\theta \\cos \\theta \\end{align} } The double-angle formula for cosine can be derived similarly, and is: $\\cos(2\\theta) = \\cos^2 \\theta - \\sin^2 \\theta$ Notice that we can apply the Pythagorean identities to get two more variations of the cosine formula: \\displaystyle{ \\begin{align} \\cos{\\left(2\\theta \\right)} &= \\cos^2 \\theta - \\sin^2 \\theta \\\\ &= \\left(1- \\sin^2 \\theta \\right) - \\sin^2 \\theta \\\\ &= 1- 2\\sin^2 \\theta \\end{align} } \\displaystyle{ \\begin{align} \\cos{\\left(2\\theta\\right)} &= \\cos^2 \\theta - \\sin^2 \\theta \\\\ &= \\cos^2 \\theta - \\left(1- \\cos^2 \\theta \\right) \\\\ &= 2 \\cos^2 \\theta -1 \\end{align} } Similarly, to derive the double-angle formula for tangent, replacing $\\alpha = \\beta = \\theta$ in the sum formula gives \\displaystyle{ \\begin{align} \\tan{\\left(\\alpha + \\beta \\right)} &= \\frac{ \\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta} \\\\ \\tan{\\left(\\theta + \\theta \\right)} &= \\frac{ \\tan \\theta + \\tan \\theta}{1 - \\tan \\theta \\tan \\theta} \\\\ \\tan{\\left(2\\theta \\right)} &= \\frac{ 2\\tan \\theta }{1 - \\tan^2 \\theta} \\end{align} } The double-angle formulae are summarized as follows: •", "= 2 · cos x · sin((n − 1)x) − sin((n − 2)x). This can be proved by adding formulae for sin((n − 1)x + x) and sin((n − 1)xx). Serving a purpose similar to that of the Chebyshev method, for the tangent we can write: ${\\displaystyle \\tan(nx)={\\frac {\\tan((n-1)x)+\\tan x}{1-\\tan((n-1)x)\\tan x}}\\,.}$ Tangent of an average ${\\displaystyle \\tan \\left({\\frac {\\alpha +\\beta }{2}}\\right)={\\frac {\\sin \\alpha +\\sin \\beta }{\\cos \\alpha +\\cos \\beta }}=-\\,{\\frac {\\cos \\alpha -\\cos \\beta }{\\sin \\alpha -\\sin \\beta }}}$ Setting either α or β to 0 gives the usual tangent half-angle formulae. Viète's infinite product ${\\displaystyle \\cos {\\frac {\\theta }{2}}\\cdot \\cos {\\frac {\\theta }{4}}\\cdot \\cos {\\frac {\\theta }{8}}\\cdots =\\prod _{n=1}^{\\infty }\\cos {\\frac {\\theta }{2^{n}}}={\\frac {\\sin \\theta }{\\theta }}=\\operatorname {sinc} \\theta .}$ (Refer to sinc function.) Power-reduction formulae Obtained by solving the second and third versions of the cosine double-angle formula. Sine Cosine Other ${\\displaystyle \\sin ^{2}\\theta ={\\frac {1-\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\cos ^{2}\\theta ={\\frac {1+\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\sin ^{2}\\theta \\cos ^{2}\\theta ={\\frac {1-\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{3}\\theta ={\\frac {3\\sin \\theta -\\sin(3\\theta )}{4}}}$ ${\\displaystyle \\cos ^{3}\\theta ={\\frac {3\\cos \\theta +\\cos(3\\theta )}{4}}}$ ${\\displaystyle \\sin ^{3}\\theta \\cos ^{3}\\theta ={\\frac {3\\sin(2\\theta )-\\sin(6\\theta )}{32}}}$ ${\\displaystyle \\sin ^{4}\\theta ={\\frac {3-4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\cos ^{4}\\theta ={\\frac {3+4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{4}\\theta \\cos ^{4}\\theta ={\\frac {3-4\\cos(4\\theta )+\\cos(8\\theta )}{128}}}$ ${\\displaystyle \\sin ^{5}\\theta ={\\frac {10\\sin \\theta -5\\sin(3\\theta", "\\tan(\\theta + \\pi) &= \\tan\\theta \\\\ \\end{align} From the symmetry of the respective graphs it also follows that \\begin{align} \\sin(-\\theta) &= -\\sin\\theta \\\\ \\cos(-\\theta) &= \\cos\\theta \\\\ \\tan(-\\theta) &= -\\tan\\theta \\end{align} Also, if the graph of cosine is translated to the left by $\\pi/2$ and reflected in the vertical axis the graph of sine is obtained. Hence the equation $$\\cos(-\\theta + \\frac{\\pi}{2}) = \\sin\\theta$$ is an identity. Similar identity can be found for since. Hence following identities \\begin{align} \\sin(\\frac{\\pi}{2} - \\theta) = \\cos\\theta \\\\ \\cos(-\\theta + \\frac{\\pi}{2}) = \\sin\\theta \\end{align} ## 4.3 Cosecant, secant and cotangent In right angled triangle, three known trigonometric functions are defined as $$\\sin\\theta = \\frac{\\text{opp}}{\\text{hyp}} ~~~ \\cos\\theta = \\frac{\\text{adj}}{\\text{hyp}} ~~~ \\tan\\theta = \\frac{\\text{opp}}{\\text{adj}} ~~~$$ There are three further ratios cosecant, secant, cotangent of $\\theta$: $$\\text{cosec}\\theta = \\frac{\\text{hyp}}{\\text{opp}} ~~~ \\text{sec}\\theta = \\frac{\\text{hyp}}{\\text{adj}} ~~~ \\text{cot}\\theta = \\frac{\\text{adj}}{\\text{opp}} ~~~$$ Generalized as ### Cosecant, secant and cotangent \\begin{align} \\text{cosec}\\theta &= \\frac{1}{\\sin\\theta} ~~~(\\text{provided} \\sin\\theta \\ne 0) \\\\ \\text{sec}\\theta &= \\frac{1}{\\cos\\theta} ~~~(\\text{provided} \\cos\\theta \\ne 0) \\\\ \\text{cot}\\theta &= \\frac{\\cos\\theta}{\\sin\\theta} ~~~(\\text{provided} \\sin\\theta \\ne 0) \\\\ \\end{align} Given $$\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$$ it follows that $$\\cot\\theta = \\frac{1}{\\tan\\theta} ~~~ (\\text{provided} \\sin\\theta\\ne0 ~\\text{and} \\cos\\theta\\ne0)$$ Essentially cosecant, secant and cotangent are the reciprocals of the sine, cosine and tangent functions (with same periods). Two new identities can be obtained by taking previous identity $$\\sin^2\\theta", "\\theta \\tan \\theta}{\\sin \\theta - \\cos \\theta} + \\frac{\\cos\\theta \\cot \\theta}{\\cos \\theta - \\sin \\theta} \\\\ &= \\frac{\\sin\\theta\\tan\\theta - \\cos\\theta \\cot \\theta}{\\sin \\theta - \\cos \\theta} \\\\&= \\frac{\\sin^3 \\theta - \\cos^3 \\theta}{\\sin \\theta \\cos \\theta (\\sin\\theta - \\cos\\theta)} \\\\&= \\frac{\\sin^2 \\theta + \\sin \\theta \\cos \\theta + \\cos^2\\theta }{\\sin \\theta \\cos \\theta}\\\\&= 1 + \\tan \\theta + \\cot \\theta\\\\&=\\boxed{\\text{D}} \\end{align*} - add comment \\begin{align} \\frac{\\tan(\\theta)}{1-\\cot(\\theta)}+\\frac{\\cot(\\theta)}{1-\\tan(\\theta)} &=\\frac{\\cot(\\theta)-\\tan^2{\\theta}}{1-\\tan(\\theta)}\\\\ &=\\cot(\\theta)\\frac{1-\\tan^3(\\theta)}{1-\\tan(\\theta)}\\\\ &=\\cot(\\theta)+1+\\tan(\\theta) \\end{align} - add comment" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "1 & 1 \\end{pmatrix}$", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "\\\\ 0 & 1\\end{pmatrix}.$ -", "\\begin {pmatrix} 12 & 6 & -2 \\\\ -6 & -3 & 1 \\\\ 6 & 3 & -1 \\end {pmatrix}$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "it is to compute $2\\cdot124865$; for a start, there's more input.", "6 & 0 & 5 \\end{pmatrix}$", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "1/2 \\\\ \\end{pmatrix}$$", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "& 4 \\end{pmatrix} \\begin{pmatrix} 3 & 6 & 9 \\end{pmatrix} \\begin{pmatrix} 5 & 8 \\end{pmatrix}$.", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "See all short problems arranged by curriculum topic in the short problems collection How can you get from $(10,66)$ to $(20,12)$ using only integer multiples of the vectors $\\begin{pmatrix}2\\\\-3\\end{pmatrix}$ and $\\begin{pmatrix}3\\\\2\\end{pmatrix}$? If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.", "2 } \\\\ { 20 } & { 6 } \\\\ { 0 } & { 14 } \\end{pmatrix} \\$", "{\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} \\ right | = \\ left | {\\ begin {pmatrix} -3 \\\\ - 6 \\ \\ 6 \\ end {pmatrix}} \\ right | = {\\ sqrt {81}} = 9}$.", "have that $$\\theta_{t*}(X_{(x,y)}) = \\begin{pmatrix} e^{2t} & 0 \\\\ 0 & e^{-3t} \\end{pmatrix} \\begin{pmatrix} 2x \\\\ -3y \\end{pmatrix} = \\begin{pmatrix} 2xe^{2t} \\\\ -3ye^{-3y} \\end{pmatrix} .$$ Here again the basis is understood to be $$(\\frac{\\partial}{\\partial x}, \\frac{\\partial}{\\partial y})$$. So we are done." ]

[ "5 × 2 + 6 × 8 ) = 58 {\\displaystyle c_{2,1}={\\begin{pmatrix}5&6\\end{pmatrix}}{\\begin{pmatrix}2\\\\8\\end{pmatrix}}=(5\\times 2+6\\times 8)=58} c 2 , 2 = ( 5 6 ) ( 6 7 ) = ( 5 × 6 + 6 × 7 ) = 72 {\\displaystyle c_{2,2}={\\begin{pmatrix}5&6\\end{pmatrix}}{\\begin{pmatrix}6\\\\7\\end{pmatrix}}=(5\\times 6+6\\times 7)=72} i.e. C = ( 22 32 58 72 ) {\\displaystyle C={\\begin{pmatrix}22&32\\\\58&72\\end{pmatrix}}} d 1 , 1 = ( 2 6 ) ( 3 5 ) = ( 2 × 3 + 6 × 5 ) = 36 {\\displaystyle d_{1,1}={\\begin{pmatrix}2&6\\end{pmatrix}}{\\begin{pmatrix}3\\\\5\\end{pmatrix}}=(2\\times 3+6\\times 5)=36} d 1 , 2 = ( 2 6 ) ( 2 6 ) = ( 2 × 2 + 6 × 6 ) = 40 {\\displaystyle d_{1,2}={\\begin{pmatrix}2&6\\end{pmatrix}}{\\begin{pmatrix}2\\\\6\\end{pmatrix}}=(2\\times 2+6\\times 6)=40} d 2 , 1 = ( 8 7 ) ( 3 5 ) = ( 8 × 3 + 7 × 5 ) = 59 {\\displaystyle d_{2,1}={\\begin{pmatrix}8&7\\end{pmatrix}}{\\begin{pmatrix}3\\\\5\\end{pmatrix}}=(8\\times 3+7\\times 5)=59} d 2 , 2 = ( 8 7 ) ( 2 6 ) = ( 8 × 2 + 7 × 6 ) = 58 {\\displaystyle d_{2,2}={\\begin{pmatrix}8&7\\end{pmatrix}}{\\begin{pmatrix}2\\\\6\\end{pmatrix}}=(8\\times 2+7\\times 6)=58} i.e. D = ( 36 40 59 58 ) {\\displaystyle D={\\begin{pmatrix}36&40\\\\59&58\\end{pmatrix}}} Example 2 Evaluate AB and BA where A = ( 5 17 2 7 ) {\\displaystyle A={\\begin{pmatrix}5&17\\\\2&7\\end{pmatrix}}} B = ( 7 − 17 − 2 5 ) {\\displaystyle B={\\begin{pmatrix}7&-17\\\\-2&5\\end{pmatrix}}} Solution ( 5 17 2 7 ) (", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "\\\\ 7 \\end{pmatrix} = 7 \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\end{align*} so, it is correct.", "{10}}}\\\\{-{\\sqrt {10}}}\\\\{-i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{-3}\\\\{-i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{-3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{1}\\\\{-i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{-i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{1}\\\\{i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{-3}\\\\{i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i}\\\\{\\sqrt {5}}\\\\{-i{\\sqrt {10}}}\\\\{-{\\sqrt {10}}}\\\\{i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}}\\\\S_{z}={\\frac {\\hbar }{2}}{\\begin{pmatrix}5&0&0&0&0&0\\\\0&3&0&0&0&0\\\\0&0&1&0&0&0\\\\0&0&0&-1&0&0\\\\0&0&0&0&-3&0\\\\0&0&0&0&0&-5\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\end{pmatrix}}\\\\\\end{array}}}$ 5. The generalization of these matrices for arbitrary spin s is {\\displaystyle {\\begin{aligned}\\left(S_{x}\\right)_{ab}&={\\frac {\\hbar }{2}}\\left(\\delta _{a,b+1}+\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\,\\\\\\left(S_{y}\\right)_{ab}&={\\frac {i\\hbar }{2}}\\left(\\delta _{a,b+1}-\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\\\\\left(S_{z}\\right)_{ab}&=\\hbar (s+1-a)\\delta _{a,b}=\\hbar (s+1-b)\\delta _{a,b}\\end{aligned}}} where indices ${\\displaystyle a,b}$ are integer numbers such that ${\\displaystyle 1\\leq a\\leq 2s+1}$ and ${\\displaystyle 1\\leq b\\leq 2s+1}$ Also useful in the quantum mechanics of multiparticle systems, the general Pauli group Gn is defined to consist of all n-fold tensor products of Pauli matrices. The analog formula of Euler's formula in terms of the Pauli matrices: ${\\displaystyle e^{i\\theta \\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)}=I\\cos(\\theta )+i\\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)\\sin(\\theta )}$ for higher spins is tractable, but less simple.[20] ## Parity In tables of the spin quantum number s for nuclei or particles, the spin is often followed by a \"+\" or \"−\". This refers to the parity with \"+\" for even parity (wave function unchanged by spatial inversion) and \"−\" for odd", "5 & -8 -4 & 12 + 3 \\\\ 3 \\times 2 + 5 & -6 - 4 & 9 +3\\\\7 \\times 2 + 5 & -14 - 4 & 21 + 3\\end{pmatrix} \\\\ & = \\begin{pmatrix}13 & -12 & 15 \\\\ 11 & -10 & 12 \\\\ 19 & -18 & 24\\end{pmatrix}\\end{align*} Similarly: CB = \\begin{pmatrix}23 & -2 & 3 \\\\ 0 & 0 & 0 \\\\ 29 & -7 & 4\\end{pmatrix} 1 Like", "to choose 4 from the 8 with cracks and 1 from the remaining 17. 5. Jan 19, 2010 ### exitwound b.)$$\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix} = \\frac{8!}{4!4!} * \\frac{17!}{1!16!}= 1190$$ c.) $$\\frac{\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} = \\frac{\\frac{8!}{4!4!} * \\frac{17!}{1!16!}}{\\frac{25!}{5!20!}} = .0224$$ d.) $$\\frac{\\begin{pmatrix} 8\\\\ 5 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 0 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} + AnswerToPartC = \\frac{\\frac{8!}{5!3!} * \\frac{17!}{0!17!}}{\\frac{25!}{5!20!}} + .0224 = .0235$$ Is this how it's done? 6. Jan 19, 2010 ### vela Staff Emeritus Yup, perfect.", "\\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} 1^3 - 3\\times 1^2\\end{pmatrix} - \\begin{pmatrix} \\begin{pmatrix}\\frac{1}{2} \\end{pmatrix}^3 - 3\\begin{pmatrix}\\frac{1}{2} \\end{pmatrix}^2\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} 1 - 3\\end{pmatrix} - \\begin{pmatrix} \\frac{1}{8} - 3\\times \\frac{1}{4}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} -2\\end{pmatrix} - \\begin{pmatrix} \\frac{1}{8} - \\frac{3}{4}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 - \\begin{pmatrix} \\frac{1}{8} - \\frac{6}{8}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 - \\begin{pmatrix} - \\frac{5}{8}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 + \\frac{5}{8} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} - \\frac{11}{8} \\end{bmatrix} \\\\ & = \\frac{11}{32} \\\\ P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = 0.344 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, between $$x=0.5$$ and $$x=1$$ is equal to $$0.344$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} = 0.344$$ 3. To find $$P\\begin{pmatrix}X \\geq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}X \\geq 1\\end{pmatrix} & = \\int_1^{+\\infty}f(x)dx \\\\ & = \\int_1^2 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4}\\int_1^2 x(x-2)dx \\\\ & = -\\frac{3}{4}\\int_1^2 \\begin{pmatrix} x^2 - 2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix}\\frac{x^3}{3}-x^2 \\end{bmatrix}_1^2 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix}x^3-3x^2 \\end{bmatrix}_1^2 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}2^3-3\\times 2^2\\end{pmatrix} - \\begin{pmatrix}1^3-3\\times 1^2\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}8-12\\end{pmatrix} - \\begin{pmatrix}1-3\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}-4\\end{pmatrix} - \\begin{pmatrix}-2\\end{pmatrix} \\end{bmatrix} \\\\ & =", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "\\\\ B &= \\frac1{c}\\begin{pmatrix} c\\\\ 1 \\end{pmatrix} \\\\ C &= \\frac1{b}\\begin{pmatrix} b\\\\ 1 \\end{pmatrix} \\\\ M &= \\frac1{2 b c}\\begin{pmatrix} 2 b c\\\\ b + c \\end{pmatrix} \\\\ L &= \\frac1{b^{2} c^{2} + b^{2} + c^{2} + 1}\\begin{pmatrix} b^{2} c^{2} - 1\\\\ b^{2} c + b c^{2} + b + c \\end{pmatrix} \\\\ K &= \\frac1{b^{2} c^{2} + b^{2} + 4 b c + c^{2} + 1}\\begin{pmatrix} b^{2} c^{2} + b^{2} + 2 b c + c^{2} - 1\\\\ 2 b + 2 c \\end{pmatrix} \\\\ T &= {\\scriptsize\\frac1{16 b^{4} c^{4} + 9 b^{4} c^{2} + 18 b^{3} c^{3} + 9 b^{2} c^{4} + b^{4} + 6 b^{3} c + 10 b^{2} c^{2} + 6 b c^{3} + c^{4} + b^{2} + 2 b c + c^{2}}}\\cdot\\\\ &\\phantom={\\scriptsize\\begin{pmatrix} 16 b^{4} c^{4} + 7 b^{4} c^{2} + 14 b^{3} c^{3} + 7 b^{2} c^{4} + b^{4} + 2 b^{3} c + 2 b^{2} c^{2} + 2 b c^{3} + c^{4} - b^{2} - 2 b c - c^{2}\\\\ 8 b^{4} c^{3} + 8 b^{3} c^{4} + 2 b^{4} c + 14 b^{3} c^{2} + 14 b^{2} c^{3} + 2 b c^{4} + 2 b^{3} + 6 b^{2} c + 6 b c^{2} + 2 c^{3} \\end{pmatrix}} \\end{align*} I'd like to paste equations for the circles as well, but I", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "\\end{pmatrix} + } + & + \\color{red} { + MS_{6} = + \\begin{pmatrix} ​ + 3 & 21 & 19 & 12 & 10 \\\\ + 22 & 20 & 13 & 6 & 4 \\\\ + 16 & 14 & 7 & 5 & 23 \\\\ + 15 & 18 & 1 & 24 & 17 \\\\ + 9 & 2 & 25 & 18 & 11 + \\end{pmatrix} + } + \\end{align} + + + + \\begin{align} + \\color{green} { + MS_{7} = + \\begin{pmatrix} ​ + 11 & 10 & 4 & 23 & 17 \\\\ + 18 & 12 & 6 & 5 & 24 \\\\ + 25 & 19 & 13 & 7 & 1 \\\\ + 2 & 21 & 20 & 14 & 8 \\\\ + 9 & 3 & 22 & 16 & 15 + \\end{pmatrix} + } + & + \\color{red} { + MS_{8} = + \\begin{pmatrix} ​ + 10 & 4 & 23 & 17 & 11 \\\\ + 12 & 6 & 5 & 24 & 18 \\\\ + 19 & 13 & 7 & 1 & 25 \\\\ + 21 & 20 & 14 & 8 & 2 \\\\ + 3 & 22 & 16 & 15 & 9 + \\end{pmatrix} + } + \\end{align} +", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03", "& 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 4 & 7 & 9 &10 \\end{pmatrix}\\overset{(1)}\\sim \\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 3 & 5 & 6\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\overset{(2)}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 6-\\lambda & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 0 & 3\\lambda-13\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 0 & 1 & 1 & 0\\\\ 0 & 0 & 1 & 3\\\\ 0", "&= 64 &=& 1 &\\pmod{21}.\\\\ 2^3 &= 8 \\pmod{21}, & & & & & }$ \\end{document}", "& 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\ub{12} & 3 & 16 & \\ub{4} \\end{pmatrix} \\qquad \\begin{pmatrix} \\hl{1} & 3 & 2 & \\hl{6} \\\\ 4 & 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\hl{12} & 3 & 16 & \\hl{4} \\end{pmatrix}$$ \\end{document}", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "• {BA,BD}+{BB,N3} • {BB,BD}+{BA,N3} as different selections even though they lead to exactly the same reading. - Perhaps one should add that the correct approach is to say \"4 choose 2\" then \"6 choose 2\" + \"4 choose 3\" then \"6 choose 1\" + \"4 choose 4\" then \"6 choose 0\" which leads to $$\\begin{pmatrix} 4 \\\\ 2 \\end{pmatrix}\\begin{pmatrix} 6 \\\\ 2 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} \\begin{pmatrix} 6 \\\\ 1 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix}\\begin{pmatrix} 6 \\\\ 0 \\end{pmatrix} = 115$$. – Patrick Da Silva Dec 28 '11 at 3:33 @Patrick, that appears to be implicit it in the part of the question beginning \"I understand how the answer solved it\". – Henning Makholm Dec 28 '11 at 3:34 Great! Excellent answer, your example (with the BA, BB, etc and N1 to N6) helped me see where I was double counting. Thank you so much! – oxuser Dec 28 '11 at 10:41", ". & . & . & . \\\\ . & . & . & . & . & . & 2 & . & . & . \\\\ . & . & . & . & . & . & . & 3 & . & . \\\\ . & . & . & . & . & . & . & . & 4 & . \\end{pmatrix} = \\begin{pmatrix} 1 & . & . & . & . & . & . & . & . & . \\\\ -4 & 1 & . & . & . & . & . & . & . & . \\\\ 6 & -3 & 1 & . & . & . & . & . & . & . \\\\ -4 & 3 & -2 & 1 & . & . & . & . & . & . \\\\ 1 & -1 & 1 & -1 & 1 & . & . & . & . & . \\\\ . & . & . & . & . & 1 & . & . & . & . \\\\ . & . & . & . & . & 1 & 1 & . & . & . \\\\ . & . & . & . &", "\\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "Example Using the method we just saw and without using a calculator, calculate each of the following: 1. $$101^2$$ 2. $$27^2$$ 3. $$23^2$$ 4. $$99^2$$ 5. $$104^2$$ 6. $$48^2$$ ## Solution Without Working 1. $$101^2 = 10201$$ 2. $$27^2 = 711$$ 3. $$23^2 = 529$$ 4. $$99^2 = 9801$$ 5. $$104^2 = 10816$$ 6. $$48^2 = 2304$$ ## Solution With Working 1. We calculate $$101^2$$ as follows: \\begin{aligned} 101^2 & = \\begin{pmatrix} 100+1\\end{pmatrix}^2 \\\\ & = 100^2 + 2 \\times 100 \\times 1 + 1^2 \\\\ & = 10000 + 200 + 1\\\\ 101^2 &= 10201 \\end{aligned} 2. We calculate $$27^2$$ as follows: \\begin{aligned} 27^2 & = \\begin{pmatrix} 30 - 3 \\end{pmatrix}^2 \\\\ & = 30^2 - 2 \\times 30 \\times 3 - 3^2 \\\\ & = 900 - 180 - 9 \\\\ & = 720 - 9 \\\\ 27^2 & = 711 \\end{aligned} 3. We calculate $$23^2$$ as follows: \\begin{aligned} 23^2 & = \\begin{pmatrix} 20 + 3 \\end{pmatrix}^2 \\\\ & = 20^2 + 2 \\times 20 \\times 3 + 3^2 \\\\ & = 400 + 120 + 9 \\\\ 23^2 & = 529 \\end{aligned} 4. We calculate $$99^2$$ as follows: \\begin{aligned} 99^2 & = \\begin{pmatrix} 100 - 1 \\end{pmatrix} \\\\ & = 100^2 - 2\\times 100 \\times 1 + 1^2 \\\\ & = 10000 - 200 +", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "\\\\- 0.283871 \\end{pmatrix}$ $\\begin{pmatrix}-3.95018 \\\\- 0.24367 \\end{pmatrix}$ $\\begin{pmatrix}-3.8219 \\\\- 0.0463343 \\end{pmatrix}$ $\\begin{pmatrix}-3.8219\\\\0.0463343 \\end{pmatrix}$ $\\begin{pmatrix}-3.95018\\\\0.24367\\end{pmatrix}$ $\\begin{pmatrix}-4.17269\\\\0.283871 \\end{pmatrix}$ $10$ $\\begin{pmatrix}-4.20324 \\\\- 0.275251 \\end{pmatrix}$ $\\begin{pmatrix}-3.8754 \\\\- 0.173249\\end{pmatrix}$ $\\begin{pmatrix}-3.8754\\\\0.173249\\end{pmatrix}$ $\\begin{pmatrix}-4.20324\\\\0.275251\\end{pmatrix}$ ${}^\\ast$: $\\begin{pmatrix}-3.85567\\\\- 0.14285\\end{pmatrix} \\equiv -3.85567- 0.14285 \\; i, \\; i = \\sqrt{-1}$ Table 3. Typical numerical values of $\\{\\xi, {\\lambda} \\}$ for satellite and primitive components with $q k \\le 8$ $(q, k)$ Type $\\xi$ $\\lambda$ Fig. No. $(2, 1)$ Primitive $-0.23512621166835 + 1.6020106612492319\\; i$ $-1.951962936703127 -1.7485319734836857\\; i$ 7(a) $(3, 1)$ Primitive $0.454659800907803 + 0\\; i$ $-5.219261654371707 + 0\\; i$ 7(b) $(4, 1)$ Primitive $0.341977833651053 + 0.4814672920477362\\; i$ $-3.564458921821090 + 1.646535885566244\\; i$ 7(c) $(5, 1)$ Primitive $0.757190777481740 + 0.2122872817258981\\; i$ $-4.045956427432157 + 0.346070091333350\\; i$ 7(d) $(6, 1)$ Primitive $-1.10304555596122 + 1.0059215339291120\\; i$ $-0.568308288946734 - 0.953277135441219\\; i$ 7(e) $(7, 1)$ Primitive $0.764464537114743 + 0.2097254339881594\\; i$ $-4.018513045186225 + 0.345332811593217\\; i$ 7(f) $(2, 2)$ Satellite $0.851625771743808 + 0.2369760927165886\\; i$ $-3.752943559023020 + 0.074408072396025\\; i$ 7(g) $(3, 2)$ Satellite $0.875562175989576 + 0.2566238834815281\\; i$ $-3.838763107389483 + 0.152045173864673\\; i$ 7(h) $(4, 2)$ Satellite $1.201354999600545 - 0.2912383604676560\\; i$ $-3.921140825708978 + 0.257336038244326\\; i$ 7(i) • [1] L. V. Ahlfors, Complex Analysis. An Introduction to the Theory of Analytic Functions of One Complex Variable, International Series in Pure and Applied Mathematics, McGraw-Hill Book Co., New York, 1978. [2] J. Ainsworth, M. Dawson, J. Pianta and J. Warwick, The Farey Sequence, 2003. Available", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "so $\\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ -1 \\end{pmatrix}s =\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix}(2-(1/3)s).$ 2. These two are equal. To show that $S_1\\subseteq S_2$, we check that for any t,s we can find an appropriate m,n so that these hold. $\\begin{array}{*{2}{rc}r} 4m &- &4n &= &1t+2s \\ 7m &- &2n &= &3t+1s \\ 7m &- &10n &= &1t+5s \\end{array}$ Use Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 7 &-2 &3t+1s \\ 7 &-10 &1t+5s \\end{array}\\right) &\\xrightarrow[(-7/4)\\rho_1+\\rho_3]{(-7/4)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &-3 &-(3/4)t+(6/4)s \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to conclude that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 1 \\\\ 5 \\end{pmatrix}s =\\begin{pmatrix} 4 \\\\ 7 \\\\ 7 \\end{pmatrix}((1/2)t) +\\begin{pmatrix} -4 \\\\ -2 \\\\ -10 \\end{pmatrix}((1/4)t-(1/2)s)$ and so $S_1\\subseteq S_2$. For $S_2\\subseteq S_1$, solve $\\begin{array}{*{2}{rc}r} 1t &+ &2s &= &4m-4n \\ 3t &+ &1s &= &7m-2n \\ 1t &+ &5s &= &7m-10n \\end{array}$ with Gaussian reduction $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 3 &1 &7m-2n \\ 1 &5 &7m-10n \\end{array}\\right) &\\xrightarrow[-\\rho_1+\\rho_3]{-3\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &3 &3m-6n \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get", "If I'm correct, I got $b = \\begin{pmatrix} 2\\\\ 5 \\\\ 7 \\\\ 10 \\end{pmatrix} = b_1 \\begin{pmatrix} 8.1 \\\\ 10.2 \\\\ 12.3 \\\\ 14.4 \\end{pmatrix} + b_2 \\begin{pmatrix} -6.1 \\\\ -5.2 \\\\ -5.3 \\\\ -4.4 \\end{pmatrix}$ Also can anyone explain the first question to me? Thanks, in advance! Last edited: Apr 18, 2012", "b_{21} & b_{22} \\end{pmatrix} \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix} \\Rightarrow \\\\ \\begin{cases}\\require{cancel}\\cancel{a_{11}b_{11}}+a_{12}b_{21}=\\cancel{b_{11}a_{11}}+b_{12}a_{21}\\\\ a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\\\ a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\\\ a_{21}b_{12}+\\cancel{a_{22}b_{22}}=b_{21}a_{12}+\\cancel{b_{22}a_{22}}\\end{cases}$$ From$(1)$, since$b_{12}$and$b_{21}$can be any number, in particular,$b_{12}=0$and$b_{21}\\ne 0$, we get:$a_{12}=0$. Similarly, for$b_{12}\\ne 0$and$b_{21}=0$, we get$a_{21}=0$. From$(2)$, since$a_{12}=0$and$b_{12}$is an arbitrary number, we get$a_{11}b_{12}=b_{12}a_{22} \\Rightarrow a_{11}=a_{22}\\$.", "0.200 \\\\ 0.200 \\\\ 0.200 \\end{pmatrix}, \\mathbf{p}_1 = \\begin{pmatrix} 0.064 \\\\ 0.291 \\\\ 0.291 \\\\ 0.205 \\\\ 0.149 \\end{pmatrix}, \\mathbf{p}_2 = \\begin{pmatrix} 0.055 \\\\ 0.321 \\\\ 0.284 \\\\ 0.197 \\\\ 0.143 \\end{pmatrix}, \\mathbf{p}_3 = \\begin{pmatrix} 0.054 \\\\ 0.315 \\\\ 0.289 \\\\ 0.201 \\\\ 0.141 \\end{pmatrix}$ $\\mathbf{p}_4 = \\begin{pmatrix} 0.054 \\\\ 0.315 \\\\ 0.289 \\\\ 0.203 \\\\ 0.139 \\end{pmatrix}, \\mathbf{p}_5 = \\begin{pmatrix} 0.054 \\\\ 0.315 \\\\ 0.289 \\\\ 0.203 \\\\ 0.139 \\end{pmatrix}, \\mathbf{p}_6 = \\begin{pmatrix} 0.054 \\\\ 0.314 \\\\ 0.289 \\\\ 0.203 \\\\ 0.140 \\end{pmatrix}, \\mathbf{p}_7 = \\begin{pmatrix} 0.054 \\\\ 0.315 \\\\ 0.289 \\\\ 0.202 \\\\ 0.140 \\end{pmatrix}$. The sum of absolute differences between pk+1 and pk decreases exponentially: $|\\mathbf{p}_1 - \\mathbf{p}_0| = 0.374 \\\\ |\\mathbf{p}_2 - \\mathbf{p}_1| = 0.060 \\\\ |\\mathbf{p}_3 - \\mathbf{p}_2| = 0.029 \\\\ |\\mathbf{p}_4 - \\mathbf{p}_3| = 0.013 \\\\ |\\mathbf{p}_5 - \\mathbf{p}_4| = 0.005 \\\\ |\\mathbf{p}_6 - \\mathbf{p}_5| = 0.002 \\\\ |\\mathbf{p}_7 - \\mathbf{p}_6| = 0.001$ ### The code #!/usr/bin/python3 import scipy from scipy import sparse \"\"\"Computes the PageRank for given |links|. The source or target elements of |links| are numbered from zero upwards. Gaps in the numbering represent pages having neither inbound nor outbound links. Note that with normal settings, such pages still get a nonzero PageRank. Args: links: A list of ⟨source, target⟩ pairs. Source and target must be integers >= 0. c:", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "0.0 & 1.0 \\end{array} \\right)}^2 \\begin{pmatrix}v_1\\\\v_2\\\\v_3\\\\v_4\\end{pmatrix} \\\\ &= \\left( \\begin{array}{cccc} 0.22 & 0.21 & 0.0 & 0.0 \\\\ 0.19 & 0.57 & 0.0 & 0.0 \\\\ 0.14 & 0.03 & 1.0 & 0.0 \\\\ 0.23 & 0.16 & 0.0 & 1.0 \\end{array} \\right) \\begin{pmatrix}v_1\\\\v_2\\\\v_3\\\\v_4\\end{pmatrix} \\end{aligned} . The amount in each country after $n$ days is $C^n \\mathbf{v}$ .", "for to distribute the 5 chosen circular bowls. With this correction, the sum becomes $$\\begin{pmatrix}9\\\\5\\end{pmatrix}\\begin{pmatrix}5\\\\2\\end{pmatrix}+\\begin{pmatrix}9\\\\2\\end{pmatrix}=1296.$$", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Real Multiplication Distributes over Addition $\\ds \\mathbf A \\mathbf C + \\mathbf B \\mathbf C$ $=$ $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix}$ $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + a_{12} c_{21} & a_{11} c_{12} + a_{12} c_{22} \\\\ a_{21} c_{11} + a_{22} c_{21} & a_{21} c_{12} + a_{22} c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} c_{11} + b_{12} c_{21} & b_{11} c_{12} + b_{12} c_{22} \\\\ b_{21} c_{11} + b_{22} c_{21} & b_{21} c_{12} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Product (Conventional) $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + b_{11} c_{11} + a_{12} c_{21} + b_{12} c_{21} & a_{11} c_{12} + b_{11} c_{12} + a_{12} c_{22} + b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Entrywise Addition The expression for $\\paren {\\mathbf A", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "\\\\$ What’s interesting is that you can take a step with any magnitude! If we calculated the limit of $\\Delta X$ as it approaches infinity, we would see that: $\\lim_{x \\to \\infty} \\Delta X = \\lim_{x \\to \\infty} -\\begin{pmatrix} \\frac{h}{1+h}x_0 \\\\[0.5em] \\frac{h}{1+kh}ky_0 \\end{pmatrix} = -\\begin{pmatrix} x_0 \\\\[0.5em] \\frac{1}{k}ky_0 \\end{pmatrix} \\\\[0.5em] = -\\begin{pmatrix} x_0 \\\\[0.5em] y_0 \\end{pmatrix} \\\\[0.5em]$ So $X*{new} = X_0 + -(X_0) = \\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix}$ as _h* approaches infinity! Now we can use the program I wrote to see this example. after navigating to the folder where you downloaded the code, run the following: make ./out 0.2 1000.0 100.0 100.0 where h=0.2, k=1000.0, x0=100.0 and **y0 **=100.0. You should see two files implicit.txt and explicit.txt added to the folder. In implicit.txt we see that y converges to zero! $\\begin{array}{cc} x_0 & y_0 \\\\ \\hline \\\\ 100.000000 & 100.000000 \\\\ 83.333328 & 0.497505 \\\\ 69.444443 & 0.002475 \\\\ 57.870369 & 0.000012 \\\\ 48.225307 & 0.000000 \\\\ 40.187756 & 0.000000 \\\\ 33.489796 & 0.000000 \\\\ 27.908163 & 0.000000 \\\\ 23.256802 & 0.000000 \\\\ 19.380669 & 0.000000 \\\\ \\end {array}$ But in explicit.txt it’s a completely different story: $\\begin{array}{cc} x_0 & y_0 \\\\ \\hline \\\\ 100.000000 & 100.000000 \\\\ 80.000000 & -19900.000000 \\\\ 64.000000 & 3960100.000000 \\\\ 51.200001 & -788059904.000000 \\\\ 40.959999 & 156823928832.000000 \\\\ 32.767998 &" ]

[ "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", ")}$ or $r=\\frac{ed}{(r± e \\sin \\theta)}$ where r traces out the curve as $$r= f( \\theta )$$ and e follows an ellipse if e < 1, a parabola if e = 1, a hyperbola if e > 1. Public Topic 2 years 17 weeks ago Polar Equation for Conics", "## Curvature of functions Find the curvature of the following curve at the specified point. $r=a \\sin 3\\theta$; $\\theta =\\frac{\\pi}{12}$ I'm pretty much lost on this one. Thanks for any help!", "# pedal equations Find the pedal equation of the curve: $\\frac{l}{r}=1-e\\cos\\theta$, with respect to focus.", "same. The conventional equation is $$r(\\theta) = \\frac{ed}{1-e\\sin\\theta}$$. Our new equation is $$r(\\theta) = \\frac{R(1+e)}{1-e\\sin\\theta}$$. This means that R(1 + e) = ed. Remember that d is the distance from the focus to the conventional directrix. This allows us the state the distance from the vertex of the conic to the conventional directix in terms of e and R. Specifically, that distance is R/e. This is no surprise since we can also obtain the same result using a, b, and c. Negative Eccentricity Figure 3: Negative values of e and conics (to be loaded) When e < –1, the conic is a hyperbola in blue. When –1 < e < 0, the conic is an ellipse in red. The generating circle is shown in green. The line y = 2R is in cyan. Other Forms Originally, when I worked this out, I used a circle of radius R centered at (0, R) and the line y = 2R to derive the equations. Below is what I discovered originally before I reworked out the equations presented above. The Parabola A parabola is a locus of all points such that a point on the parabola is equidistant from a given line and a fixed point, the focus. This is the classic definition of a parabola. Its eccentricity, e, is 1.", "Tangents to Polar Curve)}\\qquad \\frac{dy}{dx}=\\frac{dy/d\\theta}{dx/d\\theta}=\\frac{f'(\\theta)\\sin\\theta+f(\\theta)\\cos\\theta}{f'(\\theta)\\cos\\theta-f(\\theta)\\sin\\theta}. }", "# Asymptotes of Hyperbolas Before discussing rectangular hyperbolas, we must first understand what asymptotes are. Asymptotes : An asymptote to a curve is a straight line, to which the tangent to the curve tends as the point of contact goes to infinity. If this sounds confusing, you can think of an asymptote as follows: an asymptote to a curve is a straight line such that the perpendicular distance of a point $$P(x,\\,y)$$ on the curve from this line tends to zero as the point P goes to infinity (along some branch of the curve). For example, the line $$y=0$$ is an asymptote to the curve \\begin{align}y=\\frac{1}{1+{{x}^{2}}}\\end{align} as shown below : Consider a hyperbola \\begin{align}\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1.\\end{align} Can this have asymptotes ? If so, what will be their equations ? Let $$y=mx+c$$ be an asymptote to the given hyperbola. We need to determine m and c. Consider any point $$P(a\\sec \\theta ,\\,b\\tan \\theta )$$ on the hyperbola. The perpendicular distance of P from our supposed asymptote $$y=mx+c$$ will be given by ${{d}^{2}}=\\frac{{{(am\\sec \\theta -b\\tan \\theta +c)}^{2}}}{1+{{m}^{2}}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\ldots \\left( 1 \\right)$ Now first observe that as $$\\theta \\to \\pm \\frac{\\pi }{2},$$ the point P tends to go away to infinity along some arm of the hyperbola. Thus, we want a value of m and c, such that in (1), As\\;\\quad\\begin{align}\\theta \\to \\pm \\frac{\\pi", "# General Parametric Curves In this post, we will take a look at some of the general parametric curves in $\\mathbb{R}^{2}$ as well as $\\mathbb{R}^{3}$, viz., circles, cycloids, and helixes; explore their curves to see how and why their respective equations describe such paths. # Circles A circle can be defined as the locus of all points that satisfy the equations $$x = a\\cos{\\theta}\\quad y = a\\sin{\\theta}$$ where $x$, $y$ are the coordinates of any point on the circle, $a$ is the radius of the circle and $\\theta$ is the parameter - the angle subtended by the point at the centre of the circle. 0 0 Figure 1: Circle as a parametric curve Another way to parameterize a circle by rational functions is: $$x = \\frac{1-\\theta^{2}}{1+\\theta^{2}}\\quad y = \\frac{2\\theta}{1+\\theta^{2}}$$ It is implied that any point not on the circle does not satisfy this pair of equations. If the centre of the circle does not lie on the origin, we translate the circle from the origin to its proper location, i.e.:$$x = h + a\\cos{\\theta}\\quad y = k + a\\sin{\\theta}$$ where $(h,k)$ is the centre of the circle. # Cycloids Let $P$ be a fixed point on a circle. If the circle rolls along a straight line with uniform velocity and without slipping, the path traced out by $P$", "# Tangents of polar curves ### Tangents of polar curves In this lesson, we will learn how to find the tangent line of polar curves. Just like how we can find the tangent of Cartesian and parametric equations, we can do the same for polar equations. First, we will examine a generalized formula to taking the derivative, and apply it to finding tangents. Then we will look at a few examples to finding the first derivative. Lastly, we will do some applications which involve finding tangent lines of polar curves at a specified point. #### Lessons In order to find the tangent line to polar curves, we have to take the derivative in polar coordinates. Here is the formula to take the derivative in polar coordinates: $\\frac{dy}{dx}=\\frac{\\frac{dr}{d \\theta}\\sin \\theta+r\\;\\cos \\theta}{\\frac{dr}{d \\theta}\\cos \\theta-r\\;\\sin \\theta}$ • Introduction Tangents of Polar Curves Overview • 1. Finding the Derivative Find $\\frac{dy}{dx}$ for each of the following polar equations: a) $r=\\sin \\theta + \\theta$ b) $r= \\frac{\\sin \\theta}{\\cos \\theta}$ • 2. Finding the Tangent Line Find the tangent line with the following polar curves at the specified point: $r=\\sin (3\\theta)$ at $\\theta = \\frac{\\pi}{4}$ • 3. Finding the Tangent Line Find the tangent line with the following polar curves at the specified point: $r=\\theta \\cos \\theta$ at $\\theta =0$", "following curves at any point: 8. The circle $\\rho = a \\sin \\theta$. Ans. $R = \\frac{a}{2}$. 9. The spiral of Archimedes $\\rho = a\\theta$. $R = \\frac{ (\\rho^2 = a^2)^{\\frac{3}{2}} }{ \\rho^2 + 2a^2 }$. 10. The cardioid $\\rho = a (1 - \\cos \\theta)$. $R = \\frac{2}{3} \\sqrt{2a\\rho}$. 11. The lemniscate $\\rho^2 = a^2 \\cos 2 \\theta$. $R = \\frac{a^2}{3\\rho}$. 12. The parabola $\\rho = a \\sec^2 \\frac{\\theta}{2}$. $R = 2 a \\sec^3 \\frac{\\theta}{2}$. 13. The curve $\\rho = a sin^3 \\frac{\\theta}{3}$. 14. The trisectrix $\\rho = 2 a \\cos \\theta - a$. Ans. $R = \\frac{ a(5 - 4 \\cos \\theta)^{\\frac{3}{2}} }{ 9 - 6 \\cos \\theta }$ 15. The equilateral hyperbola $\\rho^2 \\cos 2 \\theta = a^2$. Ans. $R = \\frac{\\rho^3}{a^2}$. 16. The conic $\\rho = \\frac{a(1 - e^2)}{1 - e \\cos \\theta}$. Ans. $R = \\frac{a(1 - e^2)(1 - 2e \\cos \\theta + e^2)^{\\frac{3}{2}}}{(1 - e \\cos \\theta)^3}$. 17. The curve $\\begin{cases}x = 3 t^2, \\\\ y = 3 t - t^3.\\ t = 1.\\end{cases}$ Ans. $R = 6.$ 18. The hypocycloid $\\begin{cases}x = a \\cos^3 t, \\\\ y = a \\sin^3 t.\\ t = t_1.\\end{cases}$ Ans. $R = 3 a \\sin t_1 \\cos t_1.$ 19. The curve $\\begin{cases}x = a(\\cos t + t \\sin t), \\\\ y = a(\\sin t - t", "Polar representation of conic sections $r(\\theta)=\\frac1{1 + e \\cos\\theta}$ Consider a curve given in polar coordinates by $r(\\theta) = \\dfrac1{1 + e \\cos\\theta}$, where $e\\ge0$. a) Show that the distance of each point on this curve to the line $x=\\frac1e$ is a constant multiple of $r(\\theta)$. b) When $e>1$, show that the curve approaches two asymptotes, find them and sketch the curve. Hint: If the critical angles are $\\pm\\theta_0$, compute the vertical distance of the point of the curve at angle $\\theta = \\theta_0+h$ to the line $\\theta=\\theta_0$, and take a limit using Taylor approximations. c) Observe that the curve is bounded if and only if $e < 1$. Show that the curve is an ellipse as follows: Let $a$ be the midpoint between the two points intersecting the $x$-axis. Show that $(1-e^2)(x-a)^2 + y^2$ is constant. d) What happens when $e=1$? I'm not really sure where to begin with this question. I know that the curve is an ellipse if $e<1$, a parabola if $e=1$ and a hyperbola if $e>1$. I converted to parametric form, $x=\\frac{\\cos(\\theta)}{1+e\\cos(\\theta)}$, $y=\\frac{\\sin(\\theta)}{1+e\\cos(\\theta)}$ but I'm not sure this helps. I'm also not sure why this is on a differential equation assignment. Any input is greatly appreciated. EDIT: Okay, so I've got part a) down. Using the Cartesian coordinates, $\\left|\\frac{1}{e}-\\frac{\\cos\\theta}{1+e\\cos\\theta}\\right|=\\left|1+e\\cos\\theta-e\\cos\\theta\\right|=1$ which is constant.", "## Calculus 8th Edition The graph of $$r=\\frac{3}{1+2\\sin\\theta}$$ is a hyperbola. See the attached graph. The graph of $$r=\\frac{3}{1+2\\sin\\theta}$$ is a hyperbola. See the attached graph.", "= 5\\sec\\left(\\theta - \\frac{\\pi}{4}\\right)$ describe? This equation describes the line with point $$(5, \\pi/4)$$ closest to the origin. Show question Question What curve is described by a function of the form: $r = a\\sin\\theta ?$ A circle of radius $$a$$ centered at (0, a). Show question Question What curve is described by a function of the form: $r = a \\cos\\theta ?$ A circle of radius $$a$$ through the point $$(a, 0)$$. Show question Question What curve is described by a function of the form: $r = \\frac{b}{\\sin\\theta - m\\cos\\theta} ?$ A line with slope $$m$$ through the point $$(0, b)$$. Show question Question What curve is described by a function of the form: $r = 2a\\cos\\theta + 2b\\sin\\theta?$ A circle with radius $$\\sqrt{a^2 + b^2}$$ through the point $$(a, b)$$. Show question Question Is the equation for a circle a function? The equation for a circle can be written as a function in polar coordinates, but not in rectangular coordinates. Show question 60% of the users don't pass the Polar Coordinates Functions quiz! Will you pass the quiz? Start Quiz Study Plan Be perfectly prepared on time with an individual plan. Quizzes Test your knowledge with gamified quizzes. Flashcards Create and find flashcards in record time. Notes Create beautiful notes faster than ever before. Study Sets Have", "this direction. For an ellipse or a hyperbola we can define the pedal curve as the locus of the feet of the perpendiculars dropped from the center of the conic to its tangents. We can define a second curve as the envelope of the circles whose centers lie on an ellipse or a hyperbola and which pass through the center of the said conic. We found that these two curves differ only in scale with the later being double the former in a given dimension. If $a, b$ are the semi- major and minor axes of the parent conic, these curves have the polar equations: $\\rho^2 = a^2\\cos^2(\\theta)+b^2\\sin^2(\\theta)$ the pedal curve for an ellipse $\\rho^2 = 4\\left(a^2\\cos^2(\\theta)+b^2\\sin^2(\\theta)\\right)$ the envelope curve for an ellipse $\\rho^2 = a^2\\cos^2(\\theta)-b^2\\sin^2(\\theta)$ the pedal curve for a hyperbola $\\rho^2 = 4\\left(a^2\\cos^2(\\theta)-b^2\\sin^2(\\theta)\\right)$ the envelope curve for a hyperbola We subsequently learned that these elliptical version of the curves was termed the hippopede by the great Proclus who investigated them (Figure 6A). He was consciously one of the last in the line of great yavana investigators of curves starting from the discovery of the conics at the Platonic academy and this died with the destruction of the Greek tradition by the “Typhonic winds” of the second Abrahamism. One observes that if $a=b$ then the hippopedes become", "$r(\\theta) = \\frac{\\lambda}{\\sin \\theta}$ but it's not clear that it would not be favorable for the curve to descend below this line if it can follow the path of a circle for some rangle of angle. </p> <p>If this type of problem is known then references would also be appreciated. </p>", "15 views The equations $x=a\\cos\\theta+b\\sin\\theta$ and $y=a\\sin\\theta+b\\cos\\theta,(0\\leq\\theta\\leq2\\pi$ and $a,b$ are arbitrary constants$)$ represent 1. a circle 2. a parabola 3. an ellipse 4. a hyperbola in Geometry recategorized | 15 views 0 Here, $x$ should be $x = a\\cos\\theta - b \\sin \\theta$ I think this is the mistake in question??", "# How do you name the curve given by the conic r=6/(2+sintheta)? Aug 22, 2016 Ellipse #### Explanation: The polar equation of a conic, referred to a/the focus as the pole r = 0 and the perpendicular from the pole to the (corresponding) directrix as the initial line $\\theta = 0$, is $\\frac{l}{r} = 1 + e \\cos \\theta$, where e is the eccentricity of the conic and l is the semi latus rectum = (1/2) X (length of the chord of the conic through the focus,that is perpendicular to the initial line). The conic is named an ellipse, parabola or hyperbola according as e < = > 1. Interestingly, for the circle the focus is at the center and l = radius a and e = 0. The equation is simply r = a. Here, the equation is $\\frac{3}{r} = 1 + \\frac{1}{2} \\cos \\theta$, So, e = 1/2 < 1, and so, the conic is an ellipse. The semi major axis a is given by l = a(1-e^2)=a(1-1/4) =3a/4=3. So, a = 1/4#", "lobes of the figure. Hyperbola An east-west opening hyperbola can be represented parametrically by :$\\begin x &= a\\sec t + h \\\\ y &= b\\tan t + k \\end\\quad$ or, rationally $\\quad\\begin x &= a\\frac + h \\\\ y &= b\\frac + k \\end$ A north-south opening hyperbola can be represented parametrically as :$\\begin x = b\\tan t + h \\\\ y = a\\sec t + k \\\\ \\end\\quad$ or, rationally $\\quad\\begin x = b\\frac + h \\\\ y = a\\frac + k \\\\ \\end$ In all these formulae (''h'',''k'') are the center coordinates of the hyperbola, ''a'' is the length of the semi-major axis, and ''b'' is the length of the semi-minor axis. Hypotrochoid A hypotrochoid is a curve traced by a point attached to a circle of radius ''r'' rolling around the inside of a fixed circle of radius ''R'', where the point is at a distance ''d'' from the center of the interior circle. Image:2-circles hypotrochoid.gif| A hypotrochoid for which ''r'' = ''d'' Image:HypotrochoidOutThreeFifths.gif| A hypotrochoid for which ''R'' = 5, ''r'' = 3, ''d'' = 5 The parametric equations for the hypotrochoids are: :$\\begin x \\left(\\theta\\right) &= \\left(R - r\\right)\\cos\\theta + d\\cos\\left\\left(\\theta\\right\\right) \\\\ y \\left(\\theta\\right) &= \\left(R - r\\right)\\sin\\theta - d\\sin\\left\\left(\\theta\\right\\right) \\end$ Some sophisticated functions Other examples are shown: :$\\begin x &=- b\\cos\\left(t\\right)\\ +", "Free Version Easy # Given Slope, Find the Type of Curve APCALC-SBPLKC If the slope of a certain curve is $-\\cfrac {x}{y}$ for every point $(x, \\, y)$, then the curve is a(n) A Hyperbola. B Parabola. C Ellipse. D Circle.", "# Solve the following : Question: What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle $0 / 2$ with the horizontal? Solution: Since acceleration in horizontal direction is zero $u \\cos \\theta=\\operatorname{vcos}\\left(\\frac{\\theta}{2}\\right)$ $v=\\frac{u \\cos \\theta}{\\cos _{\\frac{\\theta}{2}}^{\\theta}}$ $\\mathrm{a}_{\\text {radial }}=\\frac{V^{2}}{R}$ $R=\\frac{\\frac{w^{2} \\cos ^{2} \\theta}{\\cos ^{2} \\theta / 2}}{g \\cos \\theta / 2}$ $R=\\frac{u^{2} \\cos ^{2} \\theta}{g \\cos ^{3}(\\theta / 2)}$" ]

[ "|z|=r$$. And we can write it in this form: $$\\large\\rm z=r(\\cos\\theta+i\\sin\\theta)$$ 8. zepdrix Differentiate this with respect to the argument, theta,$\\large\\rm \\frac{dz}{d \\theta}=r(-\\sin \\theta+i \\cos \\theta)$And then I guess this is where the tricky step comes in. We're going to factor a -1 out of each term: 9. zepdrix Woops, I mean factor an $$\\large\\rm i$$ out of each term :) $\\large\\rm \\frac{dz}{d \\theta}=r i(i\\sin \\theta+\\cos \\theta)$Recall that $$\\large\\rm -1=i\\cdot i$$ 10. zepdrix So what we've shown is that:$\\large\\rm \\frac{dz}{d \\theta}= i r(\\color{orangered}{\\cos \\theta+i \\sin \\theta})$Ooo look we've got our z showing up!$\\large\\rm \\frac{dz}{d \\theta}= i r(\\color{orangered}{z})$ 11. zepdrix Woops woops, little mistake, my bad. the r is part of my z 12. zepdrix $\\large\\rm \\frac{dz}{d \\theta}= i (\\color{orangered}{z})$ 13. anonymous I'm still here - just processing the information and looking everything up and cross referencing in my formula book ... ! 14. zepdrix Then separation of variables and integrating gives us:$\\large\\rm \\int\\limits \\frac{dz}{z}=\\int\\limits i d\\theta$Leading to,$\\large\\rm \\ln z=i \\theta+c$ 15. IrishBoy123 brill astro!! historically, Euler found this from comparing Taylor series 16. zepdrix Exponentiating gives us:$\\large\\rm z=e^{i \\theta+c}=e^{i \\theta}e^c$ $\\Large\\rm z=e^c \\color{orangered}{e^{i \\theta}}=r(\\color{orangered}{\\cos \\theta+i \\sin \\theta})$ 17. Astrophysics Thanks irishboy xD Nice one zep! 18. anonymous Okay. Working on the trig method. In order to understand (and recall old lessons) I am starting from scratch given from", "value: In [2]: v0, g, theta = sym.symbols('v0 g theta') R = v0**2*sym.sin(2*theta)/g Rdiff = sym.diff(R, theta) display(Math(sym.latex('R =') + sym.latex(R))) display(Math(sym.latex(r'\\frac{\\mathrm{d}R}{\\mathrm{d}\\theta} =') + sym.latex(Rdiff))) ang = sym.solve(Rdiff, theta) display(Math(sym.latex('Sols:') + sym.latex(ang))) display(Math(sym.latex(r'R(\\theta=\\pi/4)=') + sym.latex(sym.N(R.subs({v0: 10, g: 9.8, theta: sym.pi/4}), 4)))) plot(R.subs({v0: 10, g: 9.8}), (theta, 0, sym.pi/2), xlabel=r'$\\theta(rad)$', ylabel=r'$R$'); $$R =\\frac{v_{0}^{2}}{g} \\sin{\\left (2 \\theta \\right )}$$ $$\\frac{\\mathrm{d}R}{\\mathrm{d}\\theta} =\\frac{2 v_{0}^{2}}{g} \\cos{\\left (2 \\theta \\right )}$$ $$Sols:\\left [ \\frac{\\pi}{4}, \\quad \\frac{3 \\pi}{4}\\right ]$$ $$R(\\theta=\\pi/4)=10.2$$ ### Maximum volume of a box¶ We want to make a box from a square cardboard with side $a$ such that its volume should be maximum. What is the optimal distance where the square cardbox should be cut and folded to make a box with maximum volume? If the distance where to cut and fold the cardbox is $b$, see figure above, the volume of the box will be: $$\\begin{array}{l l} V(b) = b(a-2b)(a-2b) \\\\ \\\\ V(b) = a^2b - 4ab^2 + 4b^3 \\end{array}$$ The expression for $V$ is the cost function, $b$ is the design variable, and the constraint is that feasible values of $b$ are in the interval $[0, a/2]$. The first derivative of $V$ is: $$\\frac{\\mathrm{d}V}{\\mathrm{d}b} = a^2 - 8ab + 12b^2$$ We have to find the values for $b$ where this derivative is zero. Let's use Sympy for that: In [3]:", "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "draw(X--Y); label(\"r-1\",(X+B)/2,S); label(\"1\",(Y+C)/2,N); label(\"s\",(O+Y)/2,W); label(\"s\",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label(\"1\",(C+P)/2,NE); label(\"r\",(B+P)/2,NE); [/asy]$ (diagram by copeland?) Then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ , which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$. Subtracting $r^2-2r+1$ from both sides, $4r=4s^2$ Solving for s, we end up with $$s=\\sqrt{r}$$. Next, we can find the volume of the frustum and of the sphere. Since we know $V_{frustum}=2V_{sphere}$, we can solve for $s$ using $V_{frustum}=\\frac{\\pi*h}{3}(R^2+r^2+Rr)$ we get: $$V_{frustum}=\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)$$ Using $V_{sphere}=\\dfrac{4r^{3}\\pi}{3}$ , we get $$V_{sphere}=\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ so we have: $$\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)=2*\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ Dividing by $\\frac{2\\pi*\\sqrt{r}}{3}$, we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ $r=\\dfrac{3\\pm\\sqrt{(-3)^2-4*1*1}}{2*1}$ , so $$r=\\dfrac{3+\\sqrt{5}}{2} \\longrightarrow \\boxed{\\textbf{(E)}}$$", "equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From similar triangles, we have that $PC=\\frac{x\\sqrt{x^2+18x}}{x+9}$. Similarly, $TP=QT=\\frac{9\\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem, $BQ=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem once again, $BP=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2+(\\frac{18\\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\\sqrt{405+\\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\\sqrt{432}$, so the answer is $(\\sqrt{432})^2=\\boxed{432}$. Solution 4 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\\frac{9x}{x+9}$. Now, apply law of cosines on $\\triangle PAB$. $$BP^2=\\left(\\frac{9x^2}{x+9}\\right)^2+18^2-2(18)\\left(\\frac{9x}{x+9}\\right)\\cos(\\angle PAB).$$ We can see that $\\cos(\\angle PAB)=\\cos(180^{\\circ}-\\angle", "draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); label(\"$y = mx$\", (8,12*sqrt(221)/49*8), N); dot(\"$(9,6)$\", (9,6), NE); dot(\"$(kr,r)$\", O[1], N); dot(P,red); [/asy] Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\\frac{|r - krm|}{\\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \\neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \\neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \\frac{2k}{k^2 - 1} = \\frac{2 \\sqrt{\\frac{117}{68}}}{\\frac{117}{68} - 1} = \\boxed{\\frac{12 \\sqrt{221}}{49}}.$$", "You have $$x(0)=\\frac{f}{r}cos(\\theta ) \\mbox{ and }v(0)=\\frac{f\\omega}{r}\\sin(\\theta )$$, use that $$cos(\\theta ) = \\frac{1}{\\sqrt{1+tan^2(\\theta )}}\\mbox{, }sin(\\theta )=\\frac{tan(\\theta )}{\\sqrt{1+tan^2(\\theta )}} \\mbox{ and } tan(tan^{-1}(\\theta))=\\theta$$. ehild", "as possible: . $\\frac{\\sin[2(x-y)]}{2\\cos(x)\\cos(y)\\cos(x-y)}$ In the numerator, use: . $\\sin2\\theta \\:=\\:2\\sin\\theta\\cos\\theta$ We have: . $\\frac{{\\color{red}\\rlap{/}}2\\sin(x-y){\\color{red}\\rlap{/////////}}\\cos(x-y)}{{\\color{red}\\rlap{/}}2\\cos(x)\\cos(y){\\color{red}\\rlap{/////////}}\\cos(x-y)} \\;=\\;\\frac{\\sin(x-y)}{\\cos(x)\\cos(y)}$ . . $= \\;\\frac{\\sin(x)\\cos(y) - \\cos(x)\\sin(y)}{\\cos(x)\\cos(y)} \\;=\\;\\frac{\\sin(x){\\color{red}\\rlap{/////}}\\cos(y)}{\\cos(x){\\color{red}\\rlap{/////}}\\cos(y)} - \\frac{{\\color{red}\\rlap{/////}}\\cos(x)\\sin(y)}{{\\color{red}\\rlap{/////}}\\cos(x)\\cos(y)}$ . . $=\\;\\frac{\\sin(x)}{\\cos(x)} - \\frac{\\sin(y)}{\\cos(y)} \\;=\\;\\tan(x) - \\tan(y)$ 9. Originally Posted by Amer I prefer to give you hints and if you can't solve it ask for help for the first question $\\cos^{-1}a=0 \\Rightarrow \\cos(\\cos^{-1}a)=cos(0) \\Rightarrow a=1$ since $\\cos(\\cos^{-1}a)=a$ the same thing for $sin^{-1}(\\frac{\\pi}{6})=b$ $\\frac{\\pi}{6}=\\sin(b)$ now you have a=1 and you have sin(b) now to find the value of $\\sin(a\\pi+b)=sin(\\pi+b)$ since a=1 use $\\sin(x+y)=\\sin(x)\\cos(y)+\\sin(y)\\cos(x)$ the second question $sin(\\frac{-5\\pi}{12})=??$ first thing you should know $sin(-\\theta)=-sin(\\theta)$ you can write the angle $\\frac{5\\pi}{12}=\\frac{3\\pi}{12}+\\frac{2\\pi}{12}$ then use the formula $\\sin(x+y)=\\sin(x)\\cos(y)+\\sin(y)\\cos(x)$ I'm still struggling with #1 and #5. Could anyone help me further with them? Also, I would like to thank all of you so much, for being so kind and helpful! 10. Originally Posted by wdma I'm still struggling with #1 and #5. Could anyone help me further with them? Also, I would like to thank all of you so much, for being so kind and helpful! 5) \\begin{aligned} 2\\sin^2 x &= 1 \\\\ \\sin^2 x &= \\frac{1}{2} \\\\ \\sin x &= \\pm \\frac{\\sqrt{2}}{2} \\end{aligned} You should have recognized the ratios $\\pm\\frac{\\sqrt{2}}{2}$ (the alternate form is $\\frac{1}{\\sqrt{2}}$). Ask yourself, the sine of what angles is $\\frac{\\sqrt{2}}{2}$? (There are", "value of $\\mu$)... ok so i get my three equations as: $$\\frac{d^2 r}{dt^2}-r \\frac{d^2 \\theta}{dt^2} - r \\sin^2{\\theta} \\frac{d^2 \\phi}{dt^2}=0$$ $$\\frac{d^2 \\theta}{dt^2} + \\frac{2}{r} \\frac{dr}{dt} \\frac{d \\theta}{dt} - \\sin{\\theta} \\cos{\\theta} \\frac{d^2 \\phi}{dt^2}=0$$ $$\\frac{d^2 \\phi}{dt^2} + \\frac{2}{r} \\frac{dr}{dt} \\frac{d \\phi}{dt} + 2 \\cot{\\theta} \\frac{d \\theta}{dt} \\frac{d \\phi}{dt}=0$$ do they look correct? how on earth do i solve them? Recognitions: Homework Help Assuming you meant $\\frac{d^2 \\theta}{dt^2} +\\frac{2}{r} \\frac{dr}{dt} \\frac{d \\theta}{dt} - \\sin{\\theta} \\cos{\\theta} \\frac{d^2 \\phi}{dt^2}=0$ for the 2nd ODE, then yes those look right....Now, do you really need to solve them, in order to verify that the solutions are straight lines in Cartesian coordinates?...What is the general (parametrized in terms of $t$) form of a straight line in Cartesian coords?...What is it when you convert to Spherical coords?...Does that form satisfy the ODEs? Could there be any other solutions? in cartesian isnt it just going to be $\\vec{r}(t)=\\vec{a}+\\vec{b}t$ with $\\vec{a}, \\vec{b} \\in \\mathbb{R}^3$. i'm not sure how to convert that to sphericals though? have i used the wrong form above? Recognitions: Homework Help No, the general parametrized form of a line in 3D, is $ax(t)+by(t)+cz(t)=0$ $$\\frac{x-x_0}{a}=\\frac{y-y_0}{b}=\\frac{z-z_0}{c}$$? Recognitions: Homework Help Did you see my previous post? ok. so if that's the cartesian form i need to convert it to spherical polars: $$ar(t) \\sin{\\theta (t) } \\cos{\\phi (t) } + br(t)", "to order 1 : $g_0R_0(1-\\frac{h}{R_0}) + \\frac{\\omega^2R^2_0\\sin^2\\, \\theta}{2}(1+\\frac{2h}{R_0}) - g_0R_0(1-\\frac{h_P}{R_0}) = 0$ After simplifications, $-g_0h + \\frac{\\omega^2R^2_0\\sin^2\\, \\theta}{2} + \\omega^2R_0h\\sin^2\\, \\theta + g_0h_P = 0$ For $\\theta = pi/2$, we get : $-g_0h_E + \\frac{\\omega^2R^2_0}{2} + \\omega^2R_0h_E + g_0h_P = 0$ Except that I have a $h_E$ with a product, not possible to isolate $\\Delta h$ ... Where is my mistake? Thank you.", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "way: $$r_1e^{1\\theta_1}\\cdot r_2e^{1\\theta_2}=r_1r_2e^{i(\\theta_1+\\theta_2)},$$, $$\\frac{r_1e^{1\\theta_1}}{r_2e^{1\\theta_2}}=\\frac{r_1}{r_2}e^{i(\\theta_1-\\theta_2)}$$, z_{1}=2(cos(\\frac{pi}{3})+i sin (\\frac{pi}{3}) )=2e^{i\\frac{pi}{3}}\\\\z_{2}=1(cos(\\frac{pi}{6})-i sin (\\frac{pi}{6}) )=1(cos(\\frac{pi}{6}) In your case, $a,b,c$ and $d$ are all given so just plug in the numbers. Thanks to all of you who support me on Patreon. When dividing two complex numbers you are basically rationalizing the denominator of a rational expression. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Answer your tough homework and study questions direct sum matrices into the form =! Tileable, vertical redstone in minecraft redstone in minecraft has angle A_ANGLE_REP and radius.... To many people and Simplify i am stuck at square one, any help would great... 'S salary receipt Open in its respective personal webmail in someone else 's computer indicated operations an write final. Help would be great times cosine alpha plus i sine beta a page! Missing something LOOse '' pronounced differently are all given so just plug in the complex number all you have do! The domains *.kastatic.org and *.kasandbox.org are unblocked, just like vectors, can also be written polar. Entire Q & a library imaginary part: a + 0i 1 Sina! Of a complex number corresponds to a point", "= \\frac{1}{p_\\theta} - \\frac{1}{q_\\theta} = \\frac{1}{q_\\theta'} - \\frac{1}{p_\\theta'}.$$ Note $$\\frac{1}{r_3} = 1 - \\frac{1 - \\theta}{q_0} - \\frac{\\theta}{p_1'} \\ge 1 - (1 - \\theta) - \\theta = 0.$$ If $\\frac{1}{r_3} = 0$, then we proceed as below but using Young's inequality for the two variables $r_1$ and $r_2$ and ignoring all terms with $r_3$. Also let $$r = \\frac{q_0}{r_1}, \\hspace{0.2in} s = \\frac{p_1'}{r_2}, \\hspace{0.2in} t = \\frac{p_\\theta}{r_1},$$ $$u = \\frac{p_\\theta}{r_3}, \\hspace{0.2in} v = \\frac{q_\\theta'}{r_2}, \\hspace{0.2in} w = \\frac{q_\\theta'}{r_3}.$$ By construction, these satisfy $$\\frac{1}{r_1} + \\frac{1}{r_2} + \\frac{1}{r_3} = 1,$$ $$r + s = 1, \\hspace{0.2in} t + u = 1, \\hspace{0.2in} v + w = 1,$$ $$rr_1 = q_0, \\hspace{0.1in} sr_2 = p_1', \\hspace{0.1in} tr_1 = p_\\theta, \\hspace{0.1in} vr_2 = q_\\theta', \\hspace{0.1in} ur_3 = p_\\theta, \\hspace{0.1in} wr_3 = q_\\theta'.$$ Thus by Young's inequality for three variables (or two variables if $r_3$ is infinite), $$\\left|K(x, y) f(x) h(y)\\right| = (|K(x, y)|^{r}|f(x)|^t)(|K(x, y)|^s |h(y)|^v)(|f(x)|^u |h(y)|^w) \\le \\frac{1}{r_1} |K(x, y)|^{rr_1}|f(x)|^{tr_1} + \\frac{1}{r_2} |K(x, y)|^{sr_2} |h(y)|^{vr_2} + \\frac{1}{r_3} |f(x)|^{ur_3} |h(x)|^{wr_3}$$ $$\\le \\frac{1}{r_1} |K(x, y)|^{q_0} |f(x)|^{p_\\theta} + \\frac{1}{r_2} |K(x, y)|^{p_1'} |h(y)|^{q_\\theta'} + \\frac{1}{r_3}|f(x)|^{p_\\theta} |h(y)|^{q_\\theta'} \\hspace{0.2in} (1).$$ Using Fubini-Tonelli and the assumption that $||K(x, \\cdot)||_{L^{q_0}(Y)} \\le 1$ for almost every $x \\in X$ \\begin{eqnarray*} \\int_{Y} \\int_X |K(x, y)|^{q_0} |h(y)|^{q_\\theta'} \\, dx \\, dy & = & \\int_X \\int_Y |K(x, y)|^{q_0} |h(y)|^{q_\\theta'} \\, dy \\,", "$$\\frac{f(x,cx)}{g(x,cx)} = \\frac{cx^3}{x^4+c^2x^2} =\\frac{cx}{c^2+x^2}\\to 0 \\quad\\text{as }x\\to 0\\,,$$ unless $c=0$, but that case is easily handled directly. On the other hand, if we let $y=x^2$, then $$\\frac{f(x,x^2)}{g(x,x^2)} = \\frac{x^4}{x^4+x^4} = \\frac12 \\quad\\text{for all }x\\ne 0\\,,$$ If you want a single application of L'Hôpital to do it, let's try $$f(x,y) = x|y|^{1/2} \\quad\\text{and}\\quad g(x,y)=x^2+|y|\\,.$$ The same sort of analysis works here: Along $y=cx$, $\\dfrac{f(x,cx)}{g(x,cx)} \\to 0$, and along $y=x^2$, $\\dfrac{f(x,x^2)}{g(x,x^2)}=\\dfrac{|x|}{x} = \\pm 1$. I believe that if your limit $C\\ne 0$ and $f(r,\\theta)$ and $g(r,\\theta)$ are actually differentiable functions, then your result works, because we'll have $$f(r,\\theta) = arh(\\theta) + o(r) \\quad\\text{and}\\quad g(r,\\theta)=brk(\\theta) + o(r)\\,,$$ from which we'll have $$\\frac{f(r,\\theta)}{g(r,\\theta)} = \\frac{a r h(\\theta) + o(r)}{b r k(\\theta) + o(r)} \\to \\frac{ah(\\theta)}{bk(\\theta)}\\,;$$ if this is a constant $C$, we must have $h(\\theta) = k(\\theta)$, and it works. - I'm sorry, but I still can't figure it out ... Let's look at the second example you suggested. When moving to polar coordinates, and differentiate with respect to $\\theta$ , we get the limit of $\\frac{1.5 \\sqrt{r} cos(\\theta) \\sqrt{|sin \\theta | } }{2rcos^2 \\theta + |sin \\theta| }$ , and this limit does not exist (we can take $\\theta =r$ and then the limit is $0/0$ ... Can you please re explain why did you say that this limit is a", "r(-sin\\theta)\\frac{d\\theta}{dt}.$", "$x_\\theta = -y , y_\\theta = x$ $\\frac{df}{d\\theta} = f_x (-y) + f_y (x)$ simplify it and find the sum you will get the solution $\\sin ^2 x + \\cos ^2 x =1$ use this 7. $x= e^r \\cos \\theta$ $y=e^r \\sin \\theta$ $x_\\theta = \\frac{\\partial x}{\\partial \\theta } = e^r (-\\sin \\theta) = -e^r\\sin \\theta = -y$ $y_\\theta = \\frac{\\partial y}{\\partial \\theta } = e^r \\cos \\theta = x$ 8. er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ 9. Originally Posted by billym er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ $f_r = f_x \\cdot x_r + f_y \\cdot y_r$ $f_{rr} = (f_{xx}\\cdot x_r)x_r + f_x\\cdot x_{rr} + (f_{yy} \\cdot y_r)y_r + f_y \\cdot y_{rr}$ now $x_{rr} = x_r=x , y_{rr} = y_r=y$ $f_{rr} = f_{xx}\\cdot x^2 + f_x \\cdot x + f_{yy}\\cdot y^2 + f_y \\cdot y$ .................(1) other one I will symbol theta with t ok $f_{t} = f_x \\cdot x_t + f_y \\cdot y_t$ $f_{tt}= (f_{xx} x_t)x_t + f_x \\cdot x_{tt} + (f_{yy} \\cdot y_t)y_t + f_y \\cdot y_{tt}$ $x_t = -y , y_t = x$ and $x_{tt} = -x , y_{tt} = -y$ $f_{tt} = f_{xx} \\cdot y^2 + f_x \\cdot -x + f_{yy} \\cdot x^2 + f_y \\cdot -y$.........(2) sub x value and y value in 1,2 and", "sx + d$ $s = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{cos(\\theta)}{sin(\\theta)}$ and $d = \\frac{\\rho}{sin(\\theta)}$ so that we can represent the set of lines passing through a pixel at $(x_0, y_0)$ with the expression $\\rho = -x_0 cos(\\theta) + y_0 sin(\\theta)$ With this expression, called normal Hesse form or simply normal form, we can represent the set of lines that pass through a pixel at $(x_0, y_0)$ in the image by a sinusoidal function in the parameter space $(\\theta, \\rho)$. If two points belong to the same line in the image, their representations as sinusoidal functions in the parameter space must intersect at a certain point $(\\theta_0, \\rho_0)$. Similarly to what has been said before, the more sinusoidal curves intersect in a point $(\\theta_0, \\rho_0)$ of the parameter space, the more its corresponding line in the image ranks high enough to be elected as a real line. We can count the number of sinusoidal curves that intersect at each point of the parameter space by dividing this space into a grid of cells whose width and height depends on the angular and spatial resolution of the image. For example, if we can distinguish two lines in the image that are rotated by 1 degree and two lines that are separated by one pixel, we can set the", "\\sqrt{r} \\cos \\theta$ and $Y = \\sqrt{r} \\sin \\theta$ have a joint pdf on the disk $\\frac{1}{\\pi}$, i.e. describes a point picked uniformly at random on the disk. And $R = \\sqrt{X^2 + Y^2}$.", "call it a. a is the amount of lines I want drawn from the centre of the circle to the edge of the circle, all evenly spaced (so r is the length of each line). I'm assuming that in order to do this, I have to do this: 360 / a, which gives me a degree (I'm totally making these variable names up) which I will somehow use to evenly space the lines. This leads to the question! Given xy, r & a, how can I go about getting the x & y co-ordinates of the lines' end-points? I hope this is clear enough! Please let me know if I need to clarify anything. - ## 1 Answer Each line extends from $(x,y)$ to $(x+r \\cos \\theta, y+r \\sin \\theta)$. You are right that if you want $a$ lines to fill up the circle, they should be spaced at $\\frac {360}a$ degree intervals. So use $\\theta = \\frac {360n}a$ for $n=0,1,2,\\ldots a-1$ - The only part I don't understand is the θ=360na for n=0,1,2,…a−1. Can you clarify it? Again, I haven't done anything like this in years :) – Djentleman May 28 '12 at 10:39 @Djentleman: You want a total of $a$ lines. I was presuming you start with one at $\\theta = 0$, so if you count", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$." ]

[ "# Math Help - Curve Tracing 1. ## Curve Tracing The equation of the curve is- r^2cos2(theta)=q^2 Sorry I don't know to write this in Natural form but I just hope that I am clear enough 2. What are you supposed to do with this equation? 3. Originally Posted by stapel What are you supposed to do with this equation? Make the cure mate.. 4. That's even more confusing! And what is \"q\"? Some constant? Do you know the identity \" $cos(2\\theta)= cos^2(\\theta)- sin^2(\\theta)$\"? You can use that to write your equation as $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q$ and use the fact that $x= r cos(\\theta)$, $y= r sin(\\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me. 5. Originally Posted by HallsofIvy That's even more confusing! And what is \"q\"? Some constant? Do you know the identity \" $cos(2\\theta)= cos^2(\\theta)- sin^2(\\theta)$\"? You can use that to write your equation as $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q$ and use the fact that $x= r cos(\\theta)$, $y= r sin(\\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me. Yes q is constant..!! and the question is $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q^2$ So I think it will be a rectangular hyperbola..?? 6. Originally Posted by stapel What are you supposed to do with this equation? You have to make the curve with this.", "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", "from that of the ellipse in having −b2 instead of b2 , it will be found that many propositions for the hyperbola are derived from those for the ellipse by changing the sign of b2 . Some results for the hyperbola $\\displaystyle \\frac{x^2}{a^2} – \\frac{y^2}{b^2} = 1$ are : (i) The tangent at any point (x1, y1) on the curve is $\\displaystyle \\frac{x x_1}{a^2} – \\frac{y y_1}{b^2} = 1$ (ii) The tangent at the point ‘ θ ‘ is $\\displaystyle \\frac{x sec\\theta}{a} – \\frac{y tan\\theta}{b } = 1$ (iii) The straight line y = mx + c is a tangent to the curve , if c2 = a2 m2 − b2 . In other words , $\\displaystyle y = m x \\pm \\sqrt{a^2 m^2 – b^2}$ touches the curve, for all those values of m when m > b/a or m < − b/a. If m = ±b/a the tangents touch at infinity and are called as asymptotes. (iv) Equation of the normal at any point (x1, y1) to the curve is $\\displaystyle \\frac{x-x_1}{x_1^2 /a^2} = \\frac{y-y_1}{y_1/(-b^2)}$ (v) The equation of the chord through the points θ1 and θ2 is $\\large \\left| \\begin{array}{ccc} x & y & 1 \\\\ a sec\\theta_1 & b tan\\theta_1 & 1 \\\\ a sec\\theta_2 & b tan\\theta_2 & 1\\end{array} \\right| =0$ (vi) The", "OP.) – John Hughes Aug 2 '17 at 12:14 Let $r_1$ and $r_2$ straight lines so that: $$r_1: y =kx$$ and $$r_2: y =0.$$ Let $\\Gamma$ a circle centered at $C_0=(x_0,y_0)$ with radius $r$. Solve the following equation for $\\theta$ and you will get the point T of tangency. $$ky_0\\sin(\\theta)+kr(\\sin(\\theta))^2=kx_0\\cos(\\theta)-2y_0\\cos(\\theta)+rk(\\cos(\\theta))^2-2r\\sin(\\theta)\\cos(\\theta)$$ $$T=(x_0+r\\cos(\\theta),y_0+r\\sin(\\theta)).$$ Draw tangent lines at T, and you will get the answer of the problem. For $x_0=2$, $y_0=1$, $k=1$ and $r=0.5$, you get $\\theta=13.18$ degrees and $\\theta=142.38$ degrees. Let $AB$ just be tangent with circle, and their midpoint $M$ goes into a hyperbola which is tangent to the circle. Now we aim to find the tangent point. Not sure if possible (we have two equations but only one unknown value, so can just remove some unnecessary part)", ")}$ or $r=\\frac{ed}{(r± e \\sin \\theta)}$ where r traces out the curve as $$r= f( \\theta )$$ and e follows an ellipse if e < 1, a parabola if e = 1, a hyperbola if e > 1. Public Topic 2 years 17 weeks ago Polar Equation for Conics", "double the former in a given dimension. If $a, b$ are the semi- major and minor axes of the parent conic, these curves have the polar equations: $\\rho^2 = a^2\\cos^2(\\theta)+b^2\\sin^2(\\theta)$ the pedal curve for an ellipse $\\rho^2 = 4\\left(a^2\\cos^2(\\theta)+b^2\\sin^2(\\theta)\\right)$ the envelope curve for an ellipse $\\rho^2 = a^2\\cos^2(\\theta)-b^2\\sin^2(\\theta)$ the pedal curve for a hyperbola $\\rho^2 = 4\\left(a^2\\cos^2(\\theta)-b^2\\sin^2(\\theta)\\right)$ the envelope curve for a hyperbola We subsequently learned that these elliptical version of the curves was termed the hippopede by the great Proclus who investigated them (Figure 6A). He was consciously one of the last in the line of great yavana investigators of curves starting from the discovery of the conics at the Platonic academy and this died with the destruction of the Greek tradition by the “Typhonic winds” of the second Abrahamism. One observes that if $a=b$ then the hippopedes become circles. If $b=0$ then again we get a pair of circles. The hyperbolic versions in contrast specify $\\infty$-shaped curves that cross at origin. If the hyperbola is rectangular (i.e. $a=b$) then we get the pedal version to be: $\\rho^2 = a^2\\left(\\cos^2(\\theta)-\\sin^2(\\theta)\\right)= a^2\\cos(2\\theta)$. This curve was first studied by Jakob Bernoulli, the eldest brother of the first famous generation of the Bernoulli clan. It has an interesting property: if the ellipse is the locus of points the sum of", "Polar representation of conic sections $r(\\theta)=\\frac1{1 + e \\cos\\theta}$ Consider a curve given in polar coordinates by $r(\\theta) = \\dfrac1{1 + e \\cos\\theta}$, where $e\\ge0$. a) Show that the distance of each point on this curve to the line $x=\\frac1e$ is a constant multiple of $r(\\theta)$. b) When $e>1$, show that the curve approaches two asymptotes, find them and sketch the curve. Hint: If the critical angles are $\\pm\\theta_0$, compute the vertical distance of the point of the curve at angle $\\theta = \\theta_0+h$ to the line $\\theta=\\theta_0$, and take a limit using Taylor approximations. c) Observe that the curve is bounded if and only if $e < 1$. Show that the curve is an ellipse as follows: Let $a$ be the midpoint between the two points intersecting the $x$-axis. Show that $(1-e^2)(x-a)^2 + y^2$ is constant. d) What happens when $e=1$? I'm not really sure where to begin with this question. I know that the curve is an ellipse if $e<1$, a parabola if $e=1$ and a hyperbola if $e>1$. I converted to parametric form, $x=\\frac{\\cos(\\theta)}{1+e\\cos(\\theta)}$, $y=\\frac{\\sin(\\theta)}{1+e\\cos(\\theta)}$ but I'm not sure this helps. I'm also not sure why this is on a differential equation assignment. Any input is greatly appreciated. EDIT: Okay, so I've got part a) down. Using the Cartesian coordinates, $\\left|\\frac{1}{e}-\\frac{\\cos\\theta}{1+e\\cos\\theta}\\right|=\\left|1+e\\cos\\theta-e\\cos\\theta\\right|=1$ which is constant.", "curve defined by $x = r(\\theta - \\sin \\theta)$ and $y = r(1 - \\cos \\theta)$. First let's differentiate dx/dθ and dy/dθ to obtain dy/dx. (5) \\begin{align} \\frac{dx}{d\\theta} = r - r\\cos \\theta \\quad \\quad \\quad \\frac{dy}{d\\theta} = r\\sin \\theta \\end{align} Therefore: (6) \\begin{align} \\frac{dy}{dx} = \\frac{r \\sin \\theta}{r - r \\cos \\theta} \\\\ = \\frac{r \\sin \\theta}{r(1 - \\cos \\theta)} \\\\ = \\frac{\\sin \\theta}{1 - \\cos \\theta} \\end{align} So dx/dθ = 0 when θ = 0, π, 2π, … But notice that when θ = 0, π, 2π, …, then dy/dθ = 0, since sin(0) = 0, sin(π) = 0, …. So we need to use our knowledge of limits in order to figure out what is happening here. Let's just use 2π as an example, (7) \\begin{align} \\lim_{\\theta \\to 2\\pi} \\frac{dy}{dx} = \\lim_{\\theta \\to 2\\pi} \\frac{\\sin x}{1 - \\cos x} \\end{align} We can use L'Hospital's Rule to evaluate this limit further, first from the right side: (8) \\begin{align} \\lim_{\\theta \\to 2\\pi^+} \\frac{\\sin x}{1 - \\cos x} \\\\ = \\lim_{\\theta \\to 2\\pi^+} \\frac{\\cos x}{\\sin x} \\\\ = \\infty \\end{align} Now let's look at the limit from the left: (9) \\begin{align} \\lim_{\\theta \\to 2\\pi^-} \\frac{\\sin x}{1 - \\cos x} \\\\ = \\lim_{\\theta \\to 2\\pi^+} \\frac{\\cos x}{\\sin x} \\\\ = -\\infty \\end{align} Since 2π exists as a point on", "the significant point to notice is that there is a minimum value for r, say r0. We conclude that this minimum value of r provides an upper estimate for the size of the nucleus. A similar calculation was first performed by Lord Rutherford i in the early years of the 20th century; prior to this time very little was known about the size of the nucleus. ### Rectangular hyperbola A special case arises when a = b, for then Equation 24, $\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1$(Eqn 24) reduces to x2y2 = a2(30) and the asymptotes (from Equation 25) $y=\\pm\\dfrac ba x$(Eqn 25) are y = x and y = −x, which are perpendicular to each other. Such a curve is known as a rectangular hyperbola. Question T12 Find the equations of the tangents to the hyperbola $\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1$ that pass through the point S(a/2, 0). From Equation 29, $\\dfrac{\\cosh\\theta_1}{a}x - \\dfrac{\\sinh\\theta_1}{b}y = 1$(Eqn 29) the tangent at the point on the hyperbola corresponding to the parameter θ is $\\dfrac{\\cosh\\theta}{a}x - \\dfrac{\\sinh\\theta}{b}y = 1$, and this passes through the point (a/2, 0) if $\\dfrac{\\cosh\\theta}{a}\\left(\\dfrac a2\\right) - \\dfrac{\\sinh\\theta}{b}0 = 1$, i.e. if coshθ = 2. Since sinh2θ = cosh2θ − 1 it follows that $\\sinh\\theta = \\pm\\sqrt{\\cosh^2\\theta-1}=\\pm\\sqrt{4-1\\os} = \\sqrt{3\\os}$ The equations of the tangent lines are therefore", "following curves at any point: 8. The circle $\\rho = a \\sin \\theta$. Ans. $R = \\frac{a}{2}$. 9. The spiral of Archimedes $\\rho = a\\theta$. $R = \\frac{ (\\rho^2 = a^2)^{\\frac{3}{2}} }{ \\rho^2 + 2a^2 }$. 10. The cardioid $\\rho = a (1 - \\cos \\theta)$. $R = \\frac{2}{3} \\sqrt{2a\\rho}$. 11. The lemniscate $\\rho^2 = a^2 \\cos 2 \\theta$. $R = \\frac{a^2}{3\\rho}$. 12. The parabola $\\rho = a \\sec^2 \\frac{\\theta}{2}$. $R = 2 a \\sec^3 \\frac{\\theta}{2}$. 13. The curve $\\rho = a sin^3 \\frac{\\theta}{3}$. 14. The trisectrix $\\rho = 2 a \\cos \\theta - a$. Ans. $R = \\frac{ a(5 - 4 \\cos \\theta)^{\\frac{3}{2}} }{ 9 - 6 \\cos \\theta }$ 15. The equilateral hyperbola $\\rho^2 \\cos 2 \\theta = a^2$. Ans. $R = \\frac{\\rho^3}{a^2}$. 16. The conic $\\rho = \\frac{a(1 - e^2)}{1 - e \\cos \\theta}$. Ans. $R = \\frac{a(1 - e^2)(1 - 2e \\cos \\theta + e^2)^{\\frac{3}{2}}}{(1 - e \\cos \\theta)^3}$. 17. The curve $\\begin{cases}x = 3 t^2, \\\\ y = 3 t - t^3.\\ t = 1.\\end{cases}$ Ans. $R = 6.$ 18. The hypocycloid $\\begin{cases}x = a \\cos^3 t, \\\\ y = a \\sin^3 t.\\ t = t_1.\\end{cases}$ Ans. $R = 3 a \\sin t_1 \\cos t_1.$ 19. The curve $\\begin{cases}x = a(\\cos t + t \\sin t), \\\\ y = a(\\sin t - t", "# Epicycloid The red curve is an epicycloid traced as the small circle (radius r = 1) rolls around the outside of the large circle (radius R = 3). In geometry, an epicycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circle—called an epicycle—which rolls without slipping around a fixed circle. It is a particular kind of roulette. ## Equations If the smaller circle has radius r, and the larger circle has radius R = kr, then the parametric equations for the curve can be given by either: {\\displaystyle {\\begin{aligned}&x(\\theta )=(R+r)\\cos \\theta \\ -r\\cos \\left({\\frac {R+r}{r}}\\theta \\right)\\\\&y(\\theta )=(R+r)\\sin \\theta \\ -r\\sin \\left({\\frac {R+r}{r}}\\theta \\right)\\end{aligned}}} or: {\\displaystyle {\\begin{aligned}&x(\\theta )=r(k+1)\\cos \\theta -r\\cos \\left((k+1)\\theta \\right)\\\\&y(\\theta )=r(k+1)\\sin \\theta -r\\sin \\left((k+1)\\theta \\right)\\end{aligned}}} in a more concise and complex form[1] ${\\displaystyle z(\\theta )=r\\left(e^{i(k+1)\\theta }-(k+1)e^{i\\theta }\\right)}$ where • angle θ is in turns: ${\\displaystyle \\theta \\in [0,2\\pi ].}$ • smaller circle has radius r • the larger circle has radius kr ## Area (Assuming the initial point lies on the larger circle.) When k is a positive integer, the area of this epicycloid is ${\\displaystyle A=(k+1)(k+2)\\pi r^{2}.}$ It means that the epicycloid is ${\\displaystyle {\\frac {(k+1)(k+2)}{k^{2}}}}$ larger than the original stationary circle. If k is a positive integer, then the curve is closed, and has", "$(6) \\quad {R}(t)=e^{-x}{i}+e^x{j}$ Exercise. Find the point(s) where the ellipse $36x^2+9y^2=121$ has maximum curvature. Exercise. Find the points on the curve in the $xy$-plane described by $y=3x-x^3$ at which the curvature is maximum. Exercise. Show that the curvature of the parabola $y=x^2+3$ approaches 0 as $x\\to \\infty .$ Exercise. Show that the curvature at every point on a line is 0. Exercise. Show that the maximum curvature of a parabola occurs at the vertex. Exercise. Show that for an ellipse, the maximum and minim curvature occurs at ends of the major and minor axes, respectively. Exercise. Show that for a hyperbola, the maximum curvature occurs at the ends of the traverse axis. Exercise. Show that the curvature of the helix $${R}(t)= a\\cos t {i}+a \\sin t {j}+b t$$ is $\\kappa=a/(a^2+b^2).$ What is the largest value that $\\kappa$ can have for a given value of $b$? Exercise. Suppose a curve $C$ in the $xy$-plane described by a polar equation $r=f(\\theta).$ Show that the curvature of $C$ is $$\\kappa(\\theta) = \\frac{| \\, \\, [f(\\theta)]^2+2[f'(\\theta)]^2-f(\\theta)f”(\\theta)\\, \\, |}{\\, {[f(\\theta)]^2+[f'(\\theta)]^2\\, }^{3/2}}.$$ Exercise. Find the curvature of the polar curve at $P(r,\\theta).$ $(1) \\quad$ the cardoid $r=a(1-\\cos \\theta)$, where $0< \\theta<2\\pi.$ $(2) \\quad$ the four-leafed rose $r=\\sin 2\\theta$, where $0<\\theta < \\pi/2$ $(3) \\quad$ the spiral $r=e^{a\\theta}$ $(4) \\quad r=1-\\cos \\theta$, $\\theta=3\\pi/2$ $(5) \\quad", "# Asymptotes of Hyperbolas Before discussing rectangular hyperbolas, we must first understand what asymptotes are. Asymptotes : An asymptote to a curve is a straight line, to which the tangent to the curve tends as the point of contact goes to infinity. If this sounds confusing, you can think of an asymptote as follows: an asymptote to a curve is a straight line such that the perpendicular distance of a point $$P(x,\\,y)$$ on the curve from this line tends to zero as the point P goes to infinity (along some branch of the curve). For example, the line $$y=0$$ is an asymptote to the curve \\begin{align}y=\\frac{1}{1+{{x}^{2}}}\\end{align} as shown below : Consider a hyperbola \\begin{align}\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1.\\end{align} Can this have asymptotes ? If so, what will be their equations ? Let $$y=mx+c$$ be an asymptote to the given hyperbola. We need to determine m and c. Consider any point $$P(a\\sec \\theta ,\\,b\\tan \\theta )$$ on the hyperbola. The perpendicular distance of P from our supposed asymptote $$y=mx+c$$ will be given by ${{d}^{2}}=\\frac{{{(am\\sec \\theta -b\\tan \\theta +c)}^{2}}}{1+{{m}^{2}}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\ldots \\left( 1 \\right)$ Now first observe that as $$\\theta \\to \\pm \\frac{\\pi }{2},$$ the point P tends to go away to infinity along some arm of the hyperbola. Thus, we want a value of m and c, such that in (1), As\\;\\quad\\begin{align}\\theta \\to \\pm \\frac{\\pi", "lobes of the figure. Hyperbola An east-west opening hyperbola can be represented parametrically by :$\\begin x &= a\\sec t + h \\\\ y &= b\\tan t + k \\end\\quad$ or, rationally $\\quad\\begin x &= a\\frac + h \\\\ y &= b\\frac + k \\end$ A north-south opening hyperbola can be represented parametrically as :$\\begin x = b\\tan t + h \\\\ y = a\\sec t + k \\\\ \\end\\quad$ or, rationally $\\quad\\begin x = b\\frac + h \\\\ y = a\\frac + k \\\\ \\end$ In all these formulae (''h'',''k'') are the center coordinates of the hyperbola, ''a'' is the length of the semi-major axis, and ''b'' is the length of the semi-minor axis. Hypotrochoid A hypotrochoid is a curve traced by a point attached to a circle of radius ''r'' rolling around the inside of a fixed circle of radius ''R'', where the point is at a distance ''d'' from the center of the interior circle. Image:2-circles hypotrochoid.gif| A hypotrochoid for which ''r'' = ''d'' Image:HypotrochoidOutThreeFifths.gif| A hypotrochoid for which ''R'' = 5, ''r'' = 3, ''d'' = 5 The parametric equations for the hypotrochoids are: :$\\begin x \\left(\\theta\\right) &= \\left(R - r\\right)\\cos\\theta + d\\cos\\left\\left(\\theta\\right\\right) \\\\ y \\left(\\theta\\right) &= \\left(R - r\\right)\\sin\\theta - d\\sin\\left\\left(\\theta\\right\\right) \\end$ Some sophisticated functions Other examples are shown: :$\\begin x &=- b\\cos\\left(t\\right)\\ +", "# Math Help - hyperbolic spiral 1. ## hyperbolic spiral The hyperbolic spiral is given by the formula (in polar form), $r\\theta=a$ for $\\theta>0$ and some postive constant $a>0$. Use the fact that $\\lim_{\\theta\\rightarrow 0}\\frac{\\sin\\theta}{\\theta}=1$ to prove there is a horizontal asympotote at $y=a$ and hence sketch the curve. Basically, all i've done is a sketch of $r=\\frac{a}{\\theta}$ which isnt of great help. Ive also tried multiplying both sides by $\\sin\\theta$ to give $y=r\\sin\\theta=\\frac{a\\sin\\theta}{\\theta}$ but im unsure how complete the question and the sketch. Can someone give me some help please. 2. Originally Posted by vuze88 The hyperbolic spiral is given by the formula (in polar form), $r\\theta=a$ for $\\theta>0$ and some postive constant $a>0$. Use the fact that $\\lim_{\\theta\\rightarrow 0}\\frac{\\sin\\theta}{\\theta}=1$ to prove there is a horizontal asympotote at $y=a$ and hence sketch the curve. Basically, all i've done is a sketch of $r=\\frac{a}{\\theta}$ which isnt of great help. Ive also tried multiplying both sides by $\\sin\\theta$ to give $y=r\\sin\\theta=\\frac{a\\sin\\theta}{\\theta}$ but im unsure how complete the question and the sketch. Can someone give me some help please. Excellent! Now use the hint that they gave you: since the limit, as $\\theta$ goes to 0, of $\\frac{sin(\\theta)}{\\theta}$ is 1, $y= \\frac{a sin(\\theta)}{\\theta}$ goes to a. y= a is a horizontal asymptote. 3. hm...i was thinking that, but dont you", "How do you name the curve given by the conic r=4/(1+costheta)? Jan 13, 2018 Convert to the General Cartesian Form : $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$ Compute the determinant: $\\Delta = {B}^{2} - 4 A C$ If $\\Delta < 0$, then it is an ellipse or a circle. If $B = 0 \\mathmr{and} A = C$, then it is a circle. Otherwise, it is an ellipse. If $\\Delta = 0$, then it is a parabola. If $\\Delta > 0$, them it is a hyperbola. Given: $r = \\frac{4}{1 + \\cos \\left(\\theta\\right)}$ $r + r \\cos \\left(\\theta\\right) = 4$ Substitute $r = \\sqrt{{x}^{2} + {y}^{2}}$ and $r \\cos \\left(\\theta\\right) = x$: $\\sqrt{{x}^{2} + {y}^{2}} + x = 4$ $\\sqrt{{x}^{2} + {y}^{2}} = 4 - x$ ${x}^{2} + {y}^{2} = {x}^{2} - 8 x + 16$ ${y}^{2} + 8 x - 16 = 0$ Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, $A = B = E = 0 , C = 1 , D = 8 , \\mathmr{and} F = - 16$ $\\Delta = {0}^{2} - 4 \\left(0\\right) \\left(1\\right) = 0$ It is a parabola.", "x=4\\cos \\theta$ and $y=3\\sin\\theta$, $0\\leq \\theta \\leq 2\\pi$. $(2) \\quad x=\\sin t$ and $y=\\sin2t$, $0\\leq t \\leq 2\\pi$. $(3) \\quad C: x=t^2$, $y=t-1$; $0\\leq t \\leq 3$ $(4) \\quad C: x=t^2+1$, $y=2t^2-1$; $-2\\leq t\\leq 2$ Exercise. Sketch the graph of the parametric equations $x=\\cos (\\pi -t)$ and $y=\\sin (\\pi -t)$, then indicate the direction of increasing $t.$ Exercise. Sketch the graph of the parametric equations $x=3 t$ and $y=9t^2$, then indicate the direction of increasing $-\\infty <t<\\infty .$ Exercise. Sketch the graph of the parametric equations $x=3-3t$ and $y=2t$, then indicate the direction of increasing $0\\leq t\\leq 1.$ Exercise. Find parametric equations and a parameter interval for the motion of a particle that starts at $(a,0)$ and traces the ellipse $$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1.$$ (a) once clockwise (b) once counterclockwise (c) twice clockwise (d) twice counterclockwise. Exercise. Find a parametrization for the curve whose graph is the lower half of the parabola $x-1=y^2.$ Exercise. Find a parametrization for the curve whose graph is the ray (half-line) with initial point $(-1,2)$ that passes through the point $(0,0).$ Exercise. Find an equation for the line tangent to the curve $x=t$ and $y=\\sqrt{t}$ at $t=1/4.$ Also, find the value of $\\frac{d^2y}{dx^2}$ at this point. Exercise. Convert $\\left(x^2+y^2\\right)^2=4\\left(x^2-y^2\\right)$ to polar form to find all points on the lemniscate of Bernoulli where the tangent line is", "solid curves correspond to level lines of $H_0(x,y)$ Figure 3. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 4. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 0$, $a_0 = 1$, $a_1 = 0.8$, $b_0 = 0$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curve corresponds to $r = r(1) t^{{|\\vartheta_4|}/{2}}$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 5. The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = 1$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ Figure 6. The evolution of $(x(t),y(t))$, $R(t)$, $\\theta(t)$ for solutions of (50) with $h = 1/6$, $a_0", "# Math Help - Equation of tangent line (rec. form) to a polar curve!! 1. ## Equation of tangent line (rec. form) to a polar curve!! Hello Quesiton: Find the rectangular form of the equation of the tangent line to the polar curve $r=cos^3(\\theta)$ at the point corresponding to $\\theta=\\frac{\\pi}{4}$ How to do that? I mean finding it in RECTANGULAR FORM ! i know that $\\frac{dy}{dx}=\\frac{\\frac{dr}{d\\theta} sin(\\theta)+rcos(\\theta)}{\\frac{dr}{d\\theta} cos(\\theta) - r sin(\\theta)}$ I will calculate $\\frac{dy}{dx}$ at $\\theta=\\frac{\\pi}{4}$ , and this is easy .. The problem here is that, how can I find the equation of the tangent line in RECATNGULAR FORM ?? The equation of the tangent line is : $y-y1=m(x-x1)$ m = the slope , and this one will be calculated by using the formula .. but what about x1 and y1? 2. $r =\\cos^3{\\theta}$ $x = r\\cos{\\theta} = \\cos^4{\\theta}$ $y = r\\sin{\\theta} = \\cos^3{\\theta} \\sin{\\theta} $ sub in $\\frac{\\pi}{4}$ for $\\theta$ to determine x and y", "curve (strictly speaken: any curve which does not pass through the origin, but we can solve this by continuity). Example. It is a bit hard to write this down for the ellipse (as requested by you in a comment) because the ellipse is already hard to reparametrize to arc length (which is a requirement in your post). Therefore I will demonstrate it for the square with corners $(0,0), (1,0), (1,1)$ and $(0,1)$. An arc length parametrization $x:I\\to\\Bbb R^2$ must be defined on the interval $[0,4]$. We then have $$r(t)=\\begin{cases} t & \\text{for t\\in[0,1]} \\\\ \\sqrt{t^2-2t+2} & \\text{for t\\in[1,2]} \\\\ \\sqrt{t^2-6t+10} & \\text{for t\\in[2,3]} \\\\ 4-t & \\text{for t\\in[3,4]} \\end{cases},\\qquad \\theta(t)=\\begin{cases} 0 & \\text{for t\\in[0,1]} \\\\ \\arctan(t-1) & \\text{for t\\in[1,2]} \\\\ \\pi/2-\\arctan(3-t) & \\text{for t\\in[2,3]} \\\\ \\pi/2 & \\text{for t\\in[3,4]} \\\\ \\end{cases}.$$ Example. Note that your post requires that the curve starts in $(0,0)$, hence a full circle is not parametrized by $r=\\mathrm{const}$ (this would set the circles center at $(0,0)$ instead). But think about the unit circle around the point $(1,0)$. It can be parametrized on $I=[0,2\\pi]$ by $$r(t)=\\sqrt{2+2\\cos t},\\qquad \\theta(t)=\\arctan\\left(\\frac{\\sin(t)}{\\cos(t)+1}\\right).$$ Note: in both examples I dropped the requirement $I=[0,\\pi]$ as given in your post, but I think this is no problem as you can simply scale the curves in question (this is what the author had in" ]

[ "- k_1 / m) v_n + (1 - k_b) \\begin{pmatrix} T_n \\\\ θ_n \\\\ \\end{pmatrix}$$ where: $T$ = ...", "-4 & 1 & -3 & 0 \\\\ 3 & -3 & -2 & -3 & 3 & 5 & 0 & -3 & 0 & -1 & -2 & 0 & 3 & 2 & 6 \\end{pmatrix} \\\\ T &=", "\\\\ \\mathbf{\\hat y} \\end{pmatrix} = \\begin{pmatrix} \\cos \\phi & -\\sin \\phi \\\\ \\sin \\phi & \\cos \\phi \\end{pmatrix} \\begin{pmatrix} \\boldsymbol{\\hat \\rho} \\\\ \\boldsymbol{\\hat\\phi} \\end{pmatrix}$$", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "\\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle {\\mathbf {a} }_{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad {\\mathbf {b} }_{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle {\\mathbf {AB} }={\\begin{pmatrix}{\\mathbf {A} }{\\mathbf {b} }_{1}&{\\mathbf {A} }{\\mathbf {b} }_{2}&\\dots &{\\mathbf {A} }{\\mathbf {b} }_{p}\\end{pmatrix}}={\\begin{pmatrix}{\\mathbf {a} }_{1}{\\mathbf {B} }\\\\{\\mathbf {a} }_{2}{\\mathbf {B} }\\\\\\vdots \\\\{\\mathbf {a} }_{n}{\\mathbf {B} }\\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).{{ safesubst:#invoke:Unsubst||date=__DATE__ |B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} Outer product The outer product (also known as the dyadic product or tensor product) of two", "Now you have to find $$\\sqrt{B}$$, Let $$B=\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}^2=\\begin{pmatrix} a^2+bc & b(a+d) \\\\ c(a+d) & d^2+bc \\end{pmatrix}=\\begin{pmatrix} 4 & 3 \\\\ 5 & 6 \\end{pmatrix}~~~~(3)$$ Now solve for $$a,b,c,d$$ the set of four equations $$a^2+bc=4~~~(i),~ b(a+d)=3~~~(ii), ~c(a+d)=5~~~(iii),~ d^2+bc=6~~~(iv)~~~(4)$$ Let $$c=k$$ then from (ii) and (iii) get $$b=3k/5.$$ From (i) and (ii) have $$a^2-d^2=-2 \\Rightarrow (a-d)(a+d)=-2.$$ Next use (ii) to get $$a-d=-6k/5$$ and from (iii) we have $$(a+d)=5/k$$. Solve these two to get $$a=(5/k-6k/5)/2$$. Put this $$a$$ and $$b,c$$ in (i) to get $$(5/k-6k/3)^2/4+3k^2/5=4 ~~~~~~(5)$$ whose roots are $$k=\\pm 5/2,~~ \\pm \\frac{5}{2 \\sqrt{6}}~~~~~~~~~~(6)$$ Four $$2 \\times 2$$ matrices are possible for $$\\sqrt{B}$$. $$\\sqrt{B}=\\frac{\\pm 1}{2} \\begin{pmatrix} 1 & 3 \\\\ 5 & 3 \\end{pmatrix} \\mbox{and two more for other value of}~ k~~~~ (7)$$ The other two matrices do satisfy $$Trace(\\sqrt{B})^2=Trace(B)$$ but they do not satisfy $$(\\det |\\sqrt{B}|)^2=\\det |B|$$, hence they are rejected. from (2) you get $$A_1=\\frac{1}{2}\\begin{pmatrix} 5 & 3\\\\ 5 & 7 \\end{pmatrix},~~~A_2=\\frac{1}{2}\\begin{pmatrix} 3 & -3 \\\\ -5 & 1 \\end{pmatrix}.$$ It will be fun to check that these two matrices for A$will satisfy the original equation. Note that my answer though consistent will be different from that of the Answer of other methods. Also note that this question does not have a unique matrix as answer. • For matrices$P^2=Q$doesn't necessarily imply$P=\\pm", "b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Real Multiplication Distributes over Addition $\\ds \\mathbf A \\mathbf C + \\mathbf B \\mathbf C$ $=$ $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix}$ $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + a_{12} c_{21} & a_{11} c_{12} + a_{12} c_{22} \\\\ a_{21} c_{11} + a_{22} c_{21} & a_{21} c_{12} + a_{22} c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} c_{11} + b_{12} c_{21} & b_{11} c_{12} + b_{12} c_{22} \\\\ b_{21} c_{11} + b_{22} c_{21} & b_{21} c_{12} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Product (Conventional) $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + b_{11} c_{11} + a_{12} c_{21} + b_{12} c_{21} & a_{11} c_{12} + b_{11} c_{12} + a_{12} c_{22} + b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Entrywise Addition The expression for $\\paren {\\mathbf A", "& 1 \\\\ 1 & 0 \\end{pmatrix}$ $D = \\begin {pmatrix} -1 & 0 \\\\ 0 & 1 \\end {pmatrix}$ $P = \\begin {pmatrix} -1 & 1 \\\\ 1 & 1 \\end{pmatrix}$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "A_{22} B_{21} & A_{21} B_{12} + A_{22} B_{22} \\end{pmatrix}$$ This is how equations (51.1) and (51.2) were derived.", "$\\rho = \\begin{pmatrix} 1 & 2 & 4 \\end{pmatrix} \\begin{pmatrix} 3 & 5 & 7 & 9 \\end{pmatrix} \\begin{pmatrix} 1 & 3 & 9 \\end{pmatrix} \\begin{pmatrix} 2 & 3 & 4 & 5 & 6 & 8 \\end{pmatrix}$ Then $\\rho$ evaluates to: $\\begin{pmatrix} 1 & 5 & 6 & 8 & 4 & 7 & 9 & 2 & 3 \\end{pmatrix}$", "c= 0, d+ f= 1, a+ 2c= 0, and d+ 2f= 0, to solve for a, b, c, d, e, and f. From a+ c= 0 and a+ 2c= 0, c= 0. From d+ f= 1 and d+ 2f= 0, f= -1. Then a+ 0= 0 so a= 0 and d- 1= 1 so d= 2. a+ b+ c= 0+ b+ 0= 1 so b= 1. d+ e+ f= 2+ e- 1= 0 so e= -1. $A= \\begin{pmatrix}0 & 1 & 0 \\\\ 2 & -1 & -1 \\end{pmatrix}$.", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "these sets are equal ${\\displaystyle \\{{\\begin{pmatrix}1\\\\4\\\\1\\\\1\\end{pmatrix}}+{\\begin{pmatrix}-1\\\\0\\\\1\\\\0\\end{pmatrix}}z\\,{\\big |}\\,z\\in \\mathbb {R} \\}\\quad {\\text{and}}\\quad \\{{\\begin{pmatrix}0\\\\4\\\\2\\\\1\\end{pmatrix}}+{\\begin{pmatrix}-1\\\\0\\\\1\\\\0\\end{pmatrix}}k\\,{\\big |}\\,k\\in \\mathbb {R} \\},}$ and that both describe the solution set of this system. ${\\displaystyle {\\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\\\&&y&&&-&w&=&3\\\\x&&&+&z&+&2w&=&4\\end{array}}}$ That system reduces like this: ${\\displaystyle {\\begin{array}{rcl}&{\\xrightarrow[{}]{-\\rho _{1}+\\rho _{2}}}&{\\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\\\&&y&&&-&w&=&3\\\\&&y&&&+&w&=&5\\end{array}}\\\\&{\\xrightarrow[{}]{-\\rho _{2}+\\rho _{3}}}&{\\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\\\&&y&&&-&w&=&3\\\\&&&&&&2w&=&2\\end{array}}\\end{array}}}$ showing that ${\\displaystyle w=1}$, ${\\displaystyle y=4}$ and ${\\displaystyle x=2-z}$. This exercise is recommended for all readers. Problem 5 Decide if the sets are equal. 1. ${\\displaystyle \\{{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t\\,{\\big |}\\,t\\in \\mathbb {R} \\}}$ and ${\\displaystyle \\{{\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s\\,{\\big |}\\,s\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle \\{{\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s\\,{\\big |}\\,t,s\\in \\mathbb {R} \\}}$ and ${\\displaystyle \\{{\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n\\,{\\big |}\\,m,n\\in \\mathbb {R} \\}}$ 3. ${\\displaystyle \\{{\\begin{pmatrix}1\\\\2\\end{pmatrix}}t\\,{\\big |}\\,t\\in \\mathbb {R} \\}}$ and ${\\displaystyle \\{{\\begin{pmatrix}2\\\\4\\end{pmatrix}}m+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}n\\,{\\big |}\\,m,n\\in \\mathbb {R} \\}}$ 4. ${\\displaystyle \\{{\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}}s+{\\begin{pmatrix}-1\\\\1\\\\0\\end{pmatrix}}t\\,{\\big |}\\,s,t\\in \\mathbb {R} \\}}$ and ${\\displaystyle \\{{\\begin{pmatrix}-1\\\\1\\\\1\\end{pmatrix}}m+{\\begin{pmatrix}0\\\\1\\\\3\\end{pmatrix}}n\\,{\\big |}\\,m,n\\in \\mathbb {R} \\}}$ 5. ${\\displaystyle \\{{\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\4\\\\6\\end{pmatrix}}s\\,{\\big |}\\,t,s\\in \\mathbb {R} \\}}$ and ${\\displaystyle \\{{\\begin{pmatrix}3\\\\7\\\\7\\end{pmatrix}}t+{\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}s\\,{\\big |}\\,t,s\\in \\mathbb {R} \\}}$ For each item, we call the first set ${\\displaystyle S_{1}}$ and the other ${\\displaystyle S_{2}}$. 1. They are equal. To see that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must show that any element of the first set is in the second, that is, for any vector of the form ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t}$ there is an appropriate ${\\displaystyle s}$ such that ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s.}$ Restated, given ${\\displaystyle t}$ we must find ${\\displaystyle s}$ so that this holds. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0s&=&1+0t\\\\8&-&1s&=&2+3t\\end{array}}}$ That system reduces to ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\s&=&6-3t\\end{array}}}$ That is, ${\\displaystyle", "## Algebra 2 (1st Edition) $$\\begin{pmatrix}-2&12\\\\ -6&9\\\\ -40&24\\end{pmatrix}$$ Subtracting the matrices by subtracting corresponding elements, we find: $$\\begin{pmatrix}6-8&24-12\\\\ \\left(-1\\right)-5&3-\\left(-6\\right)\\\\ \\left(-8\\right)-32&15-\\left(-9\\right)\\end{pmatrix} \\\\ \\begin{pmatrix}-2&12\\\\ -6&9\\\\ -40&24\\end{pmatrix}$$", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03", "= x\\cdot \\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix} + y\\cdot \\begin{pmatrix} 6 & 5 & 4 \\end{pmatrix} + z\\cdot \\begin{pmatrix} 7 & 8 & 9 \\end{pmatrix}.$$", "0 & 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 1& 0 \\\\ 1 & 0 \\end{pmatrix} = B \\cdot C$$ As you said, what you can't do is to write this as $A \\cdot C^{-1} = B$ or $B^{-1} \\cdot A = C$", "where $$A= \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\mbox{ and } I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}.$$", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures." ]

[ "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "right order. So let’s look at our matrix product again: we wan’t the i-th column of the matrix product $\\mathbf{B}\\mathbf{A}$, so we look at $(\\mathbf{B}\\mathbf{A})\\mathbf{e}_i = \\mathbf{B}(\\mathbf{A}\\mathbf{e}_i) = \\mathbf{B} \\mathbf{a}_i$ exactly as claimed. As always, there’s a corresponding construction using row vectors. Using the dual basis of linear functionals (note no transpose this time!) $\\mathbf{e}^1 = \\begin{pmatrix} 1 & 0 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^2 = \\begin{pmatrix} 0 & 1 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^m = \\begin{pmatrix} 0 & 0 & \\cdots & 1 \\end{pmatrix}$ we can disassemble a matrix into its rows: $\\mathbf{A} = \\begin{pmatrix} \\mathbf{a}^1 \\\\ \\mathbf{a}^2 \\\\ \\vdots \\\\ \\mathbf{a}^m \\end{pmatrix} = \\begin{pmatrix} \\mathbf{e}^1 \\mathbf{A} \\\\ \\mathbf{e}^2 \\mathbf{A} \\\\ \\vdots \\\\ \\mathbf{e}^m \\mathbf{A} \\end{pmatrix}$ and since $\\mathbf{e}^i (\\mathbf{A} \\mathbf{B}) = (\\mathbf{e}^i \\mathbf{A}) \\mathbf{B} = \\mathbf{a}^i \\mathbf{B}$, we can write matrix products in terms of what happens to the row vectors too (though this time, we’re expanding in terms of the row vectors of the first not second factor): $\\mathbf{A} \\mathbf{B} = \\begin{pmatrix} \\mathbf{a}^1 \\mathbf{B} \\\\ \\mathbf{a}^2 \\mathbf{B} \\\\ \\vdots \\\\ \\mathbf{a}^m \\mathbf{B} \\end{pmatrix}$. ### Gluing it back together Above, the trick was to write the “slicing” step as a matrix product with one of the basis vectors – $\\mathbf{A} \\mathbf{e}_i$ and so forth. We’ll soon need to deal with the gluing step too,", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "= i, \\quad \\mathbf{A} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}$ $i\\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ij \\\\ \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & k \\\\ \\end{pmatrix} \\ne \\begin{pmatrix} -1 & 0 \\\\ 0 & -k \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ji \\\\ \\end{pmatrix} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}i\\,,$ where i, j, k are the quaternion units. The non-commutativity of quaternion multiplication prevents the transition of changing ij = +k to ji = −k. Matrix product (two matrices) Assume two matrices are to be multiplied (the generalization to any number is discussed below). General definition of the matrix product Arithmetic process of multiplying numbers (solid lines) in row i in matrix A and column j in matrix B, then adding the terms (dashed lines) to obtain entry ij in the final matrix. If A is an n × m matrix and B is an m × p matrix, $\\mathbf{A}=\\begin{pmatrix} A_{11} & A_{12} & \\cdots & A_{1m} \\\\ A_{21} & A_{22} & \\cdots & A_{2m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n1} & A_{n2} & \\cdots & A_{nm} \\\\ \\end{pmatrix},\\quad\\mathbf{B}=\\begin{pmatrix} B_{11} & B_{12} & \\cdots", "outer product is a row vector multiplied on the left by a column vector: $\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$ where $\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi:[1] $\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p", "## Non-commutative product For an arbitrary asymmetric square matrix $$\\mathbf A$$ construct a matrix $$\\mathbf B$$ s.t. their products do not commute, i.e. $$\\mathbf A\\mathbf B\\ne\\mathbf B\\mathbf A$$. Do products commute when both matrices are symmetric? • #### Hint Use facts from the previous exercise. • #### Resolution There are several solutions to this problem. If $$a_{i,j}\\ne a_{j,i}$$, we can choose $$\\mathbf B$$ s.t. $$b_{i,j}=b_{j,i}=1$$, and $$b_{k,l}=0$$ otherwise. Then we get $$(\\mathbf A\\mathbf B)_{i,i}=a_{i,j}\\ne a_{j,i}=(\\mathbf B\\mathbf A)_{i,i}$$. Another option is to choose $$\\mathbf B$$ s.t. the only nonzero element is $$b_{i,i}=1$$. Then the two products cannot agree on elements with indices $$i$$ and $$j$$. We get $$(\\mathbf A\\mathbf B)_{i,j}=0=(\\mathbf B\\mathbf A)_{i,j}=a_{i,j}$$, while $$(\\mathbf B\\mathbf A)_{j,i}=0=(\\mathbf A\\mathbf B)_{j,i}=a_{j,i}$$, but at least one of $$a_{i,j}$$ and $$a_{j,i}$$ is distinct from zero. Even on symmetric matrices the product does not commute (even the result need not to be symmetric), e.g. $$\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} = \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\ne \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\\\ \\end{pmatrix}$$", "product AB is defined by[3][4] $\\mathbf{A}\\mathbf{B} =\\begin{pmatrix} (AB)_{11} & (AB)_{12} & \\cdots & (AB)_{1p} \\\\ (AB)_{21} & (AB)_{22} & \\cdots & (AB)_{2p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (AB)_{n1} & (AB)_{n2} & \\cdots & (AB)_{np} \\\\ \\end{pmatrix}$ (with no multiplication signs or dots) where AB has entries defined by $(AB)_{ij} = \\sum_{k=1}^m A_{ik}B_{kj}.$ Treating the rows and columns in each matrix as row and column vectors respectively, this entry is also their vector dot product: $\\mathbf{a}_i=\\begin{pmatrix} A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix}\\,, \\quad \\mathbf{b}_j=\\begin{pmatrix} B_{1j} \\\\ B_{2j} \\\\ \\vdots \\\\ B_{mj} \\end{pmatrix}, \\quad (AB)_{ij} = \\mathbf{a}_i \\cdot \\mathbf{b}_j.$ (See below for further details). Usually the entries are numbers or expressions, but can even be matrices themselves (see block matrix). The matrix product can still be calculated exactly the same way. ### Illustration The figure to the right illustrates diagrammatically the product of two matrices A and B, showing how each intersection in the product matrix corresponds to a row of A and a column of B. $\\overset{4\\times 2 \\text{ matrix}}{\\begin{bmatrix} \\color{BrickRed} a_{11} & \\color{BrickRed} a_{12} \\\\ \\cdot & \\cdot \\\\ \\color{BurntOrange} a_{31} & \\color{BurntOrange} a_{32} \\\\ \\cdot & \\cdot \\\\ \\end{bmatrix}} \\overset{2\\times 3\\text{ matrix}}{\\begin{bmatrix} \\cdot & \\color{RedViolet}b_{12} & \\color{Violet}b_{13} \\\\ \\cdot & \\color{RedViolet}b_{22} & \\color{Violet}b_{23} \\\\ \\end{bmatrix}} = \\overset{4\\times 3\\text{ matrix}}{\\begin{bmatrix} \\cdot &", "R^{-1}= \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix}$$ • #### Result The given matrix can be factorized into Jordan normal form as $$\\mathbf A= \\begin{pmatrix} 1 & 1 & 1\\\\ 0 & 1 & 0\\\\ -1 & 0 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 2\\\\ 0 & 0 & -1\\\\ 1 & 1 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 1 & 0\\\\ 0 & 2 & 0\\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix} = \\mathbf R\\mathbf J\\mathbf R^{-1}$$ • #### Variant $$\\mathbf A= \\begin{pmatrix} 2 & 1 & 0 & 2\\\\ 0 & 2 & 0 & -2\\\\ 0 & 0 & 2 & 2\\\\ 0 & 0 & 0 & 2 \\end{pmatrix}$$ • #### Solution The matrix has a single eigenvalue $$\\lambda = 2$$ of algebraicmultiplicity 4, eigenvectors are nontrivial solutions of a system of equation with matrix $\\mathbf A-\\lambda \\mathbf I = \\begin{pmatrix} 0&1&0&2\\\\ 0&0&0&-2\\\\ 0&0&0&2\\\\ 0&0&0&0 \\end{pmatrix} \\sim \\begin{pmatrix} 0&1&0&0\\\\ 0&0&0&1\\\\ \\end{pmatrix}$ The dimension of $$\\ker(\\mathbf A-\\lambda \\mathbf I)$$ is 2, so the matrix $$\\mathbf A$$ has two linearly independent eigenvectors, e.g. $$\\mathbf x_1=(0{,}0,1{,}0)^T$$ and $$\\mathbf x_2=(1{,}0,0{,}0)^T$$ To get a regular matrix $$\\mathbf R$$ we", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "{b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).[citation needed] ### Outer productEdit The outer product (also known as the dyadic product or tensor product) of two vectors in matrix form is equivalent to a row vector multiplied on the left by a column vector: {\\displaystyle {\\begin{aligned}\\mathbf {a} \\otimes \\mathbf {b} &=\\mathbf {a} \\mathbf {b} ^{\\mathrm {T} }\\\\&={\\begin{pmatrix}a_{1}\\\\a_{2}\\\\\\vdots \\\\a_{n}\\end{pmatrix}}{\\begin{pmatrix}b_{1}&b_{2}&\\cdots &b_{n}\\end{pmatrix}}\\\\&={\\begin{pmatrix}a_{1}b_{1}&a_{1}b_{2}&\\cdots &a_{1}b_{n}\\\\a_{2}b_{1}&a_{2}b_{2}&\\cdots &a_{2}b_{n}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\a_{n}b_{1}&a_{n}b_{2}&\\cdots &a_{n}b_{n}\\\\\\end{pmatrix}}.\\end{aligned}}} An alternative method is to express", "Product of a 2x2 and a general matrices Determine products $$\\mathbf A\\mathbf B_i$$ and $$\\mathbf B_i\\mathbf A$$, for matrices $$\\mathbf A=\\begin{pmatrix} a_{1{,}1} & a_{1{,}2} \\\\ a_{2{,}1} & a_{2{,}2} \\end{pmatrix}$$, $$\\mathbf B_1=\\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix}$$, $$\\mathbf B_2=\\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$$ a $$\\mathbf B_3=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}$$. Which transformations of $$\\mathbf A$$ these products provide? • Resolution $$\\mathbf A\\mathbf B_1= \\begin{pmatrix} a_{1{,}1} & 0 \\\\ a_{2{,}1} & 0 \\end{pmatrix}$$ – corresponds of the choice of the first column. $$\\mathbf B_1\\mathbf A= \\begin{pmatrix} b_{1{,}1} & b_{1{,}2} \\\\ 0 & 0 \\end{pmatrix}$$ – corresponds of the choice of the first row. $$\\mathbf A\\mathbf B_2= \\begin{pmatrix} 0 & a_{1{,}1} \\\\ 0 & a_{2{,}1} \\end{pmatrix}$$ – chooses the first column and puts it as the second. $$\\mathbf B_2\\mathbf A= \\begin{pmatrix} a_{2{,}1} & a_{2{,}2} \\\\ 0 & 0 \\end{pmatrix}$$ – chooses thesecond row and puts it as the first. $$\\mathbf A\\mathbf B_3= \\begin{pmatrix} a_{1{,}2} & a_{1{,}1} \\\\ a_{2{,}2} & a_{2{,}1} \\end{pmatrix}$$ a $$\\mathbf B_3\\mathbf A= \\begin{pmatrix} a_{2{,}1} & a_{2{,}2} \\\\ a_{1{,}1} & a_{1{,}2} \\end{pmatrix}$$ – the first product swaps columns, the other rows. Products $$\\mathbf B\\mathbf A$$ correspond to the row transformations of $$\\mathbf A$$, while products $$\\mathbf A\\mathbf B$$ yield column transformations of $$\\mathbf A$$.", "\\end{pmatrix}^k = \\begin{pmatrix} A_{11}^k & 0 & \\cdots & 0 \\\\ 0 & A_{22}^k & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & A_{nn}^k \\end{pmatrix}$$ meaning its easy to raise a diagonal matrix to a power. When raising an arbitrary matrix (not necessarily a diagonal matrix) to a power, it is often helpful to exploit this property by diagonalizing the matrix first. The inner and outer products Given two column vectors a and b, the Euclidean inner product and outer product are the simplest special cases of the matrix product, by transposing the column vectors into row vectors.[5] The inner product is a column vector multiplied on the left by a row vector: $$\\mathbf{a}\\cdot \\mathbf{b} = \\mathbf{a}^\\mathrm{T}\\mathbf{b}$$ More explicitly, $$\\mathbf{a}\\cdot \\mathbf{b} = \\begin{pmatrix}a_1 & a_2 & \\cdots & a_n\\end{pmatrix} \\begin{pmatrix}b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n\\end{pmatrix} = a_1b_1+a_2b_2+\\cdots+a_nb_n = \\sum_{i=1}^n a_ib_i$$ The outer product is a row vector multiplied on the left by a column vector: $$\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$$ where $$\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots", "the matrix product in terms of the outer product. The decomposition is done the other way around, the first matrix A is decomposed into column vectors ai and the second matrix B into row vectors bi: {\\displaystyle {\\begin{aligned}\\mathbf {AB} &={\\begin{pmatrix}\\mathbf {\\bar {a}} _{1}&\\mathbf {\\bar {a}} _{2}&\\cdots &\\mathbf {\\bar {a}} _{m}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {\\bar {b}} _{1}\\\\\\mathbf {\\bar {b}} _{2}\\\\\\vdots \\\\\\mathbf {\\bar {b}} _{m}\\end{pmatrix}}\\\\&=\\mathbf {\\bar {a}} _{1}\\otimes \\mathbf {\\bar {b}} _{1}+\\mathbf {\\bar {a}} _{2}\\otimes \\mathbf {\\bar {b}} _{2}+\\cdots +\\mathbf {\\bar {a}} _{m}\\otimes \\mathbf {\\bar {b}} _{m}\\\\&=\\sum _{i=1}^{m}\\mathbf {\\bar {a}} _{i}\\otimes \\mathbf {\\bar {b}} _{i}\\end{aligned}}} where this time ${\\displaystyle \\mathbf {\\bar {a}} _{i}={\\begin{pmatrix}A_{1i}\\\\A_{2i}\\\\\\vdots \\\\A_{ni}\\end{pmatrix}}\\,,\\quad \\mathbf {\\bar {b}} _{i}={\\begin{pmatrix}B_{i1}&B_{i2}&\\cdots &B_{ip}\\end{pmatrix}}\\,.}$ This method emphasizes the effect of individual column/row pairs on the result, which is a useful point of view with e.g. covariance matrices, where each such pair corresponds to the effect of a single sample point.[citation needed] {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}{\\color {Brown}1}&{\\color {Orange}2}&{\\color {Violet}3}\\\\{\\color {Brown}4}&{\\color {Orange}5}&{\\color {Violet}6}\\\\{\\color {Brown}7}&{\\color {Orange}8}&{\\color {Violet}9}\\\\\\end{pmatrix}}{\\begin{pmatrix}{\\color {Brown}a}&{\\color {Brown}d}\\\\{\\color {Orange}b}&{\\color {Orange}e}\\\\{\\color {Violet}c}&{\\color {Violet}f}\\\\\\end{pmatrix}}&={\\begin{pmatrix}{\\color {Brown}1}\\\\{\\color {Brown}4}\\\\{\\color {Brown}7}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Brown}{a}}&{\\color {Brown}{d}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Orange}2}\\\\{\\color {Orange}5}\\\\{\\color {Orange}8}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Orange}{b}}&{\\color {Orange}{e}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Violet}3}\\\\{\\color {Violet}6}\\\\{\\color {Violet}9}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Violet}c}&{\\color {Violet}f}\\\\\\end{pmatrix}}\\\\&={\\begin{pmatrix}{\\color {Brown}{1a}}&{\\color {Brown}{1d}}\\\\{\\color {Brown}{4a}}&{\\color {Brown}{4d}}\\\\{\\color {Brown}{7a}}&{\\color {Brown}{7d}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Orange}{2b}}&{\\color {Orange}{2e}}\\\\{\\color {Orange}{5b}}&{\\color {Orange}{5e}}\\\\{\\color {Orange}{8b}}&{\\color {Orange}{8e}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Violet}{3c}}&{\\color {Violet}{3f}}\\\\{\\color {Violet}{6c}}&{\\color {Violet}{6f}}\\\\{\\color {Violet}{9c}}&{\\color {Violet}{9f}}\\\\\\end{pmatrix}}\\\\&={\\begin{pmatrix}{\\color {Brown}{1a}}+{\\color {Orange}{2b}}+{\\color {Violet}{3c}}&{\\color {Brown}{1d}}+{\\color {Orange}{2e}}+{\\color {Violet}{3f}}\\\\{\\color {Brown}{4a}}+{\\color {Orange}{5b}}+{\\color {Violet}{6c}}&{\\color {Brown}{4d}}+{\\color {Orange}{5e}}+{\\color {Violet}{6f}}\\\\{\\color {Brown}{7a}}+{\\color {Orange}{8b}}+{\\color {Violet}{9c}}&{\\color {Brown}{7d}}+{\\color {Orange}{8e}}+{\\color {Violet}{9f}}\\\\\\end{pmatrix}}.\\end{aligned}}} ## Algorithms for efficient matrix multiplicationEdit", "# Matrix Product (Conventional)/Examples/2 x 2 Real Matrices Let $\\mathbf A = \\begin {pmatrix} p & q \\\\ r & s \\end {pmatrix}$ and $\\mathbf B = \\begin {pmatrix} w & x \\\\ y & z \\end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix product of $\\mathbf A$ with $\\mathbf B$ is given by: $\\mathbf A \\mathbf B = \\begin {pmatrix} p w + q y & p x + q z \\\\ r w + s y & r x + s z \\end {pmatrix}$", "{1 \\choose 0}(1\\ \\ 0) = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix}.$$ We may then use the kronecker product to compute the direct product of two matrices $$A, B$$( where $$[A] = m\\times n$$ and $$[B]= p\\times q$$) $$\\mathbf {A} \\otimes \\mathbf {B} ={\\begin{bmatrix}a_{11}\\mathbf {B} &\\cdots &a_{1n}\\mathbf {B} \\\\\\vdots &\\ddots &\\vdots \\\\a_{m1}\\mathbf {B} &\\cdots &a_{mn}\\mathbf {B} \\end{bmatrix}}.$$ to construct the complete hamiltonian in the 2-particle hilbert space (which it looks like you did correctly).", "the product $\\mathbf A \\mathbf B$ as a column-block matrix : $\\mathbf A \\mathbf B = \\begin {pmatrix} \\ds \\sum_{s_1 \\mathop = 1}^n \\mathbf B_{s_1, 1} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\ds \\sum_{s_n \\mathop = 1}^n B_{s_n, n} \\mathbf A \\mathbf e_{s_n} \\end {pmatrix}$ Using linearity with respect to each columns, we get: $\\map \\det {\\mathbf A \\mathbf B} = \\ds \\sum_{1 \\mathop \\leqslant s_1, \\ldots, s_n \\mathop \\leqslant n} \\paren {\\prod_{i \\mathop = 1}^n \\mathbf B_{s_i, i} } \\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end {pmatrix}$ Now notice that $\\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end{pmatrix}$ is zero once two entries are the same (since $\\det$ is an alternating map), it means that if for some $k \\ne \\ell$ we have $\\mathbf A \\mathbf e_{s_k} = \\mathbf A \\mathbf e_{s_\\ell}$, then $\\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end {pmatrix} = 0$. Therefore the only nonzero summands are those one the $s_1, \\ldots, s_n$ are all distinct. In other words, the \"selector\" $s$ represents some permutation of the numbers $1, \\ldots, n$. As a result, the determinant of the product can now be expressed as a sum of precisely $n!$ terms using", "as the inner products between the rows of $$\\mathbf{A}$$ and the the vector $$\\mathbf{v}$$. Specifically, let $$\\mathbf{A}_{(1,:)}$$ and $$\\mathbf{A}_{(2,:)}$$ be column vectors that form the first and second rows of $$\\mathbf{A}$$. Then the matrix-vector product $$\\mathbf{Av}$$ can also be written as $\\mathbf{A}\\mathbf{v} = \\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\end{pmatrix} \\begin{pmatrix} v_{1} \\\\ v_{2} \\\\ v_{3} \\end{pmatrix} = \\begin{pmatrix} \\langle \\mathbf{A}_{(1,:)}, \\mathbf{v} \\rangle \\\\ \\langle \\mathbf{A}_{(2,:)}, \\mathbf{v} \\rangle \\end{pmatrix}.$ Matrix-matrix multiplication Now that we’re comfortable with matrix-vector multiplication, we can easily extend this idea to matrix-matrix multiplication! Say we have two matrices $$\\mathbf{A} \\in \\mathbb{R}^{3 \\times 2}$$ and $$\\mathbf{B} \\in \\mathbb{R}^{2 \\times 2}$$. We obtain the matrix product $$\\mathbf{AB}$$ as follows $\\mathbf{AB} = \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\\\ a_{31} & a_{32} \\end{pmatrix} \\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix} = \\begin{bmatrix} b_{11} \\begin{pmatrix} a_{11} \\\\ a_{21} \\\\ a_{31} \\end{pmatrix} + b_{21} \\begin{pmatrix} a_{12} \\\\ a_{22} \\\\ a_{32} \\end{pmatrix} & b_{12} \\begin{pmatrix} a_{11} \\\\ a_{21} \\\\ a_{31} \\end{pmatrix} + b_{22} \\begin{pmatrix} a_{12} \\\\ a_{22} \\\\ a_{32} \\end{pmatrix} \\end{bmatrix}.$ A few things to keep in mind… 1. The first column of $$\\mathbf{AB}$$ is the matrix-vector product between $$\\mathbf{A}$$ and the first column of $$\\mathbf{B}$$. Similarly, the second column of $$\\mathbf{AB}$$ is the matrix-vector product between $$\\mathbf{A}$$ and the", "elementary matrices are not too advanced, here is an answer. Technically it's possible to show the result using only row reductions, but it is rather tedious and not nearly as clear. Suppose that we have a matrix $A$ with reduced row echelon form $R$. Then there is a series of elementary matrices which bring $A$ to $R$. Let $E$ be their product so that $$EA=R$$ Now if we write the matrix column-wise, we have $$E\\begin{pmatrix}\\mathbf{a_1} & \\cdots & \\mathbf{a_n}\\end{pmatrix} = \\begin{pmatrix}E\\mathbf{a_1} & \\cdots & E\\mathbf{a_n}\\end{pmatrix} =\\begin{pmatrix}\\mathbf{r_1} & \\cdots & \\mathbf{r_n}\\end{pmatrix}$$ where $\\mathbf{a_i}$ and $\\mathbf{r_i}$ are the columns of $A$ and $R$ respectively. From this we get the equality $$E\\mathbf{a_i} = \\mathbf{r_i}$$ for $1\\le i \\le n$. Now consider the equation $$c_1\\mathbf{r_1} + \\cdots + c_n\\mathbf{r_n} = \\mathbf{0}$$ for scalars $c_i$. We rewrite this as $$c_1\\left(E\\mathbf{a_1}\\right) + \\cdots + c_n\\left(E\\mathbf{a_n}\\right) = E\\left(c_1\\mathbf{a_1} + \\cdots + c_n\\mathbf{a_n}\\right)=\\mathbf{0}$$ since $E$ is a product of elementary matrices, it is invertible. In particular this implies $$c_1\\mathbf{a_1} + \\cdots + c_n\\mathbf{a_n}\\iff c_1\\mathbf{r_1} + \\cdots + c_n\\mathbf{r_n} = \\mathbf{0}$$ And it's easy to see that this holds for any subset of the columns. What this means is that elementary row operations preserves linear relations between columns. If a set of columns vectors are linearly independent in $R$ then they will remain linearly independent after transformation" ]

[ "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "R^{-1}= \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix}$$ • #### Result The given matrix can be factorized into Jordan normal form as $$\\mathbf A= \\begin{pmatrix} 1 & 1 & 1\\\\ 0 & 1 & 0\\\\ -1 & 0 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 2\\\\ 0 & 0 & -1\\\\ 1 & 1 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 1 & 0\\\\ 0 & 2 & 0\\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix} = \\mathbf R\\mathbf J\\mathbf R^{-1}$$ • #### Variant $$\\mathbf A= \\begin{pmatrix} 2 & 1 & 0 & 2\\\\ 0 & 2 & 0 & -2\\\\ 0 & 0 & 2 & 2\\\\ 0 & 0 & 0 & 2 \\end{pmatrix}$$ • #### Solution The matrix has a single eigenvalue $$\\lambda = 2$$ of algebraicmultiplicity 4, eigenvectors are nontrivial solutions of a system of equation with matrix $\\mathbf A-\\lambda \\mathbf I = \\begin{pmatrix} 0&1&0&2\\\\ 0&0&0&-2\\\\ 0&0&0&2\\\\ 0&0&0&0 \\end{pmatrix} \\sim \\begin{pmatrix} 0&1&0&0\\\\ 0&0&0&1\\\\ \\end{pmatrix}$ The dimension of $$\\ker(\\mathbf A-\\lambda \\mathbf I)$$ is 2, so the matrix $$\\mathbf A$$ has two linearly independent eigenvectors, e.g. $$\\mathbf x_1=(0{,}0,1{,}0)^T$$ and $$\\mathbf x_2=(1{,}0,0{,}0)^T$$ To get a regular matrix $$\\mathbf R$$ we", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "= i, \\quad \\mathbf{A} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}$ $i\\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ij \\\\ \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & k \\\\ \\end{pmatrix} \\ne \\begin{pmatrix} -1 & 0 \\\\ 0 & -k \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ji \\\\ \\end{pmatrix} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}i\\,,$ where i, j, k are the quaternion units. The non-commutativity of quaternion multiplication prevents the transition of changing ij = +k to ji = −k. Matrix product (two matrices) Assume two matrices are to be multiplied (the generalization to any number is discussed below). General definition of the matrix product Arithmetic process of multiplying numbers (solid lines) in row i in matrix A and column j in matrix B, then adding the terms (dashed lines) to obtain entry ij in the final matrix. If A is an n × m matrix and B is an m × p matrix, $\\mathbf{A}=\\begin{pmatrix} A_{11} & A_{12} & \\cdots & A_{1m} \\\\ A_{21} & A_{22} & \\cdots & A_{2m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n1} & A_{n2} & \\cdots & A_{nm} \\\\ \\end{pmatrix},\\quad\\mathbf{B}=\\begin{pmatrix} B_{11} & B_{12} & \\cdots", "the product $\\mathbf A \\mathbf B$ as a column-block matrix : $\\mathbf A \\mathbf B = \\begin {pmatrix} \\ds \\sum_{s_1 \\mathop = 1}^n \\mathbf B_{s_1, 1} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\ds \\sum_{s_n \\mathop = 1}^n B_{s_n, n} \\mathbf A \\mathbf e_{s_n} \\end {pmatrix}$ Using linearity with respect to each columns, we get: $\\map \\det {\\mathbf A \\mathbf B} = \\ds \\sum_{1 \\mathop \\leqslant s_1, \\ldots, s_n \\mathop \\leqslant n} \\paren {\\prod_{i \\mathop = 1}^n \\mathbf B_{s_i, i} } \\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end {pmatrix}$ Now notice that $\\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end{pmatrix}$ is zero once two entries are the same (since $\\det$ is an alternating map), it means that if for some $k \\ne \\ell$ we have $\\mathbf A \\mathbf e_{s_k} = \\mathbf A \\mathbf e_{s_\\ell}$, then $\\det \\begin {pmatrix} \\mathbf A \\mathbf e_{s_1} & \\cdots & \\mathbf A \\mathbf e_{s_n} \\end {pmatrix} = 0$. Therefore the only nonzero summands are those one the $s_1, \\ldots, s_n$ are all distinct. In other words, the \"selector\" $s$ represents some permutation of the numbers $1, \\ldots, n$. As a result, the determinant of the product can now be expressed as a sum of precisely $n!$ terms using", "R^{2^{n}\\times 2^{n}}}$ If the matrices A, B are not of type 2n × 2n we fill the missing rows and columns with zeros. We partition A, B and C into equally sized block matrices ${\\displaystyle \\mathbf {A} ={\\begin{bmatrix}\\mathbf {A} _{1,1}&\\mathbf {A} _{1,2}\\\\\\mathbf {A} _{2,1}&\\mathbf {A} _{2,2}\\end{bmatrix}}{\\mbox{ , }}\\mathbf {B} ={\\begin{bmatrix}\\mathbf {B} _{1,1}&\\mathbf {B} _{1,2}\\\\\\mathbf {B} _{2,1}&\\mathbf {B} _{2,2}\\end{bmatrix}}{\\mbox{ , }}\\mathbf {C} ={\\begin{bmatrix}\\mathbf {C} _{1,1}&\\mathbf {C} _{1,2}\\\\\\mathbf {C} _{2,1}&\\mathbf {C} _{2,2}\\end{bmatrix}}}$ with ${\\displaystyle \\mathbf {A} _{i,j},\\mathbf {B} _{i,j},\\mathbf {C} _{i,j}\\in R^{2^{n-1}\\times 2^{n-1}}}$. The naive algorithm would be: ${\\displaystyle \\mathbf {C} _{1,1}=\\mathbf {A} _{1,1}\\mathbf {B} _{1,1}+\\mathbf {A} _{1,2}\\mathbf {B} _{2,1}}$ ${\\displaystyle \\mathbf {C} _{1,2}=\\mathbf {A} _{1,1}\\mathbf {B} _{1,2}+\\mathbf {A} _{1,2}\\mathbf {B} _{2,2}}$ ${\\displaystyle \\mathbf {C} _{2,1}=\\mathbf {A} _{2,1}\\mathbf {B} _{1,1}+\\mathbf {A} _{2,2}\\mathbf {B} _{2,1}}$ ${\\displaystyle \\mathbf {C} _{2,2}=\\mathbf {A} _{2,1}\\mathbf {B} _{1,2}+\\mathbf {A} _{2,2}\\mathbf {B} _{2,2}}$ With this construction we have not reduced the number of multiplications. We still need 8 multiplications to calculate the Ci,j matrices, the same number of multiplications we need when using standard matrix multiplication. The Strassen algorithm defines instead new matrices: ${\\displaystyle \\mathbf {M} _{1}:=(\\mathbf {A} _{1,1}+\\mathbf {A} _{2,2})(\\mathbf {B} _{1,1}+\\mathbf {B} _{2,2})}$ ${\\displaystyle \\mathbf {M} _{2}:=(\\mathbf {A} _{2,1}+\\mathbf {A} _{2,2})\\mathbf {B} _{1,1}}$ ${\\displaystyle \\mathbf {M} _{3}:=\\mathbf {A} _{1,1}(\\mathbf {B} _{1,2}-\\mathbf {B} _{2,2})}$ ${\\displaystyle \\mathbf {M} _{4}:=\\mathbf {A} _{2,2}(\\mathbf {B} _{2,1}-\\mathbf {B} _{1,1})}$ ${\\displaystyle \\mathbf", "{b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).[citation needed] ### Outer productEdit The outer product (also known as the dyadic product or tensor product) of two vectors in matrix form is equivalent to a row vector multiplied on the left by a column vector: {\\displaystyle {\\begin{aligned}\\mathbf {a} \\otimes \\mathbf {b} &=\\mathbf {a} \\mathbf {b} ^{\\mathrm {T} }\\\\&={\\begin{pmatrix}a_{1}\\\\a_{2}\\\\\\vdots \\\\a_{n}\\end{pmatrix}}{\\begin{pmatrix}b_{1}&b_{2}&\\cdots &b_{n}\\end{pmatrix}}\\\\&={\\begin{pmatrix}a_{1}b_{1}&a_{1}b_{2}&\\cdots &a_{1}b_{n}\\\\a_{2}b_{1}&a_{2}b_{2}&\\cdots &a_{2}b_{n}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\a_{n}b_{1}&a_{n}b_{2}&\\cdots &a_{n}b_{n}\\\\\\end{pmatrix}}.\\end{aligned}}} An alternative method is to express", "elementary matrices are not too advanced, here is an answer. Technically it's possible to show the result using only row reductions, but it is rather tedious and not nearly as clear. Suppose that we have a matrix $A$ with reduced row echelon form $R$. Then there is a series of elementary matrices which bring $A$ to $R$. Let $E$ be their product so that $$EA=R$$ Now if we write the matrix column-wise, we have $$E\\begin{pmatrix}\\mathbf{a_1} & \\cdots & \\mathbf{a_n}\\end{pmatrix} = \\begin{pmatrix}E\\mathbf{a_1} & \\cdots & E\\mathbf{a_n}\\end{pmatrix} =\\begin{pmatrix}\\mathbf{r_1} & \\cdots & \\mathbf{r_n}\\end{pmatrix}$$ where $\\mathbf{a_i}$ and $\\mathbf{r_i}$ are the columns of $A$ and $R$ respectively. From this we get the equality $$E\\mathbf{a_i} = \\mathbf{r_i}$$ for $1\\le i \\le n$. Now consider the equation $$c_1\\mathbf{r_1} + \\cdots + c_n\\mathbf{r_n} = \\mathbf{0}$$ for scalars $c_i$. We rewrite this as $$c_1\\left(E\\mathbf{a_1}\\right) + \\cdots + c_n\\left(E\\mathbf{a_n}\\right) = E\\left(c_1\\mathbf{a_1} + \\cdots + c_n\\mathbf{a_n}\\right)=\\mathbf{0}$$ since $E$ is a product of elementary matrices, it is invertible. In particular this implies $$c_1\\mathbf{a_1} + \\cdots + c_n\\mathbf{a_n}\\iff c_1\\mathbf{r_1} + \\cdots + c_n\\mathbf{r_n} = \\mathbf{0}$$ And it's easy to see that this holds for any subset of the columns. What this means is that elementary row operations preserves linear relations between columns. If a set of columns vectors are linearly independent in $R$ then they will remain linearly independent after transformation", "# Can this matrix group be obtained from $(\\mathbb R,+)$ and $(\\mathbb R^*,\\cdot)$? I have stumbled upon this group of matrices in an old midterm: $$G=\\{ \\begin{pmatrix} a & b \\\\ 0 & \\frac1a \\\\ \\end{pmatrix}; a,b\\in\\mathbb R, a\\ne0 \\}.$$ The students were asked to show that this is a group (under matrix multiplication). It is easy to notice that this group contains two subgroups \\begin{align*} H_1&=\\{ \\begin{pmatrix} a & 0 \\\\ 0 & \\frac1a \\\\ \\end{pmatrix}; a\\in\\mathbb R^* \\}\\\\ H_2&=\\{ \\begin{pmatrix} 1 & b \\\\ 0 & 1 \\\\ \\end{pmatrix}; b\\in\\mathbb R \\} \\end{align*} such that $H_1 \\cong (\\mathbb R^*, \\cdot)$ and $H_2\\cong (\\mathbb R,+)$. I wonder whether there is some group theoretic construction using which we can obtain $G$ from $H_1$ and $H_2$. (I.e., some kind of construction which, given the two groups $(\\mathbb R^*,\\cdot)$ and $(\\mathbb R,+)$ returns a group isomorphic to $G$.) It is not very difficult to see that: • Every element $g\\in G$ can be expressed in exactly one way as $g=h_1h_2$ where $h_1\\in H_1$ and $h_2\\in H_2$. $$\\begin{pmatrix} a & 0 \\\\ 0 & \\frac1a \\end{pmatrix} \\begin{pmatrix} 1 & b \\\\ 0 & 1 \\end{pmatrix}= \\begin{pmatrix} a & ab \\\\ 0 & \\frac1a \\end{pmatrix}$$ • Of course, the same is true for expression of the form $g=h_2h_1$ with $h_i\\in H_i$. (We", "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "right order. So let’s look at our matrix product again: we wan’t the i-th column of the matrix product $\\mathbf{B}\\mathbf{A}$, so we look at $(\\mathbf{B}\\mathbf{A})\\mathbf{e}_i = \\mathbf{B}(\\mathbf{A}\\mathbf{e}_i) = \\mathbf{B} \\mathbf{a}_i$ exactly as claimed. As always, there’s a corresponding construction using row vectors. Using the dual basis of linear functionals (note no transpose this time!) $\\mathbf{e}^1 = \\begin{pmatrix} 1 & 0 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^2 = \\begin{pmatrix} 0 & 1 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^m = \\begin{pmatrix} 0 & 0 & \\cdots & 1 \\end{pmatrix}$ we can disassemble a matrix into its rows: $\\mathbf{A} = \\begin{pmatrix} \\mathbf{a}^1 \\\\ \\mathbf{a}^2 \\\\ \\vdots \\\\ \\mathbf{a}^m \\end{pmatrix} = \\begin{pmatrix} \\mathbf{e}^1 \\mathbf{A} \\\\ \\mathbf{e}^2 \\mathbf{A} \\\\ \\vdots \\\\ \\mathbf{e}^m \\mathbf{A} \\end{pmatrix}$ and since $\\mathbf{e}^i (\\mathbf{A} \\mathbf{B}) = (\\mathbf{e}^i \\mathbf{A}) \\mathbf{B} = \\mathbf{a}^i \\mathbf{B}$, we can write matrix products in terms of what happens to the row vectors too (though this time, we’re expanding in terms of the row vectors of the first not second factor): $\\mathbf{A} \\mathbf{B} = \\begin{pmatrix} \\mathbf{a}^1 \\mathbf{B} \\\\ \\mathbf{a}^2 \\mathbf{B} \\\\ \\vdots \\\\ \\mathbf{a}^m \\mathbf{B} \\end{pmatrix}$. ### Gluing it back together Above, the trick was to write the “slicing” step as a matrix product with one of the basis vectors – $\\mathbf{A} \\mathbf{e}_i$ and so forth. We’ll soon need to deal with the gluing step too,", "{b} )\\cdot (\\mathbf {c} \\times \\mathbf {d} )=(\\mathbf {a} \\cdot \\mathbf {c} )(\\mathbf {b} \\cdot \\mathbf {d} )-(\\mathbf {a} \\cdot \\mathbf {d} )(\\mathbf {b} \\cdot \\mathbf {c} ).}$ If a = c and b = d this simplifies to the formula above. Alternative ways to compute the cross product Conversion to matrix multiplication The vector cross product also can be expressed as the product of a skew-symmetric matrix and a vector:[11] ${\\displaystyle \\mathbf {a} \\times \\mathbf {b} =[\\mathbf {a} ]_{\\times }\\mathbf {b} ={\\begin{bmatrix}\\,0&\\!-a_{3}&\\,\\,a_{2}\\\\\\,\\,a_{3}&0&\\!-a_{1}\\\\-a_{2}&\\,\\,a_{1}&\\,0\\end{bmatrix}}{\\begin{bmatrix}b_{1}\\\\b_{2}\\\\b_{3}\\end{bmatrix}}}$ ${\\displaystyle \\mathbf {a} \\times \\mathbf {b} =[\\mathbf {b} ]_{\\times }^{\\mathrm {T} }\\mathbf {a} ={\\begin{bmatrix}\\,0&\\,\\,b_{3}&\\!-b_{2}\\\\-b_{3}&0&\\,\\,b_{1}\\\\\\,\\,b_{2}&\\!-b_{1}&\\,0\\end{bmatrix}}{\\begin{bmatrix}a_{1}\\\\a_{2}\\\\a_{3}\\end{bmatrix}}}$ where superscript T refers to the transpose operation, and [a]× is defined by: ${\\displaystyle [\\mathbf {a} ]_{\\times }{\\stackrel {\\rm {def}}{=}}{\\begin{bmatrix}\\,\\,0&\\!-a_{3}&\\,\\,\\,a_{2}\\\\\\,\\,\\,a_{3}&0&\\!-a_{1}\\\\\\!-a_{2}&\\,\\,a_{1}&\\,\\,0\\end{bmatrix}}.}$ It should be noted that [a]× is a singular matrix where a is it's (right and left) null-vector. Also, if a is itself a cross product: ${\\displaystyle \\mathbf {a} =\\mathbf {c} \\times \\mathbf {d} }$ then ${\\displaystyle [\\mathbf {a} ]_{\\times }=\\mathbf {d} \\mathbf {c} ^{\\mathrm {T} }-\\mathbf {c} \\mathbf {d} ^{\\mathrm {T} }.}$ This result can be generalized to higher dimensions using geometric algebra. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector.{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B=", "must be 1. In other words, if we have a column vector $\\mathbf{v}=\\begin{pmatrix} a \\\\ b\\end{pmatrix}$, then $\\mathbf{v}$'s inner product with itself must equal 1. For two vectors $\\mathbf{v}_1=\\begin{pmatrix} a_1 \\\\ b_1\\end{pmatrix}$ and $\\mathbf{v}_2=\\begin{pmatrix} a_2 \\\\ b_2\\end{pmatrix}$, the definition of the inner product of $\\mathbf{v}_1$ and $\\mathbf{v}_2$, denoted $\\left \\langle \\mathbf{v}_1 | \\mathbf{v}_2 \\right \\rangle$ can be written in matrix notation as:$$\\left \\langle \\mathbf{v_1} | \\mathbf{v_2} \\right \\rangle=\\mathbf{v}_1^\\dagger \\mathbf{v}_2 = \\begin{pmatrix} a_1^* & b_1^*\\end{pmatrix}\\begin{pmatrix} a_2 \\\\ b_2 \\end{pmatrix}=a_1^*a_2+b_1^*b_2$$ where * denotes the complex conjugate: e.g. if a=1+2i, a*=1-2i. It should be clear in this that the symbol $\\dagger$ denotes the adjoint operation, which is the same as taking the complex conjugate and then the transpose (with a column vector, the transpose simply changes a column vector into a row vector.) Equivalently, you could write the definition of the inner product of $\\mathbf{v}_1$ and $\\mathbf{v}_2$ in vector notation as: $$\\left \\langle \\mathbf{v_1} | \\mathbf{v_2} \\right \\rangle=\\mathbf{v}_1^* \\cdot \\mathbf{v}_2 = \\begin{pmatrix} a_1^* \\\\ b_1^*\\end{pmatrix}\\cdot\\begin{pmatrix} a_2 \\\\ b_2 \\end{pmatrix}=a_1^*a_2+b_1^*b_2$$ As I stated earlier, the definition of $\\mathbf{v}=\\begin{pmatrix} a \\\\ b\\end{pmatrix}$ being normalized is that its inner product with itself is 1: $$1=\\left \\langle \\mathbf{v} | \\mathbf{v} \\right \\rangle=\\mathbf{v}^\\dagger \\mathbf{v} = \\begin{pmatrix} a^* & b^*\\end{pmatrix}\\begin{pmatrix} a \\\\ b\\end{pmatrix}=a^*a+b^*b$$ So let's see what happens if we use your suggestion that $\\mathbf{v}=\\begin{pmatrix} \\frac{1}{2} \\\\", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "We partition A, B and C into equally sized block matrices $\\mathbf{A} = \\begin{bmatrix} \\mathbf{A}_{1,1} & \\mathbf{A}_{1,2} \\\\ \\mathbf{A}_{2,1} & \\mathbf{A}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{B} = \\begin{bmatrix} \\mathbf{B}_{1,1} & \\mathbf{B}_{1,2} \\\\ \\mathbf{B}_{2,1} & \\mathbf{B}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{C} = \\begin{bmatrix} \\mathbf{C}_{1,1} & \\mathbf{C}_{1,2} \\\\ \\mathbf{C}_{2,1} & \\mathbf{C}_{2,2} \\end{bmatrix}$ with $\\mathbf{A}_{i,j}, \\mathbf{B}_{i,j}, \\mathbf{C}_{i,j} \\in R^{2^{n-1} \\times 2^{n-1}}$ then $\\mathbf{C}_{1,1} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{1,2} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,2}$ $\\mathbf{C}_{2,1} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{2,2} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,2}$ With this construction we have not reduced the number of multiplications. We still need 8 multiplications to calculate the Ci,j matrices, the same number of multiplications we need when using standard matrix multiplication. Now comes the important part. We define new matrices $\\mathbf{M}_{1} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{2,2}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{2,2})$ $\\mathbf{M}_{2} := (\\mathbf{A}_{2,1} + \\mathbf{A}_{2,2}) \\mathbf{B}_{1,1}$ $\\mathbf{M}_{3} := \\mathbf{A}_{1,1} (\\mathbf{B}_{1,2} - \\mathbf{B}_{2,2})$ $\\mathbf{M}_{4} := \\mathbf{A}_{2,2} (\\mathbf{B}_{2,1} - \\mathbf{B}_{1,1})$ $\\mathbf{M}_{5} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{1,2}) \\mathbf{B}_{2,2}$ $\\mathbf{M}_{6} := (\\mathbf{A}_{2,1} - \\mathbf{A}_{1,1}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{1,2})$ $\\mathbf{M}_{7} := (\\mathbf{A}_{1,2} - \\mathbf{A}_{2,2}) (\\mathbf{B}_{2,1} + \\mathbf{B}_{2,2})$ only using 7 multiplications (one for each Mk) instead of 8. We may now express the Ci,j in terms of Mk, like this: $\\mathbf{C}_{1,1} = \\mathbf{M}_{1} + \\mathbf{M}_{4} - \\mathbf{M}_{5} + \\mathbf{M}_{7}$ $\\mathbf{C}_{1,2} = \\mathbf{M}_{3} + \\mathbf{M}_{5}$ $\\mathbf{C}_{2,1} =", "# Thread: Left inverse of matrix 1. ## Left inverse of matrix Consider the matrix $\\mathbf{C} = \\begin{pmatrix}1 & -1\\\\ 1&1 \\\\ 0&1 \\end{pmatrix}$ Find a left-inverse of C. Attempt: $\\begin{pmatrix}x1 & y1 & z1\\\\ x2&y2 &z2 \\end{pmatrix}\\begin{pmatrix}1 & -1\\\\ 1&1\\\\ 0&1 \\end{pmatrix} = \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ $\\begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\\\ x2 + y2& -x2 + y2 + z2 \\end{pmatrix}= \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ I know that you suppose to somehow use gauss reduction, and thought about gauss reducing: $x1 + y1 = 1$ $-x1 + y1 + z1 = 0$ and $x2 + y2 = 0$ $-x2 + y2 + z2 = 1$ but didnt get any luck 2. Only square matrices have inverses. To solve a matrix equation of the form $\\mathbf{Y} = \\mathbf{AX}$ for $\\mathbf{X}$ where $\\mathbf{A}$ is not square, you need to premultiply both sides by $\\mathbf{A}^T$ first. $\\mathbf{Y} = \\mathbf{AX}$ $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$. Now $\\mathbf{A}^T\\mathbf{A}$ is a square matrix, so you can premultiply both sides by $(\\mathbf{A}^T\\mathbf{A})^{-1}$. $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = (\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{IX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{X}$. 3. Originally Posted by Prove It Only square matrices have inverses. Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both. 4. You have", "matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & 0 & 1 & 1 \\\\ 0 & 0 & 3 & 2 \\\\ 0 & 1 & 6 & 10 \\end {pmatrix}$ ### Arbitrary Matrix $3$ Let $\\mathbf A$ be the matrix: $\\mathbf A = \\begin {pmatrix} 0 & 4 & 7 & 10 \\\\ 2 & 5 & 8 & 11 \\\\ 3 & 6 & 9 & 12 \\end {pmatrix}$ The matrix $\\mathbf R$ corresponding to the row operation to clear the first column is: $\\mathbf R = \\begin {pmatrix} 0 & \\dfrac 1 2 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & -\\dfrac 3 2 & 1 \\end {pmatrix}$ and the matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & \\dfrac 5 2 & 4 & \\dfrac {11} 2 \\\\ 0 & 4 & 7 & 10 \\\\ 0 & -\\dfrac 3 2 & -3 & -\\dfrac 9 2 \\end {pmatrix}$ ## Also presented as Some sources are not concerned about making $b_{1 1}$ equal to $1$. Hence they do not use step $3$ of the algorithm.", "$\\begin {pmatrix} \\mathbf A_2 & \\mathbf b_2 \\end {pmatrix} = \\paren {\\begin {array} {ccc|c} 1 & -1 & -1 & 1 \\\\ 0 & 1 & 2 & -1 \\\\ 0 & 2 & 4 & -1 \\\\ \\end {array} }$ $e_3 := r_1 \\to r_1 + r_2$ $e_4 := r_3 \\to r_3 - 2 r_2$ Hence: $\\begin {pmatrix} \\mathbf A_4 & \\mathbf b_4 \\end {pmatrix} = \\paren {\\begin {array} {ccc|c} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 2 & -1 \\\\ 0 & 0 & 0 & 1 \\\\ \\end {array} }$ The bottom line of this augmented matrix leads to the false statement $0 = 1$. It follows that this system of simultaneous linear equations is inconsistent. $\\blacksquare$", "no effect). Second, the matrices that reduce to $\\begin{pmatrix} 1 &a \\\\ 0 &0 \\end{pmatrix}$ have the form $\\begin{pmatrix} b &ba \\\\ c &ca \\end{pmatrix}$ (where $a,b,c\\in\\mathbb{R}$, and $b$ and $c$ are not both zero). Next, the matrices that reduce to $\\begin{pmatrix} 0 &1 \\\\ 0 &0 \\end{pmatrix}$ have the form $\\begin{pmatrix} 0 &a \\\\ 0 &b \\end{pmatrix}$ (where $a,b\\in\\mathbb{R}$, and not both are zero). Finally, the matrices that reduce to $\\begin{pmatrix} 1 &0 \\\\ 0 &1 \\end{pmatrix}$ are the nonsingular matrices. That's because a linear system for which this is the matrix of coefficients will have a unique solution, and that is the definition of nonsingular. (Another way to say the same thing is to say that they fall into none of the above classes.) Problem 3 Describe all matrices in the row equivalence class of these. 1. $\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &2 \\\\ 2 &4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &1 \\\\ 1 &3 \\end{pmatrix}$ 1. They have the form $\\begin{pmatrix} a &0 \\\\ b &0 \\end{pmatrix}$ where $a,b\\in\\mathbb{R}$. 2. They have this form (for $a,b\\in\\mathbb{R}$). $\\begin{pmatrix} 1a &2a \\\\ 1b &2b \\end{pmatrix}$ 3. They have the form $\\begin{pmatrix} a &b \\\\ c &d \\end{pmatrix}$ (for $a,b,c,d\\in\\mathbb{R}$) where $ad-bc\\neq 0$. (This is the formula that determines when a $2 \\! \\times" ]

[ "intersections of the line with the unit circle are the solutions. Clearly there are two solutions for A, one positive and one negative. 3. Feb 27, 2012 ### SammyS Staff Emeritus The general solution to the equation, $\\tan(x)=A\\,,$ is $x=\\arctan(A)+k\\pi\\,,$ where k is an integer.", "Problems Problem 1: Solve: tan(arcsin 12/13) Problem 2: Find the value of x, cos(arccos 1) = cos x", "? sin2 A or cos A = ?1 ? sinA (Ignoring the negative value , since sin A is positive, Cos A will also be positive.) tan A = sin A/cos A = sin A/?1 ? sinA sec A = 1/cosA = 1/?1 ? sinA", "0.3$ since cos(u) = 0.3 , there is one solution in quad I and one solution in quad IV (cosine is positive in quads I and IV). note that the inverse cosine function (arccos) has a range from 0 to 90 degrees for positive values of cosine, so ... $u = \\arccos(0.3) \\approx 72.5^\\circ$ $u = 360 - \\arccos(0.3) \\approx 287.5^\\circ$ divide each angle above by 20 to get the values for t between 0 and 18.", "# What does -sin(arccos(2))-tan(arcsec(4)) equal? If a = arc cos 2, cos a = 2 > 1. A cosine value has to be $\\in \\left[- 1 , + 1\\right]$.", "# Arccos(x) where x>1 or x<-1 1. Sep 4, 2011 ### Philosophaie What happens if you have to take the arccos of a number greater than 1 or less than -1. I know from a right triangle if you have: [ tex ]\\Theta[ /tex ]=arccos(x) if you use x as the adjacent side, 1 as the hypotenuse and [ tex ]\\sqrt{1-x^2}[ /tex ] as the opposite. Is that where the negative sign start? I have an equation that has numbers from -1.01 to 1.01 as inputs \"x\". The output runs from 0 to 180 without the numbers 180 to 359 or any negative numbers. Where do I find the negative or the higher angles? Last edited: Sep 4, 2011 2. Sep 4, 2011 ### mathman Cos and sin of real variables are confined to the [-1,1] range. To get outside this range, you have to use complex numbers, i.e. cos(z) where z is complex. To get arccos of u, when |u| > 1, you will end up with a complex number. 3. Sep 4, 2011 ### HallsofIvy As mathman said, for any real x, cos(x) lies in the interval [-1, 1]. To go outside that you must look for a complex z. For z a complex number, $cos(z)= (e^z+ e^{-z})/2$. If y> 1, you must solve $(e^z+ e^{-z})/2=", "## anonymous 5 years ago Simplify tan(arcsin x +arccos X) for 0 =< x =< 1 (=< means less than or equal to) Let$\\theta =(\\sin^{-1} x +\\cos^{-1} x)$Make a right triangle with angle tan theta.", "right triangle problem and we know that the angle θ must be somewhere between 0 and 90° then the solution is the single value: θ = arccos (c) On the other hand, if this is a more advanced problem and we need to find all the possible angles whose cos is c then the solution is the entire set of values: θ = Arccos (c) The solutions in these two cases follow directly from the definitions of the arccos function and Arccos relation. Note that if c is greater than 1 or less than −1 then there are no real solutions. However there are complex solutions. Evaluating Arccos(c) If c is a number then the entire set of values of Arccos(c) can be found using the following procedure. Refer to the graph to the right where the dots are the desired values. • The first value (the principal value), denoted θPV , is found by evaluating arccos(c) with a calculator or with the Algebra Coach. • The second value, called θ2 , is found by using the symmetry of the Arccos curve. Notice that the two blue arrows in the graph have the same length. This means that θ2 is just as far below 2π as θPV is above zero. In formula form: θ2 = 2 π −", "For those inputs, sin() is always positive. That rules out the final answer of the negative root for every value of x. Last edited: mfb Mentor The arcsine goes from -pi/2 to +pi/2, in this range the cosine is positive. Just check the ranges to see which sign is necessary.", "they made the denominator be |x| instead of just x. My guess is that they wanted to force tan(arccos(x)) to be positive. However, tan(arccos(x)) is only positive when x is positive 25. jim_thompson5910 So it should be $\\Large \\tan(\\arccos(x)) = \\frac{\\sqrt{1-x^2}}{x}$", "– x^2). Are we sure that our solution doesn’t need a negative sign? The fact is that we don’t need it; we can show this based on the fact that arccos(x) is in quadrant 1 or 2. This is similar to what happened above with a cosecant; that, and the sine, are positive for the entire range of arccos, which keeps things simple. In other problems, this warning about the sign might be important. So we have gone through a good list of problems combining trig functions and their inverses. There are probably several more types we could look at, but this will do. This site uses Akismet to reduce spam. Learn how your comment data is processed.", "# What does sin(arccos(2))-3sec(arc tan(3)) equal? arc cos(2) is invalid arc cos function. $| \\cos x | \\le 1$. So, there is no arc cos for which cosine is 2.", "1. For those inputs, sin() is always positive. That rules out the final answer of the negative root for every value of x. Last edited: mfb Mentor The arcsine goes from -pi/2 to +pi/2, in this range the cosine is positive. Just check the ranges to see which sign is necessary. Mr Davis 97", "x+y$$ and evaluate: $\\tan (z+u)=\\frac{\\tan z+\\tan u}{1-\\tan z \\cdot \\tan u}=\\frac{-3+3}{1-9}=0$ So $$\\tan( x+y+z ) = 0$$. Hence, $$\\arctan (1)+\\arctan (2)+\\arctan (3)=x+y+z=\\pi$$. ### Intersection of a line and periodic function I A1, 2012 Find the number of real solutions to the equation $$x=99 \\sin (\\pi x)$$. Solution Since the equation is symmetric, if $$r$$ is a solution then $$-r$$ is also a solution. Hence, we restrict to finding the positive roots. Note that $$x=0$$ is also a root. The function $$99 \\sin \\pi x$$ is positive when $$2k < x < 2k+1$$. It equals $$x$$ when $$x\\leq 99$$. The positive cycle of $$\\sin\\pi x$$ meets $$y=x$$ twice, so there are 198 positive solutions. Counting the zero gives 199 solutions. ### Intersection of a line and periodic function II A3, 2011 Find the number of solutions to the equation $$4 \\sin (3x+2)=1$$ in the domain $$x\\in[0,2\\pi]$$. Solution We want the number of intersection points between the line $$y=1$$ and the periodic function $$f(x) = 4\\sin (3x+2)$$. The function $$f$$ has period $$2\\pi/3$$. Between every maxima and a minima, there is exactly one solution. At $$x=0$$ and $$x=2\\pi$$, $$f$$ is close to maxima. So all we need to do is find the number of minimas of $$f$$ in the given domain and multiply it by 2. The minimas occur at", "## Trigonometry (10th Edition) 1. x is positive 2. y is negative 3. Therefore, $\\tan\\theta=\\frac{x}{y}=\\frac{+}{-}=-$", "Don't have an account Signup Now # Introduction $sin^{–1} x,\\ cos^{–1} x,\\ tan^{–1} x$ etc. represents angles or numbers whose values of sine, cosine and tangent is ‘$x$’, provided that the value in numerical form is smallest. These can be written as $\\arcsin x,\\ arccos x\\$ etc. If two angles whose modulus is equal, in which one is positive and other is negative then we take positive sign.", "all the positive elements. How to solve:", "# Find all positive integral solutions of the equation $\\tan^{-1}x+\\cot^{-1}y=\\tan^{-1}3$ $\\begin{array}{1 1}(a)\\;x=1,y=2&(b)\\;x=0,y=1\\\\(c)\\;x=2,y=3&(d)\\;x=2,y=3\\end{array}$", "# What does 2sin(arccos(4))+tan(arctan(1)) equal? arc cos(4) is undefined $| \\cos \\left(x\\right) | \\le 1$. There is no solution for cos x - 4 > 1. 'arc cos' is used for 'the angle whose cosine is'.", "= 0 x = arcsin(0) x = n(pi) where n is an integer. When applying such maths to physics, it is necessary to choose which answer you require, which may, or may not be obvious. In your question, you assumed that the only solution to arccos(0.8) is 0.6435... when in fact, -0.6435 is an equally valid solution. (This is a consequence of cos(x) = cos(-x)). Substituting this in to your equation gives the answer provided by the book. If you are uncomfortable with having to choose which answer is correct without using educated guesswork or drawing graphs, use the equation for the velocity of the object to find out whether each solution corresponds to a positive or negative velocity. 3. Apr 29, 2003 gnome Thanks Claude, that was very helpful." ]

[ "# $\\sin(2\\arcsin(\\frac{1}{3}))=?$ A broad one I have written it as $$\\sin(2\\arcsin(\\frac{1}{3}))=2\\sin(\\arcsin(\\frac{1}{3}))\\cos(\\arcsin(\\frac{1}{3})).$$Then, solved for $$\\cos(\\arcsin(\\frac{1}{3}))=\\frac{\\sqrt{10}}{3}$$ $$\\arcsin(\\frac{1}{3})=\\theta$$$$\\sin\\theta=\\frac{1}{3}\\Rightarrow\\cos\\theta=\\frac{\\sqrt{10}}{3}.$$But noticed something weird (1) it didn't satisfy the Pythagorean theorem as $\\sqrt{10}\\approx3.1622,$ which means that the adjacent side to the angle $\\theta$ is greater than the hypotenuse, (which is just 3) why this method (using right triangle to determine $\\cos(\\theta)$) didn't work here although the answer ($\\frac{\\sqrt{10}}{3}$) satisfied the range of the cosine function, which is $R(\\cos\\theta)=\\Bbb{R}$? After it (2) used this formula$$\\arcsin(x)=\\arccos(\\sqrt{1-x^2});$$$$\\arcsin(\\frac{1}{3})=\\arccos(\\sqrt{1-\\frac{1}{9}})=\\arccos(\\frac{\\sqrt{8}}{3});$$then, checked if it satisfies or not $D(\\arccos(\\theta))=[-1;1],$ satisfied, since $\\frac{\\sqrt{8}}{3}\\approx0.942.$ Then proceeded on my problem $$2\\sin(\\arcsin(\\frac{1}{3}))\\cos(\\arcsin(\\frac{1}{3}))=\\frac{2}{3}\\cdot\\frac{\\sqrt{8}}{3}=\\frac{2\\sqrt{8}}{9}.$$ Anyway, it didn't turn out to be the right answer. Where is my mistake? • Why do you think this is not the correct answer? – kingW3 Jun 23 '17 at 21:20 • – jim Jun 23 '17 at 21:20 • @kingW3 The person found $\\cos \\theta = \\sqrt{10}/3 > 1$ – jim Jun 23 '17 at 21:22 • Because there wasn't such ōne. – user438365 Jun 23 '17 at 21:22 • @jim Yeah I agree on that one but,he said bottom that $\\frac{2\\sqrt{8}}9$ is not an answer. – kingW3 Jun 23 '17 at 21:22 Your mistake is that $\\cos\\theta=\\dfrac{2\\sqrt{2}}{3}$, not $\\dfrac{\\sqrt{10}}{3}$. $$\\sin2\\arcsin\\frac{1}{3}=2\\cdot\\frac{1}{3}\\cdot\\sqrt{1-\\frac{1}{9}}=\\frac{4}{9}\\sqrt2$$", "# Question on inverse tangent. 1. Feb 17, 2013 ### yungman $$\\Psi=\\tan^{-1}\\left(\\frac{\\cos\\omega t}{\\cos(\\omega t+\\delta)}\\right)$$ I want to find out whether $\\Psi$ increase or decrease with time t, if $\\delta$ is positive and if $\\delta$ is negative. $$\\Psi=\\tan^{-1}\\left(\\frac{\\cos\\omega t}{\\cos(\\cos\\omega t \\cos \\delta+\\sin\\omega t \\sin\\delta)}\\right)\\;=\\;\\tan^{-1}\\left(\\frac {1}{(\\cos\\delta + \\tan(\\omega t)\\sin \\delta)}\\right)$$ $$\\Rightarrow\\;\\Psi=\\tan^{-1}\\left(\\frac{1}{K_1+K_2\\tan \\omega t}\\right)$$ where $K_1$ is a constant and $K_2$ is a constant following the sign of $\\delta$. This is as far as I know how to go, what can I do to simplify it. I want to find out whether $\\Psi$ increase or decrease with time if $\\delta$ is positive and if $\\delta$ is negative. 2. Feb 17, 2013 ### Simon Bridge You'll have to use the known limits to the problem and the properties of the tangent function. I would be thinking: if $\\Psi$ is increasing, what happens to $\\tan\\Psi$ ? What are the allowed values for $\\omega t$ ? 3. Feb 17, 2013 ### yungman Thanks for the reply. I think I come up with a way and see whether you agree or not: Let $\\theta=\\frac {\\pi}{2}-\\Psi$. So if $\\tan \\Psi=\\frac y x\\;\\Rightarrow \\;\\tan \\theta = \\frac x y$ So $\\theta \\;=\\;\\tan ^{-1} \\left(\\frac{\\cos(\\omega t +\\delta}{\\cos \\omega t}\\right)$ This will be easy to solve and find out the direction. Then we know if θ increase with t,", "\\arccos x$ $\\arccsc \\frac{1}{x} = \\arcsin x$ If you only have a fragment of a sine table: $\\arccos x = \\arcsin \\sqrt{1-x^2},\\text{ if }0 \\leq x \\leq 1$ $\\arctan x = \\arcsin \\frac{x}{\\sqrt{x^2+1}}$ Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real). From the half-angle formula $\\tan \\frac{\\theta}{2} = \\frac{\\sin \\theta}{1+\\cos \\theta}$, we get: $\\arcsin x = 2 \\arctan \\frac{x}{1+\\sqrt{1-x^2}}$ $\\arccos x = 2 \\arctan \\frac{\\sqrt{1-x^2}}{1+x},\\text{ if }-1 < x \\leq +1$ $\\arctan x = 2 \\arctan \\frac{x}{1+\\sqrt{1+x^2}}$ ## Relationships between trigonometric functions and inverse trigonometric functions $\\sin (\\arccos(x)) = \\cos(\\arcsin(x)) = \\sqrt{1-x^2}$ $\\sin (\\arctan(x)) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan(x)) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arcsin(x)) = \\frac{x}{\\sqrt{1-x^2}}$ $\\tan (\\arccos(x)) = \\frac{\\sqrt{1-x^2}}{x}$ ## General solutions Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent", "here we continued to try to understand the differences in reasoning and answers. Given that $\\cos(\\theta)=-x$ can be simplified to $\\theta=\\arccos(x)+\\pi$, then it can also be simplified into $\\arccos(-x)=\\theta$ which would suggest that $$\\tan(\\arccos(-x)) = \\tan(\\arccos(x)+\\pi)$$ but putting both functions in Wolfram Alpha did not result in the same graph. So my question is why is this case? And what errors did I make in my calculations that would result in something happening like this? Wolfram Alpha: tan(arccos(x)+π) tan(arccos(-x)) Sorry if this formatting is terrible, I did this late at night. I might fix it later. • Since $\\theta$ lies in $[\\pi,2\\pi]$, $\\sin\\theta$ is always negative. So $\\tan\\theta=\\dfrac{\\sqrt{1-x^2}}{x}$ – AnotherJohnDoe Sep 19 '18 at 4:22 • Also, $\\cos\\theta=-x\\nRightarrow\\theta=cos^{-1}(-x)$, because $\\theta\\ge\\pi$ – AnotherJohnDoe Sep 19 '18 at 4:25 The range of $\\arccos x$ is always in $[0,\\pi]$ so $\\tan \\arccos x = \\frac {\\sqrt{1-x^2}}{x}$ however if you say $\\cos \\theta = x$ there are two values of $\\theta$ in $[0,2\\pi)$ that $\\theta$ could take. and $\\tan \\theta = \\pm \\frac{\\sqrt{1-x^2}}{x}$ depending on which value of $\\theta$ you choose. You need to keep track of what quadrants you might be in. This seems to be the source of your problem. The you could us the identity $\\tan x+\\pi = \\tan x$ $\\tan(\\arccos x+ \\pi) = \\tan (\\arccos x)$ $\\tan(\\arccos x+", "# Math Help - evaluate arccosx 1. ## evaluate arccosx evaluate sin (arccos x) , 0 < x< 1 2. Note that: $\\sin x = \\pm \\sqrt{1 + cos^2 x}$ $\\sin (\\arccos x) = +\\sqrt{1 - \\left[\\cos (\\arccos x)\\right]^{2}}$ (since $0 \\leq x \\leq 1 \\: \\: \\Rightarrow \\: \\: 0 \\leq \\arccos x \\leq \\frac{\\pi}{2}$) Can you finish off? 3. Hello, cruxkitty! ${\\color{red}\\rlap{////////}}\\text{Evaluate: }\\;\\sin(\\arccos x),\\quad 0 < x< 1$ . They meant to say \"Simplify\". Recall that an inverse trig expression is an angle. $\\text{We have: }\\;\\sin\\underbrace{(\\arccos x)}_{\\theta}$ . . . . . . angle whose cosine is x \"The angle whose cosine is $x$\" looks like this: Code: * * * 1 * * ____ * * √1-x² * * * θ * * * * * * * * x $\\cos\\theta \\:=\\:\\frac{x}{1} \\:=\\:\\frac{adj}{hyp} \\quad\\Rightarrow\\quad adj = x,\\;\\;hyp = 1$ From Pythagorus, we find that: . $opp = \\sqrt{1-x^2}$ Hence: . $\\sin\\theta \\:=\\:\\frac{opp}{hyp} \\:=\\:\\frac{\\sqrt{1-x^2}}{1} \\:=\\:\\sqrt{1-x^2}$ Therefore: . $\\boxed{\\sin(\\arccos x) \\;=\\;\\sqrt{1-x^2}}$ 4. $\\sin(\\arccos{x})$ let ... $\\theta = \\arccos{x}$ so ... $\\cos{\\theta} = \\frac{x}{1}$ remember the basic right triangle from geometry days? $\\cos{\\theta} = \\frac{x}{1} = \\frac{adj}{hyp}$ the adjacent side has length x and the hypotenuse has length 1. using Pythagoras, the opposite side has length = $\\sqrt{1-x^2}$ $\\sin(\\arccos{x}) = \\sin{\\theta} = \\frac{opp}{hyp} = \\frac{\\sqrt{1-x^2}}{1}$ Soroban was quicker", "derivations are longer. $\\theta$ $\\sin \\theta$ $\\cos \\theta$ $\\tan \\theta$ Diagram $\\arcsin x$ $\\sin (\\arcsin x) = x$ $\\cos (\\arcsin x) = \\sqrt{1-x^2}$ $\\tan (\\arcsin x) = \\frac{x}{\\sqrt{1-x^2}}$ $\\arccos x$ $\\sin (\\arccos x) = \\sqrt{1-x^2}$ $\\cos (\\arccos x) = x$ $\\tan (\\arccos x) = \\frac{\\sqrt{1-x^2}}{x}$ $\\arctan x$ $\\sin (\\arctan x) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan x) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arctan x) = x$ $\\arccot x$ $\\sin (\\arccot x) = \\frac{1}{\\sqrt{1+x^2}}$ $\\cos (\\arccot x) = \\frac{x}{\\sqrt{1+x^2}}$ $\\tan (\\arccot x) = \\frac{1}{x}$ $\\arcsec x$ $\\sin (\\arcsec x) = \\frac{\\sqrt{x^2-1}}{x}$ $\\cos (\\arcsec x) = \\frac{1}{x}$ $\\tan (\\arcsec x) = \\sqrt{x^2-1}$ $\\arccsc x$ $\\sin (\\arccsc x) = \\frac{1}{x}$ $\\cos (\\arccsc x) = \\frac{\\sqrt{x^2-1}}{x}$ $\\tan (\\arccsc x) = \\frac{1}{\\sqrt{x^2-1}}$ ### Relationships among the inverse trigonometric functions The usual principal values of the arcsin(x) (red) and arccos(x) (blue) functions graphed on the cartesian plane. The usual principal values of the arctan(x) and arccot(x) functions graphed on the cartesian plane. Principal values of the arcsec(x) and arccsc(x) functions graphed on the cartesian plane. Complementary angles: \\begin{align} \\arccos x &= \\frac{\\pi}{2} - \\arcsin x \\\\[0.5em] \\arccot x &= \\frac{\\pi}{2} - \\arctan x \\\\[0.5em] \\arccsc x &= \\frac{\\pi}{2} - \\arcsec x \\end{align} Negative arguments: \\begin{align} \\arcsin (-x) &= -\\arcsin x \\\\ \\arccos (-x) &= \\pi -\\arccos x \\\\ \\arctan (-x) &= -\\arctan x \\\\ \\arccot (-x) &= \\pi", "also be replaced with $$\\arctan\\left(\\sqrt{x^2-1}\\right)$$. For $$\\theta \\in \\left(\\frac{\\pi}{2}, \\pi\\right]$$, $$\\tan(\\theta) \\leq 0$$, so you get $$-\\int \\tan^2(\\theta) \\mathrm d\\theta = -\\int \\left[\\sec^2(\\theta)-1\\right] \\mathrm d\\theta = \\theta-\\tan(\\theta)+C \\longrightarrow \\sqrt{x^2-1}+\\text{arcsec}(x)+C$$ Since tangent is negative in the second quadrant, $$\\tan(\\theta) = \\tan(\\theta -\\pi) = -\\sqrt{x^2-1}$$ (remember that tangent takes arguments in the first and fourth quadrants), so the $$\\theta$$ term can also be replaced with $$\\pi-\\arctan\\left(\\sqrt{x^2-1}\\right)$$. In both cases, the anti-derivative could be re-written as $$\\sqrt{x^2-1}-\\arctan\\left(\\sqrt{x^2-1}\\right)+C$$. Basically, it just \"combines\" your other two anti-derivatives and expresses them as a single function rather than having one for each case. • Would just leaving the answer as: √(x^2-1) - arcsec(x) + C , still be correct? – user749176 Feb 18 '20 at 0:09 • For indefinite integrals, we usually assume $\\tan(\\theta)$ is positive (at least from what I've seen), so yeah, that's fine. – KM101 Feb 18 '20 at 0:16 Now, both answers are correct. They merely look different. They differ by a constant. Note 1... $$-\\sqrt{\\frac{x+1}{1-x}}+\\sqrt{\\frac{x+1}{1-x}}\\;x = -\\sqrt{\\frac{x+1}{1-x}}\\;(1-x) = -\\frac{\\sqrt{x+1}\\;(1-x)}{\\sqrt{1-x}} \\\\= -\\sqrt{x+1}\\sqrt{1-x} =-\\sqrt{(1+x)(1-x)} =-\\sqrt{1-x^2}$$ Note 2... $$2\\,\\arcsin \\left( \\frac{\\sqrt {1+x}}{\\sqrt {2}} \\right) =\\pi-2\\,\\arcsin \\left( \\frac{\\sqrt {1-x}}{\\sqrt {2}} \\right)$$ • For the second line of step, I multiplied top and bottom by √(1+x) in order to use the trig substitution of x = sinθ – user749176 Feb 17 '20 at 21:12 •", "\\theta = \\frac{x}{\\sqrt{x^2+1}} && \\csc (\\tan^{-1}x) = \\csc \\theta = \\frac{\\sqrt{x^2+1}}{x}\\\\\\cos (\\tan^{-1}x) & = \\cos \\theta = \\frac{1}{\\sqrt{x^2+1}} && \\sec (\\tan^{-1}x) = \\sec \\theta = \\sqrt{x^2+1}\\\\\\tan (\\tan^{-1}x) & = \\tan \\theta = x && \\cot (\\tan^{-1}x) = \\cot \\theta = \\frac{1}{x}$ #### Example A Find $\\sin (\\tan^{-1} 3x)$ . Solution: Instead of using $x$ in the ratios above, use $3x$ . $\\sin (\\tan^{-1} 3x) = \\sin \\theta = \\frac{3x}{\\sqrt{(3x)^2 + 1}} = \\frac{3x}{\\sqrt{9x^2+1}}$ #### Example B Find $\\sec^2 (\\tan^{-1} x)$ . Solution: This problem might be better written as $[\\sec (\\tan^{-1} x)]^2$ . Therefore, all you need to do is square the ratio above. $[\\sec (\\tan^{-1}x)]^2 = \\left ( \\sqrt{x^2+1} \\right )^2 = x^2 + 1$ You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section. #### Example C Find $\\csc^3 (\\tan^{-1} 4x)$ . Solution: This example is similar to both of the examples above. First, use $4x$ instead of $x$ in the ratios above. Second, the $\\csc^3$ is the same as taking the $\\csc$ function and cubing it. $[\\csc (\\tan^{-1}4x)]^3 = \\left ( \\frac{\\sqrt{(4x)^2+1}}{x} \\right)^3 = \\frac{(16x^2+1)^{3/2}}{x^{3}}$ ### Vocabulary Trigonometric Function: A trigonometric function is a function describing a relationship", "F^{\\prime}(x)-c G^{\\prime}(x)=\\psi (x)$$ $$F(x)-G(x)=\\frac{1}{c} \\int_{0}^{x} \\psi(s) \\mathrm{d} s+C$$ \\begin{aligned} &F(x)=\\frac{1}{2} \\theta (x)+\\frac{1}{2 c} \\int_{0}^{x} \\psi (s) \\mathrm{d} s+\\frac{C}{2} \\\\ &G(x)=\\frac{1}{2} \\theta (x)-\\frac{1}{2 c} \\int_{0}^{x} \\psi (s) \\mathrm{d} s-\\frac{C}{2} \\end{aligned} $$\\phi(x, t)=\\frac{\\theta (x+c t)+\\theta (x-c t)}{2}+\\frac{1}{2 c} \\int_{x-c t}^{x+c t} \\psi (s) \\mathrm{d} s \\quad \\text{d’Alembert’s formula}$$ $$\\phi (x,0)= \\theta (x) \\\\ \\phi_{t} (x,0)= \\psi(x)=0$$ In this case, the d’Alembert’s solution takes the form $$\\phi(x, t)=\\frac{1}{2}[\\theta (x+c t)+\\theta (x-c t)]$$ $$\\phi (x,0)= \\theta (x)=0 \\\\ \\phi_{t} (x,0)= \\psi(x)$$ For this case the d’Alembert’s solution assumes the form $$\\phi (x,t)=\\frac{1}{2 c} \\int_{x-c t}^{x+c t} \\psi (s) \\mathrm{d} s= \\frac{1}{2}[g(x+ct) − g(x−ct)]$$ ———————————————下面进入数值求解部分————————————————— #### 3. Finite differencing approximation 有限差分近似 $$u_{i+1}= u_{i} + \\left(\\frac{\\partial u}{\\partial x}\\right)_{i}(\\Delta x)+\\frac{1}{2 !}\\left(\\frac{\\partial^{2} u}{\\partial x^{2}}\\right)_{i}(\\Delta x)^{2}+\\frac{1}{3 !}\\left(\\frac{\\partial^{3} u}{\\partial x^{3}}\\right)_{i}(\\Delta x)^{3}+\\frac{1}{4 !}\\left(\\frac{\\partial^{4} u}{\\partial x^{4}}\\right)_{i}(\\Delta x)^{4} +O\\left(\\Delta x^{5}\\right)$$ $$u_{i-1}=u_{i}+\\left(\\frac{\\partial u}{\\partial x}\\right)_{i}(-\\Delta x)+\\frac{1}{2 !}\\left(\\frac{\\partial^{2} u}{\\partial x^{2}}\\right)_{i}(-\\Delta x)^{2}+\\frac{1}{3 !}\\left(\\frac{\\partial^{3} u}{\\partial x^{3}}\\right)_{i}(-\\Delta x)^{3}+\\frac{1}{4 !}\\left(\\frac{\\partial^{4} u}{\\partial x^{4}}\\right)_{i}(-\\Delta x)^{4} +O\\left(\\Delta x^{5}\\right)$$ $$\\frac{u_{i+1}-u_{i-1}}{2 h} = \\frac{u'_{i}(2h) +\\frac{1}{6} u_{i}^{'''}(2h^3)+O(h^{4})}{2h} \\qquad \\to \\ \\frac{du}{dx}=u'_{i}\\approx\\frac{u_{i+1}-u_{i-1}}{2 h} - \\frac{1}{6}u_{i}^{'''}h^{2}$$ $$\\frac{u_{i+1}-2 u_{i}+u_{i-1}}{h^{2}} = \\frac{\\frac{1}{2}u^{''}_{i}(2h^{2}) +\\frac{1}{24} u_{i}^{''''}(2h^{4})+O(h^{5})}{h^{2}}$$ $$\\to \\frac{d^{2} u}{d x^{2}}=u^{''}_{i} \\approx \\frac{u_{i+1}-2 u_{i}+u_{i-1}}{h^{2}}-\\frac{1}{12} u^{''''}_{i} h^{2}$$ $$u_{i+1, j}=u_{i, j}+\\left(\\frac{\\partial u}{\\partial x}\\right)_{i, j} \\Delta x+\\left(\\frac{\\partial^{2} u}{\\partial x^{2}}\\right)_{i, j} \\frac{(\\Delta x)^{2}}{2}+\\left(\\frac{\\partial^{3} u}{\\partial x^{3}}\\right)_{i, j} \\frac{(\\Delta x)^{3}}{6}+\\cdots$$ $$\\left(\\frac{\\partial u}{\\partial y}\\right)_{i+1, j}=\\left(\\frac{\\partial u}{\\partial y}\\right)_{i, j}+\\left(\\frac{\\partial^{2} u}{\\partial x \\partial y}\\right)_{i, j} \\Delta x+\\left(\\frac{\\partial^{3} u}{\\partial x^{2} \\partial y}\\right)_{i, j} \\frac{(\\Delta x)^{2}}{2}+\\left(\\frac{\\partial^{4} u}{\\partial x^{3} \\partial y}\\right)_{i, j} \\frac{(\\Delta", "doing these problems correctly. But it's less of an issue here because you attempted to justify your way around it. The justification behind the \"$\\pm$'s cancelling out into a negatives\" isn't sufficient. The subtlety is that you haven't convinced me between which \"$\\sqrt{\\frac{a^2}{b^2}}=\\frac{\\sqrt{a^2}}{\\sqrt{b^2}}=\\frac{\\pm a}{\\pm b}$\" or \"$\\sqrt{\\frac{a^2}{b^2}}=\\frac{\\sqrt{a^2}}{\\sqrt{b^2}}=\\frac{\\pm a}{\\mp b}$\" is the favourable one. Saying that \"observing that taking the negative one for the top one will give the desired answer\" is not convincing because it looks like you're assuming what you're trying to prove, A better justification here would be that $\\sqrt{\\frac{(\\sin\\theta-\\cos\\theta)^2}{(\\sin\\theta+\\cos\\theta)^2}} = \\left| \\frac{\\sin\\theta-\\cos\\theta}{\\sin\\theta+\\cos\\theta} \\right| = \\left| \\frac{\\tan\\theta-1}{\\tan\\theta+1} \\right| = -\\frac{\\tan\\theta-1}{\\tan\\theta+1}$, justifying the last equality here by the fact that $\\boxed{0\\leq\\theta\\leq\\frac\\pi4}$. We know that over this domain, $0 \\leq \\tan\\theta \\leq 1$. Hence on one hand, it is clear that $\\tan\\theta+1 > 0$. On the other hand, rearranging $\\tan\\theta \\leq 1$ gives $\\tan\\theta - 1\\leq 0$. Therefore $\\frac{\\tan\\theta-1}{\\tan\\theta+1} \\leq0$ (non-positive number divided by positive number), so $\\left|\\frac{\\tan\\theta-1}{\\tan\\theta+1}\\right| = -\\frac{\\tan\\theta-1}{\\tan\\theta+1}$. This now relies on the actual definition of the absolute value $|x| = \\begin{cases}x&x\\geq 0\\\\ -x&x<0\\end{cases}$ instead of intuition that cannot be justified. « Last Edit: July 27, 2019, 06:08:57 pm by RuiAce » #### mani.s_ • Posts: 17 • Respect: 0 ##### Re: 3U Maths Question Thread « Reply #4158 on: July 27, 2019, 07:08:34 pm »", "$z = |z| (\\cos \\theta + i\\sin \\theta)$ So too $e^{ix} = r(\\cos \\psi + i \\sin \\psi)$ where $r$ and $\\psi$ depend on $x$. Now (and applying the product rule) $\\frac{d}{dx}e^{ix} = ie^{ix} = \\frac{dr}{dx}(\\cos \\psi + i\\sin \\psi) + \\frac{d \\psi}{dx}r(-\\sin \\psi + i \\cos \\psi)$ So we equate the real and imaginary sides of both sides of this equality we have: $ie^{ix} = (\\cos \\psi \\frac{dr}{dx} - r \\sin \\psi \\frac{d \\psi}{dx}) + i(\\sin \\psi \\frac{dr}{dx} + r \\cos \\psi \\frac{d\\psi}{dx})$ Then, recalling $e^{ix} = r(\\cos \\psi + i \\sin \\psi)$, we have $ir (\\cos \\psi + i \\sin \\psi) = ir \\cos \\psi - r sin \\psi = ( \\cos \\psi \\frac{dr}{dx} - r \\sin \\psi \\frac{d\\psi}{dx}) + i( \\sin \\psi \\frac{dr}{dx} + r \\cos \\psi \\frac{d\\psi}{dx})$ By inspection we can see that $\\frac{dr}{dx} = 0$ and $\\frac{d\\psi}{dx} = 1$, giving us: $ie^{ix} = - r \\sin \\psi + ir \\cos \\psi$ and multiplying both sides by $i$ we have: $-e^{ix} = -r \\cos \\psi - i r \\sin \\psi$ Reversing the signs: $e^{ix} = r \\cos \\psi + i r \\sin \\psi$ But what of $r$ and $\\psi$? Well, we have $\\frac{dr}{dx} = 0$ and $\\frac{d \\psi}{dx} = 1$. So $r$ is constant with respect to $x$ while $\\psi$ varies as $x$. If", "like that? – weno Jan 23 at 11:40 • Which assumptions do you have in mind? I defined $c$ as $\\cos(\\arctan 2)$. And, no, you don't need to do it like this. If you find any other way, I'm fine with it. – José Carlos Santos Jan 23 at 11:43 Ok, so I will try to illustrate how I generally approach these by virtue of some examples. arccos(sinx) : you want $$sinx = cosy$$ for some $$y$$ . Obviously $$y = \\frac{\\pi}{2} - x$$ works so $$arccos(sinx) = \\frac{\\pi}{2} - x$$ (wherever this is properly defined, note!) cos(arcsinx) : write $$cos(arcsinx) = +-\\sqrt{1-sin(arcsinx)^2} = +-\\sqrt{1-x^2}$$ (the sign depends on your actual value for x and convention for arcsin) sin(arctanx)): use the fact that $$sinx = \\frac{2tan(\\frac{x}{2})}{1+tan(\\frac{x}{2})^2}$$ The rest of the combinations should be deducible from these. $$\\sin(\\arctan2)$$ is asking you \"the tangent of some angle is $$2$$, what is the sine ?\" Now, you can use $$4=\\tan^2\\theta=\\frac{\\sin^2\\theta}{\\cos^2\\theta}=\\frac{\\sin^2\\theta}{1-\\sin^2\\theta}.$$ Solving for $$\\sin^2\\theta$$, you get $$\\dfrac45$$. You might also know the relation $$\\sin^2\\theta=\\frac{\\tan^2\\theta}{\\tan^2\\theta+1}.$$ If you want to simplify $$\\sin(\\arccos x),$$ for example, try to use formulas to re-express the $$\\sin$$ with $$\\cos,$$ and then make use of $$\\cos(\\arccos x) = x.$$ Specifically, in this example, use $$\\sin(x)=\\sqrt{1-\\cos^2(x)}$$ to arrive at $$\\sin(\\arccos x) = \\sqrt{1-\\cos^2(\\arccos x)} = \\sqrt{1-x^2}.$$ (Skipping over issues of", "question. Let $\\alpha = \\frac{1+\\sqrt{5}}{2}$. Then $\\cos(\\theta) = \\frac{1+\\sqrt{5}}{4}$. This gives us $\\theta = \\pm \\frac{\\pi}{5} \\mod 2\\pi$. Therefore the answer to the problem is $2\\cos (400\\pi)=2$. • That was a wonderful argument. – infinitylord Dec 2 '16 at 18:01 Let $\\psi = \\frac{1+\\sqrt{5}}{2}$ $$x + \\frac{1}{x} = \\psi$$ $$x^2 + 1 = \\psi x$$ $$x^2 - \\psi x + 1 = 0$$ Use quadratic formula here to solve for $x$. Then plug that into the given expression. • Superior answer – Decaf-Math Dec 1 '16 at 18:34 • $x + \\frac{1}{x} = \\psi$ has no real solutions. – Anurag A Dec 1 '16 at 18:41 • @AnuragA Go for complex..? – Frank Dec 1 '16 at 18:53 • That might be very dificult to simplify. $(a+b\\sqrt{c})^{2000} + \\frac 1{(a+b\\sqrt{c})^{2000}}$ might not be an acceptable answer. (Then again maybe it will work out nice.) – fleablood Dec 1 '16 at 23:00 Nice try of yours. You can further continue...: $$x^{2000} + \\frac{1}{x^{2000}} = \\left(x^{1000} + \\frac{1}{x^{1000}}\\right)^2 - 2 \\\\ x^{1000} + \\frac{1}{x^{1000}} = \\left(x^{500} + \\frac{1}{x^{500}}\\right)^2 - 2 \\\\ x^{500} + \\frac{1}{x^{500}} = \\left(x^{250} + \\frac{1}{x^{250}}\\right)^2 - 2 \\\\ \\vdots \\\\ x^{n} + \\frac{1}{x^n} = \\left(x^{n/2} + \\frac{1}{x^{n/2}}\\right)^2 - 2 \\\\$$ In the case of $2000 = 2^4 \\cdot 5^3$. That is, when you are done with all", "1$$ there is a $\\psi$ such that $$\\cos(\\psi) = \\frac{A}{\\sqrt{A^2+B^2}} \\text{ and } \\sin(\\psi) = \\frac{B}{\\sqrt{A^2+B^2}}.$$ Thus we have $$\\cos(\\psi)\\sin(x) + \\sin(\\psi)\\cos(x) = \\frac{C}{\\sqrt{A^2+B^2}}$$ and using the sum formula we find: $$\\sin(x+\\psi) = \\frac{C}{\\sqrt{A^2+B^2}}.$$ In this case $A=3$ and $B=-1$ so $\\sqrt{A^2+B^2} = \\sqrt{10}$. Since $A > 0$ and $B < -1$, this means $\\psi$ is in the third quadrant. Thus $\\arctan(-1/3) = \\psi$. Finally this means $$\\sin(x + \\arctan(-1/3)) = 2/\\sqrt{10}.$$ Hint: $$3\\sin x - \\cos x = 2 \\implies 3\\tan x - 1 = 2\\sec x$$ $$9\\tan^2x - 6\\tan x + 1 = 4\\sec^2x = 4(1 + \\tan^2x)$$ $$5\\tan^2x - 6\\tan x - 3 = 0$$ Just need to be careful about any extraneous solutions If $a \\sin x + b \\cos x = c$, divide by $\\sqrt{a^2+b^2}$ to get $A \\sin x + B \\cos x = C$ where $A^2+B^2 = 1$. Choose $y$ so that $A = \\cos y$ and $B = \\sin y$ (i.e., $\\tan y = B/A$). Then $C = \\sin x \\cos y + \\cos x \\sin y = \\sin(x+y)$, so $x+y = \\sin^{-1}(C)$ or $x = \\sin^{-1}(C)-y$ . As is often the case, nothing original here. When you arrive at $$9\\sin^2x - 6\\sin x\\cos x + \\cos^2x = 4$$ just note that $4=4\\sin^2x+4\\cos^2x$ so you can rewrite your equation as", "# How to prove that $2 \\arctan\\sqrt{x} = \\arcsin \\frac{x-1}{x+1} + \\frac{\\pi}{2}$ I want to prove that$$2 \\arctan\\sqrt{x} = \\arcsin \\frac{x-1}{x+1} + \\frac{\\pi}{2}, x\\geq 0$$ I have started from the arcsin part and I tried to end to the arctan one but I failed. Can anyone help me solve it? - Define $\\displaystyle g(x)=2\\arctan(\\sqrt{x})-\\arcsin\\left(\\frac{x-1}{x+1}\\right)-\\frac{\\pi}{2}$. Differentiate to find that $g'(x)=0$. Since this was true on the connected set $[0,\\infty)$ you can conclude that $g$ is contant. Note then that $g(0)=0$ to finish. - +1. A nice way to look at it but unnecessarily complicated. – user17762 Dec 9 '11 at 0:33 Let $\\arctan{\\sqrt{x}} = \\theta$. Then we have $x = \\tan^2 (\\theta)$. Hence, $$\\frac{x-1}{x+1} = \\frac{\\tan^2 (\\theta)-1}{\\tan^2 (\\theta)+1} = \\sin^2(\\theta) - \\cos^2(\\theta) = - \\cos(2 \\theta)= \\sin \\left(2 \\theta - \\frac{\\pi}{2} \\right)$$ Hence, $$2 \\arctan{\\sqrt{x}} = 2 \\theta = \\arcsin \\left(\\frac{x-1}{x+1} \\right) + \\frac{\\pi}{2}$$ - Very nice.. One short comment: $\\frac{\\tan^2 (\\theta)-1}{\\tan^2 (\\theta)+1} = - \\cos(2 \\theta)$ is just the standard $t =\\tan(\\frac{x}{2})$ substitution :) – N. S. Dec 9 '11 at 0:41 One way would be to notice that both functions have the same derivative and then find out what the \"constant\" is by plugging in $x=0$. Here's another way. Look at $$\\frac\\pi2 - 2\\arctan\\sqrt{x} = 2\\left( \\frac\\pi4 - \\arctan\\sqrt{x} \\right) = 2\\left( \\arctan1-\\arctan\\sqrt{x} \\right).$$ Now remember the", "× # 1/sinx + 1/tanx=pi/2 Prove that 2x²+1=√5? Mar 8, 2018 #### Explanation: I hope the Question is : If, ${\\sin}^{-} 1 x + {\\tan}^{-} 1 x = \\frac{\\pi}{2}$, prove that, $2 {x}^{2} + 1 = 5$. Let ${\\tan}^{-} 1 x = \\theta \\therefore \\tan \\theta = x \\ldots \\ldots \\left(\\star 1\\right)$. Given that, ${\\sin}^{-} 1 x + {\\tan}^{-} 1 x = \\frac{\\pi}{2} \\therefore {\\sin}^{-} 1 x + \\theta = \\frac{\\pi}{2}$. $\\therefore {\\sin}^{-} 1 x = \\frac{\\pi}{2} - \\theta$. $\\therefore \\sin \\left(\\frac{\\pi}{2} - \\theta\\right) = x$. $\\therefore \\cos \\theta = x , \\mathmr{and} , \\sec \\theta = \\frac{1}{x} \\ldots \\ldots \\left(\\star 2\\right)$. $\\text{But, } {\\tan}^{2} \\theta = {\\sec}^{2} \\theta - 1$. $\\therefore {x}^{2} = \\frac{1}{x} ^ 2 - 1 , \\mathmr{and} , {x}^{4} + {x}^{2} - 1 = 0$. This is a quadr. eqn. in ${x}^{2}$. Solving it with the help of the quadr. formula, we get, ${x}^{2} = \\frac{- 1 \\pm \\sqrt{1 + 4}}{2} = \\frac{- 1 \\pm \\sqrt{5}}{2}$. Since, ${x}^{2} > 0 , {x}^{2} \\ne \\frac{- 1 - \\sqrt{5}}{2}$. $\\therefore {x}^{2} = \\frac{- 1 + \\sqrt{5}}{2}$, or what is the same as to say that, $2 {x}^{2} + 1 = \\sqrt{5}$.", "\\mathop {\\lim }\\limits_{\\theta \\to 0} \\frac{{\\sin \\theta }} {\\theta } \\cdot \\mathop {\\lim }\\limits_{\\theta\\to 0} \\frac{1} {{1 + \\dfrac{{\\tan \\theta }} {\\theta }}}\\\\ &= \\frac{1} {2}.\\end{aligned} This is equal to THP's work. While I was posting he beat me to it. (Taylor series solution will make no sense to Coco87.) 7. Originally Posted by Coco87 Hey, This is a problem in my Calculus book, but I'm pretty sure it's a trig issue. The problem is: $\\lim_{\\theta\\to0}\\frac{\\sin{\\theta}}{\\theta+\\tan{\\ theta}}$ I know the answer is: $\\frac{1}{2}$ I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer. I get as far as: $\\lim_{\\theta\\to0}\\frac{\\sin{\\theta}\\cos{\\theta}}{\\ theta\\cos{\\theta}+\\sin{\\theta}}$ Thanks! Or you could just use L'Hopital's rule. (Ducks the flying rotten fruit.) $\\lim_{\\theta \\to 0} \\frac{sin(\\theta)}{\\theta + tan(\\theta)}$ $= \\lim_{ \\theta \\to 0} \\frac{cos(\\theta)}{1 + sec^2(\\theta)}$ $= \\frac{1}{1 + 1} = \\frac{1}{2}$ -Dan 8. Originally Posted by Krizalid \\begin{aligned}\\mathop {\\lim }\\limits_{\\theta \\to 0} \\frac{{\\sin \\theta }} {{\\theta + \\tan \\theta }} &= \\mathop {\\lim }\\limits_{\\theta \\to 0} \\left\\{ {\\frac{{\\sin \\theta }} {\\theta }\\cdot \\frac{\\theta } {{\\theta +\\tan\\theta }}} \\right\\}\\\\ &= \\mathop {\\lim }\\limits_{\\theta \\to 0} \\frac{{\\sin \\theta }} {\\theta } \\cdot \\mathop {\\lim }\\limits_{\\theta\\to 0} \\frac{1} {{1 + \\dfrac{{\\tan \\theta }} {\\theta }}}\\\\ &= \\frac{1} {2}.\\end{aligned} This is equal to THP's work. While I was posting he", "\\int_{\\Omega}\\sqrt{d\\theta^{2} + \\sin^{2}\\theta d\\phi^{2}}$ AlphaNumeric, This is also a solution In differential form; if you integrate the equation below you will get the equation above: This is conditional: $d\\Omega^2 = \\frac{dA_{{\\theta}{\\phi}}}{r^2} = (d\\theta^{2} + \\sin(\\theta_{0})^{2} d\\phi^{2}) = {3}(\\frac{dg_{{\\theta}{\\phi}}}{r^3})$.$-> radians^2$ This is general: $d\\Omega^2 = \\frac{dA_{{\\theta}{\\phi}}}{r^2} = (d\\theta^{2} + \\sin(\\theta)^{2} d\\phi^{2}) = {3}(\\frac{dg_{{\\theta}{\\phi}}}{r^3})$.$-> radians^2$ Let's review the conditional Let $d\\Omega^2 = 0$, meaning that $\\Omega^2 = constant$ $d\\theta^{2} = - \\sin(\\theta_{0})^{2} d\\phi^{2}$.$-> radians^2$ which becomes $\\frac{d\\theta^{2}}{d\\phi^{2}} = - \\sin(\\theta_{0})^{2}$.$-> unit less$ or $\\frac{ds}{dr} = \\sqrt{1 + \\frac{d\\theta^{2}}{d\\phi^{2}}} = \\cos(\\theta_{0})$.$-> unit less$ $ds = (\\sqrt{1 + \\frac{d\\theta^{2}}{d\\phi^{2}}}){dr}$.$-> m$ $s(L) = \\int_{L}ds = \\int_{L}(\\sqrt{1 + \\frac{d\\theta^{2}}{d\\phi^{2}}}){dr}$.$-> m$ See link for reference: Geodesic - Arc Length Last edited: Apr 16, 2011", "this. We get $-\\arcsin(\\cos 2\\theta)$. Now we have to be careful. When $\\theta$ is small and positive, $\\arcsin(\\cos 2\\theta)$ is simply $\\pi/2-2\\theta$. But when $\\theta$ is small negative, then $\\arcsin(\\cos 2\\theta)$ is not equal to $\\pi/2-2\\theta$, it is $\\pi/2+ 2\\theta$. In all cases when $\\theta$ is not far from $0$, we have $$\\arcsin(\\cos 2\\theta)=\\pi/2-2|\\theta|,$$ and therefore $$\\arcsin\\left(\\frac{x^2-1}{x^2+1}\\right)-\\arcsin(-1)=-(\\pi/2-2|\\theta|)+ \\pi/2=2|\\theta|.$$ So it all comes down to finding $$\\lim_{\\theta\\to 0^-} \\frac{2|\\theta|}{\\tan\\theta}.$$ This is easy. if we know about the behaviour of $\\frac{\\sin x}{x}$ near $0$. Remark: The $2$ in the limits comes from a double-angle identity. The discontinuous behaviour comes from the kinky behaviour of $\\arcsin(\\cos\\phi)$ when $\\phi$ is close to $0$. There is less to this solution than meets the eye. The expression $\\frac{1-x^2}{1+x^2}$ is easy to recognize from the standard rational parametrization of the circle. It also comes up in calculus when we do the so-called Weierstrass substitution. (If I were sentimental I would have said let $x=\\tan(\\theta/2)$.) - That's a lovely answer! it's a bit long, but it does the job. thanks for you efforts! – RB14 May 27 '12 at 22:13 Since the sine function is odd and $\\cos(\\pi/2-u)=\\sin(u)$, the numerator is also $$n(x)=\\arccos\\left(\\frac{1-x^2}{1+x^2}\\right).$$ When $x\\to0$, $\\dfrac{1-x^2}{1+x^2}=1-2x^2+o(x^2)$. Since $\\cos(u)=1-\\frac12u^2+o(u^2)$ when $u\\to0$, $\\arccos(1-v)\\sim\\sqrt{2v}$ when $v\\to0^+$. Applying this to $v=2x^2$ yields $n(x)\\sim\\sqrt{4x^2}=2|x|$. Finally, $\\dfrac{n(x)}x\\to2$ when $x\\to0^+$ and $\\dfrac{n(x)}x\\to-2$ when", "- \\sqrt{2} \\sqrt[4]{3}\\; u + \\sqrt{3}\\right)$$ may be useful. I tried this one with trigonometric substitutions. The $4$ possibilities are $x=\\sin\\theta$, $x=\\cos\\theta$, $x=\\tanh\\theta$, and $x=\\text{sech}\\,\\theta$. $x=\\sin\\theta$ and $x=\\tanh\\theta$ force you to complete the square at one point, and $x=\\text{sech}\\,\\theta$ has a hole in the middle of its domain, so that leaves $x=\\cos\\theta$. Then $$\\int\\frac x{(1-x^3)\\sqrt{1-x^2}}dx=\\int\\frac{\\cos\\theta}{\\cos^3\\theta-1}d\\theta$$ Expanding by partial fractions, $$\\frac x{x^3-1}=\\frac x{\\prod_{k=0}^2(x-\\omega_k)}=\\sum_{k=0}^2\\frac{A_k}{x-\\omega_k}$$ Where $\\omega_k=e^{i\\theta_k}$, $\\theta_k=\\frac{2\\pi ik}3$. By L'Hopital's rule, $$\\lim_{x\\rightarrow\\omega_k}\\frac{(x-\\omega_k)x}{x^3-1}=\\frac{\\omega_k}{3\\omega_k^2}=\\frac{\\omega_k^2}3=A_k$$ So we can substitute $z=\\tan\\left(\\frac{\\theta}2\\right)$ to get \\begin{align}\\int\\frac x{(1-x^3)\\sqrt{1-x^2}}dx&=\\sum_{k=0}^2\\frac{\\omega_k^2}3\\int\\frac{d\\theta}{\\cos\\theta-\\omega_k}\\\\ &=\\sum_{k=0}^2\\frac{\\omega_k^2}3\\int\\frac{2dz}{1-z^2-\\omega_k(1+z^2)}\\\\ &=\\sum_{k=0}^2\\frac{2\\omega_k^2}3\\int\\frac{dz}{1-\\omega_k-(1+\\omega_k)z^2}\\end{align} At this point it can be observed that $z=\\sqrt{\\frac{1-x}{1+x}}$, just like Robert Israel recommended. Here is where completing the square for $x=\\sin\\theta$ would come in. The $k=0$ term in the sum is $$-\\frac13\\int\\frac{dz}{z^2}=\\frac1{3z}+C_1$$ and the $k=2$ term = is complex conjugate of the $k=1$ term so we only have to integrate the latter term. Some arithmetic we will need is $1-\\omega_1=\\sqrt3e^{\\frac{-i\\pi}6}$, $1+\\omega_1=e^{\\frac{i\\pi}3}$, and $\\omega_1^2=e^{\\frac{-2\\pi i}3}$. Now we can sustitute $\\sqrt{1+\\omega_1}z=\\sqrt{1-\\omega_1}u$ transforming to \\begin{align}\\frac{2\\omega_1^2}3\\int\\frac{dz}{1-\\omega_1-(1+\\omega_1)z^2}&=\\frac{2\\omega_1^2}3\\int\\frac{\\frac{\\sqrt{1-\\omega_1}}{\\sqrt{1+\\omega_1}}du}{(1-\\omega_1)(1-u^2)}\\\\ &=\\frac23\\frac{e^{\\frac{-2\\pi i}3}}{3^{1/4}e^{\\frac{-\\pi i}{12}}e^{\\frac{\\pi i}6}}\\int\\frac{du}{1-u^2}\\\\ &=\\frac2{3^{5/4}}e^{\\frac{-3\\pi i}4}\\frac12\\ln\\left(\\frac{1+u}{1-u}\\right)+C_2\\\\ &=\\frac{(-1-i)}{2^{1/2}3^{5/4}}\\ln\\left(\\frac{\\frac{3^{1/4}}{2^{1/2}}(1-i)+z}{\\frac{3^{1/4}}{2^{1/2}}(1-i)-z}\\right)+C_2\\\\ &=\\frac{(-1-i)}{2^{1/2}3^{5/4}}\\left\\{\\frac12\\ln\\left(\\left(z+\\frac{3^{1/4}}{2^{1/2}}\\right)^2+\\frac{\\sqrt3}2\\right)-\\\\ \\frac12\\ln\\left(\\left(z-\\frac{3^{1/4}}{2^{1/2}}\\right)^2+\\frac{\\sqrt3}2\\right)+\\\\ i\\,\\text{atan2}\\left(-\\frac{3^{1/4}}{2^{1/2}},\\frac{3^{1/4}}{2^{1/2}}+z\\right)\\\\ -i\\,\\text{atan2}\\left(-\\frac{3^{1/4}}{2^{1/2}},\\frac{3^{1/4}}{2^{1/2}}-z\\right)\\right\\}+C_2\\end{align} We rotate to $$\\text{atan2}\\left(-\\frac{3^{1/4}}{2^{1/2}},\\frac{3^{1/4}}{2^{1/2}}-z\\right)=-\\frac{\\pi}2+\\text{atan2}\\left(\\frac{3^{1/4}}{2^{1/2}}-z,\\frac{3^{1/4}}{2^{1/2}}\\right)=-\\frac{\\pi}2+\\tan^{-1}\\left(\\frac{\\frac{3^{1/4}}{2^{1/2}}-z}{\\frac{3^{1/4}}{2^{1/2}}}\\right)$$ $$\\text{atan2}\\left(-\\frac{3^{1/4}}{2^{1/2}},\\frac{3^{1/4}}{2^{1/2}}+z\\right)=-\\frac{\\pi}2+\\text{atan2}\\left(\\frac{3^{1/4}}{2^{1/2}}+z,\\frac{3^{1/4}}{2^{1/2}}\\right)=-\\frac{\\pi}2+\\tan^{-1}\\left(\\frac{\\frac{3^{1/4}}{2^{1/2}}+z}{\\frac{3^{1/4}}{2^{1/2}}}\\right)$$ Since we need only twice the real part of the $k=1$ term the overall integral is \\begin{align}\\int\\frac x{(1-x^3)\\sqrt{1-x^2}}dx&=\\frac1{3z}-\\\\ &\\frac1{2^{1/2}3^{5/4}}\\left\\{\\ln\\left(\\left(z+\\frac{3^{1/4}}{2^{1/2}}\\right)^2+\\frac{\\sqrt3}2\\right)-\\ln\\left(\\left(z-\\frac{3^{1/4}}{2^{1/2}}\\right)^2+\\frac{\\sqrt3}2\\right)\\right\\}+\\\\ &\\frac{2^{1/2}}{3^{5/4}}\\left\\{\\tan^{-1}\\left(\\frac{\\frac{3^{1/4}}{2^{1/2}}+z}{\\frac{3^{1/4}}{2^{1/2}}}\\right)-\\tan^{-1}\\left(\\frac{\\frac{3^{1/4}}{2^{1/2}}-z}{\\frac{3^{1/4}}{2^{1/2}}}\\right)\\right\\}+C\\end{align} Where again $z=\\sqrt{\\frac{1-x}{1+x}}$. Verified by numerical quadrature. There seemed to be a bit of an advantage in getting to Robert Israel's substitution in two steps like this because" ]

[ "\\arccos(\\sin(x)) = \\arccos(\\cos(\\pi/2 - x)) = -x + \\pi/2 $$The sine funcion on [\\pi/2, \\pi/2 + \\pi] is$$ \\sin(x) = -\\sin(x-\\pi) = \\sin(\\pi - x) $$and maps to [-1,1]. We get$$ \\arccos(\\sin(x)) = \\arccos(\\cos(\\pi/2 - ... 2 Just for some new ideas! I would reccomend a completely different method. This method uses the Gudermannian $\\text{gd}$ function. So you would substitute $x=\\text{gd}(a);\\text{d}x=\\text{sech}\\space a\\text{d}a$ That transforms the integral into: $$\\int \\frac{\\tanh a}{\\tanh a+\\text{sech}\\space a}(\\text{sech}\\space a)\\mathrm da$$ Through some hyperbolic trig ... 2 In $1^{st}$ quadrant both $\\sin \\theta$ and $\\cos \\theta$ are positive. In $2^{nd}$ quadrant from $90^{\\circ}$ to $135^{\\circ}$ degrees $\\sin \\theta > \\cos \\theta$ and $\\sin\\theta$ is positive while $\\cos\\theta$ is negative. In $3^{rd}$ quadrant both are negative. In $4^{th}$ quadrant from $315^{\\circ}$ to $360^{\\circ}$ $\\cos \\theta > \\sin \\theta$ ... 2 You want to read the alternate form assuming $x$ is real. (The one about $x>0$ is a red herring, just written there in case that was what a user wanted to know about.) If $x$ is not real but rather complex then the square root is a more complicated animal (see https://en.wikipedia.org/wiki/Square_root for more info). 2 This doesn't hold. Try $A=B=C=\\pi/3$. 2 HINT: $$\\int\\frac{1}{\\left(t+\\cos(a)\\right)^2+\\sin^2(a)}\\space\\text{d}t=$$ Substitute $u=\\cos(a)+t$ and $\\text{d}u=\\text{d}t$: $$\\int\\frac{1}{\\sin^2(a)+u^2}\\space\\text{d}u=$$ $$\\int\\frac{\\csc^2(a)}{u^2\\csc^2(a)+1}\\space\\text{d}u=$$ $$\\csc^2(a)\\int\\frac{1}{u^2\\csc^2(a)+1}\\space\\text{d}u=$$ Substitute $s=u\\csc(a)$ and ... 2 $$\\frac{1+\\tan^2 (x/2)}{1-\\tan^2 (x/2)}=\\frac{\\cos^2(x/2)+\\sin^2 (x/2)}{\\cos^2(x/2)-\\sin^2 (x/2)}=\\frac{1}{\\cos^2(x/2)-\\sin^2 (x/2)}=\\frac{1}{\\cos x}=\\sec x$$ 2 Thanks", "# Question on inverse tangent. 1. Feb 17, 2013 ### yungman $$\\Psi=\\tan^{-1}\\left(\\frac{\\cos\\omega t}{\\cos(\\omega t+\\delta)}\\right)$$ I want to find out whether $\\Psi$ increase or decrease with time t, if $\\delta$ is positive and if $\\delta$ is negative. $$\\Psi=\\tan^{-1}\\left(\\frac{\\cos\\omega t}{\\cos(\\cos\\omega t \\cos \\delta+\\sin\\omega t \\sin\\delta)}\\right)\\;=\\;\\tan^{-1}\\left(\\frac {1}{(\\cos\\delta + \\tan(\\omega t)\\sin \\delta)}\\right)$$ $$\\Rightarrow\\;\\Psi=\\tan^{-1}\\left(\\frac{1}{K_1+K_2\\tan \\omega t}\\right)$$ where $K_1$ is a constant and $K_2$ is a constant following the sign of $\\delta$. This is as far as I know how to go, what can I do to simplify it. I want to find out whether $\\Psi$ increase or decrease with time if $\\delta$ is positive and if $\\delta$ is negative. 2. Feb 17, 2013 ### Simon Bridge You'll have to use the known limits to the problem and the properties of the tangent function. I would be thinking: if $\\Psi$ is increasing, what happens to $\\tan\\Psi$ ? What are the allowed values for $\\omega t$ ? 3. Feb 17, 2013 ### yungman Thanks for the reply. I think I come up with a way and see whether you agree or not: Let $\\theta=\\frac {\\pi}{2}-\\Psi$. So if $\\tan \\Psi=\\frac y x\\;\\Rightarrow \\;\\tan \\theta = \\frac x y$ So $\\theta \\;=\\;\\tan ^{-1} \\left(\\frac{\\cos(\\omega t +\\delta}{\\cos \\omega t}\\right)$ This will be easy to solve and find out the direction. Then we know if θ increase with t,", "\\arccos x$ $\\arccsc \\frac{1}{x} = \\arcsin x$ If you only have a fragment of a sine table: $\\arccos x = \\arcsin \\sqrt{1-x^2},\\text{ if }0 \\leq x \\leq 1$ $\\arctan x = \\arcsin \\frac{x}{\\sqrt{x^2+1}}$ Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real). From the half-angle formula $\\tan \\frac{\\theta}{2} = \\frac{\\sin \\theta}{1+\\cos \\theta}$, we get: $\\arcsin x = 2 \\arctan \\frac{x}{1+\\sqrt{1-x^2}}$ $\\arccos x = 2 \\arctan \\frac{\\sqrt{1-x^2}}{1+x},\\text{ if }-1 < x \\leq +1$ $\\arctan x = 2 \\arctan \\frac{x}{1+\\sqrt{1+x^2}}$ ## Relationships between trigonometric functions and inverse trigonometric functions $\\sin (\\arccos(x)) = \\cos(\\arcsin(x)) = \\sqrt{1-x^2}$ $\\sin (\\arctan(x)) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan(x)) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arcsin(x)) = \\frac{x}{\\sqrt{1-x^2}}$ $\\tan (\\arccos(x)) = \\frac{\\sqrt{1-x^2}}{x}$ ## General solutions Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent", "# $\\sin(2\\arcsin(\\frac{1}{3}))=?$ A broad one I have written it as $$\\sin(2\\arcsin(\\frac{1}{3}))=2\\sin(\\arcsin(\\frac{1}{3}))\\cos(\\arcsin(\\frac{1}{3})).$$Then, solved for $$\\cos(\\arcsin(\\frac{1}{3}))=\\frac{\\sqrt{10}}{3}$$ $$\\arcsin(\\frac{1}{3})=\\theta$$$$\\sin\\theta=\\frac{1}{3}\\Rightarrow\\cos\\theta=\\frac{\\sqrt{10}}{3}.$$But noticed something weird (1) it didn't satisfy the Pythagorean theorem as $\\sqrt{10}\\approx3.1622,$ which means that the adjacent side to the angle $\\theta$ is greater than the hypotenuse, (which is just 3) why this method (using right triangle to determine $\\cos(\\theta)$) didn't work here although the answer ($\\frac{\\sqrt{10}}{3}$) satisfied the range of the cosine function, which is $R(\\cos\\theta)=\\Bbb{R}$? After it (2) used this formula$$\\arcsin(x)=\\arccos(\\sqrt{1-x^2});$$$$\\arcsin(\\frac{1}{3})=\\arccos(\\sqrt{1-\\frac{1}{9}})=\\arccos(\\frac{\\sqrt{8}}{3});$$then, checked if it satisfies or not $D(\\arccos(\\theta))=[-1;1],$ satisfied, since $\\frac{\\sqrt{8}}{3}\\approx0.942.$ Then proceeded on my problem $$2\\sin(\\arcsin(\\frac{1}{3}))\\cos(\\arcsin(\\frac{1}{3}))=\\frac{2}{3}\\cdot\\frac{\\sqrt{8}}{3}=\\frac{2\\sqrt{8}}{9}.$$ Anyway, it didn't turn out to be the right answer. Where is my mistake? • Why do you think this is not the correct answer? – kingW3 Jun 23 '17 at 21:20 • – jim Jun 23 '17 at 21:20 • @kingW3 The person found $\\cos \\theta = \\sqrt{10}/3 > 1$ – jim Jun 23 '17 at 21:22 • Because there wasn't such ōne. – user438365 Jun 23 '17 at 21:22 • @jim Yeah I agree on that one but,he said bottom that $\\frac{2\\sqrt{8}}9$ is not an answer. – kingW3 Jun 23 '17 at 21:22 Your mistake is that $\\cos\\theta=\\dfrac{2\\sqrt{2}}{3}$, not $\\dfrac{\\sqrt{10}}{3}$. $$\\sin2\\arcsin\\frac{1}{3}=2\\cdot\\frac{1}{3}\\cdot\\sqrt{1-\\frac{1}{9}}=\\frac{4}{9}\\sqrt2$$", "\\theta = \\frac{x}{\\sqrt{x^2+1}} && \\csc (\\tan^{-1}x) = \\csc \\theta = \\frac{\\sqrt{x^2+1}}{x}\\\\\\cos (\\tan^{-1}x) & = \\cos \\theta = \\frac{1}{\\sqrt{x^2+1}} && \\sec (\\tan^{-1}x) = \\sec \\theta = \\sqrt{x^2+1}\\\\\\tan (\\tan^{-1}x) & = \\tan \\theta = x && \\cot (\\tan^{-1}x) = \\cot \\theta = \\frac{1}{x}$ #### Example A Find $\\sin (\\tan^{-1} 3x)$ . Solution: Instead of using $x$ in the ratios above, use $3x$ . $\\sin (\\tan^{-1} 3x) = \\sin \\theta = \\frac{3x}{\\sqrt{(3x)^2 + 1}} = \\frac{3x}{\\sqrt{9x^2+1}}$ #### Example B Find $\\sec^2 (\\tan^{-1} x)$ . Solution: This problem might be better written as $[\\sec (\\tan^{-1} x)]^2$ . Therefore, all you need to do is square the ratio above. $[\\sec (\\tan^{-1}x)]^2 = \\left ( \\sqrt{x^2+1} \\right )^2 = x^2 + 1$ You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section. #### Example C Find $\\csc^3 (\\tan^{-1} 4x)$ . Solution: This example is similar to both of the examples above. First, use $4x$ instead of $x$ in the ratios above. Second, the $\\csc^3$ is the same as taking the $\\csc$ function and cubing it. $[\\csc (\\tan^{-1}4x)]^3 = \\left ( \\frac{\\sqrt{(4x)^2+1}}{x} \\right)^3 = \\frac{(16x^2+1)^{3/2}}{x^{3}}$ ### Vocabulary Trigonometric Function: A trigonometric function is a function describing a relationship", "derivations are longer. $\\theta$ $\\sin \\theta$ $\\cos \\theta$ $\\tan \\theta$ Diagram $\\arcsin x$ $\\sin (\\arcsin x) = x$ $\\cos (\\arcsin x) = \\sqrt{1-x^2}$ $\\tan (\\arcsin x) = \\frac{x}{\\sqrt{1-x^2}}$ $\\arccos x$ $\\sin (\\arccos x) = \\sqrt{1-x^2}$ $\\cos (\\arccos x) = x$ $\\tan (\\arccos x) = \\frac{\\sqrt{1-x^2}}{x}$ $\\arctan x$ $\\sin (\\arctan x) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan x) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arctan x) = x$ $\\arccot x$ $\\sin (\\arccot x) = \\frac{1}{\\sqrt{1+x^2}}$ $\\cos (\\arccot x) = \\frac{x}{\\sqrt{1+x^2}}$ $\\tan (\\arccot x) = \\frac{1}{x}$ $\\arcsec x$ $\\sin (\\arcsec x) = \\frac{\\sqrt{x^2-1}}{x}$ $\\cos (\\arcsec x) = \\frac{1}{x}$ $\\tan (\\arcsec x) = \\sqrt{x^2-1}$ $\\arccsc x$ $\\sin (\\arccsc x) = \\frac{1}{x}$ $\\cos (\\arccsc x) = \\frac{\\sqrt{x^2-1}}{x}$ $\\tan (\\arccsc x) = \\frac{1}{\\sqrt{x^2-1}}$ ### Relationships among the inverse trigonometric functions The usual principal values of the arcsin(x) (red) and arccos(x) (blue) functions graphed on the cartesian plane. The usual principal values of the arctan(x) and arccot(x) functions graphed on the cartesian plane. Principal values of the arcsec(x) and arccsc(x) functions graphed on the cartesian plane. Complementary angles: \\begin{align} \\arccos x &= \\frac{\\pi}{2} - \\arcsin x \\\\[0.5em] \\arccot x &= \\frac{\\pi}{2} - \\arctan x \\\\[0.5em] \\arccsc x &= \\frac{\\pi}{2} - \\arcsec x \\end{align} Negative arguments: \\begin{align} \\arcsin (-x) &= -\\arcsin x \\\\ \\arccos (-x) &= \\pi -\\arccos x \\\\ \\arctan (-x) &= -\\arctan x \\\\ \\arccot (-x) &= \\pi", "1$$ there is a $\\psi$ such that $$\\cos(\\psi) = \\frac{A}{\\sqrt{A^2+B^2}} \\text{ and } \\sin(\\psi) = \\frac{B}{\\sqrt{A^2+B^2}}.$$ Thus we have $$\\cos(\\psi)\\sin(x) + \\sin(\\psi)\\cos(x) = \\frac{C}{\\sqrt{A^2+B^2}}$$ and using the sum formula we find: $$\\sin(x+\\psi) = \\frac{C}{\\sqrt{A^2+B^2}}.$$ In this case $A=3$ and $B=-1$ so $\\sqrt{A^2+B^2} = \\sqrt{10}$. Since $A > 0$ and $B < -1$, this means $\\psi$ is in the third quadrant. Thus $\\arctan(-1/3) = \\psi$. Finally this means $$\\sin(x + \\arctan(-1/3)) = 2/\\sqrt{10}.$$ Hint: $$3\\sin x - \\cos x = 2 \\implies 3\\tan x - 1 = 2\\sec x$$ $$9\\tan^2x - 6\\tan x + 1 = 4\\sec^2x = 4(1 + \\tan^2x)$$ $$5\\tan^2x - 6\\tan x - 3 = 0$$ Just need to be careful about any extraneous solutions If $a \\sin x + b \\cos x = c$, divide by $\\sqrt{a^2+b^2}$ to get $A \\sin x + B \\cos x = C$ where $A^2+B^2 = 1$. Choose $y$ so that $A = \\cos y$ and $B = \\sin y$ (i.e., $\\tan y = B/A$). Then $C = \\sin x \\cos y + \\cos x \\sin y = \\sin(x+y)$, so $x+y = \\sin^{-1}(C)$ or $x = \\sin^{-1}(C)-y$ . As is often the case, nothing original here. When you arrive at $$9\\sin^2x - 6\\sin x\\cos x + \\cos^2x = 4$$ just note that $4=4\\sin^2x+4\\cos^2x$ so you can rewrite your equation as", "here we continued to try to understand the differences in reasoning and answers. Given that $\\cos(\\theta)=-x$ can be simplified to $\\theta=\\arccos(x)+\\pi$, then it can also be simplified into $\\arccos(-x)=\\theta$ which would suggest that $$\\tan(\\arccos(-x)) = \\tan(\\arccos(x)+\\pi)$$ but putting both functions in Wolfram Alpha did not result in the same graph. So my question is why is this case? And what errors did I make in my calculations that would result in something happening like this? Wolfram Alpha: tan(arccos(x)+π) tan(arccos(-x)) Sorry if this formatting is terrible, I did this late at night. I might fix it later. • Since $\\theta$ lies in $[\\pi,2\\pi]$, $\\sin\\theta$ is always negative. So $\\tan\\theta=\\dfrac{\\sqrt{1-x^2}}{x}$ – AnotherJohnDoe Sep 19 '18 at 4:22 • Also, $\\cos\\theta=-x\\nRightarrow\\theta=cos^{-1}(-x)$, because $\\theta\\ge\\pi$ – AnotherJohnDoe Sep 19 '18 at 4:25 The range of $\\arccos x$ is always in $[0,\\pi]$ so $\\tan \\arccos x = \\frac {\\sqrt{1-x^2}}{x}$ however if you say $\\cos \\theta = x$ there are two values of $\\theta$ in $[0,2\\pi)$ that $\\theta$ could take. and $\\tan \\theta = \\pm \\frac{\\sqrt{1-x^2}}{x}$ depending on which value of $\\theta$ you choose. You need to keep track of what quadrants you might be in. This seems to be the source of your problem. The you could us the identity $\\tan x+\\pi = \\tan x$ $\\tan(\\arccos x+ \\pi) = \\tan (\\arccos x)$ $\\tan(\\arccos x+", "we have $$\\theta = \\arcsin(x)$$. Taking the sine of both sides yields $$\\sin(\\theta) = x$$. We use this equation to set up the triangle to the right, $$a = x$$ and $$c = 1$$. Now we use the Pythagorean Theorem to solve for $$b$$. $$\\begin{array}{rcl} x^2 + b^2 & = & 1 \\\\ b^2 & = & 1 - x^2 \\\\ b & = & \\pm \\sqrt{1-x^2} \\end{array}$$ We choose the positive square root. Now we can write $$\\cos(\\theta) = \\sqrt{1-x^2}$$. Final Answer $$\\cos( \\arcsin(x))$$ $$= \\sqrt{1-x^2}$$ Log in to rate this practice problem and to see it's current rating. $$\\sec( \\arctan(x/3))$$ Problem Statement Evaluate (simplify) $$\\sec( \\arctan(x/3))$$ without using a calculator and give your answer in exact, simplified form. Final Answer $$\\sec( \\arctan(x/3))$$ $$\\displaystyle{ = \\frac{\\sqrt{x^2+9}}{3}}$$ Problem Statement Evaluate (simplify) $$\\sec( \\arctan(x/3))$$ without using a calculator and give your answer in exact, simplified form. Solution We start on the inside by letting $$\\theta = \\arctan(x/3)$$. Take the tangent of both sides to get $$\\tan(\\theta) = x/3$$. We use this equation to set up the triangle to the right, $$a = x, ~~ b = 3$$. Using the Pythagorean Theorem, we can find $$c$$, i.e. $$c^2 = x^2 + 3^2 ~~ \\to ~~ c = \\pm \\sqrt{x^2 + 9}$$. We choose the positive square root. Now we have", "I think, assuming that $$\\frac{-1}{\\phi} = \\frac{1-\\sqrt{5}}{2}$$ Which I seem to remember a similar property about phi... 12. Sep 13, 2010 Helios note that tan( x ) = cos( x ) = 1 / sqr( phi ) so the solution can be made in terms of arctan or arccos. 13. Sep 15, 2010 willem2 Since tan(x) increases monotonically from 0 to infinity in the domain (0, pi/1) and cos (x) decreases from 1 to 0 in the same domain, there should be a solution x between 0 and pi/2 and 0< sin(x) < 1. There's a solution in the second quadrant as well. cos(x) and tan(x) have different signs in the 3rd and fourth quadrant, so a solution where sin(x) < 0 is not possible Once you get all the signs right in the quadratic formula, you'll get $$sin(x) = \\frac {1} {\\phi} = \\phi - 1$$ from one of the solutions. The other solution has sin(x) >1 so that isn't a valid solution. }", "like that? – weno Jan 23 at 11:40 • Which assumptions do you have in mind? I defined $c$ as $\\cos(\\arctan 2)$. And, no, you don't need to do it like this. If you find any other way, I'm fine with it. – José Carlos Santos Jan 23 at 11:43 Ok, so I will try to illustrate how I generally approach these by virtue of some examples. arccos(sinx) : you want $$sinx = cosy$$ for some $$y$$ . Obviously $$y = \\frac{\\pi}{2} - x$$ works so $$arccos(sinx) = \\frac{\\pi}{2} - x$$ (wherever this is properly defined, note!) cos(arcsinx) : write $$cos(arcsinx) = +-\\sqrt{1-sin(arcsinx)^2} = +-\\sqrt{1-x^2}$$ (the sign depends on your actual value for x and convention for arcsin) sin(arctanx)): use the fact that $$sinx = \\frac{2tan(\\frac{x}{2})}{1+tan(\\frac{x}{2})^2}$$ The rest of the combinations should be deducible from these. $$\\sin(\\arctan2)$$ is asking you \"the tangent of some angle is $$2$$, what is the sine ?\" Now, you can use $$4=\\tan^2\\theta=\\frac{\\sin^2\\theta}{\\cos^2\\theta}=\\frac{\\sin^2\\theta}{1-\\sin^2\\theta}.$$ Solving for $$\\sin^2\\theta$$, you get $$\\dfrac45$$. You might also know the relation $$\\sin^2\\theta=\\frac{\\tan^2\\theta}{\\tan^2\\theta+1}.$$ If you want to simplify $$\\sin(\\arccos x),$$ for example, try to use formulas to re-express the $$\\sin$$ with $$\\cos,$$ and then make use of $$\\cos(\\arccos x) = x.$$ Specifically, in this example, use $$\\sin(x)=\\sqrt{1-\\cos^2(x)}$$ to arrive at $$\\sin(\\arccos x) = \\sqrt{1-\\cos^2(\\arccos x)} = \\sqrt{1-x^2}.$$ (Skipping over issues of", "function is positive in quadrants I and III. So, the hypotenuse is equal to square root of (5^2 + 2^2) = square root of 29. Then sin theta is equal to + 2/(sqrt of 29) and - 2/(sqrt of 29), because for sin theta, positive in quadrant I and negative in quadrant III. 6. Hello, deej813! if $\\tan \\theta = \\tfrac{2}{5}\\;\\text{ for }0< \\theta <360^o$, find the values of $\\sin \\theta$ Since $\\tan\\theta$ is positive, $\\theta$ is in Quadrant 1 or 3. We have: . $\\tan\\theta \\:=\\:\\frac{2}{5} \\:=\\:\\frac{opp}{adj}$ Assume $\\theta$ is an acute angle. Then $\\theta$ is in a right triangle with: . $opp = 2,\\;adj = 5$ Pythagorus says: . $hyp \\:=\\:\\sqrt{2^2 + 5^2} \\:=\\:\\sqrt{29}$ In Quadrant 1, we have: . $\\begin{Bmatrix}opp &=& 2 \\\\ adj &=& 5 \\\\ hyp &=& \\sqrt{29}\\end{Bmatrix}$ . . Hence: . $\\sin\\theta \\:=\\:\\frac{opp}{adj} \\;=\\;\\frac{2}{\\sqrt{29}}$ In Quadrant 3, we have: . $\\begin{Bmatrix}opp &=& \\text{-}2 \\\\ adj &=& \\text{-}5 \\\\ hyp &=& \\sqrt{29} \\end{Bmatrix}$ . . Hence: . $\\sin\\theta \\:=\\:\\frac{opp}{hyp} \\:=\\:\\frac{-2}{\\sqrt{29}}$ Therefore: . $\\boxed{\\sin\\theta \\;=\\;\\pm\\frac{2}{\\sqrt{29}}}$", "# Math Help - evaluate arccosx 1. ## evaluate arccosx evaluate sin (arccos x) , 0 < x< 1 2. Note that: $\\sin x = \\pm \\sqrt{1 + cos^2 x}$ $\\sin (\\arccos x) = +\\sqrt{1 - \\left[\\cos (\\arccos x)\\right]^{2}}$ (since $0 \\leq x \\leq 1 \\: \\: \\Rightarrow \\: \\: 0 \\leq \\arccos x \\leq \\frac{\\pi}{2}$) Can you finish off? 3. Hello, cruxkitty! ${\\color{red}\\rlap{////////}}\\text{Evaluate: }\\;\\sin(\\arccos x),\\quad 0 < x< 1$ . They meant to say \"Simplify\". Recall that an inverse trig expression is an angle. $\\text{We have: }\\;\\sin\\underbrace{(\\arccos x)}_{\\theta}$ . . . . . . angle whose cosine is x \"The angle whose cosine is $x$\" looks like this: Code: * * * 1 * * ____ * * √1-x² * * * θ * * * * * * * * x $\\cos\\theta \\:=\\:\\frac{x}{1} \\:=\\:\\frac{adj}{hyp} \\quad\\Rightarrow\\quad adj = x,\\;\\;hyp = 1$ From Pythagorus, we find that: . $opp = \\sqrt{1-x^2}$ Hence: . $\\sin\\theta \\:=\\:\\frac{opp}{hyp} \\:=\\:\\frac{\\sqrt{1-x^2}}{1} \\:=\\:\\sqrt{1-x^2}$ Therefore: . $\\boxed{\\sin(\\arccos x) \\;=\\;\\sqrt{1-x^2}}$ 4. $\\sin(\\arccos{x})$ let ... $\\theta = \\arccos{x}$ so ... $\\cos{\\theta} = \\frac{x}{1}$ remember the basic right triangle from geometry days? $\\cos{\\theta} = \\frac{x}{1} = \\frac{adj}{hyp}$ the adjacent side has length x and the hypotenuse has length 1. using Pythagoras, the opposite side has length = $\\sqrt{1-x^2}$ $\\sin(\\arccos{x}) = \\sin{\\theta} = \\frac{opp}{hyp} = \\frac{\\sqrt{1-x^2}}{1}$ Soroban was quicker", "Oct 5, 2006 6. Oct 5, 2006 ### matt grime Because arctan only returns values in the region -pi/2 to pi/2, so it is wrong to say that theta is arctan(y/x) alone. You must also use your common sense to see that the argument is actually pi, not zero. It is correct to say that tan(theta)=y/x, but that is *not* the same as theta =arctan(y/x) at all. 7. Oct 5, 2006 ### murshid_islam yeah, i can see that $$\\theta = \\pi$$ from the position of z = -1 in the argand plane. so, $$\\sqrt{z} = \\sqrt{-1} = \\sqrt{r}e^{i\\frac{\\theta}{2}} = \\sqrt{1}e^{i\\frac{\\pi}{2}} = 1.(\\cos{\\frac{\\pi}{2}} + i.\\sin{\\frac{\\pi}{2}}) = i$$ am i right? 8. Oct 7, 2006 ### Robokapp This is my assumption: $$\\sqrt{i^{2}} = \\sqrt{-1} = i$$ $$i^{2}$$ is equal to -1 and square root of -1 is defined as i. 9. Oct 8, 2006 ### Data That expression for $r$ works for all $z$, but the one for $\\theta$ only works for $x>0$, assuming you're using the usual branch of arctan. In general finding the polar representation of a complex number involves solving $$\\sin{\\theta} = \\frac{y}{r}$$ and $$\\cos{\\theta} = \\frac{x}{r}$$ simultaneously (taking $\\arctan (y/x)$ makes the possible solutions obvious though!).", "$[-1,1]$, because of $\\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\\tan\\arcsin x$. (Note: your $x\\in[-\\pi/4,\\pi/4]$ is wrong and the probable cause for the low grade.) The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\\in(0,1)$. For every root we find, also its negative will be a root. If $y=\\sin\\arctan x$, then $\\arctan x=\\arcsin y$ and so $x=\\tan\\arcsin y$ so $$x^2=\\frac{\\sin^2\\arcsin y}{1-\\sin^2\\arcsin y}=\\frac{y^2}{1-y^2}$$ whence $$y^2=\\frac{x^2}{1+x^2}$$ If $z=\\tan\\arcsin x$, then $$z^2=\\frac{\\sin^2\\arcsin x}{1-\\sin^2\\arcsin x}=\\frac{x^2}{1-x^2}$$ So your equation $\\frac{1}{y^2}-\\frac{1}{z^2}=4x^2$ becomes $$\\frac{1+x^2}{x^2}-\\frac{1-x^2}{x^2}=4x^2$$ that is, $$2x^2=4x^4$$ and, since $x\\ne0$, $x^2=\\frac{1}{2}$. The only positive root is $$x=\\frac{1}{\\sqrt{2}}$$ which indeed belongs to $(0,1)$. Thus the solutions are $$\\frac{1}{\\sqrt{2}}\\qquad\\text{and}\\qquad{-}\\frac{1}{\\sqrt{2}}$$ • Yes, ofcourse! arctan is undefined at $-\\pi/2$, $0$ and $\\pi/2$. So the domain is $(-1,0)\\cup(0,1)$, if I'm not mistaken? Which in turn means that the solutions to the equation $x\\in[\\pi/4,\\pi/4] / {0}$? Sep 18, 2017 at 20:53 • @Parseval No: $x\\in(-1,0)\\cup(0,1)$. Why should you take the arctangent? Sep 18, 2017 at 20:54 • Because it's 22:57 and I've been staring at math for 18 hours. This is the kind of blind mistakes you do in such a condition. Thank you very much for clearing up everything, I really appreciate it mate! Sep 18, 2017 at 20:57 Taking derivative of", "Aug 12th 2007, 05:44 PM Krizalid 1) You know that $\\sin x=\\sqrt{1-\\cos^2x}$, now compute $\\sin(\\arccos x)$ 2) $\\sin2x=\\sin(x+x)$, now apply the given formula. It doesn't understand very well the exponents of some identities, try to use LaTeX • Aug 12th 2007, 05:58 PM Soroban Hello, forkball42! Welcome aboard . . . Here's some help . . . Quote: 1) Express $\\sin(\\arccos x)$ in a form that contains no trigonometric functions or their inverses. Let $\\theta \\:=\\:\\arccos x\\quad\\Rightarrow\\quad \\cos\\theta \\:=\\:x \\:=\\:\\frac{x}{1} \\:=\\:\\frac{adj}{hyp}$ So $\\theta$ is an acute angle in a right triangle . . with: . $adj \\,= \\,x,\\:hyp \\,= \\,1$ Using Pythagorus, we find that: . $opp \\,=\\,\\sqrt{1-x^2}$ . . Hence: . $\\sin\\theta \\:=\\:\\frac{opp}{hyp} \\:=\\:\\frac{\\sqrt{1-x^2}}{1} \\:=\\:\\sqrt{1-x^2}$ Therefore: . $\\sin(\\arccos x) \\:=\\:\\sqrt{1-x^2}$ Quote: 2) Use the identity: . $\\sin(A+B) \\:= \\:\\sin A\\cos B + \\cos A\\sin B$ to verify that: . $\\sin2x \\:= \\:2\\sin x \\cos x$ We have: . $\\sin2x \\;=\\;\\sin(x + x) \\;= \\;\\sin x\\cos x + \\sin x\\cos x \\;=\\;2\\sin x\\cos x$ Quote: 3) Show that sin⁵ x cos⁴ x = sin⁵ x - 2sin⁷ x + sin⁹ x. As you can see, we can't read the exponents. . . But I'll take a guess . . . Is it: . $\\sin^2\\!x\\cos^4\\!x \\;=\\;\\sin^6\\!x - 2\\sin^4\\!x + \\sin^2\\!x$ ? We need the identity: . $\\sin^2\\!\\theta + \\cos^2\\!\\theta", "# How do you simplify the expression Sin(arctan(x)+arccos(x))? Jul 26, 2016 sin 2x #### Explanation: arctan x --> x arccos x --> x sin (arctan x + arccos x) = sin (x + x) = sin 2x Jul 26, 2016 .$= \\left(\\frac{1}{\\sqrt{1 + {x}^{2}}}\\right) \\left(\\sqrt{1 - {x}^{2}} \\pm {x}^{2}\\right)$, x in [-1, 1]. The negative sign is used, when $x \\in \\left[- 1 , 0\\right]$. #### Explanation: Let a = arc tan (x). The principal value of $a \\in \\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ Then x = tan a. sin a = $\\pm \\frac{x}{\\sqrt{1 + {x}^{2}}}$ and cos a = $\\frac{1}{\\sqrt{1 + {x}^{2}}}$. Let b = arc cos x. The principal value of $b \\in \\left[0 , \\pi\\right]$ Then, x = cos b and sin b = $\\sqrt{1 - {x}^{2}}$. Also, $x \\in \\left[- 1 , 1\\right]$. The given expression = $\\sin \\left(a + b\\right)$ $= \\sin a \\cos b + \\cos a \\sin b$ $= \\left(\\pm \\frac{x}{\\sqrt{1 + {x}^{2}}}\\right) \\left(x\\right) + \\left(\\frac{1}{\\sqrt{1 + {x}^{2}}}\\right) \\sqrt{1 - {x}^{2}}$ $= \\left(\\frac{1}{\\sqrt{1 + {x}^{2}}}\\right) \\left(\\sqrt{1 - {x}^{2}} \\pm {x}^{2}\\right)$, x in [-1, 1]. The negative sign is used when $x \\in \\left[- 1 , 0\\right]$.", "1. Yep. That's it. Good job. It is easy if you spot it. Those are the fun ones. 2. Here is one, have fun with the simplicity $\\displaystyle \\lim_{\\psi\\to{0^{+}}}\\frac{\\sin(\\sqrt{\\psi})\\tan(e ^{10\\psi}-1)}{\\sqrt{15\\psi^8-17\\psi^3+64\\psi}}$ 3. Originally Posted by galactus $\\displaystyle \\lim_{n\\to\\rightarrow{\\infty}}\\frac{1}{n}\\sum_{j=1 }^{n}\\sum_{k=1}^{n}\\frac{j}{j^{2}+k^{2}}$ It would seem that this is 0 by a \"power counting\" argument. But I am having some trouble justifying it Mathematically. -Dan 4. Originally Posted by Mathstud28 Here is one, have fun with the simplicity $\\displaystyle \\lim_{\\psi\\to{0^{+}}}\\frac{\\sin(\\sqrt{\\psi})\\tan(e ^{10\\psi}-1)}{\\sqrt{15\\psi^8-17\\psi^3+64\\psi}}$ hint ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 5. Originally Posted by Mathstud28 Here is one, have fun with the simplicity $\\displaystyle \\lim_{\\psi\\to{0^{+}}}\\frac{\\sin(\\sqrt{\\psi})\\tan(e ^{10\\psi}-1)}{\\sqrt{15\\psi^8-17\\psi^3+64\\psi}}$ This actually turns out to be really easy Let $\\displaystyle u(0)=0$ Then $\\displaystyle \\lim_{x\\to{0}}\\frac{\\tan(u(x))}{u(x)}=1$ This can be done by making the sub with $\\displaystyle \\psi=u(x)$ $\\displaystyle \\therefore\\tan(u(x))\\sim{u(x)}$ if $\\displaystyle u(0)$ and the sine case was listed earlier, the other case can be seen before which states that if a polynomial approaches zero then it is ~ to its lowest powered x term $\\displaystyle \\therefore\\sqrt{15x^8-17x^3+64x}\\sim{8\\sqrt{x}}$ $\\displaystyle \\therefore\\lim_{\\psi\\to{0^{+}}}\\frac{\\sin(\\sqrt{\\p si})\\tan(e^{10\\psi}-1)}{\\sqrt{15\\psi^8-17\\psi^3+64\\psi}}\\sim\\lim_{\\psi\\to{0^{+}}}\\frac{\\s qrt{\\psi}(e^{10\\psi}-1)}{8\\sqrt{\\psi}}=0$ 6. Originally Posted by galactus $\\displaystyle \\lim_{n\\to\\infty}\\frac{1}{n}\\sum_{j=1}^{n}\\sum_{k= 1}^{n}\\frac{j}{j^{2}+k^{2}}$ nice problem! the sum is a double Riemann sum of $\\displaystyle f(x,y)=\\frac{x}{x^2+y^2}$ on the unit square: $\\displaystyle \\lim_{n\\to\\infty} \\frac{1}{n}\\sum_{j=1}^n \\sum_{k=1}^n \\frac{j}{j^2 + k^2}= \\lim_{n\\to\\infty} \\frac{1}{n^2} \\sum_{j=1}^n\\sum_{k=1}^n \\frac{\\frac{j}{n}}{(\\frac{j}{n})^2 + (\\frac{k}{n})^2}=\\int_0^1 \\int_0^1 \\frac{x}{x^2 + y^2} \\ dx \\ dy$ $\\displaystyle =\\frac{1}{2}\\int_0^1 (\\ln(y^2 + 1) - 2\\ln y) \\ dy=\\frac{\\pi}{4} +", "# Taking 1/2 of arcsin/arccos/arctan • October 27th 2010, 06:21 PM kjv6675 Taking 1/2 of arcsin/arccos/arctan Could I get some help with the following: sin{1/2acrsin(-7/25)} cos(1/2arctan 8/15) tan(1/2acrcos 3/5) I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are: 1. -1/10 * 2^(1/2) 2. 4/17 * 17^(1/2) 3. 1/2 • October 27th 2010, 08:06 PM TheEmptySet Quote: Originally Posted by kjv6675 Could I get some help with the following: sin{1/2acrsin(-7/25)} cos(1/2arctan 8/15) tan(1/2acrcos 3/5) I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are: 1. -1/10 * 2^(1/2) 2. 4/17 * 17^(1/2) 3. 1/2 List of trigonometric identities - Wikipedia, the free encyclopedia This will be helpful here is the first $\\displaystyle \\sin\\left( \\frac{1}{2}\\theta\\right)=\\pm \\sqrt{\\frac{1-\\cos(\\theta)}{2}}$ This gives $\\displaystyle \\sin\\left( \\frac{1}{2}\\sin^{-1}(\\frac{-7}{25})\\right)=\\pm \\sqrt{\\frac{1-\\cos(\\sin^{-1}(\\frac{-7}{25}))}{2}}$ $-\\sqrt{\\frac{1-\\frac{24}{25}}{2}}=-\\sqrt{\\frac{1}{50}}=-\\frac{\\sqrt{2}}{10}$ • October 27th 2010, 08:24 PM sa-ri-ga-ma $Let \\frac{1}{2}arcsin(\\frac{-7}{25}) = \\theta$ $arcsin(\\frac{-7}{25}) = 2\\theta$ $(\\frac{-7}{25}) = \\sin{2 \\theta}$ $(\\frac{-7}{25}) = \\frac{2\\tan{\\theta}}{1 + \\tan^2{\\theta}}$ Simplify and solve the quadratic to find tanθ. from that find sinθ. Similarly try the other problems. • October 27th 2010, 08:25 PM sa-ri-ga-ma $Let \\frac{1}{2}arcsin(\\frac{-7}{25}) = \\theta$ $arcsin(\\frac{-7}{25}) = 2\\theta$ $(\\frac{-7}{25}) = \\sin{2 \\theta}$ $(\\frac{-7}{25}) = \\frac{2\\tan{\\theta}}{1 +", "also be replaced with $$\\arctan\\left(\\sqrt{x^2-1}\\right)$$. For $$\\theta \\in \\left(\\frac{\\pi}{2}, \\pi\\right]$$, $$\\tan(\\theta) \\leq 0$$, so you get $$-\\int \\tan^2(\\theta) \\mathrm d\\theta = -\\int \\left[\\sec^2(\\theta)-1\\right] \\mathrm d\\theta = \\theta-\\tan(\\theta)+C \\longrightarrow \\sqrt{x^2-1}+\\text{arcsec}(x)+C$$ Since tangent is negative in the second quadrant, $$\\tan(\\theta) = \\tan(\\theta -\\pi) = -\\sqrt{x^2-1}$$ (remember that tangent takes arguments in the first and fourth quadrants), so the $$\\theta$$ term can also be replaced with $$\\pi-\\arctan\\left(\\sqrt{x^2-1}\\right)$$. In both cases, the anti-derivative could be re-written as $$\\sqrt{x^2-1}-\\arctan\\left(\\sqrt{x^2-1}\\right)+C$$. Basically, it just \"combines\" your other two anti-derivatives and expresses them as a single function rather than having one for each case. • Would just leaving the answer as: √(x^2-1) - arcsec(x) + C , still be correct? – user749176 Feb 18 '20 at 0:09 • For indefinite integrals, we usually assume $\\tan(\\theta)$ is positive (at least from what I've seen), so yeah, that's fine. – KM101 Feb 18 '20 at 0:16 Now, both answers are correct. They merely look different. They differ by a constant. Note 1... $$-\\sqrt{\\frac{x+1}{1-x}}+\\sqrt{\\frac{x+1}{1-x}}\\;x = -\\sqrt{\\frac{x+1}{1-x}}\\;(1-x) = -\\frac{\\sqrt{x+1}\\;(1-x)}{\\sqrt{1-x}} \\\\= -\\sqrt{x+1}\\sqrt{1-x} =-\\sqrt{(1+x)(1-x)} =-\\sqrt{1-x^2}$$ Note 2... $$2\\,\\arcsin \\left( \\frac{\\sqrt {1+x}}{\\sqrt {2}} \\right) =\\pi-2\\,\\arcsin \\left( \\frac{\\sqrt {1-x}}{\\sqrt {2}} \\right)$$ • For the second line of step, I multiplied top and bottom by √(1+x) in order to use the trig substitution of x = sinθ – user749176 Feb 17 '20 at 21:12 •" ]

[ "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "\\\\ 0 & 0 \\end{pmatrix}} -c {\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}} \\\\ &= bE_3 - cE_2 \\end{align*} How do I continu to find $M_{E_1}$? Remember that given your basis $\\xi$, for $B \\in M_2(\\mathbb{R})$, $$[B]_\\xi = \\begin{pmatrix}B_{11} \\\\ B_{21} \\\\ B_{12} \\\\ B_{22} \\end{pmatrix};$$ in particular, for $A \\in M_2(\\mathbb{R})$, $M_A := [L_A]_{\\xi \\xi}$ is the matrix of the linear transformation $L_A : M_2(\\mathbb{R}) \\to M_2(\\mathbb{R})$ with respect to $\\xi$, and hence a $4 \\times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \\times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\\xi$, as given by the equation above. So, given $A \\in M_2(\\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$. Hint:do you know that the matrix you are looking for is $4\\times4$? Now you are looking for a $4\\times4$ matrix such that $M_{E_1}[X]_\\xi=[L_{E_1}(X)]_\\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\\xi=(0,b,-c,0)$ column vector, while $[X]_\\xi=(a,b,c,d)$ column. Do you see how to proceed?", "+0 # Long solution matrix part 2 0 237 1 Let A be a $$2 \\times 2$$ matrix. Suppose that for every two-dimensional vector v, there exists a two-dimensional vector w such that $$\\mathbf{A} \\mathbf{w} = \\mathbf{v}$$ Show that we can find a matrix B such that $$\\mathbf{A} \\mathbf{B} = \\mathbf{I}$$. Mar 13, 2019 $$\\exists w_1 \\ni A w_1 = \\begin{pmatrix}1\\\\0\\end{pmatrix}\\\\ \\exists w_2 \\ni A w_2 = \\begin{pmatrix}0\\\\1\\end{pmatrix}\\\\ B = \\begin{pmatrix}w_1 &w_2\\end{pmatrix}$$", "following (2x2) matrices. ${\\displaystyle {\\begin{pmatrix}3&5\\\\6&4\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}1&2\\\\3&4\\\\\\end{pmatrix}}}$ = ${\\displaystyle {\\begin{pmatrix}3{\\rm {x}}1+5{\\rm {x}}3&3{\\rm {x}}2+5{\\rm {x}}4\\\\6{\\rm {x}}1+4{\\rm {x}}3&6{\\rm {x}}2+4{\\rm {x}}4\\\\\\end{pmatrix}}}$ ii) Multiply the following (3x3) matrices. ${\\displaystyle {\\begin{pmatrix}38.15&-42.17&4.02\\\\4.69&0.76&-5.35\\\\-9.94&8.20&2.74\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}0.205&0.665&1\\\\0.181&0.647&1\\\\0.207&0.476&1\\\\\\end{pmatrix}}}$ You will notice that this gives a unit matrix as its product. ${\\displaystyle {\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\\\\\end{pmatrix}}}$ The first matrix is the inverse of the 2nd. Computers use the inverse of a matrix to solve simultaneous equations. If we have ${\\displaystyle {\\begin{pmatrix}y_{1}=a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}.....+a_{1n}x_{n}\\\\y_{2}=a_{21}x_{2}+a_{22}x_{2}+a_{23}x_{3}.....+a_{2n}x_{n}\\\\.....\\\\.....\\\\y_{n}=a_{n1}x_{n}+a_{n2}x_{n}+a_{n3}x_{3}.....+a_{nn}x_{n}\\\\\\end{pmatrix}}}$ In matrix form this is.... ${\\displaystyle Y=AX}$ ${\\displaystyle A^{-1}Y=X}$ In terms of work this is equivalent to the elimination method you have already employed for small equations but can be performed by computers for ${\\displaystyle 10~000}$ simultaneous equations. (Examples of large systems of equations are the fitting of reference data to 200 references molecules, dimension 200, or the calculation of the quantum mechanical gradient of the energy where there is an equation for every way of exciting 1 electron from an occupied orbital to an excited, (called virtual, orbital, (typically ${\\displaystyle 10~000}$ equations.) ## Finding the inverse How do you find the inverse... You use Maple or Matlab on your PC but if the matrix is small you can use the formula... ${\\displaystyle A^{-1}={\\frac {1}{DetA}}AdjA}$ Here Adj A is the adjoint matrix, the transposed matrix of cofactors. These strange objects are best described by example..... ${\\displaystyle {\\begin{vmatrix}1&-1&2\\\\2&1&1\\\\3&-1&1\\\\\\end{vmatrix}}}$ This determinant is equal", "{A}}x_{2}\\\\{\\textbf {A}}x_{3}\\\\...\\\\{\\textbf {A}}x_{n}\\end{pmatrix}}}$, you'll get ${\\displaystyle \\nabla f({\\textbf {x}})={\\begin{pmatrix}{\\textbf {A}}\\\\{\\textbf {A}}\\\\{\\textbf {A}}\\\\...\\\\{\\textbf {A}}\\end{pmatrix}}}$. This is just A! Indeed, that's why ${\\displaystyle {\\frac {d}{d{\\textbf {x}}}}{\\textbf {A}}{\\textbf {x}}={\\textbf {A}}}$. ### A first step towards matrices Now we return back to the other problem: ${\\displaystyle {\\frac {d}{d{\\textbf {A}}}}{\\textbf {A}}{\\textbf {x}}}$. Let's assume that A is a 2 by 2 matrix, and represent A as ${\\displaystyle {\\textbf {A}}={\\begin{pmatrix}a_{11}&a_{12}\\\\a_{21}&a_{22}\\end{pmatrix}}}$. Using the same notation for x, perform matrix multiplication: ${\\displaystyle {\\textbf {A}}{\\textbf {x}}={\\begin{pmatrix}a_{11}&a_{12}\\\\a_{21}&a_{22}\\end{pmatrix}}{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}={\\begin{pmatrix}a_{11}\\times x_{1}+a_{12}\\times x_{2}\\\\a_{21}\\times x_{1}+a_{22}\\times x_{2}\\end{pmatrix}}}$ Take the partial derivatives with respect to each of the elements of A (this is equivalent to finding the Jacobian). For ${\\displaystyle 1\\leq i,j\\leq 2}$, find ${\\displaystyle {\\frac {\\partial }{\\partial A_{ij}}}}$: ${\\displaystyle {\\frac {\\partial }{\\partial A_{11}}}={\\begin{pmatrix}x_{1}\\\\0\\end{pmatrix}}}$, ${\\displaystyle {\\frac {\\partial }{\\partial A_{12}}}={\\begin{pmatrix}x_{2}\\\\0\\end{pmatrix}}}$, ${\\displaystyle {\\frac {\\partial }{\\partial A_{21}}}={\\begin{pmatrix}0\\\\x_{1}\\end{pmatrix}}}$ and ${\\displaystyle {\\frac {\\partial }{\\partial A_{22}}}={\\begin{pmatrix}0\\\\x_{2}\\end{pmatrix}}}$ But then what do you do with that? How do you \"combine\" the result? Clearly, we are missing something. ### Dimensions of ∇f Let's take a step back and ask the question: given a vector ${\\displaystyle {\\textbf {f}}={\\textbf {A}}{\\textbf {x}}}$, what should be the dimension of ${\\displaystyle \\nabla {\\textbf {f}}}$? Consider the example above. We have two variables: A with 4 elements (2x2) and x with 2 elements, and we want to find ${\\displaystyle {\\frac {d{\\textbf {f}}}{d{\\textbf {A}}}}}$. It is straightforward to show that the", "## Matrix inverses Show that $\\mathbf{R_1} = \\begin{pmatrix}-1 & 1 \\\\ 6& -4 \\\\ 4& -3\\end{pmatrix}$ and $\\mathbf{R_2} = \\begin{pmatrix}1 & -1 \\\\ -4& 6 \\\\ -4& 5\\end{pmatrix}$ are both right-inverses of the matrix $\\mathbf{A} = \\begin{pmatrix}1 &1 &-1 \\\\ 4&0 &1 \\end{pmatrix}$. Use the right-inverses $\\mathbf{R_1}$ and $\\mathbf{R_2}$ to find two solutions $\\mathbf{x_1}$ and $\\mathbf{x_2}$ of the equation $\\mathbf{Ax = b}$, where $\\mathbf{b} =\\begin{pmatrix}0\\\\ 8\\end{pmatrix}$. By what I understand, the only way to solve Ax = b is with an inverse: $\\mathbf{A^{-1}Ax = A^{-1}b}$ $\\mathbf{x = A^{-1}b}$ and matrix $\\mathbf{A}$ doesnt have an inverse but the question asks to use the right-inverse and this is what I dont understand", "for all i, j with ${\\displaystyle \\scriptstyle 1\\leq i\\leq m}$ and ${\\displaystyle \\scriptstyle 1\\leq j\\leq n}$ For example: ${\\displaystyle {\\begin{pmatrix}1&2&3\\\\4&5&6\\end{pmatrix}}+{\\begin{pmatrix}9&16&12\\\\5&17&15\\end{pmatrix}}={\\begin{pmatrix}(1+9)&(2+16)&(3+12)\\\\(4+5)&(5+17)&(6+15)\\end{pmatrix}}={\\begin{pmatrix}10&18&15\\\\9&22&21\\end{pmatrix}}}$ ${\\displaystyle A+B=B+A}$ and associative: ${\\displaystyle A+(B+C)=(A+B)+C}$ ### Scalar multiplication Any scalar may be multiplied by any matrix. To obtain the resultant matrix, multiply each entry of the original matrix by the scalar. That is, if c is a scalar and A is an m×n matrix, then cA is an m×n matrix such that ${\\displaystyle \\scriptstyle (cA)_{i,j}=c\\times A_{i,j}}$ For example: ${\\displaystyle 3{\\begin{pmatrix}1&2&3\\\\4&5&6\\end{pmatrix}}={\\begin{pmatrix}(3\\times 1)&(3\\times 2)&(3\\times 3)\\\\(3\\times 4)&(3\\times 5)&(3\\times 6)\\end{pmatrix}}={\\begin{pmatrix}3&6&9\\\\12&15&18\\end{pmatrix}}}$ ### Matrix multiplication Two matrices A and B may be multiplied if A has as many columns as B has rows. (Otherwise, they are said to be incompatible and their product is undefined.) That is, an m×n matrix may be multiplied by an n×p matrix. Then the resultant matrix AB is m×p and the (i,j)th entry of AB is the vector dot product of the ith row of A and the jth column of B. Formally: ${\\displaystyle (AB)_{i,j}=\\sum _{k=1}^{n}A_{i,k}B_{k,j}}$ For example: ${\\displaystyle {\\begin{pmatrix}1&2\\\\3&4\\end{pmatrix}}{\\begin{pmatrix}3&4&5\\\\6&7&8\\end{pmatrix}}={\\begin{pmatrix}(1\\times 3+2\\times 6)&(1\\times 4+2\\times 7)&(1\\times 5+2\\times 8)\\\\(3\\times 3+4\\times 6)&(3\\times 4+4\\times 7)&(3\\times 5+4\\times 8)\\end{pmatrix}}={\\begin{pmatrix}15&18&21\\\\33&40&47\\end{pmatrix}}}$ Even if AB is defined, BA may not be. If both matrices are defined, they may have different dimensions. Even if they have the same dimensions, they may not be equal. Thus, matrix multiplication is clearly", "A $2\\times 2$ matrix $\\textbf{P}$ has an eigen value $\\lambda_{1}=2$ with eigenvector $x_{1}=\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}$ and another eigenvalue $\\lambda_{2}=5,$ with eigenvector $x_{2}=\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}.$ The matrix $\\textbf{P}$ is 1. $\\begin{pmatrix} 2 & 0 \\\\ 0 & 5 \\end{pmatrix}$ 2. $\\begin{pmatrix} 2 & 3 \\\\ 0 & 5 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}$ 4. $\\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$", "Free Version Easy # Find 3x3 Inverse - 3 LINALG-TGB9X1 Find the inverse of the matrix: $$\\begin{pmatrix}1&1&3\\\\\\ 4&1&6\\\\\\ 7&1&9\\end{pmatrix}$$ ...if it exists. A $\\begin{pmatrix}1&2&1\\\\\\ 26&4&-6\\\\\\ 4&2&-2\\end{pmatrix}$ B $\\begin{pmatrix}1&2&1\\\\\\ 24&4&-6\\\\\\ 4&2&-2\\end{pmatrix}$ C $\\begin{pmatrix}5&-2&3\\\\\\ -2&-4&2\\\\\\ -4&2&-\\cfrac{5}{2}\\end{pmatrix}$ D $\\begin{pmatrix}5&-2&-3\\\\\\ -2&-4&-2\\\\\\ -4&2&\\cfrac{5}{2}\\end{pmatrix}$ E This matrix is not invertible", "3$ matrix $\\mathbf{A} = \\begin{pmatrix}1&0&2\\\\3&1&0\\\\-1&1&4\\end{pmatrix}$. Workbook This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples. More Support You can get one-to-one support from Maths-Aid.", "? Free Version Easy # 3x3 Matrix Product - 1 LINALG-GKW5LH Find the product: $$\\begin{pmatrix}2&7&-7\\\\\\ -7&-9&9\\\\\\ -9&-8&6\\end{pmatrix}\\begin{pmatrix}-8&-5&-10\\\\\\ 7&5&10\\\\\\ 9&2&-3\\end{pmatrix}$$ A $\\begin{pmatrix}-30&11&-71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ B $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&8\\end{pmatrix}$ C $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ D $\\begin{pmatrix}-30&11&71\\\\\\ -74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ E This matrix product is not defined.", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "the same length: • For$m = 1$: the roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$are • (all separate)$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$, • (two pairs)$\\begin{pmatrix}2 & 5\\end{pmatrix} \\begin{pmatrix}8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8\\end{pmatrix} \\begin{pmatrix}5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 14\\end{pmatrix} \\begin{pmatrix}5 & 8\\end{pmatrix}$, • (all together)$\\begin{pmatrix}2 & 5 & 8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 5 & 14 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 14 & 5\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 5 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 8 & 5\\end{pmatrix}$. So there are$1 + 3 + 6 = 10$roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$. • For$m = 2$, the roots of$\\begin{pmatrix}1 & 7\\end{pmatrix} \\begin{pmatrix}4 & 10\\end{pmatrix} \\begin{pmatrix}6 & 18\\end{pmatrix} \\begin{pmatrix}12 & 17\\end{pmatrix}$are$6 \\times 8 = 48$in number (pick one of the$3!$orderings of the last three, then one of the$2^3$orderings of each of the three). • For$m = 3$, the roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$are • (both separate)$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$• (both together)$\\begin{pmatrix}3 & 11 & 9 & 13 & 16 & 15\\end{pmatrix}$,$\\begin{pmatrix}3 & 15 & 9 & 11 & 16 & 13\\end{pmatrix}$,$\\begin{pmatrix}3 & 13 & 9 & 15 & 16 & 11\\end{pmatrix}$. So there are$1 + 3 = 4$roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$. So totally$N(\\sigma) = 10 \\times 48 \\times 4 = 1920$. - In", "Question # Let A = $$[m \\, n] _1 \\, _\\times \\, _2$$ , B = $$\\begin{bmatrix}m\\\\ n\\end{bmatrix} \\, _2 \\, _\\times \\,_ 1$$ and C = $$[25] \\, _1 \\, _\\times \\,_1$$ be three matrices such that AB = C, where m and n are integers, then the number of ordered pairs of (m,n) are A 8 B 12 C 10 D 4 Solution ## The correct option is C 12$$m^2 \\,+ \\, n^2 \\, = \\, 25$$ $$\\therefore \\, (0 \\, \\pm \\, 5), \\, \\left ( \\pm 5,0 \\right ) \\, \\left ( \\pm 3, \\, \\pm 4 \\right ),\\, \\left ( \\pm 4 ,\\pm 3 \\right )$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More", "\\begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \\end{pmatrix}$ $\\textbf u = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3\\\\...\\\\ b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = \\begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 \\\\ b_2 \\\\b_3 \\\\ ... \\\\b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$ Notice that it only works if $m = n$ and it always results in a matrix of order $1 \\times 1$. The outer product of these two vectors is $\\textbf v \\cdot \\textbf u^T$. $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\\\ ... \\\\ a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \\end{pmatrix}$ $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\\\...& ... & ... & ... & ... \\\\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \\end{pmatrix}$ If you have a column vector of order $m \\times 1$ and a row vector of order $1 \\times n$ respectively, their outer product will be a matrix of order", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "$$n = 2$$ and $$D ((a)) = a$$ then $$D_ {11} begin {pmatrix} a + a & # 39; & b + b & # 39; \\ c & d end {pmatrix} = d$$ while $$D_ {11} begin {pmatrix} a & b \\ c & d end {pmatrix} + D_ {11} begin {pmatrix} a & # 39; & bsp; \\ c & d end {pmatrix} = d + d = 2d.$$ But the authors state $$A_ {ij} D_ {ij}$$ is $$n$$-linear." ]

[ "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} , \\quad (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We see that the numbers in the parentheses between both cycles are difference. Hence $(23)$ and $(165)$ are disjoint cycles. Also notice that: (2) \\begin{align} \\quad (23) \\circ (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 3 & 2 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We also have that: (3) \\begin{align} \\quad (165) \\circ (23) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ = \\begin{pmatrix} 1 & 2 &", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "& d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$and set$D=ad-bc$as before. Then the following are equivalent: 1.$(D,N)=1$. 2.$A$has an inverse matrix. 3. if$x$and$y$are not both$0$in$\\mathbb{Z}/N\\mathbb{Z}$, then$A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4.$A$gives a one-to-one correspondence of$(\\mathbb{Z}/N\\mathbb{Z})^2with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruenceAX\\equiv B\\mod 26$or as the matrix equation$AX=B$in$\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$$D=16-21=-5$and$(-5,26)=1$, so by above Theorem, we see that$A^{-1}$exists. In fact, we have already found$A^{-1}$in the previous example. The solution$Xis given by \\begin{align*} X&\\equiv A^{-1}B\\mod 26\\\\&\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 26\\\\ &\\equiv\\begin{pmatrix} 10\\\\ 11 \\end{pmatrix}\\mod 26 \\end{align*} 2. The matrix in parts 2 and 3 does not have an inverse matrix in\\mathbb{Z}/26\\mathbb{Z}$because$D=9-21=-12$and$(-12,26)=2\\ne 1$. However, it does have an inverse matrix in$\\mathbb{Z}/13\\mathbb{Z}and the solution is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in\\mathbb{Z}/26\\mathbb{Z}$. Since$x\\equiv 3\\mod 13$and$y\\equiv 8\\mod 13$,$x$and$y$can be written as$x=3+13k_1$and$y=8+13k_2$for some$k_1,k_2\\in\\mathbb{Z}. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only ifk_1+k_2$is even. Suppose that$k_1+k_2$is even, so that$x+3y\\equiv", "C. $B’(-1,2)$ Solution Rotation rule is given as follows. The coordinates of the image of point $(x, y)$ after being rotated counter clockwise by an angle of $\\theta$ about the center point $(a, b)$ are given by $$\\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta &-\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix} \\begin{pmatrix} x-a \\\\ y-b \\end{pmatrix} + \\begin{pmatrix} a \\\\ b \\end{pmatrix}$$For $(x, y) = (3,-2)$ and rotation of $\\theta = 90^{\\circ}$ about the center point $(-1, 1)$, we have \\begin{aligned} \\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} & = \\begin{pmatrix} \\cos 90^{\\circ} &-\\sin 90^{\\circ} \\\\ \\sin 90^{\\circ} & \\cos 90^{\\circ} \\end{pmatrix} \\begin{pmatrix} 3-(-1) \\\\-2-1 \\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 &-1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 4 \\\\-3 \\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} \\end{aligned}Thus, the coordinates of the image of point $B$ are $\\boxed{B'(2,5)}$ [collapse] Problem Number 6 The coordinates of the image of point $P(2, -3)$ under rotation that is given by $R[O,90^{\\circ}]$ are $\\cdots \\cdot$ A. $P’(3,2)$ D. $P’(-3,2)$ B. $P’(2,3)$ E. $P’(-3,-2)$ C. $P’(-2,3)$ Solution Rotation rule is given as follows. The coordinates of the image of point $(x, y)$ after being rotated counter", "\\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle {\\mathbf {a} }_{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad {\\mathbf {b} }_{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle {\\mathbf {AB} }={\\begin{pmatrix}{\\mathbf {A} }{\\mathbf {b} }_{1}&{\\mathbf {A} }{\\mathbf {b} }_{2}&\\dots &{\\mathbf {A} }{\\mathbf {b} }_{p}\\end{pmatrix}}={\\begin{pmatrix}{\\mathbf {a} }_{1}{\\mathbf {B} }\\\\{\\mathbf {a} }_{2}{\\mathbf {B} }\\\\\\vdots \\\\{\\mathbf {a} }_{n}{\\mathbf {B} }\\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).{{ safesubst:#invoke:Unsubst||date=__DATE__ |B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} Outer product The outer product (also known as the dyadic product or tensor product) of two", "& 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 3\\mod 13$ and $y\\equiv 8\\mod 13$, $x$ and $y$ can be written as $x=3+13k_1$ and $y=8+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Suppose that $k_1+k_2$ is even, so that $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 15+13(k_1+k_2)\\mod 26\\\\ &\\equiv 15\\mod 26\\\\ &\\not\\equiv 2\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are no solutions for part 2. 3. Similarly to part 2, the solution in $\\mathbb{Z}/13\\mathbb{Z}$ is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$", "(m_1 - m_2) * qInv mod p. iv. Let m = m_2 + q * h. v. If u > 2, let R = r_1 and for i = 3 to u do 1. Let R = R * r_(i-1). 2. Let h = (m_i - m) * t_i mod r_i. 3. Let m = m + R * h. 1. $m_1=c^{d_p}\\bmod p,\\ m_2=c^{d_q}\\bmod q$ 2. $h=(m_1-m_2)*q_p \\bmod p$ 3. $m=m_2+qh$ \\begin{aligned} h&=(m_1-m_2)q_p\\bmod p\\\\ &=(k_2q-k_1p)q_p\\bmod p\\\\ &=k_2q(q^{-1}\\bmod p)\\bmod p\\ -k_1pq_p\\bmod p\\\\ &=k_2 \\end{aligned} \\begin{aligned} m&=m_2+qh\\\\ &=(m-k_2q)+k_2q\\\\ &=m \\end{aligned} ## 如何尝试恢复私钥 \\begin{aligned} \\varphi(n) &= n-p-q+1\\\\ ed&\\equiv1\\pmod{\\varphi(n)}\\\\ ed_p&\\equiv 1 \\pmod{(p-1)}\\\\ ed_q&\\equiv 1\\pmod{(q-1)}\\\\ n &= pq \\end{aligned} \\begin{aligned} ed& = k\\varphi(n) + 1\\\\ ed_p& = k_p(p-1) + 1\\\\ ed_q&=k_q(q-1)+1 \\end{aligned} \\begin{aligned} \\varphi(n)&\\equiv n-p-q+1&\\pmod {16^t}\\\\ ed&\\equiv k\\varphi(n)+1&\\pmod {16^t}\\\\ ed_p&\\equiv k_p(p-1)+1&\\pmod {16^t}\\\\ ed_q&\\equiv k_q(q-1)+1&\\pmod {16^t}\\\\ n&\\equiv pq&\\pmod {16^t} \\end{aligned} 1. 枚举第t位p可能的取值 2. 因为$n\\equiv pq\\pmod{16^t}$，所以计算$p=n\\times q^{-1}\\bmod {16^t}$，检查q的已知值 3. 计算$d=(1+k(n-p-q+1))\\times e^{-1}\\bmod {16^t}$，检查d与已知部分是否符合 4. 计算$d_p = (1+d_p(p-1)) \\times e^{-1}\\bmod {16^t}$，检查$d_p$与已知部分是否符合 5. 计算$d_q=(1+d_q(q-1))\\times e^{-1}\\bmod{16^t}$，检查$d_q$与已知部分是否符合 ## Jarvis OJ GodLike RSA openssl rsautl -decrypt -inkey private.pem -keyform PEM -in flag.enc -oaep", "\\\\ Y\\end{bmatrix} \\qquad (x,y) = [X,Y]P^T$ $q(\\mathbf{v}) = (x,y)A\\begin{pmatrix}x \\\\ y\\end{pmatrix} = [X,Y]P^T AP\\begin{bmatrix}X \\\\ Y\\end{bmatrix}$ If we can find an orthogonal change of basis $(P^T = P^{-1})$ then $P$ ... Eigenvalues of $A = \\begin{pmatrix}2 & 1 \\\\ 1 & 2\\end{pmatrix}$ \\begin{aligned} 0 &= \\begin{vmatrix}\\lambda I - A \\end{vmatrix} = \\begin{vmatrix} \\lambda -2 & -1 \\\\ -1 & \\lambda -2\\end{vmatrix}= (\\lambda -2)^2 - 1 \\\\ 0 &= (\\lambda -2)^2 - 1^2 = ((\\lambda -2)-1)((\\lambda -2)+1) \\\\ 0 &= (\\lambda -3)(\\lambda -1)\\end{aligned} $\\begin{cases}\\lambda_1 = 3 \\lambda_2 = 1\\end{cases}$ eigenvalues $3X^2 + Y^2$ is $q$ in the new system, Eigenvalue $\\lambda_1 = 3$ $\\begin{pmatrix} 3-2 & -1 \\\\ -1 & 3-2 \\end{pmatrix} \\begin{pmatrix}\\alpha \\\\ \\beta\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0\\end{pmatrix} \\begin{cases} \\alpha - \\beta = 0 \\\\ -\\alpha + \\beta = 0 \\end{cases} so \\alpha = \\beta \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$ will do Note $<\\begin{pmatrix}1 \\\\ 1\\end{pmatrix},\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}> = 0$ orthogonal basis of eigenvectors $\\|\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}\\|^2 = 2 =\\|\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}\\|^2$ Corresponding orthonormal basis: $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}-\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ But another can be $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ and two more candidates... We may take $P = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} = (P^{-1})^T$ Note $det P = \\pm 1$ (rotation) Claim $\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} \\begin{pmatrix} 2 & 1", "/ a^* = e^{2 i \\phi}$ for some $\\phi$ real – reuns Oct 7 '16 at 14:24 ## 1 Answer \\begin{align*} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} &= \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} \\bar{d} & -\\bar{b} \\\\ -\\bar{c} & \\bar{a} \\end{pmatrix} \\\\ &= \\begin{pmatrix} a\\bar{d}-b\\bar{c} & b\\bar{a}-a\\bar{b} \\\\ c\\bar{d}-d\\bar{c} & d\\bar{a}-c\\bar{b} \\end{pmatrix} \\end{align*} Equating \"$0$\": \\left \\{ \\begin{align*} \\frac{a}{\\bar{a}} &= \\frac{b}{\\bar{b}} \\\\ \\frac{c}{\\bar{c}} &= \\frac{d}{\\bar{d}} \\\\ \\end{align*} \\right. Let \\left \\{ \\begin{align*} a &= Ae^{i\\theta} \\\\ b &= Be^{i\\theta} \\end{align*} \\right. and \\left \\{ \\begin{align*} c &= Ce^{i\\phi} \\\\ d &= De^{i\\phi}\\ \\end{align*} \\right. where $A,B,C,D\\in \\mathbb{R}$ Equating \"$1$\": \\left \\{ \\begin{align*} (AD-BC)e^{i(\\theta-\\phi)} &= 1 \\\\ (AD-BC)e^{i(\\phi-\\theta)} &= 1 \\\\ \\end{align*} \\right. \\implies \\left \\{ \\begin{align*} AD-BC &= 1 \\\\ \\theta &= \\phi \\\\ \\end{align*} \\right.", "k & \\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}(ae+bg)i+(af+bh)k & (ae+bg)j+(af+bh)\\ell \\\\ (ce+dg)i+(cf+dh)k & (ce+dg)j+(cf+dh)\\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\ell+bh\\ell \\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\ell+dh\\ell\\end{pmatrix}\\end{align*} \\begin{align*}A\\cdot (B\\cdot C)&=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot\\left ( \\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}\\cdot \\begin{pmatrix}i & j \\\\ k & \\ell\\end{pmatrix}\\right )= \\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot\\begin{pmatrix}ei+fk & ej+f\\ell \\\\ gi+hk & gj+h\\ell\\end{pmatrix} \\\\ & = \\begin{pmatrix}a(ei+fk)+b(gi+hk) & a(ej+f\\ell)+b(gj+h\\ell) \\\\ c(ei+fk)+d(gi+hk) & c(ej+f\\ell)+d(gj+h\\ell)\\end{pmatrix} \\\\ & = \\begin{pmatrix}aei+afk+bgi+bhk & aej+af\\ell+bgj+bh\\ell \\\\ cei+cfk+dgi+dhk & cej+cf\\ell+dgj+dh\\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\ell+bh\\ell \\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\ell+dh\\ell\\end{pmatrix}\\end{align*} The results are the same. Therefore it holds that $(A\\cdot B)\\cdot C=A\\cdot (B\\cdot C)$ which means the multiplication over $M_2(\\mathbb{R})$ is associative. 2. Let $A=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}, \\ B=\\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}$. Then we have the following: \\begin{equation*}A\\cdot B=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot \\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}=\\begin{pmatrix}ae+bg & af+bh \\\\ ce+dg & cf+dh\\end{pmatrix} \\end{equation*} Then we have the following: \\begin{equation*}B\\cdot A=\\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}\\cdot \\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}=\\begin{pmatrix}ea+fc & eb+fd \\\\ ga+hc & gb+hd\\end{pmatrix} \\end{equation*} We see that $A\\cdot B\\neq B\\cdot A$, which means that the multiplication over $M_2(\\mathbb{R})$ is not commutative. 3. The neutral element in respect of the multiplication over $M_2(\\mathbb{R})$ is the identity matrix \\begin{equation*}I_2=\\begin{pmatrix}1 & 0 \\\\ 0", "ca&=ac=b. \\end{align*} Here, $\\cdot$ is the function composition i.e. successive application of rotations. By labling the vertices of the rectangle as 1, 2, 3, 4 as seen in the picture, we can define each rotation as a permutation of $\\{1,2,3,4\\}$. \\begin{align*} e&=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 1 & 2 & 3 & 4 \\end{pmatrix},\\ a=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 & 1 & 4 &3 \\end{pmatrix},\\\\ b&=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1 \\end{pmatrix},\\ c=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 3 & 4 & 1 &2 \\end{pmatrix}. \\end{align*} For instance, we calculate $ab$: \\begin{align*} ab&=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 & 1 & 4 &3 \\end{pmatrix}\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 & 1 & 4 &3 \\end{pmatrix}\\begin{pmatrix} 4 & 3 & 2 & 1\\\\ 3 & 4 & 1 & 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 3 & 4 & 1 &2 \\end{pmatrix}\\\\ &=c. \\end{align*} Example. [The $n$-th dihedral group $D_n$, $n\\geq 3$] Consider an equilateral triangle shown in the following picture. Dihedral group D3 $\\rho$ is counterclockwise rotation about the axis coming out of the", "given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$ or $k_1=k_2=1$. Therefore, the solutions are \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 26\\ \\mbox{or}\\ \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 19\\\\ 20 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} \\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ with $(D,N)=1$ where $D=ad-bc$ be an enciphering matrix. Then it defines the enciphering transformation $C=AP$ where $P=\\begin{pmatrix} x\\\\ y \\end{pmatrix}$ and $C=\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}$ are plaintext message unit and ciphertext, respectively. That is, $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}$$ The deciphering transformation is then given by $P=A^{-1}C$, i.e. $$\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} D^{-1}d & -D^{-1}b\\\\ -D^{-1}c &", "the same variables together, we have \\begin{align*} 25a+11b &\\equiv 13 \\pmod{26}\\\\ 19a+8b &\\equiv 12 \\pmod{26} \\end{align*} \\begin{align*} 25c+11d &\\equiv 20 \\pmod{26}\\\\ 19c+8d &\\equiv 1 \\pmod{26} \\end{align*} Now, notice (and this is a nice trick) that we can write the two systems of congruences as matrix congruences. That is to say, $$\\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix} \\begin{pmatrix}a\\\\b\\end{pmatrix} \\equiv \\begin{pmatrix}13\\\\12\\end{pmatrix} \\textrm{ and } \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}\\begin{pmatrix}c\\\\d\\end{pmatrix}\\equiv\\begin{pmatrix}20\\\\1\\end{pmatrix} \\pmod{26}$$ Both of these can be solved by applying the inverse of the matrix involved to both sides of the congruence... $$\\begin{pmatrix}a\\\\b\\end{pmatrix} \\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}13\\\\12\\end{pmatrix} \\textrm{ and } \\begin{pmatrix}c\\\\d\\end{pmatrix} \\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}20\\\\1\\end{pmatrix} \\pmod{26}$$ We just need to find the inverse matrix in question. As seen previously, we can find the inverse of a matrix with the formula $$M^{-1} = \\begin{bmatrix}x & y\\\\z & w\\end{bmatrix}^{-1} = \\textrm{det}^{-1}\\begin{bmatrix}w & -y\\\\-z & x\\end{bmatrix}$$ Finding the determinant$\\pmod{26}$ is easy enough, $$\\textrm{det}=25 \\cdot 8 - 19 \\cdot 11 = -9 \\equiv 17 \\pmod{26}$$ but to find $\\textrm{det}^{-1}$, since we are working with congruences instead of equations, we need to find the multiplicative inverse of 17$\\pmod{26}$. Recall that the product of any number and its multiplicative inverse must be the multiplicative identity, 1, so we need to solve $$17m \\equiv 1 \\pmod{26}$$ This in turn, leads to solving (in integers)", "as the splitting (6), where \\begin{align} & \\mathbf{D} = \\begin{pmatrix} 6 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 5 \\end{pmatrix}, \\quad \\mathbf{U} = \\begin{pmatrix} 0 & 2 & 3 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 0 \\end{pmatrix}, \\quad \\mathbf{L} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 1 & 0 & 0 \\\\ 3 & 1 & 0 \\end{pmatrix}. \\quad (14) \\end{align} We then have \\begin{align} & \\mathbf{(D-L)^{-1}} = \\frac{1}{120} \\begin{pmatrix} 20 & 0 & 0 \\\\ 5 & 30 & 0 \\\\ 13 & 6 & 24 \\end{pmatrix}, \\end{align} \\begin{align} & \\mathbf{(D-L)^{-1}U} = \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix}, \\quad \\mathbf{(D-L)^{-1}k} = \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}. \\end{align} The Gauss-Seidel method (8) applied to the problem (10) takes the form \\bold x^{(m+1)} = \\begin{align} & \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix} \\bold x^{(m)} + \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}, \\end{align} \\quad m = 0, 1, 2, \\ldots \\quad (15) The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From", "b_{21} & b_{22} \\end{pmatrix} \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix} \\Rightarrow \\\\ \\begin{cases}\\require{cancel}\\cancel{a_{11}b_{11}}+a_{12}b_{21}=\\cancel{b_{11}a_{11}}+b_{12}a_{21}\\\\ a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\\\ a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\\\ a_{21}b_{12}+\\cancel{a_{22}b_{22}}=b_{21}a_{12}+\\cancel{b_{22}a_{22}}\\end{cases}$$ From$(1)$, since$b_{12}$and$b_{21}$can be any number, in particular,$b_{12}=0$and$b_{21}\\ne 0$, we get:$a_{12}=0$. Similarly, for$b_{12}\\ne 0$and$b_{21}=0$, we get$a_{21}=0$. From$(2)$, since$a_{12}=0$and$b_{12}$is an arbitrary number, we get$a_{11}b_{12}=b_{12}a_{22} \\Rightarrow a_{11}=a_{22}\\$.", "+ 3 x_{22}\\\\ \\end{pmatrix} \\end{align*} So our equation is $$\\begin{pmatrix} 9 & -3 \\\\ 5 & -5 \\end{pmatrix} - \\begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\\\ \\end{pmatrix}= \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ Now we write it as a system of equations: \\begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\\\ 0&= -3-(-2x_{11} +3x_{12}) \\\\ 0&=5-(-9x_{21} +8 x_{21}) \\\\ 1&=-5 - (-2x_{21}+3x_{22}) \\\\ \\end{align*} If you need help solving this system tell me. - Thank you very much for your excellent answer! – Lukas Arvidsson Mar 15 at 9:27 You are welcome :) – Dominic Michaelis Mar 15 at 9:28 First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution. - Denote $B=\\begin{pmatrix}-9&-2\\\\8&5\\end{pmatrix}$. Then we have: $$\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E=XB$$ Observe that $\\det B=-45+16\\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\\left(\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E\\right)B^{-1}$$ Do you know how to find $B^{-1}$? - Thanks for your answer! I get the final result to be: $\\left[ \\begin{array}{cc} \\frac{-64}{29} & \\frac{-43}{29}\\\\ \\frac{-48}{29} & \\frac{-54}{29}\\end{array} \\right]$ Is that correct?", "and $B=(1,3,-2)$. The line that joints them is $$C = \\begin{pmatrix} 1 \\\\ 7 \\\\ 2 \\end{pmatrix} \\times \\begin{pmatrix} 1 \\\\ 3 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -20 \\\\ 4 \\\\ -4 \\end{pmatrix}$$ with the equation $$(-20)+(4)x+(-4)y=0 \\\\ x-y-5 = 0$$ -", "\\\\ 1 && -2\\end{bmatrix}$$ I'll write the eigenvalue equation as$$\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix} = c \\begin{pmatrix} a \\\\ b\\end{pmatrix}$$. The bottom line of the matrix equation gives $a-2b=cb$. Thus $a=(c+2)b$. So we have shown that the vector $\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix}$ should be an eigenvector. So we can plug this back into the eigenvalue equation: $$c\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix}=\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix} =\\begin{pmatrix}-2cb-3b\\\\ cb \\end{pmatrix}$$ Reading across the top, we see the equation $$c(c+2)b=-2cb-3b$$Or$$0=c^2+4c+3=(c+3)(c+1)$$ so the two eigenvalues are $c=-3$ and $c=-1$. Now we plug these two values back into the eigenvalue equation to find the eigenvectors corresponding to each eigenvalue. For $c=-3$ we have: $$\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix} = -3 \\begin{pmatrix} a \\\\ b\\end{pmatrix}$$ which implies $b=-a$. This means any vector of the form $\\begin{pmatrix}a \\\\ -a \\end{pmatrix}$ is an eigenvector with eigenvalue -3. We want our states to be normalized though, so we pick $a=\\frac{1}{\\sqrt{2}}$. See if you can find the other eigenvector (the one corresponding to $c=-1$) for yourself, and then see if you can use your answers to solve the original problem with the k's and m's. Last edited: Sep 10, 2013", "1 \\pmod 3 \\\\ 20m &\\equiv 1 \\pmod 3\\\\ 2m &\\equiv 1 \\pmod 3\\\\ m &\\equiv 2 \\pmod 3 \\\\ n &\\equiv 40 \\pmod{60} \\end{align} (B) If $n \\equiv 0 \\pmod 3$ and $n \\equiv 0 \\pmod 5$, then $n \\equiv 0 \\pmod{15},$ that is $n = 15m$ for some integer $m$. So \\begin{align} n &\\equiv 1 \\pmod 4 \\\\ 15m &\\equiv 1 \\pmod 4\\\\ -m &\\equiv 1 \\pmod 4\\\\ m &\\equiv 3 \\pmod 4 \\\\ n &\\equiv 45 \\pmod{60} \\end{align} (C) If $n \\equiv 0 \\pmod 3$ and $n \\equiv 0 \\pmod 4$, then $n \\equiv 0 \\pmod{12},$ that is $n = 12m$ for some integer $m$. So \\begin{align} n &\\equiv 1 \\pmod 5 \\\\ 12m &\\equiv 1 \\pmod 5\\\\ 2m &\\equiv 1 \\pmod 5\\\\ m &\\equiv 3 \\pmod 5 \\\\ n &\\equiv 36 \\pmod{60} \\end{align}" ]

[ "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "& d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$and set$D=ad-bc$as before. Then the following are equivalent: 1.$(D,N)=1$. 2.$A$has an inverse matrix. 3. if$x$and$y$are not both$0$in$\\mathbb{Z}/N\\mathbb{Z}$, then$A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4.$A$gives a one-to-one correspondence of$(\\mathbb{Z}/N\\mathbb{Z})^2with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruenceAX\\equiv B\\mod 26$or as the matrix equation$AX=B$in$\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$$D=16-21=-5$and$(-5,26)=1$, so by above Theorem, we see that$A^{-1}$exists. In fact, we have already found$A^{-1}$in the previous example. The solution$Xis given by \\begin{align*} X&\\equiv A^{-1}B\\mod 26\\\\&\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 26\\\\ &\\equiv\\begin{pmatrix} 10\\\\ 11 \\end{pmatrix}\\mod 26 \\end{align*} 2. The matrix in parts 2 and 3 does not have an inverse matrix in\\mathbb{Z}/26\\mathbb{Z}$because$D=9-21=-12$and$(-12,26)=2\\ne 1$. However, it does have an inverse matrix in$\\mathbb{Z}/13\\mathbb{Z}and the solution is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in\\mathbb{Z}/26\\mathbb{Z}$. Since$x\\equiv 3\\mod 13$and$y\\equiv 8\\mod 13$,$x$and$y$can be written as$x=3+13k_1$and$y=8+13k_2$for some$k_1,k_2\\in\\mathbb{Z}. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only ifk_1+k_2$is even. Suppose that$k_1+k_2$is even, so that$x+3y\\equiv", "4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} , \\quad (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We see that the numbers in the parentheses between both cycles are difference. Hence $(23)$ and $(165)$ are disjoint cycles. Also notice that: (2) \\begin{align} \\quad (23) \\circ (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 3 & 2 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We also have that: (3) \\begin{align} \\quad (165) \\circ (23) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ = \\begin{pmatrix} 1 & 2 &", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "the same variables together, we have \\begin{align*} 25a+11b &\\equiv 13 \\pmod{26}\\\\ 19a+8b &\\equiv 12 \\pmod{26} \\end{align*} \\begin{align*} 25c+11d &\\equiv 20 \\pmod{26}\\\\ 19c+8d &\\equiv 1 \\pmod{26} \\end{align*} Now, notice (and this is a nice trick) that we can write the two systems of congruences as matrix congruences. That is to say, $$\\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix} \\begin{pmatrix}a\\\\b\\end{pmatrix} \\equiv \\begin{pmatrix}13\\\\12\\end{pmatrix} \\textrm{ and } \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}\\begin{pmatrix}c\\\\d\\end{pmatrix}\\equiv\\begin{pmatrix}20\\\\1\\end{pmatrix} \\pmod{26}$$ Both of these can be solved by applying the inverse of the matrix involved to both sides of the congruence... $$\\begin{pmatrix}a\\\\b\\end{pmatrix} \\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}13\\\\12\\end{pmatrix} \\textrm{ and } \\begin{pmatrix}c\\\\d\\end{pmatrix} \\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}20\\\\1\\end{pmatrix} \\pmod{26}$$ We just need to find the inverse matrix in question. As seen previously, we can find the inverse of a matrix with the formula $$M^{-1} = \\begin{bmatrix}x & y\\\\z & w\\end{bmatrix}^{-1} = \\textrm{det}^{-1}\\begin{bmatrix}w & -y\\\\-z & x\\end{bmatrix}$$ Finding the determinant$\\pmod{26}$ is easy enough, $$\\textrm{det}=25 \\cdot 8 - 19 \\cdot 11 = -9 \\equiv 17 \\pmod{26}$$ but to find $\\textrm{det}^{-1}$, since we are working with congruences instead of equations, we need to find the multiplicative inverse of 17$\\pmod{26}$. Recall that the product of any number and its multiplicative inverse must be the multiplicative identity, 1, so we need to solve $$17m \\equiv 1 \\pmod{26}$$ This in turn, leads to solving (in integers)", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "gives a few basic definitions. An $$n\\times n$$ matrix is an $$n\\times n$$ \"box'' of numbers. For example, the matrix belowis a $$2\\times 2$$ matrix: \\begin{align*} \\begin{pmatrix} 1 & -4 \\\\ 5 & 0 \\end{pmatrix}. \\end{align*} And the matrix below is a $$3\\times 3$$ matrix: \\begin{align*} \\begin{pmatrix} 1 & -4 & 6\\\\ 5 & 0 & 6\\\\ 2 & 1 & -1 \\end{pmatrix}. \\end{align*} Matrices can be \"rectangular\" (e.g. the one below is $$2\\times 3$$) \\begin{align*} \\begin{pmatrix} 1 & -2 & 8\\\\ 0 & 1 & 0 \\end{pmatrix} \\end{align*} But in ODEs square matrices are the ones that come up and so those are the ones we will deal with here. A better way to think about an $$n\\times n$$ matrix (at least for our purposes) is to think of it as $$n$$ vectors in $$\\mathbb{R}^n$$. So, the matrix: \\begin{align*} \\begin{pmatrix} 1 & -4 & 6\\\\ 5 & 0 & 6\\\\ 2 & 1 & -1 \\end{pmatrix} \\end{align*} can be thought of as an arrangement of the vectors: \\begin{align*} \\begin{pmatrix}1\\\\5\\\\2\\end{pmatrix}, \\begin{pmatrix}-4\\\\0\\\\1\\end{pmatrix}, \\begin{pmatrix}6\\\\6\\\\-1\\end{pmatrix}. \\end{align*} Note: order matters. Similarly, if: \\begin{align*} \\boldsymbol{v}_1 = \\begin{pmatrix}1\\\\5\\end{pmatrix} \\hspace{.5in} \\boldsymbol{v}_2 = \\begin{pmatrix}-4\\\\0\\end{pmatrix} \\end{align*} Then the matrix $$\\begin{pmatrix}\\boldsymbol{v}_1 & \\boldsymbol{v}_2\\end{pmatrix}$$ means: \\begin{align*} \\begin{pmatrix} 1 & -4\\\\5 & 0 \\end{pmatrix}. \\end{align*} Returning to the equation in the beginning, we write the equation as the matrix–vector equation:", "&\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}13\\\\12\\end{pmatrix}\\\\ &\\equiv \\begin{bmatrix}2 & 7\\\\5 & 3\\end{bmatrix} \\begin{pmatrix}13\\\\12\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}2 \\cdot 13 + 7 \\cdot 12\\\\5 \\cdot 13 + 3 \\cdot 12\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}110\\\\101\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}6\\\\23\\end{pmatrix} \\end{align*} So $a=6$ and $b=23$. In a similar manner, we can solve for $c$ and $d$... \\begin{align*} \\begin{pmatrix}c\\\\d\\end{pmatrix} &\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}20\\\\1\\end{pmatrix}\\\\ &\\equiv \\begin{bmatrix}2 & 7\\\\5 & 3\\end{bmatrix} \\begin{pmatrix}20\\\\1\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}2 \\cdot 20 + 7 \\cdot 1\\\\5 \\cdot 20 + 3 \\cdot 1\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}47\\\\103\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}21\\\\25\\end{pmatrix} \\end{align*} So $c=21$ and $d=25$. Now we have our (decrypting) matrix! $$D = \\begin{bmatrix}a & b\\\\c & d\\end{bmatrix} = \\begin{bmatrix}6 & 23\\\\21 & 25\\end{bmatrix}$$ Decrypting the rest of the message is now easy! We simply convert the letter blocks to vectors and apply $D$ to them, and out pops the vectors corresponding to the decrypted letter blocks! So, for example, consider the third letter block \"ZO\", which we suspect should decrypt to \"ER\". Does it? \\begin{align*} D(\\textrm{\"ZO\"}) &= \\begin{bmatrix}6 & 23\\\\21 & 25\\end{bmatrix} \\begin{pmatrix}25\\\\14\\end{pmatrix} \\\\ &= \\begin{pmatrix}6 \\cdot 25 + 23 \\cdot 14\\\\21 \\cdot 25 + 25 \\cdot 14\\end{pmatrix} \\\\ &= \\begin{pmatrix}472\\\\875\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}4\\\\17\\end{pmatrix} \\pmod{26} \\end{align*} which, is the letter block \"ER\"! Looks like our guess about the word \"NUMBER\" was right!", "/ a^* = e^{2 i \\phi}$ for some $\\phi$ real – reuns Oct 7 '16 at 14:24 ## 1 Answer \\begin{align*} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} &= \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} \\bar{d} & -\\bar{b} \\\\ -\\bar{c} & \\bar{a} \\end{pmatrix} \\\\ &= \\begin{pmatrix} a\\bar{d}-b\\bar{c} & b\\bar{a}-a\\bar{b} \\\\ c\\bar{d}-d\\bar{c} & d\\bar{a}-c\\bar{b} \\end{pmatrix} \\end{align*} Equating \"$0$\": \\left \\{ \\begin{align*} \\frac{a}{\\bar{a}} &= \\frac{b}{\\bar{b}} \\\\ \\frac{c}{\\bar{c}} &= \\frac{d}{\\bar{d}} \\\\ \\end{align*} \\right. Let \\left \\{ \\begin{align*} a &= Ae^{i\\theta} \\\\ b &= Be^{i\\theta} \\end{align*} \\right. and \\left \\{ \\begin{align*} c &= Ce^{i\\phi} \\\\ d &= De^{i\\phi}\\ \\end{align*} \\right. where $A,B,C,D\\in \\mathbb{R}$ Equating \"$1$\": \\left \\{ \\begin{align*} (AD-BC)e^{i(\\theta-\\phi)} &= 1 \\\\ (AD-BC)e^{i(\\phi-\\theta)} &= 1 \\\\ \\end{align*} \\right. \\implies \\left \\{ \\begin{align*} AD-BC &= 1 \\\\ \\theta &= \\phi \\\\ \\end{align*} \\right.", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "& 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 3\\mod 13$ and $y\\equiv 8\\mod 13$, $x$ and $y$ can be written as $x=3+13k_1$ and $y=8+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Suppose that $k_1+k_2$ is even, so that $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 15+13(k_1+k_2)\\mod 26\\\\ &\\equiv 15\\mod 26\\\\ &\\not\\equiv 2\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are no solutions for part 2. 3. Similarly to part 2, the solution in $\\mathbb{Z}/13\\mathbb{Z}$ is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "\\\\ 0 & 0 \\end{pmatrix}} -c {\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}} \\\\ &= bE_3 - cE_2 \\end{align*} How do I continu to find $M_{E_1}$? Remember that given your basis $\\xi$, for $B \\in M_2(\\mathbb{R})$, $$[B]_\\xi = \\begin{pmatrix}B_{11} \\\\ B_{21} \\\\ B_{12} \\\\ B_{22} \\end{pmatrix};$$ in particular, for $A \\in M_2(\\mathbb{R})$, $M_A := [L_A]_{\\xi \\xi}$ is the matrix of the linear transformation $L_A : M_2(\\mathbb{R}) \\to M_2(\\mathbb{R})$ with respect to $\\xi$, and hence a $4 \\times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \\times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\\xi$, as given by the equation above. So, given $A \\in M_2(\\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$. Hint:do you know that the matrix you are looking for is $4\\times4$? Now you are looking for a $4\\times4$ matrix such that $M_{E_1}[X]_\\xi=[L_{E_1}(X)]_\\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\\xi=(0,b,-c,0)$ column vector, while $[X]_\\xi=(a,b,c,d)$ column. Do you see how to proceed?", "given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$ or $k_1=k_2=1$. Therefore, the solutions are \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 26\\ \\mbox{or}\\ \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 19\\\\ 20 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} \\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ with $(D,N)=1$ where $D=ad-bc$ be an enciphering matrix. Then it defines the enciphering transformation $C=AP$ where $P=\\begin{pmatrix} x\\\\ y \\end{pmatrix}$ and $C=\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}$ are plaintext message unit and ciphertext, respectively. That is, $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}$$ The deciphering transformation is then given by $P=A^{-1}C$, i.e. $$\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} D^{-1}d & -D^{-1}b\\\\ -D^{-1}c &", "theorem without proof. Theorem. Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ and set $D=ad-bc$ as before. Then the following are equivalent: 1. $(D,N)=1$. 2. $A$ has an inverse matrix. 3. if $x$ and $y$ are not both $0$ in $\\mathbb{Z}/N\\mathbb{Z}$, then $A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4. $A$ gives a one-to-one correspondence of $(\\mathbb{Z}/N\\mathbb{Z})^2$ with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruence $AX\\equiv B\\mod 26$ or as the matrix equation $AX=B$ in $\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$ $D=16-21=-5$ and $(-5,26)=1$, so by above Theorem, we see that $A^{-1}$ exists. In fact, we have already found $A^{-1}$ in the previous example. The solution $X$ is given by \\begin{align*} X&\\equiv A^{-1}B\\mod 26\\\\&\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 26\\\\ &\\equiv\\begin{pmatrix} 10\\\\ 11 \\end{pmatrix}\\mod 26 \\end{align*} 2. The matrix in parts 2 and 3 does not have an inverse matrix in $\\mathbb{Z}/26\\mathbb{Z}$ because $D=9-21=-12$ and $(-12,26)=2\\ne 1$. However, it does have an inverse matrix in $\\mathbb{Z}/13\\mathbb{Z}$ and the solution is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6", "form: \\begin{align*} \\begin{pmatrix}t\\\\t\\end{pmatrix} = t\\begin{pmatrix}1\\\\1\\end{pmatrix} \\end{align*} So we may take $$\\boldsymbol{v}_2 = \\begin{pmatrix}1\\\\1\\end{pmatrix}.$$ ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L5E1 2. Find the eigenvalues and eigenvectors of the following matrix: \\begin{align*} \\begin{pmatrix} -2 & 1\\\\ 1 & -2 \\end{pmatrix} \\end{align*} First we find the eigenvalues: \\begin{align*} =\\det\\begin{pmatrix}-2-\\lambda & 1\\\\ 1 & -2-\\lambda\\end{pmatrix} =(\\lambda +2)^2 - 1. \\end{align*} So the eigenvalues are $$\\lambda_1 = -3$$ and $$\\lambda_2 = -1$$. Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue. First, for $$\\lambda_1$$ we find (non–zero) solutions to: \\begin{align*} \\begin{pmatrix} 1 & 1\\\\1 & 1 \\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\end{pmatrix} =\\begin{pmatrix}0\\\\0\\end{pmatrix}. \\end{align*} The solutions are $$x_1 + x_2 = 0$$ or $$x_1 = -x_2$$. That is, vectors of the form: \\begin{align*} \\begin{pmatrix}t\\\\-t\\end{pmatrix}=t\\begin{pmatrix}1\\\\-1\\end{pmatrix}. \\end{align*} So we may take $$\\boldsymbol{v}_1 = \\begin{pmatrix}1\\\\-1\\end{pmatrix}$$. Next, for $$\\lambda_2$$ we find (non–zero) solutions to: \\begin{align*} \\begin{pmatrix} -1 & 1\\\\1 & -1 \\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\end{pmatrix} =\\begin{pmatrix}0\\\\0\\end{pmatrix}. \\end{align*} The solutions are $$x_1 - x_2 = 0$$ or $$x_1 = x_2$$. That is, vectors of the form: \\begin{align*} \\begin{pmatrix}t\\\\t\\end{pmatrix} = t\\begin{pmatrix}1\\\\1\\end{pmatrix} \\end{align*} So we may take $$\\boldsymbol{v}_2 = \\begin{pmatrix}1\\\\1\\end{pmatrix}.$$ ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L5E2 3. Find the eigenvalues and eigenvectors of the following matrix:", "# Find 2 by 3 matrix M such that M $\\begin{pmatrix}a \\\\ b \\\\ c\\end{pmatrix}= \\begin{pmatrix}d \\\\ e\\end{pmatrix}$ whenever $(ax^2+bx + c)'=dx+e$ Find 2 by 3 matrix M such that M $\\begin{pmatrix}a \\\\ b \\\\ c\\end{pmatrix}= \\begin{pmatrix}d \\\\ e\\end{pmatrix}$ whenever $(ax^2+bx + c)'=dx+e$. Here is my take on it. \\begin{align*}(ax^2+bx+c)' = 2ax +b = dx+e\\end{align*} which gives $a = \\frac{d}{2}$ and $b = e$. So we have, \\begin{align*}\\begin{pmatrix}m_1 & m_2 & m_3 \\\\ m_4 & m_5 & m_6\\end{pmatrix}\\begin{pmatrix}a \\\\ b \\\\ c\\end{pmatrix} = \\begin{pmatrix}am_1+bm_2+cm_3\\\\ am_4 + bm_5 + cm_6 \\end{pmatrix}=\\begin{pmatrix}2a \\\\ b\\end{pmatrix}\\end{align*}After solving the above equation, I got $m_1 +m_4 = 2, m_2 + m_5 = 1$ and $m_3 + m_6 = 0$. I can just set $m_1, m_2$ and $m_3$ to be free variable but I don't think this is the right answer. And I am not sure if I am going to the right direction either. It will be really helpful for me if someone can point me to the right direction. • How'd you get $a=\\frac{d}{2}$ and $b=e$ – user76568 Jan 31 '18 at 1:41 • $2ax + b = dx + e$ so I figure $2a = d$ and $e = b$. – abuchay Jan 31 '18 at 1:42 • You think $x$ is fancier? :) You can't conclude $b=e$ – user76568 Jan 31", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "find the vector projection a onto b and theorthogonal projection of a onto b.1) a =< 4, 3 >, b =< 0, 5 >2) a =< 4, 3 >, b =< 0, −5 > Part A: find the vector projection b onto a and the orthogonal projection of b onto a. 1) a =< 4, 3 >, b =< 0, 5 > 2) a =< 4, 3 >, b =< 0, −5 > Part B: find the vector projection a onto b and the orthogonal projection of a onto b. 1) a =< 4, 3 >, b =< 0, 5...", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "6. vector $$\\vec{f}$$: 7. vector $$\\vec{g}$$: 8. vector $$\\vec{h}$$: Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. We find vector $$\\vec{a} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$. 2. We find vector $$\\vec{b} = \\begin{pmatrix} 5 \\\\ -1 \\end{pmatrix}$$. 3. We find vector $$\\vec{c} = \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix}$$. 4. We find vector $$\\vec{d} = \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$. 5. We find vector $$\\vec{e} = \\begin{pmatrix} -5 \\\\ 0 \\end{pmatrix}$$. 6. We find vector $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$. 7. We find vector $$\\vec{g} = \\begin{pmatrix} 4 \\\\ -6 \\end{pmatrix}$$. 8. We find vector $$\\vec{h} = \\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix}$$. Scan this QR-Code with your phone/tablet and view this page on your preferred device. ### Subscribe to Our Channel Subscribe Now and view all of our playlists & tutorials.", "+0 0 61 2 Consider the projection matrix P that satisfies Pv = Projection of v onto $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$. Find 3 vectors with integer components and magnitude no greater than 8 that satisfy Pv = $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$. Use this to help you find $$P^{-1}\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ . Dec 30, 2019 #1 0 You can take the vectors [3,4], [4,1], and [5,-2]. Dec 30, 2019 #2 0 I tries this answer and it is incorrect. Thank you though. Guest Dec 30, 2019", "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "# Orthogonal projection onto span of vectors using weighted inner product • Jul 14th 2008, 07:11 PM JCS007 Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ • Jul 14th 2008, 09:43 PM CaptainBlack Quote: Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "the vector $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ 3. the vector $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ 4. the vector $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ Note: you can download a sheet of gridded paper here. Note: this exercise can be downloaded as a worksheet to practice with: ### Solutions 1. For $$\\vec{a} = \\begin{pmatrix} 5 \\\\ 2 \\end{pmatrix}$$ we find: 2. For $$\\vec{b} = \\begin{pmatrix} -3 \\\\ 4 \\end{pmatrix}$$ we find: 3. For $$\\vec{c} = \\begin{pmatrix} 6 \\\\ -3 \\end{pmatrix}$$ we find: 4. For $$\\vec{d} = \\begin{pmatrix} 0 \\\\ 5 \\end{pmatrix}$$ we find: 5. For $$\\vec{e} = \\begin{pmatrix} -2 \\\\ -5 \\end{pmatrix}$$ we find: 6. For $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ we find: 7. For $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ we find: 8. For $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ we find: ### Tutorial: How to find a vector's components In this tutorial we learn what a vector is. We learn how to write vectors as both column and row vectors as well as how to represent vectors graphically; that is how to draw vectors. ## Exercise 2 Writing your answers as column vectors, find the coordinates of each of the vectors drawn here: 1. vector $$\\vec{a}$$: 2. vector $$\\vec{b}$$: 3. vector $$\\vec{c}$$: 4. vector $$\\vec{d}$$: 5. vector $$\\vec{e}$$:", "# How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \\operatorname{span}\\left\\{ \\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix},\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix}\\right\\}$ and the vector $v = \\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix}$ How can I find the orthogonal projection of $v$ on $V$? This is what I did so far: \\begin{align}&P_v(v)=\\langle v,v_1\\rangle v_1+\\langle v,v_2\\rangle v_2 =\\\\=& \\left\\langle\\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix},\\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix}\\right\\rangle\\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix}+\\left<\\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix},\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix}\\right>\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix} = \\begin{pmatrix}10 \\\\29 \\\\38\\end{pmatrix}\\end{align} Is this the right method to compute this? That would be the correct method...if $v_1$ and $v_2$ were orthogonal and unit length. Unfortunately, they're not. Three alternatives: 1. Compute $w = v_1 \\times v_2$, and the projection of $v$ onto $w$ -- call it $q$. Then compute $v - q$, which will be the desired projection. 2. Orthgonalize $v_1$ and $v_2$ using the gram-schmidt process, and then apply your method. 3. Write $q = av_1 + bv_2$ as the proposed projection vector. You then want $v - q$ to the orthogonal to both $v_1$ and $v_2$. This gives you two equations in the unknowns $a$ adn $b$, which you can solve. Hint You have to construct by the Gram Schmidt procedure an orthonormal basis $(e_1,e_2)$ from the given basis of $V$ and then $$P_v(V)=\\langle v,e_1\\rangle e_1+\\langle v,e_2\\rangle e_2$$ Since codimension of $V$ is one, in", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "7 \\end{pmatrix}$$ The answer-vector is $\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$ away from $\\begin{pmatrix} 12 \\\\ 6 \\end{pmatrix}$ and $\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$ is a third of the sum of the x-vector and the y-vector. So if we do $6\\frac{1}{3}$ of the x-vector and $\\frac{1}{3}$ of the y-vector we'll get the right answer. That is, $x = 6\\frac{1}{3} = \\frac{19}{3}$ and $y = \\frac{1}{3}$. As you can see, there's no \"method\" per se, but thinking of it as a vector equation can help. -" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "you perform the projection as outlined above, and then perform the rotations again to get things back to the new location of the line. \\begin{align*} X_P &= \\begin{pmatrix}1&0&0\\\\0&\\cos P&-\\sin P\\\\0&\\sin P&\\cos P\\end{pmatrix} \\\\ Y_Q &= \\begin{pmatrix}\\cos Q&0&\\sin Q\\\\0&1&0\\\\-\\sin Q&0&\\cos Q\\end{pmatrix} \\\\ T &= Y_Q\\cdot X_P\\cdot M\\cdot X_P^{-1}\\cdot Y_Q^{-1} \\\\ &= Y_Q\\cdot X_P\\cdot M\\cdot X_{-P}\\cdot Y_{-Q} \\\\ &= \\small\\begin{pmatrix} \\sin^2Q\\cdot\\cos^2 P & -\\sin P\\cdot\\sin Q\\cdot\\cos P & \\sin Q\\cdot\\cos^2P\\cdot\\cos Q \\\\ -\\sin P\\cdot\\sin Q\\cdot\\cos P & \\sin^2P & -\\sin P\\cdot\\cos P\\cdot\\cos Q \\\\ \\sin Q\\cdot\\cos^2 P\\cdot\\cos Q & -\\sin P\\cdot\\cos P\\cdot\\cos Q & \\cos^2P\\cdot\\cos^2Q \\end{pmatrix} \\end{align*} This way, you get a single matrix describing your whole operation. -", "\\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\qquad \\vec{b} = \\begin{pmatrix} 0\\\\2 \\end{pmatrix}\\\\\\\\ \\begin{array}{lll} \\text{area}=\\left|\\vec{a}_{projection}\\times\\vec{b} \\right| & \\qquad \\vec{a}_{projection} = \\left(\\vec{a}\\cdot \\vec{e}_x\\right)\\cdot \\vec{e}_x & \\qquad \\vec{e}_x = \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\\\\\\\ & \\qquad \\vec{a}_{projection} =\\left( 1.5\\cdot \\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\right)\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\left( 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )} +0\\right) \\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix} \\\\\\\\ \\text{area}=\\left|\\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix}\\times\\begin{pmatrix} 0\\\\2 \\end{pmatrix} \\right| \\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\cos{ (60\\ensurement{^{\\circ}} ) } \\qquad | \\qquad \\cos{ (60\\ensurement{^{\\circ}} ) } = \\dfrac{1}{2}\\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\dfrac{1}{2}\\\\\\\\ \\mathbf{\\text{area}=1.5 ~\\mathrm{m^2}}\\\\\\\\ \\end{array}$$ heureka Jul 6, 2015 #1 +10 I will make an assumption that the tarp is 1.5 m TALL and 2 m WIDE. One side of the shadow will remain the same if the sun is shining directly down: 2m To calculate the OTHER side of this shadow rectangle, The hypotenuse of the triangle is 1.5 m . We need the cosine of 60 dgrees times the hypotenuse to calculate the other side of the shadow COS(60) * 1.5 =.75 So you have a rectangle that is 2m x .75m Area = 2 * .75 = 1.5 sq m Guest Jul 5, 2015 #2 +19653 +10 Two math students erect a", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "then at the end my constraint about having all the entries sum up to 1 doesn't hold anymore in case I have negative entries. – user2987 Nov 8 '16 at 15:03 • If you do the update $s_{i,j}^{(k+1)} = s_{i,j}^{(k)} - \\eta \\delta f(s_{i,j}^{(k)})$ and all entries become negative, what is the projection on the feasible region? In $\\mathbb{R}$ and $\\mathbb{R^2}$ this should be easy to figure out. Alternatively, Google for \"projection on the simplex\". – LinAlg Nov 8 '16 at 17:04 • The first line of Section 2 says $X=\\mathbb{R}^n$. – LinAlg Nov 8 '16 at 22:05 You have 2 options: 1. Use direct projection onto the Unit Simplex which the intersection of the 2 sets you mentioned (See Orthogonal Projection onto the Unit Simplex). 2. Use Alternating Projection on each of the sets you defined. The projection onto the Non Negative Orthant set: $$\\operatorname{Proj}_{\\mathcal{S}} \\left( x \\right) = \\max \\left\\{ 0, x \\right\\}$$ Where $S = \\left\\{ x \\in \\mathbb{R}^{n} \\mid x \\succeq 0 \\right\\}$; The projection onto the set $\\mathcal{C} = \\left\\{ x \\in \\mathbb{R}^{n} \\mid \\boldsymbol{1}^{T} x = 1 \\right\\}$: $$\\operatorname{Proj}_{\\mathcal{C}} \\left( x \\right) = x - \\frac{\\sum_{i} {x}_{i} - 1}{n} \\boldsymbol{1}$$ Pay attention that this is not what you had in your solution. To show the 2 methods in practice one could use", "Projection - Axonometric Projection \\begin{align} \\mathtt{A} \\mathtt{T}_{\\mathrm{\\small cam}}^{-1} & = \\mathtt{A} \\, \\mathtt{T}_{r_x}(\\beta) \\, \\mathtt{T}_{r_y}(\\alpha)\\\\ & = \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0& \\cos \\beta & -\\sin \\beta& 0 \\\\ 0 & \\sin \\beta & \\cos \\beta & 0 \\\\ 0 & 0 & 0 &1 \\\\ \\end{bmatrix} \\begin{bmatrix} \\cos \\alpha & 0& \\sin \\alpha& 0 \\\\ 0 & 1 & 0 & 0 \\\\ -\\sin \\alpha & 0 &\\cos \\alpha & 0 \\\\ 0 & 0 & 0 &1 \\\\ \\end{bmatrix} \\\\ & = \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} \\cos \\alpha & 0 & \\sin \\alpha & 0 \\\\ \\sin \\beta \\sin \\alpha & \\cos \\beta & -\\sin \\beta \\cos \\alpha & 0\\\\ - \\cos \\beta \\sin \\alpha & \\sin \\beta & \\cos \\beta \\cos \\alpha & 0 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{bmatrix} \\\\ & = \\begin{bmatrix} \\cos \\alpha & 0 & \\sin \\alpha & 0", "+0 # dumb precalculus 0 45 1 For $$\\bold{v} = \\begin{pmatrix} \\phantom -2 \\\\ \\phantom -3 \\\\ -1 \\end{pmatrix} and \\bold{w} = \\begin{pmatrix} \\phantom -2 \\\\ -1 \\\\ \\phantom -0 \\end{pmatrix}$$ compute the projection of $\\mathbf v$ onto $\\mathbf w.$ i literally dont know how to solve this Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ MaxWong Apr 18, 2022", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "\\right\\rbrace }$ Example (plane through the origin as a span) Let ${\\displaystyle (5,0,0)^{T}}$ and ${\\displaystyle (0,3,0)^{T}}$ be two vectors from ${\\displaystyle \\mathbb {R} ^{3}}$. The span of these two vectors is the ${\\displaystyle xy}$-plane. The following transformation shows {\\displaystyle {\\begin{aligned}\\operatorname {span} \\left(\\left\\lbrace {\\begin{pmatrix}5\\\\0\\\\0\\end{pmatrix}},{\\begin{pmatrix}0\\\\3\\\\0\\end{pmatrix}}\\right\\rbrace \\right)&=\\left\\lbrace \\lambda \\cdot {\\begin{pmatrix}5\\\\0\\\\0\\end{pmatrix}}+\\mu \\cdot {\\begin{pmatrix}0\\\\3\\\\0\\end{pmatrix}}{\\Bigg |}\\ \\lambda ,\\mu \\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&=\\,\\left\\lbrace 5\\cdot \\lambda \\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+3\\cdot \\mu \\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}{\\Bigg |}\\ \\lambda ,\\mu \\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&\\ {\\color {OliveGreen}\\left\\downarrow {\\text{set }}{{\\tilde {\\lambda }}=5\\cdot \\lambda ,\\ {\\tilde {\\mu }}=3\\cdot \\mu }\\ {\\text{ bzw. }}\\ \\lambda ={\\frac {1}{5}}\\cdot {\\tilde {\\lambda }},\\ \\mu ={\\frac {1}{3}}\\cdot {\\tilde {\\mu }}\\right.}\\\\[0.3em]&=\\,\\left\\lbrace {\\tilde {\\lambda }}\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+\\cdot {\\tilde {\\mu }}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}{\\Bigg |}\\ {\\tilde {\\lambda }},{\\tilde {\\mu }}\\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&=\\,\\left\\{{\\begin{pmatrix}{\\tilde {\\lambda }}\\\\{\\tilde {\\mu }}\\\\0\\end{pmatrix}}{\\Bigg |}\\ {\\tilde {\\lambda }},{\\tilde {\\mu }}\\in \\mathbb {R} \\right\\}\\end{aligned}}} ## Overview: Properties of the span Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space, ${\\displaystyle M}$, ${\\displaystyle N\\subseteq V}$ subsets of ${\\displaystyle V}$ and ${\\displaystyle W\\subseteq V}$ a subspace of ${\\displaystyle V}$. Then, we have • For a vector ${\\displaystyle v\\in V}$ we have ${\\displaystyle \\operatorname {span} (\\{v\\})=\\{\\lambda \\cdot v|\\lambda \\in K\\}}$ • If ${\\displaystyle N\\subseteq M}$, then ${\\displaystyle \\operatorname {span} (N)\\subseteq \\operatorname {span} (M)}$ • From ${\\displaystyle \\operatorname {span} (N)=\\operatorname {span} (M)}$ one can usually not conclude ${\\displaystyle N=M}$ • ${\\displaystyle M\\subseteq \\operatorname {span} (M)}$ •", "# Prove that perspective projection of circle is ellipse How to prove that perspective projection of a circle is an ellipse? I start with the parametric equation of circle and ellipse: Circle: $x = r\\cos t$ $y = r\\sin t$ Ellipse: $x = a\\cos(t)$ $y = b\\sin(t)$ Then I have perspective projection matrix: $$\\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & 1 \\\\ \\end{pmatrix}$$ And obtain projected coordinates (affine transform for simplicity): $$\\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ 0 & 0 & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} u \\\\ v \\\\ 1 \\\\ \\end{pmatrix}$$ If we convert this to equations: $u = a_{11}x+a_{12}y+a_{13}$ $v = a_{21}x+a_{22}y+a_{23}$ Substituting $x,y$ with parametric equations: $a\\cos t = a_{11}\\,r\\, \\cos t + a_{12}\\, r \\,\\sin t + a_{13}$ $b \\sin t = a_{21}\\,r \\,\\cos t + a_{22}\\, r \\,\\sin t + a_{23}$ What to do next? I can't see how these equations can be equal. • Please write a_{11} for indexed variables. You could also consider removing the multiplication operator * because in modern mathematics this operator is implicitly implied if the variables are written without an operator in the middle of them. If the spacing is to", "\\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 \\end{pmatrix} = \\begin{pmatrix}0\\\\1\\\\0 \\end{pmatrix} \\\\ \\phi(|{-i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right)\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0", "the representation {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}&=a\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+b\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&=a\\cdot \\left({\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+b\\cdot \\left({\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}-{\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&={\\frac {a+2b}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {a-2b}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\end{aligned}}} Thus ${\\displaystyle (a,b,c)^{T}\\in \\operatorname {span} (\\{v_{1},v_{2},v_{3}\\})}$. Since this vector was arbitrarily chosen, every vector from ${\\displaystyle \\mathbb {R} ^{3}}$ is representable as a linear combination of the linearly independent vectors ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. Thus ${\\displaystyle \\{v_{1},v_{2},v_{3}\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{3}}$. Therefore, we cannot add another vector to ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$, so the system remains linearly independent, since every other vector from ${\\displaystyle \\mathbb {R} ^{3}}$ can be represented as a linear combination of ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. In other words, ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$ form a maximal system of linearly independent vectors. In summary, we proceeded as follows to find a maximal system of linearly independent vectors: We start with a vector that is not the zero vector. That means we should not consider the null vector space here. Then we proceed step by step: Once we have found linearly independent vectors ${\\displaystyle v_{1},...,v_{n}}$, we form the span of these vectors. If this is the entire vector space, we have found a generator and are done. We choose a vector ${\\displaystyle v_{n+1}}$ that is not in ${\\displaystyle \\operatorname {span} (\\{v_{1},...,v_{n}\\})}$. This contributes a new direction", "|u \\rangle_1 |d \\rangle_2 - |d \\rangle_1 |u \\rangle_2 \\right]$ Written in tensor format the antisymmetric wave function is $| \\Psi_F \\rangle = \\frac{1}{ \\sqrt{2}} \\left[ \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\right] = frac{1}{ \\sqrt{2}} \\left[ \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ 0 \\end{pmatrix} - \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\right] = \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix}$ $\\Psi _F = \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix}$ Now the interaction with the beam splitter yields the following result. $|Psi _{Fout} = \\frac{1}{2} \\begin{pmatrix} 1 & i & i & 1 \\\\ i & 1 & -1 & i \\\\ i & -1 & 1 & i \\\\ -1 & i & i & 1 \\end{pmatrix} \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 0 \\\\ \\frac{ \\sqrt{2}}{2} \\\\ - \\frac{ \\sqrt{2}}{2} \\\\ 0 \\end{pmatrix}$ We see by inspection or calculation that fermionic photons are never observed in the same output channel after interaction with the beam splitter. $\\left( \\left| uu^T \\Psi_{Fout} \\right| \\right)^2 \\rightarrow 0$ $\\left( \\left| dd^T \\Psi_{Fout} \\right| \\right)^2 \\rightarrow 0$ Naturally this means they are always observed", "Perspective projection To get the gist of perspective projection consider the following image: Given the Point $P(p_x, p_y, p_z)$ what is are the coordinates of the projected point $P’(p’_x,p’_y,p’_z)$? $$p'_x = d \\frac{p_x}{p_z}$$ $$p'_y = d \\frac{p_y}{p_z}$$ $$p'_z = d$$ If $d = 1$ then the problem simplifies to just dividing by the z-coordinate. And this is what the whole magic about perspective projection is all about. It is just dividing by $z$. Now you do we put this into a matrix $M_P$? $$M_P \\cdot P = P''$$ $$\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} p_x \\\\ p_y \\\\ p_z \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} p''_x \\\\ p''_y \\\\ p''_z \\\\ p''_w \\end{pmatrix}$$ Applying homogeneous division results in: $$p'_x = \\frac{p''_x}{p''_z}$$ $$p'_y = \\frac{p''_y}{p''_z}$$ $$p'_z = 1$$ Field of view What is the field of view here (FOV)? $90°$. Why? Consider the following sketch: Assuming that $d=1$ and the horizontal field of view is $\\frac{h_{FOV}}{2}$ from trigonometry we know: $$\\tan(\\frac{h_{FOV}}{2}) = \\frac{S}{1}$$ What should we do if we want to have a different FOV? We can simple scale by the inverse of S: $$M_S = \\begin{pmatrix} \\frac{1}{S} & 0 & 0" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "# Orthogonal projection onto span of vectors using weighted inner product • Jul 14th 2008, 07:11 PM JCS007 Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ • Jul 14th 2008, 09:43 PM CaptainBlack Quote: Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "Projection Matrix - Maple Programming Help Home : Support : Online Help : Tasks : Linear Algebra : Matrix Manipulations : Task/ProjectionMatrix Projection Matrix Description Generate a projection matrix that will project a vector onto the subspace spanned by a list of vectors. Enter a list of vectors. > $\\left[{[}\\begin{array}{c}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}{]}{,}{[}\\begin{array}{c}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}{]}{,}{[}\\begin{array}{c}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}{]}\\right]$ $\\left[\\left[\\begin{array}{r}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}\\right]{,}\\left[\\begin{array}{r}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}\\right]{,}\\left[\\begin{array}{r}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}\\right]\\right]$ (1) Calculate the Projection Matrix. > $\\mathrm{LinearAlgebra}\\left[\\mathrm{ProjectionMatrix}\\right]\\left(\\right)$ $\\left[\\begin{array}{cccc}{1}& {0}& {0}& {0}\\\\ {0}& \\frac{{5}}{{6}}& \\frac{{1}}{{3}}& {-}\\frac{{1}}{{6}}\\\\ {0}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}\\\\ {0}& {-}\\frac{{1}}{{6}}& \\frac{{1}}{{3}}& \\frac{{5}}{{6}}\\end{array}\\right]$ (2) Commands Used", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "# How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \\operatorname{span}\\left\\{ \\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix},\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix}\\right\\}$ and the vector $v = \\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix}$ How can I find the orthogonal projection of $v$ on $V$? This is what I did so far: \\begin{align}&P_v(v)=\\langle v,v_1\\rangle v_1+\\langle v,v_2\\rangle v_2 =\\\\=& \\left\\langle\\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix},\\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix}\\right\\rangle\\begin{pmatrix}\\frac{1}{3} \\\\\\frac{2}{3} \\\\\\frac{2}{3}\\end{pmatrix}+\\left<\\begin{pmatrix}9 \\\\0 \\\\0\\end{pmatrix},\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix}\\right>\\begin{pmatrix}1 \\\\3 \\\\4\\end{pmatrix} = \\begin{pmatrix}10 \\\\29 \\\\38\\end{pmatrix}\\end{align} Is this the right method to compute this? That would be the correct method...if $v_1$ and $v_2$ were orthogonal and unit length. Unfortunately, they're not. Three alternatives: 1. Compute $w = v_1 \\times v_2$, and the projection of $v$ onto $w$ -- call it $q$. Then compute $v - q$, which will be the desired projection. 2. Orthgonalize $v_1$ and $v_2$ using the gram-schmidt process, and then apply your method. 3. Write $q = av_1 + bv_2$ as the proposed projection vector. You then want $v - q$ to the orthogonal to both $v_1$ and $v_2$. This gives you two equations in the unknowns $a$ adn $b$, which you can solve. Hint You have to construct by the Gram Schmidt procedure an orthonormal basis $(e_1,e_2)$ from the given basis of $V$ and then $$P_v(V)=\\langle v,e_1\\rangle e_1+\\langle v,e_2\\rangle e_2$$ Since codimension of $V$ is one, in", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "# Math Help - Orthogonal projection onto span of vectors using weighted inner product 1. ## Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ 2. Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "^{2}}$ is an ${\\displaystyle \\mathbb {R} }$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ### Important special cases Exercise (The identity is a linear map) Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space. Prove that the identity ${\\displaystyle \\operatorname {id} :V\\to V}$ with ${\\displaystyle \\operatorname {id} (v)=v}$ is a linear map. Proof (The identity is a linear map) The identity is additive: Let ${\\displaystyle v,w\\in V}$, then. ${\\displaystyle \\operatorname {id} (v+w)=v+w=\\operatorname {id} (v)+\\operatorname {id} (w)}$ The identity is homogeneous: Let ${\\displaystyle \\lambda \\in K}$ and ${\\displaystyle v\\in V}$, then ${\\displaystyle \\operatorname {id} (\\lambda \\cdot v)=\\lambda \\cdot v=\\lambda \\cdot \\operatorname {id} (v)}$ Exercise (The map to zero is a linear map) Let ${\\displaystyle V,W}$ be two ${\\displaystyle K}$-vector spaces. Show that", "Projection Matrix - Maple Programming Help Projection Matrix Description Generate a projection matrix that will project a vector onto the subspace spanned by a list of vectors. Enter a list of vectors. > $\\left[{[}\\begin{array}{c}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}{]}{,}{[}\\begin{array}{c}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}{]}{,}{[}\\begin{array}{c}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}{]}\\right]$ $\\left[\\left[\\begin{array}{r}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}\\right]{,}\\left[\\begin{array}{r}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}\\right]{,}\\left[\\begin{array}{r}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}\\right]\\right]$ (1) Calculate the Projection Matrix. > $\\mathrm{LinearAlgebra}\\left[\\mathrm{ProjectionMatrix}\\right]\\left(\\right)$ $\\left[\\begin{array}{cccc}{1}& {0}& {0}& {0}\\\\ {0}& \\frac{{5}}{{6}}& \\frac{{1}}{{3}}& {-}\\frac{{1}}{{6}}\\\\ {0}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}\\\\ {0}& {-}\\frac{{1}}{{6}}& \\frac{{1}}{{3}}& \\frac{{5}}{{6}}\\end{array}\\right]$ (2) Commands Used", "the inner product between vector $$\\vec{A}$$ and the normalized basis vector $$\\vec{\\epsilon}_1/\\vert \\vec{\\epsilon}_1\\vert$$. Using the same approach with $$\\widehat{P}_2$$ and $$\\widehat{P}_3$$ gives \\begin{equation*} \\widehat{P}_2\\ket{A}=\\frac{1}{45}\\begin{pmatrix}9&18&0\\\\18&36&0\\\\0&0&0\\end{pmatrix}\\begin{pmatrix}7\\\\-3\\\\2\\end{pmatrix}=\\frac{1}{45}\\begin{pmatrix}9\\\\18\\\\0\\end{pmatrix}, \\end{equation*} which has magnitude \\begin{equation*} \\vert \\widehat{P}_2\\ket{A} \\vert=\\sqrt{\\left(\\frac{9}{45}\\right)^2+\\left(\\frac{18}{45}\\right)^2+\\left(\\frac{0}{45}\\right)^2}=0.45 \\end{equation*} and \\begin{equation*} \\widehat{P}_3\\ket{A}=\\begin{pmatrix}0&0&0\\\\0&0&0\\\\0&0&1\\end{pmatrix}\\begin{pmatrix}7\\\\-3\\\\2\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\\\2\\end{pmatrix}. \\end{equation*} which has magnitude \\begin{equation*} \\vert \\widehat{P}_3\\ket{A} \\vert=\\sqrt{0^2+0^2+2^2}=2. \\end{equation*} As described in Section 2.4, applying a projection operator to a vector produces a new vector in the direction of the vector used to form the projection operator. Application of a projection operator to a vector is illustrated in Eq. 2.42: \\begin{equation*} \\widehat{P}_1\\ket{A}=\\ket{\\epsilon_1}\\bra{\\epsilon_1|A}=A_1\\ket{\\epsilon_1}, \\end{equation*} and inserting the projection operator$$\\widehat{P}_1$$ from Problem 7 to vector$$\\vec{A}=7\\hat{\\imath}-3\\hat{\\jmath}+2\\hat{k}$$ looks like this: \\begin{equation*} \\widehat{P}_1\\ket{A}=\\frac{1}{20}\\begin{pmatrix}16&-8&0\\\\-8&4&0\\\\0&0&0\\end{pmatrix}\\begin{pmatrix}7\\\\-3\\\\2\\end{pmatrix}. \\end{equation*} Performing the matrix multiplication gives \\begin{equation*} \\widehat{P}_1\\ket{A}=\\frac{1}{20}\\begin{pmatrix}136\\\\-68\\\\0\\end{pmatrix}, \\end{equation*} which is a vector in the direction of $$\\vec{\\epsilon}_1$$ (as you can tell by the ratio of the components) and with magnitude equal to the projection of vector $$\\vec{A}$$ onto the direction of $$\\vec{\\epsilon}_1$$. That magnitude is \\begin{equation*} \\vert \\widehat{P}_1\\ket{A} \\vert=\\sqrt{\\left(\\frac{136}{20}\\right)^2+\\left(\\frac{-68}{20}\\right)^2+\\left(\\frac{0}{20}\\right)^2}=7.6 \\end{equation*} which you can verify by taking the inner product between vector $$\\vec{A}$$ and the normalized basis vector $$\\vec{\\epsilon}_1/\\vert \\vec{\\epsilon}_1\\vert$$. Using the same approach with $$\\widehat{P}_2$$ and $$\\widehat{P}_3$$ gives \\begin{equation*} \\widehat{P}_2\\ket{A}=\\frac{1}{45}\\begin{pmatrix}9&18&0\\\\18&36&0\\\\0&0&0\\end{pmatrix}\\begin{pmatrix}7\\\\-3\\\\2\\end{pmatrix}=\\frac{1}{45}\\begin{pmatrix}9\\\\18\\\\0\\end{pmatrix}, \\end{equation*} which has magnitude \\begin{equation*} \\vert \\widehat{P}_2\\ket{A} \\vert=\\sqrt{\\left(\\frac{9}{45}\\right)^2+\\left(\\frac{18}{45}\\right)^2+\\left(\\frac{0}{45}\\right)^2}=0.45 \\end{equation*} and \\begin{equation*} \\widehat{P}_3\\ket{A}=\\begin{pmatrix}0&0&0\\\\0&0&0\\\\0&0&1\\end{pmatrix}\\begin{pmatrix}7\\\\-3\\\\2\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\\\2\\end{pmatrix}. \\end{equation*} which has magnitude \\begin{equation*} \\vert \\widehat{P}_3\\ket{A} \\vert=\\sqrt{0^2+0^2+2^2}=2. \\end{equation*} So as you may have guessed by comparing the components of vector", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "# Projection of the vector $\\vec{b}=\\left(\\begin{smallmatrix}3\\\\-1\\\\4\\end{smallmatrix}\\right)$ in some given direction Find the projection of the vector $\\vec{b}=\\begin{pmatrix} 3 \\\\ -1\\\\ 4 \\end{pmatrix}$ in the direction $(a)$ perpendicular to the plane: $$2x-2y+z=5.$$ $(b)$ parallel to the plane: $$2x-2y+z=5.$$ For the part $(a)$ projecting $\\vec{b}=\\begin{pmatrix}3 \\\\ -1\\\\ 4\\end{pmatrix}$ to $\\vec{a}=\\begin{pmatrix}2 \\\\ -2\\\\ 1\\end{pmatrix}$ will give me the perpendicular projection right? For part $(b)$ I know I have to subtract what I found in a from a vector to find the projection of the vector parallel to the plane, but I'm kinda lost. • You are very correct. Just proceed as you've written. Feb 6 '18 at 5:57 • I am lost in getting the vector that I am supposed to subtract the projection from... Feb 6 '18 at 6:03 The plane $E = 2x-2y+z=5$ is defined by the normal vector $\\vec{n}=\\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix}$ and the projection of $\\vec{b}$ on $\\vec{n}$, we define it as $\\vec{l}$, is obtained by $\\,|l|\\cdot \\left(\\frac{1}{|n|}\\vec{n}\\right)\\,$ (See unit vector) It follows: $$|l| = |b|\\cdot \\cos(\\alpha) = |b|\\frac{\\vec{b}\\cdot \\vec{n}}{|b|\\cdot |n|} = \\frac{\\vec{b}\\cdot \\vec{n}}{|n|} = 4,\\quad\\text{ and }\\quad \\frac{1}{|n|}\\vec{n} = \\frac{1}{3}\\cdot\\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix}$$ $$\\vec{l} = |l|\\cdot \\left(\\frac{1}{|n|}\\cdot\\vec{n}\\right) = 4\\cdot\\frac{1}{3}\\cdot \\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix} = \\begin{pmatrix}{\\frac{8}{3} \\\\ -\\frac{8}{3}\\\\ \\frac{4}{3}}\\end{pmatrix}$$ $\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad$ And the vector $\\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\\vec{b}$" ]

[ "Dot Product ## Description The dot product of any two vectors $a = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}$ and $b=\\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix}$ is defined as (1) \\begin{align} \\mathbf a \\cdot \\mathbf b = \\sum_{i=1}^n a_i b_i \\end{align} The Sage cell below calculates the dot product for the vectors $a = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ and $b = \\begin{pmatrix} 4 \\\\ 5 \\\\ 6 \\end{pmatrix}$. ## Sage Cell #### Code a = vector( QQ, 3, [ 1, 2, 3 ] ) b = vector( QQ, 3, [ 4, 5, 6 ] ) dotproduct=a.dot_product(b) dotproduct None Primary Tags: Secondary Tags:", "# What is the dot product of <1,5,-3 > and <-8,3,-5 >? $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = - 8 + 15 + 15$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$", "the dot product for work be: $(\\frac{F.v}{llvll^{2}})v$ dotted with <10,0> ? (note; I don't know how to do the dot symbol, any tips on that would be appreciated) Last edited: Mar 16, 2014", "forgot the right definition for dot product. Just grabbed some old books and everything seems to work out now. Sorry and thanks again for all the help. $$\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}$$ $$\\begin{bmatrix}1 & -2 & 2\\end{bmatrix}\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}\\begin{bmatrix}4\\\\ -3 \\\\ 0\\end{bmatrix}$$", "Finding the dot product. Finding the dot item of $(-2w) \\cdot w$ where $w=(0,-2,-2)$ $$\\text{dot product} = \\frac{v . u}{\\| v \\| . \\| u \\|}$$ So $$-2 (0,-2,-2)=(0,4,4) \\\\ (0+4+4) = 8 \\\\ (0,-2,-2)=-4 \\\\ 8 \\times -4=-32$$ $$\\| v \\|= \\sqrt{0^2+(-2)^2+(-2)^2}=2 \\sqrt 2 \\\\ \\|-2w \\|= \\sqrt{0^2+4^2+4^2}= 4 \\sqrt 2\\\\ \\frac{-32}{2 \\sqrt 2 \\times 4 \\sqrt 2} = -64$$ Is this proper? 1 2022-07-25 20:46:29 Source Share", "\\end{pmatrix}$ such that the dot product with a vector $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ satisfying $3x - 2y + z = 0$ is zero. Calculate the dot product. Can you identify $a, b, c$? When you have found $a,b,c$, normalize the vector. - For 1, normalize $(1,2)^T$: what is its length? Then your book should have an equation for the second, if not Wikipedia should help. Can you calculate the length of $(1,2)^T$? For 2, what is g? Ignoring it, can you find a vector such that $f(\\vec{t})=0?$ Then normalize it. Just pick an $x,y$, find $z$... -", "# What is the dot product of <-8,1,2 > and <-2,-8,-1 > ? $a . b = 6$ $\\vec{a} = < - 8 , 1 , 2 > \\text{ } \\vec{b} = < - 2 , - 8 , - 1 >$ $a . b = 16 - 8 - 2$ $a . b = 6$", "# What is the dot product of <9,4,1> and <7,6,3 >? $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$ $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$", "dot product? ie: v=(4,6), u*v=38, find u", "# What is the dot product of <-1,4,7> and <9,-7,5 > ? $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$ $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$", "What is the dot product of <9,4,1> and <-5,1,7 >? Feb 5, 2016 $A . B = - 34$ Explanation: $A = \\left({A}_{i} , {A}_{j} , {A}_{k}\\right)$ $B = \\left({B}_{i} , {B}_{j} , {B}_{k}\\right)$ $A . B = \\left({A}_{i} . {B}_{i} + {A}_{j} . {B}_{j} + {A}_{k} . {B}_{k}\\right)$ $A . B = \\left(- 5.9 + 4.1 + 1.7\\right)$ $A . B = - 45 + 4 + 7$ $A . B = - 34$", "Math Help - dot product help. 1. dot product help. v = <5,6> & p = <1,10> what is v*p? 2. for $a = (x_1,y_1)$and $b = (x_2,y_2)$ $a . b = x_1\\times x_2+y_1\\times y_2$", "4 \\\\ 3 \\end{pmatrix}.$ ×", "# What is the dot product of (- 4 i - 5 j + 2) and (i + j -7k)? $- 23.$ $\\left(- 4 i - 5 j + 2 k\\right) \\cdot \\left(i + j - 7 k\\right) = \\left(- 4\\right) \\left(1\\right) + \\left(- 5\\right) \\left(1\\right) + \\left(2\\right) \\left(- 7\\right) = - 4 - 5 - 14 = - 23$.", "Free Version Easy # Dot Product: Computation with a Vector Sum and a Scalar Multiple LINALG-ZJHUGR Compute $2\\boldsymbol{u} \\cdot ( \\boldsymbol{v} + \\boldsymbol{w})$ where: $\\boldsymbol{u} = \\begin{bmatrix} -1\\\\\\ -1\\\\\\ 3\\end{bmatrix}$ $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 1\\end{bmatrix}$ $\\boldsymbol{w} = \\begin{bmatrix} 0\\\\\\ -5\\\\\\ 3\\end{bmatrix}$ A $4$ B $15$ C $14$ D $36$ E $8$ F $28$", "# How do you compute the dot product for <-4, 1>*<2, -3>? $- 11$ $\\left(a , b\\right) \\times \\left(c , d\\right) = a c + b d$", "# What is the dot product of <-4,-3,8 > and <3,-1,7 >? Feb 20, 2016 $\\vec{a} . \\vec{b}$ = $47$ #### Explanation: We know, $\\vec{a} . \\vec{b}$ = ${a}_{x} . {b}_{x} + {a}_{y} . {b}_{y} + {a}_{Z} . {b}_{z}$ = (-4) · 3 + (-3) · (-1) + 8 · 7 = $- 12 + 3 + 56$ =47", "# What is the dot product of <7,1,1 > and <-2,4,-8>? Oct 21, 2016 $< 7 , 1 , 1 > \\cdot < - 2 , 4 , - 8 >$ $= 7 \\left(- 2\\right) + 1 \\left(4\\right) + 1 \\left(- 8\\right)$ $= - 14 + 4 - 8 = - 18$", "# What is the dot product of (2i -3j + 4k) and (- 4 i - 5 j + 2)? $A \\cdot B = 15$ $A \\cdot B = 15$", "# What is the dot product of <7,1,1 > and <-2,4,-8>? $< 7 , 1 , 1 > \\cdot < - 2 , 4 , - 8 >$ $= 7 \\left(- 2\\right) + 1 \\left(4\\right) + 1 \\left(- 8\\right)$ $= - 14 + 4 - 8 = - 18$" ]

[ "# What is the dot product of <1,5,-3 > and <-8,3,-5 >? $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = - 8 + 15 + 15$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$", "Dot Product ## Description The dot product of any two vectors $a = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}$ and $b=\\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix}$ is defined as (1) \\begin{align} \\mathbf a \\cdot \\mathbf b = \\sum_{i=1}^n a_i b_i \\end{align} The Sage cell below calculates the dot product for the vectors $a = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ and $b = \\begin{pmatrix} 4 \\\\ 5 \\\\ 6 \\end{pmatrix}$. ## Sage Cell #### Code a = vector( QQ, 3, [ 1, 2, 3 ] ) b = vector( QQ, 3, [ 4, 5, 6 ] ) dotproduct=a.dot_product(b) dotproduct None Primary Tags: Secondary Tags:", "forgot the right definition for dot product. Just grabbed some old books and everything seems to work out now. Sorry and thanks again for all the help. $$\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}$$ $$\\begin{bmatrix}1 & -2 & 2\\end{bmatrix}\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}\\begin{bmatrix}4\\\\ -3 \\\\ 0\\end{bmatrix}$$", "# What is the dot product of <-8,1,2 > and <-2,-8,-1 > ? $a . b = 6$ $\\vec{a} = < - 8 , 1 , 2 > \\text{ } \\vec{b} = < - 2 , - 8 , - 1 >$ $a . b = 16 - 8 - 2$ $a . b = 6$", "# What is the dot product of <-1,4,7> and <9,-7,5 > ? $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$ $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$", "Finding the dot product. Finding the dot item of $(-2w) \\cdot w$ where $w=(0,-2,-2)$ $$\\text{dot product} = \\frac{v . u}{\\| v \\| . \\| u \\|}$$ So $$-2 (0,-2,-2)=(0,4,4) \\\\ (0+4+4) = 8 \\\\ (0,-2,-2)=-4 \\\\ 8 \\times -4=-32$$ $$\\| v \\|= \\sqrt{0^2+(-2)^2+(-2)^2}=2 \\sqrt 2 \\\\ \\|-2w \\|= \\sqrt{0^2+4^2+4^2}= 4 \\sqrt 2\\\\ \\frac{-32}{2 \\sqrt 2 \\times 4 \\sqrt 2} = -64$$ Is this proper? 1 2022-07-25 20:46:29 Source Share", "# What is the dot product of <9,4,1> and <7,6,3 >? $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$ $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$", "# What is the dot product of (- 4 i - 5 j + 2) and (i + j -7k)? $- 23.$ $\\left(- 4 i - 5 j + 2 k\\right) \\cdot \\left(i + j - 7 k\\right) = \\left(- 4\\right) \\left(1\\right) + \\left(- 5\\right) \\left(1\\right) + \\left(2\\right) \\left(- 7\\right) = - 4 - 5 - 14 = - 23$.", "Free Version Easy # Dot Product: Computation with a Vector Sum and a Scalar Multiple LINALG-ZJHUGR Compute $2\\boldsymbol{u} \\cdot ( \\boldsymbol{v} + \\boldsymbol{w})$ where: $\\boldsymbol{u} = \\begin{bmatrix} -1\\\\\\ -1\\\\\\ 3\\end{bmatrix}$ $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 1\\end{bmatrix}$ $\\boldsymbol{w} = \\begin{bmatrix} 0\\\\\\ -5\\\\\\ 3\\end{bmatrix}$ A $4$ B $15$ C $14$ D $36$ E $8$ F $28$", "# What is the dot product of <-4,-3,8 > and <3,-1,7 >? Feb 20, 2016 $\\vec{a} . \\vec{b}$ = $47$ #### Explanation: We know, $\\vec{a} . \\vec{b}$ = ${a}_{x} . {b}_{x} + {a}_{y} . {b}_{y} + {a}_{Z} . {b}_{z}$ = (-4) · 3 + (-3) · (-1) + 8 · 7 = $- 12 + 3 + 56$ =47", "\\\\ then $10x=3.\\dot{3}$ \\\\ $10x-x=3.\\dot{3}-0.\\dot{3}$ \\\\ $9x=3$ \\\\ $9x=\\frac{3}{9}$ \\item Express $0.\\dot{1}\\dot{2}$ as a fraction. \\\\ Let $x=0.\\dot{1}\\dot{2}$ \\\\ then $100x=12.\\dot{1}\\dot{2}$ \\\\ $100x-10x=12.\\dot{1}\\dot{2}-0.\\dot{1}\\dot{2}$ \\\\ $90x=12$ \\\\ $x=\\frac{12}{90}$ \\\\ $x=\\frac{2}{15}$ \\end{enumerate} \\end{document} • thanks bernard, I will read through this and try to understand! – user1448093 May 27 '14 at 17:20", "What is the dot product of <9,4,1> and <-5,1,7 >? Feb 5, 2016 $A . B = - 34$ Explanation: $A = \\left({A}_{i} , {A}_{j} , {A}_{k}\\right)$ $B = \\left({B}_{i} , {B}_{j} , {B}_{k}\\right)$ $A . B = \\left({A}_{i} . {B}_{i} + {A}_{j} . {B}_{j} + {A}_{k} . {B}_{k}\\right)$ $A . B = \\left(- 5.9 + 4.1 + 1.7\\right)$ $A . B = - 45 + 4 + 7$ $A . B = - 34$", "# What is the dot product of <7,1,1 > and <-2,4,-8>? Oct 21, 2016 $< 7 , 1 , 1 > \\cdot < - 2 , 4 , - 8 >$ $= 7 \\left(- 2\\right) + 1 \\left(4\\right) + 1 \\left(- 8\\right)$ $= - 14 + 4 - 8 = - 18$", "4 \\\\ 3 \\end{pmatrix}.$ ×", "Math Help - dot product help. 1. dot product help. v = <5,6> & p = <1,10> what is v*p? 2. for $a = (x_1,y_1)$and $b = (x_2,y_2)$ $a . b = x_1\\times x_2+y_1\\times y_2$", "& \\dots \\\\ 0 & 0 & 0 & \\dots & 0 \\\\ a_0 & a_1 & a_2 & \\dots & a_{n-1}\\end{pmatrix} .$$ – Martin Sleziak Nov 13 '12 at 11:26 This seems to be a better version: math.washington.edu/Seminars/Milliman/TrefethenDay2_handout.pdf – Martin Sleziak Nov 13 '12 at 11:33", "# What is the dot product of <5,4,-4> and <-4,1,5 >? $a \\cdot b = 4$ $a = < 5 , 4 , - 4 > \\text{ } b = < - 4 , 1 , 5 >$ $a \\cdot b = 5 \\cdot \\left(- 4\\right) + 4 \\cdot 1 + \\left(- 4\\right) \\cdot 5$ $a \\cdot b = - 20 + 4 - 20$ $a \\cdot b = 4$", "# What is the dot product of <7,1,1 > and <-2,4,-8>? $< 7 , 1 , 1 > \\cdot < - 2 , 4 , - 8 >$ $= 7 \\left(- 2\\right) + 1 \\left(4\\right) + 1 \\left(- 8\\right)$ $= - 14 + 4 - 8 = - 18$", "1. ## Dot Product Problem Assume that , , and . What is the value of ? From $|\\vec u| = 3$ you know that $\\vec u \\cdot \\vec u = 9$ $7 \\vec u \\left(4 \\vec u - 3 \\vec v \\right) = 28 \\vec u \\cdot \\vec u - 21 \\vec u \\cdot \\vec v = 28 \\cdot 9 - 21 \\cdot 3 = 189$", "Hw7 Problem 2 Return to Homework 7, Glossary, Theorems ### Problem 2 Write the following permutations as (a) a product of disjoint cycles; (b) a product of transpositions (a) $(1235)(413)$ (b) $(13256)(23)(46512)$ (c) $(12)(13)(23)(142)$ Solution (a) $\\begin{pmatrix} 1&2&3&4&5&6\\\\ 5&3&4&2&1&6\\end{pmatrix}$ Hence, the products are $(15)(234)$ and $(15)(24)(23)$. (b) $\\begin{pmatrix} 1&2&3&4&5&6\\\\ 2&4&5&1&3&6\\end{pmatrix}$ Hence, the products are $(124)(35)$ and $(14)(12)(35)$. (c) $\\begin{pmatrix} 1&2&3&4\\\\ 4&3&1&2\\end{pmatrix}$ Hence, the products are $(1423)$ and $(13)(12)(14)$. Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License" ]

[ "Dot Product ## Description The dot product of any two vectors $a = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}$ and $b=\\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix}$ is defined as (1) \\begin{align} \\mathbf a \\cdot \\mathbf b = \\sum_{i=1}^n a_i b_i \\end{align} The Sage cell below calculates the dot product for the vectors $a = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ and $b = \\begin{pmatrix} 4 \\\\ 5 \\\\ 6 \\end{pmatrix}$. ## Sage Cell #### Code a = vector( QQ, 3, [ 1, 2, 3 ] ) b = vector( QQ, 3, [ 4, 5, 6 ] ) dotproduct=a.dot_product(b) dotproduct None Primary Tags: Secondary Tags:", "4 \\\\ 3 \\end{pmatrix}.$ ×", "Finding the dot product. Finding the dot item of $(-2w) \\cdot w$ where $w=(0,-2,-2)$ $$\\text{dot product} = \\frac{v . u}{\\| v \\| . \\| u \\|}$$ So $$-2 (0,-2,-2)=(0,4,4) \\\\ (0+4+4) = 8 \\\\ (0,-2,-2)=-4 \\\\ 8 \\times -4=-32$$ $$\\| v \\|= \\sqrt{0^2+(-2)^2+(-2)^2}=2 \\sqrt 2 \\\\ \\|-2w \\|= \\sqrt{0^2+4^2+4^2}= 4 \\sqrt 2\\\\ \\frac{-32}{2 \\sqrt 2 \\times 4 \\sqrt 2} = -64$$ Is this proper? 1 2022-07-25 20:46:29 Source Share", "# What is the dot product of <1,5,-3 > and <-8,3,-5 >? $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = - 8 + 15 + 15$ $< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$", "\\end{pmatrix}$ such that the dot product with a vector $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ satisfying $3x - 2y + z = 0$ is zero. Calculate the dot product. Can you identify $a, b, c$? When you have found $a,b,c$, normalize the vector. - For 1, normalize $(1,2)^T$: what is its length? Then your book should have an equation for the second, if not Wikipedia should help. Can you calculate the length of $(1,2)^T$? For 2, what is g? Ignoring it, can you find a vector such that $f(\\vec{t})=0?$ Then normalize it. Just pick an $x,y$, find $z$... -", "forgot the right definition for dot product. Just grabbed some old books and everything seems to work out now. Sorry and thanks again for all the help. $$\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}$$ $$\\begin{bmatrix}1 & -2 & 2\\end{bmatrix}\\begin{bmatrix}6 & 4 & 0 \\\\ -2 & 0 & 0 \\\\ 0 & 4 & 0\\end{bmatrix}\\begin{bmatrix}4\\\\ -3 \\\\ 0\\end{bmatrix}$$", "Free Version Easy # Dot Product: Computation with a Vector Sum and a Scalar Multiple LINALG-ZJHUGR Compute $2\\boldsymbol{u} \\cdot ( \\boldsymbol{v} + \\boldsymbol{w})$ where: $\\boldsymbol{u} = \\begin{bmatrix} -1\\\\\\ -1\\\\\\ 3\\end{bmatrix}$ $\\boldsymbol{v} = \\begin{bmatrix} 1\\\\\\ -2\\\\\\ 1\\end{bmatrix}$ $\\boldsymbol{w} = \\begin{bmatrix} 0\\\\\\ -5\\\\\\ 3\\end{bmatrix}$ A $4$ B $15$ C $14$ D $36$ E $8$ F $28$", "of these vectors. Can we find the \"dot product\"? Yes: $$u \\dot {} v=u_1v_1+u_2v_2=cos(\\theta)*\\left || u |\\right |*\\left || v| \\right |=cos(60)*3*2=3$$ So, we got: $$u_1v_1+u_2v_2=3$$ Next, let's see what the question really wants: $$Au=\\begin{bmatrix} u_1 && u_2 \\\\ v_1 && v_2 \\end{bmatrix}$$ $$*\\begin{bmatrix} u_1 \\\\ u_2 \\end{bmatrix}$$ = $$\\begin{pmatrix} u_1^2+u_2^2\\\\ u_1v_1+u_2v_2 \\end{pmatrix}$$ = $$\\begin{pmatrix} 3^2\\\\ 3 \\end{pmatrix}=\\begin{pmatrix} 9\\\\ 3 \\end{pmatrix}$$ Now, in a similar way, find $$Av$$. Hope this helps! Jul 19, 2022 #3 0 I got Av= (9;3) Guest Jul 20, 2022 #4 0 Edit: Sorry I meant Au= 3;9 Guest Jul 20, 2022 #5 +285 +3 The dot product of u and v is equal to ||u||* ||v||*cos 60 = 3. (Prove this before you start using it!) row_1(Au) = $$\\mathbf{u \\cdot u}$$ = 9 row_2(Au) = Dot product of u and v, equaling 3 Au = <9, 3> Similarly: row_1(Av) = 3 row_2(Av) = v \\cdot v = 4 Av = <3, 4> Jul 21, 2022 #6 +1 There seems to be a notational or transcription error in the question. The statement that the first row of the matrix A is the vector u implies that u is a row vector. (Similarly, the implication is that v is a row vector.) We are not given the dimension of u and v, but assuming that", "# What is the dot product of <-1,4,7> and <9,-7,5 > ? $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$ $< - 1 , 4 , 7 > \\cdot < 9 , - 7 , 5 > = 26$", "# What is the dot product of (- 4 i - 5 j + 2) and (i + j -7k)? $- 23.$ $\\left(- 4 i - 5 j + 2 k\\right) \\cdot \\left(i + j - 7 k\\right) = \\left(- 4\\right) \\left(1\\right) + \\left(- 5\\right) \\left(1\\right) + \\left(2\\right) \\left(- 7\\right) = - 4 - 5 - 14 = - 23$.", "& 18 \\end{pmatrix}$$ and you want to find the product $$\\displaystyle AB= \\begin{pmatrix} 1 & 2 & 3\\\\ 4 & 5 & 6 \\\\7 & 8 & 9 \\end{pmatrix}\\begin{pmatrix}10 & 11 & 12 \\\\ 13 & 14 & 15 \\\\ 16 & 17 & 18 \\end{pmatrix}$$. One way to think about this is that each row of matrix A is a vector, that each column of matrix B is a vector, and you want to take the \"dot product\" of the three vectors from A with each of the three vectors from B. (you appear to have labeled row and columns with \"p\", \"q\", etc. but that seems to me more complicated and confusing than useful.) For example, the first row in A is [1, 2. 3] and the first column in B is [10, 13, 16]. Their \"dot product\" is 1(10)+ 2(13)+ 3(16)= 10+ 26+ 48= 84. The value in the first row, first column of AB is 84. The first row in A is [1, 2, 3] and the second column in B is [11, 14, 17]. Their \"dot product\" is 1(11)+ 2(14)+ 3(17)= 11+ 28+ 51= 90. The value in the first row, second column of AB is 90. The last number in the first row of AB is 1(12)+2(15)+ 3(18)= 12+ 30+ 54= 96.", "# What is the dot product of <9,4,1> and <7,6,3 >? $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$ $A . B = 9 \\cdot 7 + 4 \\cdot 6 + 1 \\cdot 3 = 90$", "# What is the dot product of <-4,-3,8 > and <3,-1,7 >? Feb 20, 2016 $\\vec{a} . \\vec{b}$ = $47$ #### Explanation: We know, $\\vec{a} . \\vec{b}$ = ${a}_{x} . {b}_{x} + {a}_{y} . {b}_{y} + {a}_{Z} . {b}_{z}$ = (-4) · 3 + (-3) · (-1) + 8 · 7 = $- 12 + 3 + 56$ =47", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "What is the dot product of <9,4,1> and <-5,1,7 >? Feb 5, 2016 $A . B = - 34$ Explanation: $A = \\left({A}_{i} , {A}_{j} , {A}_{k}\\right)$ $B = \\left({B}_{i} , {B}_{j} , {B}_{k}\\right)$ $A . B = \\left({A}_{i} . {B}_{i} + {A}_{j} . {B}_{j} + {A}_{k} . {B}_{k}\\right)$ $A . B = \\left(- 5.9 + 4.1 + 1.7\\right)$ $A . B = - 45 + 4 + 7$ $A . B = - 34$", "the dot product that allows you to take the alpha's and pull them in front: $$\\vec{a}\\cdot \\alpha\\vec{b}= \\alpha\\vec{a}\\cdot \\vec{b}= \\alpha(\\vec{a}\\cdot \\vec{b})$$ P: 492 $$||\\vec{y}||^2=\\alpha^2 \\vec{a} \\cdot \\vec{a}$$?", "# What is the dot product of <7,1,1 > and <-2,4,-8>? Oct 21, 2016 $< 7 , 1 , 1 > \\cdot < - 2 , 4 , - 8 >$ $= 7 \\left(- 2\\right) + 1 \\left(4\\right) + 1 \\left(- 8\\right)$ $= - 14 + 4 - 8 = - 18$", "1. ## Dot Product Problem Assume that , , and . What is the value of ? From $|\\vec u| = 3$ you know that $\\vec u \\cdot \\vec u = 9$ $7 \\vec u \\left(4 \\vec u - 3 \\vec v \\right) = 28 \\vec u \\cdot \\vec u - 21 \\vec u \\cdot \\vec v = 28 \\cdot 9 - 21 \\cdot 3 = 189$", "therefore a must-know topic. In this guide, we will explore the dot product in depth! Definition. # Dot product Suppose we have the following two vectors $\\boldsymbol{v}$ and $\\boldsymbol{w}$: $$\\boldsymbol{v}=\\begin{pmatrix} v_1\\\\ v_2\\\\ v_3\\\\ \\end{pmatrix},\\;\\;\\;\\; \\boldsymbol{w}=\\begin{pmatrix} w_1\\\\ w_2\\\\ w_3\\\\ \\end{pmatrix}$$ The dot product is defined as: $$\\boldsymbol{v}\\cdot{\\boldsymbol{w}}= v_1w_1+ v_2w_2+ v_3w_3$$ Note that $\\boldsymbol{v}$ and $\\boldsymbol{w}$ are in $\\mathbb{R}^3$, but the dot product returns a scalar. The dot product is also sometimes referred to as the scalar product or the inner product. Example. ## Computing the dot product between two vectors Compute the dot product $\\boldsymbol{v}\\cdot{\\boldsymbol{w}}$ where $\\boldsymbol{v}$ and $\\boldsymbol{w}$ are defined as: $$\\boldsymbol{v}= \\begin{pmatrix} 1\\\\2 \\end{pmatrix},\\;\\;\\;\\; \\boldsymbol{w}= \\begin{pmatrix} 3\\\\4 \\end{pmatrix}$$ Solution. The dot product $\\boldsymbol{v}\\cdot\\boldsymbol{w}$ is: \\begin{align*} \\boldsymbol{v}\\cdot{\\boldsymbol{w}}&= (1)(3)+(2)(4)\\\\ &=11 \\end{align*} Theorem. # Using dot product to compute the angle between two vectors Suppose we have the following two vectors $\\boldsymbol{v}$ and $\\boldsymbol{w}$: Let the angle between $\\boldsymbol{v}$ and $\\boldsymbol{w}$ be $\\theta$. The dot product of $\\boldsymbol{v}$ and $\\boldsymbol{w}$ is: $$\\boldsymbol{v}\\cdot{\\boldsymbol{w}}= \\Vert{\\boldsymbol{v}}\\Vert\\Vert{\\boldsymbol{w}}\\Vert \\cos{(\\theta)}$$ Here, $\\Vert{\\boldsymbol{v}}\\Vert$ is the magnitude of $\\boldsymbol{v}$. Proof. Suppose we have the following: Here, the vector $\\boldsymbol{w}-\\boldsymbol{v}$ points from $A$ to $B$ because of the geometric interpretation of vector additionlink. The lengths of the sides are: Using the cosine rule from elementary trigonometry, we have that: $$$$\\label{eq:xF6GwOYW3PvcsraNGkP} \\Vert{\\boldsymbol{w}-\\boldsymbol{v}}\\Vert^2 =\\Vert\\boldsymbol{v}\\Vert^2+ \\Vert\\boldsymbol{w}\\Vert^2-2\\Vert{\\boldsymbol{v}}\\Vert \\Vert{\\boldsymbol{w}}\\Vert\\cos(\\theta)$$$$ Let's assume $\\boldsymbol{w}$ and $\\boldsymbol{v}$", "vector and B is column vector, then the dot product is defined as follows; ${\\displaystyle A\\cdot B=|A||B|cos\\theta }$ or ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} ={\\begin{pmatrix}a_{1}&a_{2}&\\cdots &a_{n}\\end{pmatrix}}{\\begin{pmatrix}b_{1}\\\\b_{2}\\\\\\vdots \\\\b_{n}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}+\\cdots +a_{n}b_{n}=\\sum _{i=1}^{n}a_{i}b_{i}}$ Suppose ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}a_{1}&a_{2}&a_{3}\\end{pmatrix}}}$ and ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}b_{1}\\\\b_{2}\\\\b_{3}\\end{pmatrix}}}$ The dot product is ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} ={\\begin{pmatrix}a_{1}&a_{2}&a_{3}\\end{pmatrix}}{\\begin{pmatrix}b_{1}\\\\b_{2}\\\\b_{3}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}$ ### Example Suppose ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}2\\\\1\\\\3\\end{pmatrix}}}$ and ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}7\\\\5\\\\4\\end{pmatrix}}}$ ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} ={\\begin{pmatrix}2&1&3\\end{pmatrix}}{\\begin{pmatrix}7\\\\5\\\\4\\end{pmatrix}}}$ ${\\displaystyle =2\\cdot 7+1\\cdot 5+3\\cdot 4}$ ${\\displaystyle =14+5+12}$ ${\\displaystyle =31}$ ### Practice problems (1) ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}3\\\\2\\\\5\\end{pmatrix}}}$ and ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}1\\\\4\\\\3\\end{pmatrix}}}$ ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} =}$ (2) ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}1\\\\0\\\\3\\end{pmatrix}}}$ and ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}6\\\\9\\\\2\\end{pmatrix}}}$ ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} =}$ ## Cross product Cross product is defined as follows: ${\\displaystyle A\\times B=|A||B|sin\\theta }$ Or, using detriment, ${\\displaystyle \\mathbf {A\\times B} ={\\begin{vmatrix}e_{x}&e_{y}&e_{z}\\\\a_{x}&a_{y}&a_{z}\\\\b_{x}&b_{y}&b_{z}\\\\\\end{vmatrix}}=(a_{y}b_{z}-a_{z}b_{y},a_{z}b_{x}-a_{x}b_{z},a_{x}b_{y}-a_{y}b_{x})}$ where ${\\displaystyle e}$ is unit vector. ## References 1. Sourced from Matrix (mathematics), Wikipedia, 28th March 2013. 2. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013. 3. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013. # Lagrange Equations ## Lagrange Equation ${\\displaystyle {\\frac {d}{dt}}\\left({\\frac {\\partial L}{\\partial {\\dot {x}}}}\\right)-{\\frac {\\partial L}{\\partial x}}=0}$ where ${\\displaystyle {\\dot {x}}={\\frac {dx}{dt}}}$ The equation above is called Lagrange Equation. Let the kinetic energy of the point mass be ${\\displaystyle T}$ and the potential energy be ${\\displaystyle U}$." ]

[ "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "# How do you find the value of $\\sum_{r=0}^{44} \\tan^2(2r+1)$? Problem: Find the value of $$\\sum_{r=0}^{44} \\tan^2(2r+1)$$ Note: The angles here are in degrees. I don't know how to solve this question because trigonometric simplifications didn't get me anywhere. I think there was a method to solve this question using complex numbers which I no longer remember. Any hint/help will be appreciated. $$\\tan90x=\\dfrac{\\binom{90}1t-\\binom{90}3t^3+\\cdots+\\binom{90}{89}t^{89}}{\\binom{90}0-\\binom{90}2t^2+\\cdots+\\binom{90}{90}t^{90}}$$ where $t=\\tan x$ If $\\tan 90x=\\tan90^\\circ=\\infty$ $90x=180^\\circ n+90^\\circ=90^\\circ(2n+1)$ where $n$ is any integer $\\implies x=(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ So, the roots of $$t^{90}-\\binom{90}{88}t^{88}+\\cdots=0$$ are $\\tan(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ Now as $-\\tan(2n+1)^\\circ=\\tan\\{180^\\circ-(2n+1)^\\circ\\}=\\tan\\{2(89-n)+1)^\\circ\\}$ $$\\implies\\sum_{r=0}^{89}\\tan^2(2n+1)^\\circ=2\\sum_{r=0}^{44}\\tan^2(2n+1)^\\circ$$ and the roots of $$s^{45}-\\binom{90}{88}s^{44}+\\cdots=0$$ are $\\tan^2(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ Using Vieta's formula, $$\\sum_{r=0}^{89}\\tan^2(2n+1)^\\circ=\\binom{90}{88}=\\binom{90}{90-88}=?$$ Can you take it from here? Using an idea similar to this answer, note that the function $$\\frac{90/z}{z^{90}-1}$$ has residue $1$ for $z=e^{k\\pi i/45}$ and residue $-90$ at $z=0$. On $|z|=1$, $$\\tan(\\theta/2)=-i\\frac{z-1}{z+1}$$ Integrating $$f(z)=-\\left(\\frac{z-1}{z+1}\\right)^2\\frac{90/z}{z^{90}-1}$$ around a large circle is $0$ since the integrand is approximately $|z|^{-91}$. Thus, the sum of residues is $$2\\sum_{k=0}^{44}\\tan^2\\left(\\frac{2k\\pi}{180}\\right)+\\operatorname*{Res}_{z=0}f(z)+\\operatorname*{Res}_{z=-1}f(z)=0$$ Since $\\operatorname*{Res}\\limits_{z=0}f(z)=90$ and $\\operatorname*{Res} \\limits_{z=-1}f(z)=-\\frac{8102}{3}$, we get \\begin{align} \\sum_{k=0}^{44}\\tan^2\\left(\\frac{2k\\pi}{180}\\right) &=\\frac12\\left(\\frac{8102}{3}-90\\right)\\\\ &=\\frac{3916}{3} \\end{align} The sum in the question is the difference between the value in answer cited above and this value, that is \\begin{align} \\sum_{k=0}^{44}\\tan^2\\left(\\frac{(2k+1)\\pi}{180}\\right) &=\\frac{15931}{3}-\\frac{3916}{3}\\\\ &=4005 \\end{align}", "## tan A = tan B + tan C + tan D $\\tan \\, 84^\\circ \\; = \\; tan \\, 78^\\circ \\; + \\; tan \\, 72^\\circ \\; + \\; tan \\, 60^\\circ$ $\\tan \\, 80^\\circ \\; = \\; tan \\, 70^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 50^\\circ$ Find other 4-tuples $(a, b, c, d)$ of distinct integers between $0^\\circ$ and $90^\\circ$ that satisfy the relation $tan \\, A^\\circ\\; = \\; tan \\, B^\\circ \\; + \\; tan \\, C^\\circ \\; + \\; tan \\, D^\\circ$ Other solutions: $\\tan \\, 72^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 36^\\circ \\; + \\; tan \\, 6^\\circ$ $\\tan \\, 70^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 40^\\circ \\; + \\; tan \\, 10^\\circ$ $\\tan \\, 72^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 24^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 36^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 72^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 36^\\circ$ How many such equations can you produce if we allow repeated angles? Hint: there are 49 such equations Determine", "mentioned above, you shouldn't say that $\\tan(90^\\circ)$ is infinite but you should really say that it's undefined (careful, though, if a curious student enters $\\tan(\\pi/2)$ into a certain free online calculator, they'll get something like $1.63317787 \\times 10^{16}$). • Using the double angle prove that $\\sin(90^\\circ)=1$ • Using the Pythagorean identity, prove $\\cos(90^\\circ)=0$. • By evaluating $\\tan(x)$ using the definition at $x=89^\\circ,\\, 89.9^\\circ,\\, 89.99^\\circ,\\dots$ and at $x=91^\\circ,\\, 90.1^\\circ,\\,90.01^\\circ,\\dots$ try to explain a little more about they sense in which $\\tan(90^\\circ)$ is undefined. Now try to explain it using the unit circle.", "$$1 - \\sin 10 ^ \\circ - 2 \\sin^2 10^\\circ = \\sin N^\\circ$$, where $$0 \\leq N \\leq 90$$. What is $$N$$?", "\\pm {42.0038}^{\\circ} + {360}^{\\circ} k \\quad$integer $k$", "Volume 23 , Issue 3 1987 Going from one place to another can be a difficult problem. Have you ever wondered how your calculator or computer finds quantities such as $\\sin(63^\\circ)$ or $\\tan(17^\\circ)$? In a recent edition of Mathematical Spectrum (the English equivalent of Parabola) the following famous problem was discussed: Q.720 When the initial digit of a whole number $x$ is deleted, the number decreases by a factor of 13. Find all possible values of $x$. Q.708 A four digit number $abcd$ has the property that $a+b = c\\times d$ and also $a\\times b = c+d$. Find all possibilities.", "## Trigonometry (11th Edition) Clone The solution set is $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$ $$\\tan\\theta+1=0$$ 1) Solve the equation over the interval $[0^\\circ,360^\\circ)$ $$\\tan\\theta+1=0$$ $$\\tan\\theta=-1$$ Over the interval $[0^\\circ, 360^\\circ)$, there are two values of $\\theta$ where $\\tan\\theta=-1$, which are $135^\\circ$ and $315^\\circ$. Therefore, $$\\theta=\\{135^\\circ, 315^\\circ\\}$$ 2) Solve the equation for all solutions To find all solutions, we add the integer multiples of the period of tangent function, which is $180^\\circ$, to each solution found in step 1. - There are two solutions found in step 1, $\\{135^\\circ, 315^\\circ\\}$, so it will be written as $\\theta=135^\\circ+180^\\circ n$ and $\\theta=315^\\circ+180^\\circ n$ where $n\\in Z$. - However, a closer look reveals that $135^\\circ+180^\\circ n$ in fact covers all the possible values of $315^\\circ+180^\\circ n$, and vice versa. In other words, they are technically the same. So the solution set can be shortened to $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$", "are $\\tan(2r+1)^\\circ,0\\le r\\le 44$ or $45\\le r\\le89$ $$\\implies\\sum_{r=0}^{44}\\tan^2(2r+1)^\\circ=\\frac{\\binom{90}{88}}1=\\binom{90}2=\\cdots$$", "# How do you tell whether the value of tan 90 degrees is positive, negative, zero or undefined? Mar 7, 2018 Undefined #### Explanation: We have: $\\tan {90}^{\\circ}$ Know that, $\\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta}$ So here, we get $\\tan {90}^{\\circ} = \\frac{\\sin {90}^{\\circ}}{\\cos {90}^{\\circ}}$ Looking at the trigonometric table, we see that $\\sin {90}^{\\circ} = 1 , \\cos {90}^{\\circ} = 0$. Therefore, $\\tan {90}^{\\circ} = \\frac{1}{0}$ Since anything divided by $0$ (except $0$ itself) is considered to be undefined, we get that $\\tan {90}^{\\circ}$ is undefined as well.", "by $2(\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ))$. However, if we write these roots as $r_1,...,r_6$, then we also have $$\\sum_i r_i^2 = \\left(\\sum_i r_i\\right)^2 - 2\\sum_{i < j} r_ir_j$$ But $\\sum_i r_i$ is the coefficient of $x^5$ in the above equation, namely zero, and $\\sum_{i < j} r_ir_j$ is the coefficient of $x^4$, namely $-33$. So you get $$2(\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ)) = -2\\times-33$$ So we conclude that $$\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ) = 33$$", "# How do you find the value of tan 90? $\\tan 90 = \\infty$ $\\tan 90 = \\sin \\frac{90}{\\cos} 90 = \\frac{1}{0} = \\infty$", "# Tan on the floor! Level pending A positive integer $$n \\in S$$, the set of tanny integers, if and only if for some value of $$\\theta \\in (0,\\frac{\\pi}{2})$$, $$\\lfloor \\tan^2{\\theta} \\rfloor + \\tan{\\theta} = n$$. When the elements of $$S$$ are arranged in increasing order, let $$N_{T_{2014}}$$ denote the $$2014$$th tanny integer. Find the last three digits of $$N_{T_{2014}}$$. ×", "## Where n(n – 1) and n(n + 1) are both pandigital Find an integer $n$ such that both $n \\,(n - 1)$ and $n \\,(n + 1)$ are pandigital Antonio Roldán found n = 55152", "7 '18 at 10:08 All possibilities are wrong! Try $$A=270^{\\circ}$$, $$B=-90^{\\circ}$$ and $$C=0^{\\circ}.$$", "x}{120} ~~~~~~~~~~~ x > 30^{\\circ}$$ From this you can derive $$\\cos x = \\sin (90-x)$$ and $$\\tan x \\approx 0.017\\ x$$ With an error from $5$ to $8$ percent (for the tangent one). • How did you arrive at this approximation formula ? – user399078 Jan 10 '17 at 11:11 • @QUANTUM it is an approximate version of the answer above. Jan 10 '17 at 12:24", "+0 # Mod arithmetic 0 157 1 FInd the integer n, 0 <= n <= 11, such that n = 1089 (mod 12). Apr 23, 2022 #1 +9461 0 Note that $$12 \\times 90 = 1080$$. Then, $$1089 = 12\\times 90 + 9$$. Taking mod 12 on both sides results in $$1089 \\equiv 9 \\pmod{12}$$. Apr 23, 2022", "for some integer n. now substitute these two equations into the expression 3q+r $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ We know m(1+n) is an integer, since m and n are integers solet s=m(1+n) then 3q+r = ps therefore by the definiton of divides $$p | (3q + r)$$ QED", "$sin(140^\\circ) = 0.6428$, $cos(140^\\circ) = -0.7660$, and $tan(140^\\circ) = -0.8391$.", "70 \\cos^4(10) \\sin^4(10) - 28 \\cos^2 (10) \\sin^6(10) + \\sin^8 (10)} = \\tan (80)$$" ]

[ "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "NZ Level 7 (NZC) Level 2 (NCEA) Period changes for tangents Lesson We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\\sin kx$sinkx and $\\cos kx$coskx. The fact that the sine and cosine functions both have period $360^\\circ$360° is expressed by the statements $\\sin(x+360^\\circ n)=\\sin x$sin(x+360°n)=sinx and $\\cos(x+360^\\circ n)=\\cos x$cos(x+360°n)=cosx for all integers $n$n. So it must be true that since $\\tan x=\\frac{\\sin x}{\\cos x}$tanx=sinxcosx, then $\\tan(x+360^\\circ n)=\\tan x$tan(x+360°n)=tanx. However, $360^\\circ$360° is not the smallest interval at which the tangent function repeats. A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\\circ$180°. That is, $\\tan(x+180^\\circ n)=\\tan x$tan(x+180°n)=tanx for all integers $n$n. The fact that the tangent function repeats at intervals of $180^\\circ$180° can be verified by considering the unit circle diagram. If $180^\\circ$180° is added to an angle $\\alpha$α, then the diagram shows that $\\sin(\\alpha+180^\\circ)$sin(α+180°) has the same magnitude as $\\sin\\alpha$sinα but opposite sign. The same relation holds between $\\cos(\\alpha+180^\\circ)$cos(α+180°) and $\\cos\\alpha$cosα. We make use of the definition: $\\tan\\alpha=\\frac{\\sin\\alpha}{\\cos\\alpha}$tanα=sinαcosα $\\tan(\\alpha+180^\\circ)$tan(α+180°) $=$= $\\frac{\\sin(\\alpha+180^\\circ)}{\\cos(\\alpha+180^\\circ)}$sin(α+180°)cos(α+180°)​ $=$= $\\frac{-\\sin\\alpha}{-\\cos\\alpha}$−sinα−cosα​ $=$= $\\tan\\alpha$tanα We are now in a position to determine the period of the function $\\tan kx$tankx where $k$k is any number. We can define a new", "n\\cdot 180^\\circ;$$ Other angles are $$=20^\\circ,200^\\circ,380^\\circ,460^\\circ$$ We can add or subtract $$n\\cdot \\pi$$ ( $$n$$ positive or negative integer) to $$20^\\circ$$ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding $$180^\\circ$$. You suggest we use that $$\\tan \\theta = \\tan(\\theta - 180^\\circ).$$ Equivalently, if we set $$\\phi = \\theta - 180^\\circ$$, the above equation reads $$\\tan(\\phi+180^\\circ) = \\tan\\phi.$$ To summarize: the tangent function is $$180^\\circ$$-periodic: it repeats every $$180^\\circ$$. So the possible answers are $$\\dots,-340^\\circ,-160^\\circ, 20^\\circ, 200^\\circ, 380^\\circ, \\dots$$ Narasimham's answer shows how to express this in a single equation with $$n$$ varying, ABCD does the same in the language of radians ($$180^\\circ = \\pi$$ radians).", "# 2015-19 Sum of tangent functions Evaluate $\\sum_{k=0}^{n} 2^k \\tan \\frac{\\pi}{4\\cdot 2^{n-k}}.$ GD Star Rating loading... 2015-19 Sum of tangent functions, 3.9 out of 5 based on 7 ratings", "tangent is $\\pi$-periodic (or $180°$-periodic if you prefer), $\\tan 225° = \\tan 45°=1$, so the previous equality holds and your sum is actually $1$. -", "Mathematics » Functions II » The Tangent Function # Sketching Tangent Graphs ## Example ### Question Sketch the graph of $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$ for $$-\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}$$. ### Examine the form of the equation From the equation we see that $$0 < k < 1$$, therefore the branches of the graph will be less steep than the standard tangent graph $$y = \\tan \\theta$$. We also notice that $$p < 0$$ so the graph will be shifted to the right on the $$x$$-axis. ### Determine the period The period for $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$ is: \\begin{align*} \\text{Period} &= \\cfrac{\\text{180}\\text{°}}{|k|} \\\\ &= \\dfrac{\\text{180}\\text{°}}{\\cfrac{1}{2}} \\\\ &= \\text{360}\\text{°} \\end{align*} ### Determine the asymptotes The standard tangent graph, $$y = \\tan \\theta$$, for $$-\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}$$ is undefined at $$\\theta = -\\text{90}\\text{°}$$ and $$\\theta = \\text{90}\\text{°}$$. Therefore we can determine the asymptotes of $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$: • $$\\cfrac{-\\text{90}\\text{°}}{\\text{0.5}} + \\text{30}\\text{°} = -\\text{150}\\text{°}$$ • $$\\cfrac{\\text{90}\\text{°}}{\\text{0.5}} + \\text{30}\\text{°} = \\text{210}\\text{°}$$ The asymptote at $$\\theta = \\text{210}\\text{°}$$ lies outside the required interval. ### Plot the points and join with a smooth curve Period: $$\\text{360}\\text{°}$$ Domain: $$\\{ \\theta: -\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}, \\theta \\ne -\\text{150}\\text{°} \\}$$ Range: $$(-\\infty;\\infty)$$ $$y$$-intercepts: $$(\\text{0}\\text{°};-\\text{0.27})$$ $$x$$-intercept: $$(\\text{30}\\text{°};0)$$ Asymptotes: $$\\theta = -\\text{150}\\text{°}$$ Do you want to suggest a correction", "# Difference between revisions of \"1996 AIME Problems/Problem 10\" ## Problem Find the smallest positive integer solution to $\\tan{19x^{\\circ}}=\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$. ## Solution $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-45^{\\circ})}} =$ $\\dfrac{2\\sin{141^{\\circ}}\\cos{45^{\\circ}}}{2\\cos{141^{\\circ}}\\sin{45^{\\circ}}} = \\tan{141^{\\circ}}$. The period of the tangent function is $180^\\circ$, and the tangent function is one-to-one over each period of its domain. Thus, $19x \\equiv 141 \\pmod{180}$. Since $19^2 \\equiv 361 \\equiv 1 \\pmod{180}$, multiplying both sides by $19$ yields $x \\equiv 141 \\cdot 19 \\equiv (140+1)(18+1) \\equiv 0+140+18+1 \\equiv 159 \\pmod{180}$. Therefore, the smallest positive solution is $x = \\boxed{159}$. ## Solution 2 $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\dfrac{1 + \\tan{96^{\\circ}}}{1-\\tan{96^{\\circ}}}$ which is the same as $\\dfrac{\\tan{45^{\\circ}} + \\tan{96^{\\circ}}}{1-\\tan{45^{\\circ}}\\tan{96^{\\circ}}} = \\tan{141{^\\circ}}$. So $19x = 141 +180n$, for some integer $n$. Multiplying by $19$ gives $x \\equiv 141 \\cdot 19 \\equiv 2679 \\equiv 159 \\pmod{180}$. The smallest positive solution of this is $x = \\boxed{159}$ ## Solution 3 (Only sine and cosine sum formulas) It seems reasonable to assume that $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\tan{\\theta}$ for some angle $\\theta$. This means $$\\dfrac{\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}})}{\\alpha (\\cos{96^{\\circ}}-\\sin{96^{\\circ}})} = \\frac{\\sin{\\theta}}{\\cos{\\theta}}$$ for some constant $\\alpha$. We can set $\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}}) = \\alpha \\sin{\\theta}$.Note that if we have $\\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\\sin{45^{\\circ}} = \\cos{45^{\\circ}} = \\tfrac{\\sqrt{2}}{2}$,", "# Difference between revisions of \"1996 AIME Problems/Problem 10\" ## Problem Find the smallest positive integer solution to $\\tan{19x^{\\circ}}=\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$. ## Solution $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-45^{\\circ})}} =$ $\\dfrac{2\\sin{141^{\\circ}}\\cos{45^{\\circ}}}{2\\cos{141^{\\circ}}\\sin{45^{\\circ}}} = \\tan{141^{\\circ}}$. The period of the tangent function is $180^\\circ$, and the tangent function is one-to-one over each period of its domain. Thus, $19x \\equiv 141 \\pmod{180}$. Since $19^2 \\equiv 361 \\equiv 1 \\pmod{180}$, multiplying both sides by $19$ yields $x \\equiv 141 \\cdot 19 \\equiv (140+1)(18+1) \\equiv 0+140+18+1 \\equiv 159 \\pmod{180}$. Therefore, the smallest positive solution is $x = \\boxed{159}$. ## Solution 2 $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$ So $19x = 141 +180n$, for some integer $n$. Then $0 \\equiv 8 + 9n \\pmod{19}$, from which $n \\equiv 16 \\pmod{19}$ after some manipulation. The smallest suitable value of $n$ is therefore 16 from which $x = \\boxed{159}$", "Trigonometry (11th Edition) Clone $\\tan{1} \\gt \\tan{2}$ Recall: (1) The tangent function's period is $\\pi$ or around $3.14$; (2) The tangent function is increasing in the intervals $(-1.57, 1.57), (1.57, 4.71), (4.71, 7.85), ...$ (3) The value of the tangent function is $0$ when $x=0, \\pi, 2\\pi, 3\\pi, ...$ or, in approximate decimal form, $x=0, 3.14, 6.28, 9.42, ...$. (4) The value of the tangent function is negative in the intervals $(-1.57, 0), (1.57, 3.14), (4.71, 6.28), ....$ (5) The value of the tangent function is positive in the intervals $(0, 1.57), (3.14, 4.71), (6.28, 7.85), ....$ Thus, the value of $\\tan{1}$ is positive, and the value of $\\tan{2}$ is negative. Therefore, $\\tan{1} \\gt \\tan{2}$.", "Question If $\\tan ({\\text{A + B) = }}\\dfrac{1}{{\\sqrt 3 }}$ and $\\tan ({\\text{A - B) = }}\\dfrac{1}{{\\sqrt 3 }}$, ${0^0}$< A + B < ${90^0}$. Find A and B. Hint: Compare the given two trigonometric functions and use the value of tan${30^0}$ is $\\dfrac{1}{{\\sqrt 3 }}$ and period of tangent is ${180^0}$. Given, $\\tan ({\\text{A + B) = }}\\dfrac{1}{{\\sqrt 3 }}$………………………………………………………..(1) $\\tan ({\\text{A - B) = }}\\dfrac{1}{{\\sqrt 3 }}$………………………………………………………….(2) So, the period of tan is ${180^0}$. Neither A nor B can be near this angle because we have stated in the question that the sum of A and B must be less than 90$^0$. So, from equation (1) and (2) tan(A+B) = tan(A – B) Equating the angles of tangents , we get A + B = A – B A + B – A + B = 0 2B = 0 B = 0$^0$ Putting the value of B in equation (1), we get tan(A + 0) = $\\dfrac{1}{{\\sqrt 3 }}$ tanA = tan${30^0}$ A = ${30^0}$. Hence, A = ${30^0}$ and B = 0$^0$ is the answer. Note: In these types of questions, the periodicity of the trigonometric function matters. Periodicity is the property of the function such that it repeats itself after a fixed interval of time called period of the function.", "simply 1, and in any case $50^\\circ$ is closer to $45^\\circ$ than to $60^\\circ$, so it makes more sense to use $45^\\circ$ as the special angle. 9) Leaving calculators in radian mode will yield the wrong answers $-0.9820$ and $45.1831$. 10) Students may end up thinking a little too hard about this problem. All they’re supposed to conjecture is that the two expressions are not equal, so if any of them struggle with this problem, find out if they’ve got that much figured out and reassure them they can stop there. 1) Use a calculator to find the tangent of $86^\\circ, 87^\\circ, 88^\\circ,$ and $89^\\circ$. Then, find the tangent of $94^\\circ, 93^\\circ, 92^\\circ,$ and $91^\\circ$. Now make a conjecture about the behavior of the tangent function as $x$ approaches $90^\\circ.$ 1) The values of the tangent function are as follows: $& X && \\tan x\\\\& 86^\\circ && 14.3007\\\\& 87^\\circ && 19.0811\\\\& 88^\\circ && 28.6363\\\\& 89^\\circ && 57.2900\\\\& 90^\\circ && \\text{undefined}\\\\& 91^\\circ && -57.2900\\\\& 92^\\circ && -28.6363\\\\& 93^\\circ && -19.0811\\\\& 94^\\circ && -14.3007$ The tangent function approaches infinity as $x$ approaches $90^\\circ$ from below, and approaches negative infinity as $x$ approaches $90^\\circ$ from above. ## Relating Trigonometric Functions In-Text Examples 4) Demonstrating that $\\cot \\theta = \\frac{\\cos \\theta}{\\sin \\theta}$ because cot is the reciprocal of tan and $\\tan \\theta", "$0 < x < 180$ as was stated in your original post. 7. Jun 15, 2015 ### Tangeton Ahh okay I forgot about that bit.. I guess I would have to remember to use the CAST diagram for my answer. Thanks for all help, everyone. 8. Jun 15, 2015 ### ehild You know that the tangent function is periodic with period of 180°. And you need to find solution in the range 0<θ<180°. Adding 180° to your result is also solution, just in the desired range.", "From $180^\\circ$ to $360^\\circ$, you remember, it was always negative, and you can readily see, it has a minimum value of $-1$ at $270^\\circ$. Beyond $360^\\circ$, of course, it just repeats. The function is what we call an oscillatory function, and in your studies in physics, you will find it most important to appreciate this property, particularly in the study of waves and alternating current in electricity and electronics. How does the cosine function behave? Panel 10 again shows you. At $0^\\circ$, $q = p$ and so the value is $1$. At $90^\\circ$, $p = 0$, and so $\\cos(90^\\circ) = 0$. This is shown in Panel 10b. You can work out the rest of it for yourself. You see that the cosine function is exactly the same as the sine function if you slide the sine function graph back $90^\\circ$. This says that $\\sin(90^\\circ + q) = \\cos (\\theta)$. The tangent curve looks quite different. Panel 11 shows that $\\tan(0) = r/p = 0$, but $\\tan(90^\\circ) = r/p = \\infty$ since $r$ is finite but $p$ has gone to $0$. Panel 11 shows this function. The tangent function is a repeating one but not oscillatory. One last point which you will find used many times are the approximate values of these functions when $\\theta$ is very small. Look", "c) $$sin \\; 390^{\\circ}$$ d) $$sin \\; (-30^{\\circ})^{\\circ}$$ e) $$cos \\; 60^{\\circ}$$ • 1, Chapter 4.2, Tangent Function, ### Tangent Function The last function to consider here is $$tan \\; x$$, which has rather different properties from $$sin \\; x$$ and $$cos \\; x$$. We can use a graphic calculator or computer to sketch the function $$y=tan \\; x$$ for $$x$$ in any range. The function is shown below. The period of the function is the distance between any two consecutive asymptotes (dotted lines); the period for $$y=tan \\; x$$ is $$180^{\\circ}$$. Unlike the sine and cosine functions, tan $$x$$ is not bounded by $$\\pm 1$$. In fact, as $$x$$ increases to $$90^{\\circ}$$, $$tan \\; x$$ increases without limit. • 1, Chapter 4.3, Remember, ### Remember Here is a useful reminder of some of the values of the trig functions • 1, Chapter 4, Task 2, Worked Example 4, ### Worked Example • #### Worked Example Using the values: $$tan \\; 45^{\\circ} = 1 \\;$$ and $$tan \\; 30^{\\circ} = \\frac{1}{\\sqrt{3}}$$ find the following without a calculator: a) $$tan \\; 135^{\\circ}$$ b) $$tan \\; (-45^{\\circ})$$ c) $$tan \\; 315^{\\circ}$$ d) $$tan \\; (-30^{\\circ})$$ e) $$tan \\; 150^{\\circ}$$ f) $$tan \\; 60^{\\circ}$$ • 0, Chapter 5, Solving Trigonometrical Equations, ### Solving Trigonometrical Equations Trigonometric functions can be used to", "period of $6\\pi$ . 1. The period is $\\frac{\\pi}{\\frac{3}{2}}=\\pi \\cdot \\frac{2}{3}=\\frac{2\\pi}{3}$ . 2. The zeros are where $y$ is zero. $0 &=-4\\tan \\frac{3}{2}x \\\\0 &=\\tan \\frac{3}{2}x \\\\\\frac{3}{2}x &=\\tan ^{-1}0=0, \\pi, 2\\pi, 3\\pi \\\\x &=\\frac{2}{3}(0, \\pi, 2\\pi, 3\\pi ) \\\\x &=0, \\frac{2\\pi}{3}, \\frac{4\\pi}{3}, 2\\pi$ 3. The general equation is $y=a\\tan bx$ . We know that $a = 8$ . Let’s use the period to solve for the frequency, or $b$ . $\\frac{\\pi}{b} &=6\\pi \\\\b &=\\frac{\\pi}{6\\pi}=\\frac{1}{6}$ The equation is $y=8\\tan \\frac{1}{6}x$ . ### Vocabulary Tangent Function Defined by the coordinates $(\\theta, \\tan \\theta)$ , where $\\theta$ is the central angle from the unit circle and tangent is the ratio of the sine and cosine functions. ### Practice Graph the following tangent functions over $[0, 4\\pi]$ . Determine the period, domain, and range. 1. $y=2\\tan x$ 2. $y=-\\frac{1}{3}\\tan x$ 3. $y=-\\tan 3x$ 4. $y=4\\tan 2x$ 5. $y=\\frac{1}{2}\\tan 4x$ 6. $y=-\\tan \\frac{1}{2}x$ 7. $y=4+\\tan x$ 8. $y=-3+\\tan 3x$ 9. $y=1+\\frac{2}{3}\\tan \\frac{1}{2}x$ 10. Find the zeros of the function from #1. 11. Find the zeros of the function from #3. 12. Find the zeros of the function from #5. Write the equation of the tangent function, in the form $y=a\\tan bx$ , with the given amplitude and period. 1. Amplitude: 3 Period: $\\frac{3\\pi}{2}$ 2. Amplitude: $\\frac{1}{4}$ Period: $2\\pi$ 3. Amplitude: -2.5 Period: 8", "function $\\tan u$, is cyclical with a period of $\\pi$. Substituting $u$ with an expression about $x$ gives $3 \\cdot x = k \\cdot \\pi$ $x = \\frac{k}{3} \\cdot \\pi$ Thus coordinates of $x$ -intercepts of this function shall fit into the general expression $\\left(\\frac{k}{3} \\cdot \\pi , 0\\right)$ Taking $k = - 1$, $k = 0$, and $k = 1$ gives $x$ -intersects $\\left(- \\frac{k}{3} \\cdot \\pi , 0\\right)$, $\\left(0 , 0\\right)$, and $\\left(\\frac{k}{3} \\cdot \\pi , 0\\right)$. 2. Monotonies The tangent function always increases as the angle grows, as seen from a unit circle. Therefore the graph of $\\tan x$ should slope upwards and extend to the upper-right corner of the Cartesian plane. This observation is also the case for a composite tangent function like $\\tan u$ as long as the inner function $u = f \\left(x\\right)$ is increasing. 3.Asymptotes The tangent function is not defined at the sum of $\\frac{\\pi}{2}$ any integer multiple of $\\pi$ and shows asymptotic behavior at each of these values of $x$. That is $x = \\frac{\\pi}{2} + k \\cdot \\pi$ where $k$ is an integer. For the composite tangent function here, the general expression for all the asymptotes would be $3 \\cdot x = u = \\frac{\\pi}{2} + k \\cdot \\pi$ and $x = \\left(\\frac{1}{6} + \\frac{k}{3}\\right) \\cdot \\pi$ Evaluating", "blue segment) is the tangent of $$(180{}^\\circ-\\theta)$$. $$\\tan (\\theta \\pm 90{}^\\circ)$$ = $$-1/\\tan \\theta$$. The picture below illustrates the geometry of this identity when $$\\theta$$ is in the first quadrant. The line formed at $$\\theta + 90{}^\\circ$$ is perpendicular to OC and $$\\triangle COB\\sim \\triangle ODB$$. Thus $$\\dfrac{BD}{OB}=\\dfrac{OB}{BC}$$, and with appropriate signs, $$\\tan (\\theta + 90{}^\\circ)$$ = $$-1/\\tan \\theta$$. Since $$\\tan \\theta$$=$$\\tan (\\theta+180{}^\\circ)$$, $$\\tan (\\theta +90{}^\\circ)=\\tan(\\theta-90{}^\\circ)$$. The applet below shows the geometry in all quadrants, and it gives a dynamic sense of the relationship between $$\\tan\\theta$$ and $$\\tan(-\\theta)$$. Again, move the blue point: ## Special Bonus: The Secant Function The signed length of the segment OC is called the secant function, $$\\sec\\theta$$. Using similar triangles, we see that $$\\sec \\theta = \\dfrac{1}{\\cos \\theta}$$. The Pythagorean Theorem applied to $$\\triangle COB$$ shows that $$\\tan^2\\theta+1=\\sec^2 \\theta$$. When the tangent function is big, so is the secant function, and when the tangent function is small, so is the secant function. Also $$\\sec \\theta$$ is close to $$\\pm 1$$ when $$\\theta$$ is close to the x-axis and when $$\\tan \\theta$$ is close to 0. The graphs of the two functions look nice together: ### 9 Responses 1. ##### Michael Enright I was doing something today with a student and it dawned on me that the tangent was called that because it really", "even $\\cos(-x) = \\cos x$, we can simplify the above equation to: $$\\tan(-x) = -\\left(\\frac{\\sin x }{\\cos x}\\right) = -\\tan x$$ which means that the tangent function is an odd function.", "? Free Version Easy # Converting Tangent to Cotangent TRIG-AGNXTA Given $\\tan { { 85 }^{ \\circ } }$, find the $\\cot { { 85 }^{ \\circ } }$ (rounded to the nearest thousandth). A $0.088$ B $0.087$ C $1.004$ D $11.474$", "# Chapter 3 - Review Exercises - Page 136: 41 $\\tan{1} \\gt \\tan{2}$ #### Work Step by Step Recall: (1) The tangent function's period is $\\pi$ or around $3.14$; (2) The tangent function is increasing in the intervals $(-1.57, 1.57), (1.57, 4.71), (4.71, 7.85), ...$ (3) The value of the tangent function is $0$ when $x=0, \\pi, 2\\pi, 3\\pi, ...$ or, in approximate decimal form, $x=0, 3.14, 6.28, 9.42, ...$. (4) The value of the tangent function is negative in the intervals $(-1.57, 0), (1.57, 3.14), (4.71, 6.28), ....$ (5) The value of the tangent function is positive in the intervals $(0, 1.57), (3.14, 4.71), (6.28, 7.85), ....$ Thus, the value of $\\tan{1}$ is positive, and the value of $\\tan{2}$ is negative. Therefore, $\\tan{1} \\gt \\tan{2}$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]

[ "Mathematics » Functions II » The Tangent Function # Sketching Tangent Graphs ## Example ### Question Sketch the graph of $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$ for $$-\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}$$. ### Examine the form of the equation From the equation we see that $$0 < k < 1$$, therefore the branches of the graph will be less steep than the standard tangent graph $$y = \\tan \\theta$$. We also notice that $$p < 0$$ so the graph will be shifted to the right on the $$x$$-axis. ### Determine the period The period for $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$ is: \\begin{align*} \\text{Period} &= \\cfrac{\\text{180}\\text{°}}{|k|} \\\\ &= \\dfrac{\\text{180}\\text{°}}{\\cfrac{1}{2}} \\\\ &= \\text{360}\\text{°} \\end{align*} ### Determine the asymptotes The standard tangent graph, $$y = \\tan \\theta$$, for $$-\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}$$ is undefined at $$\\theta = -\\text{90}\\text{°}$$ and $$\\theta = \\text{90}\\text{°}$$. Therefore we can determine the asymptotes of $$f(\\theta) = \\tan \\cfrac{1}{2}(\\theta – \\text{30}\\text{°})$$: • $$\\cfrac{-\\text{90}\\text{°}}{\\text{0.5}} + \\text{30}\\text{°} = -\\text{150}\\text{°}$$ • $$\\cfrac{\\text{90}\\text{°}}{\\text{0.5}} + \\text{30}\\text{°} = \\text{210}\\text{°}$$ The asymptote at $$\\theta = \\text{210}\\text{°}$$ lies outside the required interval. ### Plot the points and join with a smooth curve Period: $$\\text{360}\\text{°}$$ Domain: $$\\{ \\theta: -\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}, \\theta \\ne -\\text{150}\\text{°} \\}$$ Range: $$(-\\infty;\\infty)$$ $$y$$-intercepts: $$(\\text{0}\\text{°};-\\text{0.27})$$ $$x$$-intercept: $$(\\text{30}\\text{°};0)$$ Asymptotes: $$\\theta = -\\text{150}\\text{°}$$ Do you want to suggest a correction", "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "n\\cdot 180^\\circ;$$ Other angles are $$=20^\\circ,200^\\circ,380^\\circ,460^\\circ$$ We can add or subtract $$n\\cdot \\pi$$ ( $$n$$ positive or negative integer) to $$20^\\circ$$ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding $$180^\\circ$$. You suggest we use that $$\\tan \\theta = \\tan(\\theta - 180^\\circ).$$ Equivalently, if we set $$\\phi = \\theta - 180^\\circ$$, the above equation reads $$\\tan(\\phi+180^\\circ) = \\tan\\phi.$$ To summarize: the tangent function is $$180^\\circ$$-periodic: it repeats every $$180^\\circ$$. So the possible answers are $$\\dots,-340^\\circ,-160^\\circ, 20^\\circ, 200^\\circ, 380^\\circ, \\dots$$ Narasimham's answer shows how to express this in a single equation with $$n$$ varying, ABCD does the same in the language of radians ($$180^\\circ = \\pi$$ radians).", "## Algebra 2 (1st Edition) For $\\theta=90^{\\circ}$, $\\cos \\theta=\\cos 90^{\\circ}=0$ Thus, the function tangent is undefined. AND For $\\theta=90^{\\circ}$, $\\sin \\theta=\\sin 90^{\\circ}=1 \\neq 0$ Thus, the function cotangent is defined. Since, $\\tan \\theta=\\dfrac{Opposite}{Adjacent}=\\dfrac{y}{x}$ and $\\tan \\theta=\\dfrac{\\sin \\theta }{\\cos \\theta}$ when $\\cos \\theta=0$, then the tangent function will be undefined. For $\\theta=90^{\\circ}$, $\\cos \\theta=\\cos 90^{\\circ}=0$ Thus, the function tangent is undefined. Also, $\\cot \\theta=\\dfrac{1}{\\dfrac{\\sin \\theta }{\\cos \\theta}}$ Thus, $\\cot \\theta=\\dfrac{\\cos \\theta }{\\sin \\theta}$ For $\\theta=90^{\\circ}$, $\\sin \\theta=\\sin 90^{\\circ}=1 \\neq 0$ Thus, the function cotangent is defined.", "simply 1, and in any case $50^\\circ$ is closer to $45^\\circ$ than to $60^\\circ$, so it makes more sense to use $45^\\circ$ as the special angle. 9) Leaving calculators in radian mode will yield the wrong answers $-0.9820$ and $45.1831$. 10) Students may end up thinking a little too hard about this problem. All they’re supposed to conjecture is that the two expressions are not equal, so if any of them struggle with this problem, find out if they’ve got that much figured out and reassure them they can stop there. 1) Use a calculator to find the tangent of $86^\\circ, 87^\\circ, 88^\\circ,$ and $89^\\circ$. Then, find the tangent of $94^\\circ, 93^\\circ, 92^\\circ,$ and $91^\\circ$. Now make a conjecture about the behavior of the tangent function as $x$ approaches $90^\\circ.$ 1) The values of the tangent function are as follows: $& X && \\tan x\\\\& 86^\\circ && 14.3007\\\\& 87^\\circ && 19.0811\\\\& 88^\\circ && 28.6363\\\\& 89^\\circ && 57.2900\\\\& 90^\\circ && \\text{undefined}\\\\& 91^\\circ && -57.2900\\\\& 92^\\circ && -28.6363\\\\& 93^\\circ && -19.0811\\\\& 94^\\circ && -14.3007$ The tangent function approaches infinity as $x$ approaches $90^\\circ$ from below, and approaches negative infinity as $x$ approaches $90^\\circ$ from above. ## Relating Trigonometric Functions In-Text Examples 4) Demonstrating that $\\cot \\theta = \\frac{\\cos \\theta}{\\sin \\theta}$ because cot is the reciprocal of tan and $\\tan \\theta", "## Trigonometry (11th Edition) Clone The solution set is $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$ $$\\tan\\theta+1=0$$ 1) Solve the equation over the interval $[0^\\circ,360^\\circ)$ $$\\tan\\theta+1=0$$ $$\\tan\\theta=-1$$ Over the interval $[0^\\circ, 360^\\circ)$, there are two values of $\\theta$ where $\\tan\\theta=-1$, which are $135^\\circ$ and $315^\\circ$. Therefore, $$\\theta=\\{135^\\circ, 315^\\circ\\}$$ 2) Solve the equation for all solutions To find all solutions, we add the integer multiples of the period of tangent function, which is $180^\\circ$, to each solution found in step 1. - There are two solutions found in step 1, $\\{135^\\circ, 315^\\circ\\}$, so it will be written as $\\theta=135^\\circ+180^\\circ n$ and $\\theta=315^\\circ+180^\\circ n$ where $n\\in Z$. - However, a closer look reveals that $135^\\circ+180^\\circ n$ in fact covers all the possible values of $315^\\circ+180^\\circ n$, and vice versa. In other words, they are technically the same. So the solution set can be shortened to $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$", "# Difference between revisions of \"1996 AIME Problems/Problem 10\" ## Problem Find the smallest positive integer solution to $\\tan{19x^{\\circ}}=\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$. ## Solution $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-45^{\\circ})}} =$ $\\dfrac{2\\sin{141^{\\circ}}\\cos{45^{\\circ}}}{2\\cos{141^{\\circ}}\\sin{45^{\\circ}}} = \\tan{141^{\\circ}}$. The period of the tangent function is $180^\\circ$, and the tangent function is one-to-one over each period of its domain. Thus, $19x \\equiv 141 \\pmod{180}$. Since $19^2 \\equiv 361 \\equiv 1 \\pmod{180}$, multiplying both sides by $19$ yields $x \\equiv 141 \\cdot 19 \\equiv (140+1)(18+1) \\equiv 0+140+18+1 \\equiv 159 \\pmod{180}$. Therefore, the smallest positive solution is $x = \\boxed{159}$. ## Solution 2 $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\dfrac{1 + \\tan{96^{\\circ}}}{1-\\tan{96^{\\circ}}}$ which is the same as $\\dfrac{\\tan{45^{\\circ}} + \\tan{96^{\\circ}}}{1-\\tan{45^{\\circ}}\\tan{96^{\\circ}}} = \\tan{141{^\\circ}}$. So $19x = 141 +180n$, for some integer $n$. Multiplying by $19$ gives $x \\equiv 141 \\cdot 19 \\equiv 2679 \\equiv 159 \\pmod{180}$. The smallest positive solution of this is $x = \\boxed{159}$ ## Solution 3 (Only sine and cosine sum formulas) It seems reasonable to assume that $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\tan{\\theta}$ for some angle $\\theta$. This means $$\\dfrac{\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}})}{\\alpha (\\cos{96^{\\circ}}-\\sin{96^{\\circ}})} = \\frac{\\sin{\\theta}}{\\cos{\\theta}}$$ for some constant $\\alpha$. We can set $\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}}) = \\alpha \\sin{\\theta}$.Note that if we have $\\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\\sin{45^{\\circ}} = \\cos{45^{\\circ}} = \\tfrac{\\sqrt{2}}{2}$,", "# Difference between revisions of \"1996 AIME Problems/Problem 10\" ## Problem Find the smallest positive integer solution to $\\tan{19x^{\\circ}}=\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$. ## Solution $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-45^{\\circ})}} =$ $\\dfrac{2\\sin{141^{\\circ}}\\cos{45^{\\circ}}}{2\\cos{141^{\\circ}}\\sin{45^{\\circ}}} = \\tan{141^{\\circ}}$. The period of the tangent function is $180^\\circ$, and the tangent function is one-to-one over each period of its domain. Thus, $19x \\equiv 141 \\pmod{180}$. Since $19^2 \\equiv 361 \\equiv 1 \\pmod{180}$, multiplying both sides by $19$ yields $x \\equiv 141 \\cdot 19 \\equiv (140+1)(18+1) \\equiv 0+140+18+1 \\equiv 159 \\pmod{180}$. Therefore, the smallest positive solution is $x = \\boxed{159}$. ## Solution 2 $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$ So $19x = 141 +180n$, for some integer $n$. Then $0 \\equiv 8 + 9n \\pmod{19}$, from which $n \\equiv 16 \\pmod{19}$ after some manipulation. The smallest suitable value of $n$ is therefore 16 from which $x = \\boxed{159}$", "c) $$sin \\; 390^{\\circ}$$ d) $$sin \\; (-30^{\\circ})^{\\circ}$$ e) $$cos \\; 60^{\\circ}$$ • 1, Chapter 4.2, Tangent Function, ### Tangent Function The last function to consider here is $$tan \\; x$$, which has rather different properties from $$sin \\; x$$ and $$cos \\; x$$. We can use a graphic calculator or computer to sketch the function $$y=tan \\; x$$ for $$x$$ in any range. The function is shown below. The period of the function is the distance between any two consecutive asymptotes (dotted lines); the period for $$y=tan \\; x$$ is $$180^{\\circ}$$. Unlike the sine and cosine functions, tan $$x$$ is not bounded by $$\\pm 1$$. In fact, as $$x$$ increases to $$90^{\\circ}$$, $$tan \\; x$$ increases without limit. • 1, Chapter 4.3, Remember, ### Remember Here is a useful reminder of some of the values of the trig functions • 1, Chapter 4, Task 2, Worked Example 4, ### Worked Example • #### Worked Example Using the values: $$tan \\; 45^{\\circ} = 1 \\;$$ and $$tan \\; 30^{\\circ} = \\frac{1}{\\sqrt{3}}$$ find the following without a calculator: a) $$tan \\; 135^{\\circ}$$ b) $$tan \\; (-45^{\\circ})$$ c) $$tan \\; 315^{\\circ}$$ d) $$tan \\; (-30^{\\circ})$$ e) $$tan \\; 150^{\\circ}$$ f) $$tan \\; 60^{\\circ}$$ • 0, Chapter 5, Solving Trigonometrical Equations, ### Solving Trigonometrical Equations Trigonometric functions can be used to", "Trigonometry (11th Edition) Clone $\\tan{1} \\gt \\tan{2}$ Recall: (1) The tangent function's period is $\\pi$ or around $3.14$; (2) The tangent function is increasing in the intervals $(-1.57, 1.57), (1.57, 4.71), (4.71, 7.85), ...$ (3) The value of the tangent function is $0$ when $x=0, \\pi, 2\\pi, 3\\pi, ...$ or, in approximate decimal form, $x=0, 3.14, 6.28, 9.42, ...$. (4) The value of the tangent function is negative in the intervals $(-1.57, 0), (1.57, 3.14), (4.71, 6.28), ....$ (5) The value of the tangent function is positive in the intervals $(0, 1.57), (3.14, 4.71), (6.28, 7.85), ....$ Thus, the value of $\\tan{1}$ is positive, and the value of $\\tan{2}$ is negative. Therefore, $\\tan{1} \\gt \\tan{2}$.", "$$k > 1$$, the period of the tangent function decreases. For $$0 < k < 1$$, the period of the tangent function increases. • For $$k < 0$$: For $$-1 < k < 0$$, the graph is reflected about the $$y$$-axis and the period increases. For $$k < -1$$, the graph is reflected about the $$y$$-axis and the period decreases. Negative angles: $\\tan (-\\theta) = -\\tan \\theta$ Calculating the period: To determine the period of $$y = \\tan k\\theta$$ we use, $\\text{Period} = \\cfrac{\\text{180}\\text{°}}{|k|}$ where $$|k|$$ is the absolute value of $$k$$. $$k > 0$$ $$k < 0$$ ## Example ### Question 1. Sketch the following functions on the same set of axes for $$-\\text{180}\\text{°} \\leq \\theta \\leq \\text{180}\\text{°}$$. 1. $$y_1 = \\tan \\theta$$ 2. $$y_2 = \\tan \\cfrac{3\\theta}{2}$$ 2. For each function determine the following: • Period • Domain and range • $$x$$- and $$y$$-intercepts • Asymptotes ### Examine the equations of the form $$y = \\tan k\\theta$$ Notice that $$k > 1$$ for $$y_2 = \\tan \\cfrac{3\\theta}{2}$$, therefore the period of the graph decreases. ### Complete a table of values θ $$-\\text{180}$$$$\\text{°}$$ $$-\\text{135}$$$$\\text{°}$$ $$-\\text{90}$$$$\\text{°}$$ $$-\\text{45}$$$$\\text{°}$$ $$\\text{0}$$$$\\text{°}$$ $$\\text{45}$$$$\\text{°}$$ $$\\text{90}$$$$\\text{°}$$ $$\\text{135}$$$$\\text{°}$$ $$\\text{180}$$$$\\text{°}$$ $$\\tan \\theta$$ $$\\text{0}$$ $$\\text{1}$$ undef $$-\\text{1}$$ $$\\text{0}$$ $$\\text{1}$$ undef $$-\\text{1}$$ $$\\text{0}$$ $$\\tan \\cfrac{3\\theta}{2}$$ undef $$-\\text{0.41}$$ $$\\text{1}$$ $$-\\text{2.41}$$ $$\\text{0}$$ $$\\text{2.41}$$ $$-\\text{1}$$ $$\\text{0.41}$$ undef ### Complete the table", "1.39 Exercise 1.39: A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert: $tan x = \\frac{x}{1 - \\frac {x^2}{3 - \\frac{x^2}{5-\\cdots}}}$ where x is in radians. Define a procedure (tan-cf x k) that computes an approximation to the tangent function based on Lambert’s formula. k specifies the number of terms to compute, as in Exercise 1.37. Let's first review the general continued fraction: $f = \\frac{N_1}{D_1 + \\frac{N_2}{D_2 + \\frac{N_3}{D_3 + \\cdots}}}$ Here, $f = tan, f(x) = tan(x)$. And $D_1 = 1, D_2 = 3, D_3 = 5, \\cdots$, while $N_1(x) = x^1, N_2(x)=-x^2, N_3(x)=-x^2, \\cdots$. Firstly we can define d as: (define (d i) (- (* 2 i) 1)) (d 3) Then we can define n as: (define (n i x) (if (= i 1) x (- (* x x)))) (n 1 3) (n 2 3) (n 3 3) (n 4 3) Now, let's define tan-cf based on cont-frac: (define (cont-frac n d k x) (define (g i k) (if (= i k) (/ (n k x) (d k)) (/ (n i x) (+ (d i) (g (+ i 1) k))) ) ) (g 1 k) ) (define (tan-cf x k) (cont-frac n d k x) ) (tan-cf 3 30) Compare the result with (tan 3)", "blue segment) is the tangent of $$(180{}^\\circ-\\theta)$$. $$\\tan (\\theta \\pm 90{}^\\circ)$$ = $$-1/\\tan \\theta$$. The picture below illustrates the geometry of this identity when $$\\theta$$ is in the first quadrant. The line formed at $$\\theta + 90{}^\\circ$$ is perpendicular to OC and $$\\triangle COB\\sim \\triangle ODB$$. Thus $$\\dfrac{BD}{OB}=\\dfrac{OB}{BC}$$, and with appropriate signs, $$\\tan (\\theta + 90{}^\\circ)$$ = $$-1/\\tan \\theta$$. Since $$\\tan \\theta$$=$$\\tan (\\theta+180{}^\\circ)$$, $$\\tan (\\theta +90{}^\\circ)=\\tan(\\theta-90{}^\\circ)$$. The applet below shows the geometry in all quadrants, and it gives a dynamic sense of the relationship between $$\\tan\\theta$$ and $$\\tan(-\\theta)$$. Again, move the blue point: ## Special Bonus: The Secant Function The signed length of the segment OC is called the secant function, $$\\sec\\theta$$. Using similar triangles, we see that $$\\sec \\theta = \\dfrac{1}{\\cos \\theta}$$. The Pythagorean Theorem applied to $$\\triangle COB$$ shows that $$\\tan^2\\theta+1=\\sec^2 \\theta$$. When the tangent function is big, so is the secant function, and when the tangent function is small, so is the secant function. Also $$\\sec \\theta$$ is close to $$\\pm 1$$ when $$\\theta$$ is close to the x-axis and when $$\\tan \\theta$$ is close to 0. The graphs of the two functions look nice together: ### 9 Responses 1. ##### Michael Enright I was doing something today with a student and it dawned on me that the tangent was called that because it really", "# 2015-19 Sum of tangent functions Evaluate $\\sum_{k=0}^{n} 2^k \\tan \\frac{\\pi}{4\\cdot 2^{n-k}}.$ GD Star Rating loading... 2015-19 Sum of tangent functions, 3.9 out of 5 based on 7 ratings", "tangent, ${\\tan}^{-} 1$ to find the value of $A$.", "to Euclid's formula) may be generated sequentially with $$k_{n+1}=k_n+\\sqrt{2k_n^2+(-1)^{k_n}}.$$ These are pell numbers $$\\{1,2,5,12,\\cdots\\}$$ to be used in pairs like $$\\space(2,1),\\space (5,2),\\space (12,5)\\cdots.\\space$$ For $$\\tan\\theta<1,\\space$$ we use these steps 1. Convert tangent to a $$1$$-to-$$3$$ digit fraction to identify the A:B ratio. 2. Solve the tangent function for $$k$$ 3. Test a range of $$m$$-values to see which yields a $$k$$-value closest to an integer, 4. Use $$m$$ and the rounded value of $$k$$ to generate the triple with Euclid's formula. $$$$\\tan\\theta=\\dfrac{A}{B} = \\dfrac{m^2-k^2}{2mk}\\\\ \\\\ 2mkA=(m^2-k^2)B\\\\ \\\\ B k^2+2 A k m - B m^2 = 0 \\\\ \\implies k = \\dfrac{\\sqrt{A^2 m^2 + B^2 m^2} - A m}{B}\\\\ \\quad \\text{ for }\\quad 2 \\le m \\le 50$$$$ This range is chosen to accommodate fractions up to $$3$$ digits. Example $$\\tan39^\\circ\\approx 0.80978\\approx \\dfrac{149}{184} \\implies A=149\\quad B=184$$ $$k = \\dfrac{\\sqrt{149^2 m^2 + 184^2 m^2} - 149 m}{184} \\quad \\text{ for } 2\\le m \\le 30 \\\\\\text{and we find the best fit is } \\quad m=21\\quad k\\approx 10.016\\approx 10 \\\\ f(21,10)=(341,420,541)\\\\ \\tan\\theta\\approx\\dfrac{341}{420}\\approx 0.81190476 \\\\ \\tan^{-1} 0.81190476\\approx 39.07^\\circ$$ For any positive integer $$m>n$$, $$\\begin{gather} z=m+ni= r\\angle \\phi\\\\ z^2=(m+ni)^2=(m^2-n^2)+2mni=r^2\\angle 2\\phi \\end{gather}$$ where $$r=\\sqrt{m^2+n^2}$$ and $$\\tan \\phi=\\frac{n}{m}$$. Thus $$\\{m^2+n^2,m^2-n^2,2mn\\}$$ is a Pythagorean triplet. In your case $$2\\phi=56.25^\\circ$$ or $$\\phi=28.125^\\circ$$. Find the simplest $$m>n$$ such that $$\\tan 28.125^\\circ=\\frac{n}{m}$$. \\begin{align} \\tan 28.125^\\circ&=\\frac{n}{m}\\\\ 0.5&\\approx\\frac{n}{m}\\\\ \\end{align} $$m=2$$", "# How do you find the value of $\\sum_{r=0}^{44} \\tan^2(2r+1)$? Problem: Find the value of $$\\sum_{r=0}^{44} \\tan^2(2r+1)$$ Note: The angles here are in degrees. I don't know how to solve this question because trigonometric simplifications didn't get me anywhere. I think there was a method to solve this question using complex numbers which I no longer remember. Any hint/help will be appreciated. $$\\tan90x=\\dfrac{\\binom{90}1t-\\binom{90}3t^3+\\cdots+\\binom{90}{89}t^{89}}{\\binom{90}0-\\binom{90}2t^2+\\cdots+\\binom{90}{90}t^{90}}$$ where $t=\\tan x$ If $\\tan 90x=\\tan90^\\circ=\\infty$ $90x=180^\\circ n+90^\\circ=90^\\circ(2n+1)$ where $n$ is any integer $\\implies x=(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ So, the roots of $$t^{90}-\\binom{90}{88}t^{88}+\\cdots=0$$ are $\\tan(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ Now as $-\\tan(2n+1)^\\circ=\\tan\\{180^\\circ-(2n+1)^\\circ\\}=\\tan\\{2(89-n)+1)^\\circ\\}$ $$\\implies\\sum_{r=0}^{89}\\tan^2(2n+1)^\\circ=2\\sum_{r=0}^{44}\\tan^2(2n+1)^\\circ$$ and the roots of $$s^{45}-\\binom{90}{88}s^{44}+\\cdots=0$$ are $\\tan^2(2n+1)^\\circ$ where $n\\equiv0,1,2,\\cdots,88,89\\pmod{90}$ Using Vieta's formula, $$\\sum_{r=0}^{89}\\tan^2(2n+1)^\\circ=\\binom{90}{88}=\\binom{90}{90-88}=?$$ Can you take it from here? Using an idea similar to this answer, note that the function $$\\frac{90/z}{z^{90}-1}$$ has residue $1$ for $z=e^{k\\pi i/45}$ and residue $-90$ at $z=0$. On $|z|=1$, $$\\tan(\\theta/2)=-i\\frac{z-1}{z+1}$$ Integrating $$f(z)=-\\left(\\frac{z-1}{z+1}\\right)^2\\frac{90/z}{z^{90}-1}$$ around a large circle is $0$ since the integrand is approximately $|z|^{-91}$. Thus, the sum of residues is $$2\\sum_{k=0}^{44}\\tan^2\\left(\\frac{2k\\pi}{180}\\right)+\\operatorname*{Res}_{z=0}f(z)+\\operatorname*{Res}_{z=-1}f(z)=0$$ Since $\\operatorname*{Res}\\limits_{z=0}f(z)=90$ and $\\operatorname*{Res} \\limits_{z=-1}f(z)=-\\frac{8102}{3}$, we get \\begin{align} \\sum_{k=0}^{44}\\tan^2\\left(\\frac{2k\\pi}{180}\\right) &=\\frac12\\left(\\frac{8102}{3}-90\\right)\\\\ &=\\frac{3916}{3} \\end{align} The sum in the question is the difference between the value in answer cited above and this value, that is \\begin{align} \\sum_{k=0}^{44}\\tan^2\\left(\\frac{(2k+1)\\pi}{180}\\right) &=\\frac{15931}{3}-\\frac{3916}{3}\\\\ &=4005 \\end{align}", "Move the blue point below; the tangent is the length of the red segment. (If a label is getting in the way, right click and toggle “show label” from the menu). The circle definition of the tangent function leads to geometric illustrations of many standard properties and identities. (If this were my class, I would stop here and tell you to explore on your own and with others). ## Some things to notice: $$\\left| \\tan \\theta \\right|$$ gets big as $$\\theta$$ approaches $$\\pm 90{}^\\circ$$. $$\\tan (\\pm 90{}^\\circ)$$ is undefined, because at these angles, $$\\overline{OA}$$ is parallel to x=1, so the two lines don’t intersect, and point C doesn’t exist. $$\\tan 90{}^\\circ$$ tends toward $$+\\infty$$, $$\\tan (-90{}^\\circ)$$ tends toward $$-\\infty$$. $$\\tan \\theta$$ is positive in the first and third quadrants, negative in the second and fourth quadrants. $$\\tan \\theta$$=$$tan (\\theta+180{}^\\circ)$$ — the angles $$\\theta$$ and $$\\theta +180{}^\\circ$$ form the same line. Thus the period of the tangent function is $$180 {}^\\circ = \\pi$$ radians. $$\\tan \\theta$$ = $$- \\tan (-\\theta)$$. Moving from $$\\theta$$ to $$-\\theta$$ reflects $$OC$$ about the x-axis. $$\\tan \\theta$$ is equal to the slope of OA (rise = $$\\tan \\theta$$ , run =1), which is also equal to $$\\dfrac{\\sin\\theta}{\\cos\\theta}$$, as well as Opposite over Adjacent for angle $$\\theta$$ in right triangle CBO. $$\\tan (45{}^\\circ)=1$$. When $$\\theta=45{}^\\circ$$,", "Trigonometry (11th Edition) Clone The solution set is $$\\{60^\\circ+180^\\circ n, n\\in Z\\}$$ $$\\tan\\theta\\csc\\theta-\\sqrt3\\csc\\theta=0$$ 1) Solve the equation over the interval $[0^\\circ,360^\\circ)$ $$\\tan\\theta\\csc\\theta-\\sqrt3\\csc\\theta=0$$ $$\\csc\\theta(\\tan\\theta-\\sqrt3)=0$$ $$\\csc\\theta=0\\hspace{1cm}\\text{or}\\hspace{1cm}\\tan\\theta=\\sqrt3$$ - For $\\csc\\theta=0$ The range of a cosectant function is $(-\\infty,-1]\\cup[1,\\infty)$. Since $0\\notin(-\\infty,-1]\\cup[1,\\infty)$, there cannot be any values of $\\theta\\in[0^\\circ, 360^\\circ)$ that have $\\csc\\theta=0$. - For $\\tan\\theta=\\sqrt3$ Over the interval $[0^\\circ, 360^\\circ)$, there are 2 values of $\\theta$ where $\\tan\\theta=\\sqrt3$, which are $60^\\circ$ and $240^\\circ$. Therefore, overall, $$\\theta=\\{60^\\circ, 240^\\circ\\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the tangent function is $180^\\circ$. - We apply it to each solution found in step 1, which would lead to $\\theta=60^\\circ+180^\\circ n$ and $\\theta=240^\\circ+180^\\circ n$ $(n\\in Z)$ - However, both $60^\\circ+180^\\circ n$ and $240^\\circ+180^\\circ n$ refer to the same set of points, so we only need to include one in the solution set. The solution set is $$\\{60^\\circ+180^\\circ n, n\\in Z\\}$$", "= \\frac {2}{3}\\\\ \\tan\\theta = \\frac {2}{3}\\\\ \\theta = \\arctan (\\frac {2}{3})$ This is not one of your special angles, so you will need a calculator. $\\theta = 33.69^\\circ$ $\\tan \\theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\\circ)$. There is second answer in the interval $[0,360^\\circ)$. $\\tan \\theta = \\tan (\\theta + 180)$ $213.69^\\circ$ is also in the solution set. The next one might even be a little bit trickier. $\\cos 3\\theta = -0.5\\\\ 3\\theta = \\arccos (-0.5)\\\\ 3\\theta = 120^\\circ$ but again, this isn't the only answer. $\\theta = \\arccos (-0.5)$ has two answers in the 360 degree interval, and $3\\theta = \\arccos (-0.5)$ will have $6.$ $\\cos \\theta = \\cos (-\\theta)$ and $\\cos \\theta = \\cos (360+\\theta)$ using these identities we find other values of theta. $3\\theta = 240^\\circ, 480^\\circ, 600^\\circ, 840^\\circ, 960^\\circ$ are all viable, as well. $\\theta \\in \\{40, 80, 160, 200, 280, 320\\}$ a) Reduce to $$\\tan\\theta=\\frac23.$$ Then as the tangent function has period $180°$ and is monotonic in a period, $$\\tan\\theta=\\tan(\\theta+k180°)$$ and all solutions are of the form $$\\theta+k180°=\\arctan\\frac23.$$ Find all $k$ such that $0\\le\\theta\\le360°$. b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that $$\\cos3\\theta=\\cos(k360°\\pm3\\theta).$$ Then all solutions are of the form $$k120°\\pm\\theta=\\frac13\\arccos(-0.5).$$ Again, find all $k$" ]

[ "\\frac{w}{\\alpha r}$. Hence $\\alpha = \\frac{w}{\\rho r}$ (radians). - 1. Compute tangents intersections between the object and the circle. How you do this depends a lot on how your objects are represented. 2. Compute the angle corresponding to this pair of points. An aran2 kind of function might help in this step. 3. Divide that angle by the ratio you're given, i.e. by 0.5 for 50%. -", "Calculus 8th Edition $\\frac{2}{3}$ radians We use the formula $s=θr$ We re-arrange the formula for $θ$: $θ=\\frac{s}{r}$ We plug in the given values and simplify: $θ=\\frac{1}{1.5}=\\frac{2}{3}$ radians", "\\frac{2}{\\sqrt{\\theta}}\\right) = 0$ $\\implies \\theta = 2$ radian Hence option (D) is correct", "of an arc whose measure is $$\\frac{7\\pi}{12} \\text{ radians}$$?", "must be expressed in radians. Returns the arcsine (the number whose sine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arccosine (the number whose cosine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arctangent (the number whose tangent is Number) in radians.", "Browse Questions # If $1^{\\circ}=\\alpha$ radians then the approximate value of $\\cos(60^{\\circ}1')$ is $(a)\\;\\frac{1}{2}+\\frac{\\alpha \\sqrt 3}{120} \\quad (b)\\;\\frac{1}{2}-\\frac{\\alpha}{120} \\quad (c)\\;\\frac{1}{2}-\\frac{\\alpha \\sqrt 3}{120} \\quad (d)\\;\\frac{1}{2}+\\frac{\\alpha}{120}$ Can you answer this question?", "Question To calculate: The simplification of the expression \\frac{7}{\\sqrt{z}} Conic sections To calculate: The simplification of the expression $$\\displaystyle{\\frac{{{7}}}{{\\sqrt{{{z}}}}}}$$", "Express your answer in radians per second to three significant figures.", "x$, so $\\sin 1^\\circ=\\sin (\\frac \\pi{180})$radians $\\approx \\frac \\pi{180}$", "it gives (up to a constant) the same expression as $\\frac{1}{\\sqrt{5}}\\arcsin(\\frac{5x}{3}-\\frac{2}{3})$.", "Question # To calculate: The simplification of the expression , \\frac{4}{\\sqrt{y}} Conic sections To calculate: The simplification of the expression , $$\\displaystyle{\\frac{{{4}}}{{\\sqrt{{{y}}}}}}$$", "+0 help 0 59 1 Find all numbers in the range of $$f(x) = \\arctan x + \\arctan \\frac{1 - x}{1 + x},$$ expressed in radians. Enter all the numbers, separated by commas. Sep 17, 2019", "### Home > PC3 > Chapter 2 > Lesson 2.3.3 > Problem2-154 2-154. Evaluate each expression (in radians) and leave your answers in fraction form when appropriate. Remember the restricted ranges for the functions. 1. $\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)$ 1. $\\sin^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(-\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(\\frac{\\sqrt{3}}{2}\\right)$ The domain of inverse sine is $\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$. The domain of inverse cosine is $[0,\\pi]$.", "Computes the arctangent of one value divided by another value (y/x). Computes both the sine and cosine of a specified value (x), where x is in radians.", "Question # To calculate: The simplification of the expression \\frac{7}{\\sqrt[4]{z}} Conic sections To calculate: The simplification of the expression $$\\frac{7}{\\sqrt[4]{z}}$$", "= x$. Then $I = x \\, \\arcsin (\\sqrt{x}) - \\frac{1}{2} \\int \\frac{\\sqrt{x}}{\\sqrt{1 - x}} \\, dx$. Now refer to post #2.", "α = $$\\frac{g}{r (1 + \\frac{1}{2} (\\frac{R}{r})^2 }$$ we look for the angle, remember that they must be measured in radians θ = s / r in this case we approximate the arc to the distance s = y θ = y / r we substitute w = $$\\sqrt{2 \\frac{g}{ r( \\frac{1}{2} (\\frac{R}{r})^2 } \\frac{y}{r} }$$ w = $$\\sqrt{\\frac{2gy}{r^2 + \\frac{1}{2} R^2 } }$$ for the simple case where r = R w = $$\\sqrt{ \\frac{2gy}{ \\frac{3}{2} R^2 } }$$ w = $$\\sqrt{ \\frac{4}{3} \\frac{gy}{R^2} }$$", "## Calculus 10th Edition $\\int_0^7\\frac{1}{\\sqrt {100-x^{2}}}dx=arcsin\\frac{7}{10}$ $\\int_0^7\\frac{1}{\\sqrt {100-x^{2}}}dx$ $=\\int_0^7\\frac{1}{\\sqrt {(10)^{2}-x^{2}}}dx$ Use arcsine formula: $arcsin\\frac{u}{a}+C=\\int \\frac{1}{\\sqrt {a^{2}-u^{2}}}du$ $=arcsin\\frac{x}{10}|_0^7$ $=arcsin\\frac{7}{10}-arcsin0$ $=arcsin\\frac{7}{10}-0$ $=arcsin\\frac{7}{10}$", "## Algebra 2 (1st Edition) $\\frac{\\sqrt3}{2}$ Using the calculator in radian mode we get: $\\cos{\\frac{\\pi}{6}}=\\frac{\\sqrt3}{2}$", "Question # To calculate: The simplification of the expression , \\frac{\\sqrt{12}}{\\sqrt{x+1}} Conic sections To calculate: The simplification of the expression , $$\\displaystyle{\\frac{{\\sqrt{{{12}}}}}{{\\sqrt{{{x}+{1}}}}}}$$" ]

[ "= \\boxed{\\frac{\\pi}{4}.}$$ $$\\int_{0}^{\\frac{\\pi}{2}}\\frac{{cos}^{n}\\left(x\\right)\\:dx}{{cos}^{n}\\left(x\\right)+ {sin}^{n}\\left(x\\right)} .$$", "as $\\cos x=\\sin(\\frac\\pi2-x)$ and $\\arcsin(\\sin z)=z$ Hence, $$\\arcsin \\sqrt {\\cos2t}=\\frac\\pi2-y=\\frac\\pi2-\\arcsin(\\sqrt 2\\sin t)$$ -", "2. $\\int{\\frac{dt}{\\sqrt{1-{{t}^{2}}}}}=\\arcsin t+k.$ Now, if you want to derive this, put $t=\\sin u.$", "## Trigonometry (11th Edition) Clone Since $\\frac{5\\pi}{6} \\gt \\frac{\\pi}{2}$, then $arcsin(sin~\\frac{5\\pi}{6}) \\neq \\frac{5\\pi}{6}$ $arcsin(sin~\\frac{5\\pi}{6}) = arcsin(0.5) = \\frac{\\pi}{6}$ The domain of arcsin is $[-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$ Since $\\frac{5\\pi}{6} \\gt \\frac{\\pi}{2}$, then $arcsin(sin~\\frac{5\\pi}{6}) \\neq \\frac{5\\pi}{6}$", "$-1\\le t\\le 1.$ and then $\\frac{-\\pi}{2}\\le\\arcsin(t)\\le\\frac{\\pi}{2}.$", "^2 x) = \\sqrt 2 x - \\frac{{\\sqrt 2 }}{{12}}x^3 + O(x^5 )\\;\\; {\\rm as} \\; x \\to 0^+.$$ From Wikipedia, $$\\arccos x = 2\\arctan \\frac{{\\sqrt {1 - x^2 } }}{{1 + x}},\\;\\; - 1 < x \\le 1.$$ Hence $$\\arccos (\\cos ^2 x) = 2\\arctan \\frac{{\\sqrt {1 - \\cos ^4 x} }}{{1 + \\cos ^2 x}} = 2\\arctan \\frac{{\\sin x}}{{\\sqrt {1 + \\cos ^2 x} }} = 2\\arctan \\frac{{\\sin x}}{{\\sqrt {2 - \\sin ^2 x} }}.$$ Next, from $$\\frac{1}{{\\sqrt {2 - x} }} = \\frac{1}{{\\sqrt 2 }} + \\frac{1}{{4\\sqrt 2 }}x + O(x^2 )$$ we get $$\\frac{{\\sin x}}{{\\sqrt {2 - \\sin ^2 x} }} = \\bigg(\\frac{1}{{\\sqrt 2 }} + \\frac{1}{{4\\sqrt 2 }}\\sin ^2 x + O(x^4 )\\bigg)\\sin x.$$ Then, from $$\\sin x = x - \\frac{{x^3 }}{6} + O(x^5 )$$ we get $$\\frac{{\\sin x}}{{\\sqrt {2 - \\sin ^2 x} }} = \\frac{1}{{\\sqrt 2 }}\\bigg(x - \\frac{{x^3 }}{6}\\bigg) + \\frac{1}{{4\\sqrt 2 }}x^3 + O(x^5 ) = \\frac{1}{{\\sqrt 2 }}x + \\frac{1}{{12\\sqrt 2 }}x^3 + O(x^5 ).$$ Next, from $$\\arctan x = x - \\frac{{x^3 }}{3} + O(x^5 )$$ we get $$\\arctan \\frac{{\\sin x}}{{\\sqrt {2 - \\sin ^2 x} }} = \\frac{1}{{\\sqrt 2 }}x + \\frac{1}{{12\\sqrt 2 }}x^3 - \\frac{1}{{3\\sqrt 2 ^3 }}x^3 + O(x^5 ) = \\frac{1}{{\\sqrt 2 }}x - \\frac{1}{{12\\sqrt 2 }}x^3 + O(x^5 ).$$ Finally, $$\\arccos", "It is in agreement with the fact that $\\arcsin(\\frac{1}{\\sqrt{2}})= \\frac{\\pi}{4}.$ Fig. 1 In Fig. 2, we compute $\\arcsin(x)$ from repeatedly solving $\\sin(y) = x$ for $y$ where $x=-1+i \\cdot\\frac{1-(-1)}{n}, i=1,2,...n.$ Fig. 2 Since $\\sin(y)$ is a periodic function, $\\sin(y) = x$ has infinitely many solutions. It is possible that the solution obtained by Newton’s method lies outside of $[-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$, the range of $\\arcsin(x)$ by its definition. Such solution cannot be considered the value of $\\arcsin(x).$ Fig. 3 Exercise-1 Compute $\\arccos(x)$ by solving $\\cos(y) = x$ for $y.$ Exercise-2 Explain Fig. 3. # A Mathematical Allegory We have defined function $\\arcsin$ as a set: $\\{(x, y) | \\sin(y) =x, \\frac{-\\pi}{2} \\le y \\le \\frac{\\pi}{2}\\}.$ By definition, $\\arcsin(-1) = \\frac{-\\pi}{2}, \\arcsin(0)=0, \\arcsin(1) = \\frac{\\pi}{2}$ and $\\frac{d}{dx} \\arcsin(x) = \\frac{1}{\\sqrt{1-x^2}}.$ It means that $\\arcsin(x)$ is the unique solution of $\\frac{dy}{dx} = \\frac{1}{\\sqrt{1-x^2}}\\quad\\quad\\quad(\\star)$ where $y(-1)=-\\frac{\\pi}{2}, y(0)=0$ and $y(1)=\\frac{\\pi}{2}.$ To compute $\\arcsin(x)$, we solve $(\\star)$ for $y(x)$ as follows: Integrate $\\frac{dy}{dx} = \\frac{1}{\\sqrt{1-x^2}}$ from $-1$ to $x$ gives $\\displaystyle\\int\\limits_{-1}^{x}\\frac{dy}{dx} \\,dx=\\displaystyle\\int\\limits_{-1}^{x}\\frac{1}{\\sqrt{1-\\xi^2}}\\; d\\xi\\overset{\\textbf{FTC}}{\\implies}y(x) - y(-1) = \\displaystyle\\int\\limits_{-1}^{x} \\frac{1}{\\sqrt{1-\\xi^2}}\\, d\\xi.$ Therefore, $y(x) = \\displaystyle\\int\\limits_{-1}^{x}\\frac{1}{\\sqrt(1-\\xi^2}\\,d\\xi + y(-1) \\overset{y(-1)=\\frac{-\\pi}{2}}{=} \\displaystyle\\int\\limits_{-1}^{x}\\frac{1}{\\sqrt{1-\\xi^2}}\\,d\\xi - \\frac{\\pi}{2}.$ That is, $\\arcsin(x) = \\displaystyle\\int\\limits_{-1}^{x} \\frac{1}{\\sqrt{1-\\xi^2}}\\;d\\xi - \\frac{\\pi}{2}.$ To obtain $\\arcsin(x), -1 < x < 1$, we numerically evaluate $\\int\\limits_{-1}^{x}\\frac{1}{\\sqrt{1-\\xi^2}}\\,d\\xi$, using function ‘quad_qags’. Fig. 1 The result is visually validated in Fig. 2. Fig. 2", "It is as simple as that. This is followed by the substitution $$v(x)=\\frac{\\sin x+\\cos x}{\\sqrt{2}}$$. \\begin{align} \\frac{d}{dx}\\arcsin(v(x)) & =\\frac{1}{\\sqrt{1-v^2(x)}}v'(x) \\\\ &= \\frac{1}{\\sqrt{1-\\Biggl(\\frac{\\sin x+\\cos x}{\\sqrt{2}}\\Biggl)^2}}\\Biggl(\\frac{\\cos x-\\sin x}{\\sqrt{2}}\\Biggl) \\\\ &= \\frac{\\sqrt{2}}{\\sqrt{(\\cos x-\\sin x)^2}}\\frac{\\cos x-\\sin x}{\\sqrt{2}} \\\\ &= \\frac{\\cos x-\\sin x}{|\\cos x-\\sin x|} \\\\ &= \\frac{\\cos(x+\\frac{\\pi}{4})}{|\\cos(x+\\frac{\\pi}{4})|} \\\\ \\end{align}", "you let $x=\\frac{3\\pi}{4}$ , then $sin(x)=\\frac{1}{\\sqrt{2}}$ But $arcsin(\\frac{1}{\\sqrt{2}})=\\frac{\\pi}{4}$ , not $\\frac{3\\pi}{4}$ This is why it is only true when $x\\in [-\\frac{\\pi}{2},\\frac{\\pi}{2}]$ 8. Awesome,thank you", "x\\sqrt{4-x^2}+2\\int \\frac{1}{\\sqrt{1-(x/2)^2}}\\,d(x/2)\\\\\\\\ &=\\frac12 x\\sqrt{4-x^2}+2\\arcsin(x/2)+C \\end{align}", "r sin(Θ/2)$$ $$Θ=2 arcsin{\\frac{Y}{2r}}$$ $$\\frac{dΘ}{dY}=\\frac{2}{r\\sqrt{4-\\frac{Y^2}{r^2}}}$$, $$f(θ)=1/2π$$ so $$f_Y(y)=\\frac{1}{r\\pi\\sqrt{4-\\frac{Y^2}{r^2}}}$$", "square is $r\\sqrt{2}$ and $p=\\frac{S_{square}}{S_{circle}}=\\frac{2 r^2}{\\pi r^2}=\\frac{2}{\\pi}$", "bound is $2$. Furthermore, we note that the outer radius is $x^2$ and the inner radius is $1$, therefore we get $A_{\\mathrm{outer disk}} = \\pi(x^2)^2 = \\pi x^4$, and $A_{\\mathrm{inner disk}} = \\pi (1)^2 = \\pi$ (4) \\begin{align} V = \\int_{1}^{2} \\pi [ x^4 - 1] \\: dx \\\\ V = \\left ( \\frac{x^5}{5} - x \\right ) \\biggr \\rvert_{1}^{2} \\\\ V = \\left ( \\frac{32}{5} - 2 \\right ) - \\left ( \\frac{1}{5} - 1 \\right ) \\\\ V = \\frac{31}{5} - 1 \\\\ V = \\frac{26}{5} \\end{align}", "# Find the value of $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})$ Find the value of $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})$ Let $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})=x$ $\\arcsin\\frac{\\sqrt{63}}{8}=4\\arcsin x$ $\\arcsin\\frac{\\sqrt{63}}{8}=\\arcsin(4x\\sqrt{1-x^2})(2x^2-1)$ $\\frac{\\sqrt{63}}{8}=(4x\\sqrt{1-x^2})(2x^2-1)$ $\\frac{63}{64}=16x^2(1-x^2)(2x^2-1)^2$ Let $x^2=t$ $\\frac{63}{64}=16t(1-t)(2t-1)^2$ $64t^4-128t^3+80t^2-16t+\\frac{63}{64}=0$ Now solving this fourth degree equation is getting really difficult,even after using rational roots theorem. Is there a better method possible?The answer given is $x=\\frac{\\sqrt2}{4}$.Please help me. $$x = \\arcsin{\\frac{\\sqrt{63}}{8}} = \\arccos{\\frac18}$$ $$\\sin{\\frac{x}{4}} = \\sqrt{\\frac{1-\\cos{(x/2)}}{2}} = \\sqrt{\\frac{1-\\sqrt{\\frac{1+\\cos{(x)}}{2}} }{2}}$$ In this case, $\\cos{x} = 1/8$ so that $$\\sin{\\left (\\frac14 \\arcsin{\\frac{\\sqrt{63}}{8}} \\right )} = \\sqrt{\\frac{1-\\sqrt{\\frac{9/8}{2}} }{2}} = \\frac1{2 \\sqrt{2}}$$", "= x$. Then $I = x \\, \\arcsin (\\sqrt{x}) - \\frac{1}{2} \\int \\frac{\\sqrt{x}}{\\sqrt{1 - x}} \\, dx$. Now refer to post #2.", "• # question_answer A circular arc AB of thin wire frame of radius R and mass M makes an angle of $90{}^\\circ$ at the origin. The centre of mass of the arc lies at A) $\\left[ 0,\\,\\left( \\frac{2}{\\pi } \\right)R \\right]$ B) $\\left[ 0,\\,\\left( \\frac{\\sqrt{2}}{\\pi } \\right)R \\right]$ C) $\\left[ 0,\\,\\left( \\frac{2\\sqrt{2}}{\\pi } \\right)R \\right]$ D) $\\left[ 0,\\,\\left( \\frac{4}{\\pi } \\right)R \\right]$ By symmetry, ${{X}_{CM}}=0$ As, $dm=\\lambda R\\,\\,d\\theta$ $\\therefore$ ${{y}_{CM}}=\\frac{\\int{y\\,\\,dm}}{\\int{dm}}=\\frac{1}{m}\\,\\int{R\\,\\cos \\,\\theta \\lambda R\\,\\,d\\theta }$ $=\\frac{{{R}^{2}}\\lambda }{m}\\int_{-\\pi 4}^{\\pi /4}{\\cos \\,\\theta \\,d\\theta }$ $=\\frac{{{R}^{2}}\\lambda }{R\\times \\lambda \\times \\left( \\frac{\\pi }{2} \\right)}\\left[ 2\\,\\sin \\,\\frac{\\pi }{4} \\right]$ $=\\frac{2R}{\\pi }\\,2\\frac{1}{\\sqrt{2}}=\\frac{2\\sqrt{2}\\,R}{\\pi }$", "## Trigonometry (11th Edition) Clone $y=-\\frac{\\pi}{2}$ RECALL: $y=\\arcsin{x} \\longrightarrow \\sin{y}=x$, $y$ is in the interval $[-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$ Note that $\\sin {(\\frac{\\pi}{2})}=1$. Since sine is an odd function, $\\sin{(-x)}=-\\sin{x}$. Thus, $\\sin{(-\\frac{\\pi}{2})}=-\\sin{(\\frac{\\pi}{2})}=-1$ Therefore, $y=\\arcsin{(-1)}\\longrightarrow y=-\\frac{\\pi}{2}$", "Sir, please could you explain how you got this step: $$\\int_{0}^{1}\\frac{1}{a^{2}+1-a^{2}v^{2}}dv=\\frac{1}{a+1}\\int_{0}^{1}\\frac{1}{1-\\frac{a^{2}v^{2}}{a+1}}dv$$ According to me, it seems to be $a^2+1$ instead of $a+1$, Sir. – Ishan Jun 10 '15 at 19:35 HINT: You might want to use the fact that: $$\\dfrac{1}{\\sqrt{1-x^2}}=\\dfrac{\\partial \\arcsin(x)}{\\partial x}=-\\dfrac{\\partial \\arccos(x)}{\\partial x}$$", "{x^{2}}{6\\,a}}{\\sqrt {{\\frac {1}{a^{2}\\,x^{2}}}+1}}\\,+\\,C}$ ${\\displaystyle \\int x^{m}\\,\\operatorname {arcsch} (a\\,x)dx={\\frac {x^{m+1}\\operatorname {arcsch} (a\\,x)}{m+1}}\\,+\\,{\\frac {1}{a(m+1)}}\\int {\\frac {x^{m-1}}{\\sqrt {{\\frac {1}{a^{2}\\,x^{2}}}+1}}}\\,dx\\quad (m\\neq -1)}$", "to $$\\arccos\\frac1{1+(n-1)h^2/2}\\;,$$ and for small $h$ this is approximately $$\\arccos\\frac1{1+(n-1)h^2/2}\\approx\\arccos\\left(1-(n-1)h^2/2\\right)\\approx\\sqrt{n-1}h\\;.$$ (Again the covering radius is half of that.) 0 2022-07-25 22:23:40 Source" ]

[ "$\\frac{3 \\pi}{4}$ radians $\\tan \\left(\\pi - \\frac{\\pi}{4}\\right) = \\tan \\left(3 \\frac{\\pi}{4}\\right)$ = -1. Hence $\\arctan \\left(- 1\\right) = \\frac{3 \\pi}{4}$ radians", "$-1\\le t\\le 1.$ and then $\\frac{-\\pi}{2}\\le\\arcsin(t)\\le\\frac{\\pi}{2}.$", "\\arcsin{(\\frac{d_a}{d_b}\\sin{\\alpha)}}.$ Other than a numerical approach, it is plausible to assume $\\alpha$ is small especially when $d_a \\gg r$. This might justify linearizing both the sines and arcsines to get something like: $\\alpha \\approx \\frac{\\alpha' + \\beta' - \\pi}{1 + \\frac{d_a}{d_b} - 2 \\frac{d_a}{r}}.$", "Browse Questions # If $1^{\\circ}=\\alpha$ radians then the approximate value of $\\cos(60^{\\circ}1')$ is $(a)\\;\\frac{1}{2}+\\frac{\\alpha \\sqrt 3}{120} \\quad (b)\\;\\frac{1}{2}-\\frac{\\alpha}{120} \\quad (c)\\;\\frac{1}{2}-\\frac{\\alpha \\sqrt 3}{120} \\quad (d)\\;\\frac{1}{2}+\\frac{\\alpha}{120}$ Can you answer this question?", "x$, so $\\sin 1^\\circ=\\sin (\\frac \\pi{180})$radians $\\approx \\frac \\pi{180}$", "= x$. Then $I = x \\, \\arcsin (\\sqrt{x}) - \\frac{1}{2} \\int \\frac{\\sqrt{x}}{\\sqrt{1 - x}} \\, dx$. Now refer to post #2.", "# How do you simplify cos[Arcsin(-2/5)-Arctan3]? Jun 29, 2016 $\\frac{\\sqrt{21} - 6}{5 \\sqrt{10}}$, against principal values of arc sin and arc tan. $\\frac{\\pm \\sqrt{2} \\pm 6}{5 \\sqrt{10}}$, against general values.. #### Explanation: Let a = arc sin (-2/5)3. Then, $\\sin a = \\left(- \\frac{2}{5}\\right) < 0$. So, $\\cos a = \\frac{\\sqrt{21}}{5}$ for principal a and $\\pm \\frac{\\sqrt{21}}{5}$ for general a. Let b = arc tan 3. Then, $\\tan b = 3 > 0$. So, $\\sin b = \\frac{3}{\\sqrt{10}}$, for principal b and $\\pm \\frac{3}{\\sqrt{10}}$, for general b.. $\\cos b = \\frac{1}{\\sqrt{10}}$, for principal b and $\\pm \\frac{1}{\\sqrt{10}}$, for general b.. Note that sin b and cos b are of the same sign , when tan b >0.. Now, the given expression is $\\cos \\left(a - b\\right) = \\cos a \\cos b + \\sin a \\sin . b$ $\\frac{\\sqrt{21} - 6}{5 \\sqrt{10}}$, against principal values of arc sin and arc tan. $\\frac{\\pm \\sqrt{2} \\pm 6}{5 \\sqrt{10}}$, against general values..", "## anonymous 3 years ago Evaluate exactly in radians: arctan(1/sqrt(3)). I know it equals tan(x)=sqrt(3)/3 and the answer is pi/6 but I do not know why. Can someone explain? 1. myininaya $\\arctan(\\frac{1}{\\sqrt{3}})=\\arctan(\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}})$ The range of arctan( ) is -pi/2 to pi/2 For when do you have from -pi/2 to pi/2 that $\\tan( )=\\frac{\\sin( )}{\\cos( )}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}$ 2. myininaya Use unit circle. 3. anonymous Ah, your explanation is exactly what I needed to see. I never though about the fact that the sine and cosine would have common denominators and therefore cancel. Thank you 4. myininaya np :) Find more explanations on OpenStudy", "\\pi}{6}$ 3. $\\frac{9 \\pi}{2}$ 4. $-\\frac{3 \\pi}{4}$ 5. $\\frac{5 \\pi}{6}$ For problems 11-15, find two coterminal angles (one positive, one negative) and the reference angle for each angle in radians. 1. $\\frac{8 \\pi}{3}$ 2. $\\frac{11 \\pi}{4}$ 3. $-\\frac{\\pi}{6}$ 4. $\\frac{4 \\pi}{3}$ 5. $-\\frac{17 \\pi}{6}$ ### Vocabulary Language: English subtended arc subtended arc A subtended arc is the part of the circle in between the two rays that make the central angle.", "\\frac{w}{\\alpha r}$. Hence $\\alpha = \\frac{w}{\\rho r}$ (radians). - 1. Compute tangents intersections between the object and the circle. How you do this depends a lot on how your objects are represented. 2. Compute the angle corresponding to this pair of points. An aran2 kind of function might help in this step. 3. Divide that angle by the ratio you're given, i.e. by 0.5 for 50%. -", "+0 # Didn't understand concept plz help with HW 0 151 1 +24 1) Compute $$sin^{-1} \\left(\\sin \\left(\\frac{7\\pi}6\\right)\\right).$$ 2) Compute $$\\sin\\left(\\arccos\\left(\\frac{\\sqrt{5}}3\\right)\\right).$$ 3) Compute $$\\arccos\\left(\\frac{\\sqrt{3}}{2}\\right).$$ Oct 5, 2019 #1 +1 1 - Do what is inside brackets first. Sin(7pi/6) = - 1/2 - This is in radians. Then take the arcsin of - 1/2 = - pi/6 2 - Do the same thing here. Arccos(sqrt(5) / 3) = 0.729727656- This is in radians. Take its sin(0.729727656) = 2/3 3 - This is straightforward. Arccos(sqrt(3) / 2) =pi / 6 Oct 5, 2019", "Browse Questions If $1^{\\circ}=\\alpha$ radians then the approximate value of $\\cos(60^{\\circ}1')$ is $(a)\\;\\frac{1}{2}+\\frac{\\alpha \\sqrt 3}{120} \\quad (b)\\;\\frac{1}{2}-\\frac{\\alpha}{120} \\quad (c)\\;\\frac{1}{2}-\\frac{\\alpha \\sqrt 3}{120} \\quad (d)\\;\\frac{1}{2}+\\frac{\\alpha}{120}$", "# How do you evaluate arcsin(sqrt 2/2)? May 24, 2015 $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}$ is the length of one side of the isoceles right angled triangle with sides $\\frac{\\sqrt{2}}{2}$, $\\frac{\\sqrt{2}}{2}$ and $1$, which has internal angles $\\frac{\\pi}{4}$, $\\frac{\\pi}{4}$ and $\\frac{\\pi}{2}$. ($\\frac{\\pi}{4}$ radians = ${45}^{o}$ and $\\frac{\\pi}{2}$ radians = ${90}^{o}$ if you prefer) To show this is right angled, check with Pythagoras: ${\\left(\\frac{\\sqrt{2}}{2}\\right)}^{2} + {\\left(\\frac{\\sqrt{2}}{2}\\right)}^{2}$ $= {\\sqrt{2}}^{2} / {2}^{2} + {\\sqrt{2}}^{2} / {2}^{2}$ $= \\frac{2}{4} + \\frac{2}{4} = \\frac{1}{2} + \\frac{1}{2} = 1 = {1}^{2}$ So since $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}$ and $\\frac{\\pi}{4}$ is in the required range for $\\arcsin$ viz $- \\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}$, we find $\\arcsin \\left(\\frac{\\sqrt{2}}{2}\\right) = \\frac{\\pi}{4}$", "How do you evaluate Sin(2 arctan(sqrt 2))? May 7, 2016 $\\frac{\\sqrt{2}}{3}$. Explanation: Let $a = a r c \\tan \\sqrt{2}$. Then $\\tan a = \\sqrt{2}$. Tangent is positive. Both sine and cosine have the same sign. So, either $\\sin a = \\frac{\\sqrt{2}}{\\sqrt{3}} \\mathmr{and} \\cos a = \\frac{1}{\\sqrt{3}}$ or $\\sin a = - \\frac{\\sqrt{2}}{\\sqrt{3}} \\mathmr{and} \\cos a = - \\frac{1}{\\sqrt{3}}$. Now, $\\sin \\left(2 a r c \\tan \\sqrt{2}\\right) = \\sin 2 a = 2 \\sin a \\cos a = \\frac{\\sqrt{2}}{3}$, in both the cases.", "## Calculus 10th Edition $\\int_0^7\\frac{1}{\\sqrt {100-x^{2}}}dx=arcsin\\frac{7}{10}$ $\\int_0^7\\frac{1}{\\sqrt {100-x^{2}}}dx$ $=\\int_0^7\\frac{1}{\\sqrt {(10)^{2}-x^{2}}}dx$ Use arcsine formula: $arcsin\\frac{u}{a}+C=\\int \\frac{1}{\\sqrt {a^{2}-u^{2}}}du$ $=arcsin\\frac{x}{10}|_0^7$ $=arcsin\\frac{7}{10}-arcsin0$ $=arcsin\\frac{7}{10}-0$ $=arcsin\\frac{7}{10}$", "# Compute $2\\arctan{2}+\\arcsin{\\frac{4}{5}}.$ Proceeding as you guys have tought me on my previous posts by setting the whole thing equal to $X$ taking the tangent of the whole expression and then using the additon formula for tangent: $$\\tan{X}=\\tan{\\left(2\\arctan{2}+\\arcsin{\\frac{4}{5}}\\right)}=\\frac{\\tan{(2\\arctan{2})}+\\tan{\\left(\\arcsin{\\frac{4}{5}}\\right)}}{1-\\tan{(2\\arctan{2})}\\cdot\\tan{\\left(\\arcsin{\\frac{4}{5}}\\right)}}.$$ So lets split this into two sub-computations like this: i): $$N=\\tan{(2\\arctan{2})}=\\tan{(\\arctan{2}+\\arctan{2})}=\\frac{2\\tan{(\\arctan{2})}}{1-\\tan^2{(\\arctan{2})}}=\\frac{2\\cdot 2}{1-2^2}=-\\frac{4}{3}.$$ ii): $$M=\\tan{\\left(\\arcsin{\\frac{4}{5}}\\right)}=\\frac{\\sin{\\left(\\arcsin{\\frac{4}{5}}\\right)}}{\\cos{\\left(\\arcsin{\\frac{4}{5}}\\right)}}=\\frac{\\sin{\\left(\\arcsin{\\frac{4}{5}}\\right)}}{\\sqrt{1-\\sin^2{\\left(\\arcsin{\\frac{4}{5}}\\right)}}}=\\frac{\\frac{4}{5}}{\\sqrt{1-\\left(\\frac{4}{5}\\right)^2}}=\\frac{\\frac{4}{5}}{\\frac{3}{5}}=\\frac{4}{3}.$$ So, $$\\tan{X}=\\frac{N+M}{1-MN}=\\frac{-\\frac{4}{3}+\\frac{4}{3}}{1+\\frac{4}{3}\\cdot\\frac{4}{3}}=0\\Longrightarrow X=\\arctan{0}=0.$$ The book says $X=\\pi.$ • $\\pi$ is the right answer Sep 16 '17 at 21:56 • Solutions of the equation $\\tan x = c$ are determined only up to addition by integer multiples of $\\pi$. Moreover, without any further input, all of them equally makes sense. In your case, you may write $$X = \\pi - 2\\arctan\\left(\\frac{1}{2}\\right) + \\arctan\\left(\\frac{4}{3}\\right)$$ and then proceed. Sep 16 '17 at 21:57 • @SangchulLee - I'm still not sure where my solution is wrong. Sep 16 '17 at 21:59 • Your mistake is the part where you concluded $X = 0$ from $\\tan X = 0$. As I explained, $\\tan X = 0$ only tells you that $X = n \\pi$ for some $n \\in \\mathbb{Z}$. You still need to determine which $n$ fits into your $X$. This can be done if you carefully examine $X$. For instance, Chappers explains one possible way. Sep 16 '17 at 22:00 • Yes, I forgot to add the periodicity of tangent", "$$\\frac{\\pi}{3}$$ radian. Consequently, m ∠EVO = $$\\frac{2\\pi}{3}$$. Looking at triangle EVO , the law of sines gives $$\\frac{6}{\\sin \\left(\\frac{2 \\pi}{3}\\right)}$$ = $$\\frac{1.5}{\\sin (m \\angle V E O)}$$, and so sin ⁡(m∠VEO ) = $$\\frac{\\sin \\left(\\frac{2 \\pi}{3}\\right)}{4}$$ =$$\\frac{\\sqrt{3}}{8}$$. Thus, m ∠VEO = sin–1 ($$\\frac{\\sqrt{3}}{8}$$) radians. Since ∠EVO is obtuse, ∠VEO is acute, and we must work with the value of sin–1 ($$\\frac{\\sqrt{3}}{8}$$) that corresponds to the measure of an acute angle. It follows that m ∠a = $$\\frac{\\pi}{3}$$ – sin–1 ($$\\frac{\\sqrt{3}}{8}$$) ≈ 0.6 radians (which is about 34° ).", "it gives (up to a constant) the same expression as $\\frac{1}{\\sqrt{5}}\\arcsin(\\frac{5x}{3}-\\frac{2}{3})$.", "# Find the value of $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})$ Find the value of $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})$ Let $\\sin(\\frac{1}{4}\\arcsin\\frac{\\sqrt{63}}{8})=x$ $\\arcsin\\frac{\\sqrt{63}}{8}=4\\arcsin x$ $\\arcsin\\frac{\\sqrt{63}}{8}=\\arcsin(4x\\sqrt{1-x^2})(2x^2-1)$ $\\frac{\\sqrt{63}}{8}=(4x\\sqrt{1-x^2})(2x^2-1)$ $\\frac{63}{64}=16x^2(1-x^2)(2x^2-1)^2$ Let $x^2=t$ $\\frac{63}{64}=16t(1-t)(2t-1)^2$ $64t^4-128t^3+80t^2-16t+\\frac{63}{64}=0$ Now solving this fourth degree equation is getting really difficult,even after using rational roots theorem. Is there a better method possible?The answer given is $x=\\frac{\\sqrt2}{4}$.Please help me. $$x = \\arcsin{\\frac{\\sqrt{63}}{8}} = \\arccos{\\frac18}$$ $$\\sin{\\frac{x}{4}} = \\sqrt{\\frac{1-\\cos{(x/2)}}{2}} = \\sqrt{\\frac{1-\\sqrt{\\frac{1+\\cos{(x)}}{2}} }{2}}$$ In this case, $\\cos{x} = 1/8$ so that $$\\sin{\\left (\\frac14 \\arcsin{\\frac{\\sqrt{63}}{8}} \\right )} = \\sqrt{\\frac{1-\\sqrt{\\frac{9/8}{2}} }{2}} = \\frac1{2 \\sqrt{2}}$$", "### Home > PC3 > Chapter 2 > Lesson 2.3.3 > Problem2-154 2-154. Evaluate each expression (in radians) and leave your answers in fraction form when appropriate. Remember the restricted ranges for the functions. 1. $\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)$ 1. $\\sin^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(-\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(\\frac{\\sqrt{3}}{2}\\right)$ The domain of inverse sine is $\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$. The domain of inverse cosine is $[0,\\pi]$." ]

[ "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "to the one of your original line: $- 3$ $s l o p e = \\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ $- 3 = \\frac{y - 2}{x - 7}$ $- 3 x + 21 = y - 2$ and finally your line: $y + 3 x = 23$", "get $u_2 =\\begin{pmatrix}\\frac{1}{\\sqrt{2}}\\\\-\\frac{1}{\\sqrt{2}}\\end{pmatrix} \\;\\;\\; \\text{and} \\;\\;\\; u_3 =\\begin{pmatrix}\\frac{3}{5}\\\\-\\frac{-4}{5}\\end{pmatrix}$", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "get $d=\\begin{vmatrix}\\large\\frac{(8\\hat i+8\\hat j+4\\hat k).(-10\\hat i-2\\hat j-3\\hat k)}{12}\\end{vmatrix}$ $d=\\begin{vmatrix}\\large\\frac{-80-16-12}{12}\\end{vmatrix}$ $\\;\\;\\;=\\large\\frac{108}{12}$ $\\;\\;\\;=9$ units. Hence the shortest distance between the given two lines is $9$ units.", "A line segment $P_1P_2$ can be represented in parametric form by \\begin{aligned} x &= x_1 + (x_2 - x_1)t \\\\ y &= y_1 + (y_2 - y_1)t \\\\ & 0 \\leq t \\leq 1 \\end{aligned} Two line segments $P_1P_2$ and $P_3P_4$ intersect if any only if the numbers $$s = \\frac{ \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_1 & y_3 - y_1 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} ~~ \\text{and} ~~ t = \\frac{ \\begin{vmatrix} x_3 - x_1 & y_3 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}}$$ satisfy $0 \\leq s \\leq 1$ and $0 \\leq t \\leq 1$.", "same except it has three lines and three columns $$W=\\left |\\pmatrix{f_1 &f_2 & f_3 \\\\f_1' &f_2' & f_3' \\\\f_1'' &f_2'' & f_3'' }\\right |=\\left |\\pmatrix{2t-3 &2t^2+1 & 3t^2+t\\\\2 &.. & ... \\\\0 & 4 & 6 }\\right |$$ Note that: $$f_3(t) = \\frac 3 2 f_2+\\frac 12f_1$$", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "\\frac xr \\neq \\frac rx.$$ Rather, if that relation does hold then it implies that $x$ is independent of $v$ so that $\\frac{\\partial x}{\\partial v} = 0$. anuttarasammyak Gold Member Further to my post #2 $$A= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix}$$ $$B= \\begin{pmatrix} x_u & x_v \\\\ y_u & y_v \\\\ \\end{pmatrix}$$ $$AB= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix} \\begin{pmatrix} x_u & x_v \\\\ y_u & y_v \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ It meets with post #3. $$A=B^{-1}$$ $$\\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix} =\\frac{1}{x_uy_v-x_vy_u} \\begin{pmatrix} y_v & -x_v \\\\ -y_u & x_u \\\\ \\end{pmatrix}$$ For an example (1,1) component says $$u_x=\\frac{1}{x_uy_v-x_vy_u}y_v$$ You see in order ##u_x =\\frac{1}{x_u}## as you expect, ##x_vy_u=0 ## and ##y_v \\neq 0## is required. Last edited:", "$$ln(x)$$, and some other function $$u(x)$$ then the chain rule will always lead to the following result: $\\text{if} \\quad f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ $\\text{then} \\quad f'(x) = \\frac{u'(x)}{u(x)}$ This special case of the chain rule is further explained and illustrated in Tutorial 3. Spend a few minutes to watch it now. ## Exercise 3 Differentiate each of the following: 1. $$f(x) = ln \\begin{pmatrix} x^2 - 3 \\end{pmatrix}$$ 2. $$y = 3.ln \\begin{pmatrix} x^3-3x^2 + 6 \\end{pmatrix}$$ 3. $$f(x) = -2.ln \\begin{pmatrix} 4x+ 5 \\end{pmatrix}$$ 4. $$y = 4.ln \\begin{pmatrix} \\sqrt{x} \\end{pmatrix}$$ 5. $$f(x) = ln \\begin{pmatrix} \\frac{1}{x} \\end{pmatrix}$$ 1. $$f'(x) = \\frac{2x}{x^2-3}$$ 2. $$\\frac{dy}{dx} = \\frac{9x^2 - 18x}{x^3 - 3x^2 + 6}$$ 3. $$f'(x) = - \\frac{8}{4x+5}$$ 4. $$\\frac{dy}{dx} = \\frac{2}{x}$$ 5. $$f'(x) = - \\frac{1}{x}$$ 1. The function $$f(x) = ln \\begin{pmatrix} x^2 - 3\\end{pmatrix}$$ can be written: $f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = x^2 - 3$ and $u'(x) = 2x$ Using the formula we saw above we can state: \\begin{aligned} f'(x) & = \\frac{u'(x)}{u(x)} \\\\ f'(x) & = \\frac{2x}{x^2 - 3} \\end{aligned} 2. The function $$y = 3. ln \\begin{pmatrix} x^3 - 3x^2 + 6\\end{pmatrix}$$ can be written: $y = 3 ln \\begin{pmatrix} u(x)\\end{pmatrix}$ Where $u(x) = x^3 - 3x^2 + 6$ and $u'(x) = 3x^2 -6x$ Using the formula", "$$f(x)$$. That's $$y_1=f(x_1)$$, $$y_2 = f(x_2)$$ ... . This is further explained and illustrated in the following tutorial. ## Tutorial - Finding Stationary Points In the following tutorial we review the ... . ## Exercise 1 Using the method, described above, find the coordinates of any stationary points on each of the following curves: 1. $$y = 2x^2 - 8x + 7$$ 2. $$y = 2x^3 - 15x^2 + 2x + 1$$ 3. $$y=-x^3-3x^2+çx +1$$ 4. $$y = x + \\frac{1}{x}$$ 5. $$y = \\frac{x^2}{2} - ln(x)+4$$ 1. One stationary point at $$\\begin{pmatrix}2,-1\\end{pmatrix}$$. 2. Two stationary points: $$\\begin{pmatrix}1,-1 \\end{pmatrix}$$ and $$\\begin{pmatrix}4,3\\end{pmatrix}$$. 3. Two stationary points: $$\\begin{pmatrix}-1,0\\end{pmatrix}$$ and $$\\begin{pmatrix}1,0\\end{pmatrix}$$.", "$= \\lim_{x \\to 6} \\frac{x-6}{-6x(-6+x)}$ $= \\lim_{x \\to 6} \\frac{1}{-6x} = -\\frac {1}{36}$ fixed the next-to-last line", "How to draw the figure of $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$-type straight line Discussion in 'Mathematics' started by Ordinary users68, Aug 2, 2020 at 2:25 AM. 1. Ordinary users68Guest I'd like to draw the straight line that is expressed in symmetrical form directly: $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$ ContourPlot3D[(x - 1)/2 == (y - 1)/3 == (z - 2)/3, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}, BoxRatios -> Automatic] But I can't get the correct result with the above code. How can I draw this line directly? In addition, $\\frac{x-1}{2}=\\frac{y-1}{3}=2$ should represent a vertical line, which I hope can be drawn by built-in or custom functions.", "equation becomes $\\frac{d}{dt}\\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix} 0 & 1 \\\\ - 3 & -4\\end{pmatrix}\\begin{pmatrix}y \\\\ x \\end{pmatrix}$ $= \\begin{pmatrix} x \\\\-3y- 4x\\end{pmatrix}$ Which then reduces to the two scalar equations $\\frac{dy}{dt}= x$ and $\\frac{dx}{dt}= -3y- 4x$ with the intial conditions x(0)= -2, y(0)= -4. Differentiate the first equation twice to get $\\frac{d^2y}{dt^2}= \\frac{dx}{dt}$ and from the second equation that is equal to -3y- 4x: $\\frac{d^2y}{dt^2}= -3y- 4x$. From the first equation $x= \\frac{dy}{dt}$ so we can replace it by that: $\\frac{d^2y}{dt^2}= -3y- 4\\frac{dy}{dt}$ or $\\frac{d^2y}{dt^2}+ 3y+ 4= 0$. Since x(0)= -2 and x= y', y(0)= -4, y'(0)= -2. To find the solution to the intial problem means, of course, to find the vector function $Y(t)= \\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}y(t)\\\\ y'(t)\\end{pmatrix}$ (since x= y'). Since you have found that $y(t)= e^{-3t}+ e^{-t}$ (NOT \" $e^{-3x}+ e^{-x}$\" because y is a function of t, not x!), $x(t)= y'= -3e^{-3t}- e^{-t}$ and $Y(t)= \\begin{pmatrix}y(t)\\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}e^{-3t}+ e^{-t} \\\\ -3e^{-3t}- e^{-t}\\end{pmatrix}$.", "as it should help develop a much better sense of what we're actually doing when we use the chain rule. If you'd prefer to try the exercises right away, by all means do so but do make sure to watch Tutorial 2 afterwards. ## Exercise 1 1. Differentiate the function defined as $$f(x) = sin\\begin{pmatrix} x^3-2 \\end{pmatrix}$$. 2. Given $$y = \\sqrt{3x^2-4x+7}$$, find an expression for $$\\frac{dy}{dx}$$. 3. Differentiate the function $$f(x) = \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^7$$. 4. Find $$\\frac{dy}{dx}$$, given $$y = e^{3x^2-1}$$. 5. Differentiate the function $$f(x) = ln \\begin{pmatrix} x^2 + x \\end{pmatrix}$$. 6. Given $$y = cos \\begin{pmatrix} 2x + 1 \\end{pmatrix}$$, find $$\\frac{dy}{dx}$$. 1. $$f'(x) = 3x^2.cos \\begin{pmatrix} x^3 - 2 \\end{pmatrix}$$ 2. $$\\frac{dy}{dx} = \\frac{3x-2}{\\sqrt{3x^2 - 4x + 7}}$$ 3. $$f'(x) = 14x \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6$$ 4. $$\\frac{dy}{dx} = 6x.e^{3x^2-1}$$ 5. $$f'(x) = \\frac{2x+1}{x^2+x}$$ 6. $$\\frac{dy}{dx} = -2.sin \\begin{pmatrix} 2x + 1 \\end{pmatrix}$$ 1. The function $$f(x) = sin \\begin{pmatrix} x^3 - 2 \\end{pmatrix}$$ can be written $$f(x) = g \\begin{bmatrix} h(x) \\end{bmatrix}$$, where: $g(x) = sin(x) \\quad \\text{and} \\quad h(x) = x^3 - 2$ and $g'(x) = cos(x) \\quad \\text{and} \\quad h'(x) = 3x^2$ We can now use the chain rule to write: $f'(x) = h'(x) \\times g'\\begin{bmatrix} h(x) \\end{bmatrix}$ That's: $f'(x) = 3x^2.cos \\begin{pmatrix} x^3-2\\end{pmatrix}$ 2. We" ]

[ "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "case 2 If $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, then $z=t$, a free parameter, which can be any value. Second row in (1) now gives \\begin{align*} \\frac{4}{3}y+\\left( k-\\frac{7}{3}\\right) t & =0\\\\ y & =-\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\end{align*} And first row gives \\begin{align*} 3x+2y+z & =0\\\\ 3x & =-2\\left( -\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\right) -t\\\\ x & =\\frac{1}{2}t\\left( k-3\\right) \\end{align*} Hence in this case, the NullSpace is $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} =t \\begin{pmatrix} \\frac{1}{2}\\left( k-3\\right) \\\\ -\\frac{3}{4}\\left( k-\\frac{7}{3}\\right) \\\\ 1 \\end{pmatrix}$$ Since $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, the solution is $k=5,k=4$. Substituting these values into the above gives the NullSpace as $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = t\\left\\{ \\begin{pmatrix} 1\\\\ -2\\\\ 1 \\end{pmatrix} , \\begin{pmatrix} \\frac{1}{2}\\\\ -\\frac{5}{4}\\\\ 1 \\end{pmatrix} \\right\\}$$ For any $t$. For only $t=0$, this gives the result gives by Mathematica. So Wolfram Mathematica only considered the case 1 above, and did not consider case 2. Wolfram Alpha, which has additional A.I. engine, seems to have looked more into this. • Here's how to use Mathematica for constructing the reduced form: UpperTriangularize[First[LUDecomposition[{{3, 2, 1}, {7, 6, k}, {k, 4, 3}}]]]. Mar 27 '17 at 4:50 • @J.M. Thank you. Mar 27 '17 at 5:04 • @J.M. btw, this only works for square matrices. For non-square matrices, I had to write an M function to do this. Mar 27 '17 at 5:20 • Indeed, it is unfortunate that Mathematica", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "correct indeed, by a direct check, for $$t=-\\frac 6{11}$$ we obtain $$N=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} -\\frac 6{11}\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=\\begin{pmatrix} \\frac {18}{11} \\\\ \\frac {1}{11} \\\\ \\frac {40}{11} \\end{pmatrix}$$ $$\\overrightarrow{CN}=\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix}$$ $$\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}= -\\frac {21}{11}+\\frac {42}{11}-\\frac {21}{11}=0$$", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "1 \\end{pmatrix}$$. ### Solution 1. To find $$P\\begin{pmatrix}X \\leq 1.5\\end{pmatrix}$$, we use write: \\begin{aligned} P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = \\int_{-\\infty}^{1.5} f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}-\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}\\begin{pmatrix} x^2 - 2x \\end{pmatrix}dx \\\\ & = -\\frac{3}{4} \\begin{bmatrix}\\frac{x^3}{3} - x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}x^3 - 3x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}\\begin{pmatrix} \\frac{3}{2}\\end{pmatrix}^3 - 3 \\times \\begin{pmatrix} \\frac{3}{2} \\end{pmatrix}^2 - 0 \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\frac{27}{8} - \\frac{27}{4} \\end{bmatrix} \\\\ & = - \\frac{1}{4} \\begin{bmatrix} -\\frac{27}{8} \\end{bmatrix} \\\\ & = \\frac{27}{32} \\\\ P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = 0.844 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, from $$-\\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix} X \\leq 1.5\\end{pmatrix} = 0.844$$ 2. To find $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = \\int_{0.5}^1f(x)dx \\\\ & = \\int_{0.5}^1 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{0.5}^1 x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{\\frac{1}{2}}^1 \\begin{pmatrix} x^2-2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix} \\frac{x^3}{3} - x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "# Let me know your opinion about simple formula to find $\\begin{pmatrix} n \\\\ k \\end{pmatrix}$ [closed] I often say to my students: There is a simple way to find binomial formula ,and the way is :$$\\begin{pmatrix} n \\\\ k \\end{pmatrix} =\\dfrac{\\underbrace{n.(n-1).(n-2)...}_{\\text{k-terms decreasing}}}{\\underbrace{1.2.3...k}_{\\text{k-terms increasing}}}$$ for example: $$\\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix}= \\frac{5!}{3!(5-3)!} \\to \\\\ \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix}= \\frac{5.4.3}{1.2.3}=5.2=10$$ for example $$\\begin{pmatrix} n \\\\ 2 \\end{pmatrix}= \\frac{n(n-1)}{1.2} =\\frac{n(n-1)}{2}\\\\\\begin{pmatrix} n \\\\ 3 \\end{pmatrix}= \\frac{\\underbrace{n(n-1)(n-2)}_{\\text{3-terms decreasing}}}{\\underbrace{1.2.3}_{\\text{3-terms increasing}}}$$ or $$\\begin{pmatrix} 11 \\\\ 7 \\end{pmatrix}\\underbrace{=}_{\\text{w.r.t. to }\\begin{pmatrix} n \\\\ k \\end{pmatrix}=\\begin{pmatrix} n \\\\ n-k \\end{pmatrix}} \\begin{pmatrix} 11 \\\\ 4 \\end{pmatrix}=\\frac{11.10.9.8}{1.2.3.4}$$ I tel this for speedy work with binomial to k-11,k-12 students.Is this a good way to teach binomial calculation for the students ? (instead of formal calculation $\\frac{n!}{r!(n-r)!}$) .If not tell my why ? ## closed as unclear what you're asking by Lord Shark the Unknown, Cameron Buie, José Carlos Santos, Simply Beautiful Art, NamasteAug 3 '17 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. • i think you should move", "All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $cos{\\theta}=\\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}|}{\\sqrt{9} \\sqrt{49}}=\\frac{16}{21}$ $\\theta = {40.4}^{\\circ}$ (ii) Using Graphing Calculator, equation of line, $r = \\begin{pmatrix}{-\\frac{1}{6}}\\\\{-\\frac{2}{3}}\\\\0\\end{pmatrix} + \\lambda \\begin{pmatrix}7\\\\10\\\\6\\end{pmatrix}, \\lambda \\in \\mathbb{R}$ (iii) Distance A to $P_1 = \\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} - 1|}{\\sqrt{9}} = \\frac{|1+c|}{3}$ Distance A to $P_1 = \\frac{|\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} + 1|}{\\sqrt{49}} = \\frac{|2c-14|}{7}$ $\\frac{|1+c|}{3} = \\frac{|2c-14|}{7}$ $49(1+c)^{2} = 9(2c-14)^{2}$ $c = -49 \\mathrm{~or~} \\frac{35}{13}$", "equation becomes $\\frac{d}{dt}\\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix} 0 & 1 \\\\ - 3 & -4\\end{pmatrix}\\begin{pmatrix}y \\\\ x \\end{pmatrix}$ $= \\begin{pmatrix} x \\\\-3y- 4x\\end{pmatrix}$ Which then reduces to the two scalar equations $\\frac{dy}{dt}= x$ and $\\frac{dx}{dt}= -3y- 4x$ with the intial conditions x(0)= -2, y(0)= -4. Differentiate the first equation twice to get $\\frac{d^2y}{dt^2}= \\frac{dx}{dt}$ and from the second equation that is equal to -3y- 4x: $\\frac{d^2y}{dt^2}= -3y- 4x$. From the first equation $x= \\frac{dy}{dt}$ so we can replace it by that: $\\frac{d^2y}{dt^2}= -3y- 4\\frac{dy}{dt}$ or $\\frac{d^2y}{dt^2}+ 3y+ 4= 0$. Since x(0)= -2 and x= y', y(0)= -4, y'(0)= -2. To find the solution to the intial problem means, of course, to find the vector function $Y(t)= \\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}y(t)\\\\ y'(t)\\end{pmatrix}$ (since x= y'). Since you have found that $y(t)= e^{-3t}+ e^{-t}$ (NOT \" $e^{-3x}+ e^{-x}$\" because y is a function of t, not x!), $x(t)= y'= -3e^{-3t}- e^{-t}$ and $Y(t)= \\begin{pmatrix}y(t)\\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}e^{-3t}+ e^{-t} \\\\ -3e^{-3t}- e^{-t}\\end{pmatrix}$.", "because the first column tells us $$e^{5it}=cos 5t+isin5t,$$ and the second tells us $$e^{-5it}=cos5t-isin5t$$. We can then solve $$pmatrix{10\\8}=Mpmatrix{A\\B}$$ as $$pmatrix{A\\B}=M^{-1}pmatrix{10\\8}=pmatrix{1&1\\i&-i}^{-1}pmatrix{10\\8}$$ $$=dfrac{pmatrix{-i&-1\\-i&1}}{-2i}pmatrix{10\\8}=pmatrix{frac12&-frac i2\\frac12&frac i2}pmatrix{10\\8}=pmatrix{5-4i\\5+4i}.$$", "We already know the value of $$\\mu$$. We now calculate $$E\\begin{pmatrix}X^2\\end{pmatrix}$$: \\begin{aligned} E\\begin{pmatrix}X^2\\end{pmatrix} &= \\int_{-\\infty}^{+\\infty}x^2.f(x)dx \\\\ &= \\int_{-\\infty}^{+\\infty} x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\int_0^3 x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x-3)^2dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x^2-6x+9)dx \\\\ &= \\frac{1}{9}\\int_0^3 (x^4-6x^3+9x^2)dx \\\\ &= \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{6}{4}x^4+\\frac{9}{3}x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{3}{2}x^4+3x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{3^5}{5}-\\frac{3}{2}\\times 3^4+3\\times 3^3 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{243}{5}-\\frac{243}{2}+81 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{486}{10} - \\frac{1215}{10} + \\frac{810}{10}\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{81}{10} \\end{bmatrix} \\\\ & = \\frac{81}{9\\times 10} \\\\ & = \\frac{9}{10} \\\\ E\\begin{pmatrix}X^2\\end{pmatrix} & = 0.9 \\end{aligned} We can now go back to the definition of variance and calculate: \\begin{aligned} Var\\begin{pmatrix}X \\end{pmatrix} &= E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 0.9 - 0.75^2 \\\\ Var\\begin{pmatrix}X \\end{pmatrix} & = 0.3375 \\end{aligned} 3. We now calculate the standard deviation using the definition: $\\sigma = \\sqrt{Var\\begin{pmatrix} X\\end{pmatrix}}$ Since $$Var\\begin{pmatrix}X \\end{pmatrix} = 0.3375$$ we find: $\\sigma = \\sqrt{0.3375} = 0.581$ The standard deviation is $$\\sigma = 0.581$$. This tells that, on average amongst all the different cell phones, we can expect the time taken for the app to launch to be $$0.581$$ seconds more, or less, than the mean value $$\\mu = 0.75$$ seconds. ## Exercise 1 A continuous random variable $$X$$ has probability density function defined as: $f(x) = \\begin{cases} \\frac{x^2}{9}, \\quad 0\\leq", "how to prove the formula. – làntèrn Aug 6 '15 at 16:38 • What if $a=0$?. – Michael Galuza Aug 6 '15 at 16:43 The fast approach, as I see it is (in case $a \\neq 0$): you have $ax+by = c$ , just plug in $y=0+ta$, and you get $$ax + bta = c$$ $$x= \\frac ca - bt$$. alternatively, in case you don't have the solution in advance, use same approach and you will get: $$x = \\frac ca - \\frac ba y$$, so the general solution will be $$(x,y) = (\\frac ca - \\frac bat , t)$$ $$(x,y) = (\\frac ca, 0) + (-\\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result. • So, using $t$ here is redundant? – làntèrn Aug 6 '15 at 16:58 • I've edited my answer, I should have you $t$ ,, $y$ is somewhat confusing. – d_e Aug 6 '15 at 17:03 • So $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} -b \\\\ a \\\\ \\end{pmatrix}$ would be same with $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} \\frac{-b}{a} \\\\ 1 \\\\ \\end{pmatrix}$ – làntèrn Aug 6 '15 at 17:27" ]

[ "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "equation becomes $\\frac{d}{dt}\\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix} 0 & 1 \\\\ - 3 & -4\\end{pmatrix}\\begin{pmatrix}y \\\\ x \\end{pmatrix}$ $= \\begin{pmatrix} x \\\\-3y- 4x\\end{pmatrix}$ Which then reduces to the two scalar equations $\\frac{dy}{dt}= x$ and $\\frac{dx}{dt}= -3y- 4x$ with the intial conditions x(0)= -2, y(0)= -4. Differentiate the first equation twice to get $\\frac{d^2y}{dt^2}= \\frac{dx}{dt}$ and from the second equation that is equal to -3y- 4x: $\\frac{d^2y}{dt^2}= -3y- 4x$. From the first equation $x= \\frac{dy}{dt}$ so we can replace it by that: $\\frac{d^2y}{dt^2}= -3y- 4\\frac{dy}{dt}$ or $\\frac{d^2y}{dt^2}+ 3y+ 4= 0$. Since x(0)= -2 and x= y', y(0)= -4, y'(0)= -2. To find the solution to the intial problem means, of course, to find the vector function $Y(t)= \\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}y(t)\\\\ y'(t)\\end{pmatrix}$ (since x= y'). Since you have found that $y(t)= e^{-3t}+ e^{-t}$ (NOT \" $e^{-3x}+ e^{-x}$\" because y is a function of t, not x!), $x(t)= y'= -3e^{-3t}- e^{-t}$ and $Y(t)= \\begin{pmatrix}y(t)\\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}e^{-3t}+ e^{-t} \\\\ -3e^{-3t}- e^{-t}\\end{pmatrix}$.", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "\\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {11}{12} & \\frac {-5}{12}\\\\ \\frac {22}{9} & \\frac {5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {11 - 5}{12}\\\\ \\frac {22 + 5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {6}{12}\\\\ \\frac {27}{9}\\\\ \\end {pmatrix}$$ ∴X = $$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ i.e.$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, 4x + $$\\frac y5$$ = 7 or, $$\\frac {20x + y}{5}$$ = 7 or, 20x + y = 35.........................................(1) 3x + $$\\frac y4$$ = 5 or, $$\\frac {12x + y}{4}$$ = 5 or, 12x + y = 20..........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 20 & 1\\\\ 12 &", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "correct indeed, by a direct check, for $$t=-\\frac 6{11}$$ we obtain $$N=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} -\\frac 6{11}\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=\\begin{pmatrix} \\frac {18}{11} \\\\ \\frac {1}{11} \\\\ \\frac {40}{11} \\end{pmatrix}$$ $$\\overrightarrow{CN}=\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix}$$ $$\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}= -\\frac {21}{11}+\\frac {42}{11}-\\frac {21}{11}=0$$", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "how to prove the formula. – làntèrn Aug 6 '15 at 16:38 • What if $a=0$?. – Michael Galuza Aug 6 '15 at 16:43 The fast approach, as I see it is (in case $a \\neq 0$): you have $ax+by = c$ , just plug in $y=0+ta$, and you get $$ax + bta = c$$ $$x= \\frac ca - bt$$. alternatively, in case you don't have the solution in advance, use same approach and you will get: $$x = \\frac ca - \\frac ba y$$, so the general solution will be $$(x,y) = (\\frac ca - \\frac bat , t)$$ $$(x,y) = (\\frac ca, 0) + (-\\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result. • So, using $t$ here is redundant? – làntèrn Aug 6 '15 at 16:58 • I've edited my answer, I should have you $t$ ,, $y$ is somewhat confusing. – d_e Aug 6 '15 at 17:03 • So $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} -b \\\\ a \\\\ \\end{pmatrix}$ would be same with $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} \\frac{-b}{a} \\\\ 1 \\\\ \\end{pmatrix}$ – làntèrn Aug 6 '15 at 17:27", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "(-9) - 4 × 3 = -18 - 12 = -30 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-30}$$$$\\begin {pmatrix} -9 & -3\\\\ -4 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {9}{30} & \\frac {3}{30}\\\\ \\frac {4}{30} & \\frac {-2}{30}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {9}{30} & \\frac {3}{30}\\\\ \\frac {4}{30} & \\frac {-2}{30}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {9×2}{30} - \\frac {3×1}{30}\\\\ \\frac {4×2}{30} + \\frac {2×1}{30}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {18-3}{30}\\\\ \\frac {8+2}{30}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {15}{30}\\\\ \\frac {10}{30}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} \\frac 12\\\\ \\frac 13\\\\ \\end {pmatrix}$$ i.e.$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac 12\\\\ \\frac 13\\\\ \\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 $$\\frac 1y$$ = $$\\frac 13$$ ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac {3x + 5y}{4}$$ =4, $$\\frac {7x + 3y}{5}$$ = 4 3x + 5y = 16..............................(1) 7x + 3y = 20..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 3 & 5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix}", "case 2 If $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, then $z=t$, a free parameter, which can be any value. Second row in (1) now gives \\begin{align*} \\frac{4}{3}y+\\left( k-\\frac{7}{3}\\right) t & =0\\\\ y & =-\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\end{align*} And first row gives \\begin{align*} 3x+2y+z & =0\\\\ 3x & =-2\\left( -\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\right) -t\\\\ x & =\\frac{1}{2}t\\left( k-3\\right) \\end{align*} Hence in this case, the NullSpace is $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} =t \\begin{pmatrix} \\frac{1}{2}\\left( k-3\\right) \\\\ -\\frac{3}{4}\\left( k-\\frac{7}{3}\\right) \\\\ 1 \\end{pmatrix}$$ Since $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, the solution is $k=5,k=4$. Substituting these values into the above gives the NullSpace as $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = t\\left\\{ \\begin{pmatrix} 1\\\\ -2\\\\ 1 \\end{pmatrix} , \\begin{pmatrix} \\frac{1}{2}\\\\ -\\frac{5}{4}\\\\ 1 \\end{pmatrix} \\right\\}$$ For any $t$. For only $t=0$, this gives the result gives by Mathematica. So Wolfram Mathematica only considered the case 1 above, and did not consider case 2. Wolfram Alpha, which has additional A.I. engine, seems to have looked more into this. • Here's how to use Mathematica for constructing the reduced form: UpperTriangularize[First[LUDecomposition[{{3, 2, 1}, {7, 6, k}, {k, 4, 3}}]]]. Mar 27 '17 at 4:50 • @J.M. Thank you. Mar 27 '17 at 5:04 • @J.M. btw, this only works for square matrices. For non-square matrices, I had to write an M function to do this. Mar 27 '17 at 5:20 • Indeed, it is unfortunate that Mathematica", "f'(x) & = - \\frac{1}{x} \\end{aligned} ## Special Case 3 If a function is composed of a function $$h(x)$$ and a power function $$x^n$$, then the chain rule leads to the following result: $\\text{if} \\quad f(x) = \\begin{bmatrix} h(x) \\end{bmatrix}^n$ $\\text{then} \\quad f'(x) = h'(x)\\times n \\begin{bmatrix}h(x) \\end{bmatrix}^{n-1}$ We further illustrate and explain this sepcial case in Tutorial 4. Spend a few minutes to watch it now. ## Exercise 4 Differentiate each of the following: 1. $$f(x)=\\begin{pmatrix} 4x^2-7\\end{pmatrix}^5$$ 2. $$y = sin^5(x)$$ 3. $$f(x) = \\begin{pmatrix}2cos(x)\\end{pmatrix}^3$$ 4. $$y= \\begin{bmatrix} \\frac{ln(x)}{2} \\end{bmatrix}^6$$ 5. $$f(x) = tan^5(x)$$ 1. $$f'(x) = 40x\\begin{pmatrix}4x^2-7 \\end{pmatrix}^4$$ 2. $$\\frac{dy}{dx} = 5.cos(x).sin^4(x)$$ 3. $$f'(x) = -24.sin(x).cos^2(x)$$ 4. $$\\frac{dy}{dx}=\\frac{3.ln^5(x)}{32x}$$ 5. Two answers possible: $f'(x)=5.sec^2(x).tan^4(x)$ or $f'(x)=5.tan^4(x)+5tan^6(x)$ ## Solution With Working 1. We can see that $$f(x) = \\begin{pmatrix} 4x^2 - 7 \\end{pmatrix}^5$$ can be written $$f(x)=\\begin{bmatrix}h(x)\\end{bmatrix}^5$$, where: $$h(x)=4x^2-7$$ and $$h'(x)=8x$$ So using the special case 3, we can write: \\begin{aligned} f'(x) & = h'(x)\\times 5\\begin{bmatrix}h(x) \\end{bmatrix}^4 \\\\ & = 8x \\times 5 \\begin{bmatrix} 4x^2 - 7 \\end{bmatrix}^4 \\\\ f'(x) & = 40x \\begin{pmatrix} 4x^2 - 7 \\end{pmatrix}^4 \\end{aligned} 2. We can see that $$y=sin^5(x)$$ can be written $$y=\\begin{bmatrix}h(x)\\end{bmatrix}^5$$, where: $$h(x)=sin(x)$$ and $$h'(x)=cos(x)$$. Using the special case 3 we find the derivative function: \\begin{aligned} \\frac{dy}{dx} & = h'(x)\\times 5 \\begin{bmatrix}h(x)\\end{bmatrix}^4 \\\\ & = cos(x)\\times 5 \\begin{bmatrix} sin(x) \\end{bmatrix}^4", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$" ]

[ "line segment $$PQ$$ has an end-point $$P\\begin{pmatrix}-3,4\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}2,-1\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$Q$$. 5. A line segment $$AB$$ has an end-point $$A\\begin{pmatrix}1,-3\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}5,2\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$B$$. Note: this exercise can be downloaded as a worksheet to practice with: 1. We find $$B\\begin{pmatrix}4,13\\end{pmatrix}$$. 2. We find $$Q\\begin{pmatrix}-5,-1\\end{pmatrix}$$. 3. We find $$T\\begin{pmatrix}6,-2\\end{pmatrix}$$. 4. We find $$Q\\begin{pmatrix}7,-6\\end{pmatrix}$$. 5. We find $$B\\begin{pmatrix}9,7\\end{pmatrix}$$.", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "A line segment $P_1P_2$ can be represented in parametric form by \\begin{aligned} x &= x_1 + (x_2 - x_1)t \\\\ y &= y_1 + (y_2 - y_1)t \\\\ & 0 \\leq t \\leq 1 \\end{aligned} Two line segments $P_1P_2$ and $P_3P_4$ intersect if any only if the numbers $$s = \\frac{ \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_1 & y_3 - y_1 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} ~~ \\text{and} ~~ t = \\frac{ \\begin{vmatrix} x_3 - x_1 & y_3 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}}$$ satisfy $0 \\leq s \\leq 1$ and $0 \\leq t \\leq 1$.", "segment is drawn to connect the last point in the path to the first point in the new line segment. ## ◆ add_lines() [1/2] void xtd::drawing::drawing2d::graphics_path::add_lines ( const std::vector< xtd::drawing::point > & points ) Appends a series of connected line segments to the end of this xtd::drawing::drawing2d::graphics_path. Parameters points An array of xtd::drawing::point structures that represents the points that define the line segments to add. Remarks If there are previous lines or curves in the figure, a line is added to connect the endpoint of the previous segment the starting point of the line. The points parameter specifies an array of endpoints. The first two specify the first line. Each additional point specifies the endpoint of a line segment whose starting point is the endpoint of the previous line. ## ◆ add_lines() [2/2] void xtd::drawing::drawing2d::graphics_path::add_lines ( const std::vector< xtd::drawing::point_f > & points ) Appends a series of connected line segments to the end of this xtd::drawing::drawing2d::graphics_path. Parameters points An array of xtd::drawing::point_f structures that represents the points that define the line segments to add. Remarks If there are previous lines or curves in the figure, a line is added to connect the endpoint of the previous segment the starting point of the line. The points parameter specifies an array of endpoints. The first two specify the first line.", "General question about parameterized equations for lines I'm supposed to find the equation of a line perpendicular to and intersecting another line. The original line was parameterized in t. I got the answer as L = (0, 1, 2) + s <3/2, -1/2, -1> The book gave it as L = (0, 1 ,2> + t <3, -1, -2> Since both equations describe the same line, is my answer okay? Thanks!", "of [:y= [$line]:]. [:y=:][___________]{$mutians} When [:x=[$x1]:], predict the value of [:y:]. [:y=:][____]{$multians} When [:y=[$y2]:], predict the value of [:x:]. [:x=:][____]{$multians} END_PGML ENDDOCUMENT();", "diagram: The way to determine if two finite lines cross is first to work out if the infinite versions of the lines would cross, then work out if the crossing point is within the finite range of the lines that we’re interested in. ## Crossing Place Let’s get to the maths. We have a checkpoint line which goes from (xCheckStart, yCheckStart) to (xCheckEnd, yCheckEnd), which we can express in the following form: $\\begin{pmatrix} \\text{x} \\\\ \\text{y} \\end{pmatrix} = \\begin{pmatrix} \\text{xCheckStart} \\\\ \\text{yCheckStart} \\end{pmatrix} + \\text{checkDist} \\times \\begin{pmatrix} \\text{xCheckEnd} - \\text{xCheckStart} \\\\ \\text{yCheckEnd} - \\text{yCheckStart} \\end{pmatrix}$ We also have the bus’s motion in an identical form: $\\begin{pmatrix} \\text{x} \\\\ \\text{y} \\end{pmatrix} = \\begin{pmatrix} \\text{xBusStart} \\\\ \\text{yBusStart} \\end{pmatrix} + \\text{busDist} \\times \\begin{pmatrix} \\text{xBusEnd} - \\text{xBusStart} \\\\ \\text{yBusEnd} - \\text{yBusStart} \\end{pmatrix}$ Each of these expressions defines an (x, y) point on the respective lines. Finding out whether the lines cross is the same as checking if there is an (x, y) point that lies on both lines. For this, we set the line equations equal to each other: $\\begin{pmatrix} \\text{xCheckStart} \\\\ \\text{yCheckStart} \\end{pmatrix} + \\text{checkDist} \\times \\begin{pmatrix} \\text{xCheckEnd} - \\text{xCheckStart} \\\\ \\text{yCheckEnd} - \\text{yCheckStart} \\end{pmatrix} =\\\\ \\begin{pmatrix} \\text{xBusStart} \\\\ \\text{yBusStart} \\end{pmatrix} + \\text{busDist} \\times \\begin{pmatrix} \\text{xBusEnd} - \\text{xBusStart} \\\\ \\text{yBusEnd} - \\text{yBusStart} \\end{pmatrix}$ ## A Little Help From Our Friends To", "- y_1)t \\\\ & 0 \\leq t \\leq 1 \\end{aligned} Two line segments P1P2 and P3P4 intersect if any only if the numbers $$s = \\frac{ \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_1 & y_3 - y_1 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} ~~ \\text{and} ~~ t = \\frac{ \\begin{vmatrix} x_3 - x_1 & y_3 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}} { \\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\\\ x_3 - x_4 & y_3 - y_4 \\end{vmatrix}}$$ satisfy $$0 \\leq s \\leq 1$$ and $$0 \\leq t \\leq 1$$", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "to vector form $$y = \\frac{x}{3} + 2$$ We want this in the form $$\\underline{r} =$$ 1st point + direction Using the cartesian equation we can find the direction as we know that the gradient of a line is: $$m = \\frac{\\text{Change in y}}{\\text{Change in x}}$$ We can see that for every time y is increased by 1 x is increases by 1/3 as x is divided by 3 giving us: $$m = \\frac{1}{(\\frac{1}{3})}=\\frac{3}{1}$$ This means we can write the direction as: $$\\underline{r} =$$ 1st point + $$λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ Now we juust need to find the 1st point to do this we simply set the value of λ=0 as this is where the line starts From the equation $$y = \\frac{x}{3} + 2$$ We can see that when x=0 y=2 so this is where our line starts Therefore the vector equation can be written as: $$\\underline{r} = \\begin{pmatrix}0\\\\\\ 2\\end{pmatrix} + λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ #### Intersection of two lines What about if we need to find the exact point that two lines intersect knowing only the vector equations of these two lines? $$\\underline{r_1} = \\begin{pmatrix}2\\\\\\ 3\\end{pmatrix} + λ\\begin{pmatrix}1\\\\\\ 2\\end{pmatrix}$$ $$\\underline{r_2} = \\begin{pmatrix}6\\\\\\ 1\\end{pmatrix} + μ\\begin{pmatrix}1\\\\\\ -3\\end{pmatrix}$$ For the two lines to intercept there must be a point when the value of the two equations are the same this means we", "last point in the path to the first point in the new line segment. ## ◆ add_line() [3/4] void xtd::drawing::drawing2d::graphics_path::add_line ( int32_t x1, int32_t y1, int32_t x2, int32_t y2 ) Appends a line segment to this xtd::drawing::drawing2d::graphics_path. Parameters x1 The x-coordinate of the starting point of the line. y1 The y-coordinate of the starting point of the line. x2 The x-coordinate of the endpoint of the line. y2 The y-coordinate of the endpoint of the line. Remarks This method adds the line segment defined by the specified points to the end of this xtd::drawing::drawing2d::graphics_path. If there are previous lines or curves in the xtd::drawing::drawing2d::graphics_path, a line segment is drawn to connect the last point in the path to the first point in the new line segment. ## ◆ add_line() [4/4] void xtd::drawing::drawing2d::graphics_path::add_line ( float x1, float y1, float x2, float y2 ) Appends a line segment to this xtd::drawing::drawing2d::graphics_path. Parameters x1 The x-coordinate of the starting point of the line. y1 The y-coordinate of the starting point of the line. x2 The x-coordinate of the endpoint of the line. y2 The y-coordinate of the endpoint of the line. Remarks This method adds the line segment defined by the specified points to the end of this xtd::drawing::drawing2d::graphics_path. If there are previous lines or curves in the xtd::drawing::drawing2d::graphics_path, a line", "mirror images with respect to a line through $\\mathbf{P}_1$ and $\\mathbf{P}_2$); • $f_S$ which arc between $\\mathbf{P}_1$ and $\\mathbf{P}_2$ should be considered (i.e. the longer or the shorter one). So, starting from the parameters above, we have the goal to find: • the center of the ellipse $\\mathbf{C}$; • the parameter $t_1$ corresponding to point $\\mathbf{P}_1$; • the parameter $t_2$ corresponding to point $\\mathbf{P}_2$; • the parameters $t_{begin}$ and $t_{end}$ from $t_1$ and $t_2$ so that $[t_{begin}, t_{end}]$ is a contiguous interval representing the arc of ellipse we are after. As you might have already guessed, our next move when we have these parameters will be to find the minimum and maximum values for the two coordinates when $t$ varies in the $[t_{begin}, t_{end}]$ range, and this will give us the bounding box. Stay tuned!", "# Approximating a sine curve with cubic Bézier splines ## Thomas W. We approximate the interval $[0,\\pi /2]$ with a cubic Bézier curve. We need four control points, $P_0$, $P_1$, $P_2$ and $P_3$. As we're approximating the interval $[0,\\pi /2]$ and want the boundary points to be accurate, we know $P_0$ and $P_3$: \\begin{align*} P_0 &= \\begin{pmatrix} 0 \\\\sin(0)\\end{pmatrix} = \\begin{pmatrix} 0 \\\\0\\end{pmatrix}\\\\ P_3 &= \\begin{pmatrix} 0 \\\\sin(\\pi/2)\\end{pmatrix} = \\begin{pmatrix}\\pi/2\\\\1\\end{pmatrix}. \\end{align*} We want the tangents to be accurate in the endpoints. The slope in the start point $P_0$ is the slope of the vector $$P_1 - P_0 = P_1 - \\begin{pmatrix}0\\\\0\\end{pmatrix}=P_1=\\begin{pmatrix}x_1\\\\y_1\\end{pmatrix}.$$ The slope in the end point $P_3$ is the slope of the vector $$P_3 - P_2 = \\begin{pmatrix}\\pi/2\\\\1\\end{pmatrix} - \\begin{pmatrix}x_2\\\\y_2\\end{pmatrix}.$$ We know that the slope at $x=0$ (i.e. in $P_1$) is $1$, therefore \\begin{align*} y_1/x_1 = 1 \\Rightarrow y_1 = x_1\\\\ \\Rightarrow P_1 = \\begin{pmatrix}x_1\\\\x_1\\end{pmatrix}. \\end{align*} The slope at $x=\\pi/2$ (i.e. in point $P_2$) is 0, therefore \\begin{align*} {{y_2 - 1}\\over{x_2 - \\pi/2}} = 0 \\Rightarrow y_2 = 1\\\\ \\Rightarrow P_2 = \\begin{pmatrix}x_2\\\\1\\end{pmatrix}. \\end{align*} To find $x_1$ and $x_2$, we require the curvature of the approximating Bézier curve to be accurate in the boundary points. The curvature of a graph of function $f$ is $$k(x) = \\frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$$ For the $sin$ function, it's therefore $$k(x) = \\frac{-sin(x)}{(1+cos(x)^2)^{3/2}}.$$ We", "+0 # Let O = (0, 0, 0) as usual. Consider the line consisting of all points Q such that for some real value of t and the line consisting of all -1 52 1 +82 Let O = (0, 0, 0) as usual. Consider the line consisting of all points Q such that $$\\overrightarrow{OQ}= \\begin{pmatrix} 7 \\\\ -3 \\\\ 1 \\end{pmatrix} + \\begin{pmatrix} -2 \\\\ 5 \\\\ 1 \\end{pmatrix} t$$for some real value of t and the line consisting of all points R such that $$\\overrightarrow{OR} = \\begin{pmatrix} 8 \\\\ -1 \\\\ -1 \\end{pmatrix} + \\begin{pmatrix} 1 \\\\ -4 \\\\ 0 \\end{pmatrix} u$$ for some real value of u. Enter the point of intersection of the two lines in the usual (a, b, c) format. Any help would be appreciate, thank you so much! Mar 2, 2020 #1 +29249 +3 If they intersect at the point (x, y, z) then you have: 7 - 2x = 8 + x -3 + 5y = -1 - 4y 1 + z = -1 I'm sure you can take it from here. Mar 2, 2020", "& 0 \\leq t \\leq 1 \\end{aligned} Two line segments P1P2 and P3P4 intersect if any only if the numbers $$s = \\frac{ \\begin{vmatrix} x_2 – x_1 & y_2 – y_1 \\\\ x_3 – x_1 & y_3 – y_1 \\end{vmatrix}} { \\begin{vmatrix} x_2 – x_1 & y_2 – y_1 \\\\ x_3 – x_4 & y_3 – y_4 \\end{vmatrix}} ~~ \\text{and} ~~ t = \\frac{ \\begin{vmatrix} x_3 – x_1 & y_3 – y_1 \\\\ x_3 – x_4 & y_3 – y_4 \\end{vmatrix}} { \\begin{vmatrix} x_2 – x_1 & y_2 – y_1 \\\\ x_3 – x_4 & y_3 – y_4 \\end{vmatrix}}$$ satisfy $$0 \\leq s \\leq 1$$ and $$0 \\leq t \\leq 1$$", "{ \\sqrt{\\begin{vmatrix} b_1 & c_1 \\\\ b_2 & c_2 \\end{vmatrix}^2 + \\begin{vmatrix} c_1 & a_1 \\\\ c_2 & a_2 \\end{vmatrix}^2 + \\begin{vmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}^2 }}$$ The two lines intersect if: $$\\begin{vmatrix} x_2-x_1 & y_2 - y_1 & z_2 - z_1 \\\\ a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\end{vmatrix} = 0$$", "parameter in order to be considered to be part of the room. Another possibility is that the end point of a line lies on the contour of a room. In order to determine if this is happening the closest distance of this end point to the line spanned by the two nearby corners should be determined. This can be done using simple vector calculus. The shortest distance between a line the origin is expressed in the following equation: $\\left(\\begin{pmatrix}x_{p_{1}}\\\\y_{p_{1}}\\\\\\alpha_{p_{1}}\\end{pmatrix}+t\\begin{pmatrix}x_{p_{1}}-x_{p_{2}}\\\\y_{p_{1}}-y_{p_{2}}\\\\\\alpha_{p_{1}}-\\alpha_{p_{2}}\\end{pmatrix}\\right)\\cdot\\begin{pmatrix}x_{p_{1}}-x_{p_{2}}\\\\y_{p_{1}}-y_{p_{2}}\\\\\\alpha_{p_{1}}-\\alpha_{p_{2}}\\end{pmatrix}=0$ In this equation p1 and p2 are calculated by substracting the end point of the line under consideration from the corners of the room. t is calculated, and substituted in the following equation to obtain the vector spanned between the end point and the line: $\\begin{pmatrix}x_{p_{1}}\\\\y_{p_{1}}\\\\\\alpha_{p_{1}}\\end{pmatrix}+t\\begin{pmatrix}x_{p_{1}}-x_{p_{2}}\\\\y_{p_{1}}-y_{p_{2}}\\\\\\alpha_{p_{1}}-\\alpha_{p_{2}}\\end{pmatrix}$ The length of this vector is also compared with the fixed parameter used to test the coincidence of the end points and the corners. ### Localisation Because the lines are created in a certain way, it can be derived to which of the four walls they belong. The start- and end points of lines are defined in a counter-clockwise manner around the coordinate system, and the angle of the lines is calculated with respect to the x-axis of a this coordinate system. The lines corresponding to the rightmost wall would thus have an", "their bisector have equation $w+z=k$ for some $k$, i.e. $$\\begin{pmatrix} 1 & 1 \\end{pmatrix} \\begin{pmatrix} w \\\\ z \\end{pmatrix} = k$$ in the original coordinates the line is $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix} 3 & 4 \\\\ 4 & -3 \\end{pmatrix} \\left( \\begin{pmatrix} x \\\\ y \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix}\\right) = k$$ or $$7x+y =k'$$ and since the line passes through $(1,2)$ we must have $k'=9$", "parameterization gives, your method almost works. so r(t) = [t , sqrt(-t2-2t+3) , t-2] r'(t) = (1 , 0.5(-t2-2t+3)-0.5(-2t-2) , 1) at point (1,0,-1) my t = 1 therefore r'(t) = (1 , 0 , 1) this is a vector parallel to the line right? Are you ready for this? At t= 1, $(-t^2- 2t+ 3)^{1/2}= 0$ but you have the negative 1/2 power. That is not 0, it \"blows up\" to infinity. This particular parameterization does not work at this particular point! therefore the line parameterization at (1,0,-1) is x = 1 + t y = 0 z = -1 + t is this correct? somethings wrong..... i solve it using a different method and got a diff ans :( ∇f = (2x+1, 2y , 1) ∇g = (-1 , 0, 1) ∇f(1,0,-1) = (3,0,1) ∇g(1,0,-1) = (-1,0,1) so ∇f x ∇g gives me a vector parallel to the line at (1,0,-1) right? so parameterized line is x = 1 y= -4t z = -1 ??? which is totally different from my 1st post answer the answer is x=1, y=2t, z=-1 does anyone know what went wrong? thanks Yes, that works. However, one line may have many different parameterizations. For example, the first method above, using the parameterization $x= 2cos(\\theta)- 1$, $y= 2sin(\\theta)$, $z= 2cos(\\theta)- 3$ gives", "## Derivation of Formula For Distance Between Parallel Lines If two lines are parallel, the coefficients in the Cartesian equations are the same, since the coefficients of the Cartesian equation are the components of the normal, which is the same for parallel lines. Suppose we want to find the distance between the parallel lines $ax+by=d_1$ (1) $ax+by=d_1$ (2) Suppose a normal connects points $(x_1,y_1)$ in line (1) to point $(x_2,y_2)$ in line (2). Then $ax_1+by_1=d_1$ (3) $ax_2+by_2=d_2$ (4) (4)-(3) gives $a(x_2-x_1)+b(y_2-y_1)=d_2-d_1$ (5) Let the equation of the normal from line (1) to (2) be $\\mathbf{r}(t)= \\begin{pmatrix}x_1\\\\y_\\\\end{pmatrix} + \\begin{pmatrix}a\\\\b\\end{pmatrix} t$ . Then $\\begin{pmatrix}x_2\\\\y_2\\end{pmatrix}= \\begin{pmatrix}x_1\\\\y_1\\end{pmatrix} + \\begin{pmatrix}a\\\\b\\end{pmatrix} t_0 \\rightarrow \\begin{pmatrix}x_2-x_1\\\\y_2-y_1\\end{pmatrix}= \\begin{pmatrix}a\\\\b\\end{pmatrix} t_0$ Substitution into (5) gives $a^2t_0+b^2t_0=d_2-d_1 \\rightarrow t_0=\\frac{d_2-d_1}{a^2+b^2}$ $\\begin{pmatrix}x_2-x_1\\\\y_2-y_1-\\end{pmatrix}-= \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} \\frac{d_2-d_1}{a^2+b^2}$ The distance between the lines is then $\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\\sqrt{a^2+b^2}\\frac{d_2-d_1}{a^2+b^2+c^2}= \\frac{d_2-d_1}{\\sqrt{a^2+b^2}$" ]

[ "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "\\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {11}{12} & \\frac {-5}{12}\\\\ \\frac {22}{9} & \\frac {5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {11 - 5}{12}\\\\ \\frac {22 + 5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {6}{12}\\\\ \\frac {27}{9}\\\\ \\end {pmatrix}$$ ∴X = $$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ i.e.$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, 4x + $$\\frac y5$$ = 7 or, $$\\frac {20x + y}{5}$$ = 7 or, 20x + y = 35.........................................(1) 3x + $$\\frac y4$$ = 5 or, $$\\frac {12x + y}{4}$$ = 5 or, 12x + y = 20..........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 20 & 1\\\\ 12 &", "\\\\\\dfrac 5 6 &0\\end{pmatrix}\\\\ R = \\begin{pmatrix}\\dfrac 1 2 &0 \\\\0 &\\dfrac 1 6\\end{pmatrix}\\\\ N = \\left(I_2 - Q\\right)^{-1} = \\begin{pmatrix} \\dfrac{12}{7} & \\dfrac{6}{7} \\\\ \\dfrac{10}{7} & \\dfrac{12}{7} \\\\ \\end{pmatrix}\\\\ B = NR = \\begin{pmatrix} \\dfrac{6}{7} & \\dfrac{1}{7} \\\\ \\dfrac{5}{7} & \\dfrac{2}{7} \\\\ \\end{pmatrix}$ From $B$ we can directly read off that $P[\\text{MW|start with MP}] = \\dfrac 6 7$ and $P[\\text{MW|start with SP}] = \\dfrac 5 7$ OH thank you so much", "correct indeed, by a direct check, for $$t=-\\frac 6{11}$$ we obtain $$N=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} -\\frac 6{11}\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=\\begin{pmatrix} \\frac {18}{11} \\\\ \\frac {1}{11} \\\\ \\frac {40}{11} \\end{pmatrix}$$ $$\\overrightarrow{CN}=\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix}$$ $$\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}= -\\frac {21}{11}+\\frac {42}{11}-\\frac {21}{11}=0$$", "(-9) - 4 × 3 = -18 - 12 = -30 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-30}$$$$\\begin {pmatrix} -9 & -3\\\\ -4 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {9}{30} & \\frac {3}{30}\\\\ \\frac {4}{30} & \\frac {-2}{30}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {9}{30} & \\frac {3}{30}\\\\ \\frac {4}{30} & \\frac {-2}{30}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {9×2}{30} - \\frac {3×1}{30}\\\\ \\frac {4×2}{30} + \\frac {2×1}{30}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {18-3}{30}\\\\ \\frac {8+2}{30}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {15}{30}\\\\ \\frac {10}{30}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} \\frac 12\\\\ \\frac 13\\\\ \\end {pmatrix}$$ i.e.$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac 12\\\\ \\frac 13\\\\ \\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 $$\\frac 1y$$ = $$\\frac 13$$ ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac {3x + 5y}{4}$$ =4, $$\\frac {7x + 3y}{5}$$ = 4 3x + 5y = 16..............................(1) 7x + 3y = 20..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 3 & 5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix}", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "\\begin{pmatrix} yuv & xyu \\\\ 2x & 8v \\end{pmatrix} = -\\frac{1}{16} \\det\\begin{pmatrix} -2 & -2 \\\\ 2 & 8 \\end{pmatrix} = -\\frac{1}{16} (-16+4) = -\\frac{-12}{16} = \\frac{3}{4} \\\\ \\frac{\\partial u}{\\partial y} & = -\\frac{1}{J} \\frac{\\partial (F, G)}{\\partial (y, v)} = -\\frac{1}{16} \\det\\begin{pmatrix} xuv & xyu \\\\ 4y & 8v \\end{pmatrix} = -\\frac{1}{16} \\det\\begin{pmatrix} 2 & -2 \\\\ -4 & 8 \\end{pmatrix} = -\\frac{1}{16} (16 - 8) = -\\frac{8}{16} = -\\frac{1}{2} \\\\ \\frac{\\partial v}{\\partial x} & = -\\frac{1}{J} \\frac{\\partial (F, G)}{\\partial (u, x)} = -\\frac{1}{16} \\det\\begin{pmatrix} xyv & yvu \\\\ 6u & 2x \\end{pmatrix} = -\\frac{1}{16} \\det\\begin{pmatrix} -1 & -2 \\\\ 12 & 2 \\end{pmatrix} = -\\frac{1}{16} (-2 + 24) = -\\frac{22}{16} = -\\frac{3}{8} \\\\ \\frac{\\partial v}{\\partial y} & = -\\frac{1}{J} \\frac{\\partial (F, G)}{\\partial (u, y)} = -\\frac{1}{16} \\det\\begin{pmatrix} xyv & xuv \\\\ 6u & 4y \\end{pmatrix} = -\\frac{1}{16} \\det\\begin{pmatrix} -1 & 2 \\\\ 12 & -4 \\end{pmatrix} = -\\frac{1}{16} (4 - 24) = -\\frac{-20}{16} = \\frac{5}{4} \\end{align*} Consequently, the matrix we're looking for is $$\\begin{pmatrix} \\frac{3}{4} & -\\frac{3}{8} \\\\ -\\frac{1}{2} & \\frac{5}{4} \\end{pmatrix}$$ Maybe ## 2Question 2¶ Lagrange multipliers aren't in the syllabus this year I believe ## 3Question 3¶ Let $R$ be the region in the first quadrant of the $xy$-plane bounded by the lines $y=x$ and $y=2x$ and by the hyperbolas $xy=1$ and $xy=3$. Evaluate $\\displaystyle", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "& d\\end{pmatrix}\\in\\text{SU}(2)$$ It is true that $$A^{-1}=A^{*}\\;\\text{and}\\;\\text{det}(A)=1$$: $$A^{-1}=\\begin{pmatrix}d & -b \\\\ -c & a\\end{pmatrix}$$ and $$A^{*}=\\begin{pmatrix}\\overline{a} & \\overline{c} \\\\ \\overline{b} & \\overline{d}\\end{pmatrix}$$ hence $$A^{-1}=A^*$$ implies that $$d=\\overline{a}\\;\\text{and}\\; b=-\\overline{c}.$$ Then $$A=\\begin{pmatrix}a & -\\overline{c} \\\\ c & \\overline{a}\\end{pmatrix}.$$", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "{10}}}\\\\{-{\\sqrt {10}}}\\\\{-i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{-3}\\\\{-i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{-3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{1}\\\\{-i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{-i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{1}\\\\{i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{-3}\\\\{i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i}\\\\{\\sqrt {5}}\\\\{-i{\\sqrt {10}}}\\\\{-{\\sqrt {10}}}\\\\{i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}}\\\\S_{z}={\\frac {\\hbar }{2}}{\\begin{pmatrix}5&0&0&0&0&0\\\\0&3&0&0&0&0\\\\0&0&1&0&0&0\\\\0&0&0&-1&0&0\\\\0&0&0&0&-3&0\\\\0&0&0&0&0&-5\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\end{pmatrix}}\\\\\\end{array}}}$ 5. The generalization of these matrices for arbitrary spin s is {\\displaystyle {\\begin{aligned}\\left(S_{x}\\right)_{ab}&={\\frac {\\hbar }{2}}\\left(\\delta _{a,b+1}+\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\,\\\\\\left(S_{y}\\right)_{ab}&={\\frac {i\\hbar }{2}}\\left(\\delta _{a,b+1}-\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\\\\\left(S_{z}\\right)_{ab}&=\\hbar (s+1-a)\\delta _{a,b}=\\hbar (s+1-b)\\delta _{a,b}\\end{aligned}}} where indices ${\\displaystyle a,b}$ are integer numbers such that ${\\displaystyle 1\\leq a\\leq 2s+1}$ and ${\\displaystyle 1\\leq b\\leq 2s+1}$ Also useful in the quantum mechanics of multiparticle systems, the general Pauli group Gn is defined to consist of all n-fold tensor products of Pauli matrices. The analog formula of Euler's formula in terms of the Pauli matrices: ${\\displaystyle e^{i\\theta \\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)}=I\\cos(\\theta )+i\\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)\\sin(\\theta )}$ for higher spins is tractable, but less simple.[20] ## Parity In tables of the spin quantum number s for nuclei or particles, the spin is often followed by a \"+\" or \"−\". This refers to the parity with \"+\" for even parity (wave function unchanged by spatial inversion) and \"−\" for odd" ]

[ "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "parameters, which means there are 2 free variables $y,z$ and an infinite amount of solutions. A general solution for $(O)$ looks like: $(-2t+s, t, s)$ where $t,s\\in \\mathbb{R}$. General solution for $(M)$ $A\\underline{x}=\\underline0, \\ \\ \\ A\\underline{c}=\\underline{b}\\ \\ \\ \\ \\rightarrow \\ \\ \\ A\\underline{x}+A\\underline{c}=\\underline0+\\underline{b}=\\underline{b} \\ \\ \\ \\ \\rightarrow \\ \\ \\ A(\\underline{x}+\\underline{c})=\\underline{b}$ Hence, the general solution for $(M)$ is: ${\\{(2, -3, 1)+(-2t+s, t, s)|t,s \\in \\mathbb{R}\\}}$ Another way to write the general solution: $\\begin{pmatrix}2\\\\-3\\\\1\\end{pmatrix}+t\\begin{pmatrix}-2\\\\1\\\\0\\end{pmatrix}+s\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}$, $t,s \\in \\mathbb{R}$", "1 \\\\1\\end{pmatrix}\\\\ P_2 &= \\begin{pmatrix} 1 \\\\1\\end{pmatrix}\\\\ P_3 &= \\begin{pmatrix}\\pi/2\\\\1\\end{pmatrix} \\end{align*} These control points can be transformed to piece together a spline interpolation of the sine curve. Wednesday, 19 December 2012 07:51 GMT", "= \\frac{1}{\\textbf{h}^{3T}\\textbf{x}}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\textbf{h}^{1T}\\textbf{x}}{(\\textbf{h}^{3T}\\textbf{x})^2} \\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Substituting $x’ = \\textbf{h}^{1T}\\textbf{x}$ and $w’ = \\textbf{h}^{3T}\\textbf{x}$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{x’}{(w’)^2}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Finally, substituting $\\tilde{x’} = x’/w’$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\tilde{x’} }{w’}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ $$= \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’} \\textbf{x}^T \\end{pmatrix}$$ Performing a similar process for $\\tilde{y’}$, we get $$\\partial\\tilde{y}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} 0 & \\textbf{x}^T & -\\tilde{y’} \\textbf{x}^T \\end{pmatrix}$$ Hence, we have established that $$\\partial{\\tilde{\\textbf{x}}}’ / \\partial{\\textbf{h}} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’}\\textbf{x}^T \\\\ 0 & \\textbf{x}^T & -\\tilde{y’}\\textbf{x}^T\\end{pmatrix}$$ #### VI. Sampson error with non-isotropic error distributions. The derivation of Sampson error in section 4.2.6(p98) assumed that points were measured with circular error distributions. In the case where the point $\\textbf{X} = (x, y, x’, y’)$ is measured with covariance matrix $\\Sigma_\\textbf{x}$ it is appropriate instead to minimize the Mahalanobis norm $\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\boldsymbol{\\delta}^T_\\textbf{x}\\Sigma^{-1}_\\textbf{x}\\boldsymbol{\\delta}_\\textbf{x}$. Show that in this case the formulae corresponding to (4.11-p99) and (4.12-p99) are $$\\boldsymbol{\\delta}_\\textbf{x} = -\\Sigma_\\textbf{x}J^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ and $$\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\epsilon^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ Note that if the measurements in the two images are independent, then the covariance matrix $\\Sigma_\\textbf{x}$ will be block-diagonal with two $2\\times2$ diagonal blocks", "point c) intersect in a line d) intersect in a plane We have: . $\\begin{pmatrix}1 & 3 & \\text{-}1 & | & 4\\\\ 2 & 8 & 0 & | & 18\\\\ 1 & \\text{-}1 & \\text{-}3 & | & \\text{-}6\\end{pmatrix}$ . . . . $\\begin{array}{cccc} \\\\ \\frac{1}{2}R_2\\\\ \\\\\\end{array}\\;\\begin{pmatrix}1 & 3 & \\text{-}1 & | & 4\\\\1 & 4 & 0 & | & 9 \\\\ 1 & \\text{-}1 & \\text{-}3 & | & \\text{-}6\\end{pmatrix}$ $\\begin{array}{ccc} \\\\ R_2-R_1\\\\ R_3 - R_1\\end{array}\\;\\begin{pmatrix}1 & 3 & \\text{-}1 & | & 4\\\\ 0 & 1 & 1 & | & 5\\\\0 & \\text{-}4 & \\text{-}2 & | &\\text{-}10\\end{pmatrix}$ $\\begin{array}{ccc}R_1-3R_2\\\\ \\\\ R_3+4R_2\\end{array}\\;\\begin{pmatrix}1 & 0 & \\text{-}4 & | & \\text{-}11 \\\\ 0 & 1 & 1 & | & 5\\\\ 0 & 0 & 2 & | & 10\\end{pmatrix}$ . . . . $\\begin{array}{ccc} \\\\ \\\\ \\frac{1}{2}R_3\\end{array}\\;\\begin{pmatrix}1 & 0 & \\text{-}4 & | & \\text{-}11\\\\0 & 1 & 1 & | & 5\\\\ 0 & 0 & 1 & | & 5\\end{pmatrix}$ . $\\begin{array}{cccc} R_1+4R_3 \\\\ R_2-R_3 \\\\ \\\\ \\end{array}\\;\\begin{pmatrix}1 & 0 & 0 & | & 9 \\\\ 0 & 1 & 0 & | & 0\\\\0 & 0 & 1 & | & 5\\end{pmatrix}$ Therefore: (b) The planes intersect in a point, $\\bf{(9,0,5)}$", "& d\\end{pmatrix}\\in\\text{SU}(2)$$ It is true that $$A^{-1}=A^{*}\\;\\text{and}\\;\\text{det}(A)=1$$: $$A^{-1}=\\begin{pmatrix}d & -b \\\\ -c & a\\end{pmatrix}$$ and $$A^{*}=\\begin{pmatrix}\\overline{a} & \\overline{c} \\\\ \\overline{b} & \\overline{d}\\end{pmatrix}$$ hence $$A^{-1}=A^*$$ implies that $$d=\\overline{a}\\;\\text{and}\\; b=-\\overline{c}.$$ Then $$A=\\begin{pmatrix}a & -\\overline{c} \\\\ c & \\overline{a}\\end{pmatrix}.$$", "\\\\ 5, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 16, & 3 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{3--7}{16-5} && \\text{Fill in the values}\\\\& m =\\frac{3+7}{16-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{10}{11} && \\text{Calculate the value of the numerator and the denominator}$ ii) (–6, –7) and (–1, –4) $& \\begin{pmatrix} x_1, & y_1 \\\\ -6, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -1, & -4 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-4--7}{-1--6} && \\text{Fill in the values}\\\\& m =\\frac{-4+7}{-1+6} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{3}{5} && \\text{Calculate the value of the numerator and the denominator}$ iii) (5, –12) and (0, –6) $& \\begin{pmatrix} x_1, & y_1 \\\\ 5, & -12 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 0, & -6 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-6--12}{0-5} && \\text{Fill in the values}\\\\& m =\\frac{-6+12}{0-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{6}{-5} && \\text{Calculate the value of the numerator and the denominator}\\\\& m =\\frac{-6}{5} && \\text{Do not leave the negative sign in the denominator}$ 3. i) You are not given the", "diagram: The way to determine if two finite lines cross is first to work out if the infinite versions of the lines would cross, then work out if the crossing point is within the finite range of the lines that we’re interested in. ## Crossing Place Let’s get to the maths. We have a checkpoint line which goes from (xCheckStart, yCheckStart) to (xCheckEnd, yCheckEnd), which we can express in the following form: $\\begin{pmatrix} \\text{x} \\\\ \\text{y} \\end{pmatrix} = \\begin{pmatrix} \\text{xCheckStart} \\\\ \\text{yCheckStart} \\end{pmatrix} + \\text{checkDist} \\times \\begin{pmatrix} \\text{xCheckEnd} - \\text{xCheckStart} \\\\ \\text{yCheckEnd} - \\text{yCheckStart} \\end{pmatrix}$ We also have the bus’s motion in an identical form: $\\begin{pmatrix} \\text{x} \\\\ \\text{y} \\end{pmatrix} = \\begin{pmatrix} \\text{xBusStart} \\\\ \\text{yBusStart} \\end{pmatrix} + \\text{busDist} \\times \\begin{pmatrix} \\text{xBusEnd} - \\text{xBusStart} \\\\ \\text{yBusEnd} - \\text{yBusStart} \\end{pmatrix}$ Each of these expressions defines an (x, y) point on the respective lines. Finding out whether the lines cross is the same as checking if there is an (x, y) point that lies on both lines. For this, we set the line equations equal to each other: $\\begin{pmatrix} \\text{xCheckStart} \\\\ \\text{yCheckStart} \\end{pmatrix} + \\text{checkDist} \\times \\begin{pmatrix} \\text{xCheckEnd} - \\text{xCheckStart} \\\\ \\text{yCheckEnd} - \\text{yCheckStart} \\end{pmatrix} =\\\\ \\begin{pmatrix} \\text{xBusStart} \\\\ \\text{yBusStart} \\end{pmatrix} + \\text{busDist} \\times \\begin{pmatrix} \\text{xBusEnd} - \\text{xBusStart} \\\\ \\text{yBusEnd} - \\text{yBusStart} \\end{pmatrix}$ ## A Little Help From Our Friends To", "the observer). \\begin{aligned} R_x(\\theta)&=\\begin{pmatrix}1 &0 &0\\\\ 0 &\\cos\\theta &-\\sin\\theta\\\\ 0 &\\sin\\theta &\\cos\\theta \\end{pmatrix}\\\\ R_y(\\theta)&=\\begin{pmatrix}\\cos\\theta &0 &\\sin\\theta\\\\ 0 &1 &0\\\\ -\\sin\\theta &0 &\\cos\\theta \\end{pmatrix}\\\\ R_z(\\theta)&=\\begin{pmatrix} \\cos\\theta &-\\sin\\theta &0\\\\ \\sin\\theta &\\cos\\theta &0\\\\ 0 &0 &1 \\end{pmatrix} \\end{aligned} ## Example 1 Rotating $\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} ## Example 2 Rotating $\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}-\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} Posted in Uncategorized | 1 Comment ## 十二平均律 12-tone Equal Temperament 《十二平均律》 是明朝 ＂布衣王子＂ 朱载堉 (1536年－1610年) 发明的，由当时在中国传教的意大利人 利马窦 (Matteo Ricci, 传给欧洲的法国数学家 Marin Mersenne (Mersenne Prime ) 。现代音乐之父 巴哈 Bach 第一个采用，制作世界第一架钢琴有12黑白键，并作曲 《Bach 12-tone Equal Temperament》。 Side Note: 1977年法国大学数学教授在课堂好奇地问我，你们祖先如何解代数？是用算盘吗？当时计算机还不流行，复杂的算法只能用Log Table 或 计算尺 (Slide Rule) 。 Bach 12-tone Equal Temperament https://zh.m.wikipedia.org/wiki/%E5%8D%81%E4%BA%8C%E5%B9%B3%E5%9D%87%E5%BE%8B View original post ## Richard Dedekind Julius Wilhelmina Richard Dedekind (6 Oct 1831 – 12 Feb 1916) – Last student of Gauss at Göttingen – Student and closed friend of Dirichlet who influenced his Mathematical education – Introduced the word Field (Körper) – Gave the first university course on Galois Theory – Developed Real Number ‘Dedekind Cut‘ in 1872 – Accomplished musician – Never married, lived with", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded", "## Dividing a Line Segment in a Given Ratio Suppose the line between $A(2,3)$ and $B((9,20)$ is to be divided in the ratio $2:3$ . Ket the point that divides the line segment in this ratio be $P$ . The vector from $A$ to $B$ is $\\vec{AB}=\\vec{OB}=\\vec{OA}= \\begin{pmatrix}9\\\\20\\end{pmatrix}-\\begin{pmatrix}2\\end{pmatrix}=\\begin{pmatrix}7\\\\17\\end{pmatrix}$ . The vector $\\vec{AP}=\\frac{2}{5} \\vec{AB}=\\frac{2}{5}\\begin{pmatrix}7\\\\17\\end{pmatrix}=\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}$ . Then $\\vec{OP}=\\vec{OA}+ \\vec{AP}=\\begin{pmatrix}2\\\\3\\end{pmatrix}+\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}=\\begin{pmatrix}24/5\\\\49/5\\end{pmatrix}$ . The point $P$ is $(24/5,49/5)$ .", "we saw above we can write: \\begin{aligned} \\frac{dy}{dx} & = 3 \\times \\frac{u'(x)}{u(x)} \\\\ & = 3 \\times \\frac{3x^2 -6x}{x^3 - 3x^2 + 6} \\\\ & = \\frac{3\\begin{pmatrix} 3x^2 - 6x \\end{pmatrix}}{x^3 - 3x^2 + 6} \\\\ \\frac{dy}{dx} & = \\frac{ 9x^2 - 18x}{x^3 - 3x^2 + 6} \\end{aligned} 3. The function $$f(x) = -2 ln \\begin{pmatrix} 4x+5 \\end{pmatrix}$$ can be written: $f(x) = -2ln \\begin{pmatrix} u(x) \\end{pmatrix}$ where $u(x) = 4x+5$ and $u'(x) = 4$ So \\begin{aligned} f'(x) & = -2 \\times \\frac{u'(x)}{u(x)} \\\\ & = -2 \\times \\frac{4}{4x+5} \\\\ & = - \\frac{2\\times 4}{4x+5} \\\\ f'(x) & = - \\frac{-8}{4x+5} \\end{aligned} 4. The function $$y = 4.ln \\begin{pmatrix} \\sqrt{x} \\end{pmatrix}$$ can be written: $y =4.ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = \\sqrt{x}$ and $u'(x) = \\frac{1}{2 \\sqrt{x}}$ So \\begin{aligned} \\frac{dy}{dx} & = 4 \\times \\frac{u'(x)}{u(x)} \\\\ & = 4 \\times \\frac{ \\frac{1}{2\\sqrt{x}}}{\\sqrt{x}} \\\\ & = 4\\times \\frac{1}{2\\sqrt{x}} \\times \\frac{1}{\\sqrt{x}} \\\\ & = \\frac{4}{2 \\sqrt{x} \\times \\sqrt{x}} \\\\ & = \\frac{4}{2x} \\\\ \\frac{dy}{dx} &= \\frac{2}{x} \\end{aligned} 5. The function $$f(x) = ln \\begin{pmatrix} \\frac{1}{x} \\end{pmatrix}$$ can be written: $f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = \\frac{1}{x}$ and $u'(x) = - \\frac{1}{x^2}$ So \\begin{aligned} f'(x) & = \\frac{u'(x)}{u(x)} \\\\ & = \\frac{ - \\frac{1}{x^2}}{\\frac{1}{x}} \\\\ & = - \\frac{1}{x^2}\\times x \\\\ & = - \\frac{x}{x^2} \\\\", "list of related equations, and the \"cases\" bracket was only there because someone wanted to group them together visually.) Also $$\\underline y :=\\begin{pmatrix}y_1\\\\y_2\\\\\\vdots\\\\y_n\\end{pmatrix}$$ and we can first use this to simplify the notation for the $f_i$: $$f_i(t, \\underline y) := f_i(t, y_1, y_2, \\dots, y_n)$$ And $\\underline f$ is defined by $$\\underline f(t,\\underline y) :=\\begin{pmatrix}f_1(t, \\underline y)\\\\f_2(t, \\underline y)\\\\\\vdots\\\\f_n(t, \\underline y)\\end{pmatrix}$$ Now, $$\\underline{\\dot y} =\\begin{pmatrix}\\dot y_1\\\\\\dot y_2\\\\\\vdots\\\\\\dot y_n\\end{pmatrix} = \\begin{pmatrix}f_1(t, \\underline y)\\\\f_2(t, \\underline y)\\\\\\vdots\\\\f_n(t, \\underline y)\\end{pmatrix} = \\underline f(t,\\underline y)$$", "\\cos \\theta\\\\r \\sin \\theta\\end{pmatrix} \\to \\begin{pmatrix}s \\cos \\phi\\\\s \\sin \\phi\\end{pmatrix}} rh(\\theta)\\\\ &=\\left(\\lim_{\\begin{pmatrix}r \\cos \\theta\\\\r \\sin \\theta\\end{pmatrix} \\to \\begin{pmatrix}s \\cos \\phi\\\\s \\sin \\phi\\end{pmatrix}} r\\right)\\left( \\lim_{\\begin{pmatrix}r \\cos \\theta\\\\r \\sin \\theta\\end{pmatrix} \\to \\begin{pmatrix}s \\cos \\phi\\\\s \\sin \\phi\\end{pmatrix}} h(\\theta)\\right)\\\\ &= s\\cdot h(\\phi) \\text{ (by continuity)}\\\\ &=f\\begin{pmatrix}s\\cos\\phi \\\\ s \\sin \\phi\\end{pmatrix}. \\end{align*} Now if you are approaching the origin (which corresponds to $s=0$), $\\theta$ need not approach a limit. To show this function is continuous at $(0,0)$, you need to show $$\\lim_{r \\to 0}f\\begin{pmatrix}r\\cos\\theta\\\\r\\sin\\theta \\end{pmatrix}=f\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix}$$ regardless of $\\theta$. For this you should use the squeeze theorem. Note that since $h$ is continuous and periodic with period $2\\pi$, $h$ is bounded on $[0,2\\pi]$ and thus on all of $\\mathbb{R}$. • Thank you. I think I'd got confused by the way $f$ is defined. – GoetheGrimm Oct 5 '16 at 20:17 • (a) Observe \\begin{align} |f(r\\cos\\theta, r\\sin\\theta) - f(s\\cos\\phi, s\\sin\\phi)| = |rh(\\theta)-sh(\\phi)| = |h'(\\xi)(r-s)|\\leq M|r-s| \\end{align} since $h$ is periodic and $C^1$. • (c) By the definition, you need to show \\begin{align} \\lim_{t\\rightarrow 0} \\frac{f(tm, tk)-f(0, 0)-t[\\partial_xf(0,0)m+\\partial_yf(0, 0)k]}{t} =0 \\end{align} for all $(m, k)= (a\\cos\\phi, a\\sin\\phi)\\neq(0, 0)$. Using the given relation, we have that \\begin{align} &\\frac{f(tm, tk)-f(0, 0) - t[\\partial_xf(0,0)h+\\partial_yf(0, 0)k]}{t} =\\ \\frac{at h(\\phi)-t[\\partial_xf(0,0)m+\\partial_yf(0, 0)k]}{t}\\\\ &=\\ ah(\\phi) -a[\\partial_xf(0,0)\\cos\\phi+\\partial_yf(0, 0)\\sin\\phi] =0. \\end{align} Then you need to show the above holds for all", "${\\displaystyle P}$ and ${\\displaystyle P^{-1}}$ to change bases from ${\\displaystyle B}$ to ${\\displaystyle D}$ and back again. One possible choice of the bases is ${\\displaystyle B=\\langle {\\begin{pmatrix}1\\\\2\\end{pmatrix}},{\\begin{pmatrix}-1\\\\1\\end{pmatrix}}\\rangle \\qquad D={\\mathcal {E}}_{2}=\\langle {\\begin{pmatrix}1\\\\0\\end{pmatrix}},{\\begin{pmatrix}0\\\\1\\end{pmatrix}}\\rangle }$ (this ${\\displaystyle B}$ is suggested by the map description). To find the matrix ${\\displaystyle T={\\rm {Rep}}_{B,B}(t)}$, solve the relations ${\\displaystyle c_{1}{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+c_{2}{\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\qquad {\\hat {c}}_{1}{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\hat {c}}_{2}{\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\2\\end{pmatrix}}}$ to get ${\\displaystyle c_{1}=1}$, ${\\displaystyle c_{2}=-2}$, ${\\displaystyle {\\hat {c}}_{1}=1/3}$ and ${\\displaystyle {\\hat {c}}_{2}=4/3}$. ${\\displaystyle {\\rm {Rep}}_{B,B}(t)={\\begin{pmatrix}1&1/3\\\\-2&4/3\\end{pmatrix}}}$ Finding ${\\displaystyle {\\rm {Rep}}_{D,D}(t)}$ involves a bit more computation. We first find ${\\displaystyle t({\\vec {e}}_{1})}$. The relation ${\\displaystyle c_{1}{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+c_{2}{\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\end{pmatrix}}}$ gives ${\\displaystyle c_{1}=1/3}$ and ${\\displaystyle c_{2}=-2/3}$, and so ${\\displaystyle {\\rm {Rep}}_{B}({\\vec {e}}_{1})={\\begin{pmatrix}1/3\\\\-2/3\\end{pmatrix}}_{B}}$ making ${\\displaystyle {\\rm {Rep}}_{B}(t({\\vec {e}}_{1}))={\\begin{pmatrix}1&1/3\\\\-2&4/3\\end{pmatrix}}_{B,B}{\\begin{pmatrix}1/3\\\\-2/3\\end{pmatrix}}_{B}={\\begin{pmatrix}1/9\\\\-14/9\\end{pmatrix}}_{B}}$ and hence ${\\displaystyle t}$ acts on the first basis vector ${\\displaystyle {\\vec {e}}_{1}}$ in this way. ${\\displaystyle t({\\vec {e}}_{1})=(1/9)\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}-(14/9)\\cdot {\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}5/3\\\\-4/3\\end{pmatrix}}}$ The computation for ${\\displaystyle t({\\vec {e}}_{2})}$ is similar. The relation ${\\displaystyle c_{1}{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+c_{2}{\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}0\\\\1\\end{pmatrix}}}$ gives ${\\displaystyle c_{1}=1/3}$ and ${\\displaystyle c_{2}=1/3}$, so ${\\displaystyle {\\rm {Rep}}_{B}({\\vec {e}}_{1})={\\begin{pmatrix}1/3\\\\1/3\\end{pmatrix}}_{B}}$ making ${\\displaystyle {\\rm {Rep}}_{B}(t({\\vec {e}}_{1}))={\\begin{pmatrix}1&1/3\\\\-2&4/3\\end{pmatrix}}_{B,B}{\\begin{pmatrix}1/3\\\\1/3\\end{pmatrix}}_{B}={\\begin{pmatrix}4/9\\\\-2/9\\end{pmatrix}}_{B}}$ and hence ${\\displaystyle t}$ acts on the second basis vector ${\\displaystyle {\\vec {e}}_{2}}$ in this way. ${\\displaystyle t({\\vec {e}}_{2})=(4/9)\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}-(2/9)\\cdot {\\begin{pmatrix}-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}2/3\\\\2/3\\end{pmatrix}}}$ Therefore ${\\displaystyle {\\rm {Rep}}_{D,D}(t)={\\begin{pmatrix}5/3&2/3\\\\-4/3&2/3\\end{pmatrix}}}$ and these are the change of basis matrices. ${\\displaystyle P={\\rm {Rep}}_{B,D}({\\mbox{id}})={\\begin{pmatrix}1&-1\\\\2&1\\end{pmatrix}}\\qquad P^{-1}={\\bigl (}{\\rm {Rep}}_{B,D}({\\mbox{id}}){\\bigr )}^{-1}={\\begin{pmatrix}1&-1\\\\2&1\\end{pmatrix}}^{-1}={\\begin{pmatrix}1/3&1/3\\\\-2/3&1/3\\end{pmatrix}}}$ The check of these computations is routine. ${\\displaystyle {\\begin{pmatrix}1&-1\\\\2&1\\end{pmatrix}}{\\begin{pmatrix}1&1/3\\\\-2&4/3\\end{pmatrix}}{\\begin{pmatrix}1/3&1/3\\\\-2/3&1/3\\end{pmatrix}}={\\begin{pmatrix}5/3&2/3\\\\-4/3&2/3\\end{pmatrix}}}$ Problem 6 Explain Example 1.3 in terms of maps. The only representation", "\\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7\\end{pmatrix}$ $= \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Now we set up an equation system and solve it. 1. $10+9r=4+s$ 2. $9+8r=4+s$ 3. $7+7r=6+2s$ Equation method: Equate I. and II. $4+s=4+s$ $10+9r=9+8r\\quad|-8r$ $10+r=9\\quad|-10$ $r=-1$ Compute $s$ with e.g. I. (for that insert $r=-1$) $10+9\\cdot(-1)=4+s$ $1=4+s\\quad|-4$ $s=-3$ For validation insert $r=-1$ and $s=-3$ into III. $7+7\\cdot(-1)=6+2\\cdot(-3)$ $0=0$ $r=-1$ and $s=-3$ result from all equations => The lines intersect. 3. #### Determining point of intersection To determine the point of intersection we insert $r=-1$ into $g$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + (-1) \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $=\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ => The lines intersect in point $S(1|1|0)$." ]

[ "Find $\\mathbf{a} \\cdot \\mathbf{b}$ if $|\\mathbf{a}| = 6$, $|\\mathbf{b}| = 6$, and the angle between $\\mathbf{a}$ and $\\mathbf{b}$ is $\\pi/2$.", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "$$c(\\mathbf{x}) = c(\\mathbf{x}+2\\mathbf{u}) = c(\\mathbf{x}+2\\mathbf{v})$$ or $$c(\\mathbf{x}) = c(\\mathbf{x}+4\\mathbf{u}) = c(\\mathbf{x}+4\\mathbf{v})$$ And C2 and C3 contradict C1, hence no such coloring is possible. Proof: C1$\\implies$C2: Either $c(\\mathbf{o}+2m\\mathbf{u})=c(\\mathbf{o}),c(\\mathbf{o}+(2m+1)\\mathbf{u})=c(\\mathbf{o}+\\mathbf{u})$ for all $m$, or else there is some $\\mathbf{x} = \\mathbf{o}+m_0\\mathbf{u}$ with $c(\\mathbf{x}),c(\\mathbf{x}+\\mathbf{u}),c(\\mathbf{x}+2\\mathbf{u})$ three different colors, wlog $A,B,C$. In that case the constraints of C1 uniquely determine $c(\\mathbf{x}+n\\mathbf{v})$. Specifically $c(\\mathbf{x}+2n\\mathbf{v})=A$ and $c(\\mathbf{x}+(2n+1)\\mathbf{v})=C$. In a picture, starting with $\\mathbf{x}$ in the top left with $\\mathbf{u}$ going right and $\\mathbf{v}$ going down: $$\\begin{array}{cccc} \\cdots & A & B & C & \\cdots \\\\ \\cdots & C & D & A & \\cdots \\\\ \\cdots & A & B & C & \\cdots \\\\ \\cdots & C & D & A & \\cdots \\\\ \\cdots & A & B & C & \\cdots \\\\ & \\vdots & \\vdots & \\vdots \\end{array}$$ Hence either the row containing $\\mathbf{o}$ or the column containing $\\mathbf{x}$ is periodic with period $2\\mathbf{u}$ or $2\\mathbf{v}$ respectively, and it's easy to see that either implies that the rest of the lattice is periodic with the same period. C1 and C2 $\\implies$ C3: From C2 wlog assume coloring on the $\\mathbf{o},\\mathbf{u},\\mathbf{v}$ lattice has period $2\\mathbf{u}$. Either there is some $\\mathbf{x}$ with $c(\\mathbf{x}+2\\mathbf{v})=c(\\mathbf{x})=c(\\mathbf{x}+2\\mathbf{u})$, satisfying the first condition of C3, or else $c(\\mathbf{x}+2\\mathbf{v})\\ne c(\\mathbf{x})$ for every $\\mathbf{x}$ in the lattice. In that case wlog", "b = \\mathbf x$, but that is not what you need to prove. You need to find the distance of $\\mathbf x$ to $\\frac 12 \\mathbf b$ and proof that it is $\\frac 12 b$. That is, can you prove: $(\\mathbf x - \\frac 12 \\mathbf b)^2 \\overset{?}{=} (\\frac 12 b)^2$​ $(\\mathbf{x} -\\frac{1}{2}\\mathbf{b})^2 = \\mathbf{x}\\cdot\\mathbf{x} - \\mathbf{x}\\cdot\\mathbf{b} + \\frac{1}{4}\\mathbf{b}\\cdot\\mathbf{b}$ How do I go from here? Klaas van Aarsen MHB Seeker Staff member $(\\mathbf{x} -\\frac{1}{2}\\mathbf{b})^2 = \\mathbf{x}\\cdot\\mathbf{x} - \\mathbf{x}\\cdot\\mathbf{b} + \\frac{1}{4}\\mathbf{b}\\cdot\\mathbf{b}$ How do I go from here? What do you get from $(\\mathbf{x} - \\mathbf{b})\\cdot\\mathbf{x} = 0$?", "when indeed $~X=\\mathbf 1_{X=1}~$ and $~(1-X) = \\mathbf 1_{X=0}~$.", "0 ### Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $|\\mathbf{b}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{c}|$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $|\\mathbf{a}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{b}|$ • if $|\\mathbf{b}\\cdot\\mathbf{c}| = (\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $|\\mathbf{a}\\cdot\\mathbf{c}| = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c} = 0$ • if $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|) = (\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{a}$$2|\\mathbf{a}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{b}|$", "(\\mathbf{a}\\cdot\\mathbf{b})(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{c}\\cdot\\mathbf{a})$. If T > 0, the cell is of type I, if T ≤ 0 it is of type II. The conditions for a primitive cell to be a reduced cell can all be stated analytically as follows. Type-I cell Main conditions • $\\mathbf{a}\\cdot\\mathbf{a}$$\\mathbf{b}\\cdot\\mathbf{b}$$\\mathbf{c}\\cdot\\mathbf{c}$ • $|\\mathbf{b}\\cdot\\mathbf{c}|$$(\\mathbf{b}\\cdot\\mathbf{b})/2$ • $|\\mathbf{a}\\cdot\\mathbf{c}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $|\\mathbf{a}\\cdot\\mathbf{b}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $\\mathbf{b}\\cdot\\mathbf{c} > 0$ • $\\mathbf{a}\\cdot\\mathbf{c} > 0$ • $\\mathbf{a}\\cdot\\mathbf{b} > 0$ Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $\\mathbf{b}\\cdot\\mathbf{c}$$\\mathbf{a}\\cdot\\mathbf{c}$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $\\mathbf{a}\\cdot\\mathbf{c}$$\\mathbf{a}\\cdot\\mathbf{b}$ • if $\\mathbf{b}\\cdot\\mathbf{c} = (\\mathbf{b}\\cdot\\mathbf{b})$/2 then $\\mathbf{a}\\cdot\\mathbf{b}$$2\\mathbf{a}\\cdot\\mathbf{c}$ • if $\\mathbf{a}\\cdot\\mathbf{c} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b}$$2\\mathbf{b}\\cdot\\mathbf{c}$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c}$$2\\mathbf{b}\\cdot\\mathbf{c}$ Type-II cell Main conditions • $\\mathbf{a}\\cdot\\mathbf{a}$$\\mathbf{b}\\cdot\\mathbf{b}$$\\mathbf{c}\\cdot\\mathbf{c}$ • $|\\mathbf{b}\\cdot\\mathbf{c}|$$(\\mathbf{b}\\cdot\\mathbf{b})/2$ • $|\\mathbf{a}\\cdot\\mathbf{c}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $|\\mathbf{a}\\cdot\\mathbf{b}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|)$$(\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ • $\\mathbf{b}\\cdot\\mathbf{c}$ ≤ 0 • $\\mathbf{a}\\cdot\\mathbf{c}$ ≤ 0 • $\\mathbf{a}\\cdot\\mathbf{b}$ ≤ 0 Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $|\\mathbf{b}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{c}|$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $|\\mathbf{a}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{b}|$ • if $|\\mathbf{b}\\cdot\\mathbf{c}| = (\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $|\\mathbf{a}\\cdot\\mathbf{c}| = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c} = 0$ • if $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|) = (\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{a}$$2|\\mathbf{a}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{b}|$ Geometrical meaning of the reduced cell The main conditions express the following two requirements: • Of all lattice vectors, none is shorter than a; of those not directed along a, none is shorter than b; of those not lying in", "≤ 0 ### Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $|\\mathbf{b}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{c}|$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $|\\mathbf{a}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{b}|$ • if $|\\mathbf{b}\\cdot\\mathbf{c}| = (\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $|\\mathbf{a}\\cdot\\mathbf{c}| = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c} = 0$ • if $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|) = (\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{a}$$2|\\mathbf{a}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{b}|$ ## Geometrical meaning of the reduced cell The main conditions express the following two requirements: • Of all lattice vectors, none is shorter than a; of those not directed along a, none is shorter than b; of those not lying in the ab plane, none is shorter than c. • The three angles between basis vectors are either all acute (type I) or all non-acute (type II).", "= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. (1) But as cab, we also have $\\mathbf{c} \\cdot \\mathbf{c} = (\\mathbf{a} - \\mathbf{b}) \\cdot (\\mathbf{a} - \\mathbf{b}) \\;$, which, according to the distributive law, expands to $\\mathbf{c} \\cdot \\mathbf{c} = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) \\;$. (2) Merging the two c · c equations, (1) and (2), we obtain $\\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. Subtracting a · a + b · b from both sides and dividing by −2 leaves $\\mathbf{a} \\cdot \\mathbf{b} = ab \\cos\\theta \\;$,", "$\\mathbf{a}\\cdot\\mathbf{b} > 0$ Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $\\mathbf{b}\\cdot\\mathbf{c}$$\\mathbf{a}\\cdot\\mathbf{c}$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $\\mathbf{a}\\cdot\\mathbf{c}$$\\mathbf{a}\\cdot\\mathbf{b}$ • if $\\mathbf{b}\\cdot\\mathbf{c} = (\\mathbf{b}\\cdot\\mathbf{b})$/2 then $\\mathbf{a}\\cdot\\mathbf{b}$$2\\mathbf{a}\\cdot\\mathbf{c}$ • if $\\mathbf{a}\\cdot\\mathbf{c} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b}$$2\\mathbf{b}\\cdot\\mathbf{c}$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c}$$2\\mathbf{b}\\cdot\\mathbf{c}$ Type-II cell Main conditions • $\\mathbf{a}\\cdot\\mathbf{a}$$\\mathbf{b}\\cdot\\mathbf{b}$$\\mathbf{c}\\cdot\\mathbf{c}$ • $|\\mathbf{b}\\cdot\\mathbf{c}|$$(\\mathbf{b}\\cdot\\mathbf{b})/2$ • $|\\mathbf{a}\\cdot\\mathbf{c}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $|\\mathbf{a}\\cdot\\mathbf{b}|$$(\\mathbf{a}\\cdot\\mathbf{a})/2$ • $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|)$$(\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ • $\\mathbf{b}\\cdot\\mathbf{c}$ ≤ 0 • $\\mathbf{a}\\cdot\\mathbf{c}$ ≤ 0 • $\\mathbf{a}\\cdot\\mathbf{b}$ ≤ 0 Special conditions • if $\\mathbf{a}\\cdot\\mathbf{a} = \\mathbf{b}\\cdot\\mathbf{b}$ then $|\\mathbf{b}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{c}|$ • if $\\mathbf{b}\\cdot\\mathbf{b} = \\mathbf{c}\\cdot\\mathbf{c}$ then $|\\mathbf{a}\\cdot\\mathbf{c}|$$|\\mathbf{a}\\cdot\\mathbf{b}|$ • if $|\\mathbf{b}\\cdot\\mathbf{c}| = (\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $|\\mathbf{a}\\cdot\\mathbf{c}| = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{b} = 0$ • if $\\mathbf{a}\\cdot\\mathbf{b} = (\\mathbf{a}\\cdot\\mathbf{a})$/2 then $\\mathbf{a}\\cdot\\mathbf{c} = 0$ • if $(|\\mathbf{b}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{c}|+|\\mathbf{a}\\cdot\\mathbf{b}|) = (\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b})/2$ then $\\mathbf{a}\\cdot\\mathbf{a}$$2|\\mathbf{a}\\cdot\\mathbf{c}|+ |\\mathbf{a}\\cdot\\mathbf{b}|$", "of $$\\mathbf{3}$$ is $$\\mathbf{53}$$ heureka Sep 18, 2018", "\\in \\textsf {V} \\hspace {12pt} 1 \\cdot \\mathbf {}{x} = \\mathbf {x}$$ • 5. We also have a kind of “multiplicative distributivity” $$a(b\\mathbf {x}) = (ab) \\mathbf {x}$$ Not exactly the most exciting stuff, but we can’t build castles without foundations! I’m not a structural engineer, but I’m pretty sure this is a true statement.", "\\mathbf{a}_2(\\mathbf{b}_2+\\mathbf{c}_2), ..., \\mathbf{a}_d(\\mathbf{b}_d+\\mathbf{c}_d)]\\\\&= [\\mathbf{a}_1\\mathbf{b}_1+\\mathbf{a}_1\\mathbf{c}_1, \\mathbf{a}_2\\mathbf{b}_2+\\mathbf{a}_2\\mathbf{c}_2, ..., \\mathbf{a}_d\\mathbf{b}_d+\\mathbf{a}_d\\mathbf{c}_d]\\\\&= [\\mathbf{a}_1\\mathbf{b}_1, \\mathbf{a}_2\\mathbf{b}_2, ..., \\mathbf{a}_d\\mathbf{b}_d]+[\\mathbf{a}_1\\mathbf{c}_1,\\mathbf{a}_2\\mathbf{c}_2, ..., \\mathbf{a}_d\\mathbf{c}_d]\\\\&= [\\mathbf{a}_1, \\mathbf{a}_2, ..., \\mathbf{a}_d]\\cdot[\\mathbf{b}_1, \\mathbf{b}_2, ..., \\mathbf{b}_d]+[\\mathbf{a}_1, \\mathbf{a}_2, ..., \\mathbf{a}_d]\\cdot(\\mathbf{c}_1, \\mathbf{c}_2, ..., \\mathbf{c}_d]\\\\&=\\mathbf{a}\\cdot\\mathbf{b}+\\mathbf{a}\\cdot\\mathbf{c}\\end{array}$ ## Bilinear $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Vector Projection Given two vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ we want to determine the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b}$$, which is the vector $$\\mathbf{a}'$$ of length $$a'$$ parallel to the vector $$\\mathbf{b}$$. The length $$a'$$ is called the Scalar Projection and is determined by $a' = |\\mathbf{a}|\\cos\\theta = |\\mathbf{a}|\\underbrace{\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}|\\cdot|\\mathbf{b}|}}_{\\cos\\theta}=\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{b}|}=\\mathbf{a}\\cdot\\frac{\\mathbf{b}}{|\\mathbf{b}|}=\\mathbf{a}\\cdot\\hat{\\mathbf{b}}$ Having the Scalar Projection, we can then calculate the Vector Projection: $\\mathbf{a}' = a'\\cdot\\hat{\\mathbf{b}} = (\\mathbf{a}\\cdot\\hat{\\mathbf{b}})\\cdot\\hat{\\mathbf{b}}=\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{b}|}\\cdot\\frac{\\mathbf{b}}{|\\mathbf{b}|}=\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{b}|^2}\\mathbf{b}=\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{\\mathbf{b}\\cdot\\mathbf{b}}\\mathbf{b}=:proj_{\\mathbf{b}}(\\mathbf{a})$ Scalar Projection Percentage: The length of the distance $$a'$$ to the total length of $$\\mathbf{b}$$ can be expressed in percent as follows: $\\frac{a'}{|\\mathbf{b}|} = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{b}|\\cdot |\\mathbf{b}|} =\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{b}|^2}=\\frac{\\mathbf{a}\\cdot\\mathbf{b}}{\\mathbf{b}\\cdot\\mathbf{b}}$ Which is interesting. It says that the Vector Projection is just a scaling of vector $$\\mathbf{b}$$ by the proportion of $$\\mathbf{a}'$$ to $$\\mathbf{b}$$. ### Direction Cosines In three dimensional space, a vector $$\\mathbf{a}$$ will form the angle $$\\alpha$$ with the x-axis, $$\\beta$$ with the y-axis and $$\\gamma$$ with the z-axis. These angles are", "Select Page 1. Let $\\mathbf{a_1,a_2,\\cdots, a_n }$ and $\\mathbf{ b_1,b_2,\\cdots, b_n }$ be two permutations of the numbers $\\mathbf{1,2,\\cdots, n }$. Show that $\\mathbf{\\sum_{i=1}^n i(n+1-i) \\le \\sum_{i=1}^n a_ib_i \\le \\sum_{i=1}^n i^2 }$ 1. Let a,b,c,d be distinct digits such that the product of the 2-digit numbers $\\mathbf{\\overline{ab}}$ and $\\mathbf{\\overline{cb}}$ is of the form $\\mathbf{\\overline{ddd}}$. Find all possible values of a+b+c+d. 1. Let $\\mathbf{I_1, I_2, I_3}$ be three open intervals of $\\mathbf{\\mathbb{R}}$ such that none is contained in another. If $\\mathbf{I_1\\cap I_2 \\cap I_3}$ is non-empty, then show that at least one of these intervals is contained in the union of the other two. 1. A real valued function f is defined on the interval (-1,2). A point $\\mathbf{x_0}$ is said to be a fixed point of f if $\\mathbf{f(x_0)=x_0}$. Suppose that f is a differentiable function such that f(0)>0 and f(1)=1. Show that if f'(1)>1, then f has a fixed point in the interval (0,1). 1. Let A be the set of all functions $\\mathbf{f:\\mathbb{R} to \\mathbb{R}}$ such that f(xy)=xf(y) for all $\\mathbf{x,y \\in \\mathbb{R}}$.(a) If $\\mathbf{f \\in A}$ then show that f(x+y)=f(x)+f(y) for all x,y $\\mathbf{\\in \\mathbb{R}}$(b) For $\\mathbf{g,h \\in A}$, define a function $$\\mathbf{g \\circ h}$$ by $\\mathbf{(g \\circ h)(x)=g(h(x))}$ for $\\mathbf{x \\in \\mathbb{R}}$. Prove that $\\mathbf{g \\circ h}$ is in A and is equal to", "length $||\\mathbf{v}||\\neq 0$. Thus, we must have $c_1=0$. We repeat the above argument with $\\mathbf{v}_1$ replaced by $\\mathbf{v}_2$ as follows. Since $c_1=0$, the (*) becomes $c_2\\mathbf{v}_2+c_3\\mathbf{v}_3=\\mathbf{0}.$ We have \\begin{align*} \\mathbf{0}&=\\mathbf{v}_2\\cdot \\mathbf{0}=\\mathbf{v}_2\\cdot (c_2\\mathbf{v}_2+c_3\\mathbf{v}_3)\\\\ &=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2+c_3\\mathbf{v}_2\\cdot\\mathbf{v}_3. \\end{align*} Since $\\mathbf{v}_2\\cdot\\mathbf{v}_3=0$, we have $\\mathbf{0}=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2=c_2||\\mathbf{v}_2||^2.$ Since $\\mathbf{x}\\neq \\mathbf{0}$, we have $||\\mathbf{v}_2||\\neq 0$, and thus $c_2=0$. Hence (*) is now just $c_3\\mathbf{v}_3=\\mathbf{0}$, and since $\\mathbf{x}_3\\neq \\mathbf{0}$, we conclude that $c_3=0$. Therefore, we have shown that $c_1=c_2=c_3=0$, and thus the set $\\{\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3\\}$ is linearly independent. ## Comment. These are Quiz 3 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. ### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017 There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. ### 3 Responses 1. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 2. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 3. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] ##### Find a Nonsingular Matrix Satisfying Some Relation Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A,$ where $B$ is the following matrix. If such a nonsingular... Close", "^ * mathbf {u}} right)$$ and $$mathbf {H} ^ * mathbf {H} = mathbf {I}$$but I do not see how $$mathbf {H} = mathbf {H ^ {- 1}}$$, I even understand that $$mathbf {H} ^ {- 1} mathbf {e} _1 = mathbf {x}$$what should be expected. Any help is greatly appreciated.", "in $\\mathbb{R}^n$: $$\\mathbf{x} = \\sum_{j=1}^n (\\mathbf{x} \\cdot \\mathbf{u}_i) \\mathbf{u}_i$$ where $\\mathbf{u}_1 = \\mathbf{u}$. Now subtract the first term from both sides of this equality. -", "### pasmith The result is true for all $t \\in \\mathbb{R}$, so in particular it is true for $t = 1$. Hence $$\\mathbf{x} \\cdot \\nabla f = n 1^{n-1} f(\\mathbf{x}) = n f(\\mathbf{x})$$ as required.", "$\\mathbf{F} = y \\mathbf{j}$ then $\\mathbf{\\nabla} \\cdot \\mathbf{F} =1$ and $V= \\int \\int_S y \\: dx \\: dz$ Take $\\mathbf{F} = y \\mathbf{i}$ then $\\mathbf{\\nabla} \\cdot \\mathbf{F} =1$ and $V= \\int \\int_S y \\: dx \\: dz$ Hence $V= \\frac{1}{3} \\int \\int_S x \\: dy \\: dz + y \\: dx \\: dz + z \\: dx \\: dy$", "= \\sqrt{(\\mathbf{x} - \\mathbf{y})^T (\\mathbf{x} - \\mathbf{y})} = \\varepsilon(\\mathbf x, \\mathbf y)$." ]

[ "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "and further the mathematical expression for dot product using vector coordinates. Suppose $\\mathbf{u} = [u_1, u_2, \\cdots, u_n]$ and $\\mathbf{v} = [v_1, v_2, \\cdots, v_n]$, we have the dot product expression \\begin{align} \\mathbf{u} \\cdot \\mathbf{v} &= (u_1 \\mathbf{i_1} + u_2 \\mathbf{i_2} + \\cdots + u_n \\mathbf{i_n}) \\cdot (v_1 \\mathbf{i_1} + v_2 \\mathbf{i_2} + \\cdots + v_n \\mathbf{i_n}) \\\\ &= u_1 v_1 + u_2 v_2 + \\cdots + u_n v_n \\\\ \\end{align} It happens to be the same to the definition of inner product in real space. For 3D space, we just have $\\mathbf{u} = [u_1, u_2, u_3]$, $\\mathbf{v} = [v_1, v_2, v_3]$, and \\begin{align} \\mathbf{u} \\cdot \\mathbf{v} &= u_1 v_1 + u_2 v_2 + u_3 v_3 \\end{align} By the way, it is a common mistake that people are using the mathematical expression for dot product to prove the dot product product properties, since they overlooked the fact that we first have the definition and properties, and then we have the mathematical expression. The length of the projected vector $\\mathbf{u}$ on $\\mathbf{v}$ is \\begin{align} \\| \\mathbf{u} \\| \\cos \\theta &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{\\| \\mathbf{v} \\|} \\end{align} The projected vector is \\begin{align} \\| \\mathbf{u} \\| \\cos \\theta \\frac{\\mathbf{v}}{\\| \\mathbf{v} \\|} &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{\\| \\mathbf{v} \\|} \\frac{\\mathbf{v}}{\\| \\mathbf{v} \\|} \\\\ &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{\\| \\mathbf{v} \\| \\| \\mathbf{v} \\|} \\mathbf{v}", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "Using the $\\mathbf{x}_{ijk}$ shorthand again, the initial expansion gives \\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( { \\boldsymbol{\\partial} \\wedge T } \\right)=d^4 u\\left( {\\mathbf{x}_{123} \\cdot \\partial_4 T - \\mathbf{x}_{124} \\cdot \\partial_3 T + \\mathbf{x}_{134} \\cdot \\partial_2 T - \\mathbf{x}_{234} \\cdot \\partial_1 T} \\right).\\end{aligned} \\hspace{\\stretch{1}}(1.0.153) Applying lemma 4 to expand the inner products within the braces we have \\begin{aligned}\\begin{aligned}\\mathbf{x}_{123} \\cdot \\partial_4 T-&\\mathbf{x}_{124} \\cdot \\partial_3 T+\\mathbf{x}_{134} \\cdot \\partial_2 T-\\mathbf{x}_{234} \\cdot \\partial_1 T \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\underbrace{\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_1 T } \\right) } \\right)}_{\\text{Apply cyclic permutations}}\\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_1 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\partial_1 T } \\right) } \\right) \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot\\left( {\\mathbf{x}_3 \\cdot \\partial_4 T-\\mathbf{x}_4 \\cdot \\partial_3 T} \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3", "\\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{a} = 1,\\end{aligned} \\hspace{\\stretch{1}}(1.174) and \\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{b} = 0.\\end{aligned} \\hspace{\\stretch{1}}(1.175) \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}.\\end{aligned} \\hspace{\\stretch{1}}(1.176) Dotting with $\\mathbf{a}$ we have \\begin{aligned}\\mathbf{a} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right)=\\mathbf{a} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right)^2 - \\mathbf{b}^2 \\mathbf{a}^2,\\end{aligned} \\hspace{\\stretch{1}}(1.177) but dotting with $\\mathbf{b}$ yields zero \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right) &= \\mathbf{b} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right) \\\\ &= \\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}^2 - \\mathbf{b}^2 \\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right) \\\\ &= 0.\\end{aligned} \\hspace{\\stretch{1}}(1.178) To complete the proof, we note that the product in eq. 1.177 is just the wedge squared \\begin{aligned}\\left( { \\mathbf{a} \\wedge \\mathbf{b}} \\right)^2 &= \\left\\langle{{\\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)^2}}\\right\\rangle \\\\ &= \\left\\langle{{\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)}}\\right\\rangle \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\mathbf{a} \\mathbf{b} - 2 \\left( {\\mathbf{a} \\cdot \\mathbf{b}} \\right) \\mathbf{a} \\mathbf{b}}}\\right\\rangle+\\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right)^2 \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\left( { -\\mathbf{b} \\mathbf{a} + 2 \\mathbf{a} \\cdot \\mathbf{b} }", "initial expansion gives \\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( { \\boldsymbol{\\partial} \\wedge T } \\right)=d^4 u\\left( {\\mathbf{x}_{123} \\cdot \\partial_4 T - \\mathbf{x}_{124} \\cdot \\partial_3 T + \\mathbf{x}_{134} \\cdot \\partial_2 T - \\mathbf{x}_{234} \\cdot \\partial_1 T} \\right).\\end{aligned} \\hspace{\\stretch{1}}(1.0.153) Applying lemma 4 to expand the inner products within the braces we have \\begin{aligned}\\begin{aligned}\\mathbf{x}_{123} \\cdot \\partial_4 T-&\\mathbf{x}_{124} \\cdot \\partial_3 T+\\mathbf{x}_{134} \\cdot \\partial_2 T-\\mathbf{x}_{234} \\cdot \\partial_1 T \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\underbrace{\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_1 T } \\right) } \\right)}_{\\text{Apply cyclic permutations}}\\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_1 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\partial_1 T } \\right) } \\right) \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot\\left( {\\mathbf{x}_3 \\cdot \\partial_4 T-\\mathbf{x}_4 \\cdot \\partial_3 T} \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left(", "initial expansion gives \\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( { \\boldsymbol{\\partial} \\wedge T } \\right)=d^4 u\\left( {\\mathbf{x}_{123} \\cdot \\partial_4 T - \\mathbf{x}_{124} \\cdot \\partial_3 T + \\mathbf{x}_{134} \\cdot \\partial_2 T - \\mathbf{x}_{234} \\cdot \\partial_1 T} \\right).\\end{aligned} \\hspace{\\stretch{1}}(1.0.153) Applying lemma 4 to expand the inner products within the braces we have \\begin{aligned}\\begin{aligned}\\mathbf{x}_{123} \\cdot \\partial_4 T-&\\mathbf{x}_{124} \\cdot \\partial_3 T+\\mathbf{x}_{134} \\cdot \\partial_2 T-\\mathbf{x}_{234} \\cdot \\partial_1 T \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\underbrace{\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_1 T } \\right) } \\right)}_{\\text{Apply cyclic permutations}}\\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_1 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\partial_1 T } \\right) } \\right) \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot\\left( {\\mathbf{x}_3 \\cdot \\partial_4 T-\\mathbf{x}_4 \\cdot \\partial_3 T} \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left(", "Inner Product As we continue to study vectors, we will be looking at the concept of distance and angles between vectors. In order to do so, we need to define an inner product on a vector space. Over the powers of real numbers, we were able to use the dot product as an inner product; however, this is not the only possibility. As such, we will look at examples with dot products and with a different inner product. Over $$\\mathbb{R}^{3}$$ Let $$\\mathbf{u}=(1,1,2)$$ and $$\\mathbf{v}=(-1,3,0)$$ in the vector space $$\\mathbb{R}^{3}$$, then find the following • $$||\\mathbf{u}||$$, • $$d(\\mathbf{u},\\mathbf{v})$$, • $$\\mathbf{u}\\cdot \\mathbf{v}$$ and • the angle between $$\\mathbf{u}$$ and $$\\mathbf{v}$$. $$||\\mathbf{u}||$$ We first note that the magnitude of a vector in a vector space with a defined inner product is given by \\begin{align*} ||\\mathbf{u}||=\\sqrt{\\mathbf{u}\\cdot \\mathbf{u}}. \\end{align*} As such, we just need to find the dot product of $$\\mathbf{u}$$ with itself. This is then found at \\begin{align*} ||\\mathbf{u}||&=\\sqrt{\\mathbf{u}\\cdot \\mathbf{u}} \\\\ &=\\sqrt{(1,1,2) \\cdot (1,1,2)} \\\\ &=\\sqrt{1*1+1*1+2*2} \\\\ &=\\sqrt{6}. \\end{align*} $$d(\\mathbf{u},\\mathbf{v})$$ In order to find the distance between the two vectors, we need to find \\begin{align*} d(\\mathbf{u},\\mathbf{v})&=||\\mathbf{u}-\\mathbf{v}|| \\\\ &=\\sqrt{((1,1,2)-(-1,3,0))\\cdot((1,1,2-(-1,3,0))} \\\\ &=\\sqrt{(2,-2,2) \\cdot (2,-2,2)} \\\\ &=\\sqrt{2*2+(-2)(-2)+2*2} \\\\ &=\\sqrt{12}. \\end{align*} $$\\mathbf{u} \\cdot \\mathbf{v}$$ In order to find the dot product of the two vectors, we need only use the definition to find that \\begin{align*} \\mathbf{u}", "their dot product, by summing the products of their respective components (for example, if u = (a, b) and v = (c, d), then their dot product u·v = ac + bd). If the dot product is zero, the two vectors are said to be orthogonal to each other. Some properties of the dot product aid understanding of how W-CDMA works. If vectors a and b are orthogonal, then $\\scriptstyle\\mathbf{a}\\cdot\\mathbf{b} \\,=\\, 0$ and: \\begin{align}\\mathbf{a}\\cdot(\\mathbf{a}+\\mathbf{b}) &= \\|\\mathbf{a}\\|^2 &\\quad\\mathrm{since}\\quad \\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{a}\\cdot\\mathbf{b} &= \\|a\\|^2+0 \\\\\\mathbf{a}\\cdot(-\\mathbf{a}+\\mathbf{b}) &= -\\|\\mathbf{a}\\|^2 &\\quad\\mathrm{since}\\quad -\\mathbf{a}\\cdot\\mathbf{a}+\\mathbf{a}\\cdot\\mathbf{b} &= -\\|a\\|^2+0 \\\\\\mathbf{b}\\cdot(\\mathbf{a}+\\mathbf{b}) &= \\|\\mathbf{b}\\|^2 &\\quad\\mathrm{since}\\quad \\mathbf{b}\\cdot\\mathbf{a}+\\mathbf{b}\\cdot\\mathbf{b} &= 0+\\|b\\|^2 \\\\\\mathbf{b}\\cdot(\\mathbf{a}-\\mathbf{b}) &= -\\|\\mathbf{b}\\|^2 &\\quad\\mathrm{since}\\quad \\mathbf{b}\\cdot\\mathbf{a}-\\mathbf{b}\\cdot\\mathbf{b} &= 0-\\|b\\|^2\\end{align} Each user in synchronous CDMA uses a code orthogonal to the others' codes to modulate their signal. An example of four mutually orthogonal digital signals is shown in the figure. Orthogonal codes have a cross-correlation equal to zero; in other words, they do not interfere with each other. In the case of IS-95 64 bit Walsh codes are used to encode the signal to separate different users. Since each of the 64 Walsh codes are orthogonal to one another, the signals are channelized into 64 orthogonal signals. The following example demonstrates how each user's signal can be encoded and decoded. ### Example An example of four mutually orthogonal digital signals. Start with a set of vectors that", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "of trivector Using the $\\mathbf{x}_{ijk}$ shorthand again, the initial expansion gives \\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( { \\boldsymbol{\\partial} \\wedge T } \\right)=d^4 u\\left( {\\mathbf{x}_{123} \\cdot \\partial_4 T - \\mathbf{x}_{124} \\cdot \\partial_3 T + \\mathbf{x}_{134} \\cdot \\partial_2 T - \\mathbf{x}_{234} \\cdot \\partial_1 T} \\right).\\end{aligned} \\hspace{\\stretch{1}}(1.0.153) Applying lemma 4 to expand the inner products within the braces we have \\begin{aligned}\\begin{aligned}\\mathbf{x}_{123} \\cdot \\partial_4 T-&\\mathbf{x}_{124} \\cdot \\partial_3 T+\\mathbf{x}_{134} \\cdot \\partial_2 T-\\mathbf{x}_{234} \\cdot \\partial_1 T \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\underbrace{\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_1 T } \\right) } \\right)}_{\\text{Apply cyclic permutations}}\\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot \\partial_4 T } \\right) } \\right)-\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\partial_3 T } \\right) } \\right) \\\\ &\\quad +\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_1 \\cdot \\partial_2 T } \\right) } \\right)-\\mathbf{x}_3 \\cdot \\left( { \\mathbf{x}_4 \\cdot \\left( { \\mathbf{x}_2 \\cdot \\partial_1 T } \\right) } \\right) \\\\ &=\\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_2 \\cdot\\left( {\\mathbf{x}_3 \\cdot \\partial_4 T-\\mathbf{x}_4 \\cdot \\partial_3 T} \\right) } \\right) \\\\", "$(\\mathbf{a}+\\mathbf{b})\\cdot(\\mathbf{c}+\\mathbf{d}) = \\mathbf{a}\\cdot\\mathbf{c} + \\mathbf{a}\\cdot\\mathbf{d} + \\mathbf{b}\\cdot\\mathbf{c} + \\mathbf{b}\\cdot\\mathbf{d}$ ## Mixed associative law The dot product has a homogeneity $\\alpha (\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot \\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Bilinear Often the mixed associative law and distributive law are combined to show that the dot product is bilinear: $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Cauchy-Schwarz Inequality $|\\mathbf{a}\\cdot\\mathbf{b}|\\leq|\\mathbf{a}||\\mathbf{b}|$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Fast calculation of Cosine The definition of the dot-product is used extensively in computer graphics since it speeds up computations in many circumstances by avoiding inefficient trigonometric functions. Additionally, working with normalized vectors gives the cosine between these two vectors by taking the dot product of them: $\\cos\\theta = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}||\\mathbf{b}|} = \\hat{\\mathbf{a}}\\cdot\\hat{\\mathbf{b}}$ ### Making perpendicular vectors Since two vectors are perpendicular when $$\\mathbf{a}\\cdot\\mathbf{b} = 0$$, we can easily construct a perpendicular vector by zeroing all components except 2, flipping those two and reversing the sign of one of them. For example with $$\\mathbf{a} = (\\mathbf{a}_1, \\mathbf{a}_2, \\dots, \\mathbf{a}_n)$$ the vectors $$\\mathbf{a}' = (-\\mathbf{a}_2, \\mathbf{a}_1, 0, \\dots, 0)$$ or $$\\mathbf{a}'' = (0, -\\mathbf{a}_3, \\mathbf{a}_2, 0, \\dots, 0)$$, etc are all perpendcular to $$\\mathbf{a}$$,", "of \\mathbf{A} parallel to \\mathbf{B}}) \\end{align*} The dot product has some of the properties possessed by ordinary multiplication of numbers, \\begin{align*} \\mathbf{A}\\cdot\\mathbf{B} &= \\mathbf{B}\\cdot\\mathbf{A} \\\\ \\mathbf{A}\\cdot(\\mathbf{B}+\\mathbf{C}) &= \\mathbf{A}\\cdot\\mathbf{B}+\\mathbf{A}\\cdot\\mathbf{C} \\\\ \\left(c\\mathbf{A}\\right)\\cdot\\mathbf{B} &= c\\left(\\mathbf{A}\\cdot\\mathbf{B}\\right) , \\end{align*} but it lacks one other: the ability to undo multiplication by dividing. If you know the components of two vectors, you can easily calculate their dot product as follows: $\\begin{equation*} \\mathbf{A}\\cdot\\mathbf{B} = A_xB_x+A_yB_y+A_zB_z . \\end{equation*}$ (This can be proved by first analyzing the special case where each vector has only an $$x$$ component, and the similar cases for $$y$$ and $$z$$. We can then use the rule $$\\mathbf{A}\\cdot(\\mathbf{B}+\\mathbf{C}) = \\mathbf{A}\\cdot\\mathbf{B}+\\mathbf{A}\\cdot\\mathbf{C}$$ to make a generalization by writing each vector as the sum of its $$x$$, $$y$$, and $$z$$ components. See homework problem 17.) ##### Example 14: Magnitude expressed with a dot product If we take the dot product of any vector b with itself, we find \\begin{align*} \\mathbf{b}\\cdot\\mathbf{b} &= \\left( b_{x}\\hat{\\mathbf{x}}+ b_{y}\\hat{\\mathbf{y}}+ b_z\\hat{\\mathbf{z}}\\right) \\cdot \\left( b_{x}\\hat{\\mathbf{x}}+ b_{y}\\hat{\\mathbf{y}}+ b_z\\hat{\\mathbf{z}}\\right) \\\\ &= b_{x}^2+ b_{y}^2+ b_z^2 ,\\\\ \\text{so its magnitude can be expressed as} |\\mathbf{b}| &= \\sqrt{\\mathbf{b}\\cdot\\mathbf{b}} . \\end{align*} We will often write $$b^2$$ to mean $$\\mathbf{b}\\cdot\\mathbf{b}$$, when the context makes it clear what is intended. For example, we could express kinetic energy as $$\\text{(1/2)} m|\\mathbf{v}|^2$$, $$\\text{(1/2)} m\\mathbf{v}\\cdot\\mathbf{v}$$, or $$\\text{(1/2)} m v^2$$. In the third version, nothing but", "\\dot{\\mathbf{x}}) = 0\\end{aligned} \\hspace{\\stretch{1}}(3.26) By evaluating this, we can eventually show that we can construct a four vector equation. Doing this we have \\begin{aligned}\\frac{d}{dt} (\\gamma \\mathbf{v}) &=\\frac{d}{dt} \\left( \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\mathbf{v} \\right) \\\\ &=-2 (-1/2) \\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}})/c^2 \\left(1 - \\mathbf{v}^2/c^2\\right)^{-3/2} + \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\dot{\\mathbf{v}} \\\\ &=\\gamma \\left( \\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\right)\\end{aligned} Or \\begin{aligned}\\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} = 0\\end{aligned} \\hspace{\\stretch{1}}(3.27) Clearly $\\dot{\\mathbf{v}} = 0$ is a solution, but is it the only solution? By dotting this with $\\mathbf{v}$ we have \\begin{aligned}0 &= \\frac{\\mathbf{v}^2 (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\cdot \\mathbf{v} \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\left( 1 + \\frac{\\mathbf{v}^2}{c^2 - \\mathbf{v}^2} \\right) \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\frac{c^2}{c^2 - \\mathbf{v}^2} \\end{aligned} This implies that $\\dot{\\mathbf{v}} = 0$ (a contraction) or that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$. To examine the perpendicularity question, let’s take cross products. This gives \\begin{aligned}0 =\\frac{(\\mathbf{v} \\times \\mathbf{v}) (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\times \\mathbf{v}\\end{aligned} \\hspace{\\stretch{1}}(3.28) We have found that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$ and $\\mathbf{v} \\times \\dot{\\mathbf{v}} = 0$. This can only mean that $\\dot{\\mathbf{v}} = 0$, contradicting the assumption that is non-zero. We conclude that $\\dot{\\mathbf{v}} = 0$ is the only solution to 3.27.", "be made explicit, introducing dot and wedge product notation respectively \\begin{aligned}\\mathbf{a} \\cdot \\mathbf{b} &= \\frac{1}{{2}}( \\mathbf{a} \\mathbf{b} + \\mathbf{b} \\mathbf{a}) = x_1 y_1 + x_2 y_2 \\\\ \\mathbf{a} \\wedge \\mathbf{b} &= \\frac{1}{{2}}( \\mathbf{a} \\mathbf{b} - \\mathbf{b} \\mathbf{a}) = \\mathbf{e}_1 \\mathbf{e}_2 (x_1 y_y - x_2 y_1).\\end{aligned} \\hspace{\\stretch{1}}(3.7) so that the vector product can be written as \\begin{aligned}\\mathbf{a} \\mathbf{b} = \\mathbf{a} \\cdot \\mathbf{b} + \\mathbf{a} \\wedge \\mathbf{b}.\\end{aligned} \\hspace{\\stretch{1}}(3.9) # Pseudoscalar In many contexts it is useful to introduce an ordered product of all the unit vectors for the space is called the pseudoscalar. In our 2D case this is \\begin{aligned}i = \\mathbf{e}_1 \\mathbf{e}_2,\\end{aligned} \\hspace{\\stretch{1}}(4.10) a quantity that we find behaves like the complex imaginary. That can be shown by considering its square \\begin{aligned}(\\mathbf{e}_1 \\mathbf{e}_2)^2&=(\\mathbf{e}_1 \\mathbf{e}_2)(\\mathbf{e}_1 \\mathbf{e}_2) \\\\ &=\\mathbf{e}_1 (\\mathbf{e}_2 \\mathbf{e}_1) \\mathbf{e}_2 \\\\ &=-\\mathbf{e}_1 (\\mathbf{e}_1 \\mathbf{e}_2) \\mathbf{e}_2 \\\\ &=-(\\mathbf{e}_1 \\mathbf{e}_1) (\\mathbf{e}_2 \\mathbf{e}_2) \\\\ &=-1^2 \\\\ &= -1\\end{aligned} Here the anticommutation of normal vectors property has been used, as well as (for the first time) the associative multiplication axiom. In a 3D context, you’ll see the pseudoscalar in many places (expressing the normals to planes for example). It also shows up in a number of fundamental relationships. For example, if one writes \\begin{aligned}I = \\mathbf{e}_1 \\mathbf{e}_2 \\mathbf{e}_3\\end{aligned} \\hspace{\\stretch{1}}(4.11) for the 3D pseudoscalar, then it’s also possible to show \\begin{aligned}\\mathbf{a} \\mathbf{b} = \\mathbf{a}", "\\cdot \\mathbf{A}_\\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum. Momentum Now move on to (16). For the factors other than $\\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a \\mathbf{a} (\\mathbf{k} \\wedge \\mathbf{c}) }}\\right\\rangle&=\\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) - \\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) \\\\ &=\\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}) - \\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}),\\end{aligned} and the other \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a (\\mathbf{k} \\wedge \\mathbf{c}) \\mathbf{a} }}\\right\\rangle&=\\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) - \\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) \\\\ &=\\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}).\\end{aligned} The momentum $\\mathbf{P} = \\int T(\\gamma^0) \\wedge \\gamma_0$ in this volume follows by computation of \\begin{aligned}&\\sigma_a \\left\\langle{{ \\sigma_a\\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} - i \\mathbf{k} \\wedge {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\right) \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} + i \\mathbf{k} \\wedge \\mathbf{A}_\\mathbf{k} \\right) }}\\right\\rangle \\\\ &= i \\mathbf{A}_\\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{k} \\right) - i \\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{A}_\\mathbf{k} \\right) \\\\ &- i \\mathbf{k} \\left( \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} \\right) \\cdot {{\\mathbf{A}_\\mathbf{k}}}^{*} \\right) + i {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\left( \\left(", "the normal dot product convention is followed using the following rules. \\begin {align*} \\mathbf {ii\\cdot i} & \\mathbf {=i}\\left ( \\mathbf {i\\cdot i}\\right ) =1\\\\ \\mathbf {ij\\cdot i} & \\mathbf {=i}\\left ( \\mathbf {j\\cdot i}\\right ) =0\\\\ \\mathbf {ik\\cdot i} & \\mathbf {=i}\\left ( \\mathbf {k\\cdot i}\\right ) =0\\\\ \\mathbf {ji\\cdot i} & \\mathbf {=j}\\left ( \\mathbf {i\\cdot i}\\right ) =1\\\\ & etc\\cdots \\end {align*} Performing the multiplication gives \\begin {multline*} \\mathbf {i\\ }dx_{1}+\\mathbf {j\\ }dx_{2}+\\mathbf {k\\ }dx_{3}=\\\\ \\left ( \\mathbf {ii}\\frac {\\partial f_{1}}{\\partial X_{1}}+\\mathbf {ij}\\frac {\\partial f_{1}}{\\partial X_{2}}+\\mathbf {ik}\\frac {\\partial f_{1}}{\\partial X_{3}}+\\mathbf {ji}\\frac {\\partial f_{2}}{\\partial X_{1}}+\\mathbf {jj}\\frac {\\partial f_{2}}{\\partial X_{2}}+\\mathbf {jk}\\frac {\\partial f_{2}}{\\partial X_{3}}+\\mathbf {ki}\\frac {\\partial f_{3}}{\\partial X_{1}}+\\mathbf {kj}\\frac {\\partial f_{3}}{\\partial X_{2}}+\\mathbf {kk}\\frac {\\partial f_{3}}{\\partial X_{3}}\\right ) \\cdot \\mathbf {i\\ }dX_{1}\\\\ +\\left ( \\mathbf {ii}\\frac {\\partial f_{1}}{\\partial X_{1}}+\\mathbf {ij}\\frac {\\partial f_{1}}{\\partial X_{2}}+\\mathbf {ik}\\frac {\\partial f_{1}}{\\partial X_{3}}+\\mathbf {ji}\\frac {\\partial f_{2}}{\\partial X_{1}}+\\mathbf {jj}\\frac {\\partial f_{2}}{\\partial X_{2}}+\\mathbf {jk}\\frac {\\partial f_{2}}{\\partial X_{3}}+\\mathbf {ki}\\frac {\\partial f_{3}}{\\partial X_{1}}+\\mathbf {kj}\\frac {\\partial f_{3}}{\\partial X_{2}}+\\mathbf {kk}\\frac {\\partial f_{3}}{\\partial X_{3}}\\right ) \\cdot \\mathbf {j\\ }dX_{2}\\\\ +\\left ( \\mathbf {ii}\\frac {\\partial f_{1}}{\\partial X_{1}}+\\mathbf {ij}\\frac {\\partial f_{1}}{\\partial X_{2}}+\\mathbf {ik}\\frac {\\partial f_{1}}{\\partial X_{3}}+\\mathbf {ji}\\frac {\\partial f_{2}}{\\partial X_{1}}+\\mathbf {jj}\\frac {\\partial f_{2}}{\\partial X_{2}}+\\mathbf {jk}\\frac {\\partial f_{2}}{\\partial X_{3}}+\\mathbf {ki}\\frac {\\partial f_{3}}{\\partial X_{1}}+\\mathbf {kj}\\frac {\\partial f_{3}}{\\partial X_{2}}+\\mathbf {kk}\\frac {\\partial f_{3}}{\\partial X_{3}}\\right ) \\cdot \\mathbf {k\\ }dX_{3} \\end {multline*} The dot multiplication is simplified using", "isn’t particularly illuminating, but can be done for completeness without too much trouble \\begin{aligned}\\mathbf{R}^{*} = ( \\rho \\cos\\phi - a \\cos t_r + a k R \\sin t_r, \\rho \\sin\\phi - a \\sin t_r - a k R \\cos t_r, z )\\end{aligned} \\hspace{\\stretch{1}}(1.20) In one sense the problem could be considered solved, since we have all the pieces of the puzzle. The outstanding question is whether or not the resulting mess can be simplified at all. Let’s see if the cross product reduces at all. Using \\begin{aligned}\\mathbf{R} \\times (\\mathbf{R}^{*} \\times \\dot{\\mathbf{v}}/c^2) =\\mathbf{R}^{*} (\\mathbf{R} \\cdot \\dot{\\mathbf{v}}/c^2) - \\frac{\\dot{\\mathbf{v}}}{c^2}(\\mathbf{R} \\cdot \\mathbf{R}^{*})\\end{aligned} \\hspace{\\stretch{1}}(1.21) Perhaps one or more of these dot products can be simplified? One of them does reduce nicely \\begin{aligned}\\mathbf{R}^{*} \\cdot \\mathbf{R} &= ( \\mathbf{R} - R \\mathbf{v}/c ) \\cdot \\mathbf{R} \\\\ &= R^2 - (\\mathbf{R} \\cdot \\mathbf{v}/c) R \\\\ &= R^2 - R a k \\rho \\sin(\\phi - k c t_r) \\\\ &= R(R - a k \\rho \\sin(\\phi - k c t_r))\\end{aligned} \\begin{aligned}\\mathbf{R} \\cdot \\dot{\\mathbf{v}}/c^2&=\\Bigl(z \\mathbf{e}_3 + \\mathbf{e}_1 (\\rho e^{i \\phi} - a e^{i \\omega t_r} ) \\Bigr) \\cdot(-a k^2 \\mathbf{e}_1 e^{i \\omega t_r} ) \\\\ &=- a k^2 \\left\\langle{{\\mathbf{e}_1 (\\rho e^{i \\phi} - a e^{i \\omega t_r} ) \\mathbf{e}_1 e^{i \\omega t_r} ) }}\\right\\rangle \\\\ &=- a k^2 \\left\\langle{{(\\rho e^{i \\phi} - a e^{i \\omega t_r} )", "them. # Appendix ## Question: Expand volume elements in coordinates Show that the coordinate representation for the volume element dotted with the curl can be represented as a sum of antisymmetric terms. That is • (a)Prove eq. 1.0.144a • (b)Prove eq. 1.0.144b • (c)Prove eq. 1.0.144c • (d)Prove eq. 1.0.144d • (e)Prove eq. 1.0.144e • (f)Prove eq. 1.0.144f ### (a) Two parameter volume, curl of vector \\begin{aligned}d^2 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge \\mathbf{f} \\right) &= d^2 u\\Bigl( { \\left( \\mathbf{x}_1 \\wedge \\mathbf{x}_2 \\right) \\cdot \\mathbf{x}^i } \\Bigr) \\cdot \\partial_i \\mathbf{f} \\\\ &= d^2 u \\left( \\mathbf{x}_1 \\cdot \\partial_2 \\mathbf{f}-\\mathbf{x}_2 \\cdot \\partial_1 \\mathbf{f} \\right) \\\\ &= d^2 u\\left( \\partial_2 f_1-\\partial_1 f_2 \\right) \\\\ &= - d^2 u \\epsilon^{ab} \\partial_{a} f_{b}. \\qquad\\square\\end{aligned} \\hspace{\\stretch{1}}(1.149) ### (b) Three parameter volume, curl of vector \\begin{aligned}d^3 \\mathbf{x} \\cdot \\left( { \\boldsymbol{\\partial} \\wedge \\mathbf{f} } \\right) &= d^3 u\\Bigl( { \\left( { \\mathbf{x}_1 \\wedge \\mathbf{x}_2 \\wedge \\mathbf{x}_3 } \\right) \\cdot \\mathbf{x}^i } \\Bigr) \\cdot \\partial_i \\mathbf{f} \\\\ &= d^3 u\\Bigl( { \\left( { \\mathbf{x}_1 \\wedge \\mathbf{x}_2 } \\right) \\cdot \\partial_3 \\mathbf{f}+\\left( { \\mathbf{x}_3 \\wedge \\mathbf{x}_1 } \\right) \\cdot \\partial_2 \\mathbf{f}+\\left( { \\mathbf{x}_2 \\wedge \\mathbf{x}_3 } \\right) \\cdot \\partial_1 \\mathbf{f}} \\Bigr) \\\\ &= d^3 u\\Bigl( {\\left( { \\mathbf{x}_1 \\partial_3 \\mathbf{f} \\cdot \\mathbf{x}_2 -\\mathbf{x}_2 \\partial_3 \\mathbf{f} \\cdot \\mathbf{x}_1 } \\right)+\\left( { \\mathbf{x}_3 \\partial_2 \\mathbf{f} \\cdot \\mathbf{x}_1 -\\mathbf{x}_1", "... + b_m = n$. We will now expand the lefthand side of the equation. We first expand the first two factors to get: (4) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} = \\frac{n!}{b_1! \\cdot (n - b_1)!} \\cdot \\frac{(n - b_1)!}{b_2! \\cdot (n - b_1 - b_2)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\end{align} • We then expand this with the third factor to get: (5) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} \\cdot \\binom{n - b_1 - b_2}{b_3} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\binom{n - b_1 - b_2}{b_3} \\\\ = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\frac{(n - b_1 - b_2)!}{b_3! \\cdot (n - b_1 - b_2 - b_3)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot b_3! \\cdot (n - b_1 - b_2 - b_3)!} \\end{align} • Inductively, by expanding the $m^{\\mathrm{th}}$ factor of the product and noting that $n - b_1 - b_2 - ... - b_m = 0$, we see that: (6) \\begin{align} \\quad \\binom{b_1 + b_2 + ... + b_m}{b_1} \\cdot \\binom{b_2 + b_3 + ... + b_m}{b_2} \\cdot ... \\cdot \\binom{b_{m-1}}{b_m} \\\\ = \\binom{b_1 + b_2 + ... + b_m}{b_1, b_2, ..., b_m} = \\frac{n!}{b_1! \\cdot b_2! \\cdot ... \\cdot b_{m-1}! (n - b_1 - b_2 - ... - b_m)!}" ]

[ "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "\\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{a} = 1,\\end{aligned} \\hspace{\\stretch{1}}(1.174) and \\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{b} = 0.\\end{aligned} \\hspace{\\stretch{1}}(1.175) \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}.\\end{aligned} \\hspace{\\stretch{1}}(1.176) Dotting with $\\mathbf{a}$ we have \\begin{aligned}\\mathbf{a} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right)=\\mathbf{a} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right)^2 - \\mathbf{b}^2 \\mathbf{a}^2,\\end{aligned} \\hspace{\\stretch{1}}(1.177) but dotting with $\\mathbf{b}$ yields zero \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right) &= \\mathbf{b} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right) \\\\ &= \\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}^2 - \\mathbf{b}^2 \\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right) \\\\ &= 0.\\end{aligned} \\hspace{\\stretch{1}}(1.178) To complete the proof, we note that the product in eq. 1.177 is just the wedge squared \\begin{aligned}\\left( { \\mathbf{a} \\wedge \\mathbf{b}} \\right)^2 &= \\left\\langle{{\\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)^2}}\\right\\rangle \\\\ &= \\left\\langle{{\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)}}\\right\\rangle \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\mathbf{a} \\mathbf{b} - 2 \\left( {\\mathbf{a} \\cdot \\mathbf{b}} \\right) \\mathbf{a} \\mathbf{b}}}\\right\\rangle+\\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right)^2 \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\left( { -\\mathbf{b} \\mathbf{a} + 2 \\mathbf{a} \\cdot \\mathbf{b} }", "B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2 B \\right) \\\\ &\\qquad - \\mathbf{x}_2 \\left( \\mathbf{x}_{34} \\cdot \\partial_1 B \\right) - \\mathbf{x}_3 \\left( \\mathbf{x}_{42} \\cdot \\partial_1 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{23} \\cdot \\partial_1 B \\right)} \\Bigr) \\\\", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "\\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2", "+ (-a_1 - b_1i) = (a_1 - a_1) + (b_1 - b_1)i = 0 + 0i = 0$. • FM1: $z_1 \\cdot z_2 = (a_1 + b_1i)(a_2 + b_2i) = a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i$ and so $(z_1 \\cdot z_2) \\in \\mathbb{C}$. • FM2: $z_1 \\cdot z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i = (a_2a_1 - b_2b_1) + (a_2b_1 + b_2a_1)i = z_2 \\cdot z_1$. • FM3: (1) \\begin{align} z_1 \\cdot (z_2 \\cdot z_3) \\\\ = (a_1 + b_1i) [ (a_2a_3 - b_2b_3) + (a_2b_3 + b_2a_3)i ] \\\\ = (a_1a_2a_3 - a_1b_2b_3) + (a_1a_2b_3 + a_1b_2a_3)i + (b_1a_2a_3 - b_1b_2b_3)i + (b_1a_2b_3 + b_1b_2a_3)i^2 \\\\ \\quad = [(a_1a_2a_3 - a_1b_2b_3) - (b_1a_2b_3 + b_1b_2a_3)] + [(a_1a_2b_3 + a_1b_2a_3) + (b_1a_2a_3 - b_1b_2b_3)]i \\\\ = [ (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i ] (a_3 + b_3i) \\\\ = (z_1 \\cdot z_2) \\cdot z_3 \\end{align} • FM4: The multiplicative identity is $1 = 1 + 0i$. That is $z_1 \\cdot 1 = (a_1 + b_1)(1 + 0i) = (a_1 + a_1 \\cdot 0i) + (b_1 + b_1 \\cdot 0)i = a_1 + b_1i = z_1$. • FM5: The multiplicative inverse for $z_1$ where $z_1 \\neq 0$ is $z_1^{-1} = \\frac{a_1 -b_1i}{a_1^2 + b_1^2}$. That is: (2)", "\\dot{\\mathbf{x}}) = 0\\end{aligned} \\hspace{\\stretch{1}}(3.26) By evaluating this, we can eventually show that we can construct a four vector equation. Doing this we have \\begin{aligned}\\frac{d}{dt} (\\gamma \\mathbf{v}) &=\\frac{d}{dt} \\left( \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\mathbf{v} \\right) \\\\ &=-2 (-1/2) \\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}})/c^2 \\left(1 - \\mathbf{v}^2/c^2\\right)^{-3/2} + \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\dot{\\mathbf{v}} \\\\ &=\\gamma \\left( \\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\right)\\end{aligned} Or \\begin{aligned}\\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} = 0\\end{aligned} \\hspace{\\stretch{1}}(3.27) Clearly $\\dot{\\mathbf{v}} = 0$ is a solution, but is it the only solution? By dotting this with $\\mathbf{v}$ we have \\begin{aligned}0 &= \\frac{\\mathbf{v}^2 (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\cdot \\mathbf{v} \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\left( 1 + \\frac{\\mathbf{v}^2}{c^2 - \\mathbf{v}^2} \\right) \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\frac{c^2}{c^2 - \\mathbf{v}^2} \\end{aligned} This implies that $\\dot{\\mathbf{v}} = 0$ (a contraction) or that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$. To examine the perpendicularity question, let’s take cross products. This gives \\begin{aligned}0 =\\frac{(\\mathbf{v} \\times \\mathbf{v}) (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\times \\mathbf{v}\\end{aligned} \\hspace{\\stretch{1}}(3.28) We have found that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$ and $\\mathbf{v} \\times \\dot{\\mathbf{v}} = 0$. This can only mean that $\\dot{\\mathbf{v}} = 0$, contradicting the assumption that is non-zero. We conclude that $\\dot{\\mathbf{v}} = 0$ is the only solution to 3.27.", "of $$X$$, $$f_{X}(x)$$, is then \\begin{align*} f_{X}(0) &= \\mathbf{P}(X = 0) \\\\ &= \\mathbf{P}(\\mathrm{Hit\\:E})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 81/625 \\\\ f_{X}(1) &= \\mathbf{P}(X = 1) \\\\ &= 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 126/625 \\\\ f_{X}(2) &= \\mathbf{P}(X = 2) \\\\ &= \\begin{align*}& 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:C})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &+ \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:D}) \\end{align*} \\\\ &= 139/625 \\end{align*} and so on. From (2), the expected value of $$X$$ is $$\\mathbf{E}(X) = 0.81/625 + 1 \\cdot 126/625 + 2 \\cdot 139/625 + 3 \\cdot 124/625 + ... = 2.4$$. From (4), the variance of $$X$$ is $$\\mathrm{Var}(X) = 2.4^{2} \\cdot 81/625 + 1.4^{2} \\cdot 126/625 + 0.4^{2} \\cdot 139/625 + 0.6^{2} \\cdot 124/625 + ... = 2.72$$. The distribution function of $$X$$, $$F_{X}(x)$$, is then \\begin{align*} F_{X}(0) &= \\mathbf{P}(X \\le 0) \\\\ &= f_{X}(0) \\\\ F_{X}(1) &= \\mathbf{P}(X \\le 1) \\\\ &= f_{X}(1) + f_{X}(0) \\\\ F_{X}(2) &= \\mathbf{P}(X \\le 2) \\\\ &= f_{X}(2) + f_{X}(1) + f_{X}(0) \\end{align*} and so on. The distribution function $$F_{X}(x)$$ is shown in Figure 4. Figure 4: The distribution function $$F_{X}$$ of $$X$$ for the archery target.", "$(\\mathbf{a}+\\mathbf{b})\\cdot(\\mathbf{c}+\\mathbf{d}) = \\mathbf{a}\\cdot\\mathbf{c} + \\mathbf{a}\\cdot\\mathbf{d} + \\mathbf{b}\\cdot\\mathbf{c} + \\mathbf{b}\\cdot\\mathbf{d}$ ## Mixed associative law The dot product has a homogeneity $\\alpha (\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot \\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Bilinear Often the mixed associative law and distributive law are combined to show that the dot product is bilinear: $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Cauchy-Schwarz Inequality $|\\mathbf{a}\\cdot\\mathbf{b}|\\leq|\\mathbf{a}||\\mathbf{b}|$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Fast calculation of Cosine The definition of the dot-product is used extensively in computer graphics since it speeds up computations in many circumstances by avoiding inefficient trigonometric functions. Additionally, working with normalized vectors gives the cosine between these two vectors by taking the dot product of them: $\\cos\\theta = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}||\\mathbf{b}|} = \\hat{\\mathbf{a}}\\cdot\\hat{\\mathbf{b}}$ ### Making perpendicular vectors Since two vectors are perpendicular when $$\\mathbf{a}\\cdot\\mathbf{b} = 0$$, we can easily construct a perpendicular vector by zeroing all components except 2, flipping those two and reversing the sign of one of them. For example with $$\\mathbf{a} = (\\mathbf{a}_1, \\mathbf{a}_2, \\dots, \\mathbf{a}_n)$$ the vectors $$\\mathbf{a}' = (-\\mathbf{a}_2, \\mathbf{a}_1, 0, \\dots, 0)$$ or $$\\mathbf{a}'' = (0, -\\mathbf{a}_3, \\mathbf{a}_2, 0, \\dots, 0)$$, etc are all perpendcular to $$\\mathbf{a}$$,", "coefficients in the $i$th column of the matrix. When we compute $B\\mathbf{v}_1,\\ldots,B\\mathbf{v}_k$, each of these will lie in $E_{\\lambda}$. Therefore, each of these can be expressed as a linear combination of $\\mathbf{v}_1,\\ldots,\\mathbf{v}_k$ (since they form a basis for $E_{\\lambda}$. So, to express them as linear combinations of $\\beta$, we just add $0$s; we will have: \\begin{align*} B\\mathbf{v}_1 &= b_{11}\\mathbf{v}_1 + b_{21}\\mathbf{v}_2+\\cdots+b_{k1}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n\\\\ B\\mathbf{v}_2 &= b_{12}\\mathbf{v}_1 + b_{22}\\mathbf{v}_2 + \\cdots +b_{k2}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n\\\\ &\\vdots\\\\ B\\mathbf{v}_k &= b_{1k}\\mathbf{v}_1 + b_{2k}\\mathbf{v}_2 + \\cdots + b_{kk}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n \\end{align*} where $b_{ij}$ are some scalars (some possibly equal to $0$). So the matrix of $B$ relative to $\\beta$ would start off something like: $$\\left(\\begin{array}{ccccccc} b_{11} & b_{12} & \\cdots & b_{1k} & * & \\cdots & *\\\\ b_{21} & b_{22} & \\cdots & b_{2k} & * & \\cdots & *\\\\ \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ b_{k1} & b_{k2} & \\cdots & b_{kk} & * & \\cdots & *\\\\ 0 & 0 & \\cdots & 0 & * & \\cdots & *\\\\ \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & 0 & * & \\cdots & * \\end{array}\\right).$$ So, now suppose that you have a basis for", "length $||\\mathbf{v}||\\neq 0$. Thus, we must have $c_1=0$. We repeat the above argument with $\\mathbf{v}_1$ replaced by $\\mathbf{v}_2$ as follows. Since $c_1=0$, the (*) becomes $c_2\\mathbf{v}_2+c_3\\mathbf{v}_3=\\mathbf{0}.$ We have \\begin{align*} \\mathbf{0}&=\\mathbf{v}_2\\cdot \\mathbf{0}=\\mathbf{v}_2\\cdot (c_2\\mathbf{v}_2+c_3\\mathbf{v}_3)\\\\ &=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2+c_3\\mathbf{v}_2\\cdot\\mathbf{v}_3. \\end{align*} Since $\\mathbf{v}_2\\cdot\\mathbf{v}_3=0$, we have $\\mathbf{0}=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2=c_2||\\mathbf{v}_2||^2.$ Since $\\mathbf{x}\\neq \\mathbf{0}$, we have $||\\mathbf{v}_2||\\neq 0$, and thus $c_2=0$. Hence (*) is now just $c_3\\mathbf{v}_3=\\mathbf{0}$, and since $\\mathbf{x}_3\\neq \\mathbf{0}$, we conclude that $c_3=0$. Therefore, we have shown that $c_1=c_2=c_3=0$, and thus the set $\\{\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3\\}$ is linearly independent. ## Comment. These are Quiz 3 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. ### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017 There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. ### 3 Responses 1. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 2. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 3. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] ##### Find a Nonsingular Matrix Satisfying Some Relation Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A,$ where $B$ is the following matrix. If such a nonsingular... Close", "&\\qquad+ \\partial_1 \\left( { \\mathbf{x}_2 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2", "\\cdot \\mathbf{A}_\\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum. Momentum Now move on to (16). For the factors other than $\\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a \\mathbf{a} (\\mathbf{k} \\wedge \\mathbf{c}) }}\\right\\rangle&=\\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) - \\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) \\\\ &=\\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}) - \\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}),\\end{aligned} and the other \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a (\\mathbf{k} \\wedge \\mathbf{c}) \\mathbf{a} }}\\right\\rangle&=\\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) - \\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) \\\\ &=\\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}).\\end{aligned} The momentum $\\mathbf{P} = \\int T(\\gamma^0) \\wedge \\gamma_0$ in this volume follows by computation of \\begin{aligned}&\\sigma_a \\left\\langle{{ \\sigma_a\\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} - i \\mathbf{k} \\wedge {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\right) \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} + i \\mathbf{k} \\wedge \\mathbf{A}_\\mathbf{k} \\right) }}\\right\\rangle \\\\ &= i \\mathbf{A}_\\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{k} \\right) - i \\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{A}_\\mathbf{k} \\right) \\\\ &- i \\mathbf{k} \\left( \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} \\right) \\cdot {{\\mathbf{A}_\\mathbf{k}}}^{*} \\right) + i {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\left( \\left(", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "(for example, if u = (a, b) and v = (c, d), then their dot product u·v = ac + bd). If the dot product is zero, the two vectors are said to be orthogonal to each other. Some properties of the dot product aid understanding of how W-CDMA works. If vectors a and b are orthogonal, then ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =0}$ and: ${\\displaystyle \\mathbf {a} \\cdot (\\mathbf {a} +\\mathbf {b} )=\\|\\mathbf {a} \\|^{2},\\ {\\text{since}}\\ \\mathbf {a} \\cdot \\mathbf {a} +\\mathbf {a} \\cdot \\mathbf {b} =\\|\\mathbf {a} \\|^{2}+0,}$ ${\\displaystyle \\mathbf {a} \\cdot (-\\mathbf {a} +\\mathbf {b} )=-\\|\\mathbf {a} \\|^{2},\\ {\\text{since}}\\ {-\\mathbf {a} }\\cdot \\mathbf {a} +\\mathbf {a} \\cdot \\mathbf {b} =-\\|\\mathbf {a} \\|^{2}+0,}$ ${\\displaystyle \\mathbf {b} \\cdot (\\mathbf {a} +\\mathbf {b} )=\\|\\mathbf {b} \\|^{2},\\ {\\text{since}}\\ \\mathbf {b} \\cdot \\mathbf {a} +\\mathbf {b} \\cdot \\mathbf {b} =0+\\|\\mathbf {b} \\|^{2},}$ ${\\displaystyle \\mathbf {b} \\cdot (\\mathbf {a} -\\mathbf {b} )=-\\|\\mathbf {b} \\|^{2},\\ {\\text{since}}\\ \\mathbf {b} \\cdot \\mathbf {a} -\\mathbf {b} \\cdot \\mathbf {b} =0-\\|\\mathbf {b} \\|^{2}.}$ Each user in synchronous CDMA uses a code orthogonal to the others' codes to modulate their signal. An example of 4 mutually orthogonal digital signals is shown in the figure below. Orthogonal codes have a cross-correlation equal to zero; in other words, they do not interfere with each other. In the case", "of $$X$$. For $$X$$ and $$Y$$ independent, \\begin{align*} \\mathbf{P}(X = x \\text{ and } Y = y) &= \\mathbf{P}\\left((X = x) \\cap (Y = y)\\right) \\\\ &= \\mathbf{P}(X = x) \\cdot \\mathbf{P}(Y = y), \\end{align*} $$\\mathbf{P}(X = x \\text{ and } Y = y) = \\mathbf{P}\\left((X = x) \\cap (Y = y)\\right) = \\mathbf{P}(X = x) \\cdot \\mathbf{P}(Y = y),$$ so that $$f_{X,Y}(x, y) = f_{X}(x) \\cdot f_{Y}(y)$$. Therefore, \\begin{align*} \\mathbf{E}(X, Y) &= \\sum_{x} \\sum_{y} xy \\cdot f_{X, Y}(x, y) \\\\ &= \\sum_{x} \\sum_{y} xy \\cdot f_{X}(x) f_{Y}(y) \\\\ &= \\sum_{x} \\Big\\{ x \\cdot f_{X}(x) \\cdot \\sum_{y} y \\cdot f_{Y}(y) \\Big\\} \\\\ &= \\sum_{x} \\{ x \\cdot f_{X}(x) \\cdot \\mathbf{E}Y \\} \\\\ &= \\mathbf{E}X \\cdot \\mathbf{E}Y. \\end{align*} For $$X$$ and $$Y$$ continuous, the proof begins $$\\mathbf{E}(XY) = \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} xy \\cdot f_{X, Y}(x, y) \\mathrm{d}x\\mathrm{d}y$$ where $$f_{X,Y} : \\real^2 \\to [0, \\infty)$$ is the joint density function of $$X$$ and $$Y$$. The proof then proceeds in a similar way to the discrete case with summations replaced by integrations. For the discrete or continuous random variable $$X$$, $$\\sigma^{2} = \\mathbf{E}\\left(\\left(X - \\mathbf{E}(X)\\right)^{2}\\right) = \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}. \\tag{16}$$ $$\\sigma^{2} = \\mathbf{E}\\left(\\left(X - \\mathbf{E}(X)\\right)^{2}\\right) = \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}.$$ $$(16)$$ The proof of (16) holds for $$X$$ discrete or continuous. As a shorthand notation, let $$\\mu = \\mathbf{E}(X)$$. Then, $$\\mathbf{E}\\left(\\left(X - \\mu\\right)^{2}\\right)", "(\\text{mod} \\: m)$, the following hold: \\boxed{\\begin{align*} a + c &\\equiv b + d \\: (\\text{mod} \\: m) \\\\ a \\cdot c &\\equiv b \\cdot d \\: (\\text{mod} \\: m) \\end{align*} } Let’s prove each of these properties. #### Addition If $a \\equiv b \\: (\\text{mod} \\: m)$ and $c \\equiv d \\: (\\text{mod} \\: m)$, then $b = a + mk_1$ and $d = c + mk_2$. \\begin{align*}b + d &= a + c + mk_1 + mk_2 \\\\ &= (a + c) + m(k_1 + k_2) \\\\ &= (a + c) + m \\cdot (\\text{some integer}) \\\\ \\implies b + d &\\equiv a + c \\: (\\text{mod} \\: m) \\end{align*} #### Multiplication Multiplying our expressions for $b$ and $d$: \\begin{align*} bd &= (a + mk_1)(c + mk_2) \\\\ &= ac + ak_2m + ck_1m + k_1k_2m^2 \\\\ &= ac + m(ak_2 + ck_1 + k_1k_2m) \\\\ &= ac + m \\cdot (\\text{some integer}) \\\\ \\implies bd &\\equiv ac \\: (\\text{mod} \\: m)\\end{align*} The key takeaway of these proofs is that we can “simplify $\\text{mod} \\: m$” as we go, in any order. ### Exponentiation Suppose we want to evaluate $2^{15} \\: (\\text{mod} \\: 9)$. We could find $2^{15} = 32768$, and divide this number by $9$ and find the remainder, but there’s a better way. $2^{15}", "Find $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ for $$\\textbf{u} = (1, 2, 4)$$, $$\\textbf{v} = (2, 2, 0)$$, $$\\textbf{w} = (1, 3, 0)$$. Solution Since $$\\textbf{u} \\cdot \\textbf{v} = 6$$ and $$\\textbf{u} \\cdot \\textbf{w} = 7$$, then \\begin{align*} \\textbf{u} \\times (\\textbf{v} \\times \\textbf{w}) &= (\\textbf{u} \\cdot \\textbf{w})\\textbf{v} - (\\textbf{u} \\cdot \\textbf{v})\\textbf{w}\\\\ &= 7\\,(2, 2, 0) - 6\\,(1, 3, 0) = (14, 14, 0) - (6, 18, 0)\\\\ &= (8, -4, 0) \\end{align*} Note that $$\\textbf{v}$$ and $$\\textbf{w}$$ lie in the $$xy$$-plane, and that $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ also lies in that plane. Also, $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ is perpendicular to both $$\\textbf{u}$$ and $$\\textbf{v} \\times \\textbf{w} = (0, 0, 4)$$ (see Figure 1.4.8). Figure 1.4.8 For vectors $$\\textbf{v} = v_{1}\\textbf{i} + v_{2}\\textbf{j} + v_{3}\\textbf{k}$$ and $$\\textbf{w} = w_{1}\\textbf{i} + w_{2}\\textbf{j} + w_{3}\\textbf{k}$$ in component form, the cross product is written as: $$\\textbf{v} \\times \\textbf{w} = (v_{2}w_{3} - v_{3}w_{2})\\textbf{i} + (v_{3}w_{1} - v_{1}w_{3})\\textbf{j} + (v_{1}w_{2} - v_{2}w_{1})\\textbf{k}$$. It is often easier to use the component form for the cross product, because it can be represented as a $$\\textit{determinant}$$. We will not go too deeply into the theory of determinants; we will just cover what is essential for our purposes. A 2 $$\\times$$ 2 matrix} is an array of two rows and two columns of scalars, written as $$\\nonumber", "{ \\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot" ]

[ "# Inspired by Aakash Khandelwal Geometry Level 5 If $$a_0=\\sin^2\\left (\\dfrac{\\pi}{45} \\right)$$ and $$a_{n+1}=4a_{n}(1-a_{n})$$ for $$n\\geq 0$$, find the smallest positive integer $$n$$ such that $$a_n=a_0$$. If you come to the conclusion that no such $$n$$ exists, enter 666. ×", "The Least Positive Integer N Such that ( 2 I 1 + I ) N is a Positive Integer, Is. - Mathematics MCQ The least positive integer n such that $\\left( \\frac{2i}{1 + i} \\right)^n$ is a positive integer, is. • 16 • 8 • 4 • 2 Solution $8$ $\\text { Let } z = \\left( \\frac{2i}{1 + i} \\right)$ $\\Rightarrow z = \\frac{2i}{1 + i} \\times \\frac{1 - i}{1 - i}$ $\\Rightarrow z = \\frac{2i\\left( 1 - i \\right)}{1 - i^2}$ $\\Rightarrow z = \\frac{2i\\left( 1 - i \\right)}{1 + 1} \\left[ \\because i^2 = - 1 \\right]$ $\\Rightarrow z = \\frac{2i\\left( 1 - i \\right)}{2}$ $\\Rightarrow z = i - i^2$ $\\Rightarrow z = i + 1$ $\\text { Now }, z^n = \\left( 1 + i \\right)^n$ $\\text { For } n = 2,$ $z^2 = \\left( 1 + i \\right)^2$ $= 1 + i^2 + 2i$ $= 1 - 1 + 2i$ $= 2i . . . (1)$ $\\text { Since this is not a positive integer },$ $\\text { For } n = 4,$ $z^4 = \\left( 1 + i \\right)^4$ $= \\left[ \\left( 1 + i \\right)^2 \\right]^2$ $= \\left( 2i \\right)^2 \\left[ \\text { Using } (1) \\right]$ $= 4 i^2$ $= - 4 . . . (2)$ $\\text {", "Find the smallest positive integer value of m for which Question: Find the smallest positive integer value of $m$ for which $\\frac{(1+i)^{n}}{(1-i)^{n-2}}$ is a real number. Solution: $\\frac{(1+i)^{m}}{(1-i)^{m-2}}$ $=\\frac{(1+i)^{m}}{(1-i)^{m}} \\times(1-i)^{2}$ $=\\left(\\frac{1+i}{1-i} \\times \\frac{1+i}{1+i}\\right)^{m} \\times\\left(1+i^{2}-2 i\\right)$ $=\\left(\\frac{1+i^{2}+2 i}{1-i^{2}}\\right)^{m} \\times(1-1-2 i)$ $=\\left(\\frac{1-1+2 i}{1+1}\\right)^{m} \\times(-2 i)$ $=-2 i\\left(i^{m}\\right)$ $=-2(i)^{m+1}$ For this to be real, the smallest positive value of $m$ will be 1 . Thus, $i^{1+1}=i^{2}=-1$, which is real.", "# The least positive integer n such that Question: The least positive integer $n$ such that $\\left(\\frac{2 i}{1+i}\\right)^{n}$ is a positive integer, is (a) 16 (b) 8 (c) 4 (d) 2 Solution: (b) 8 Let $z=\\left(\\frac{2 i}{1+i}\\right)$ $\\Rightarrow z=\\frac{2 i}{1+i} \\times \\frac{1-i}{1-i}$ $\\Rightarrow z=\\frac{2 i(1-i)}{1-i^{2}}$ $\\Rightarrow z=\\frac{2 i(1-i)}{1+1} \\quad\\left[\\because i^{2}=-1\\right]$ $\\Rightarrow z=\\frac{2 i(1-i)}{2}$ $\\Rightarrow z=i-i^{2}$ $\\Rightarrow z=i+1$ Now, $z^{n}=(1+i)^{n}$ For $n=2$, $z^{2}=(1+i)^{2}$ $=1+i^{2}+2 i$ $=1-1+2 i$ $=2 i \\quad \\ldots(1)$ Since this is not a positive integer, For $n=4$, $z^{4}=(1+i)^{4}$ $=\\left[(1+i)^{2}\\right]^{2}$ $=(2 i)^{2}$ [Using (1)] $=4 i^{2}$ $=-4 \\quad \\ldots(2)$ This is a negative integer. For $n=8$, $z^{8}=(1+i)^{8}$ $=\\left[(1+i)^{4}\\right]^{2}$ $=(-4)^{2}$ = 16 This is a positive integer. Thus, $z=\\left(\\frac{2 i}{1+i}\\right)^{n}$ is positive for $n=8$. Therefore, 8 is the least positive integer such that $\\left(\\frac{2 i}{1+i}\\right)^{n}$ is a positive integer.", "• # question_answer If $n$ is a positive integer, then ${{\\left( \\frac{1+i}{1-i} \\right)}^{4n+1}}$= A) 1 B) - 1 C) $i$ D) $-i$ Since $\\frac{1+i}{1-i}=\\frac{(1+i)(1+i)}{(1-i)(1+i)}=i$ Therefore ${{\\left( \\frac{1+i}{1-i} \\right)}^{4n+1}}={{i}^{4n+1}}=i{{i}^{4n}}=i\\,\\,\\,\\,\\,(\\because {{i}^{4n}}=1)$", "For a positive integer n, find the value of Question: For a positive integer $n$, find the value of $(1-i)^{n}\\left(1-\\frac{1}{i}\\right)^{n}$. Solution: $(1-i)^{n}\\left(1-\\frac{1}{i}\\right)^{n}=(1-i)^{n}\\left(1-\\frac{i^{4}}{i}\\right)^{n}$ $\\left[\\because i^{4}=1\\right]$ $=(1-i)^{n}\\left(1-i^{3}\\right)^{n}$ $=(1-i)^{n}(1+i)^{n}$ $\\left[\\because i^{3}=-i\\right]$ $=[(1-i)(1+i)]^{n}$ $=\\left(1-i^{2}\\right)^{n}$ $=2^{n} \\quad\\left[\\because i^{2}=-1\\right]$ Thus, the value of $(1-i)^{n}\\left(1-\\frac{1}{i}\\right)^{n}$ is $2^{n}$.", "+0 # Trigonometry (helppppp) 0 100 2 Find the smallest positive integer x for which the expression $$\\log\\left(\\sin^2{x}-\\frac{1}{2}\\right)$$ is not defined. (The sine function is evaluated in terms of radians.) Jul 11, 2022 #1 0 Using a calculator, 1, 2, and 3 work, but not 4, so the answer is 4. Jul 11, 2022 #2 0 Actually, 4 works too Guest Jul 11, 2022", "• # question_answer The least positive integer $n$ which will reduce ${{\\left( \\frac{i-1}{i+1} \\right)}^{n}}$ to a real number, is [Roorkee 1998] A) 2 B) 3 C) 4 D) 5 ${{\\left( \\frac{i-1}{i+1}\\times \\frac{i-1}{i-1} \\right)}^{n}}={{\\left( \\frac{-2i}{-2} \\right)}^{n}}={{i}^{n}}$ Hence, to make the real number the least positive integer is 2.", "# Let's minimise sin! Geometry Level 5 $P_n=\\prod_{k=0}^{n-1}\\left(1-2\\sin\\left(\\frac{2k\\pi}{n}\\right)\\right)$ Find the minimal value of $$P_n$$, where $$n$$ is a positive integer. If you come to the conclusion that no minimum is attained, enter 666 as your answer. ×", "find the smallest positive integer $m$, and $0$ is not a positive integer. – MJD Nov 9 '14 at 2:32 $$\\left(\\sqrt{3}+i\\right)^m=\\left(\\sqrt{3}-i\\right)^m$$ $$2^m\\left(\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)^m=2^m\\left(\\frac{\\sqrt{3}}{2}-\\frac{i}{2}\\right)^m$$ $$e^{i\\theta}=\\cos\\theta+i\\sin\\theta$$ Using above formula we can write $$2^m(e^{i\\pi/6})^m=2^m(e^{-i\\pi/6})^m$$ $$e^{mi\\pi/6}=e^{-mi\\pi/6}$$ $$e^{2mi\\pi/6}=e^{mi\\pi/3}=1$$ $$e^{mi\\pi/3}=e^{2ik\\pi}$$ $$\\frac{mi\\pi}{3}=2ik\\pi$$ $$\\frac{m}{3}=2k$$ $$m=6k$$ where $k\\in\\mathbb Z$", "+0 Pls help 0 118 2 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 8, 2021 #1 0 The smallest n that works is 2019. Aug 8, 2021 #2 +115804 0 I don't get this question. It does not make sense to me. If $$\\displaystyle \\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.\\\\ then\\\\ \\displaystyle \\sum_{k = 1}^1 a_1 \\left( \\frac{1}{1} \\right)^2 = 1.\\\\ so\\\\a_1=1$$ Since the sum of any number of terms must equal 1 $$a_k=0\\quad \\text{Where k is an integer greater than 1}$$ Obviously, I am not understanding something. Aug 8, 2021", "you are using (if any). Mr F says: Find the smallest integer value of r that solves $\\left( \\frac{1}{6} \\right)^r > \\frac{1}{2}$. Using trial and error would be best for you. WOULD BE REALLY GRATEFUL IF SOMEONE COULD POST THE ANSWERS TO THESE QUESTIONS!!!! ..", "argument of $\\frac{1+i}{1-i}$? If your question meant \"Find the smallest positive integer $n$ such that $\\left(\\frac{1+n}{1-n}\\right)^n = 1$\", then the answer is \"there are no solutions\". Clearly we'd need $\\frac{1+n}{1-n} = 1$ if this were to work, but then we're forced to have $n = 0$. If your question meant \"Find the smallest positive integer $n$ such that $\\left(1+ \\frac{n}{1-n}\\right)^n = 1$\", you can use the same logic as above to arrive at the answer.", "Question # The smallest positive integer n for which ${\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}}$ is:$A.{\\text{ }}1 \\\\ B.{\\text{ 2}} \\\\ C.{\\text{ 3}} \\\\ D.{\\text{ 4}} \\\\$ Hint- In order to evaluate the smallest positive integer, try to bring the given terms in simpler form in order to use the property associated with even power of complex numbers. Complete step-by-step solution - Given equation is: ${\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}}$ As we know the rule of product of powers ${\\left( x \\right)^{a \\times b}} = {\\left( {{x^a}} \\right)^b}$ Applying the same on either side of the equation we obtain: $\\Rightarrow {\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}} \\\\ \\Rightarrow {\\left[ {{{\\left( {1 + i} \\right)}^2}} \\right]^n} = {\\left[ {{{\\left( {1 - i} \\right)}^2}} \\right]^n} \\\\$ Since, we know the algebraic formula for square of sum of numbers i.e. ${\\left( {a + b} \\right)^2} = {a^2} + 2ab + {b^2} \\\\ {\\left( {a - b} \\right)^2} = {a^2} - 2ab + {b^2} \\\\$ So, using the formula in the above equation, we get $\\Rightarrow {\\left[ {{{\\left( {1 + i} \\right)}^2}} \\right]^n} = {\\left[ {{{\\left( {1 - i} \\right)}^2}} \\right]^n} \\\\ \\Rightarrow {\\left[ {{{\\left( 1 \\right)}^2} + 2 \\times \\left( 1 \\right) \\times \\left( i", "$frac12 int_0^pi f\\left(x\\right)sin\\left(x\\right),dx= frac12 bigl\\left(F\\text{'}\\left(x\\right)sin x - F\\left(x\\right)cos xbigr\\right)Big|_\\left\\{x=0\\right\\}^\\left\\{x=pi\\right\\}.!$ Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1. Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 2 and 3 show that F(0) is a positive integer. Since $x\\left(pi -x\\right) = Bigl\\left(fracpi2Bigr\\right)^2-Bigl\\left(x-fracpi2Bigr\\right)^2leBigl\\left(fracpi2Bigr\\right)^2,quad xinmathbb\\left\\{R\\right\\}!$ and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have $frac12 int_0^pi f\\left(x\\right)sin\\left(x\\right),dxle frac\\left\\{b^n\\right\\}\\left\\{n!\\right\\}Bigl\\left(fracpi2Bigr\\right)^\\left\\{2n+1\\right\\}!$ which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0). ### Analysis of Niven's proof The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula begin\\left\\{align\\right\\} frac12int_0^pi f(x)sin(x),dx &=frac12sum_{j=0}^n (-1)^j bigl(f^{(2j)}(pi)+f^{(2j)}(0)bigr) &qquad+frac{(-1)^{n+1}}2int_0^pi f^{(2n+2)}(x)sin(x),dx, end{align} which is obtained by 2n + 2 partial integrations. Claim 3 essentially establishes this formula, where the use of F hides the iterated partial integrations. The last integral vanishes because f (2n + 2) is the zero polynomial. Claims 1 and 2 show that the remaining", "# Math Help - Prove that there exists a positive integer n 1. ## Prove that there exists a positive integer n 1. Suppose delta > 0. (a) Prove that there exists a positive integer n such that 0 < 1/((4n+1)pi/2) < delta and sin ((4n+1)pi/2) = 1 (b) Prove that there exists a positive integer m such that 0 < 1/((4m+3)pi/2) < delta and sin ((4m+3)pi/2) = -1 see if anyone can do it, 2. Originally Posted by 450081592 1. Suppose delta > 0. (a) Prove that there exists a positive integer n such that 0 < 1/((4n+1)pi/2) < delta and sin ((4n+1)pi/2) = 1 (b) Prove that there exists a positive integer m such that 0 < 1/((4m+3)pi/2) < delta and sin ((4m+3)pi/2) = -1 see if anyone can do it, For the first one take $n> \\frac{1}{4} (\\frac{1}{ \\delta \\pi /2} -1 )$ (just isolate $n$) and note that $\\sin ((4n+1) \\pi /2)=\\sin (2n\\pi ) \\cos (\\pi /2) + \\sin (\\pi /2) \\cos (2n \\pi )=\\sin (\\pi /2)=1$. For the second one do exactly the same thing, the only difference being $\\sin (3\\pi /2)=-1$ 3. what does this imply? I cant see the connection between this and n to be an positive integer, can you explain it please", "This is a negative integer }.$ $\\text { For } n = 8,$ $z^8 = \\left( 1 + i \\right)^8$ $= \\left[ \\left( 1 + i \\right)^4 \\right]^2$ $= \\left( - 4 \\right)^2 \\left[ \\text { Using } (2) \\right]$ $= 16$ $\\text { This is a positive integer } .$ $\\text { Thus }, z = \\left( \\frac{2i}{1 + i} \\right)^n\\text { is positive for } n = 8 .$ $\\text { Therefore, 8 is the least positive integer such that } \\left( \\frac{2i}{1 + i} \\right)^n\\text { is a positive integer } .$ Is there an error in this question or solution? APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 13 Complex Numbers Q 13 | Page 64", "Courses Courses for Kids Free study material Free LIVE classes More The smallest positive integer n for which ${\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}}$ is:A. $1$B. $2$C. $3$D. $4$ Last updated date: 23rd Mar 2023 Total views: 306k Views today: 6.83k Verified 306k+ views Hint: In order to evaluate the smallest positive integer, try to bring the given terms in simpler form in order to use the property associated with even power of complex numbers. Complete step-by-step solution: Given equation is: ${\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}}$ As we know the rule of product of powers ${\\left( x \\right)^{a \\times b}} = {\\left( {{x^a}} \\right)^b}$ Applying the same on either side of the equation we obtain: $\\Rightarrow {\\left( {1 + i} \\right)^{2n}} = {\\left( {1 - i} \\right)^{2n}}$ $\\Rightarrow {\\left[ {{{\\left( {1 + i} \\right)}^2}} \\right]^n} = {\\left[ {{{\\left( {1 - i} \\right)}^2}} \\right]^n}$ Since, we know the algebraic formula for square of sum of numbers i.e. ${\\left( {a + b} \\right)^2} = {a^2} + 2ab + {b^2}$ ${\\left( {a - b} \\right)^2} = {a^2} - 2ab + {b^2}$ So, using the formula in the above equation, we get $\\Rightarrow {\\left[ {{{\\left( {1 + i} \\right)}^2}} \\right]^n} = {\\left[ {{{\\left( {1 - i} \\right)}^2}} \\right]^n}$ $\\Rightarrow {\\left[ {{{\\left(", "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "+0 # plz help +1 102 3 Find the smallest positive solution to the equation $\\sin(x)=\\cos\\left(2x+\\dfrac{\\pi}{3}\\right)$. Nov 11, 2019 edited by Guest Nov 13, 2019 ### 3+0 Answers #1 0 Do you really think we're going to do all your homework for you? Nov 11, 2019 #2 0 Someone usually does. Guest Nov 11, 2019 #3 0 Well we are not doing it for you even if you cant Nov 11, 2019" ]

[ "since $$f'(x)=6Lx-6x^2$$, we see that $$f'(0)=f'(L)=0$$. Since $$f'''(x)=-12$$, we see from ( Equation \\ref{eq:11.3.4}) that if $$n\\ge1$$ then \\begin{aligned} a_n&=-{24L^2\\over n^3\\pi^3}\\int_0^L\\sin{n\\pi x\\over L}\\,dx =\\left.{24L^3\\over n^4\\pi^4}\\cos{n\\pi x\\over L}\\right|_{0}^{L}={24L^3\\over n^4\\pi^4}[(-1)^n-1]\\\\ &= \\left\\{\\begin{array}{cl} -{48L^3\\over(2m-1)^4\\pi^4}&\\mbox{if n=2m-1},\\\\ 0&\\mbox{if n=2m.} \\end{array}\\right.\\end{aligned}\\nonumber Therefore $C(x)={L^3\\over2}-{48L^3\\over\\pi^4}\\sum_{n=1}^\\infty{1\\over (2n-1)^4}\\cos{(2n-1)\\pi x\\over L}.\\nonumber$ Example $$\\PageIndex{6}$$ Find the Fourier sine expansion of $$f(x)=x(x^2-3Lx+2L^2)$$ on $$[0,L]$$. Solution Since $$f(0)=f(L)=0$$ and $$f''(x)=6(x-L)$$, we see from ( Equation \\ref{eq:11.3.5}) that \\begin{aligned} b_n&=- {12L\\over n^2\\pi^2}\\int_0^L(x-L)\\sin{n\\pi x\\over L}\\,dx\\\\ &=\\left.{12L^2\\over n^3\\pi^3}\\left[(x-L)\\cos{n\\pi x\\over L}\\right|_{0}^{L} -\\int_0^L\\cos{n\\pi x\\over L}\\,dx\\right]\\\\ &=\\left.{12L^2\\over n^3\\pi^3}\\left[L-\\frac{L}{n\\pi}\\sin\\frac{n\\pi x}{L}\\right|_{0}^{L}\\right] ={12L^3\\over n^3\\pi^3}.\\end{aligned}\\nonumber Therefore $S(x)=\\frac{12L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{n^{3}}\\sin\\frac{n\\pi x}{L}\\nonumber$ Example $$\\PageIndex{7}$$ Find the mixed Fourier cosine expansion of $$f(x)=3x^3-4Lx^2+L^3$$ on $$[0,L]$$. Solution Since $$f'(0)=f(L)=0$$ and $$f''(x)=2(9x-4L)$$, we see from ( Equation \\ref{eq:11.3.6}) that \\begin{aligned} c_n&= -{16L\\over(2n-1)^2\\pi^2} \\int_0^L(9x-4L)\\cos{(2n-1)\\pi x\\over2L}\\,dx\\\\ &=\\left.-{32L^2\\over(2n-1)^3\\pi^3}\\left[(9x-4L)\\sin{(2n-1)\\pi x\\over2L} \\right|_{0}^{L}-9\\int_0^L\\sin{(2n-1)\\pi x\\over2L}\\right]\\,dx\\\\ &=\\left.-{32L^2\\over(2n-1)^3\\pi^3} \\left[(-1)^{n+1}5L+{18L\\over(2n-1)\\pi}\\cos{(2n-1)\\pi x\\over2L} \\right|_{0}^{L}\\right] \\\\ &={32L^3\\over(2n-1)^3\\pi^3} \\left[(-1)^n5+{18\\over(2n-1)\\pi}\\right].\\end{aligned}\\nonumber Therefore $C_{M}(x)=\\frac{32L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{(2n-1)^{3}}\\left[(-1)^{n}5+\\frac{18}{(2n-1)\\pi } \\right]\\cos\\frac{(2n-1)\\pi x}{2L}\\nonumber$ Example $$\\PageIndex{8}$$ Find the mixed Fourier sine expansion of $f(x)=x(2x^2-9Lx+12L^2)\\nonumber$ on $$[0,L]$$. Solution Since $$f(0)=f'(L)=0$$, and $$f''(x)=6(2x-3L)$$, we see from ( Equation \\ref{eq:11.3.7}) that \\begin{aligned} d_{n}&=-\\frac{48L}{(2n-1)^{2}\\pi ^{2}}\\int_{0}^{L}(2x-3L)\\sin\\frac{(2n-1)\\pi x}{2L}dx \\\\ &=\\left.\\frac{96L^{2}}{(2n-1)^{3}\\pi ^{3}}\\left[(2x-3L)\\cos\\frac{(2n-1)\\pi x}{2L}\\right|_{0}^{L} - 2\\int_{0}^{L}\\cos\\frac{(2n-1)\\pi x}{2L}dx \\right] \\\\ &=\\left. \\frac{96L^{2}}{(2n-1)^{3}\\pi ^{3}}\\left[3L-\\frac{4L}{(2n-1)\\pi }\\sin\\frac{(2n-1)\\pi x}{2L}\\right|_{0}^{L} \\right] \\\\&=\\frac{96L^{3}}{(2n-1)^{3}\\pi ^{3}}\\left[3+(-1)^{n}\\frac{4}{(2n-1)\\pi } \\right] \\end{aligned}\\nonumber Therefore $S_{M}(x)=\\frac{96L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{(2n-1)^{3}}\\left[3+(-1)^{n}\\frac{4}{(2n-1)\\pi } \\right]\\sin\\frac{(2n-1)\\pi x}{2L}\\nonumber$", "$x = \\tan(t)$. We then get that \\begin{align} I(n) & = \\int_0^{\\infty} \\left(\\dfrac{2x}{1+x^2} \\right)^n dx = \\int_0^{\\pi/2} \\sin^n(2t) \\sec^2(t) dt = 2^n \\int_0^{\\pi/2} \\sin^n(t) \\cos^{n-2}(t) dt\\\\ & = 4\\int_0^{\\pi/2}\\sin^2(t) \\sin^{n-2}(2t)dt \\tag{$\\star$} \\end{align} Replacing $t$ by $\\pi/2-t$, we get $$I(n) = 4 \\int_0^{\\pi/2} \\cos^2(t) \\sin^{n-2}(2t) dt \\tag{\\perp}$$ Adding $(\\star)$ and $(\\perp)$, we get that $$2I(n) = 4 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt \\implies I(n) = 2 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt$$ I trust you can take it from here, using this post, which evaluates $\\displaystyle \\int_0^{\\pi} \\sin^{k}(t) dt$ and using Stirling (or) Wallis formula. - How do you go from $cos^2(t)\\sin^{n-2}(2t)$ to $\\sin^{n-2}(2t)$? – TCL May 7 '13 at 1:17 @TCL Updated the answer. Add $(\\star)$ and $(\\perp)$ to get it. – user17762 May 7 '13 at 1:19 Probably not how you want this done, but a useful and elegant method nonetheless: Note that $$f(x)=\\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\\to\\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$. $f^n \\equiv \\exp(n \\log f)$ and $n\\log(f(1+u))=-n\\frac{1}{2}u^2+n\\mathcal O(u^3)$. Then letting $v=\\sqrt{n} u$, $$\\int_0^\\infty n^{1/2}\\mathrm d x \\exp(-\\frac{1}{2}v^2+\\mathcal O(n^{-1/2})) \\approx \\int_{-\\infty}^\\infty \\mathrm d v\\, e^{-v^2/2} = \\sqrt{2\\pi}$$ Verification: WolframAlpha. Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals.", "find that \\begin{align}&c_3=1,\\ c_5=-4,\\ c_7=22,\\ c_{11}=1816,\\ c_{13}=-5056, \\\\&c_{17}=-2676224,\\ c_{19}=58473280. \\end{align} Conjecture 2. Let $p$ be an odd prime. Then both $$s_p:=\\frac{2^{(p-1)/2}}{\\sqrt p}\\text{per}\\left[\\sin2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}$$ and $$r_p:=\\frac{\\sqrt p}{2^{(p-1)/2}}\\text{per}\\left[\\csc2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}$$ are integers. Moreover, $$s_p\\equiv(-1)^{(p+1)/2}\\pmod p\\ \\ \\text{and}\\ \\ r_p\\equiv1\\pmod p.$$ Conjecture 3. Let $p$ be an odd prime. Then $$a_p:=\\frac1{2^{(p-1)/2}}\\text{per}\\left[\\sec2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}\\in\\mathbb Z.$$ If $p>3$ and $p\\equiv3\\pmod 4$, then $$\\text{per}\\left[\\sec2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le (p-1)/2}\\equiv \\text{per}\\left[\\cos2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le (p-1)/2} \\equiv-1\\pmod p.$$ Remark 2. Using Galois theory, I have shown that $a_p,c_p,r_p,s_p,t_p$ are rational numbers for any odd prime $p$. ### Enumerating all partitions induced by Voronoi diagrams for clustering Math Overflow Recent Questions - Fri, 01/18/2019 - 10:05 a classical results by M. Inaba et al. in \"Applications of Weighted Voronoi Diagrams and Randomization to Variance-Based k-CLustering\" (Theorem 3) says The number of Voronoi partitions of $n$ points by the Euclidean Voronoi diagram generated by $k$ points in $d$-dimensional space is $\\mathcal{O}(n^{dk})$, and all the Voronoi partitions can be enumerated in $\\mathcal{O}(n^{dk+1})$. They basically divide the $d$-dimensional space into equivalence classes where two sets of center $\\mu^1$ and $\\mu^2$ are equivalent if they lead to the same Voronoi Diagram. Then they show that the arrangement of the $nk(k-1)/2$ surfaces $$||x_i-\\mu_j||^2- ||x_i-\\mu_{j'}||^2 = 0$$ for each point $x_i$ and two cluster center $\\mu_j$ and $\\mu_{j'}$ coincides with the equivalence relation from Voronoi partitions. Next they argue that", "\\end{array}$$ (And remember $m=\\bigl\\lfloor \\frac n2 \\bigr\\rfloor$.) - You got to this expression: $$(1+i)^n=\\left(\\sqrt{2}\\right)^n \\left(\\cos\\left(n\\frac{\\pi}{4}\\right)+i\\sin\\left(n\\frac{\\pi}{4}\\right)\\right)$$ Now, you should note that this expression (without $\\left(\\sqrt{2}\\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look at $n=0,1,...,7$ and write the answer in the form: $$(1+i)^n=\\left(\\sqrt{2}\\right)^n\\left\\{ \\begin{array}{cc} 1 & n\\equiv0(\\mod8)\\\\ \\vdots & \\end{array}\\right.$$ - Like your way here (+1). – Basil R Sep 19 '12 at 9:19 Use: $$\\cos(n\\theta)=\\cos^n(\\theta)-\\frac{n(n-1)}{2!}\\cos^{n-2}(\\theta)\\sin^2(\\theta)+\\frac{n(n-1)(n-2)(n-3)}{4!}\\cos^{n-4}(\\theta)\\sin^4(\\theta)-...$$ and $$\\sin(n\\theta)=n\\cos^{n-1}(\\theta)\\sin(\\theta)-\\frac{n(n-1)(n-2)}{3!}\\cos^{n-3}(\\theta)\\sin^3(\\theta)-...$$ - Nice! well-done, my friend, and hoping you're reading this as your new day begins! – amWhy Mar 21 at 1:48 @amWhy: Thanks; So GN to me. ;-) – Babak S. Mar 21 at 20:25 You have a closed form already if you do just a little more work: \\begin{align*} (1+i)^n&=\\left(\\sqrt{2}\\right)^n\\left(\\cos\\frac{n\\pi}4+i\\sin\\frac{n\\pi}4\\right)\\\\ &=2^{n/2}\\left(\\cos\\frac{n\\pi}4+i\\sin\\frac{n\\pi}4\\right)\\\\ &=\\begin{cases} 2^{n/2},&\\text{if }n\\equiv 0\\pmod 8\\\\ 2^{n/2}\\left(\\frac{\\sqrt2}2+\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 1\\pmod 8\\\\ 2^{n/2}i,&\\text{if }n\\equiv 2\\pmod 8\\\\ 2^{n/2}\\left(\\frac{\\sqrt2}2-\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 3\\pmod 8\\\\ -2^{n/2},&\\text{if }n\\equiv 4\\pmod 8\\\\ -2^{n/2}\\left(\\frac{\\sqrt2}2+\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 5\\pmod 8\\\\ -2^{n/2}i,&\\text{if }n\\equiv 6\\pmod 8\\\\ -2^{n/2}\\left(\\frac{\\sqrt2}2-\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 7\\pmod 8\\;. \\end{cases}\\\\ \\end{align*} A function defined by cases is still a closed form. You can do better, though, if you allow the exponential form of the complex number: $1+i=\\sqrt2 e^{i\\pi/4}$, so $$(1+i)^n=2^{n/2}e^{in\\pi/4}\\;.$$ -", "Now to determine whether -1 is a square mod 467, we note that $467 \\equiv 3 \\mod{4}$. This means that -1 is not a square. On the other hand, to determine whether 2 is a square mod 467, we note that $467 \\equiv 3 \\mod{8}$. This means that 2 is also not a square mod 467. Hence we have (4) \\begin{align} \\left(\\frac{-2}{467}\\right) = \\left(\\frac{-1}{467}\\right)\\left(\\frac{2}{467}\\right) = (-1)(-1) = 1, \\end{align} and so we see that -2 is a square mod 467. Similarly we have (5) \\begin{split} \\left\\{\\begin{array}{cc}569 \\equiv 1 \\mod{4}\\\\569 \\equiv 1 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{569}\\right) = \\left(\\frac{-1}{569}\\right)\\left(\\frac{2}{569}\\right) = (1)(1) = 1\\\\ \\left\\{\\begin{array}{cc}821 \\equiv 1 \\mod{4}\\\\821 \\equiv 5 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{821}\\right) = \\left(\\frac{-1}{821}\\right)\\left(\\frac{2}{821}\\right) = (1)(-1) = -1\\\\ \\left\\{\\begin{array}{cc}383 \\equiv 3 \\mod{4}\\\\383 \\equiv 7 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{383}\\right) = \\left(\\frac{-1}{383}\\right)\\left(\\frac{2}{383}\\right) = (-1)(1) = -1.\\\\ \\end{split} $\\square$ # Gauss' Lemma Having practiced the rules we learned last time, let's prove the result concerning the Legendre symbol $\\left(\\frac{2}{p}\\right)$. To do so, we'll need the following Gauss' Lemma: For p an odd prime and $p \\nmid a$, let $\\displaystyle n=\\#\\left\\{1 \\leq j \\leq \\frac{p-1}{2}: \\begin{array}{cc}\\mbox{ the least non-neg residue of }\\\\ja \\mbox{ is greater than}\\frac{p}{2}\\end{array}\\right\\}.$ Then $\\left(\\frac{a}{p}\\right) = (-1)^n$. Before we prove this result, let's see it in practice. #### Example: Is 5 a square mod 11? Let's try to compute $\\left(\\fra{5}{11}\\right)$. For this, we're", "(2x)}}{{\\cos (2x)}} & = - 1\\\\ \\tan \\left( {2x} \\right) & = - 1\\end{align*} So, solving $$\\sin (2x) = - \\cos (2x)$$ is the same as solving $$\\tan (2x) = - 1$$.At some level we didn’t need to do this for this problem as all we’re looking for is angles in which sine and cosine have the same value, but opposite signs. However, for other problems this won’t be the case and we’ll want to convert to tangent. Looking at our trusty unit circle it appears that the solutions will be, \\begin{align*}2x = \\frac{{3\\pi }}{4} + 2\\pi n,& \\quad n = 0, \\pm 1, \\pm 2, \\ldots \\\\ 2x = \\frac{{7\\pi }}{4} + 2\\pi n,& \\quad n = 0, \\pm 1, \\pm 2, \\ldots \\end{align*} Or, upon dividing by the 2 we get the solutions \\begin{align*}x = \\frac{{3\\pi }}{8} + \\pi n,& \\quad n = 0, \\pm 1, \\pm 2, \\ldots \\\\ x = \\frac{{7\\pi }}{8} + \\pi n,& \\quad n = 0, \\pm 1, \\pm 2, \\ldots \\end{align*} No interval was given so we’ll stop here. 9. $$2\\sin \\left( \\theta \\right)\\cos \\left( \\theta \\right) = 1$$ Show Solution Again, we need to do a little work to get this equation into a form we can handle. The easiest way to do this one is to recall one", "same as finding the limit of $sin(x)/x$ times $x^a\\ sin(1/x)$, which is correct of both limits exists, but which is incorrect when they do not Nov 22 '17 at 19:47 • @Skyris It's just applying the limit product rule. See the addition. Nov 22 '17 at 22:06 So we have this function $f$: \\begin{align} f(x)=x^\\alpha \\sin(x)\\sin\\left( \\frac 1 x\\right) \\end{align} Okay, let's show that the limit of $x\\to 0$ does not exist for $\\alpha\\leq -1$ using your thoughts with finding sequences. Let us use $a_n= \\frac{2}{\\pi n}$ (only one sequence is sometimes enough!). Now we have: \\begin{align} f(a_n)= \\left(\\frac{\\sin\\left(\\frac{2}{\\pi n}\\right)}{\\frac{2}{\\pi n}} \\right)\\left(\\frac{2}{\\pi n}\\right)^{\\alpha+1} \\sin\\left(\\frac{\\pi n}{2}\\right) \\end{align} So we can write: \\begin{align} f(a_n) = b_n \\left(\\frac{2}{\\pi n}\\right)^{\\alpha+1} \\sin\\left(\\frac{\\pi n}{2}\\right) \\end{align} Where $b_n \\to 1$ (why?). Now look at the following: \\begin{align} \\limsup_{n\\to\\infty}f(a_n) = \\begin{cases} 1 & \\text{ if } &\\alpha=-1\\\\ \\infty & \\text{ if } & \\alpha<-1 \\end{cases} \\end{align} Similarly we have: \\begin{align} \\liminf_{n\\to\\infty}f(a_n) = \\begin{cases} -1 & \\text{ if } &\\alpha=-1\\\\ -\\infty & \\text{ if } & \\alpha<-1 \\end{cases} \\end{align} You see that the limit does not even exist for one sequence. Therefore the limit cannot exist. • Thank you though, I finally get it. Nov 22 '17 at 20:03 • @Skyris yes I see it, I removed the remark that was only true for real numbers. Nov", "4\\pi^2n^2$ – zhw. Jul 25 '15 at 18:36 Hint: For any $n \\in \\mathbb{N}$ and any $x \\in [0,a]$, we have: $|f_n(x)| = \\left|\\sin\\left(\\sqrt{4\\pi^2n^2+x}\\right)\\right|$ $= \\left|\\sin\\left(\\sqrt{4\\pi^2n^2+x} - 2\\pi n\\right)\\right|$ $\\le \\left|\\sqrt{4\\pi^2n^2+x}-2\\pi n\\right|$ $= \\sqrt{4\\pi^2n^2+x}-2\\pi n$ $= \\dfrac{x}{\\sqrt{4\\pi^2n^2+x}+2\\pi n}$. Can you continue the above work to get a uniform bound on $|f_n(x)|$ over all $x \\in [0,a]$? EDIT Spoiler: Continuing from above: $|f_n(x)| \\le \\dfrac{x}{\\sqrt{4\\pi^2n^2+x}+2\\pi n} \\le \\dfrac{a}{\\sqrt{4\\pi^2n^2}+2\\pi n} = \\dfrac{a}{4\\pi n}$ for all $n \\in \\mathbb{N}$ and all $x \\in [0,a]$. So for any $\\epsilon > 0$, choose an integer $N > \\dfrac{a}{4\\pi\\epsilon}$. Then for all $n \\ge N$ and all $x \\in [0,a]$, you have $|f_n(x)| \\le \\dfrac{a}{4\\pi n} < \\epsilon$. Hence, $f_n \\to 0$ uniformly on $[0,a]$. • +1. Why the two downvotes, though?! – triple_sec Jul 25 '15 at 6:45 I think the Arzelà–Ascoli theorem is an overkill here. Let me show you a proof based on more elementary methods. Claim 1: Define, for each $n\\in\\mathbb N$, $g_n:[0,a]\\to\\mathbb R$ as $$g_n(x)\\equiv\\sqrt{4\\pi^2 n^2+x}-\\sqrt{4\\pi^2 n^2}\\quad\\text{for every x\\in[0,a]}.$$ Then, $(g_n)_{n\\in\\mathbb N}$ converges to $0$ uniformly on $[0,a]$. Proof: First, let me show pointwise convergence. For any $x\\in[0,a]$, \\begin{align*} g_n(x)=&\\,\\frac{\\left(\\sqrt{4\\pi^2 n^2+x}-\\sqrt{4\\pi^2 n^2}\\right)\\left(\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}\\right)}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\\\ =&\\,\\frac{(4\\pi^2n^2+x)-(4\\pi^2 n^2)}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\\\ =&\\,\\frac{x}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\to0\\quad\\text{as $n\\to\\infty$,} \\end{align*} given that the denominator diverges. Uniform converges follows from the fact that $0\\leq g_n(x)\\leq", "of the time - for instance, how might you solve the case where $n=0$? $n=1$? $n=2$? Can you generalise the approach, either for all $n$, all integer $n$, or all even (or odd) integer $n$? - \\begin{align} {\\cal I}_{n} &= {1 \\over n + 1}\\int_{x\\ =\\ 0}^{x\\ =\\ \\pi/2}\\cos\\pars{x} \\,\\dd\\bracks{\\sin^{n + 1}\\pars{x}} = -\\,{1 \\over n+ 1}\\int_{0}^{\\pi/2}\\sin^{n + 1}\\pars{x}\\bracks{-\\sin\\pars{x}}\\,\\dd x \\\\[3mm]&={1 \\over n+ 1}\\int_{0}^{\\pi/2} \\sin^{n}\\pars{x}\\bracks{1 -\\cos^{2}\\pars{x}}\\,\\dd x ={1 \\over n+ 1}\\int_{0}^{\\pi/2} \\sin^{n}\\pars{x}\\,\\dd x - {1 \\over n + 1}\\,{\\cal I}_{n} \\end{align} \\begin{align} {\\cal I}_{n}&={1 \\over n + 2}\\int_{0}^{\\pi/2}\\sin^{n}\\pars{x}\\,\\dd x ={n \\over n + 2}\\bracks{{1 \\over n}\\int_{0}^{\\pi/2}\\sin^{n - 2}\\pars{x}\\,\\dd x} - {1 \\over n + 2}\\int_{0}^{\\pi/2}\\sin^{n - 2}\\pars{x}\\cos^{2}\\pars{x}\\,\\dd x \\\\[3mm]&= {n \\over n + 2}\\,{\\cal I}_{n - 2} - {1 \\over n + 2}\\,{\\cal I}_{n - 2} = {n - 1 \\over n + 2}\\,{\\cal I}_{n - 2} \\end{align} $$\\color{#0000ff}{\\large% {\\cal I}_{n} = \\int_{0}^{\\pi/2}\\sin^{n}\\pars{x}\\cos^{2}\\pars{x} \\,\\dd x = {n - 1 \\over n + 2}\\,{\\cal I}_{n - 2}\\,,\\quad n \\geq 2\\,,\\qquad \\left\\vert% \\begin{array}{rcl} {\\cal I}_{0} & = & {\\pi \\over 4} \\\\[2mm] {\\cal I}_{1} & = & {1 \\over 3} \\end{array}\\right.}$$ - (+1) nice answer. If you solve the recurrence relation you got and compare it with the solution using the beta function you will have the same answer. – Mhenni Benghorbal Jan 12 '14 at 6:23 @MhenniBenghorbal Also, the", "## Precalculus (6th Edition) Blitzer The solution on the interval $[0,2\\pi )$, correct to four decimal places is $0.4636$ , $2.6780$, $3.6052$ or $5.8195$. Let us consider the provided expression: $5{{\\sin }^{2}}x-1=0$ Add 1 on both sides: \\begin{align} & 5{{\\sin }^{2}}x-1+1=0+1 \\\\ & 5{{\\sin }^{2}}x=1 \\end{align} And divide by 5 on both sides: \\begin{align} & \\frac{5{{\\sin }^{2}}x}{5}=\\frac{1}{5} \\\\ & {{\\sin }^{2}}x=\\frac{1}{5} \\end{align} And put the square root on both sides: \\begin{align} & \\sin x=\\pm \\sqrt{\\frac{1}{5}} \\\\ & \\approx \\pm 0.4472 \\end{align} Since, the value of $\\sin x\\approx +0.4472$ $x\\approx {{\\sin }^{-1}}\\left( +0.4472 \\right)$ And the value of $x$ from $\\sin x\\approx -0.4472$ is: $x\\approx {{\\sin }^{-1}}\\left( -0.4472 \\right)$ We have to calculate the equation with the help of a calculator in radian mode: $\\theta ={{\\sin }^{-1}}\\left( +0.4472 \\right)\\approx 0.4636$ $\\theta ={{\\sin }^{-1}}\\left( -0.4472 \\right)\\approx -0.4636$ So, $x\\approx 0.4636$ Also, the sine value is positive in the II quadrant and the value of $\\pi$ is $3.14159$. Therefore, \\begin{align} & x=\\pi -0.4636 \\\\ & =3.14159-0.4636 \\\\ & \\approx 2.6780 \\end{align} The sine value is negative in the III and IV quadrants and the value of $\\pi$ is $3.14159$. So, \\begin{align} & x=\\pi +0.4636 \\\\ & =3.14159+0.4636 \\\\ & \\approx 3.6052 \\end{align} Or, \\begin{align} & x=2\\pi -0.4636 \\\\ & =2\\left( 3.14159 \\right)-0.4636 \\\\ & =6.28318-0.4636 \\\\ & \\approx 5.8195 \\end{align} Thus,", "# Finding the $n^{\\text{th}}$ term of $-1,-1,-1,-1,1,1,1,1,...$ as a repeating 8-block In my work I came across that sequence $-1,-1,-1,-1,1,1,1,1,\\dots$ and repeating this 8-block so on forever Now I cant find an ( e.g. trigonometric/complex ) expression $f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence starting with $n=2,3,4,5,6,7,8,9,…$ and so on forever. ## 7 Answers Consider the function $f(n)$ such that $$f(n) = \\left \\{ \\begin{array}[ll] (-1 & \\textrm{ if } n \\textrm{ mod } 8 \\equiv 0,1,2,3 \\\\ 1 & \\textrm{ if } n \\textrm{ mod } 8 \\equiv 4,5,6,7 \\end{array} \\right.$$ • The most unambiguous way. Nov 26 '15 at 17:04 I will assume that you really are starting at $2$. If the start is different, modification is easy. Let $g(n)=\\left\\lfloor\\frac{n+2}{4}\\right\\rfloor$. Here $\\lfloor x\\rfloor$ is the \"floor\" function, the greatest integer $\\le x$. Then your sequence is given by $(-1)^{g(n)}$. By symmetry, we can assume $$f(n) = a_1 \\sin\\frac{\\pi n}{4} +a_3 \\sin\\frac{3\\pi n}{4} +b_1 \\cos\\frac{\\pi n}{4} +b_3 \\cos\\frac{3\\pi n}{4}.$$ Then \\begin{aligned} f(1) &= \\frac{a_1}{\\sqrt{2}} +\\frac{a_3}{\\sqrt{2}} + \\frac{b_1}{\\sqrt{2}} -\\frac{b_3}{\\sqrt{2}} = -1\\\\ f(2) &= a_1 - a_3 = -1\\\\ f(3) &= \\frac{a_1}{\\sqrt{2}} +\\frac{a_3}{\\sqrt{2}} - \\frac{b_1}{\\sqrt{2}} +\\frac{b_3}{\\sqrt{2}} = -1 \\\\ f(4) &= -b_1 - b_3 = -1. \\end{aligned} So \\begin{aligned} a_1 &= -\\frac{1}{2} -\\frac{1}{\\sqrt 2}, \\\\ a_3 &= +\\frac{1}{2} -\\frac{1}{\\sqrt 2}, \\\\ b_1 &= b_3 =", "} n \\text { and } a_{n}$$ (vii) Given $$a=8, a_{n}=62, s_{n}=210, \\text { find } n \\text { and } d$$ (viii) Given $$a_{n}=4, d=2, S_{n}=-14, \\text { find } n \\text { and } a$$ (ix) Given $$a=3, n=8, S=192, \\text { find } d$$ (x) Given $$I=28, S=144 \\text { and there are total } 9$$ find a Ans : (i) Given that $$a=5, d=3, a_{n}=50$$ $$\\begin{array}{l}{\\text { As } a_{n}=a+(n-1) d} \\\\ {\\therefore 50=5+(n-1) 3} \\\\ {45=(n-1) 3} \\\\ {15=n-1} \\\\ {n=16}\\end{array}$$ \\begin{aligned} S_{n} &=\\frac{n}{2}\\left[a+a_{n}\\right] \\\\ S_{16} &=\\frac{16}{2}[5+50] \\\\ &=8 \\times 55 \\\\ &=440 \\end{aligned} (ii) Given that $$a=7, a_{13}=35$$ $$\\begin{array}{l}{\\text { As } a_{n}=a+(n-1) d_{r}} \\\\ {\\therefore a_{13}=a+(13-1) d} \\\\ {35=7+12 d} \\\\ {35-7=12 d} \\\\ {28=12 d} \\\\ {d=\\frac{7}{3}}\\end{array}$$ \\begin{aligned} S_{n} &=\\frac{n}{2}\\left[a+a_{n}\\right] \\\\ S_{13} &=\\frac{n}{2}\\left[a+a_{13}\\right] \\\\ &=\\frac{13}{2}[7+35] \\\\ &=\\frac{13 \\times 42}{2}=13 \\times 21 \\\\ &=273 \\end{aligned} (iii) Given that $$a_{12}=37, d=3$$ $$\\begin{array}{l}{\\text { As } a_{n}=a+(n-1) d} \\\\ {a_{12}=a+(12-1) 3} \\\\ {37=a+33} \\\\ {a=4}\\end{array}$$ $$\\begin{array}{l}{S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]} \\\\ {S_{n}=\\frac{12}{2}[4+37]} \\\\ {S_{n}=6(41)} \\\\ {S_{n}=246}\\end{array}$$ (iv) Given that $$a_{3}=15, S_{10}=125$$ $$\\begin{array}{l}{\\text { As } a_{n}=a+(n-1) d} \\\\ {a_{3}=a+(3-1) d} \\\\ {15=a+2 d(i)} \\\\ {S_{n}=\\frac{n}{2}[2 a+(n-1) d]} \\\\ {S_{19}=\\frac{10}{2}[2 a+(10-1) d]} \\\\ {125=5(2 a+9 d)} \\\\ {25=2 a+9 d}\\end{array}$$ On multiplying equation (I) by 2, we obtain 30 = 2a + 4d ...(iii) on subtracting equation (iii)", "and so the angle is between $(2^{-1}+2^{-4}+2^{-6})\\pi$ and $(2^{-1}+2^{-4}+2^{-6}+2^{-8})\\pi$ In degrees you replace $\\pi$ with $180$, so those are $104.06\\ldots$ and $104.77\\ldots$ The recurrence follows from the addition formula : $2\\cos(2x) = 2\\cos^2(x)-2\\sin^2(x) = 4\\cos^2(x)-2 = (2\\cos(x))^2-2$ Suppose you call $a_n \\in [0 ; \\pi]$ the angle whose cosine is $2x_n$. If $x_n\\ge 0$ then $a_n \\in [0 ; \\pi/2]$ and then $a_{n+1} = 2a_n$, so the binary digits of $a_n/\\pi$ are $.0$ followed with the binary digits of $a_{n+1}/\\pi$ If $x_n \\le 0$ then $a_n \\in [\\pi/2 ; \\pi]$ and then $a_{n+1} = 2\\pi-2a_n$, so the binary digits of $a_n/\\pi$ are $.1$ followed with the inverted binary digits of $a_{n+1}/\\pi$ Thus $a_{n+1} = \\pm 2 a_n \\mod {2\\pi}$, and by induction, $a_n = \\pm 2^n a_0 \\pmod {2\\pi}$ where the sign depends on the parity of the number of negative $x_k$ encountered for $0 \\le k < n$. The $n$th digit is $0$ if and only if $2^n a_0 \\in [0 ; \\pi] \\pmod {2\\pi}$, which means $\\pm a_n \\in [0;\\pi] \\pmod {2\\pi}$ with the same sign. But since $a_n \\in [0;\\pi]$, the digit is $0$ if the sign was $+$ and it is $1$ is the sign was $-$. And so the $n$th binary digit correspond to the parity of the number of negative cosines encountered", "X'' = 0$ with $X(0)=0$, $X(\\ell) = 0$ are $$X(x) = e^x \\sin(n \\pi x/\\ell), \\; k = -1 - n^2 \\pi^2/\\ell^2$$ for positive integers $n$. Thus you should get $$u(x,t) = \\sum_{n=1}^\\infty c_n e^{x - (1 + n^2 \\pi^2/\\ell^2) t} \\sin(n \\pi x/\\ell)$$ The $c_n$ must then be determined to satisfy your initial condition (Hint: Fourier series). • That was $u(l,t) = 0$ – ANUPAM BISHT Oct 14 '16 at 18:38 • Ah, it looked like a $1$. I'll edit. – Robert Israel Oct 14 '16 at 19:56 \\begin{align} \\varphi\\pars{x,t} & = \\sum_{k\\ \\in\\ \\Omega}A_{k}\\pars{t}\\sin\\pars{kx}\\,,\\qquad \\Omega \\equiv \\braces{k \\in \\mathbb{R}\\ \\mid\\ k = n\\,{\\pi \\over \\ell}\\,,\\ n = 1,2,3,\\ldots}\\label{1}\\tag{1} \\end{align} Expression \\eqref{1}, for $\\ds{\\varphi\\pars{x,t}}$, already satisfies the boundary conditions $\\ds{\\pars{~\\varphi\\pars{0,t} = \\varphi\\pars{\\ell,t} = 0\\,,\\ \\forall\\ t~}}$. Moreover, \\begin{align} \\sum_{q\\ \\in\\ \\Omega}\\braces{\\totald{A_{q}\\pars{t}}{t} + q^{2}}\\sin\\pars{qx} = 0 \\end{align} Multiply both sides by $\\ds{\\left.\\sin\\pars{kx}\\right\\vert_{\\ k\\ \\in\\ \\Omega}}$ and integrate over $\\ds{\\pars{0,\\ell}}$. It leads to $$0 = \\totald{A_{k}\\pars{t}}{t} + k^{2}A_{k}\\pars{t}\\implies A_{k}\\pars{t} = A_{k}\\pars{0}\\expo{-k^{2}t}$$ $\\ds{\\varphi\\pars{x,t}}$ is reduced to: \\begin{align} \\varphi\\pars{x,t} & = \\sum_{k\\ \\in\\ \\Omega}A_{k}\\pars{0}\\expo{-k^{2}t}\\sin\\pars{kx}\\implies x = \\varphi\\pars{x,0} = \\sum_{k\\ \\in\\ \\Omega}A_{k}\\pars{0}\\sin\\pars{kx} \\\\[5mm] \\implies \\int_{0}^{\\ell}x\\sin\\pars{kx}\\,\\dd x & = \\sum_{q\\ \\in\\ \\Omega} A_{q}\\pars{0}\\int_{0}^{\\ell}\\sin\\pars{kx}\\sin\\pars{qx}\\,\\dd x = {\\ell \\over 2}\\,A_{k}\\pars{0} \\\\[5mm] \\implies A_{k}\\pars{0} & = {2 \\over \\ell}\\int_{0}^{\\ell}x\\sin\\pars{kx}\\,\\dd x = \\pars{-1}^{n + 1}\\,{2 \\over n\\pi/\\ell}\\,,\\qquad k = n\\,{\\pi \\over \\ell} \\end{align} Then, $$\\varphi\\pars{x,t} = {2\\ell", "# Difference Equation $y_{n+3} − 3 y_{n+1} + 2 y_n = (−2)^n$ I get the solution to be $y_n = A(-2)^n + Bn + C + \\frac{1}{9}n(-2)^{n-1}$ but wolfram alpha gets $y(n) = c_1 (-2)^n+c_2+c_3 n+\\frac{1}{27} 2^{n-1} e^{i \\pi n} (4-3 n)$ Explanation required. Help is much appreciated. - Observe that $e^{i\\pi n}=\\cos(\\pi n)+i\\sin(\\pi n)=\\cos(\\pi n)=(-1)^n$. Hence, you have: \\begin{align*} y(n)& = c_1 (-2)^n+c_2+c_3 n+\\frac{1}{27} 2^{n-1} e^{i \\pi n} (4-3 n) \\\\ & = c_1 (-2)^n+c_2+c_3 n+\\frac{1}{27} (-2)^{n-1} (3n-4) \\\\ & = c_1 (-2)^n+c_2+c_3 n+\\frac{1}{9} n (-2)^{n-1} - \\frac{4}{27} (-2)^{n-1} \\\\ & = \\left( c_1+\\frac{2}{27}\\right)(-2)^n+c_3 n+c_2+\\frac{1}{9} n (-2)^{n-1} \\end{align*} The last line is your solution. - Thank you very much! – Alan Jan 16 '12 at 15:39 @Alan: you should accept answers that you like, so people would know that you got what you were looking for. – Dennis Gulko Jan 16 '12 at 15:44 HINT $\\$ yours $\\rm\\: =\\: \\left<f_1,\\:f_2,\\:f_3,\\:f_4\\right>\\: =\\: \\left<f_1,\\:f_2,\\:f_3,\\:f_4+c\\ f_1\\right>\\: =\\:$ alpha, for $\\rm\\ c = 2/27\\:.$ -", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of \"2017 AIME I Problems/Problem 9\" ## Problem 9 Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$. ## Solution 1 Writing out the recursive statement for $a_n, a_{n-1}, \\dots, a_{10}$ and summing them gives $$a_n+\\dots+a_{10}=100(a_{n-1}+\\dots+a_{10})+n+\\dots+10$$ Which simplifies to $$a_n=99(a_{n-1}+\\dots+a_{10})+\\frac{1}{2}(n+10)(n-9)$$ Therefore, $a_n$ is divisible by 99 if and only if $\\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\\boxed{045}$. ## Solution 2 $$a_n \\equiv a_{n-1} + n \\pmod {99}$$ By looking at the first few terms, we can see that $$a_n \\equiv 10+11+12+ \\dots + n \\pmod {99}$$ This implies $$a_n \\equiv \\frac{n(n+1)}{2} - \\frac{10*9}{2} \\pmod {99}$$ Since $a_n \\equiv 0 \\pmod {99}$, we", "apparent – Torsten Hĕrculĕ Cärlemän Jul 15 '13 at 11:25 $$\\prod_{k=1}^{n-1}\\sin\\left(\\frac{k\\pi}{n}\\right)=\\frac{n}{2^{n-1}}$$ Here, let $$x=\\prod_{k=1}^{999}\\sin\\left(\\frac{k\\pi}{1999}\\right) \\tag{1}$$ Since $\\sin t=\\sin (\\pi-t)$, therefore, $$x=\\prod_{k=1}^{999}\\sin\\left(\\frac{(1999-k)\\pi}{1999}\\right)=\\prod_{k=1000}^{1998}\\sin\\left(\\frac{k\\pi}{1999}\\right) \\tag{2}$$ Multiplying equation $(1)$ by equation $(2)$ gives, $$x^2=\\prod_{k=1}^{1998}\\sin\\left(\\frac{k\\pi}{1999}\\right)=\\frac{1999}{2^{1998}}$$ $$\\implies x=\\frac{\\sqrt{1999}}{2^{999}}$$ We took $x>0$ because all angles are in $1^{st}$ and $2^{nd}$ quadrant. • Thank you so much for your concern. But how can I prove the first indentity: $$\\prod_{k=1}^{n}\\sin\\left(\\frac{k\\pi}{n}\\right)=\\frac{n}{2^{n-1}}$$ – Young Mar 8 '13 at 6:51 • That's pretty neat! – Inceptio Mar 8 '13 at 6:51 • Can you please explain how u got $\\prod_{k=1}^{n-1}\\sin\\big(\\frac{k\\pi}{n}\\big)=\\frac{n}{2^{n-1}}$ – Phani Raj Mar 8 '13 at 6:53 • @PhaniRaj: I updated my answer. Have a look. – Aang Mar 8 '13 at 6:58 • @Avatar Nice answer! – Start wearing purple Jul 15 '13 at 11:17 I have an even more precise bound: the product is less than $$(999)! \\left(\\frac{\\pi}{1999}\\right)^{999}$$ Use Stirling approximation to get that this in turn is less than $$\\sqrt{999 e} \\left ( \\frac{999 \\pi}{1999 e}\\right )^{999} \\approx 4 \\times 10^{-237}$$ The first digits of your number are: 0.00 To get this estimate notice that all terms of the product are less than $1$ and that $|\\sin(x)| \\le |x|$. To be more precise your number is between $0$ and $10^{-80}$. • Hows this helpful? – Inceptio Mar 8 '13 at 6:34 • @Inceptio: I know more", "\\end{align} Similarly for $\\sin B$, we get \\begin{align} & {{\\sin }^{2}}B+{{\\cos }^{2}}B=1 \\\\ & {{\\sin }^{2}}B=1-{{\\cos }^{2}}B \\\\ & \\sin B=\\sqrt{1-{{\\left( \\dfrac{12}{13} \\right)}^{2}}} \\\\ & \\sin B=\\dfrac{5}{13} \\\\ \\end{align} Now, we know that $\\cos \\left( A+B \\right)=\\cos A.\\cos B-\\sin A.\\sin B$ Putting values $\\cos A=\\dfrac{4}{5},\\cos B=\\dfrac{3}{5},\\sin A\\dfrac{3}{5}\\,and\\,\\sin B=\\dfrac{5}{13}$ in the above expression we get, $\\cos \\left( A+B \\right)=\\dfrac{4}{5}\\times \\dfrac{12}{13}-\\dfrac{3}{5}\\times \\dfrac{5}{13}$ \\begin{align} & \\cos \\left( A+B \\right)=\\dfrac{48}{65}-\\dfrac{3}{13} \\\\ & \\cos \\left( A+B \\right)=\\dfrac{48-15}{65} \\\\ & \\cos \\left( A+B \\right)=\\dfrac{33}{65}=\\cos C \\\\ \\end{align} Hence we proved that $\\cos \\left( A+B \\right)=\\cos C$ i.e. $A+B=C$ and hence, we proved ${{\\cos }^{-1}}\\left( \\dfrac{4}{5} \\right)+{{\\cos }^{-1}}\\left( \\dfrac{12}{13} \\right)={{\\cos }^{-1}}\\left( \\dfrac{33}{65} \\right)$ Note: Whenever you are solving questions related to the inverse trigonometric functions try to think of ${{\\cos }^{-1}}\\theta ,{{\\sin }^{-1}}\\theta$ etc. as the value of angle $\\theta$ in the radians. And always try to convert inverse trigonometric functions to simple trigonometric functions by applying sine, cosine etc. on both sides of the equation. You also need to remember all basic trigonometric and inverse trigonometric identities in order to solve these kinds of questions and after that you will be able to easily figure out which identity to apply and when to apply it.", "\\times \\dfrac {3}{5} \\text { or } \\dfrac {48}{65}-\\dfrac {15}{65} \\\\ \\sin (\\alpha+\\beta)=\\dfrac {33}{65} \\end{array}$$ 3. Find the exact value of $$\\sin 15^{\\circ}$$ Recall that there are multiple angles that add or subtract to equal any angle. Choose whichever formula that you feel more comfortable with. \\begin{aligned} \\sin 15^{\\circ} &=\\sin \\left(45^{\\circ}-30^{\\circ}\\right) \\\\ &=\\sin 45^{\\circ} \\cos 30^{\\circ}+\\cos 45^{\\circ} \\sin 30^{\\circ} \\\\ \\sin 15^{\\circ} &=(.707) \\times(.866)+(.707) \\times(.5) \\\\ &=(.612262) \\times(.3535) \\\\ &=.2164 \\end{aligned} Example $$\\PageIndex{1}$$ Earlier, you were given a problem about Using the sine sum formula Solution With the sine sum formula, you can break the sine into easier to calculate quantities: \\begin{aligned} \\sin \\dfrac {7 \\pi}{12} &=\\sin \\left(\\dfrac {4 \\pi}{12}+\\dfrac {3 \\pi}{12}\\right) \\\\ &=\\sin \\left(\\dfrac {\\pi}{3}+\\dfrac {\\pi}{4}\\right) \\\\ &=\\sin \\left(\\dfrac {\\pi}{3}\\right) \\cos \\left(\\dfrac {\\pi}{4}\\right)+\\cos \\left(\\dfrac {\\pi}{3}\\right) \\sin \\left(\\dfrac {\\pi}{4}\\right) \\\\ &=\\left(\\dfrac {\\sqrt{3}}{2}\\right)\\left(\\dfrac {\\sqrt{2}}{2}\\right)+\\left(\\dfrac {1}{2}\\right)\\left(\\dfrac {\\sqrt{2}}{2}\\right) \\\\ &=\\dfrac {\\sqrt{6}}{4}+\\dfrac {\\sqrt{2}}{4} \\\\ &=\\dfrac {\\sqrt{6}+\\sqrt{2}}{4} \\end{aligned} Example $$\\PageIndex{2}$$ Find the exact value for $$\\sin 345^{\\circ}$$ Solution \\begin{aligned} \\sin 345^{\\circ} &=\\sin \\left(300^{\\circ}+45^{\\circ}\\right)=\\sin 300^{\\circ} \\cos 45^{\\circ}+\\cos 300^{\\circ} \\sin 45^{\\circ} \\\\ &=-\\dfrac {\\sqrt{3}}{2} \\cdot \\dfrac {\\sqrt{2}}{2}+\\dfrac {1}{2} \\cdot \\dfrac {\\sqrt{2}}{2}=-\\dfrac {\\sqrt{6}}{4}+\\dfrac {\\sqrt{2}}{4}=\\dfrac {\\sqrt{2}-\\sqrt{6}}{4} \\end{aligned} Example $$\\PageIndex{3}$$ Find the exact value for $$\\sin \\dfrac{17\\pi}{12}$$ Solution \\begin{aligned} \\sin \\dfrac {17 \\pi}{12} &=\\sin \\left(\\dfrac {9 \\pi}{12}+\\dfrac {8 \\pi}{12}\\right)=\\sin \\left(\\dfrac {3 \\pi}{4}+\\dfrac {2 \\pi}{3}\\right)=\\sin \\dfrac {3 \\pi}{4} \\cos \\dfrac {2 \\pi}{3}+\\cos \\dfrac {3 \\pi}{4} \\sin \\dfrac {2 \\pi}{3}", "i\\pi/4}$, $(1+\\mathrm i)^n$ is $(\\sqrt2)^n$ times $\\mathrm e^{\\mathrm in\\pi/4}$. • The real part of $\\mathrm e^{\\mathrm in\\pi/4}$ is $\\cos(n\\pi/4)$. Thus, $\\color{red}{x_n=\\ldots}$ - Thanks, so I think $2\\Re((1+i)^n)=2(\\sqrt{2}^n)\\cos(\\frac{n\\pi}{4})$, which can be evaluated depending on $n\\pmod{8}$? – hagar Jan 17 '12 at 16:59 Yes, depending on whether $n=0\\pmod 8$ or $n=\\pm1\\pmod 8$ or $n=\\pm2\\pmod 8$ or $n=\\pm3\\pmod 8$ or $n=4\\pmod 8$. – Did Jan 17 '12 at 17:02 This is a neat solution, that generalizes nicely : $a^n + \\bar{a}^n = 2 \\left|a\\right|^n \\cos( n \\arg (a) )$. – Michael Anderson Jan 18 '12 at 7:25 First, notice that $1+i = \\sqrt2(\\cos\\frac{\\pi}{4} + i \\sin\\frac{\\pi}{4}$) and $1-i = \\sqrt2(\\cos(-\\frac{\\pi}{4}) + i \\sin(-\\frac{\\pi}{4}))$. Using de Moivre's formula and the fact that $\\sin$ is odd and $\\cos$ is even, we get \\begin{align*} (1+i)^n + (1-i)^n &= 2(\\sqrt2)^n\\cos\\frac{n\\pi}{4} \\end{align*} - $(1+i)^n+(1-i)^n = 2 \\Re ((1+i)^n)$. Now expand $(1+i)^n$ using the binomial theorem. The real part is formed by the odd-numbered terms. Another approach is to note that $(1+i)/\\sqrt{2}$ is an 8-th root of unity and so $(1+i)^n$ depends only on $n \\bmod 8$ except for a power of $\\sqrt{2}$. - Thank you, lhf. – hagar Jan 17 '12 at 17:00 This is not a particularly elegant solution, but an alternative route is to simply note that $(1 \\pm i)^2 = \\pm 2i$. I" ]

[ "find that \\begin{align}&c_3=1,\\ c_5=-4,\\ c_7=22,\\ c_{11}=1816,\\ c_{13}=-5056, \\\\&c_{17}=-2676224,\\ c_{19}=58473280. \\end{align} Conjecture 2. Let $p$ be an odd prime. Then both $$s_p:=\\frac{2^{(p-1)/2}}{\\sqrt p}\\text{per}\\left[\\sin2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}$$ and $$r_p:=\\frac{\\sqrt p}{2^{(p-1)/2}}\\text{per}\\left[\\csc2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}$$ are integers. Moreover, $$s_p\\equiv(-1)^{(p+1)/2}\\pmod p\\ \\ \\text{and}\\ \\ r_p\\equiv1\\pmod p.$$ Conjecture 3. Let $p$ be an odd prime. Then $$a_p:=\\frac1{2^{(p-1)/2}}\\text{per}\\left[\\sec2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le(p-1)/2}\\in\\mathbb Z.$$ If $p>3$ and $p\\equiv3\\pmod 4$, then $$\\text{per}\\left[\\sec2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le (p-1)/2}\\equiv \\text{per}\\left[\\cos2\\pi\\frac{jk}p\\right]_{1\\le j,k\\le (p-1)/2} \\equiv-1\\pmod p.$$ Remark 2. Using Galois theory, I have shown that $a_p,c_p,r_p,s_p,t_p$ are rational numbers for any odd prime $p$. ### Enumerating all partitions induced by Voronoi diagrams for clustering Math Overflow Recent Questions - Fri, 01/18/2019 - 10:05 a classical results by M. Inaba et al. in \"Applications of Weighted Voronoi Diagrams and Randomization to Variance-Based k-CLustering\" (Theorem 3) says The number of Voronoi partitions of $n$ points by the Euclidean Voronoi diagram generated by $k$ points in $d$-dimensional space is $\\mathcal{O}(n^{dk})$, and all the Voronoi partitions can be enumerated in $\\mathcal{O}(n^{dk+1})$. They basically divide the $d$-dimensional space into equivalence classes where two sets of center $\\mu^1$ and $\\mu^2$ are equivalent if they lead to the same Voronoi Diagram. Then they show that the arrangement of the $nk(k-1)/2$ surfaces $$||x_i-\\mu_j||^2- ||x_i-\\mu_{j'}||^2 = 0$$ for each point $x_i$ and two cluster center $\\mu_j$ and $\\mu_{j'}$ coincides with the equivalence relation from Voronoi partitions. Next they argue that", "Now to determine whether -1 is a square mod 467, we note that $467 \\equiv 3 \\mod{4}$. This means that -1 is not a square. On the other hand, to determine whether 2 is a square mod 467, we note that $467 \\equiv 3 \\mod{8}$. This means that 2 is also not a square mod 467. Hence we have (4) \\begin{align} \\left(\\frac{-2}{467}\\right) = \\left(\\frac{-1}{467}\\right)\\left(\\frac{2}{467}\\right) = (-1)(-1) = 1, \\end{align} and so we see that -2 is a square mod 467. Similarly we have (5) \\begin{split} \\left\\{\\begin{array}{cc}569 \\equiv 1 \\mod{4}\\\\569 \\equiv 1 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{569}\\right) = \\left(\\frac{-1}{569}\\right)\\left(\\frac{2}{569}\\right) = (1)(1) = 1\\\\ \\left\\{\\begin{array}{cc}821 \\equiv 1 \\mod{4}\\\\821 \\equiv 5 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{821}\\right) = \\left(\\frac{-1}{821}\\right)\\left(\\frac{2}{821}\\right) = (1)(-1) = -1\\\\ \\left\\{\\begin{array}{cc}383 \\equiv 3 \\mod{4}\\\\383 \\equiv 7 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{383}\\right) = \\left(\\frac{-1}{383}\\right)\\left(\\frac{2}{383}\\right) = (-1)(1) = -1.\\\\ \\end{split} $\\square$ # Gauss' Lemma Having practiced the rules we learned last time, let's prove the result concerning the Legendre symbol $\\left(\\frac{2}{p}\\right)$. To do so, we'll need the following Gauss' Lemma: For p an odd prime and $p \\nmid a$, let $\\displaystyle n=\\#\\left\\{1 \\leq j \\leq \\frac{p-1}{2}: \\begin{array}{cc}\\mbox{ the least non-neg residue of }\\\\ja \\mbox{ is greater than}\\frac{p}{2}\\end{array}\\right\\}.$ Then $\\left(\\frac{a}{p}\\right) = (-1)^n$. Before we prove this result, let's see it in practice. #### Example: Is 5 a square mod 11? Let's try to compute $\\left(\\fra{5}{11}\\right)$. For this, we're", "since $$f'(x)=6Lx-6x^2$$, we see that $$f'(0)=f'(L)=0$$. Since $$f'''(x)=-12$$, we see from ( Equation \\ref{eq:11.3.4}) that if $$n\\ge1$$ then \\begin{aligned} a_n&=-{24L^2\\over n^3\\pi^3}\\int_0^L\\sin{n\\pi x\\over L}\\,dx =\\left.{24L^3\\over n^4\\pi^4}\\cos{n\\pi x\\over L}\\right|_{0}^{L}={24L^3\\over n^4\\pi^4}[(-1)^n-1]\\\\ &= \\left\\{\\begin{array}{cl} -{48L^3\\over(2m-1)^4\\pi^4}&\\mbox{if n=2m-1},\\\\ 0&\\mbox{if n=2m.} \\end{array}\\right.\\end{aligned}\\nonumber Therefore $C(x)={L^3\\over2}-{48L^3\\over\\pi^4}\\sum_{n=1}^\\infty{1\\over (2n-1)^4}\\cos{(2n-1)\\pi x\\over L}.\\nonumber$ Example $$\\PageIndex{6}$$ Find the Fourier sine expansion of $$f(x)=x(x^2-3Lx+2L^2)$$ on $$[0,L]$$. Solution Since $$f(0)=f(L)=0$$ and $$f''(x)=6(x-L)$$, we see from ( Equation \\ref{eq:11.3.5}) that \\begin{aligned} b_n&=- {12L\\over n^2\\pi^2}\\int_0^L(x-L)\\sin{n\\pi x\\over L}\\,dx\\\\ &=\\left.{12L^2\\over n^3\\pi^3}\\left[(x-L)\\cos{n\\pi x\\over L}\\right|_{0}^{L} -\\int_0^L\\cos{n\\pi x\\over L}\\,dx\\right]\\\\ &=\\left.{12L^2\\over n^3\\pi^3}\\left[L-\\frac{L}{n\\pi}\\sin\\frac{n\\pi x}{L}\\right|_{0}^{L}\\right] ={12L^3\\over n^3\\pi^3}.\\end{aligned}\\nonumber Therefore $S(x)=\\frac{12L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{n^{3}}\\sin\\frac{n\\pi x}{L}\\nonumber$ Example $$\\PageIndex{7}$$ Find the mixed Fourier cosine expansion of $$f(x)=3x^3-4Lx^2+L^3$$ on $$[0,L]$$. Solution Since $$f'(0)=f(L)=0$$ and $$f''(x)=2(9x-4L)$$, we see from ( Equation \\ref{eq:11.3.6}) that \\begin{aligned} c_n&= -{16L\\over(2n-1)^2\\pi^2} \\int_0^L(9x-4L)\\cos{(2n-1)\\pi x\\over2L}\\,dx\\\\ &=\\left.-{32L^2\\over(2n-1)^3\\pi^3}\\left[(9x-4L)\\sin{(2n-1)\\pi x\\over2L} \\right|_{0}^{L}-9\\int_0^L\\sin{(2n-1)\\pi x\\over2L}\\right]\\,dx\\\\ &=\\left.-{32L^2\\over(2n-1)^3\\pi^3} \\left[(-1)^{n+1}5L+{18L\\over(2n-1)\\pi}\\cos{(2n-1)\\pi x\\over2L} \\right|_{0}^{L}\\right] \\\\ &={32L^3\\over(2n-1)^3\\pi^3} \\left[(-1)^n5+{18\\over(2n-1)\\pi}\\right].\\end{aligned}\\nonumber Therefore $C_{M}(x)=\\frac{32L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{(2n-1)^{3}}\\left[(-1)^{n}5+\\frac{18}{(2n-1)\\pi } \\right]\\cos\\frac{(2n-1)\\pi x}{2L}\\nonumber$ Example $$\\PageIndex{8}$$ Find the mixed Fourier sine expansion of $f(x)=x(2x^2-9Lx+12L^2)\\nonumber$ on $$[0,L]$$. Solution Since $$f(0)=f'(L)=0$$, and $$f''(x)=6(2x-3L)$$, we see from ( Equation \\ref{eq:11.3.7}) that \\begin{aligned} d_{n}&=-\\frac{48L}{(2n-1)^{2}\\pi ^{2}}\\int_{0}^{L}(2x-3L)\\sin\\frac{(2n-1)\\pi x}{2L}dx \\\\ &=\\left.\\frac{96L^{2}}{(2n-1)^{3}\\pi ^{3}}\\left[(2x-3L)\\cos\\frac{(2n-1)\\pi x}{2L}\\right|_{0}^{L} - 2\\int_{0}^{L}\\cos\\frac{(2n-1)\\pi x}{2L}dx \\right] \\\\ &=\\left. \\frac{96L^{2}}{(2n-1)^{3}\\pi ^{3}}\\left[3L-\\frac{4L}{(2n-1)\\pi }\\sin\\frac{(2n-1)\\pi x}{2L}\\right|_{0}^{L} \\right] \\\\&=\\frac{96L^{3}}{(2n-1)^{3}\\pi ^{3}}\\left[3+(-1)^{n}\\frac{4}{(2n-1)\\pi } \\right] \\end{aligned}\\nonumber Therefore $S_{M}(x)=\\frac{96L^{3}}{\\pi ^{3}}\\sum_{n=1}^{\\infty}\\frac{1}{(2n-1)^{3}}\\left[3+(-1)^{n}\\frac{4}{(2n-1)\\pi } \\right]\\sin\\frac{(2n-1)\\pi x}{2L}\\nonumber$", "\\end{array}$$ (And remember $m=\\bigl\\lfloor \\frac n2 \\bigr\\rfloor$.) - You got to this expression: $$(1+i)^n=\\left(\\sqrt{2}\\right)^n \\left(\\cos\\left(n\\frac{\\pi}{4}\\right)+i\\sin\\left(n\\frac{\\pi}{4}\\right)\\right)$$ Now, you should note that this expression (without $\\left(\\sqrt{2}\\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look at $n=0,1,...,7$ and write the answer in the form: $$(1+i)^n=\\left(\\sqrt{2}\\right)^n\\left\\{ \\begin{array}{cc} 1 & n\\equiv0(\\mod8)\\\\ \\vdots & \\end{array}\\right.$$ - Like your way here (+1). – Basil R Sep 19 '12 at 9:19 Use: $$\\cos(n\\theta)=\\cos^n(\\theta)-\\frac{n(n-1)}{2!}\\cos^{n-2}(\\theta)\\sin^2(\\theta)+\\frac{n(n-1)(n-2)(n-3)}{4!}\\cos^{n-4}(\\theta)\\sin^4(\\theta)-...$$ and $$\\sin(n\\theta)=n\\cos^{n-1}(\\theta)\\sin(\\theta)-\\frac{n(n-1)(n-2)}{3!}\\cos^{n-3}(\\theta)\\sin^3(\\theta)-...$$ - Nice! well-done, my friend, and hoping you're reading this as your new day begins! – amWhy Mar 21 at 1:48 @amWhy: Thanks; So GN to me. ;-) – Babak S. Mar 21 at 20:25 You have a closed form already if you do just a little more work: \\begin{align*} (1+i)^n&=\\left(\\sqrt{2}\\right)^n\\left(\\cos\\frac{n\\pi}4+i\\sin\\frac{n\\pi}4\\right)\\\\ &=2^{n/2}\\left(\\cos\\frac{n\\pi}4+i\\sin\\frac{n\\pi}4\\right)\\\\ &=\\begin{cases} 2^{n/2},&\\text{if }n\\equiv 0\\pmod 8\\\\ 2^{n/2}\\left(\\frac{\\sqrt2}2+\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 1\\pmod 8\\\\ 2^{n/2}i,&\\text{if }n\\equiv 2\\pmod 8\\\\ 2^{n/2}\\left(\\frac{\\sqrt2}2-\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 3\\pmod 8\\\\ -2^{n/2},&\\text{if }n\\equiv 4\\pmod 8\\\\ -2^{n/2}\\left(\\frac{\\sqrt2}2+\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 5\\pmod 8\\\\ -2^{n/2}i,&\\text{if }n\\equiv 6\\pmod 8\\\\ -2^{n/2}\\left(\\frac{\\sqrt2}2-\\frac{\\sqrt2}2i\\right),&\\text{if }n\\equiv 7\\pmod 8\\;. \\end{cases}\\\\ \\end{align*} A function defined by cases is still a closed form. You can do better, though, if you allow the exponential form of the complex number: $1+i=\\sqrt2 e^{i\\pi/4}$, so $$(1+i)^n=2^{n/2}e^{in\\pi/4}\\;.$$ -", "that 2 is also not a square mod 467. Hence we have (3) \\begin{align} \\left(\\frac{-2}{467}\\right) = \\left(\\frac{-1}{467}\\right)\\left(\\frac{2}{467}\\right) = (-1)(-1) = 1, \\end{align} and so we see that -2 is a square mod 467. Similarly we have (4) \\begin{split} \\left\\{\\begin{array}{cc}569 \\equiv 1 \\mod{4}\\\\569 \\equiv 1 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{569}\\right) = \\left(\\frac{-1}{569}\\right)\\left(\\frac{2}{569}\\right) = (1)(1) = 1\\\\ \\left\\{\\begin{array}{cc}821 \\equiv 1 \\mod{4}\\\\821 \\equiv 5 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{821}\\right) = \\left(\\frac{-1}{821}\\right)\\left(\\frac{2}{821}\\right) = (1)(-1) = -1\\\\ \\left\\{\\begin{array}{cc}383 \\equiv 3 \\mod{4}\\\\383 \\equiv 7 \\mod{8}\\end{array}\\right\\} &\\Longrightarrow \\left(\\frac{-2}{383}\\right) = \\left(\\frac{-1}{383}\\right)\\left(\\frac{2}{383}\\right) = (-1)(1) = -1.\\\\ \\end{split} $\\square$ Gauss' Lemma Having practiced the rules we learned last time, let's prove the result concerning the Legendre symbol $\\left(\\frac{2}{p}\\right)$. To do so, we'll need the following Gauss' Lemma: For p an odd prime and $p \\nmid a$, let $\\displaystyle n=\\#\\left\\{1 \\leq j \\leq \\frac{p-1}{2}: \\begin{array}{cc}\\mbox{ the least non-neg residue of }\\\\ja \\mbox{ is greater than}\\frac{p}{2}\\end{array}\\right\\}.$ Then $\\left(\\frac{a}{p}\\right) = (-1)^n$. Before we prove this result, let's see it in practice. Example: Is 5 a square mod 11? Let's try to compute $\\left(\\fra{5}{11}\\right)$. For this, we're told we need to compute $5j$ for values of j between 1 and $\\frac{11-1}{2} = 5$. So let's do it: (5) \\begin{split} 5\\cdot 1 &\\equiv 5 \\mod{11}\\\\ 5\\cdot 2 &\\equiv 10 \\mod{11}\\\\ 5\\cdot 3 &\\equiv 15 \\equiv 4 \\mod{11}\\\\ 5\\cdot 4 &\\equiv 20 \\equiv 9 \\mod{11}\\\\ 5\\cdot 5 &\\equiv", "We always have $f_1(x)$ in there (the first and the $N/2$-th term) and all other terms are $\\pi/N$ shifted versions of it. Mathematically: \\begin{align} & \\sum_{n=0}^{N-1} \\sin^2\\left( x+ \\frac{\\pi}{N} n \\right) \\\\ = & \\sum_{n=0}^{N/2-1} \\sin^2\\left( x+ \\frac{\\pi}{N} n \\right) + \\sin^2\\left( x+ \\pi + \\frac{\\pi}{N} n \\right) \\\\ = & \\sum_{n=0}^{N/2-1} \\sin^2\\left( x+ \\frac{\\pi}{N} n \\right) + \\cos^2\\left( x+ \\frac{\\pi}{N} n \\right) \\\\ = & \\sum_{n=0}^{N/2-1} f_1\\left(x + \\frac{\\pi}{N} n\\right) \\\\ = & \\sum_{n=0}^{N/2-1} 1 = \\frac{N}{2}. \\end{align} Piece of cake. This is nice but here is the trouble: for odd $N$ it’s not so easy. It already breaks down for $N=3$ since $f_1(x)$ does not even appear there. Even if you prove it for $N=3$, you can use this to prove it for all integer multiples of $3$ but it will not work for $N=5$. In other words, you would have to do this for every prime number separately. Not very elegant. In essence, in case you were thinking of induction here, this will not work. Increasing $N$ by one changes all terms in the sum so you cannot go from $N$ to $N+1$ (only from $N$ to $2N$). So, to prove it for arbitrary $N$ we need something else. What is there? # Proof idea 2: Power series Wikipedia suggests that the original TPI can", "we look at the first 6 multiples of 7 namely: $7, 14, 21, 28, 35, 42.$ The least positive residues $\\pmod{13}$ are $7, 1, 8, 2, 9, 3 .$ The number of them that exceed $13/2$ is 3 namely: $7, 8, 9.$ Therefore, $$\\left(\\frac{7}{13}\\right)=(-1)^3=-1.$$ Since $(53-1)/2=26$ we look at the first 26 multiples of 9 (mod 53) namely $$\\begin{array}{ccccccccccccc} 9 & 18 & 27 & 36 & 45 & 1 & 10 & 19 & 28 & 37 & 46 & 2 & 11 \\\\ 20 & 29 & 38 & 47 & 3 & 12 & 21 & 30 & 39 & 48 & 4 & 13 & 22. \\end{array}$$ The number of them that exceeds $53/2$ is 12. Therefore, $$\\left(\\frac{9}{53}\\right)=(-1)^{12}=1$$ as desired. Example. (Quadratic Characters) Show that if $p$ is an odd prime, then $$\\left(\\frac{-1}{p}\\right)=\\begin{cases}1 & p\\equiv 1 \\pmod{4} \\ -1 & p\\equiv 3 \\pmod{4}. \\end{cases}$$ $$\\left(\\frac{2}{p}\\right)=\\begin{cases} 1 & p\\equiv 1 \\text{ or } 7\\pmod{8}\\ -1 & p\\equiv 3 \\text{ or } 5 \\pmod{8}. \\end{cases}$$ $$\\left(\\frac{3}{p}\\right)=\\begin{cases}1 & p\\equiv 1 \\text{ or } 11 \\pmod{12} \\ -1 & p\\equiv 5 \\text{ or } 7 \\pmod{12}. \\end{cases}$$ Solution. To verify the first, notice every odd prime is of the form $p=4k+1$ (that is $p\\equiv 1 \\pmod{4}$) or of the form $p=4k+3$ (that is $p\\equiv -1 \\pmod{4}$). In the", "4\\pi^2n^2$ – zhw. Jul 25 '15 at 18:36 Hint: For any $n \\in \\mathbb{N}$ and any $x \\in [0,a]$, we have: $|f_n(x)| = \\left|\\sin\\left(\\sqrt{4\\pi^2n^2+x}\\right)\\right|$ $= \\left|\\sin\\left(\\sqrt{4\\pi^2n^2+x} - 2\\pi n\\right)\\right|$ $\\le \\left|\\sqrt{4\\pi^2n^2+x}-2\\pi n\\right|$ $= \\sqrt{4\\pi^2n^2+x}-2\\pi n$ $= \\dfrac{x}{\\sqrt{4\\pi^2n^2+x}+2\\pi n}$. Can you continue the above work to get a uniform bound on $|f_n(x)|$ over all $x \\in [0,a]$? EDIT Spoiler: Continuing from above: $|f_n(x)| \\le \\dfrac{x}{\\sqrt{4\\pi^2n^2+x}+2\\pi n} \\le \\dfrac{a}{\\sqrt{4\\pi^2n^2}+2\\pi n} = \\dfrac{a}{4\\pi n}$ for all $n \\in \\mathbb{N}$ and all $x \\in [0,a]$. So for any $\\epsilon > 0$, choose an integer $N > \\dfrac{a}{4\\pi\\epsilon}$. Then for all $n \\ge N$ and all $x \\in [0,a]$, you have $|f_n(x)| \\le \\dfrac{a}{4\\pi n} < \\epsilon$. Hence, $f_n \\to 0$ uniformly on $[0,a]$. • +1. Why the two downvotes, though?! – triple_sec Jul 25 '15 at 6:45 I think the Arzelà–Ascoli theorem is an overkill here. Let me show you a proof based on more elementary methods. Claim 1: Define, for each $n\\in\\mathbb N$, $g_n:[0,a]\\to\\mathbb R$ as $$g_n(x)\\equiv\\sqrt{4\\pi^2 n^2+x}-\\sqrt{4\\pi^2 n^2}\\quad\\text{for every x\\in[0,a]}.$$ Then, $(g_n)_{n\\in\\mathbb N}$ converges to $0$ uniformly on $[0,a]$. Proof: First, let me show pointwise convergence. For any $x\\in[0,a]$, \\begin{align*} g_n(x)=&\\,\\frac{\\left(\\sqrt{4\\pi^2 n^2+x}-\\sqrt{4\\pi^2 n^2}\\right)\\left(\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}\\right)}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\\\ =&\\,\\frac{(4\\pi^2n^2+x)-(4\\pi^2 n^2)}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\\\ =&\\,\\frac{x}{\\sqrt{4\\pi^2 n^2+x}+\\sqrt{4\\pi^2 n^2}}\\to0\\quad\\text{as $n\\to\\infty$,} \\end{align*} given that the denominator diverges. Uniform converges follows from the fact that $0\\leq g_n(x)\\leq", "between 0 and π, since $(a-bx)^n>0$ when $0. Since sin(x)>0 when 0<x<π as well, it follows that $0<\\int_0^\\pi f(x)sin(x)\\,dx$. Finally, using the fundamental theorem of calculus, $\\begin{array}{ll}\\int_0^\\pi f(x)sin(x)\\,dx\\\\[2mm]= \\int_0^\\pi \\frac{d}{dx}\\left[g^\\prime(x)sin(x) - g(x)cos(x)\\right]\\,dx\\\\[2mm]= \\left(g^\\prime(\\pi)sin(\\pi) - g(\\pi)cos(\\pi)\\right) - \\left(g^\\prime(0)sin(0) - g(0)cos(0)\\right)\\\\[2mm]= \\left(g^\\prime(\\pi)(0) - g(\\pi)(-1)\\right) - \\left(g^\\prime(0)(0) - g(0)(1)\\right)\\\\[2mm] = g(\\pi)+g(0)\\end{array}$ But this is an integer, and since f(x)sin(x)>0, this integer is at least 1. Therefore, $\\int_0^\\pi f(x)sin(x)\\,dx>1$. But check this out: if $0, then $sin(x)f(x)=sin(x)\\frac{x^n(a-bx)^n}{n!}\\le\\frac{x^n(a-bx)^n}{n!}\\le\\frac{\\pi^n(a-bx)^n}{n!}\\le\\frac{\\pi^na^n}{n!}$. Therefore, $\\int_0^\\pi f(x)sin(x)\\,dx<\\int_0^\\pi \\frac{\\pi^na^n}{n!}\\,dx=\\frac{\\pi^{n+1}a^n}{n!}$. But here’s the thing; we can choose n to be any positive integer we’d like. Each one creates a slightly different version of f(x), but everything up to this point works the same for each of them. While the numerator, π(πa)n, grows exponentially fast, the denominator, n!, grows much much faster for large values of n. This is because each time n increases by one, the numerator is multiplied by πa (which is always the same), but the denominator is multiplied by n (which keeps getting bigger). Therefore, for a large enough value of n we can always force this integral to be smaller and smaller. In particular, for n large enough, $\\int_0^\\pi f(x)sin(x)\\,dx<\\frac{\\pi^{n+1}a^n}{n!}<1$. Keep in mind that $a$ is assumed to be some definite number, so it can’t “race against n”, which means that this fraction always becomes smaller and", "$x = \\tan(t)$. We then get that \\begin{align} I(n) & = \\int_0^{\\infty} \\left(\\dfrac{2x}{1+x^2} \\right)^n dx = \\int_0^{\\pi/2} \\sin^n(2t) \\sec^2(t) dt = 2^n \\int_0^{\\pi/2} \\sin^n(t) \\cos^{n-2}(t) dt\\\\ & = 4\\int_0^{\\pi/2}\\sin^2(t) \\sin^{n-2}(2t)dt \\tag{$\\star$} \\end{align} Replacing $t$ by $\\pi/2-t$, we get $$I(n) = 4 \\int_0^{\\pi/2} \\cos^2(t) \\sin^{n-2}(2t) dt \\tag{\\perp}$$ Adding $(\\star)$ and $(\\perp)$, we get that $$2I(n) = 4 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt \\implies I(n) = 2 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt$$ I trust you can take it from here, using this post, which evaluates $\\displaystyle \\int_0^{\\pi} \\sin^{k}(t) dt$ and using Stirling (or) Wallis formula. - How do you go from $cos^2(t)\\sin^{n-2}(2t)$ to $\\sin^{n-2}(2t)$? – TCL May 7 '13 at 1:17 @TCL Updated the answer. Add $(\\star)$ and $(\\perp)$ to get it. – user17762 May 7 '13 at 1:19 Probably not how you want this done, but a useful and elegant method nonetheless: Note that $$f(x)=\\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\\to\\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$. $f^n \\equiv \\exp(n \\log f)$ and $n\\log(f(1+u))=-n\\frac{1}{2}u^2+n\\mathcal O(u^3)$. Then letting $v=\\sqrt{n} u$, $$\\int_0^\\infty n^{1/2}\\mathrm d x \\exp(-\\frac{1}{2}v^2+\\mathcal O(n^{-1/2})) \\approx \\int_{-\\infty}^\\infty \\mathrm d v\\, e^{-v^2/2} = \\sqrt{2\\pi}$$ Verification: WolframAlpha. Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals.", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of \"2017 AIME I Problems/Problem 9\" ## Problem 9 Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$. ## Solution 1 Writing out the recursive statement for $a_n, a_{n-1}, \\dots, a_{10}$ and summing them gives $$a_n+\\dots+a_{10}=100(a_{n-1}+\\dots+a_{10})+n+\\dots+10$$ Which simplifies to $$a_n=99(a_{n-1}+\\dots+a_{10})+\\frac{1}{2}(n+10)(n-9)$$ Therefore, $a_n$ is divisible by 99 if and only if $\\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\\boxed{045}$. ## Solution 2 $$a_n \\equiv a_{n-1} + n \\pmod {99}$$ By looking at the first few terms, we can see that $$a_n \\equiv 10+11+12+ \\dots + n \\pmod {99}$$ This implies $$a_n \\equiv \\frac{n(n+1)}{2} - \\frac{10*9}{2} \\pmod {99}$$ Since $a_n \\equiv 0 \\pmod {99}$, we", "# Value of $\\limsup$ of $x_n=\\sup\\{x\\in [0,x_{n-1}):\\sin(1/x)=0\\},n\\ge 2$ A sequence is defined like the following $$x_1=1,$$ $$x_n=\\sup\\{x\\in [0,x_{n-1}):\\sin(1/x)=0\\},n\\ge 2$$ What is $\\limsup x_n$? I shall be highly obliged to those who will write the solution for me in detail. Thank you, I am very weak in $\\limsup$ - Well, I would write a quite explicit formula for $x_n$ if I were you: surely you know the zeroes of the sine function, hence... – Did Jul 28 '12 at 21:46 I haven't written anything down, but if I calculated this right in my head it seems the sequence $x_n$ converges, so lim sup is the limit. – Matt Jul 29 '12 at 0:02 If this is homework, you should tag it as such. – Martin Argerami Jul 29 '12 at 4:21 Note that $\\sin(\\frac{1}{x}) = 0$ iff $x = \\frac{1}{n \\pi}$ for some $n \\geq 1$. Now consider the problem $\\alpha_T = \\sup \\{x \\in [0,T)\\, | \\ x \\in \\{ \\frac{1}{n \\pi} \\}_n \\} = \\sup \\{ \\frac{1}{n \\pi} \\}_n \\cap [0,T)$. Then we want to find the smallest $n\\geq 1$ such that $\\frac{1}{n \\pi} < T \\leq \\frac{1}{(n-1) \\pi}$ (ignoring the right hand upper bound if $n=1$). A small amount of work shows that $n = \\lfloor \\frac{1}{\\pi T} \\rfloor +1$. This gives $\\alpha_T = \\frac{1}{(\\lfloor \\frac{1}{\\pi", "# For which $n$ is $2^{2x+2} \\equiv 2\\pmod n \\quad\\text{and}\\quad 2^{2x+2} \\equiv 4\\pmod {n-1}$ We begin with \\begin{align*} 2(2^p+n^2-1-2n) \\equiv 0\\pmod n \\quad&\\text{and}\\quad 2(2^p+n^2-1-2n) \\equiv 0\\pmod {n-1} \\\\ \\end{align*} Which we reduce to \\begin{align*} 2^p\\equiv1\\pmod{\\frac n{2}} \\quad&\\text{and}\\quad 2^{p-1}\\equiv1\\pmod{\\frac{n-1}{4}}. \\end{align*} We are looking to prove that $p$ cannot be odd, considering that $n \\ge 3$ So we let $p=2x+1$ (looking to prove by contradiction) \\begin{align*} 2^{2x+2} \\equiv 2\\pmod n \\quad&\\text{and}\\quad 2^{2x+2} \\equiv 4\\pmod {n-1} \\\\ \\end{align*} I know that $a^b \\equiv a\\pmod b$, so I am assuming that I should go somewhere in that direction. And one more thing, $$\\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ where $s$ is an integer and $s \\ge 3$. I have reason to believe that the only time that $p$ can be odd is when $n=2, s=2^p-1$ Also all 3 variables must be positive integers ($s,n,p$) EDIT: It seems that there are solutions even when $n$ does not equal $2$, but the question still remains, for which $n$ can it hold true • Any ideas on how to solve this? – redelectrons Nov 7 '15 at 16:45 EDIT: For a counterexample, try $p=11$ and $n=23$ We have $2(2^p+n^2−1−2n)=5060=10 \\cdot 23 \\cdot 22$. INCORRECT ANSWER: I think I found a counterexample. Set $p = 3$ and $n = 7$. Then, $2(2^p+n^2−1−2n) =", "# Rational sequence with $a_{n+1}=2a_n^2-1$ Suppose we start with a rational number $a_0$, and define $a_{n+1}=2a_n^2-1$ for $n\\geq 0$. For what $a_0$ will it be the case that $a_i=a_j$ for some $i\\neq j$? We can start with something like $a_0=1$, then $a_1=1$ so $a_0=a_1$. If $a_0=0$, we get $0, -1, 1, 1, \\ldots$ Likewise if $a_0=-1$, we get $-1,1,1,\\ldots$. But how can we find all $a_0$? • Well, for example for $\\;a_0= 1\\implies a_1=a_0\\;$ ... – DonAntonio Nov 12 '13 at 19:37 Let $a_{0} = \\cos \\theta$. Then it is easy to check that $a_{n} = \\cos (2^{n}\\theta)$. So if $a_{i} = a_{j}$ for some $i \\neq j$, then we must have Thus the problem reduces to find the condition of $(i, j, n, \\pm)$ such that $$\\cos \\left( \\frac{2n\\pi}{2^{i} \\pm 2^{j}} \\right) \\in \\Bbb{Q}.$$ Referring to this posting, this is possible if and only if \\begin{align*} \\theta \\equiv 0, \\pm \\frac{\\pi}{3}, \\pm \\frac{\\pi}{2}, \\pm \\frac{2\\pi}{3} \\pmod{2\\pi} \\end{align*} This corresponds to $a_{0} \\in \\{0, \\pm \\frac{1}{2}, \\pm 1 \\}$. • Why did you decide that $\\;|a_0|\\le 1\\;$ , @sos440 ? This follows from my answer, of course, but I can't see any reason why in yours... – DonAntonio Nov 13 '13 at 4:26 • @DonAntonio, This solution works if $\\theta$ is allowed to have complex values. But a", "n!\\frac{n(n+1)(2n+1)(50 n^3 + 39 n^2 - 11 n + 12)}{64800}$$ Note that these are not always integers. It also follows that the coefficients grows with $$n$$ as $$[x^{n+2m}] \\sim \\frac{S_2^m(n)}{m!6^m}\\sim \\frac{n! n^{3m}}{m!18^m}$$. • Very interesting answer. I was unable to get anything better. Thanks. – Gevorg Hmayakyan Oct 12 '19 at 9:39 • We can use Bell polynomials for better result. – Gevorg Hmayakyan Oct 12 '19 at 22:40 In terms of the q-Pochhammer symbol, we have \\begin{align}\\prod_{k=1}^n\\sin(kt)&=\\prod_{k=1}^n\\frac{e^{ikt}-e^{-ikt}}{2i}\\\\&=\\frac{e^{in(n+1)t/2}}{(2i)^n}\\prod_{k=1}^n(1-e^{-2ikt})\\\\&=\\frac{e^{in(n+1)t/2}}{(2i)^n}(e^{-2it};e^{-2it})_n\\end{align} from which one can produce various identities. As $$n\\to\\infty$$, one can show that the product tends to $$0$$. We can bound how fast it goes to zero by considering the following: $$|\\sin(x)\\sin(x+t)|\\le\\max\\{\\sin^2(t/2),\\cos^2(t/2)\\}$$ which gives us: \\begin{align}\\left|\\frac{\\sin((n+1)t)}{\\sin(t)}\\right|\\prod_{k=1}^n\\sin^2(kt)&=\\prod_{k=1}^n|\\sin(kt)\\sin(kt+t)|\\\\&\\le\\prod_{k=1}^n\\max\\{\\sin^2(t/2),\\cos^2(t/2)\\}\\\\&=\\max\\{\\sin^{2n}(t/2),\\cos^{2n}(t/2)\\}\\end{align} and thus, $$\\prod_{k=1}^n|\\sin(kt)|\\le\\alpha^n\\cdot\\sqrt{\\left|\\frac{\\sin(t)}{\\sin((n+1)t)}\\right|}$$ where $$\\alpha=\\max\\{|\\sin(t/2)|,|\\cos(t/2)|\\}$$, provided that $$\\sin(kt)\\ne0$$ for any natural $$k$$. In the event that $$\\sin(t/2)$$ or $$\\cos(t/2)$$ are $$1$$, then the product trivially reduces down to $$0$$. Since the irrationality measure of $$\\pi$$ has an upper bound of $$7.6063$$, we have $$\\prod_{k=1}^n|\\sin(kt)|\\in\\mathcal O\\left(n^{-3.3031}\\alpha^n\\right)$$ as $$n\\to\\infty$$. • Thanks, for the answer,but this does not allow efficient calculation for big n-s. – Gevorg Hmayakyan Sep 14 '19 at 8:27 • I doubt there is much that can be done concerning $t\\to\\infty$ as there are wild oscillations, but as $n\\to\\infty$ the product converges to $0$. – Simply Beautiful Art Sep 14 '19 at 21:56", "&= -1, &F_8 &= 21,&F_7&=13, \\\\ (\\tfrac{11}{5})& = +1, &F_{10}& = 55, &F_{11}&=89. \\end{align} It is not known whether there exists a prime p such that $F_{p-\\left(\\frac{p}{5}\\right)} \\equiv 0 \\pmod{p^2}.$ Such primes (if there are any) would be called Wall–Sun–Sun primes. Also, if p ≠ 5 is an odd prime number then:[45] $5F^2_{\\frac{p \\pm 1}{2}} \\equiv \\begin{cases} \\tfrac{1}{2} \\left (5\\left(\\frac{p}{5}\\right)\\pm 5 \\right ) \\pmod p & \\textrm{if}\\;p \\equiv 1 \\pmod 4\\\\ \\tfrac{1}{2} \\left (5\\left(\\frac{p}{5}\\right)\\mp 3 \\right ) \\pmod p & \\textrm{if}\\;p \\equiv 3 \\pmod 4. \\end{cases}$ Example 1. p = 7, in this case p ≡ 3 (mod 4) and we have: $(\\tfrac{7}{5}) = -1: \\qquad \\tfrac{1}{2}\\left (5(\\tfrac{7}{5})+3 \\right ) =-1, \\quad \\tfrac{1}{2} \\left (5(\\tfrac{7}{5})-3 \\right )=-4.$ $F_3=2 \\text{ and } F_4=3.$ $5F_3^2=20\\equiv -1 \\pmod {7}\\;\\;\\text{ and }\\;\\;5F_4^2=45\\equiv -4 \\pmod {7}$ Example 2. p = 11, in this case p ≡ 3 (mod 4) and we have: $(\\tfrac{11}{5}) = +1: \\qquad \\tfrac{1}{2}\\left (5(\\tfrac{11}{5})+3 \\right )=4, \\quad \\tfrac{1}{2} \\left (5(\\tfrac{11}{5})- 3 \\right )=1.$ $F_5=5 \\text{ and } F_6=8.$ $5F_5^2=125\\equiv 4 \\pmod {11} \\;\\;\\text{ and }\\;\\;5F_6^2=320\\equiv 1 \\pmod {11}$ Example 3. p = 13, in this case p ≡ 1 (mod 4) and we have: $(\\tfrac{13}{5}) = -1: \\qquad \\tfrac{1}{2}\\left (5(\\tfrac{13}{5})-5 \\right ) =-5, \\quad \\tfrac{1}{2}\\left (5(\\tfrac{13}{5})+ 5 \\right )=0.$ $F_6=8 \\text{ and } F_7=13.$ $5F_6^2=320\\equiv -5 \\pmod {13} \\;\\;\\text{", "+0 # Let $n$ be the inverse of $2\\pmod{17}$. 0 24 1 +122 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. Sort: #1 +19376 +1 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. $$\\begin{array}{|rcll|} \\hline 2n &\\equiv& 1 \\pmod {17} \\quad & | \\quad : 2 \\\\\\\\ n &\\equiv& \\frac12 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{\\equiv}& \\mathbf{ 2^{-1} \\pmod {17}} \\\\ \\hline \\end{array}$$ $$\\text{ Solve 2^{-1} \\pmod {17} }$$ $$\\text{Because gcd(2,17)=1 we can continue.}$$ $$\\begin{array}{|rcll|} \\hline 2^{-1} \\pmod {17} &\\equiv& 2^{\\varphi(17)-1} \\pmod {17} \\quad & | \\quad \\varphi(p) = p - 1 \\quad \\text{p is a prime number} \\\\\\\\ && \\varphi(17) = 16 \\\\\\\\ &\\equiv& 2^{16-1} \\pmod {17} \\\\ &\\equiv& 2^{15} \\pmod {17} \\\\ &\\equiv& 32768 \\pmod {17} \\\\ 2^{-1} \\pmod {17}&\\equiv& 9 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{=}& \\mathbf{9} \\\\ \\hline \\end{array}$$ $$Proof:\\\\ 2\\cdot 9 \\equiv 1 \\pmod{17} \\ \\checkmark$$ $$\\begin{array}{|rcll|} \\hline && (2^n)^2 - 2 \\pmod{17} \\quad & | \\quad n =", "i\\pi/4}$, $(1+\\mathrm i)^n$ is $(\\sqrt2)^n$ times $\\mathrm e^{\\mathrm in\\pi/4}$. • The real part of $\\mathrm e^{\\mathrm in\\pi/4}$ is $\\cos(n\\pi/4)$. Thus, $\\color{red}{x_n=\\ldots}$ - Thanks, so I think $2\\Re((1+i)^n)=2(\\sqrt{2}^n)\\cos(\\frac{n\\pi}{4})$, which can be evaluated depending on $n\\pmod{8}$? – hagar Jan 17 '12 at 16:59 Yes, depending on whether $n=0\\pmod 8$ or $n=\\pm1\\pmod 8$ or $n=\\pm2\\pmod 8$ or $n=\\pm3\\pmod 8$ or $n=4\\pmod 8$. – Did Jan 17 '12 at 17:02 This is a neat solution, that generalizes nicely : $a^n + \\bar{a}^n = 2 \\left|a\\right|^n \\cos( n \\arg (a) )$. – Michael Anderson Jan 18 '12 at 7:25 First, notice that $1+i = \\sqrt2(\\cos\\frac{\\pi}{4} + i \\sin\\frac{\\pi}{4}$) and $1-i = \\sqrt2(\\cos(-\\frac{\\pi}{4}) + i \\sin(-\\frac{\\pi}{4}))$. Using de Moivre's formula and the fact that $\\sin$ is odd and $\\cos$ is even, we get \\begin{align*} (1+i)^n + (1-i)^n &= 2(\\sqrt2)^n\\cos\\frac{n\\pi}{4} \\end{align*} - $(1+i)^n+(1-i)^n = 2 \\Re ((1+i)^n)$. Now expand $(1+i)^n$ using the binomial theorem. The real part is formed by the odd-numbered terms. Another approach is to note that $(1+i)/\\sqrt{2}$ is an 8-th root of unity and so $(1+i)^n$ depends only on $n \\bmod 8$ except for a power of $\\sqrt{2}$. - Thank you, lhf. – hagar Jan 17 '12 at 17:00 This is not a particularly elegant solution, but an alternative route is to simply note that $(1 \\pm i)^2 = \\pm 2i$. I", "of the time - for instance, how might you solve the case where $n=0$? $n=1$? $n=2$? Can you generalise the approach, either for all $n$, all integer $n$, or all even (or odd) integer $n$? - \\begin{align} {\\cal I}_{n} &= {1 \\over n + 1}\\int_{x\\ =\\ 0}^{x\\ =\\ \\pi/2}\\cos\\pars{x} \\,\\dd\\bracks{\\sin^{n + 1}\\pars{x}} = -\\,{1 \\over n+ 1}\\int_{0}^{\\pi/2}\\sin^{n + 1}\\pars{x}\\bracks{-\\sin\\pars{x}}\\,\\dd x \\\\[3mm]&={1 \\over n+ 1}\\int_{0}^{\\pi/2} \\sin^{n}\\pars{x}\\bracks{1 -\\cos^{2}\\pars{x}}\\,\\dd x ={1 \\over n+ 1}\\int_{0}^{\\pi/2} \\sin^{n}\\pars{x}\\,\\dd x - {1 \\over n + 1}\\,{\\cal I}_{n} \\end{align} \\begin{align} {\\cal I}_{n}&={1 \\over n + 2}\\int_{0}^{\\pi/2}\\sin^{n}\\pars{x}\\,\\dd x ={n \\over n + 2}\\bracks{{1 \\over n}\\int_{0}^{\\pi/2}\\sin^{n - 2}\\pars{x}\\,\\dd x} - {1 \\over n + 2}\\int_{0}^{\\pi/2}\\sin^{n - 2}\\pars{x}\\cos^{2}\\pars{x}\\,\\dd x \\\\[3mm]&= {n \\over n + 2}\\,{\\cal I}_{n - 2} - {1 \\over n + 2}\\,{\\cal I}_{n - 2} = {n - 1 \\over n + 2}\\,{\\cal I}_{n - 2} \\end{align} $$\\color{#0000ff}{\\large% {\\cal I}_{n} = \\int_{0}^{\\pi/2}\\sin^{n}\\pars{x}\\cos^{2}\\pars{x} \\,\\dd x = {n - 1 \\over n + 2}\\,{\\cal I}_{n - 2}\\,,\\quad n \\geq 2\\,,\\qquad \\left\\vert% \\begin{array}{rcl} {\\cal I}_{0} & = & {\\pi \\over 4} \\\\[2mm] {\\cal I}_{1} & = & {1 \\over 3} \\end{array}\\right.}$$ - (+1) nice answer. If you solve the recurrence relation you got and compare it with the solution using the beta function you will have the same answer. – Mhenni Benghorbal Jan 12 '14 at 6:23 @MhenniBenghorbal Also, the", "# Finding the $n^{\\text{th}}$ term of $-1,-1,-1,-1,1,1,1,1,...$ as a repeating 8-block In my work I came across that sequence $-1,-1,-1,-1,1,1,1,1,\\dots$ and repeating this 8-block so on forever Now I cant find an ( e.g. trigonometric/complex ) expression $f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence starting with $n=2,3,4,5,6,7,8,9,…$ and so on forever. ## 7 Answers Consider the function $f(n)$ such that $$f(n) = \\left \\{ \\begin{array}[ll] (-1 & \\textrm{ if } n \\textrm{ mod } 8 \\equiv 0,1,2,3 \\\\ 1 & \\textrm{ if } n \\textrm{ mod } 8 \\equiv 4,5,6,7 \\end{array} \\right.$$ • The most unambiguous way. Nov 26 '15 at 17:04 I will assume that you really are starting at $2$. If the start is different, modification is easy. Let $g(n)=\\left\\lfloor\\frac{n+2}{4}\\right\\rfloor$. Here $\\lfloor x\\rfloor$ is the \"floor\" function, the greatest integer $\\le x$. Then your sequence is given by $(-1)^{g(n)}$. By symmetry, we can assume $$f(n) = a_1 \\sin\\frac{\\pi n}{4} +a_3 \\sin\\frac{3\\pi n}{4} +b_1 \\cos\\frac{\\pi n}{4} +b_3 \\cos\\frac{3\\pi n}{4}.$$ Then \\begin{aligned} f(1) &= \\frac{a_1}{\\sqrt{2}} +\\frac{a_3}{\\sqrt{2}} + \\frac{b_1}{\\sqrt{2}} -\\frac{b_3}{\\sqrt{2}} = -1\\\\ f(2) &= a_1 - a_3 = -1\\\\ f(3) &= \\frac{a_1}{\\sqrt{2}} +\\frac{a_3}{\\sqrt{2}} - \\frac{b_1}{\\sqrt{2}} +\\frac{b_3}{\\sqrt{2}} = -1 \\\\ f(4) &= -b_1 - b_3 = -1. \\end{aligned} So \\begin{aligned} a_1 &= -\\frac{1}{2} -\\frac{1}{\\sqrt 2}, \\\\ a_3 &= +\\frac{1}{2} -\\frac{1}{\\sqrt 2}, \\\\ b_1 &= b_3 =" ]

[ "h=22.5 we have a phase shift of 22.5 left. (A phase shift of `pi/8~~.393` when using radians.) The graph of `y=cos(theta)` and the transformed graph:", "How do you graph y = cos(x- pi/6)? It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "# How do you graph y = cos(x- pi/6)? Sep 18, 2015 #### Answer: It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ #### Explanation: The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "either that the phase shift is $+ \\pi$, or that the phase shift is $- \\pi$, and the graph would look the same. $y = \\sin \\left(x\\right)$ $y = - \\frac{1}{2} \\cdot \\sin \\left(x\\right)$ • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago • 43 seconds ago • 2 minutes ago • 4 minutes ago • 4 minutes ago • 4 minutes ago • 11 minutes ago • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago", "think of the phase shift is that it's just a new start to the graph. You can treat the phase shift as a place to actually begin your graph. So.. with all this information, we can now actually graph the equation of $y = 5 \\cos \\left(x - \\pi\\right) + 1$ . Here are some visual cues. My art is pretty bad, sographing the graph on a website should be much more different. I hope this helps.", "The value of $$C$$ is the phase shift (horizontal shift) of the sinusoidal function. The graph is shifted to the right if $$C > 0$$ and shifted to the left if $$C < 0$$. • The value of $$D$$ is the vertical shift of the sinusoid. The horizontal line $$y = D$$ is the so-called center line for the graph of the sinusoidal function. • The important notes about sinusoids on page 103.", "# How do you find the phase shift of y=sin(theta-3 pi/4)? Aug 2, 2016 $$ Horizontal shift =3pi/4 #### Explanation: $y = \\sin \\left(\\theta - 3 \\frac{\\pi}{4}\\right)$ we have $a = 1$ $b = 1$ $c = 3 \\frac{\\pi}{4}$ $$ A phase shift is nothing but horizontal shift. Horizontal shift =3pi/4", "graph in a vertical direction either side of the x-axis by a factor of $5$. 3. ah icic cool thxs. so for the phase shift would be -4pi/5 correct? 4. Originally Posted by S2Krazy ah icic cool thxs. so for the phase shift would be -4pi/5 correct? Correct!", "makes no difference to the phase-shift: it merely 'stretches' the graph in a vertical direction either side of the x-axis by a factor of $5$. • October 17th 2009, 10:23 AM S2Krazy ah icic cool thxs. so for http://www.webassign.net/www38/latex...d954a06db3.gif the phase shift would be -4pi/5 correct? • October 17th 2009, 10:54 AM Quote: Originally Posted by S2Krazy ah icic cool thxs. so for http://www.webassign.net/www38/latex...d954a06db3.gif the phase shift would be -4pi/5 correct? Correct!", "Phase Shift: ( to the right) Vertical Shift:", "2, 2005 ### amcavoy At first glance it looks OK. The only thing I found is that for a phase shift to the right, it would be x-pi/3 rather than x+pi/3.", "# Laplace transform of phase-shifted sinusoid ## Homework Statement Find the laplace transform of [TEX]f(t)=sin(4t+\\frac{\\pi}{3})u(t)[/TEX]. http://img822.imageshack.us/img822/6996/codecogseqna.gif [Broken] ## Homework Equations Definition of laplace transform, and properties: http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems\" ## The Attempt at a Solution I have a feeling that I need to use the time-shift property of the transform, but I need just a bit more direction on the strategy. Last edited by a moderator: $$f(t) = \\left[ sin(4t)cos(\\frac{\\pi}{3}) + cos(4t)sin(\\frac{\\pi}{3}) \\right] u(t).$$", "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "So far I have y=10sin(pi/20,000(x-d))+70 How do I get d? Can someone clearly explain to me how to get the phase shift? thanks let the year $2000$ be $t = 0$ $T_{avg} = 10\\cos\\left(\\frac{\\pi t}{20000}\\right) + 70$", "# How do I find the phase shift of a sinusoidal graph? Jun 19, 2015 Find the value of $x$ at which we find $\\sin \\left(0\\right)$. #### Explanation: Examples: $y = 3 + 2 \\sin \\left(3 x - \\frac{\\pi}{4}\\right)$ Solve: $3 x - \\frac{\\pi}{4} = 0$ So $x = \\frac{\\pi}{12}$ is the shift. $y = 5 - \\sin \\left(4 x + \\frac{\\pi}{2}\\right)$ Solve: $4 x + \\frac{\\pi}{2} = 0$ So $x = - \\frac{\\pi}{8}$ is the shift. $y = - 3 \\sin \\left(\\pi x + \\frac{3 \\pi}{4}\\right)$ Solve: $\\pi x + \\frac{3 \\pi}{4} = 0$ So $x = - \\frac{3}{4}$ is the shift. Some textbooks use: $y = a \\sin \\left(b x + c\\right)$ has phase shift $- \\frac{c}{b}$ Other textbooks use: $y = a \\sin \\left(b x - c\\right)$ has phase shift $\\frac{c}{b}$ For me (having taught out of both kinds of textbook), it's easier just to solve the equation.", "# How do you graph sin(x/2)? It is easy to graph y= $\\sin \\left(\\frac{x}{2}\\right)$. The graph would be similar to the graph of sin x, the only difference being that its period would be $4 \\pi$, instead of $2 \\pi$. Amplitude would remain 1. There would be no phase shift. It would appear to be like this:", "The phase shift of a graph determines if the graph is going to be shifted left or right on the x-plane of the graph. $$Asin[B(x-\\frac{C}{B})]+D$$ From the example above the phase shift of the graph would be. $$\\frac{C}{B}$$ Let's do a short example of how the phase shifts would happen to a basic sin(x) function. $$sin(x)$$ If you take the function and shift it to the left $\\frac{\\pi}{2}$ length. $$sin(x+\\frac{\\pi}{2})$$ You can also go the other way and shift it to the right $\\frac{\\pi}{2}$ length. $$sin(x-\\frac{\\pi}{2})$$ Let's take a look at all three transformations.", "# Trying to read phase shift constant from the graphs I have an equation $y(x,y) = y_0 \\sin (\\omega t \\pm kx \\pm \\phi)$ and there are two graphs. One is $y(x,t=0)$ in units $mm(mm)$ and the other is $y(t,x=0)$ in units $mm(s)$. Now I have to find all of the constants inside of the equation by looking at graphs and it goes well until I am to determine phase shift $\\phi$. If I calculate it by looking at $y(x=0,t=0)$ it works out well and I get right solution, but when I try to read $\\phi$ from graph $y(x,t=0)$ or $y(t,x=0)$ I get two wrong solutions. The idea for reading $\\phi$ from graphs was to express some distance $x'$ or time $t'$ in radians but it looks like I can't get it right. The sign in front of $\\phi$ is not a problem - if my sinus (in red) is shifted right it is minus otherwise it is plus. Can anyone clear things out for me and explain why I get wrong results which also differ from one-another? • Your result using $y(x, t=0)$ appears correct, it agrees to within measurement error. Your result for $\\phi$ from $y(x=0, t)$ should be replaced with $\\pi - \\phi$, since there is a relative minus sign between $\\omega t$ and $kx$", "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "phase shift – a full period ($2\\pi$) ..." ]

[ "yields $$\\Delta p(x,y) = - a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) - b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right)$$ and $$Q = - \\Delta p(x,y) = a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) + b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right).$$ ## Input files The project file is square_1e3_poisson_sin_x_sin_y.prj. It describes the process to be solved and the related process variable together with their initial, boundary conditions and source terms. The definition of the source term $Q$ is in a Python script. The script for setting the source terms is referenced in the project file as follows: <python_script>sin_x_sin_y_source_term.py</python_script> In the source term description <source_term> <type>Python</type> <source_term_object>sinx_siny_source_term</source_term_object> </source_term> the domain is specified by the mesh-tag. The function $Q$ is defined by the Python object sinx_sinx_source_term that is created in the last line of the Python script: import OpenGeoSys from math import pi, sin a = 2.0*pi b = 2.0*pi def solution(x, y): return - sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # - laplace(solution) = source term def laplace_solution(x, y): return a*a * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) + b*b * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # source term for the benchmark class SinXSinYSourceTerm(OpenGeoSys.SourceTerm): def getFlux(self, t, coords, primary_vars): x, y, z = coords value = laplace_solution(x,y) Jac = [ 0.0 ] return (value, Jac) # instantiate source term object referenced in OpenGeoSys' prj file sinx_siny_source_term", "layers. In order for my approach to work as intended, the layer settings window must look like this: However, quite annoyingly, when you close and reopen this saved document, the layer settings window looks like this: So, it must be changed back each time to allow trouble-free math drawing. # Finding Reciprocal Functions SEC, CSC, COT on TI-30XA First, you may want to refresh your memory of the regular trig functions: Now, let’s find $\\csc \\frac{5\\pi }{4}$. If we look it up on WolframAlpha, we find that it equals $-\\sqrt{2}$, or, roughly, $-1.4142$. To get this answer on the TI-30XA, first remember that $\\csc$ is equal to $\\frac{1}{\\sin }$. So, what we’re looking for could be expressed as $\\frac{1}{\\sin \\frac{5\\pi }{4}}$. Once we know this, we can type $5\\cdot \\pi$ into the TI-30XA. This gives us ~15.7. (Make sure you’re in radian mode. If not, press the DRG button — located by the 2nd button, upper-left — until you see RAD appear in small letters on the screen.) Now, we can divide 15.7 by 4. This gives us 3.925. If we take the $\\sin$ of this (by pressing the “sin” button while 3.925 is displayed), we get $-0.7$. IF we divide 1 by $-0.7$, we get $-1.43$. As you can see, this is close to the correct value.", "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "what translation to apply, however, we first factorise the function into the form $y=-2\\sin\\left(3\\left(x+\\frac{\\pi}{12}\\right)\\right)$y=2sin(3(x+π12)). In this form, we can see that the $x$x-values are increased by $\\frac{\\pi}{12}$π12 inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $\\frac{\\pi}{12}$π12 smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\\frac{\\pi}{12}$π12 units. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=−2sin(3x+π4​) Lastly, we translate the graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=2sin(3x+π4) upwards by $2$2 units, to obtain the final graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=2sin(3x+π4)+2. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=−2sin(3x+π4​)+2 Careful! When we geometrically apply each transformation to the graph of $y=\\sin x$y=sinx, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting about the $x$x-axis, we would have needed to translate the graph downwards first. ### The general case In the example above we were transforming the graph of $y=\\sin x$y=sinx. The particular function $y=\\sin x$y=sinx was not important, however. We could have just as easily transformed the graph of $y=\\cos x$y=cosx, or even a non-trigonometric function, using the same method! Consider a function $y=f\\left(x\\right)$y=f(x). Then we can obtain the graph of $y=af\\left(b\\left(x-c\\right)\\right)+d$y=af(b(xc))+d, where $a,b,c,d$a,b,c,d are constants, by applying a series of transformations to the graph of $y=f\\left(x\\right)$y=f(x). These transformations are summarised below. Summary To obtain the graph", "it yields $\\Delta p(x,y) = - a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) - b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right)$ and $Q = - \\Delta p(x,y) = a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) + b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right).$ ## Input files The project file is square_1e3_poisson_sin_x_sin_y.prj. It describes the process to be solved and the related process variable together with their initial, boundary conditions and source terms. The definition of the source term $Q$ is in a Python script. The script for setting the source terms is referenced in the project file as follows: <python_script>sin_x_sin_y_source_term.py</python_script> In the source term descripition <source_term> <mesh>square_1x1_quad_1e3_entire_domain</mesh> <type>Python</type> <source_term_object>sinx_siny_source_term</source_term_object> </source_term> the domain is specified by the mesh-tag. The function $Q$ is defined by the Python object sinx_sinx_source_term that is created in the last line of the Python script: import OpenGeoSys from math import pi, sin a = 2.0*pi b = 2.0*pi def solution(x, y): return - sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # - laplace(solution) = source term def laplace_solution(x, y): return a*a * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) + b*b * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # source term for the benchmark class SinXSinYSourceTerm(OpenGeoSys.SourceTerm): def getFlux(self, t, coords, primary_vars): x, y, z = coords value = laplace_solution(x,y) Jac = [ 0.0 ] return (value, Jac) # instantiate source term object referenced in OpenGeoSys' prj", "# ISRO2009-46 2.2k views The shift operator $E$ is defined as $E [f(x_i)] = f (x_i+h)$ and $E'[f(x_i)]=f (x_i -h)$ then $\\triangle$ (forward difference) in terms of $E$ is 1. $E-1$ 2. $E$ 3. $1-E^{-1}$ 4. $1-E$ Forward difference operator(delta (D)) D(f(x)) = f(x + h) - f(x) Shift operator , Ef(x) = f(x+h) = f(x+h) - f(x) + f(x) = (1 + D)f(x) so E= 1 + D gives D = E - 1 selected by ## Related questions 1 1.3k views The formula $P_k = y_0 + k \\triangledown y_0+ \\frac{k(k+1)}{2} \\triangledown ^2 y_0 + \\dots + \\frac{k \\dots (k+n-1)}{n!} \\triangledown ^n y_0$ is Newton's backward formula Gauss forward formula Gauss backward formula Stirling's formula The cubic polynomial $y(x)$ which takes the following values: $y(0)=1, y(1)=0, y(2)=1$ and $y(3)=10$ is $x^3 +2x^2 +1$ $x^3 +3x^2 -1$ $x^3 +1$ $x^3 -2x^2 +1$ The formula $\\int\\limits_{x0}^{xa} y(n) dx \\simeq h/2 (y_0 + 2y_1 + \\dots +2y_{n-1} + y_n) - h/12 (\\triangledown y_n - \\triangle y_0)$ $- h/24 (\\triangledown ^2 y_n + \\triangle ^2 y_0) -19h/720 (\\triangledown ^3 y_n - \\triangle ^3 y_0) \\dots$ is called Simpson rule Trapezoidal rule Romberg's rule Gregory's formula A root $\\alpha$ of equation $f(x)=0$ can be computed to any degree of accuracy if a 'good' initial approximation $x_0$ is chosen for which $f(x_0)", "sure so I experimented, I decided to use a sine graph with a wavelength of 20 y=sin nx the wavelength is 2pi/n = 20 so n = pi/10 so my experimenting graph is $$y=sin(\\frac{\\pi }{10}x)\\\\ y=sin(\\frac{\\pi }{10}x)=sin(\\frac{\\pi }{10}(x-20))\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=sin(\\frac{\\pi }{10}\\frac{x}{5})=sin(\\frac{\\pi }{10}(\\frac{x}{5}-20))=sin(\\frac{\\pi }{10}(\\frac{x-100}{5}))$$ So the smallest a that we know with certainty is 100. Here is the graph i was playing with. https://www.desmos.com/calculator/oonutt20vz ------------------- Let me see if I can do this more simply. $$y=f(x)=f(x-20)\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=f(\\frac{x}{5})=f(\\frac{x}{5}-20)=f(\\frac{x-100}{5})$$ There you go, that was simple. The right shift is 100. Mar 27, 2018 #5 +100025 +1 2. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So \"(1,-3); (2,3); (5,-7)\" - without the quotes - is a valid answer format.) $$x^3-x^2+x+1 = x^3+x^2+x-1\\\\ -x^2+1 = x^2-1\\\\ 2=2x^2\\\\ x^2=1\\\\ x=\\pm1\\\\ f(1)=1-1+1+1=2\\\\ f(-1)=-1-1-1+1=-2\\\\$$ so the points of intersection are (-1,-2); (1,2) check: Mar 27, 2018", "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "Solved Example. Graph$\\,y=\\mathrm{arccos}\\,x\\,$and state the domain and range of the function. How to find the base point given public and private key and EC parameters except the base point. The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). Please help me answering this problem! Post-multiply both sides of … If you want to share a math-problem you need to explain what you're arguments are. For multiplication, it's division. Show … We can check … Are there any Pokémon that lose overall base stats when they evolve? Given a “special” input value, evaluate an inverse trigonometric function. [/latex], If$\\,x\\text{ is in }\\left[0,\\pi \\right],\\,$then$\\,{\\mathrm{sin}}^{-1}\\left(\\mathrm{cos}\\,x\\right)=\\frac{\\pi }{2}-x. The Inverse Park Transform block converts the time-domain direct, quadrature, and zero components in a rotating reference frame to the components of a three-phase system in an abc reference frame. Figure 11. If y = f(x) = ab x, then we may solve for x in terms of y using logarithms: .$, If$\\,x\\text{ is not in}\\,\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right],\\,$then find another angle$\\,y\\text{ in }\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right]\\,$such that$\\,\\mathrm{sin}\\,y=\\mathrm{sin}\\,x.$. prove that $\\det(ABC) = \\det(A) \\det(B) \\det(C)$ [for any $n×n$ matrices $A$, $B$, and $C$]. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical", "SYSTEM label(\"$O$\", (0, 0, 0), NW); Label Lx = Label(\"$x$\", EndPoint, W); Label Ly = Label(\"$y$\", EndPoint, S); Label Lz = Label(\"$z$\", EndPoint, W); draw(Lx, O--2X, Arrow3(emissive(black))); draw(Ly, O--2Y, Arrow3(emissive(black))); draw(Lz, O--2Z, Arrow3(emissive(black))); // SURFACE triple S(pair uv) { real x = cos(uv.x)*sin(uv.y); real y = sin(uv.x)*sin(uv.y); real z = cos(uv.y); return (x-(1-z**2-y**2), y-0.3*sin(z*pi), z); } // BOUNDARY triple dS(real u) { return S((u, pi/2)); } // TANGENT VECTORS real e = 0.001; // *this... is... infinitesimal!* triple U(pair t) { return (S((e,0)+t)-S(t))/e; } triple V(pair t) { return (S((0,e)+t)-S(t))/e; } // NORMAL VECTOR triple N(pair uv) { return cross(U(uv),V(uv))/length(cross(U(uv),V(uv))); } // GRAPHICAL ELEMENTS surface s = surface(S, (0, 0), (2pi, pi/2), 64, 64, Spline); path3 ds = graph(dS, 0, 2pi); pair t = (-pi/8, pi/8); path3 n = S(t) -- shift(-N(t))*S(t); label(\"$\\Sigma$\", shift(0,0,0.1)*shift(adj)*S((pi/2,pi/5)), NE); Label LdS = Label(\"$\\partial\\Sigma$\", Relative(0.95), S); Label Ln = Label(\"$\\mathbf n$\", EndPoint, NW); draw(s, surfacepen=material(diffusepen=gray(0.6),emissivepen=gray(0.3),specularpen=gray(0.1))); draw(LdS, ds, Arrow3(emissive(black))); draw(shift(4*unit(currentprojection.camera))*ds, dashed); draw(Ln, n, Arrow3(emissive(black))); REQUESTED APPENDIX: on dashed lines I wanted to draw the dashed border over the surface. Since this is an orthographic projection, the easiest way is translating the object along the camera axis towards it (instead of checking what segments are visible, possibly complicated): draw(shift(4*unit(currentprojection.camera))*ds, dashed); unit(currentprojection.camera) is a versor parallel to the camera axis and pointing towards it, so it", "# Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$ Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$. First $$f'(x)=\\frac{d\\sin{\\pi}x}{d({\\pi}x)}\\frac{d({\\pi}x)}{dx} ={\\pi}\\cos{\\pi}x$$ Second $$f''(x)=-{\\pi^2}\\sin{\\pi}x$$ Third $$f'''(x)= -{\\pi^3}\\cos {\\pi}x$$ These are the steps the teacher has written down on the blackboard, I have trouble understanding \"How it is solved\". Would you please provide the steps? - LaTeX tip: There are preexisting operators \\sin and \\cos for $\\sin x$ and $\\cos x$. There is a quick reference guide here. – Aeolian Mar 6 '13 at 7:50 \\begin{align*} f(x) = \\sin\\pi x \\end{align*} Here, note that $\\pi$ is a constant. So treat it like a coefficient of $x$. Progressively, what I will do is to: • Isolate the coefficient since it is not involved in the differentiation • Apply chain rule by differentiating the trigonometric function, followed by the function \"inside\" the trigo function - the $\\pi x$. Note that $\\frac{d}{dx}\\begin{pmatrix}\\pi x\\end{pmatrix} = \\pi$. \\begin{align*} f(x) &= \\sin\\pi x \\\\ f'(x) &= (\\cos \\pi x) \\times \\pi \\space\\space\\tag{by chain rule}\\\\ &= \\pi \\cos \\pi x \\\\\\\\\\\\f''(x) &= \\pi (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^2 \\sin \\pi x \\\\\\\\\\\\f'''(x) &= \\pi^2 (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^3 \\cos \\pi x \\end{align*} And that is how you do it. First, understand that $\\cos$ is a trigonometric function. I would recommend you to write $f(x)$ and $f'(x)$ as a composite", "# Math Help - Moving a Trigonometric function 1. ## Moving a Trigonometric function y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation. 2. Originally Posted by IDontunderstand y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my", "Math Help - Order of Operations with Trig Function 1. Order of Operations with Trig Function I'm trying to make a table of values for the trigonometric function $y = 2sin3(x-30) +1$, though I don't know exactly how to punch this into my calculator (confused with the order of operations). All I know is that from the parent function, $y = Sinx$, the following transformations have been made: Phase angle shift: $30deg$ right. Amplitude: V.E. by $2$. Therefore, $2$. Period: H.C. by $3$. Therefore, $120deg$. Vertical shift: $1$ up. How would I solve for y if $x = 0deg$? 2. Do you mean $y=2\\sin(3(x-30))+1$, or $y=2\\sin(3)(x-30)+1$? Putting careful parentheses around all function arguments is a very good way to enhance clarity of expressions. What kind of calculator do you have? If it's HP with the reverse Polish, let some other (poor) fellow help you out. If it's a more normal calculator, especially if it allows parentheses, you should be able to enter it more or less as it shows up on the page. Also, I would highly recommend you double-check to see if the calculator is in degrees or radian mode. I'm assuming the 30 is degrees? 3. That's the problem here, I don't know where the parenthesis are supposed to be; that's how much teacher typed it.", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(X+\\pi/4)$ Using window coordinates $-6.5 ≤ x ≤ 6.5$ $-1.25 ≤ y ≤ 1.25,$ we get the following: $Y_1 = \\sin(X)$ is shown in red, and $Y_2 = \\sin(X+\\pi/4)$ is in yellow. We notice that addition of $\\pi/4$ to the argument has shifted the graph to the left by $\\pi/4$ units.In general: Replacing $x$ by $x+c$ shifts the graph to the left $c$ units. How would we shift the graph to the right $\\pi/4$ units? (c) Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(3*X)$ Using the same window coordinates and color coding as in part (b), we obtain the following graph. Notice that the graph of $\\sin(3*x)$ oscillates three times as fast as the graph of $\\sin(x).$ Thus, every $2\\pi$ units, the graph of $\\sin(3*x)$ oscillates three times. In general: Multiplying $x$ by $b$ multiplies the rate of oscillation by $b.$ (d) $Y_1 = \\sin(\\pi*X)$ $Y_2 = 2\\sin(\\pi(X+3))$ The graphs are shown in the window $-3.2 ≤ x ≤ 3.2$ $-2.5 ≤ y ≤ 2.5$ Notice several things: • The red graph, $y = \\sin(\\pix),$ crosses the $x-$axis at integral values $0 ±1, ±2, ...$ For instance, when $x = 1,$ we get $\\sin(\\pix) = \\sin(\\pi) = 0.$ • The yellow graph, $y", "Default: False ## sin(x) Result of the operation in pi radians, which uses fewer resources on the FPGA. This output assumes the same numeric representation as x. When x is of the form x = a + bi, that is, when x is complex, the following equation defines sin: $\\mathrm{sin}\\left(x\\right)=\\mathrm{sin}\\left(a\\right)*\\mathrm{cosh}\\left(b\\right)+i\\left(\\mathrm{cos}\\left(a\\right)*\\mathrm{sinh}\\left(b\\right)\\right)$ ## cos(x) Result of the operation in pi radians, which uses fewer resources on the FPGA. This output assumes the same numeric representation as x. When x is of the form x = a + bi, that is, when x is complex, the following equation defines cos: $\\mathrm{cos}\\left(x\\right)=\\mathrm{cos}\\left(a\\right)*\\mathrm{cosh}\\left(b\\right)+i\\left(-\\mathrm{sin}\\left(a\\right)*\\mathrm{sinh}\\left(b\\right)\\right)$ ## output valid Boolean value that indicates whether this node computes a result that downstream nodes can use. Wire this output to the input valid input of a downstream node to transfer data from the node to the downstream node. True Downstream nodes can use the result this node computes. False This node returns an undefined value that downstream nodes cannot use. Note This node may return different undefined values when executed in simulation mode versus when executed on hardware. ## ready for input Boolean value that indicates whether this node is ready to accept new input data. Use Feedback Node to wire this output to the ready for output output of an upstream node. True The node is ready", "$$\\pi k$$ (the c’s (cot and csc) have the “easier” ones). To get the new asymptotes, set the trig argument to asymptotes, and solve for x. For example, if we have $$\\tan \\left( {3x+\\pi } \\right)\\,\\,(\\text{which would be}\\,\\,\\tan \\left( {3\\left( {x+\\frac{\\pi }{3}} \\right)} \\right))$$, we would solve: $$\\displaystyle 3x+\\pi =\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,\\,\\,3x=\\left( {\\frac{\\pi }{2}-\\pi } \\right)+\\pi k;$$ $$\\displaystyle 3x=-\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,x=\\frac{{-\\frac{\\pi }{2}}}{3}+\\frac{{\\pi k}}{3};\\,\\,\\,\\,x=-\\frac{\\pi }{6}+\\frac{{\\pi k}}{3}$$. 3. The new range will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 4. Draw asymptotes first; phase shift will take care of itself. (To get an asymptote starting point, you can set k = 0 in your new asymptote equation, k = –1 for one to the left, k = 1 for one to the right). Right in between the asymptotes (you can take average of the x’s), draw the middle of the graph (but make sure it is shifted up or down according to the vertical shift, or d). 5. New period will be $$\\frac{\\pi }{b}$$ (since tan and cot have periods of ). This should be the distance between asymptotes (the coefficient of “k” in the asymptote equations). 6. If trig function is negative, make sure you flip the graph. 7. New domain will be all x values except for the asymptotes. 8. To get two more points on either side", "# Graphing sin, finding phase shift, period and transformations I need to graph $$y=2 \\cos\\left(x - \\frac{\\pi}{3}\\right)$$ and I am having trouble figuring out what points will be on the graph. The book tells me to split it into four parts, which doesn't make sense since they split it into five parts anyways. I know that the points will be between $\\pi/3$ and $(\\pi/3 + 2\\pi)$ so that gives me $\\pi/3 < x < 7\\pi/3$ The only problem is I have no idea how to split that up into equal parts to make graphing easier. - Indeed, it is useful to split into four parts, as follows: $$\\bigg[\\frac{{2\\pi }}{6},\\frac{{5\\pi }}{6}\\bigg],\\bigg[\\frac{{5\\pi }}{6},\\frac{{8\\pi }}{6}\\bigg],\\bigg[\\frac{{8\\pi }}{6},\\frac{{11\\pi }}{6}\\bigg],\\bigg[\\frac{{11\\pi }}{6},\\frac{{14\\pi }}{6}\\bigg].$$ The endpoints are $\\pi/3$ and $7\\pi / 3$. The middle point is then given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{7\\pi }}{3}\\bigg)\\bigg /2 = \\frac{{8\\pi }}{6}.$$ Next, the middle point between $\\pi/3$ and $8\\pi/6$ is given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\bigg(\\frac{{2\\pi }}{6} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{5\\pi }}{6},$$ and the middle point between $8\\pi/6$ and $7\\pi/3$ is given by $$\\bigg(\\frac{8\\pi }{6} + \\frac{{7\\pi }}{3}\\bigg)\\bigg/2 = \\bigg(\\frac{{8\\pi }}{6} + \\frac{{14\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{11\\pi }}{6}.$$ Adam: $\\pi/3 = 2\\pi/6$, $7\\pi/3 = 14\\pi/6$..., the book just reduced the fractions (Shai used 6 in the denominator to help establish the four parts, you can simply" ]

[ "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "New Zealand Level 7 - NCEA Level 2 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-60^\\circ\\right)$g(x)=tan(x60°). ### Outcomes #### M7-2 Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs #### 91257 Apply graphical methods in solving problems", "sure so I experimented, I decided to use a sine graph with a wavelength of 20 y=sin nx the wavelength is 2pi/n = 20 so n = pi/10 so my experimenting graph is $$y=sin(\\frac{\\pi }{10}x)\\\\ y=sin(\\frac{\\pi }{10}x)=sin(\\frac{\\pi }{10}(x-20))\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=sin(\\frac{\\pi }{10}\\frac{x}{5})=sin(\\frac{\\pi }{10}(\\frac{x}{5}-20))=sin(\\frac{\\pi }{10}(\\frac{x-100}{5}))$$ So the smallest a that we know with certainty is 100. Here is the graph i was playing with. https://www.desmos.com/calculator/oonutt20vz ------------------- Let me see if I can do this more simply. $$y=f(x)=f(x-20)\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=f(\\frac{x}{5})=f(\\frac{x}{5}-20)=f(\\frac{x-100}{5})$$ There you go, that was simple. The right shift is 100. Mar 27, 2018 #5 +100025 +1 2. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So \"(1,-3); (2,3); (5,-7)\" - without the quotes - is a valid answer format.) $$x^3-x^2+x+1 = x^3+x^2+x-1\\\\ -x^2+1 = x^2-1\\\\ 2=2x^2\\\\ x^2=1\\\\ x=\\pm1\\\\ f(1)=1-1+1+1=2\\\\ f(-1)=-1-1-1+1=-2\\\\$$ so the points of intersection are (-1,-2); (1,2) check: Mar 27, 2018", "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "# Phase shift in frequency domain Hello, I'using Matlab to simulate phase shift in frequency domain (FD). I have got real and imaginary parts of the signal after FFT. I'd like to use phase shift in FD. This works: Y=fft(y); YY=Y.exp(-i*2*pi*nk/N*samples_delay); result=ifft(YY); But in my DSP I can't use the formula above and I need to use real and imaginary parts from fft signal. I supposed exp(-i*2*pi*nk/N*samples_delay) can be transfered to this: Shifted real part = re*cos(2*pi*nk/N*samples_delay) Shifted imaginary part = im*sin(-2*pi*nk/N*samples_delay) Then I transfered it back to time domain but result is inccorect. Could anybody help me? Thank you Related Other Physics Topics News on Phys.org DrGreg Gold Member Think about what you are doing. Is it true that$$(a + ib) (x + iy) = ax + iby?$$ sophiecentaur blue_leaf77 Homework Helper Shifted real part = re*cos(2*pi*nk/N*samples_delay) Shifted imaginary part = im*sin(-2*pi*nk/N*samples_delay) From what I understand, you actually want to calculate the real and imaginary parts of YY, am I right? Then you have to calculate Re[Y.exp(-i*2*pi*nk/N*samples_delay)] and Im[Y.exp(-i*2*pi*nk/N*samples_delay)]. Consider this $$Ze^{i\\theta} = (Re[Z] + i Im[Z])(\\cos(\\theta)+i\\sin(\\theta) )$$ Calculate the last expression to see which are the right real and imaginary parts of ##Ze^{i\\theta}##.", "graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). 1. The point $A$A on the graph of $f\\left(x\\right)$f(x) has the coordinates $\\left(0,0\\right)$(0,0). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 2. The point $B$B on the graph of $f\\left(x\\right)$f(x) has the coordinates $\\left(\\frac{\\pi}{4},1\\right)$(π4,1). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 3. The graph of $f\\left(x\\right)$f(x) has an asymptote passing through point $C$C with coordinates $\\left(\\frac{\\pi}{2},0\\right)$(π2,0). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 4. Using the answers from the previous parts, apply a phase shift to the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx to draw the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). Question 2 The graph of $f\\left(x\\right)=\\tan x$f(x)=tanx has been drawn below. On the same set of axes draw the graph of $g\\left(x\\right)=\\tan\\left(x+\\frac{\\pi}{2}\\right)$g(x)=tan(x+π2). Question 3 The graph of a function in the form $f\\left(x\\right)=\\tan\\left(x-h\\right)$f(x)=tan(xh), where $0\\le h<\\pi$0h<π, is drawn below. 1. Compared to the graph of $y=\\tan x$y=tanx, state the phase shift of the graph of $f\\left(x\\right)$f(x).", "$$\\pi k$$ (the c’s (cot and csc) have the “easier” ones). To get the new asymptotes, set the trig argument to asymptotes, and solve for x. For example, if we have $$\\tan \\left( {3x+\\pi } \\right)\\,\\,(\\text{which would be}\\,\\,\\tan \\left( {3\\left( {x+\\frac{\\pi }{3}} \\right)} \\right))$$, we would solve: $$\\displaystyle 3x+\\pi =\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,\\,\\,3x=\\left( {\\frac{\\pi }{2}-\\pi } \\right)+\\pi k;$$ $$\\displaystyle 3x=-\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,x=\\frac{{-\\frac{\\pi }{2}}}{3}+\\frac{{\\pi k}}{3};\\,\\,\\,\\,x=-\\frac{\\pi }{6}+\\frac{{\\pi k}}{3}$$. 3. The new range will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 4. Draw asymptotes first; phase shift will take care of itself. (To get an asymptote starting point, you can set k = 0 in your new asymptote equation, k = –1 for one to the left, k = 1 for one to the right). Right in between the asymptotes (you can take average of the x’s), draw the middle of the graph (but make sure it is shifted up or down according to the vertical shift, or d). 5. New period will be $$\\frac{\\pi }{b}$$ (since tan and cot have periods of ). This should be the distance between asymptotes (the coefficient of “k” in the asymptote equations). 6. If trig function is negative, make sure you flip the graph. 7. New domain will be all x values except for the asymptotes. 8. To get two more points on either side", "# If y = sin x° then find $$\\frac{{dy}}{{dx}}$$? 1. cos x 2. 0 3. -cos x 4. None of the above ## Answer (Detailed Solution Below) Option 4 : None of the above Free CRPF Head Constable Ministerial 22 Feb 2023 (Shift 1) Memory-Based Test 1.3 Lakh Users 100 Questions 100 Marks 90 Mins ## Detailed Solution Concept: Suppose that we have two functions f(x) and g(x) and they are both differentiable. • Chain Rule: $$\\frac{{\\rm{d}}}{{{\\rm{dx}}}}\\left[ {{\\rm{f}}\\left( {{\\rm{g}}\\left( {\\rm{x}} \\right)} \\right)} \\right] = {\\rm{\\;f'}}\\left( {{\\rm{g}}\\left( {\\rm{x}} \\right)} \\right){\\rm{g'}}\\left( {\\rm{x}} \\right)$$ Calculation: Given: y = sin x° We know that, 180° = π radian ∴ 1° = $$\\frac{\\pi}{180}$$ radian Now, x° = $$\\frac{\\pi x}{180}$$ radian ⇒ y = $$\\sin \\left( {\\frac{{\\rm{\\pi \\cdot x }}}{{180}}} \\right){\\rm{\\;}}$$ Differenatiate with respect to x, we get $$\\frac{{{\\rm{dy}}}}{{{\\rm{dx}}}} = {\\rm{\\;}}\\frac{{\\rm{\\pi }}}{{180}}{\\rm{\\;}} \\times \\cos \\left( {\\frac{{\\rm{\\pi \\cdot x}}}{{180}}} \\right){\\rm{\\;}}$$ Latest Airforce Group X Updates Last updated on Nov 11, 2022 Airforce Group X Phase 2 Admit Card Out. The Indian Air Force (IAF) had released the official notification for the IAF Group X (01/2023) on 7th November 2022. The selection of the candidates will depend on three stages which are Phase 1 (Online Written Test), Phase 2 ( DV, Physical Fitness Test, Adaptability Test I & II), and Phase 3 (Medical Examination). The written", "# Get slopes of integer shift of an image in radians I tried to reproduce the result from this paper, However I'm stuck at the Section III of the paper which involve integer shift. Two images which their phase correlation $\\theta(k_1,k_2)$ was calculated at pixel-accuracy(integer shift). I get their phase component as shown(from -$\\pi$ to $\\pi$). It can be represented by $\\theta(k_1,k_2)=\\theta_1(k_1,k_2)-\\theta_2(k_1,k_2)$. Since the two images shifted in parallel, it can be approximated as $\\theta(k_1,k_2) \\approx ak_1+bk_2$ where $a$ and $b$ is the slopes obtained through least square method, $k_1$ and $k_2$ are the value $MxN$ of the image. Plotting $\\theta(k_1,k_2)$: And to get phase component with only sub-pixel accuracy(decimal shift), the phase component is subtracted with integer shift. The equation is $\\theta''(k_1,k_2)=\\theta(k_1,k_2)-(a'k_1+b'k_2)$ However, as you can see from the second figure, the value of $\\theta''(k_1,k_2)$ ranges wider than -$\\pi$ to $\\pi$(as compared to Figure 4 in the paper), which I suspect the slope($a'$,$b'$) was obtained inaccurately. The paragraph above the Equation (14) mentioned that they used conventional POC to obtain the slopes. However I'm not sure did I understood it correctly. The code I implemented: function [ output_args ] = phasecorrlsa( refIm, shifIm ) F=(fft2(double(refIm))); G=(fft2(double(shifIm))); [m, n]=size(refIm); [M,N] = meshgrid(1:m,1:n); X = [M(:), N(:)]; R=(F.*conj(G))./abs(F.*conj(G)); r=(ifft2(R)); % surf(r) [apx, bpx, rhat]=lsa(angle(r)); [~,w] = max(r(:)); [del_hat2p, del_hat1p]", "maliaseth0 2022-01-25 Is the phase shift of $-3\\mathrm{sin}\\left(2x+\\frac{\\pi }{2}\\right)+1$ equal to $-\\frac{\\pi }{4}$ or just $-\\frac{\\pi }{2}$? trnovitom06 Expert The definition of \"phase shift\" actually depends on the context. For example, in a context where $A\\mathrm{sin}\\left(\\omega x-\\varphi \\right)$ is a sine wave or signal and x is time, one often calls the phase shift. To be more precise, this could be called an angular phase shift, since this parameter shifts the argument of the sine function (an angle) away from its original angle. On the other hand, in other contexts one would rather write the same sine wave as $A\\mathrm{sin}\\left(\\omega \\left(x-\\frac{\\varphi }{\\omega }\\right)\\right)$, which highlights a different kind of shift, namely a temporal shift of $\\frac{\\varphi }{\\omega }$ seconds. If you graph the sine wave as a function of time, then this quantity is the proper horizontal shift that you see on the graph. marzembreax Expert The phase shift of $-3\\mathrm{sin}\\left(2x+\\frac{\\pi }{2}\\right)+1=-3\\mathrm{sin}\\left(2\\left(x+\\frac{\\pi }{4}\\right)\\right)+1$ relative to", "How do you graph y = cos(x- pi/6)? It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "$t$ then the term $-vt$ is just some phase-shift and we can see the graph is just a horizontal shift of the sine wave. In particular we shift the $y = A \\sin(x)$ graph $vt$-units to right to form $y = A \\sin(x-vt)$. This is a right-moving wave if we assume $v>0$. To see it, I think setting $0 \\leq x-vt \\leq 2\\pi$ might be helpful, then $vt \\leq x \\leq 2\\pi+vt$ and if we animate $t$ we can actually see the wave travel. Other more complicated wave graphs could be studied. I happen to have this graph of a solution to the wave equation sitting around. Of course, adding graphs has interesting interpretations in terms of constructive and destructive interference. Or, if we add time I think you can get beats. There is much to explore here. Edit: as Dave Renfro pointed out, I am using forbidden trigonometry. To, give a less exciting, but very much applied example consider: $$y = y_o- \\frac{g}{2}t^2$$ If we start with $y=t^2$ then multiplying b $-g/2$ stretches the graph and flips it over. The significance is that $g$ is the acceleration due to gravity. Then, adding $y_o$ shifts the graph $y=-gt^2/2$ up by $y_o$ and in this discussion $y_o$ has the significance of being the initial condition. A more ambitious formula", "We’re rewarding the question askers & reputations are being recalculated! Read more. # Tag Info 0 You are really close! Change your signal and steering vectors to be complex. Specifically for the steering vectors, these coefficients are meant to act as phase shifts. Using a real sinusoid will introduce a phase shift term in the opposite angle direction, which you don't want. Doing this alone you will see an improvement in your pseudospectrum. In regards ... 4 You're almost there; you just need to connect a few dots. Let your $\\frac{10}{5 + i 2 \\pi f} = G(f)$. Then $g(t) = 10 u(t) e^{-5 t}$. Now we get into that exponent part. Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}... 0 The principal idea behind my recommended approach is to identify characteristic points in your cycle. You can then make a mapping from time within the cycle (as it is occuring) and the location of that point within your representative cycle. Once you have a set of mapping points, create a best fit function. This will then give you a mapping function of ... 1 You", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "cos((-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3))]; % Make the steering vectors complex exp((1i*(-2*pi*ant_spacings(1)/lambda)*sin(theta(i)*(pi/180) - pi/3))); exp((1i*(-2*pi*ant_spacings(2)/lambda)*sin(theta(i)*(pi/180) - pi/2))); exp((1i*(-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3)))]; spectrum_func(i) = 1/((steering_vec'*noise_vecs)*noise_vecs'*steering_vec); end % grab theta corresponding to highest peak, round to nearest half integer [mx,argmax] = max(spectrum_func); estimated_AoA = floor(2*theta(argmax)+.5)/2; fprintf(\"Estimated AoA: %.1f \\t Hard-coded AoA: %.1f.\\n\", estimated_AoA, AoA) plot(theta,20*log10(abs((spectrum_func)))) title('MUSIC Psuedospectrum') xlabel('Angle of Arrival') $$$$ • Ah, I thought this might have had something to do with it. Is there any way you could elaborate on the 'using a real sinusoid will introduce a phase shift term in the opposite angle direction' aspect? – OGBerglemir Nov 19 '19 at 20:00 • @OGBerglemir When you multiply two complex exponentials, you can add the general phase terms together. If you recall, a real sinusoid is composed of two complex exponentials, which rotate in different directions. Multiplying by a real sinusoid uses both components, so you will shift a signal's phase using both, and not one as intended. – Envidia Nov 19 '19 at 20:07 • Ah, understood. – OGBerglemir Nov 19 '19 at 21:59", "import numpy as np sr = 22050 ts = 1.0/sr duration = 2 t = np.arange(0,duration,ts) t_change = 1 freq1 =4.20 phase1 = 2*np.pi*freq1* t[t<=t_change] freq2 = 6.66 phase2 = 2*np.pi*freq2*t[t>t_change] + 2*np.pi*(freq1-freq2)*t_change total_phase = np.concatenate((phase1, phase2)) out = np.cos(total_phase) Considering more generalized cases with multiple steps, the phase offset to add for any given segment is: $$\\theta_n = \\sum_{m=1}^{n-1}2\\pi f_m t_m - \\sum_{m=1}^{n-1}2\\pi f_{m+1} t_m \\tag{1} \\label{1}$$ With the resulting phase ramp proceeding starting at $$t_n$$ as: $$\\theta(t) = 2\\pi f_n t + \\theta_n$$ This is demonstrated with a 3 segment ramp with reference to the figure below, where we note that each next phase ramp will start at the completed phase from the previous frequency. For example, the second phase offset given as $$\\theta_2$$ below is similarly determined as above by solving for the intersection of linear line equations to be: $$\\theta_2 = \\theta(t_2)-2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + \\theta_1 - 2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + 2\\pi (f_1-f_2)t_1 - 2\\pi f_3 t_2$$ Which is consistent with the generalized form given in $$\\ref{1}$$ as: $$\\theta_2 = 2\\pi f_1 t_1 + 2\\pi f_2 t_2 -2\\pi f_2 t_1 - 2\\pi f_3 t_2$$ With this, the third segment plotted below is given as: $$\\theta(t) = 2\\pi f_3 t + \\theta_2$$ Summary: In summary I have", "layers. In order for my approach to work as intended, the layer settings window must look like this: However, quite annoyingly, when you close and reopen this saved document, the layer settings window looks like this: So, it must be changed back each time to allow trouble-free math drawing. # Finding Reciprocal Functions SEC, CSC, COT on TI-30XA First, you may want to refresh your memory of the regular trig functions: Now, let’s find $\\csc \\frac{5\\pi }{4}$. If we look it up on WolframAlpha, we find that it equals $-\\sqrt{2}$, or, roughly, $-1.4142$. To get this answer on the TI-30XA, first remember that $\\csc$ is equal to $\\frac{1}{\\sin }$. So, what we’re looking for could be expressed as $\\frac{1}{\\sin \\frac{5\\pi }{4}}$. Once we know this, we can type $5\\cdot \\pi$ into the TI-30XA. This gives us ~15.7. (Make sure you’re in radian mode. If not, press the DRG button — located by the 2nd button, upper-left — until you see RAD appear in small letters on the screen.) Now, we can divide 15.7 by 4. This gives us 3.925. If we take the $\\sin$ of this (by pressing the “sin” button while 3.925 is displayed), we get $-0.7$. IF we divide 1 by $-0.7$, we get $-1.43$. As you can see, this is close to the correct value.", "you would use closed graph theorem, in fact, you would need either \\mathcal{S}=\\mathcal{S}(\\mathbb{R}^d) to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied (\\mathcal{S} is normable or compact, at least I don't think so). ... 0 The shift is defined by g_a(x) = f(x-a). Then you write$$F[g_a](\\xi) = \\int_{\\Bbb R}g_a(x)\\exp(-ix\\xi)dx = \\int_{\\Bbb R}f(x-a)\\exp(-ix\\xi)dx.Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent. 0 Let g_0 = \\frac{1}{2\\pi}\\int_{0}^{2\\pi}g(x)dx and f_0=\\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)dx. Then \\begin{align} \\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)g(nx)dx &=\\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)(g(nx)-g_0)dx+f_0g_0 \\\\ &=\\frac{1}{2\\pi}\\sum_{k=1}^{n}\\int_{\\frac{2\\pi}{n}(k-1)}^{\\frac{2\\pi}{n}k}f(x)(g(nx)-g_0)dx+f_0g_0 \\\\ &=\\frac{1}{... 2 Cosine is an even function; sine is odd. Using negative indices for these functions adds no information because, for example,5\\sin 2x+ 4\\sin(-2x) = \\sin 2x$$On the other hand, e^{-2ix} is not just a constant multiple of e^{2ix}. And we need both of these to express \\sin 2x, as in$$\\sin 2x = \\frac1{2i}(e^{2ix} - e^{-2ix})$$A more precise ... 0 After a simple classification I have figured it out. We could solve it by induction on the degree of the polynomial P(x). If deg P(x)=1, then c_1 \\in \\mathbb{R} \\setminus \\mathbb{Q} and apparently <c_1n+c_0> is equidistributed in [0,1). Assume deg P(x)=m and at least one of its coefficient c_1,……,c_m is irrational, then <P(n)> ...", "## anonymous 4 years ago Write an equation this is equivalent to f(x)=-5sin(x-pi/2) - 8 using trig.function of x 1. richyw ok so you have $f(x)=-5\\sin\\left(x-\\frac{\\pi}{2}\\right)-8$ For now just focus on $g(x)=\\sin\\left(x-\\frac{\\pi}{2}\\right)$ what does this do to the graph? 2. richyw where $$x=\\pi/2$$, what is $\\sin\\left(x-\\frac{\\pi}{2}\\right)$also brb. just gotta hand in something quick... 3. anonymous okay.. will it be sin(x-pi/2)=0?? 4. anonymous i dont really understand the question itself 5. richyw sorry open study died on me. 6. richyw You are correct that $$g(\\pi/2)=0$$ You can repeat this with some other key x-values $$x=\\pi/2,\\quad x=\\pi ,\\quad x=0$$ and find that what you get is $\\sin\\left(x-\\frac{\\pi}{2}\\right)=-\\cos\\pi$ 7. richyw you will eventually just memorize these phase shifts, but if you ever forget on a test, you can always make a quick sketch using the method I explained. So now remember that we called $g(x)=\\sin\\left(x-\\frac{\\pi}{2}\\right)=-\\cos x$ Going back to the original function you have $f(x)=-5\\sin\\left(x-\\frac{\\pi}{2}\\right)-8$$f(x)=-5g(x)-8$So the hard part is already done. The 8 is all by itself so it just shifts the up and down, and the -5 just makes the amplitude of the wave greater. All you have to do now is plug in the expression you worked out for g(x). $f(x)=-5(-\\cos x)-8$$f(x)=5\\cos x -8$ 8. richyw Oops made a mistake, in the second last post. It should say :" ]

[ "degree, the angles that a diagonal of a box with dimensions $10 \\mathrm{~cm} \\times 11 \\mathrm{~cm} \\times 25 \\mathrm{~cm}$ makes ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal.Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is ortho ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ ... close Show that if $\\mathbf{v}$ is orthogonal to both $\\mathbf{w}_{1}$ and $\\mathbf{w}_{2}$, then $\\mathbf{v}$ is orthogonal to $k_{1} \\mathbf{w}_{1}+k_{2} \\mathbf{w}_{2}$ for all scalars $k_{1}$ and $k_{2}$.Show that if $\\mathbf{v}$ is orthogonal to both $\\mathbf{w}_{1}$ and $\\mathbf{w}_{2}$, then $\\mathbf{v}$ is orthogonal to $k_{1} \\mathbf{w}_{1}+k_{2} ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Which of the following is the unit vector in direction of$a \\times", "# Math Help - Vector geometry question 1. ## Orthogonal vectors Let $\\mathbf{a,b,c}$ be three vectors in $\\mathbb{R}^3$ which satisfy the relations $\\mathbf{b=c \\times a}\\ and\\ \\mathbf{c=a \\times b}$ 1) Show that $\\mathbf{a, b\\ and\\ c}$ are a set of mutually orthogonal vectors 2) Show that $\\mathbf{b\\ and\\ c}$ are of equal length and that if $\\mathbf{b \\neq 0}$, then $\\mathbf{a}$ is a unit vector. 2. Originally Posted by acevipa Let $\\mathbf{a,b,c}$ be three vectors in $\\mathbb{R}^3$ which satisfy the relations $\\mathbf{b=c \\times a}\\ and\\ \\mathbf{c=a \\times b}$ 1) Show that $\\mathbf{a, b\\ and\\ c}$ are a set of mutually orthogonal vectors 2) Show that $\\mathbf{b\\ and\\ c}$ are of equal length and that if $\\mathbf{b \\neq 0}$, then $\\mathbf{a}$ is a unit vector. This is pretty straight forward isn't it? $\\mathbf{b= c\\times a}$ is, by definition of \"cross product\", orthogonal to both $\\mathbf{a}$ and $mathbf{c}$ and $\\mathbf{c= a\\times b}$ is orthogonal to a. Further, since $\\mathbf{a}$ is orthogonal to $\\mathbf{c}$, $|\\mathbf{b}|= |\\mathbf{a}\\times\\mathbf{c}|= |\\mathbf{a}||\\mathbf{c}\\sin(\\pi/2)$ so that $1= (|\\mathbf{a}|)(1)(1)$.", "# Solve this following Question: If $\\hat{\\mathrm{a}}$ and $\\hat{\\mathrm{b}}$ are unit vectors such that $(\\hat{\\mathrm{a}}+\\mathrm{b})$ is a unit vector, what is the angle between $\\hat{\\mathrm{a}}$ and $\\hat{\\mathrm{b}}$ ? Solution:", "+0 # Pls help me +1 32 1 +57 The vectors $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}$ satisfy $\\|\\mathbf{a}\\| = \\|\\mathbf{b}\\| = 1,$ $\\|\\mathbf{c}\\| = 2,$ and $\\mathbf{a} \\times (\\mathbf{a} \\times \\mathbf{c}) + \\mathbf{b} = \\mathbf{0}.$If $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{c},$ then find all possible values of $\\theta,$ in degrees. Apr 25, 2021", "How do you find all unit vectors orthogonal to both the vectors given below: (1,2,-1) and (3,3,-4)? Jul 5, 2016 $\\pm \\frac{1}{\\sqrt{35}} \\left(- 5 , 1 , - 3\\right)$ Explanation: The unit vectors (in opposite directions) that are orthogonal to both $a \\mathmr{and} b$ is $\\pm \\frac{a X b}{|} a X b |$. Here $a = \\left(1 , 2 , - 1\\right) \\mathmr{and} b = \\left(3 , 3 , - 4\\right) ,$ $a X b = \\left(\\left(2\\right) \\left(- 4\\right) - \\left(3\\right) \\left(- 1\\right) , \\left(- 1\\right) \\left(3\\right) - \\left(- 4\\right) \\left(1\\right) , \\left(1\\right) \\left(3\\right) - \\left(3\\right) \\left(2\\right)\\right)$ $= \\left(- 5 , 1 , - 3\\right) \\mathmr{and} | \\left(- 5 , 1 , - 3\\right) | = \\sqrt{35}$. So, the answer is $\\pm \\frac{1}{\\sqrt{35}} \\left(- 5 , 1 , - 3\\right)$", "vectors are orthogonal then we know that the angle between them is 90 degrees.", "Note: You need to know that the vector product ${\\bf a} \\times {\\bf b}$ is the product of the magnitudes of the vectors times the sine of the angle between the vectors and it is a vector perpendicular to ${\\bf a}$ and ${\\bf b}$.", "set of vectors $$\\nu_e, \\nu_\\mu, \\nu_\\tau$$ at right angles to each other. The other is the mass basis: a set of not-flavors $$\\nu_\\text{light}, \\nu_\\text{middle}, \\nu_\\text{heavy}$$ which are also at right angles to each other. To describe the orientation of one set of axes in terms of the other takes three rotations; with two angles you can get your head above the water but you might be stuck facing away from the beach.", "unit vector. $\\bf edit:$ to find all the unit vectors, we need to find an orthogonal basis first. if you project $(3,0,-2)$ onto $(2,1,0)$ you get $(4,2,0).$ that is we have split $(3,0,-2) = (4,2,0) + (-1,2,-2).$ therefore an orthogonal basis is $\\{(2,1,0), (-1,2,-2)\\}.$ we make them orthonormal by dividing by their lengths so that $$\\{u = (2/\\sqrt 5, 1/\\sqrt 5), v = (-1/3, 2/3, 2/3)\\}$$ finally, an arbitrary unit orthogoanal vector is $$u\\cos t + v \\sin t, t \\text{ any real number.}$$ • you do find all vectors orthogonal as the linear combinations of the two vectors in the post. they are linearly independent, so it is a basis for the space orthogonal to the line $(2t, -4t, 3t)$ – abel Feb 20 '15 at 1:41", "unit magnitude,acting at a point ,is of unit magnitude, then angle between these forces, is • A ${ 45 }^{ \\circ }$ • B ${ 60 }^{ \\circ }$ • C ${ 90 }^{ \\circ }$ • D ${ 120 }^{ \\circ }$", "14 views Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy$$ \\cos \\ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$. Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$.Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector ... Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors.Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. ... Let $\\vec{v}$ and $\\vec{w}$ be vectors in $\\mathbb{R}^{n}$. If $\\|\\vec{v}\\|=\\|\\vec{w}\\|$, show there is an orthogonal matrix $R$ with $R \\vec{v}=\\vec{w}$ and $R \\vec{w}=\\vec{v}$.Let $\\vec{v}$ and $\\vec{w}$ be vectors in $\\mathbb{R}^{n}$. If $\\|\\vec{v}\\|=\\|\\vec{w}\\|$, show there is an orthogonal matrix $R$ with \\(R \\v ...", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "the set of standard unit vectors in R^{3} . Show that each ordered pair of vectors in S is orthogonal. 1 answer 6 views Let S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\} be the set of standard unit vectors in R^{3} . Show that each ordered pair of vectors in S is orthogonal.Let S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\} be the set of standard unit vectors in R^{3} . Show that each ordered pair of vectors in S is ortho ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 How do I determine the equation of a normal in R^2 or R^3? 1 answer 9 views How do I determine the equation of a normal in R^2 or R^3?How do I determine the equation of a normal in R^2 or R^3? ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 What are Orthogonal Projections? 1 answer 7 views What are Orthogonal Projections?What are Orthogonal Projections? ... close 1 answer 5 views close 1 answer 5 views Find the orthogonal projections of the vectors \\mathbf{e}_{1}=(1,0) and \\mathbf{e}_{2}=(0,1) on the line L that makes an angle \\theta with the positive x-axis in R^{2}.Find the orthogonal projections of the vectors \\mathbf{e}_{1}=(1,0) and \\mathbf{e}_{2}=(0,1) on the line L that makes an angle \\theta with the ... close 1 answer 5 views If \\mathbf{u} and \\mathbf{v}", "$S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal.Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is ortho ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that two nonzero vectors $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ in $R^{3}$ are orthogonal if and only if their direction cosines satisfy Show that two nonzero vectors $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ in $R^{3}$ are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ in $R^{3}$ are orthogonal if and only if their direction cosines satisfy \\cos \\ ... Let $U, V, W$ be orthogonal vectors and let $Z=a U+b V+c W$, where $a, b, c$ are scalars. Let $U, V, W$ be orthogonal vectors and let $Z=a U+b V+c W$, where $a, b, c$ are scalars.Let $U, V, W$ be orthogonal vectors and let $Z=a U+b V+c W$, where $a, b, c$ are scalars. a) (Pythagoras) Show that \\(\\|Z\\|^{2}=a^{2}\\|U\\|^{2}+b ...", "perpendicular to $${\\bf T}$$; the vector $${\\bf N}={\\bf T}'/|{\\bf T}'|$$ is thus a unit vector perpendicular to $${\\bf T}$$, called the {\\dfont unit normal\\/}\\index{unit normal}\\index{normal} to the curve. Occasionally of use is the {\\dfont unit binormal\\/}\\index{unit binormal}\\index{binormal} $${\\bf B}={\\bf T}\\times{\\bf N}$$, a unit vector perpendicular to both $${\\bf T}$$ and $${\\bf N}$$. ### Contributors • Integrated by Justin Marshall.", "How can we find all the unit vectors that are orthogonal to the vectors?? – Mary Star Feb 20 '15 at 1:33 • As Abel notes, $\\mathbf v = \\frac{1}{\\sqrt{13}}(3, 0, 2)$ is one such vector. Taking the cross product with $(2, -4, 3)$ and normalizing gives $\\mathbf w = \\frac{1}{\\sqrt{233}} (8, -5, -12)$, and $\\mathbf v$ and $\\mathbf w$ are perpendicular, just like the unit vectors on the $x$- and $y$-axes of the plane. If you form all cosine/sine combinations of these (i.e., $\\cos(t) \\mathbf v + \\sin(t) \\mathbf w$), you get all possible unit vectors perpendicular to the ones given in your problem. – John Hughes Feb 20 '15 at 1:47 • Ok... Thanks!!! :-) – Mary Star Mar 1 '15 at 3:28 As everyone has pointed out, $\\mathbf{a}$ and $\\mathbf{b}$ are parallel, so we only need to use one of them. Using $\\mathbf{a}$, the vecotrs orthogonal to it are all vectors $(x,y,z)$ such that $2x-4y+3z=0$. We can eliminate one of these variables, say $x$, so all the orthogonal vectors are those of the form $(2y-\\frac{3}{2}z, y, z)$. To make them unit vectors, you can divide by their length, which is $\\sqrt{(2y-\\frac{3}{2}z)^2 + y^2 +z^2}$. So all the orthogonal unit vectors are the vectors of the form $$\\frac{(2y-\\frac{3}{2}z, y, z)}{\\sqrt{(2y-\\frac{3}{2}z)^2 + y^2 +z^2}}$$ • In that", "## Precalculus (6th Edition) Blitzer There can be many vectors orthogonal to a given vector $\\mathbf{v}$. Two possible answers are $\\mathbf{u}=5\\mathbf{i}+2\\mathbf{j}$ and $\\mathbf{u}=10\\mathbf{i}+4\\mathbf{j}$. Let the vector orthogonal to $\\mathbf{v}$ be $\\mathbf{u}$. And it is expressed as $\\mathbf{u}=a\\mathbf{i}+b\\mathbf{j}$ …… (1) For $\\mathbf{u}$ and $\\mathbf{v}$ to be orthogonal, their dot product must be zero, therefore $\\mathbf{u}\\cdot \\mathbf{v}=-2a+5b$ …… (2) Keeping the dot product obtained in equation (2) as zero, we get $5b=2a$ …… (3) Equation (2) is a relation between $b$ and $a$ ; for different sets of $b$ and $a$, there will be different vectors orthogonal to $\\mathbf{v}$. Let us put $a=5$ in equation (3) to get \\begin{align} & a=5 \\\\ & b=2 \\end{align} So, substitute these values of $b$ and $a$ in equation (1) to get $\\mathbf{u}=5\\mathbf{i}+2\\mathbf{j}$ Hence, The vector $\\mathbf{u}=5\\mathbf{i}+2\\mathbf{j}$ is orthogonal to $\\mathbf{v}=-2\\mathbf{i}+5\\mathbf{j}$. Let us put $a=10$ in equation (3) to get \\begin{align} & a=10 \\\\ & b=4 \\end{align} So, substitute these values of $b$ and $a$ in equation (1) to get $\\mathbf{u}=10\\mathbf{i}+4\\mathbf{j}$ Hence, The vector $\\mathbf{u}=10\\mathbf{i}+4\\mathbf{j}$ is orthogonal to $\\mathbf{v}=-2\\mathbf{i}+5\\mathbf{j}$.", "# ML Wiki ## Vector Orthogonality In geometry, we call two vectors $\\mathbf x$ and $\\mathbf y$ orthogonal of the angle between then is 90 - i.e. they are perpendicular. ### Inner Product Test • If $\\mathbf x$ and $\\mathbf y$ are perpendicular, then we can use the Pythagoras theorem • $\\| \\mathbf x \\|^2 + \\| \\mathbf y \\|^2 = \\| \\mathbf x + \\mathbf y* \\|^2$ • is there an easier way to tell if 2 vectors are orthogonal? • yes! if their Inner Product is zero, then they are: $\\mathbf x^T \\mathbf y = \\sum x_i y_i = 0 \\Rightarrow \\mathbf x \\, \\bot \\, \\mathbf y$ Why? • $\\| \\mathbf x \\|^2 = \\left(\\sqrt{\\sum x_i^2 } \\right)^2 = \\mathbf x^T \\mathbf x$ • let's expand the Pythagoras theorem: • $\\| \\mathbf x \\|^2 + \\| \\mathbf y \\|^2 = \\| \\mathbf x + \\mathbf y* \\|^2$ • $\\mathbf x^T \\mathbf x + \\mathbf y^T \\mathbf y = (\\mathbf x + \\mathbf y)^T (\\mathbf x + \\mathbf y) = \\mathbf x^T \\mathbf x + \\mathbf x^T \\mathbf y + \\mathbf y^T \\mathbf x + \\mathbf y^T \\mathbf y$ • or $\\mathbf x^T \\mathbf y + \\mathbf y^T \\mathbf x = 0$ • note that $\\mathbf x^T \\mathbf y = \\mathbf y^T \\mathbf x$, so we", "$\\mathbf{u}$ orthogonal to a. Let $\\mathbf{u}=(2,-1,3)$ and $\\mathbf{a}=(4,-1,2)$. Find the vector component of $\\mathbf{u}$ along a and the vector component of $\\mathbf{u}$ orthogonal to a.Let $\\mathbf{u}=(2,-1,3)$ and $\\mathbf{a}=(4,-1,2)$. Find the vector component of $\\mathbf{u}$ along a and the vector component of $\\mathbf{u}$ orthog ... Find the orthogonal projections of the vectors $\\mathbf{e}_{1}=(1,0)$ and $\\mathbf{e}_{2}=(0,1)$ on the line $L$ that makes an angle $\\theta$ with the positive $x$-axis in $R^{2}$.Find the orthogonal projections of the vectors $\\mathbf{e}_{1}=(1,0)$ and $\\mathbf{e}_{2}=(0,1)$ on the line $L$ that makes an angle $\\theta$ with the ..." ]

[ "3. You should know that $\\mathbf{a}\\cdot\\mathbf{b} = |\\mathbf{a}||\\mathbf{b}|\\cos{\\theta}$, where $\\theta$ is the angle between them. If $\\mathbf{x} = 6\\mathbf{i} + 3\\mathbf{j} -2\\mathbf{k}$ then $|\\mathbf{x}| = \\sqrt{6^2 + 3^2 + (-2)^2}$ $= \\sqrt{36 + 9+ 4}$ $= \\sqrt{49}$ $= 7$. If $\\mathbf{y} = -2\\mathbf{i} + t\\mathbf{j} - 4\\mathbf{k}$ then $|\\mathbf{y}| = \\sqrt{(-2)^2 + t^2 + (-4)^2}$ $= \\sqrt{4 + t^2 + 16}$ $= \\sqrt{t^2 + 20}$. $\\mathbf{x}\\cdot \\mathbf{y} = (6\\mathbf{i} + 3\\mathbf{j} -2\\mathbf{k})\\cdot (-2\\mathbf{i} + t\\mathbf{j} - 4\\mathbf{k})$ $= -12 + 3t + 8$ $= 3t - 4$. Since $\\mathbf{x} \\cdot \\mathbf{y} = |\\mathbf{x}||\\mathbf{y}|\\cos{\\theta}$ $3t - 4 = 7\\sqrt{t^2 + 20}\\cos{\\left(\\arccos{\\frac{4}{21}}\\right)}$ $3t - 4 = 7\\sqrt{t^2 + 20}\\cdot \\frac{4}{21}$ $3t - 4 = \\frac{4}{3}\\sqrt{t^2 + 20}$ $\\frac{3}{4}(3t - 4) = \\sqrt{t^2 + 20}$ $\\frac{9}{4}t - 3 = \\sqrt{t^2 + 20}$ $\\left(\\frac{9}{4}t - 3\\right)^2 = t^2 + 20$ $\\frac{81}{16}t^2 - \\frac{27}{2}t + 9 = t^2 + 20$ $\\frac{65}{16}t^2 - \\frac{27}{2}t - 11 = 0$. Solve for $t$. 5. Originally Posted by Paymemoney Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds? P.S Seeing as what you have posted is a unit of speed and not a unit of time, you can't convert it into minutes OR seconds. 6. Originally Posted by Prove It The force is the vector that has magnitude $84$ and is", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "# If A = <2 ,4 ,-3 >, B = <3 ,4 ,1 > and C=A-B, what is the angle between A and C? Mar 24, 2018 The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${63.23}^{0}$ #### Explanation: $\\vec{C} = \\vec{A} - \\vec{B} = \\left(< 2 , 4 , - 3 >\\right) - \\left(< 3 , 4 , 1 >\\right)$ $= \\left(< \\left(2 - 3\\right) , \\left(4 - 4\\right) , \\left(- 3 - 1\\right) >\\right) = < - 1 , 0 , - 4 >$ $\\vec{A} = < 2 , 4 , - 3 > \\mathmr{and} \\vec{C} = < - 1 , 0 , - 4 >$ Let $\\theta$ be the angle between them ; then we know $\\cos \\theta = \\frac{\\vec{A} \\cdot \\vec{C}}{| | \\vec{A} | | \\cdot | | \\vec{C} | |}$ =(2* (-1)+(4*0)+(-3*(-4)))/(sqrt(2^2+4^2+(-3)^2)* (sqrt((-1)^2+0^2+(-4)^2)) $= \\frac{10}{\\sqrt{29} \\cdot \\sqrt{17}} = \\frac{10}{22.2} \\approx 0.4504$ $\\therefore \\theta = {\\cos}^{-} 1 \\left(0.4504\\right) \\approx {63.23}^{0}$ The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${63.23}^{0}$ [Ans]", "= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. (1) But as cab, we also have $\\mathbf{c} \\cdot \\mathbf{c} = (\\mathbf{a} - \\mathbf{b}) \\cdot (\\mathbf{a} - \\mathbf{b}) \\;$, which, according to the distributive law, expands to $\\mathbf{c} \\cdot \\mathbf{c} = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) \\;$. (2) Merging the two c · c equations, (1) and (2), we obtain $\\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. Subtracting a · a + b · b from both sides and dividing by −2 leaves $\\mathbf{a} \\cdot \\mathbf{b} = ab \\cos\\theta \\;$,", "# If A = <4 ,-9 ,2 >, B = <1 ,1 ,-4 > and C=A-B, what is the angle between A and C? Apr 13, 2017 The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${19.60}^{0}$ #### Explanation: Let $\\vec{A} = < 4 , - 9 , 2 > , \\vec{B} = < 1 , 1 , - 4 > , \\vec{C} = \\vec{A} - \\vec{B} = < 3 , - 10 , 6 >$ ; Let $\\theta$ be the angle between $\\vec{A} \\mathmr{and} \\vec{C}$ ; then we know $\\cos \\theta = \\frac{\\vec{A} \\cdot \\vec{C}}{| | \\vec{A} | | \\cdot | | \\vec{C} | |}$ =((4*3)+(-9*-10)+(2*6))/(sqrt(4^2+(-9)^2+2^2)* (sqrt(3^2+(-10)^2+6^2)) $= \\frac{114}{\\sqrt{101} \\cdot \\sqrt{145}} = 0.94202 \\therefore \\theta = {\\cos}^{-} 1 \\left(0.94202\\right) \\approx {19.60}^{0}$[Ans]", "# If A = <2 ,4 ,-7 >, B = <3 ,5 ,2 > and C=A-B, what is the angle between A and C? Apr 30, 2018 The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${41.13}^{0}$ #### Explanation: $\\vec{C} = \\vec{A} - \\vec{B} = \\left(< 2 , 4 , - 7 >\\right) - \\left(< 3 , 5 , 2 >\\right)$ $= \\left(< 2 - 3 , 4 - 5 , - 7 - 2 >\\right) = < - 1 , - 1 , - 9 >$ vec A =<2,4, -7> and vec C =<-1,-1,-9> ; theta be the angle between them ; then we know $\\cos \\theta = \\frac{\\vec{A} \\cdot \\vec{C}}{| | \\vec{A} | | \\cdot | | \\vec{C} | |}$ =((2* -1)+(4* -1)+(-7* -9))/(sqrt(2^2+4^2+(-7)^2)* (sqrt((-1)^2+(-1)^2+(-9)^2)) $= \\frac{57}{\\sqrt{69} \\cdot \\sqrt{83}} = \\frac{57}{75.68} . \\approx 0.7532$ $\\therefore \\theta = {\\cos}^{-} 1 \\left(0.7532\\right) \\approx {41.13}^{0}$ The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${41.13}^{0}$ [Ans]", "\\mathbf{c} \\cdot (\\mathbf{a} + \\mathbf{b})(\\mathbf{a} \\times \\mathbf{b})} {2\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})}. \\,$ The Euler line relationships are: $\\mathbf{G} = \\mathbf{M} + \\frac{1}{2} (\\mathbf{O}-\\mathbf{M})\\,$ $\\mathbf{T} = \\mathbf{M} + \\frac{1}{3} (\\mathbf{O}-\\mathbf{M})\\,$ where T is twelve-point center. Also: $\\mathbf{a} \\cdot \\mathbf{O} = \\frac {\\mathbf{a^2}}{2} \\quad\\quad \\mathbf{b} \\cdot \\mathbf{O} = \\frac {\\mathbf{b^2}}{2} \\quad\\quad \\mathbf{c} \\cdot \\mathbf{O} = \\frac {\\mathbf{c^2}}{2}\\,$ and: $\\mathbf{a} \\cdot \\mathbf{M} = \\frac {\\mathbf{a} \\cdot (\\mathbf{b} + \\mathbf{c})}{2} \\quad\\quad \\mathbf{b} \\cdot \\mathbf{M} = \\frac {\\mathbf{b} \\cdot (\\mathbf{c} + \\mathbf{a})}{2} \\quad\\quad \\mathbf{c} \\cdot \\mathbf{M} = \\frac {\\mathbf{c} \\cdot (\\mathbf{a} + \\mathbf{b})}{2}.\\,$ ## Geometric relations A tetrahedron is a 3-simplex. Unlike the case of the other Platonic solids, all the vertices of a regular tetrahedron are equidistant from each other (they are the only possible arrangement of four equidistant points in 3-dimensional space). A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual. A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are (+1, +1, +1); (−1, −1, +1); (−1, +1, −1); (+1, −1, −1). This yields a tetrahedron with edge-length $\\scriptstyle 2 \\sqrt{2}$, centered at the origin. For the other tetrahedron (which is", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "given as follows, where where K is the cross product matrix of ω. That is, $\\mathbf{K} \\mathbf{v} = \\mathbf{\\omega}\\times\\mathbf{v}$ for all vectors $\\mathbf{v} \\in \\Bbb{R}$: $R = \\exp(\\theta \\mathbf{K}) = \\sum_{k=0}^\\infty\\frac{(\\theta \\mathbf{K})^k}{k!} = I + \\mathbf{K} \\theta + \\frac{1}{2!}(\\theta \\mathbf{K})^2 + \\frac{1}{3!}(\\theta \\mathbf{K})^3 + \\cdots$ Because K is skew-symmetric, and the sum of the squares of its above-diagonal entries is 1, the characteristic polynomial P(t) of K is P(t) = det(K – t I) = –(K3 + K). Since by the Hamilton-Cayley theorem we have P(K) = 0, this implies that K3 = –K. As a result, K4 = –K2, K5 = K, K6 = K2, K7 = –K. This pattern continues indefinitely, and so all higher powers of K can be expressed in terms of K and K2. Thus from the above equation we get: $R = I + \\left(\\theta - \\frac{\\theta^3}{3!} + \\frac{\\theta^5}{5!} - \\cdots\\right) \\mathbf{K} + \\left(\\frac{\\theta^2}{2!} - \\frac{\\theta^4}{4!} + \\frac{\\theta^6}{6!} - \\cdots\\right) \\mathbf{K}^2$ $R = I + \\sin(\\theta) \\mathbf{K} + (1-\\cos(\\theta)) \\mathbf{K}^2.$ This is an advanced derivation, compared to the geometric one in the article Rodrigues' rotation formula. Due to the existence of the above mentioned exponential map, the unit vector ω representing the rotation axis, and the angle θ are sometimes called the exponential coordinates of the rotation matrix R. ### Log map", "of $B C$ be ${l}_{2}$. It is given that ${l}_{1} / {l}_{2} = \\frac{2}{3}$ $\\to$ 1. Also, observe that $B D = F E$ and $B F = D E$. $\\to$ 2. We have to find $\\angle C A E$ which will give us elevation of C from A. Let $\\angle C A E$ be $\\theta$. From $\\triangle A B F$ we observe that:- $\\frac{B F}{A B} = \\sin {25}^{o} \\implies B F = A B \\cdot \\sin {25}^{o} = 0.423 \\cdot {l}_{1}$ $\\therefore$ using 2., $D E = 0.423 \\cdot {l}_{1}$ $\\to$ 3. $\\frac{A F}{A B} = \\cos {25}^{o} \\implies A F = A B \\cdot \\cos {25}^{o} = 0.906 \\cdot {l}_{1}$ $\\to$ 4. From $\\triangle B C D$ we find:- $\\frac{C D}{B C} = \\sin {40}^{o} \\implies C D = B C \\cdot \\sin {40}^{o} = 0.643 \\cdot {l}_{2}$ $\\to$ 5. $\\frac{B D}{B C} = \\cos {40}^{0} \\implies B D = B C \\cdot \\cos {40}^{o} = 0.766 \\cdot {l}_{2}$ $\\therefore$ form 2., $F E = 0.766 \\cdot {l}_{2}$ $\\to$ 6. Now, $\\tan \\theta = \\frac{C E}{A E} = \\frac{C D + D E}{A F + F E}$ Substituting values of $D E , A F , C D , F E$ from 3., 4., 5., 6. $\\tan \\theta = \\frac{0.643 \\cdot {l}_{2} + 0.423 \\cdot", "}$, which option is closest to the exact value of $\\mathbf{u}\\cdot \\mathbf{v}$ ? Exactly one option must be correct) a) 4.5 b) 6.2 c) 4.2 d) 5.1 Choice (a) is incorrect Choice (b) is incorrect Choice (c) is correct! Since $\\mathbf{u}\\cdot \\mathbf{v}=||\\mathbf{u}||\\phantom{\\rule{0.3em}{0ex}}||\\mathbf{v}||cos\\theta$, where $\\theta$ is the angle between the vectors when placed tail to tail, we have $\\mathbf{u}\\cdot \\mathbf{v}=2×3×cos4{5}^{\\circ }\\approx 4.24.$ Choice (d) is incorrect What is the approximate angle between $\\mathbf{a}$ and $\\mathbf{b}$ if $\\mathbf{a}\\cdot \\mathbf{b}=3$, $||\\mathbf{a}||=2$, and $||\\mathbf{b}||=2.6$? Exactly one option must be correct) Choice (a) is incorrect Choice (b) is incorrect Choice (c) is incorrect Choice (d) is correct! If $\\theta$ is the required angle, then $cos\\theta =\\frac{\\mathbf{a}\\cdot \\mathbf{b}}{||\\mathbf{a}||\\phantom{\\rule{0.3em}{0ex}}||\\mathbf{b}||}=\\frac{3}{5.2}\\approx 0.577$ and hence $\\theta \\approx 0.955$ radians. Find non-zero scalars $\\alpha$, $\\beta$ such that for all vectors $\\mathbf{a}$ and $\\mathbf{b}$, $\\alpha \\left(\\mathbf{a}+2\\mathbf{b}\\right)-\\beta \\mathbf{a}+\\left(4\\mathbf{b}-\\mathbf{a}\\right)=\\mathbf{0}.$ Exactly one option must be correct) a) $\\alpha =2,\\phantom{\\rule{0.3em}{0ex}}\\beta =1$ b) $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =-3$ c) $\\alpha =1,\\phantom{\\rule{0.3em}{0ex}}\\beta =3$ d) $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =3$ Choice (a) is incorrect Choice (b) is correct! If the vector equation is simplified, we get $\\left(\\alpha -\\beta -1\\right)\\mathbf{a}+\\left(2\\alpha +4\\right)\\mathbf{b}=\\mathbf{0}.$ Since this holds for all $\\mathbf{a}$ and $\\mathbf{b}$, it will hold when $\\mathbf{a}$ and $\\mathbf{b}$ are set equal to $\\mathbf{0}$ in turn. This gives the two conditions $\\alpha -\\beta -1=0$ and $2\\alpha +4=0$, whose solution is $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =-3.$ Choice (c)", "\\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2", "= \\mathbf{a} \\cdot \\mathbf{c} + \\mathbf{a} \\cdot \\mathbf{d} + \\mathbf{b} \\cdot \\mathbf{c} + \\mathbf{b} \\cdot \\mathbf{d}$ Consider a triangle $OAB$ with $\\mathbf{a}, \\mathbf{b}$ the position vectors of points $A$ and $B$. Let $\\theta$ be the angle $AOB$. By the cosine rule, $AB^2 = OA^2 + OB^2 - 2 \\, OA \\cdot OB \\cos \\theta$. $\\begin{gather} | \\mathbf{b}-\\mathbf{a} |^2 = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta \\tag{4} \\end{gather}$ By using formula 1, $|\\mathbf{b}-\\mathbf{a}|^2 = (\\mathbf{b}-\\mathbf{a}) \\cdot (\\mathbf{b}-\\mathbf{a})$. Expanding using formula 3 gives us $\\mathbf{b} \\cdot \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} - \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{a}$. The use of formulas 1 and 2 simplifies the expression to $|\\mathbf{b}|^2 - 2 \\mathbf{a}\\cdot\\mathbf{b} + |\\mathbf{a}|^2$. Plugging this into equation 4 gives us $|\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 \\mathbf{a} \\cdot \\mathbf{b} = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta$ $-2\\mathbf{a} \\cdot \\mathbf{b} = -2 |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$. ### Angles and perpendicular vectors The dot product formula is especially handy in two situations. The first is in the computation of angles. Given two vectors $\\mathbf{a}$ and $\\mathbf{b}$, the values of $\\mathbf{a} \\cdot \\mathbf{b}$, $|\\mathbf{a}|$ and $|\\mathbf{b}|$ are all relatively simple to compute. Substituting them into our formula allows us to calculate the angle between them. The second situation is when vectors", "\\Gamma \\mathbf u \\frac{(\\mathbf u \\cdot \\mathbf r)}{c^{2}} - \\gamma \\mathbf u t = \\mathbf r + \\Gamma \\mathbf u \\frac{(\\mathbf u \\cdot \\mathbf r)}{c^{2}} \\quad(t = 0) \\quad \\Rightarrow r'^{2} = r^{2} + \\frac{(\\mathbf u \\cdot \\mathbf r)^{2}\\gamma^{2}}{c^{2}},$$ $$(\\mathbf u \\cdot \\mathbf r') = \\gamma (\\mathbf u \\cdot \\mathbf r), \\quad (\\mathbf v' \\cdot \\mathbf r') = \\frac{(\\mathbf r \\cdot \\mathbf v)}{\\gamma (1 - \\frac{(\\mathbf v \\cdot \\mathbf u)}{c^{2}})} - \\gamma (\\mathbf r \\cdot \\mathbf u),$$ $$\\frac{\\mathbf F}{\\gamma(1 - \\frac{(\\mathbf v \\cdot \\mathbf u)}{c^{2}})} = \\mathbf F' + \\gamma \\frac{\\mathbf u (\\mathbf F' \\cdot \\mathbf v')}{c^{2}} + \\Gamma \\mathbf u \\frac{(\\mathbf u \\cdot \\mathbf F')}{c^{2}} \\qquad (.1).$$ Secondly, use (.1) for Coulomb's law. You can make it, because it doesn't have an information about speed of interaction, which can be proved by thought experiment with two charges, binded by the stiff spring at the rest state. So, $$\\frac{\\mathbf F}{\\gamma(1 - \\frac{(\\mathbf v \\cdot \\mathbf u)}{c^{2}})} = \\frac{Qq}{r'^{3}}\\!{\\left[\\mathbf r' + \\gamma \\mathbf u \\frac{(\\mathbf r' \\cdot \\mathbf v')}{c^{2}} + \\Gamma \\mathbf u \\frac{(\\mathbf u \\cdot \\mathbf r')}{c^{2}}\\right]} =$$ $$= \\frac{Qq}{r'^{3}}\\!{\\left[\\mathbf r + \\Gamma \\mathbf u \\frac{(\\mathbf u \\cdot \\mathbf r)}{c^{2}} + \\gamma \\frac{\\mathbf u}{c^{2}} \\left(\\frac{(\\mathbf u \\cdot \\mathbf r)}{\\gamma(1 - \\frac{(\\mathbf u \\cdot \\mathbf v)}{c^{2}})} - \\gamma (\\mathbf u \\cdot \\mathbf r)\\right) + \\Gamma \\gamma \\mathbf u \\frac{(\\mathbf", "+0 2° Functions, Delta +5 794 3 ax^2 +bx +c has 2 solutions, x1 and x2 x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? Dec 13, 2015 #2 +23278 +10 ax^2 +bx +c has 2 solutions, x1 and x2 x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? YES $$\\small{ \\begin{array}{lrcll} (1) & x_1+x_2 &=& k & \\qquad \\rightarrow \\qquad x_2 = k-x_1\\\\ (2) & \\frac{x_1}{x_2} &=& k & \\qquad \\rightarrow \\qquad x_1 = k \\cdot x_2 \\qquad \\rightarrow \\qquad x_1 = k \\cdot (k-x_1)\\\\ \\\\ \\hline \\\\ (2) & x_1 &=& k \\cdot (k-x_1) \\\\ & x_1 &=& k^2 -k \\cdot x_1 \\\\ & x_1 +k \\cdot x_1 &=& k^2 \\\\ & x_1(1 +k ) &=& k^2 \\\\ & \\mathbf{x_1} & \\mathbf{=} & \\mathbf{ \\frac{k^2}{1 +k} }\\\\ \\\\ \\hline \\\\ (1) &\\frac{x_1}{x_2} &=& k \\\\ & x_2 &=& \\frac{x_1}{k} \\\\ & x_2 &=& \\frac{\\frac{k^2}{1 +k}}{k} \\\\ & \\mathbf{x_2} & \\mathbf{=} & \\mathbf{ \\frac{k}{1 +k} } \\\\ \\\\ \\hline \\end{array} }$$ $$\\small{ \\begin{array}{rcll} (x-x_1)(x-x_2) &=& 0 \\\\ (x - \\frac{k^2}{ 1 + k } )(x - \\frac{k}{ 1 + k } ) &=& 0 \\\\ x^2-x\\cdot( \\frac{k}{ 1+k } + \\frac{k^2}{ 1+k } ) + \\frac{k^3}{ (1+k)^2 } &=& 0 \\\\ x^2-x\\cdot[ \\frac{k}{ 1+k }( 1+k ) ] + \\frac{k^3}{ (1+k)^2", "form a triangle. By the law of cosines $$15^2 = 5^2 + 12^2 - 2(5)(12)\\cos \\theta\\\\ \\frac {225 - 144 - 25}{120} = -\\cos\\theta\\\\ \\frac {56}{120} = \\cos\\theta\\\\ \\theta = \\arccos -\\frac {7}{15}$$ But that is the angle between $$u,v$$ when they are head to tail. When they are tail to tail you will get the supplement of that angle. $$\\arccos \\frac 7{15}$$ You could also do this with $$\\|\\mathbb u + \\mathbb v\\| = \\|\\mathbb w\\|$$ b) $$(\\mathbb u+\\mathbb v)\\cdot \\mathbb u = 0\\\\ \\mathbb u\\cdot \\mathbb u + \\mathbb u\\cdot \\mathbb v = 0\\\\ \\|\\mathbb u\\|^2 + \\|\\mathbb u\\|\\|\\mathbb v\\|\\cos\\theta = 0\\\\ 9 + 15\\cos\\theta = 0\\\\ \\theta = \\arccos-\\frac 35$$ For question $$1$$, answer would be $$90^\\circ$$, if the question were $$5, 12,$$ and $$1\\color{red}3$$. For question $$2$$, the answer is not $$90^\\circ$$ nor $$180^\\circ.$$ If $$\\vec u\\cdot (\\vec u+\\vec v)=0,$$ then $$\\vec u\\cdot \\vec u=-\\vec u\\cdot \\vec v$$, so $$|\\vec u|^2=-|\\vec u||\\vec v|\\cos\\theta,$$ and you are given $$|\\vec u|$$ and $$|\\vec v|$$; you should be able to find $$\\theta$$ from here. An approach could be as follows (I'm assuming you are working with real spaces). You know that: $$u\\cdot v = ||u||||v||\\cos\\theta$$ and you want to find out the value of $$\\theta$$. For this, you need to calculate $$u\\cdot v$$. But note that, because", "A) $\\frac{1}{2|b|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ B) $\\frac{1}{2|a|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ C) $\\frac{1}{2|d|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ D) None of these • question_answer139) The value of $'a'$ for which the function $f(x)=(4a-3)(x+\\log 5)+2(a-7)$ $\\cot \\frac{x}{2}\\cdot {{\\sin }^{2}}\\frac{x}{2}$does not possess critical points is A) $(-\\infty ,\\,\\,2)$ B) $(-\\infty ,\\,\\,-1)$ C) $[1,\\,\\,\\infty )$ D) $\\left( -\\infty ,\\,\\,\\frac{-4}{3} \\right)\\cup (2,\\,\\,\\infty )$ • question_answer140) $\\mathbf{a}$and $\\mathbf{c}$ are unit vectors and$|\\mathbf{b}|\\,\\,=4$. If angle between $\\mathbf{b}$ and $\\mathbf{c}$ is and ${{\\cos }^{-1}}\\left( \\frac{1}{4} \\right)$$\\text{and}$$\\mathbf{a}\\times \\mathbf{b}=2\\mathbf{a}\\times \\mathbf{c}$, then$\\mathbf{b}=\\lambda \\mathbf{a}+2\\mathbf{c}$, where$\\lambda$is equal to A) $\\pm \\frac{1}{4}$ B) $\\pm \\frac{1}{2}$ C) $\\pm \\,\\,1$ D) None of the above • question_answer141) If the planes$x-cy-bz=0$, $cx-y+az=0$and$bx+ay-z=0$ pass through a straight line, then the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$is A) $-1$ B) $2$ C) $1$ D) $0$ • question_answer142) The plane $ax+by=0$ is rotated through an angle $\\alpha$ about its line of intersection with the plane$z=0$, then the equation to the plane in new position is A) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ B) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ C) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ D) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ • question_answer143) If$y=a\\cos (\\log x)-b\\sin (\\log x)$, then the value of${{x}^{2}}\\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\\frac{dy}{dx}+y$is A) $0$ B) $1$ C) $2$ D) $3$ • question_answer144) The inverse of the function$f(x)=\\log ({{x}^{2}}+3x+1),\\,\\,x\\in [1,\\,\\,3]$, assuming it to be an onto function, is A) $\\frac{-3+\\sqrt{5+4{{e}^{x}}}}{2}$ B) $\\frac{-3\\pm \\sqrt{5+4{{e}^{x}}}}{2}$ C) $\\frac{-3-\\sqrt{5+4{{e}^{x}}}}{2}$ D) None of the above • question_answer145) $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{\\sqrt{1-\\cos", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then" ]

[ "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "refers to the dot product, denoted with a $\\cdot$. The magnitude, or length, of any dimensional vector is the square root of the sum of the elements squared. $$|(a_1, a_2, a_3)| \\equiv \\sqrt{a_1^2 + a_2^2 + a_3^2}$$ Normalizing a vector, to make it unit length and keep just its direction is done by scaling the components to divide by the length (the bold $\\mathbf{a}$ represents a vector): $$\\hat{\\mathbf{a}} \\equiv \\frac{\\mathbf{a}}{|\\mathbf{a}|}$$ ## Dot Product The dot product, also called the scalar or inner product, gives a scalar. It is not easily relatable geometrically but introduces the cosine of the angle between vectors which has some very useful properties. $$(a_1, a_2, a_3) \\cdot (b_1, b_2, b_3) \\equiv a_1 b_1 + a_2 b_2 + a_3 b_3$$ $$\\mathbf{a} \\cdot \\mathbf{b} = |\\mathbf{a}| |\\mathbf{b}| \\cos(\\theta)$$ This is less interesting until the magnitudes $|\\mathbf{a}|$ and $|\\mathbf{b}|$ are removed. An immediate relation to the angle $\\theta$ can be seen. Due to the expense of trig functions, this is not always that useful. In many cases, explicit angles can be avoided entirely. $$\\cos(\\theta) = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{|\\mathbf{a}| |\\mathbf{b}|} = \\hat{\\mathbf{a}} \\cdot \\hat{\\mathbf{b}}$$ $$\\theta = \\operatorname{acos}(\\hat{\\mathbf{a}} \\cdot \\hat{\\mathbf{b}})$$ $\\cos(\\theta)$ is quite useful to determine if two vectors are pointing in the same direction, being 1 for the same direction, 0 for right angles and -1 for opposite.", "\\end{array}$ Therefore $\\theta = 1.806$. ### Resolving a Vector into Components Exercise 19 $\\mathbf{a} \\cdot \\mathbf{i} = a_1$ $\\mathbf{a} \\cdot \\mathbf{j} = a_2$ $\\mathbf{a} \\cdot \\mathbf{k} = a_3$ The component of a vector $$\\mathbf{a}$$ in the direction of the unit vector $$\\mathbf{\\hat{u}}$$ is $$\\mathbf{a} \\cdot \\mathbf{\\hat{u}}$$. Exercise 20 1. $\\mathbf{a} \\cdot \\mathbf{c} = - 2 - 3 + 3 = -2$ $\\mathbf{a} \\cdot \\mathbf{d} = - 4 + 1 = -3$ $\\mathbf{a} \\cdot \\mathbf{e} = -2 + 3 -1 = 0$ Therefore only $$\\mathbf{a}$$ and $$\\mathbf{e}$$ are perpendicular. 2. Let $$\\mathbf{u} = \\mathbf{i} + \\mathbf{j} + \\mathbf{k}$$. Then $$\\mathbf{\\hat{u}} = \\frac{1}{\\sqrt{3}} (\\mathbf{i} + \\mathbf{j} + \\mathbf{k})$$. The component of $$\\mathbf{a} + 2\\mathbf{b}$$ along $$\\mathbf{\\hat{u}}$$ is $\\begin{array}{ll} (\\mathbf{a} + 2 \\mathbf{b}) \\cdot \\mathbf{\\hat{u}} &= (2\\mathbf{i} − 3 \\mathbf{j} + \\mathbf{k} + 2(-\\mathbf{i} + 2\\mathbf{j} + 4\\mathbf{k})) \\cdot (\\frac{1}{\\sqrt{3}} (\\mathbf{i} + \\mathbf{j} + \\mathbf{k})) \\\\ &= (2\\mathbf{i} − 3 \\mathbf{j} + \\mathbf{k} - 2\\mathbf{i} + 4\\mathbf{j} + 8\\mathbf{k}) \\cdot (\\frac{1}{\\sqrt{3}} (\\mathbf{i} + \\mathbf{j} + \\mathbf{k})) \\\\ &= (\\mathbf{j} + 9\\mathbf{k}) \\cdot (\\frac{1}{\\sqrt{3}} (\\mathbf{i} + \\mathbf{j} + \\mathbf{k})) \\\\ &= \\frac{1}{\\sqrt{3}} + \\frac{9}{\\sqrt{3}} \\\\ &= \\frac{10}{\\sqrt3} \\end{array}$ Exercise 21 $$\\mathbf{p}$$ makes the angle $$\\pi - \\alpha$$ with $$\\mathbf{i}$$ and $$\\frac{\\pi}{2} - \\alpha$$ with $$\\mathbf{j}$$. Therefore the $$\\mathbf{i}$$-component of $$\\mathbf{p}$$ is $$-\\lvert\\mathbf{p}\\rvert \\cos \\alpha = - 2.5 \\cos \\alpha$$, and the $$\\mathbf{j}$$-component", "shorthand $\\mathbf{\\hat v}$. The vector normal $\\mathbf{N}$ to this then is $\\frac{\\mathbf{T}'(t)}{|\\mathbf{v}(t)|}$. We can verify this by taking the dot product $\\mathbf{T} \\cdot \\mathbf{N} = 0$ Also note that $|\\mathbf{v}(t)| = {ds \\over dt}$ and $\\mathbf{T}(t) = \\frac{v}{|v|} = \\frac{\\frac{dr}{dt}}{\\frac{ds}{dt}} = {dr \\over ds}$ and $\\mathbf{N} = \\frac{\\mathbf{T}'(t)}{|\\mathbf{v}(t)|} = \\frac{\\frac{dT}{dt}}{\\frac{ds}{dt}} = {dT \\over ds}$ Then we can actually verify: ${d \\over ds} ( \\mathbf{T} \\cdot \\mathbf{T} )= {d \\over ds} (1)$ ${dT \\over ds} \\cdot \\mathbf{T} + \\mathbf{T} \\cdot {dT \\over ds} = 0$ $2 * \\mathbf{T} \\cdot {dT \\over ds} = 0$ $\\mathbf{T} \\cdot {dT \\over ds} = 0$ $\\mathbf{T} \\cdot \\mathbf{N} = 0$ Therefore $\\mathbf{N}$ is perpendicular to $\\mathbf{T}$ What this gives rise to is the Unit Normal Vector $\\frac{\\frac{dT}{ds}}{|\\frac{dT}{ds}|}$ of which the top-most vector is the Normal vector, but the bottom half $(|\\frac{dT}{ds}|)^{-1}$ is known as the curvature. Since the Normal vector points toward the inside of a curve, the sharper a turn, the Normal vector has a large magnitude, therefore the curvature has a small value, and is used as an index in civil engineering to reflect the sharpness of a curve (clover-leaf highways, for instance). The only other thing not mentioned is the Binormal that occurs in 3-d curves $\\mathbf{T} \\times \\mathbf{N} = \\mathbf{B}$, which is useful in creating planes parallel to the", "# On the integral $\\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$ Let $a, b$ be arbitrary vectors in $\\mathbb{R}^3$ and define $\\partial B_1(0):=\\{r\\in\\mathbb{R}^3:|r|=1\\}$ as the boundary of the three-dimensional unit sphere. I am trying to evaluate the following integral: $$I = \\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$$ So far, I've said that we can, given a vector $a$ and a fixed vector $b$, write the following: $$a = c b + x$$ where $c = \\frac{a\\cdot b}{|b|}$ and $x\\in\\mathbb{R}^3$ such that $x$ orthogonal to $b$. Then, if we choose our coordinate system so that the positive $z$-axis runs parallel to the vector $b$, then, considering spherical coordinates, the angle between $r$ and $b$ is given by $\\theta$, and we have: \\begin{align} I = \\int_0^\\pi d\\theta\\sin\\theta\\int_0^{2\\pi}d\\phi\\frac{1}{\\left(1+|a|\\cos\\theta\\right)\\left(1+c\\cos\\theta + x\\cdot r\\right)} \\end{align} Now, I know that, since $x\\cdot b = 0$, $x$ must lie in the $x-y-$plane, so we can express it as $(\\rho\\cos\\zeta, \\rho\\sin\\zeta, 0)$ for some modulus $\\rho$ and some polar angle $\\zeta$. Thus: $$x\\cdot r = \\rho\\sin\\theta\\left(\\cos\\zeta\\cos\\phi+\\sin\\zeta\\sin\\phi\\right) = \\rho\\sin\\theta\\cos\\left(\\phi-\\zeta\\right)$$ Letting $\\phi-\\zeta=:\\xi$, we have $d\\xi = d\\phi$ and thus, since all periods of the cosine function are equal, we have: \\begin{align} I&=\\int_0^\\pi d\\theta\\frac{\\sin\\theta}{1+|a|\\cos\\theta}\\int_{-\\zeta}^{2\\pi-\\zeta}\\frac{d\\xi}{1+c\\cos\\theta+\\rho\\sin\\theta\\cos\\xi}\\\\ &=\\int_{-1}^1 \\frac{du}{1+|a|u}\\int_{0}^{2\\pi}\\frac{d\\xi}{1+cu+\\rho\\sqrt{1-u^2}\\cos\\xi} \\end{align} And this is where I'm stumped. I've been thinking of maybe converting the second integral into a contour integral over the unit circle, which reduces to evaluating $$\\oint_{|z|=1}\\frac{dz}{z^2+\\alpha", "given as follows, where where K is the cross product matrix of ω. That is, $\\mathbf{K} \\mathbf{v} = \\mathbf{\\omega}\\times\\mathbf{v}$ for all vectors $\\mathbf{v} \\in \\Bbb{R}$: $R = \\exp(\\theta \\mathbf{K}) = \\sum_{k=0}^\\infty\\frac{(\\theta \\mathbf{K})^k}{k!} = I + \\mathbf{K} \\theta + \\frac{1}{2!}(\\theta \\mathbf{K})^2 + \\frac{1}{3!}(\\theta \\mathbf{K})^3 + \\cdots$ Because K is skew-symmetric, and the sum of the squares of its above-diagonal entries is 1, the characteristic polynomial P(t) of K is P(t) = det(K – t I) = –(K3 + K). Since by the Hamilton-Cayley theorem we have P(K) = 0, this implies that K3 = –K. As a result, K4 = –K2, K5 = K, K6 = K2, K7 = –K. This pattern continues indefinitely, and so all higher powers of K can be expressed in terms of K and K2. Thus from the above equation we get: $R = I + \\left(\\theta - \\frac{\\theta^3}{3!} + \\frac{\\theta^5}{5!} - \\cdots\\right) \\mathbf{K} + \\left(\\frac{\\theta^2}{2!} - \\frac{\\theta^4}{4!} + \\frac{\\theta^6}{6!} - \\cdots\\right) \\mathbf{K}^2$ $R = I + \\sin(\\theta) \\mathbf{K} + (1-\\cos(\\theta)) \\mathbf{K}^2.$ This is an advanced derivation, compared to the geometric one in the article Rodrigues' rotation formula. Due to the existence of the above mentioned exponential map, the unit vector ω representing the rotation axis, and the angle θ are sometimes called the exponential coordinates of the rotation matrix R. ### Log map", "- \\cdots (|x| < 1)$ Trigonometric Expansions: Note: All $\\theta$ are in radians. a. $\\sin \\theta = \\theta - \\dfrac{\\theta ^3}{3!} + \\dfrac{\\theta ^5}{5!} - \\cdots$ b. $\\cos \\theta = 1 - \\dfrac{\\theta ^2}{2!} + \\dfrac{\\theta ^4}{4!} - \\cdots$ c. $\\tan \\theta = \\theta + \\dfrac{\\theta ^3}{3!} + \\dfrac{2 \\theta ^5}{15!} + \\cdots$ Cramer’s Rule: Two simultaneous equations in unknown x and y, $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$ Have the solutions: $x = \\dfrac{\\begin{vmatrix}c_1 & b_1 \\\\ c_2 & b_2 \\end{vmatrix}}{\\begin{vmatrix}a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}} = \\dfrac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}$ & $y = \\dfrac{\\begin{vmatrix}a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix}}{\\begin{vmatrix}a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}} = \\dfrac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}$ Product of Vectors: If $\\hat{i}$ , $\\hat{j}$ & $\\hat{k}$ be unit vectors in the x , y and z directions , Then: $\\hat{i} . \\hat{i} = \\hat{j} . \\hat{j} = \\hat{k} . \\hat{k} = 1$ , $\\hat{i} . \\hat{j} = \\hat{j} . \\hat{k} = \\hat{k} . \\hat{i} = o$ And: $\\hat{i} \\times \\hat{i} = \\hat{j} \\times \\hat{j} = \\hat{k} \\times \\hat{k} = 0$ & $\\hat{i} \\times \\hat{j} = \\hat{k}$ , $\\hat{j} \\times \\hat{k} = \\hat{i}$ , $\\hat{k} \\times \\hat{i} = \\hat{j}$ If", "d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our integral can be rewritten as", "$$0$$) and so the momentum observed by a locally inertial observer will not simply scale as $$1/a$$. I guess maybe the angle between $$\\mathbf{c}$$ and $$\\mathbf{x}$$ changes as $$\\mathbf{x}^2$$ changes in just such a way as to keep the term in the square root constant but I don't see how to show that. Or do I have the entire logic of the exercise wrong? It turns out that last factor in the square root is actually a constant -- so everything works nicely and momentum decays as $$1/a$$ for any object moving on a geodesic as it's supposed to. I took the derivative of the term in the square root: $$\\frac{\\mathbf{c}^2-K(\\mathbf{c}\\cdot \\mathbf{x})^2}{1-K\\mathbf{x}^2} \\equiv A.$$ This introduces the terms $$\\mathbf{c}\\cdot \\mathbf{x}'$$ and $$\\mathbf{x}\\cdot \\mathbf{x}'$$, which we can get rid of using some of the formulas in the original post: \\begin{align} \\mathbf{c}\\cdot \\mathbf{x}' &= \\frac{\\mathbf{c}^2-K(\\mathbf{c}\\cdot \\mathbf{x})^2}{a^2\\sqrt{1-K\\mathbf{x}^2}} = \\frac{\\sqrt{1-K\\mathbf{x}^2}}{a^2}A \\\\ \\mathbf{x}\\cdot \\mathbf{x}' &= \\frac{1}{a^2}\\sqrt{1-K\\mathbf{x}^2}(\\mathbf{c}\\cdot \\mathbf{x}) \\end{align} \\begin{align} \\frac{dA}{d\\tau} &=\\frac{-2K (\\mathbf{c}\\cdot \\mathbf{x})(\\mathbf{c}\\cdot \\mathbf{x}')}{1-K\\mathbf{x}^2} + \\frac{\\left(\\mathbf{c}^2-K(\\mathbf{c}\\cdot \\mathbf{x})^2\\right)2K(\\mathbf{x}\\cdot \\mathbf{x}')}{(1-K\\mathbf{x}^2)^2}\\\\ &= \\frac{-2K (\\mathbf{c}\\cdot \\mathbf{x})}{1-K\\mathbf{x}^2}\\frac{\\sqrt{1-K\\mathbf{x}^2}}{a^2}A + \\frac{A}{1-K\\mathbf{x}^2} \\frac{2K}{a^2}\\sqrt{1-K\\mathbf{x}^2}(\\mathbf{c}\\cdot \\mathbf{x})\\\\ &= 0. \\end{align} So $$|\\mathbf{p}_\\mathrm{obs}| = \\frac{m}{a}\\sqrt{A}= \\frac{\\text{const}}{a}.$$", "# How do you find two unit vectors that make an angle of 60° with v = ‹3, 4›? Jul 31, 2016 Tthe reqd. unit vectors are, $\\left(\\frac{3}{10} - \\frac{2}{5} \\sqrt{3} , \\frac{2}{5} + 3 \\frac{\\sqrt{3}}{10}\\right)$, or, $\\left(\\frac{3}{10} + 2 \\frac{\\sqrt{3}}{5} , \\frac{2}{5} - 3 \\frac{\\sqrt{3}}{10}\\right)$. #### Explanation: Let $\\vec{u} = \\left(x , y\\right)$ be the reqd. unit vector. $\\therefore | | \\vec{u} | | = 1 \\Rightarrow {x}^{2} + {y}^{2} = 1. \\ldots \\ldots \\ldots \\ldots \\ldots . \\left(1\\right)$. Given that, Angle btwn. $\\vec{u} \\mathmr{and} \\vec{v}$ is $\\frac{\\pi}{3}$, we take these vectors' Dot Product, to get, $\\vec{u} \\cdot \\vec{v} = | | u | | | | v | | \\cos \\left(\\hat{\\vec{u} , \\vec{v}}\\right)$ $\\therefore \\left(x , y\\right) \\cdot \\left(3 , 4\\right) = 1 \\left(\\sqrt{{3}^{2} + {4}^{2}}\\right) \\cos \\left(\\frac{\\pi}{3}\\right)$ $\\therefore 3 x + 4 y = 1 \\cdot 5 \\cdot \\frac{1}{2} = \\frac{5}{2} \\Rightarrow 3 x = \\frac{5}{2} - 4 y$ $\\Rightarrow x = \\frac{1}{3} \\left(\\frac{5}{2} - 4 y\\right) \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . . \\left(2\\right)$. Using $\\left(2\\right)$ in $\\left(1\\right)$, we get, $\\frac{1}{9} {\\left(\\frac{5}{2} - 4 y\\right)}^{2} + {y}^{2} = 1 \\Rightarrow \\frac{25}{4} - 20 y + 16 {y}^{2} + 9 {y}^{2} = 9$ $\\Rightarrow 25 {y}^{2} - 20 y = 9 - \\frac{25}{4}$. To make the $L . H . S .$ complete", "&= I_{1 2 \\cdots (k-1)} I \\left\\lvert {d\\mathbf{x}_1 \\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I}", "are better thought of as directions that are invariant under the effect of the linear map. As directions, they do not have unique magnitudes. 2) I think this is poorly phrased but valid. I think part of the issue you're having is that, by calling these vectors$u'$and so on, it seems like it must already be proven that$u'$is part of the mutually orthonormal set the problem asks for. This is not yet proven, but the aim of the problem is to prove that property. It would be quite valid merely to define a vector$a = A(u)/\\lambda$and so on and prove the relevant properties later. I think this is a clearer progression of logic. 3) The problem asks for an orthonormal set. Setting out to prove that$a$is unit seems like a natural thing to do. 4) You would try to find$|a|^2$because it follows more easily from the idea of using$A^2$. Observe: $$|a|^2 = \\frac{1}{\\lambda^2} A(u) \\cdot A(u) = \\frac{1}{\\lambda^2} A^TA(u) \\cdot u = -\\frac{1}{\\lambda^2} A^2(u) \\cdot u$$ But you know that$A^2(u) = -\\lambda^2 u$, and you get that$|a|^2 = 1$, which implies that$|a| = 1$. 5) Remember that$A^2$is symmetric. In particular, for any$b, c$: $$A^2(b) \\cdot c = A(b) \\cdot A^T(c) = (A^T)^2(c) \\cdot b = (-1)(-1) A^2(c) \\cdot b = A^2(c) \\cdot b$$ That proves symmetry. Since$u, w$are", "\\cdot (c_2, -c_1,0) = (b_1, b_2, 0) \\cdot (c_1,c_2,c_3)$, which is obvious. $\\square$ Now, let $X$, $f$, $Y$, $g$ and $\\theta$ be as above. We write $n(t)$ for the common normal to $T_{f(t)} X$ and $T_{g(t)} Y$. Write $f(t) = |f(t)| u(t)$ and $g(t) = |g(t)| v(t)$, so $u$ and $v$ are unit vectors. From our previous computations about geodesic curvature, we want to show $$\\frac{d \\theta(t)}{dt} = \\det(v(t), v'(t), n(t)) - \\det(u(t), u'(t), n(t))$$ where prime is differentiation with respect to $t$. (Note that it was important to work out the formula for geodesic curvature without assuming we have an arc length parametrization, since it is unlikely that both $f(t)$ and $g(t)$ are unit speed.) Now, since $u$ and $v$ are unit vectors, $\\theta = \\cos^{-1}(u \\cdot v)$. We have $$\\frac{d \\theta}{dt} = - \\ \\frac{d (u \\cdot v)/dt}{\\sin \\theta} = - \\ \\frac{u' \\cdot v + u \\cdot v'}{\\sin \\theta}$$ so we must show that $$u' \\cdot v + u \\cdot v' = - \\det(v, v', (\\sin \\theta) n) + \\det(u, u', (\\sin \\theta) n). \\quad (\\ast)$$ But $u$ and $v$ are unit vectors and $n$ is normal to both of them, so $(\\sin \\theta) n = u \\times v$ and we can write the right hand side of $(\\ast)$ as $$\\det(v, v', v \\times u)", "I've reassembled the vectors in the proper positions for vector subtraction. They form an isosceles triangle with interior angles $$\\theta$$, $$\\eta$$, and $$\\eta$$. (Eta, $$\\eta$$, is my favorite Greek letter.) The law of sines gives $\\begin{equation*} \\frac{|\\Delta\\mathbf{v}|}{\\sin\\theta} = \\frac{|\\mathbf{v}|}{\\sin\\eta} . \\end{equation*}$ This tells us the magnitude of $$\\Delta \\mathbf{v}$$, which is one of the two ingredients we need for calculating the magnitude of $$\\mathbf{a}=\\Delta\\mathbf{v}/\\Delta t$$. The other ingredient is $$\\Delta t$$. The time required for the object to move through the angle $$\\theta$$ is $\\begin{equation*} \\Delta t = \\frac{\\text{length of arc}}{|\\mathbf{v}|} . \\end{equation*}$ Now if we measure our angles in radians we can use the definition of radian measure, which is $$(\\text{angle})=(\\text{length of arc})/(\\text{radius})$$, giving $$\\Delta t=\\theta r/|\\mathbf{v}|$$. Combining this with the first expression involving $$|\\Delta v|$$ gives \\begin{align*} |\\mathbf{a}| &= |\\Delta \\mathbf{v}|/\\Delta t \\\\ &= \\frac{|\\mathbf{v}|^2}{r} \\: \\cdot \\: \\frac{\\sin\\theta}{\\theta} \\: \\cdot \\: \\frac{1}{\\sin\\eta} . \\end{align*} When $$\\theta$$ becomes very small, the small-angle approximation $$\\sin \\theta\\approx \\theta$$ applies, and also $$\\eta$$ becomes close to 90°, so $$\\sin \\eta \\approx 1$$, and we have an equation for $$|\\mathbf{a}|$$: $\\begin{equation*} |\\mathbf{a}| = \\frac{|\\mathbf{v}|^2}{r} . \\shoveright{\\text{[uniform circular motion]}} \\end{equation*}$ ##### Example 4: Force required to turn on a bike $$\\triangleright$$ A bicyclist is making a turn along an arc of a circle with radius 20 m, at a speed of", "+0 # Let $\\theta$ be the angle between $\\bold{a}$ and $\\bold{b}$, where $\\bold{a}$ and $\\bold{b}$ are the vectors defined in part (a). Find \\$\\cos 0 171 2 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$. Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\cos(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{\\vec{a}\\cdot \\vec{b}}{|\\vec{a}|\\cdot |\\vec{b}|} } \\quad | \\quad |\\vec{a}| = \\sqrt{2^2+(-6)^2+5^2}=\\sqrt{65},\\ |\\vec{b}| = \\sqrt{(-1)^2+(-2)^2+0^2}=\\sqrt{5} \\\\\\\\ \\cos(\\theta) &=& \\dfrac{\\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix}\\cdot \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix} } {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{2\\cdot (-1)+(-6)\\cdot(-2)+5\\cdot 0} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{10} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {65\\cdot 5 } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {325 } \\\\\\\\ \\mathbf{\\cos^2(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{4}{13}} \\\\ \\hline \\end{array}$$ Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix}", "\\times \\mathbf{a}) + (\\mathbf{a} \\times \\mathbf{b})|} \\,$ and the radius of the circumsphere is given by: $R= \\frac {|\\mathbf{a^2}(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b^2}(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c^2}(\\mathbf{a} \\times \\mathbf{b})|} {12V} \\,$ which gives the radius of the twelve-point sphere: $r_T= \\frac {|\\mathbf{a^2}(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b^2}(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c^2}(\\mathbf{a} \\times \\mathbf{b})|} {36V} \\,$ where: $6V= |\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})|. \\,$ In the formulas throughout this section, the scalar a2 represents the inner vector product a·a; similarly b2 and c2. The vector positions of various centers are as follows: The centroid $\\mathbf{G} = \\frac{\\mathbf{a} + \\mathbf{b} + \\mathbf{c}}{4}. \\,$ The incenter $\\mathbf{I}= \\frac{ \\mathbf{a} \\cdot |\\mathbf{b}\\times \\mathbf{c}| + \\mathbf{b} \\cdot |\\mathbf{c}\\times \\mathbf{a}| + \\mathbf{c} \\cdot |\\mathbf{a}\\times \\mathbf{b}| }{ |\\mathbf{b}\\times \\mathbf{c}| + |\\mathbf{c}\\times \\mathbf{a}| + |\\mathbf{a}\\times \\mathbf{b}| + |\\mathbf{b}\\times \\mathbf{c} + \\mathbf{c}\\times \\mathbf{a} + \\mathbf{a}\\times \\mathbf{b}| }. \\,$ The circumcenter $\\mathbf{O}= \\frac {\\mathbf{a^2}(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b^2}(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c^2}(\\mathbf{a} \\times \\mathbf{b})} {2\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})}. \\,$ The Monge point $\\mathbf{M} = \\frac {\\mathbf{a} \\cdot (\\mathbf{b} + \\mathbf{c})(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b}\\cdot (\\mathbf{c} + \\mathbf{a})(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c} \\cdot (\\mathbf{a} + \\mathbf{b})(\\mathbf{a} \\times \\mathbf{b})} {2\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})}. \\,$ The Euler line relationships are: $\\mathbf{G} = \\mathbf{M} + \\frac{1}{2} (\\mathbf{O}-\\mathbf{M})\\,$ $\\mathbf{T} = \\mathbf{M} + \\frac{1}{3} (\\mathbf{O}-\\mathbf{M})\\,$ where T is twelve-point center. Also: $\\mathbf{a} \\cdot", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "dual covariant basis defined as $\\mathbf{e}_i= \\frac{\\partial\\mathbf{x}}{\\partial{u^i}} \\to \\mathbf{e}_i\\cdot\\mathbf{e}^j = \\delta_{i}^{j}~,$ and therefore relates to the contravariant vectors as $\\mathbf{e}_i = \\frac{\\mathbf{e}^j\\times\\mathbf{e}^k}{\\mathbf{e}^i\\cdot\\mathbf{e}^j\\times\\mathbf{e}^k} = \\sqrt{g}\\;\\mathbf{e}^j\\times\\mathbf{e}^k ~,$ where $(i,j,k)$ are cyclic permutations of $(1,2,3)$ and we have used the notation $(u^1, u^2, u^3) = (\\psi,\\theta,\\phi)$. The Jacobian $\\sqrt{g}$ is defined below. Similarly $\\mathbf{e}^i = \\frac{\\mathbf{e}_j\\times\\mathbf{e}_k}{\\sqrt{g}} ~.$ Any vector field $\\mathbf{B}$ can be represented as $\\mathbf{B} = (\\mathbf{B}\\cdot\\mathbf{e}^i)\\mathbf{e}_i = B^i\\mathbf{e}_i$ or $\\mathbf{B} = (\\mathbf{B}\\cdot\\mathbf{e}_i)\\mathbf{e}^i = B_i\\mathbf{e}^i ~.$ In particular any basis vector $\\mathbf{e}_i = (\\mathbf{e}_i\\cdot\\mathbf{e}_j)\\mathbf{e}^j$. The metric tensor is defined as $g_{ij} = \\mathbf{e}_i\\cdot\\mathbf{e}_j \\; ; \\; g^{ij} = \\mathbf{e}^i\\cdot\\mathbf{e}^j \\; ; \\; g^j_i = \\mathbf{e}_i\\cdot\\mathbf{e}^j = \\delta_i^j ~.$ The metric tensors can be used to raise or lower indices. Take $\\mathbf{B} = B_i\\mathbf{e}^i = B_i g^{ij}\\mathbf{e}_j = B^j\\mathbf{e}_j~,$ so that $B^j = g^{ij} B_i~.$ ### Jacobian The Jacobian of the coordinate transformation $\\mathbf{x}(\\psi, \\theta, \\phi)$ is defined as $J = \\det\\left(\\frac{\\partial(x,y,z)}{\\partial(\\psi,\\theta,\\phi)}\\right) = \\frac{\\partial\\mathbf{x}}{\\partial{\\psi}}\\cdot\\frac{\\partial\\mathbf{x}}{\\partial{\\theta}} \\times \\frac{\\partial\\mathbf{x}}{\\partial{\\phi}}$ and that of the inverse transformation $J^{-1} = \\det\\left(\\frac{\\partial(\\psi,\\theta,\\phi)}{\\partial(x,y,z)}\\right) = \\nabla{\\psi}\\cdot\\nabla{\\theta} \\times \\nabla{\\phi}$ It can be seen that [1] $g \\equiv \\det(g_{ij}) = J^2 \\Rightarrow J = \\sqrt{g}$ ### Some surface elements Consider a surface defined by a constant value of $\\phi$. Then, the surface element is $d{\\mathbf S}_\\phi = \\mathbf{e}_\\psi\\times\\mathbf{e}_\\theta d\\psi d\\theta = \\sqrt{g}\\, \\nabla\\phi d\\psi d\\theta .$ As for a surface defined by", "\\textbf{S} = \\frac{1}{\\mu _{0}} \\nabla \\cdot ( \\textbf{E} \\times \\textbf{B})$ $=\\frac{1}{\\mu _{0}}( \\textbf{B} \\cdot (\\nabla \\times \\textbf{E}) - \\textbf{E} \\cdot (\\nabla \\times \\textbf{B})$ $= \\frac{1}{\\mu _{0}} (\\frac{- \\partial}{\\partial t}(\\textbf{B} \\cdot \\textbf{B}) - \\frac{\\partial}{\\partial t}(\\frac{1}{v^{2}} ( \\textbf{E} \\cdot \\textbf{E})$ $= \\frac{- \\partial}{\\partial t} \\frac{1}{\\mu _{0}} (B + \\frac{1}{v^{2}} E)$ But I don't know how to get the vector lengths squared...nor do I know how to implement epsilon and mu...and how do I show this specifically for plane waves?? 4. Mar 28, 2013 marcusl You have left out the complex conjugation. Note that $$\\textbf{S} = \\frac{1}{\\mu _{0}}( \\textbf{E} \\times \\textbf{B}^*)$$ and so on. Then when you form B dot B you get the amplitude squared. In fact, it's usually defined in terms of H instead of B. Many authors also include a factor of 1/2 to give power units with the peak amplitudes E0 and B0. Then it's $$\\textbf{S} = \\frac{1}{2}( \\textbf{E} \\times \\textbf{H}^*)$$ Last edited: Mar 28, 2013 5. Mar 28, 2013 TSny $\\textbf{B} \\cdot (\\nabla \\times \\textbf{E}) = \\textbf{B} \\cdot( \\frac{- \\partial}{\\partial t}\\textbf{B})=-\\frac{1}{2}\\frac{ \\partial}{\\partial t}(\\textbf{B} \\cdot \\textbf{B}) = -\\frac{1}{2}\\frac{ \\partial}{\\partial t}B^2$, etc. Your approach here is going to get the result in a general way rather than specifically for plane waves. If you want to show the result for plane waves, then use the plane wave expressions for" ]

[ "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "4 \\\\ 3 \\end{pmatrix}.$ ×", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "6 & 0 & 5 \\end{pmatrix}$", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "0 & 1 & 0 & 1 & 1 \\end{pmatrix}.$$ -", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "\\begin {pmatrix} 12 & 6 & -2 \\\\ -6 & -3 & 1 \\\\ 6 & 3 & -1 \\end {pmatrix}$", "1 & -2 & -5 \\end{pmatrix},$$ $$(A-6I)(A - 12 I) = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix},$$", "& 4 \\end{pmatrix} \\begin{pmatrix} 3 & 6 & 9 \\end{pmatrix} \\begin{pmatrix} 5 & 8 \\end{pmatrix}$.", "0 & 0 & 0\\\\ 0 & 3 + i\\sqrt3 & 1\\\\ 0 & 0 & 3 + i\\sqrt3 \\end{pmatrix}.$$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "4 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ \\end{pmatrix}$$.", "0 & -1 & 0 \\\\ 0 & 0 & -4 \\end{pmatrix}$", "0 & 0\\cr 0 & 0 & 0 & 3 & 0 & 0\\cr 0 & 0 & 0 & 0 & 3 & 0\\cr 0 & 0 & 0 & 0 & 0 & 3\\end{pmatrix}$$", "$$\\begin{pmatrix} 1&1&1 \\\\1&0&0.5 \\\\ 0&1&0.5 \\end{pmatrix}$$" ]

[ "-4 & 1 & -3 & 0 \\\\ 3 & -3 & -2 & -3 & 3 & 5 & 0 & -3 & 0 & -1 & -2 & 0 & 3 & 2 & 6 \\end{pmatrix} \\\\ T &=", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "4 \\\\ 3 \\end{pmatrix}.$ ×", ". & . & . & . \\\\ . & . & . & . & . & . & 2 & . & . & . \\\\ . & . & . & . & . & . & . & 3 & . & . \\\\ . & . & . & . & . & . & . & . & 4 & . \\end{pmatrix} = \\begin{pmatrix} 1 & . & . & . & . & . & . & . & . & . \\\\ -4 & 1 & . & . & . & . & . & . & . & . \\\\ 6 & -3 & 1 & . & . & . & . & . & . & . \\\\ -4 & 3 & -2 & 1 & . & . & . & . & . & . \\\\ 1 & -1 & 1 & -1 & 1 & . & . & . & . & . \\\\ . & . & . & . & . & 1 & . & . & . & . \\\\ . & . & . & . & . & 1 & 1 & . & . & . \\\\ . & . & . & . &", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "5 & -8 -4 & 12 + 3 \\\\ 3 \\times 2 + 5 & -6 - 4 & 9 +3\\\\7 \\times 2 + 5 & -14 - 4 & 21 + 3\\end{pmatrix} \\\\ & = \\begin{pmatrix}13 & -12 & 15 \\\\ 11 & -10 & 12 \\\\ 19 & -18 & 24\\end{pmatrix}\\end{align*} Similarly: CB = \\begin{pmatrix}23 & -2 & 3 \\\\ 0 & 0 & 0 \\\\ 29 & -7 & 4\\end{pmatrix} 1 Like", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "&=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&-7\\\\ 0&0&0&7\\end{pmatrix}\\tag{R_3=R_3-R_1}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&7\\tag{R_2=-\\frac17 R_2}\\end{pmatrix}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&0\\tag{R_3=R_3 -7R_2}\\end{pmatrix} \\end{align}", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "\\end{pmatrix} + } + & + \\color{red} { + MS_{6} = + \\begin{pmatrix} ​ + 3 & 21 & 19 & 12 & 10 \\\\ + 22 & 20 & 13 & 6 & 4 \\\\ + 16 & 14 & 7 & 5 & 23 \\\\ + 15 & 18 & 1 & 24 & 17 \\\\ + 9 & 2 & 25 & 18 & 11 + \\end{pmatrix} + } + \\end{align} + + + + \\begin{align} + \\color{green} { + MS_{7} = + \\begin{pmatrix} ​ + 11 & 10 & 4 & 23 & 17 \\\\ + 18 & 12 & 6 & 5 & 24 \\\\ + 25 & 19 & 13 & 7 & 1 \\\\ + 2 & 21 & 20 & 14 & 8 \\\\ + 9 & 3 & 22 & 16 & 15 + \\end{pmatrix} + } + & + \\color{red} { + MS_{8} = + \\begin{pmatrix} ​ + 10 & 4 & 23 & 17 & 11 \\\\ + 12 & 6 & 5 & 24 & 18 \\\\ + 19 & 13 & 7 & 1 & 25 \\\\ + 21 & 20 & 14 & 8 & 2 \\\\ + 3 & 22 & 16 & 15 & 9 + \\end{pmatrix} + } + \\end{align} +", "to choose 4 from the 8 with cracks and 1 from the remaining 17. 5. Jan 19, 2010 ### exitwound b.)$$\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix} = \\frac{8!}{4!4!} * \\frac{17!}{1!16!}= 1190$$ c.) $$\\frac{\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} = \\frac{\\frac{8!}{4!4!} * \\frac{17!}{1!16!}}{\\frac{25!}{5!20!}} = .0224$$ d.) $$\\frac{\\begin{pmatrix} 8\\\\ 5 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 0 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} + AnswerToPartC = \\frac{\\frac{8!}{5!3!} * \\frac{17!}{0!17!}}{\\frac{25!}{5!20!}} + .0224 = .0235$$ Is this how it's done? 6. Jan 19, 2010 ### vela Staff Emeritus Yup, perfect.", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "& 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 4 & 7 & 9 &10 \\end{pmatrix}\\overset{(1)}\\sim \\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 3 & 5 & 6\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\overset{(2)}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 6-\\lambda & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 0 & 3\\lambda-13\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 0 & 1 & 1 & 0\\\\ 0 & 0 & 1 & 3\\\\ 0", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} , \\quad (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We see that the numbers in the parentheses between both cycles are difference. Hence $(23)$ and $(165)$ are disjoint cycles. Also notice that: (2) \\begin{align} \\quad (23) \\circ (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 3 & 2 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We also have that: (3) \\begin{align} \\quad (165) \\circ (23) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ = \\begin{pmatrix} 1 & 2 &", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "& 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\ub{12} & 3 & 16 & \\ub{4} \\end{pmatrix} \\qquad \\begin{pmatrix} \\hl{1} & 3 & 2 & \\hl{6} \\\\ 4 & 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\hl{12} & 3 & 16 & \\hl{4} \\end{pmatrix}$$ \\end{document}", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "4/9 & 0 & 5/9 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 5/9 & 0 & 4/9 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 6/9 & 0 & 3/9 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & 7/9 & 0 & 2/9 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8/9 & 0 & 1/9 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix}P_0\\\\P_1\\\\P_2\\\\P_3\\\\P_4\\\\P_5\\\\P_6\\\\P_7\\\\P_8\\\\P_9\\end{pmatrix}$$ Hence $$\\begin{pmatrix}P_0\\\\P_1\\\\P_2\\\\P_3\\\\P_4\\\\P_5\\\\P_6\\\\P_7\\\\P_8\\\\P_9\\end{pmatrix} =\\begin{pmatrix}0 \\\\ 35/83 \\\\ 315/664 \\\\ 325/664 \\\\ 165/332 \\\\ 167/332 \\\\ 339/664 \\\\ 349/664 \\\\ 48/83 \\\\ 1\\end{pmatrix}.$$ Therefore the required probability is $P_6=\\frac{339}{664}$ is just barely greater than $1/2$.", "&=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&-7\\\\ 0&0&0&7\\end{pmatrix}\\tag{R_3=R_3-R_1}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&7\\tag{R_2=-\\frac17 R_2}\\end{pmatrix}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&0\\tag{R_3=R_3 -7R_2}\\end{pmatrix} \\end{align}", "\\\\ 0 & 0 & -1 & 0 \\\\ 1 & 1 & 1-\\mathrm{i} & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix},$$ $$A^{(4)}_6= \\begin{pmatrix} 1 & 0 & 1 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_7= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_8= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix}.$$ $A_0^{(4)}$ $A_1^{(4)}$ $A_2^{(4)}$ $A_3^{(4)}$ $A_4^{(4)}$ $A_5^{(4)}$ $A_6^{(4)}$ $A_7^{(4)}$ $A_8^{(4)}$", "\\\\- 0.283871 \\end{pmatrix}$ $\\begin{pmatrix}-3.95018 \\\\- 0.24367 \\end{pmatrix}$ $\\begin{pmatrix}-3.8219 \\\\- 0.0463343 \\end{pmatrix}$ $\\begin{pmatrix}-3.8219\\\\0.0463343 \\end{pmatrix}$ $\\begin{pmatrix}-3.95018\\\\0.24367\\end{pmatrix}$ $\\begin{pmatrix}-4.17269\\\\0.283871 \\end{pmatrix}$ $10$ $\\begin{pmatrix}-4.20324 \\\\- 0.275251 \\end{pmatrix}$ $\\begin{pmatrix}-3.8754 \\\\- 0.173249\\end{pmatrix}$ $\\begin{pmatrix}-3.8754\\\\0.173249\\end{pmatrix}$ $\\begin{pmatrix}-4.20324\\\\0.275251\\end{pmatrix}$ ${}^\\ast$: $\\begin{pmatrix}-3.85567\\\\- 0.14285\\end{pmatrix} \\equiv -3.85567- 0.14285 \\; i, \\; i = \\sqrt{-1}$ Table 3. Typical numerical values of $\\{\\xi, {\\lambda} \\}$ for satellite and primitive components with $q k \\le 8$ $(q, k)$ Type $\\xi$ $\\lambda$ Fig. No. $(2, 1)$ Primitive $-0.23512621166835 + 1.6020106612492319\\; i$ $-1.951962936703127 -1.7485319734836857\\; i$ 7(a) $(3, 1)$ Primitive $0.454659800907803 + 0\\; i$ $-5.219261654371707 + 0\\; i$ 7(b) $(4, 1)$ Primitive $0.341977833651053 + 0.4814672920477362\\; i$ $-3.564458921821090 + 1.646535885566244\\; i$ 7(c) $(5, 1)$ Primitive $0.757190777481740 + 0.2122872817258981\\; i$ $-4.045956427432157 + 0.346070091333350\\; i$ 7(d) $(6, 1)$ Primitive $-1.10304555596122 + 1.0059215339291120\\; i$ $-0.568308288946734 - 0.953277135441219\\; i$ 7(e) $(7, 1)$ Primitive $0.764464537114743 + 0.2097254339881594\\; i$ $-4.018513045186225 + 0.345332811593217\\; i$ 7(f) $(2, 2)$ Satellite $0.851625771743808 + 0.2369760927165886\\; i$ $-3.752943559023020 + 0.074408072396025\\; i$ 7(g) $(3, 2)$ Satellite $0.875562175989576 + 0.2566238834815281\\; i$ $-3.838763107389483 + 0.152045173864673\\; i$ 7(h) $(4, 2)$ Satellite $1.201354999600545 - 0.2912383604676560\\; i$ $-3.921140825708978 + 0.257336038244326\\; i$ 7(i) • [1] L. V. Ahlfors, Complex Analysis. An Introduction to the Theory of Analytic Functions of One Complex Variable, International Series in Pure and Applied Mathematics, McGraw-Hill Book Co., New York, 1978. [2] J. Ainsworth, M. Dawson, J. Pianta and J. Warwick, The Farey Sequence, 2003. Available", "0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\\\end{pmatrix}$ 4 $(1,2,4,3)$ 2413 $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ 4 $(1,2,4)$ 2431 $\\begin{pmatrix} 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 0 \\\\\\end{pmatrix}$ 3 $(1,3,2)$ 3124 $\\begin{pmatrix}0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 0 \\\\", "0 \\\\ \\sin\\theta & 0 & \\cos\\theta \\end{pmatrix} \\label{9.6b}$ $R_{z}(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta & 0 \\\\ \\sin\\theta & \\cos\\theta & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\label{9.6c}$", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\stackrel{E_7}{\\sim} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} \\stackrel{E_6E_5E_4E_3E_2E_1}{\\sim} L \\end{eqnarray*} \\begin{eqnarray*} E_7= \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} =E_7^{-1} \\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}", "0 & 0 & 0 & 1 \\\\ \\end{pmatrix}$ $\\begin{pmatrix} 0 & 1 & 0 & 1 \\\\ 1 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\\\ \\end{pmatrix} || \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 & 1 \\\\ \\end{pmatrix}$ $\\begin{pmatrix} 0 & 1 & 0 & 1 \\\\ 1 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ \\end{pmatrix} || \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 1 \\\\ \\end{pmatrix}$ P: 5 If it's not too much trouble, could you work through a full example? I'm having trouble getting", "• {BA,BD}+{BB,N3} • {BB,BD}+{BA,N3} as different selections even though they lead to exactly the same reading. - Perhaps one should add that the correct approach is to say \"4 choose 2\" then \"6 choose 2\" + \"4 choose 3\" then \"6 choose 1\" + \"4 choose 4\" then \"6 choose 0\" which leads to $$\\begin{pmatrix} 4 \\\\ 2 \\end{pmatrix}\\begin{pmatrix} 6 \\\\ 2 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} \\begin{pmatrix} 6 \\\\ 1 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix}\\begin{pmatrix} 6 \\\\ 0 \\end{pmatrix} = 115$$. – Patrick Da Silva Dec 28 '11 at 3:33 @Patrick, that appears to be implicit it in the part of the question beginning \"I understand how the answer solved it\". – Henning Makholm Dec 28 '11 at 3:34 Great! Excellent answer, your example (with the BA, BB, etc and N1 to N6) helped me see where I was double counting. Thank you so much! – oxuser Dec 28 '11 at 10:41", "1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\\\end{pmatrix}$ 3 $(1,3)(2,4)$ 3412 $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\\\end{pmatrix}$ 2 $(1,3,2,4)$ 3421 $\\begin{pmatrix} 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\\\end{pmatrix}$ 4 $(1,4,3,2)$ 4123 $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\\\end{pmatrix}$ $\\begin{pmatrix} 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0", "\\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}" ]

[ "# Arccos(x) Calculator An online calculator to calculate the inverse cosine function arccos(x) in radians and degrees. ## How to use the arccos (x) calculator 1 - Enter x as a real number, within the domain of arccos function such that -1 ≤ x ≤ 1 and the number of decimal places desired then press \"enter\". Two answers are displayed: one in radians and the second in degrees. x = 1 Decimal Places = 3 arccos(x)= radians arccos(x) = degrees You may use it to explore the domain and range of arccos(x). ## More References and Links Math Calculators and Solvers.", "+0 help 0 59 1 Find all numbers in the range of $$f(x) = \\arctan x + \\arctan \\frac{1 - x}{1 + x},$$ expressed in radians. Enter all the numbers, separated by commas. Sep 17, 2019", "Function ArcCos(x) # ArcCos ArcCos( x ! (input) numerical expression ) ## Arguments x A scalar numerical expression in the range $$[-1,1]$$. ## Return Value The ArcCos function returns the arccosine of x in the range 0 to $$\\pi$$ radians. Note • A run-time error results if x is outside the range $$[-1,1]$$. • The function ArcCos can be used in constraints of nonlinear mathematical programs. The functions ArcSin, ArcTan, Cos. Arithmetic functions are discussed in full detail in Arithmetic Functions of the Language Reference.", ", use parentheses: tan^2 ( x sec^3! And the calculator is divided into two sections the scientific calculator interface on the.. Is normally represented by arccos ( x ) ) , use parentheses: tan x.: arccos ( 0.866 ) = 30° ( or 0.52 in radians and degrees on is you. As functions of the cosine of the cosine of the inverse cosine functions are the inverses the!", "(cot 1x) = x , by Equation 1.8.8. In other words, the range of cos-1 is restricted to [0, 180] or [0, ].Note: arccos refers to \"arc cosine\", or the radian measure of the arc on a circle corresponding to a given value of cosine.", "given argument button press. Radians and degrees multivalued functions that are analytic, except at a finite number of points after we get input! Compute something or you have identified an error, double-check your Expression, add and... Free functions inverse step-by-step logarithm the below solved Example problem may help you understand the mathematical function of arc. Divided into two sections the scientific calculator arccos of 0.5 is 60°: arccos ( 0.5 =. Arcosine formula ’ re in fraction to Decimal Hexadecimal scientific Notation Distance Weight Time table below below! The domain of the given value in radians and degrees cotangent, and... C. cosine C = x a triangle C = x Library of Python Language! Are analytic, except at a finite number of points when you have identified an,... Calculate an angle may help you understand the mathematical function of the adjacent side to the opposite over adjacent.", "Find the positive value of $x$ that satisfies $x = 1.900 \\sin(x)$. Give the answer to four places of accuracy. $x\\approx$ Remember to calculate the trig functions in radian mode.", "angle. Function will appear here right after we get that β = arccos ( x sec^3! Two sides of a triangle by 10 y = tan ( x ) (., pi ] , the arccos is the same as the inverse Laplace Transform, and consult table! Display tells you what mode you ’ re in, select degrees ( ° ) or (! A scientific calculator interface on the left and the calculator will find the inverse function we get input. It is normally represented by arccos ( θ ) the radians and degrees most calculators! And i 'll get something like cos A=48.735 analytic, except at a finite number of points,... { 2 } \\right ) } = 60 ^0 calculator shows values of cosine! In a market given value in radians and degrees x is a cosine value of x Unit., use parentheses: tan^2 ( x ) sec^3 ( x ) , the range is [... -2, -1,0,1,2,... } find functions inverse calculator - calculate matrix inverse calculator - find functions calculator! You skip parentheses or a multiplication sign, so 5x is to. To adjacent sides of that angle arccos is the same as saving inverse cosine of a number.... Will be parsed as tan ( x ) sec^3 ( x ) ) instead sinx... Calculate arccos", "0$$ $$\\cos(x + \\arccos(\\frac{2}{\\sqrt{5}})) = 0$$ I guess you can take it from here", "# How do you find the derivative of the function f(x)=arcsin(4x)+arccos(4x)? May 14, 2015 Since the derivative of $y = \\arcsin 4 x$ is $y ' = \\frac{1}{\\sqrt{1 - 16 {x}^{2}}} \\cdot 4$ and the derivative of $y = \\arccos 4 x$ is $y ' = - \\frac{1}{\\sqrt{1 - 16 {x}^{2}}} \\cdot 4$, The derivative of the function is $0$.", "must be expressed in radians. Returns the arcsine (the number whose sine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arccosine (the number whose cosine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arctangent (the number whose tangent is Number) in radians.", "higher terms become insignificant, for values of x close to zero, resulting in the following quite useful approximation: ### Properties The derivative of the arccos function is: The integral of the arccos function is given by: The following properties are also valid for the arccos function:", "logarithm its wrong, cause it tends to $-\\infty$ as the inside of it tends to zero. That is correct. So what is the range? 6. So would the range be (-infinity, pi/2)? 7. Consider the fact that $\\arctan$ is a bounded function: $\\displaystyle \\left( {\\left( { - \\frac{\\pi } {2}} \\right)^3 ,\\left( {\\frac{\\pi } {2}} \\right)^3 } \\right)$", "this one, to calculate the maximum range for angle. $\\theta=arccos(\\sqrt{\\frac{2gy_{0}+v^{2}}{2gy_{0}+2v^{2}}})$ It is almost what I am looking for but it doesnt take into account the distance. -", "to trigonometric functions can be viewed as real numbers (radian measure) or degrees. For example, $\\,\\cos \\pi = -1\\,$: here, $\\,\\pi\\,$ is radian measure. Equivalently, $\\,\\cos 180^\\circ = -1\\,$: here, $\\,180^\\circ\\,$ is degree measure. Here's what the definition of arccosine looks like, using degree measure instead of radian measure: $$y = \\arccos x\\ \\ \\ \\ \\ \\text{if and only if}\\ \\ \\ \\ \\bigl(\\ \\cos y = x\\ \\ \\text{AND}\\ \\ 0^\\circ \\le y\\le 180^\\circ\\ \\bigr)$$ $\\,\\arccos x\\,$ is the angle in the interval $\\,[0^\\circ,180^\\circ]\\,$ whose cosine is $\\,x\\,$ • CALCULATOR SKILLS: If a calculator is in degree mode, then $\\,\\arccos x\\,$ is reported in degrees. If a calculator is in radian mode, then $\\,\\arccos x\\,$ is reported in radians. • GRAPH OF THE ARCCOSINE FUNCTION: For a one-to-one function $\\,f\\,$, the graph of its inverse, $\\,f^{-1}\\,$, is found by reflecting the graph of $\\,f\\,$ about the line $\\,y = x\\,$. Below, this technique is used to construct the graph of the arccosine function: Here's the piece of the cosine curve that is used to define the arccosine function: domain: $\\,[0,\\pi]\\,$ range: $\\,[-1,1]\\,$ Here's the same curve, together with its reflection about the line $\\,\\color{red}{y = x}\\,$ The graph of the arccosine function domain: $\\,[-1,1]\\,$ range: $\\,[0,\\pi]\\,$ Notice that the domain and range of a function and", "distance between and is . Divide by . The period of the function is so values will repeat every radians in both directions. , for any integer Solve for x sin(x)+csc(x)=2 Scroll to top", "$$f(x) = \\tan g(x)$$ $$f'(x) = \\sec^2 g(x) \\times g'(x)$$ $$\\tan x \\Rightarrow \\sec^2 x$$ Cosecant Function $$f(x) = \\csc x$$ $$f'(x) = \\text{-} \\csc x \\cot x$$use radians Secant Function $$f(x) = \\sec x$$ $$f'(x) = \\sec x \\tan x$$use radians Cotangent Function $$f(x) = \\cot x$$ $$f'(x) = \\text{-} \\csc^2 x$$use radians arcSine Function $$f(x) = \\arcsin x$$ $$f'(x) = \\frac{1}{\\sqrt{1-x^2}}$$ arcCosine Function $$f(x) = \\arccos x$$ $$f'(x) = \\text{-} \\frac{1}{\\sqrt{1-x^2}}$$ arcTangent Function $$f(x) = \\arctan x$$ $$f'(x) = \\frac{1}{x^2+1}$$ $$m_{\\text{tangent}} = \\text{IROC} = \\lim \\limits_{h \\to 0} {f(a+h)-f(a) \\over h} = f'(a)$$ $$m_{\\text{secant}} = \\text{AROC} = {f(x_{2})-f(x_{1}) \\over x_{2} - x_{1} }$$", "occurs at $-5, 5.639^{c}$ and a maximum value occurs at $5, 2.496^{c}$ maximum function value of $y = 5$ at $\\theta = \\arctan\\left(-\\frac{3}{4}\\right) + \\pi \\approx 2.498$ RADIANS minimum function value of $y = -5$ at $\\theta = \\arctan\\left(-\\frac{3}{4}\\right) + 2\\pi \\approx 5.640$ RADIANS", "just find the resulting angle from the defined range of $y = \\arccos{x}$ and think about where it would be in another interval on the unit circle. for $-\\pi \\le x \\le 0$ $y = -\\arccos{x}$", "f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. D. is the correct answer, and it can be found without calculus (but with some trigonometry), because $4\\sin x-3\\cos x=5\\cdot\\left(\\frac{4}{5}\\cdot \\sin x-\\frac{3}{5}\\cdot\\cos x\\right)=-5\\cos(x+\\varphi)$ for $\\varphi := \\arccos(3/5)$. Now, since x varies over a whole period of the function, the maximum value of f(x) must be 5." ]

[ "$P$ can be written as: ${x}_{P}=cos\\left(\\alpha \\right)$ and ${y}_{P}=sin\\left(\\alpha \\right)$ . Using Pythagoras' theorem you can check that: ${\\left(sin\\left(\\alpha \\right)\\right)}^{2}+{\\left(cos\\left(\\alpha \\right)\\right)}^{2}=1$ To decrease the number of brackets you can use: ${sin}^{2}\\left(\\alpha \\right)+{cos}^{2}\\left(\\alpha \\right)=1$ There are some values that are easy to use: • $cos\\left(0\\right)=1$ • $cos\\left(\\frac{1}{6}\\pi \\right)=\\frac{1}{2}\\sqrt{3}$ • $cos\\left(\\frac{1}{4}\\pi \\right)=\\frac{1}{2}\\sqrt{2}$ • $cos\\left(\\frac{1}{3}\\pi \\right)=\\frac{1}{2}$ • $cos\\left(\\frac{1}{2}\\pi \\right)=0$ and vice versa: • $arccos\\left(0\\right)=\\frac{1}{2}\\pi$ • $arccos\\left(\\frac{1}{2}\\right)=\\frac{1}{3}\\pi$ • $arccos\\left(\\frac{1}{2}\\sqrt{2}\\right)=\\frac{1}{4}\\pi$ • $arccos\\left(\\frac{1}{2}\\sqrt{3}\\right)=\\frac{1}{6}\\pi$ • $arccos\\left(1\\right)=1$ Use these values, when exact answers are required.", "x}{3}+24cos\\frac{\\pi x}{3}\\equiv 26cos(\\frac{\\pi x}{3}-\\alpha)$ $26cos(\\frac{\\pi x}{3}-\\arctan\\frac{24}{10})=13$ 4. Originally Posted by arze $10sin\\frac{\\pi x}{3}+24cos\\frac{\\pi x}{3}=13$ $10sin\\frac{\\pi x}{3}+24cos\\frac{\\pi x}{3}\\equiv R(sin\\frac{\\pi x}{3}cos\\alpha+cos\\frac{\\pi x}{3}sin\\alpha)\\equiv Rcos(\\frac{\\pi x}{3}- \\alpha)$ Oops! That should be $24cos\\frac{\\pi x}{3}+10sin\\frac{\\pi x}{3}\\equiv R({\\color{red}cos}\\frac{\\pi x}{3}cos\\alpha+{\\color{red}sin}\\frac{\\pi x}{3}sin\\alpha)\\equiv Rcos(\\frac{\\pi x}{3}- \\alpha)$. (Remember that cos(a - b) = cos a cos b + sin a sin b.) So $Rcos\\alpha={\\color{red}24}$ and $Rsin\\alpha={\\color{red}10}$. Looks like the only thing that changes is $26cos(\\frac{\\pi x}{3}-\\arctan{\\color{red}\\frac{10}{24}})=13$ $26cos(\\frac{\\pi x}{3}-\\arctan\\frac{5}{12})=13$ $cos(\\frac{\\pi x}{3}-\\arctan\\frac{5}{12})=\\frac{1}{2}$ ... 01 5. Oops! That should be 01 Thanks! now i got that part. so $\\alpha=\\arctan\\frac{10}{24}$ Now what about the other part? I can't understand why 6. Originally Posted by arze $10sin\\frac{\\pi x}{3}+24cos\\frac{\\pi x}{3}=13$ $10sin\\frac{\\pi x}{3}+24cos\\frac{\\pi x}{3}\\equiv R(sin\\frac{\\pi x}{3}cos\\alpha+cos\\frac{\\pi x}{3}sin\\alpha)\\equiv R {\\color{red}cos} (\\frac{\\pi x}{3}- \\alpha)$ Mr F says: That red cos should be a sin .... $Rcos\\alpha=10$ $Rsin\\alpha=24$ $R=\\sqrt{10^2+24^2}=\\sqrt{676}=26$ $\\alpha=\\arctan\\frac{24}{10}$ So $10sin\\frac{\\pi x}{3}+24cos\\frac{\\pi x}{3}\\equiv 26cos(\\frac{\\pi x}{3}-\\alpha)$ $26cos(\\frac{\\pi x}{3}-\\arctan\\frac{24}{10})=13$ To get the solution the book gives, you should have: $10 \\sin \\frac{\\pi x}{3}+24 \\cos\\frac{\\pi x}{3} \\equiv R \\left({\\color{red}\\cos \\frac{\\pi x}{3}} \\cos\\alpha + {\\color{red}\\sin\\frac{\\pi x}{3}} \\sin\\alpha \\right) \\equiv R cos \\left(\\frac{\\pi x}{3}- \\alpha \\right)$ where $R = 26$ and $\\alpha= \\arctan \\frac{10}{24} = \\frac{5}{12}$. Then: $26 \\cos \\left(\\frac{\\pi x}{3}- \\arctan \\frac{5}{12} \\right) = 13$ $\\Rightarrow \\cos \\left(\\frac{\\pi x}{3}- \\arctan \\frac{5}{12} \\right) = \\frac{1}{2}$. Case 1: $\\frac{\\pi x}{3} - \\arctan \\frac{5}{12} = \\frac{\\pi}{3} + 2 n \\pi$ $\\Rightarrow \\pi x - 3", "edited the post above you by the way. 11. Jun 6, 2017 ### McFluffy My bad. I had edited the post above you by the way. 12. Jun 6, 2017 ### ehild The angle which tangent is given as m, can have multiple values, so it is θ=arctan(m) ± k pi, where k is any integer. Choose k so as 0<alpha<pi. 13. Jun 6, 2017 ### McFluffy Alright! It seems that I've figured it out thanks to you. So the solution should be a piecewise function: $\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right) & \\text{ if } \\arccos{(\\frac{z}{c})}<\\beta<\\pi \\end{cases}$ The reason why it's defined as a piecewise function is because of the original solution which is: $\\alpha(\\beta)=\\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ is defined to be from $(-)$ to $(+)$ during the $0<\\beta<\\frac{\\pi}{2}$ interval. I tried finding at what value of $\\beta$ when $\\alpha(\\beta)$ starts to change from being $(-)$ to $(+)$. I noticed that the term: $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ can be undefined for a certain value of $\\beta$ that makes the denominator equal to 0. Thus, setting ${c\\cos \\beta-z}=0$ to had enabled me to conclude that $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ is undefined when $\\beta=\\arccos{(\\frac{z}{c})}$. What's useful from this is that I", "Алгебра # Вычислить arccos(-√2/2)-arcsin(1) Решить уравнения: а) cos x=-√2/2 б) cos(2x-π/4)=√2/2 c)3ctgx+2=0 $1)$ $arccos(- \\frac{ \\sqrt{2} }{2})-arcsin1= \\frac{3 \\pi }{4} - \\frac{ \\pi }{2} = \\frac{3 \\pi }{4} - \\frac{2 \\pi }{4} =\\frac{ \\pi }{4}$ P. S. $arccos(- \\frac{ \\sqrt{2} }{2})= \\pi - arccos \\frac{ \\sqrt{2} }{2}= \\pi - \\frac{ \\pi }{4} = \\frac{3 \\pi }{4}$ $2)$ $a)$ $cosx= - \\frac{ \\sqrt{2} }{2}$ $x=бarccos(- \\frac{ \\sqrt{2} }{2} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б( \\pi -arccos\\frac{ \\sqrt{2} }{2} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б( \\pi - \\frac{ \\pi }{4} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б \\frac{3 \\pi }{4} +2 \\pi n,$ $n$ ∈ $Z$ $b)$ $cos(2x- \\frac{ \\pi }{4} )= \\frac{ \\sqrt{2} }{2}$ $2x- \\frac{ \\pi }{4} =бarccos \\frac{ \\sqrt{2} }{2} +2 \\pi k,$ $k$ ∈ $Z$ $2x- \\frac{ \\pi }{4} =б \\frac{ \\pi }{4} +2 \\pi k,$ $k$ ∈ $Z$ $2x=б \\frac{ \\pi }{4} + \\frac{ \\pi }{4} +2 \\pi k,$ $k$ ∈ $Z$ $x=б \\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi k,$ $k$ ∈ $Z$ $x= \\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi k,$ $k$ ∈ $Z$ или $x=-\\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi n,$ $n$ ∈ $Z$ $x=\\frac{ \\pi }{4} + \\pi k,$ $k$ ∈ $Z$ или $x= \\pi n,$ $n$ ∈ $Z$ $c)$ $3ctgx+2=0$ $3ctgx=-2$", "# What does arccos(cos ((-2pi)/3)) equal? Dec 24, 2015 $\\frac{2 \\pi}{3}$ #### Explanation: It would look strange how is that possible! A question usually which pops up isn't $\\arccos \\left(\\cos \\left(A\\right)\\right) = A$. To understand this we can use $\\cos \\left(- \\theta\\right) = \\cos \\left(\\theta\\right)$ Therefore $\\cos \\left(- \\frac{2 \\pi}{3}\\right) = \\cos \\left(\\frac{2 \\pi}{3}\\right)$ Following it up with $\\arccos \\left(\\cos \\left(- \\frac{2 \\pi}{3}\\right)\\right) = \\arccos \\left(\\cos \\left(\\frac{2 \\pi}{3}\\right)\\right)$ That leads us to our answer $\\frac{2 \\pi}{3}$. Let us understand the same in a different manner. The range of $\\arccos \\left(x\\right)$ is $\\left[0 , \\pi\\right]$. $\\cos \\left(\\frac{- 2 \\pi}{3}\\right) = - \\frac{1}{2}$ The angle $\\frac{- 2 \\pi}{3}$ is not in the range of the function. So we select the angle in the range $\\left[0 , \\pi\\right]$ which gives cos(x) = -1/2 that works out to $\\frac{2 \\pi}{3}$.", "$= \\frac{56}{65}$. If you want to write this without the names $\\alpha$ and $\\beta$ 9Or something like that, you could write: $\\sin \\left(\\arcsin \\left(\\frac{4}{5}\\right) + \\arctan \\left(\\frac{12}{5}\\right)\\right)$ $= \\sin \\left(\\arcsin \\left(\\frac{4}{5}\\right)\\right) \\cos \\left(\\arctan \\left(\\frac{12}{5}\\right)\\right) + \\cos \\left(\\arcsin \\left(\\frac{4}{5}\\right)\\right) \\sin \\left(\\arctan \\left(\\frac{12}{5}\\right)\\right)$ But I, personally do not think that is more clear. (I do think it is good for students to see it this way, however.)", "θ= pi - arcsin(m). Last edited: Jun 6, 2017 8. Jun 6, 2017 ### McFluffy Why did you define $\\theta=\\frac{a}{b}$? Are you talking about the $\\arctan$ function? Because I think if the $\\frac{a}{b}$ you are talking about are the triangle side ratios then it should be $\\arctan$ because it maps side ratios to the appropriate angle. After taking some time thinking about the terms in the equation, I noticed that $\\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)$ will always be $(+)$ for $0<\\beta<\\pi$ but for $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ however, it can be $(+)$ or $(-)$ for $0<\\beta<\\pi$ which I'm having problem interpreting it because of the $-z$ term in the denominator of the argument. Because of that $-z$ term, when $0<\\beta<\\frac{\\pi}{2}$, $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ can be $(+)$ or $(-)$ because as $\\beta$ → $\\frac{\\pi}{2}$, $c\\cos \\beta$ becomes smaller and smaller and it'll eventually get overpowered by the $-z$ term. But for $\\frac{\\pi}{2}<\\beta<\\pi$ it's less complicated because it will always of the form of $\\arctan{(\\frac{a}{-b})}$. So my problem is how do I adjust $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ for $0<\\beta<\\pi$ such that it doesn't make $\\alpha(\\beta)$ become negative? Last edited: Jun 6, 2017 9. Jun 6, 2017 ### ehild I did not. Read my previous post. It was m=a/b = tan(θ). 10. Jun 6, 2017 ### McFluffy My bad. I had", "determine $$x$$, $$y$$ and $$r$$. In the first question we found that $$\\cos \\beta = \\frac{-4}{5}$$ and so we can use this to find $$x$$ and $$r$$. \\begin{align*} \\cos \\beta & = \\frac{-4}{5} \\\\ \\frac{x}{r} & = \\frac{-4}{5} \\end{align*} Therefore $$x = -4$$ and $$r = 5$$. \\begin{align*} y^{2} & = r^{2} - x^{2} \\\\ & = (5)^{2} - (-4)^{2} \\\\ & = 25 - 16 \\\\ y & = \\pm 3 \\end{align*} We are told that $$180^{\\circ} < \\beta < 360^{\\circ}$$. Therefore the angle is in the third quadrant and $$y = -3$$. We can draw a rough sketch of the angle: We can now find $$\\frac{3}{\\tan{\\beta}} + 2 \\sin^2{\\beta}$$: \\begin{align*} \\frac{3}{\\tan{\\beta}} + 2 \\sin^2{\\beta} & = \\frac{3}{\\frac{y}{x}} + 2\\left(\\frac{y}{r}\\right)^{2} \\\\ & = \\frac{3x}{y} + \\frac{2y^{2}}{r^{2}} \\\\ & = \\frac{3(-4)}{-3} + \\frac{2(-3)^{2}}{(5)^{2}} \\\\ & = \\frac{-12}{3} + \\frac{18}{25} \\\\ & = -4 + \\frac{18}{25} \\\\ & = \\frac{-100 + 18}{25} \\\\ & = \\frac{-82}{25} \\end{align*} If $$\\sin{\\theta} = -\\dfrac{15}{17}$$ and $$\\cos \\theta < 0$$ find the following, without the use of a calculator: $$\\cos \\theta$$ We first need to determine $$x$$, $$y$$ and $$r$$. We are given $$\\sin{\\theta} = -\\frac{15}{17}$$ and so we can use this to find $$y$$ and $$r$$. \\begin{align*} \\sin \\theta & = -\\frac{15}{17} \\\\ \\frac{y}{r} & = -\\frac{15}{17} \\end{align*} Therefore $$y = -15$$", "$\\cos \\alpha = \\displaystyle \\sqrt{1 - \\left( \\frac{x-2}{2} \\right)^2 } = \\displaystyle \\frac{\\sqrt{4x-x^2}}{2}$ Fourth, $\\sin 2\\beta = 2 \\sin \\beta \\cos \\beta$. Using the above identity again, we find $\\sin 2\\beta = \\displaystyle 2 \\left( \\frac{ \\sqrt{x} }{2} \\right) \\sqrt{ 1 - \\left( \\frac{ \\sqrt{x} }{2} \\right)^2 }$ $= \\sqrt{x} \\sqrt{1 - \\displaystyle \\frac{x}{4}}$ $= \\displaystyle \\frac{\\sqrt{4x-x^2}}{2}$ Combining the above, we find $\\sin(\\alpha - 2 \\beta) = \\displaystyle -\\left( \\frac{x-2}{2} \\right)^2 - \\left( \\frac{\\sqrt{4x-x^2}}{2} \\right)^2$ $\\sin(\\alpha - 2 \\beta) = \\displaystyle \\frac{-(x^2 - 4x + 4) - (4x - x^2)}{4}$ $\\sin(\\alpha - 2 \\beta) = -1$ $\\alpha - 2 \\beta = \\displaystyle -\\frac{\\pi}{2} + 2\\pi n$ for some integer $n$ Also, since $-\\pi/2 \\le \\alpha \\le \\pi/2$ and $0 \\le -2\\beta \\le \\pi$, we see that $-\\pi/2 \\le \\alpha - 2 \\beta \\le 3\\pi/2$. (From its definition, $\\beta$ is the arcsine of a positive number and therefore must be nonnegative.) Therefore, $\\alpha - 2\\beta = -\\pi/2$. In other words, $\\sin^{-1} \\left( \\displaystyle \\frac{x-2}{2} \\right)$ and $2 \\sin^{-1} \\left( \\displaystyle \\frac{\\sqrt{x}}{2} \\right)$ differ by a constant, thus showing that the two antiderivatives are equivalent.", "# Solve algebraically and graphically: $\\arcsin(x) + \\arccos\\left(\\frac{x}{2}\\right) = \\frac{5\\pi}{6}$ Solve: $$\\arcsin(x) + \\arccos\\left(\\frac{x}{2}\\right) = \\frac{5\\pi}{6}$$ I think the algebraic solution should start like: $$\\arcsin(x) = \\frac{5\\pi}{6} - \\arccos\\left(\\frac{x}{2}\\right)$$ $$x = \\sin\\left(\\frac{5\\pi}{6} - \\arccos\\left(\\frac{x}{2}\\right)\\right)$$ at that stage probably I should use some trigonometric relation or property of $\\arccos(x)$ to convert it to $\\arcsin(x)$, however I can't figure it out. As for the solution by graph I can't even think what steps should I follow to build it. $$\\arcsin x + \\arccos\\frac{x}{2} = \\frac{5\\pi}{6}$$ $$\\arcsin x = \\frac{5\\pi}{6} - \\arccos\\frac{x}{2}$$ $$x = \\sin{\\left(\\frac{5\\pi}{6} - \\arccos\\frac{x}{2}\\right)}$$ $$x=\\sin\\frac{5\\pi}6\\cos \\left(\\arccos\\frac{x}{2}\\right)-\\cos\\frac{5\\pi}6 \\sin\\left(\\arccos\\frac{x}{2}\\right)$$ $$x=\\frac12\\cdot\\frac x2+\\frac{\\sqrt3}{2} \\cdot\\sqrt{1-\\left(\\frac x2\\right)^2}$$ Answer: $x=1$ First you have the sine of one thing minus another. $$\\sin(\\alpha-\\beta) = \\sin\\alpha\\cos\\beta - \\cos\\alpha\\sin\\beta.$$ After applying that, you will have $$\\sin\\arccos w = \\sqrt{1-w^2} \\tag 1$$ $$\\cos\\arccos w = w$$ To get the identity $(1)$, just draw a triangle. If $\\cos\\alpha =w$ then you have $\\text{adjacent} = w$ and $\\text{hypotenuse} = 1$, so by the Pythagorean theorem you have $\\text{opposite} = \\sqrt{1-w^2}.$ Then recall that $\\sin=\\dfrac{\\text{opposite}}{\\text{hypotenuse}}.$ HINT: $$\\arccos\\dfrac x2-\\dfrac\\pi3=\\dfrac\\pi2-\\arcsin x=\\arccos x$$ Applying cosine $$\\dfrac x2\\cdot\\cos\\dfrac\\pi3+\\sin\\dfrac\\pi3\\sqrt{1-\\left(\\dfrac x2\\right)^2}=x$$ $$\\implies x=1$$", "can now determine at which interval does $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ starts to change its sign. I managed to construct a sign diagram and conclude that $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)>0$ in $(0, \\arccos{(\\frac{z}{c}}))$ interval and $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)<0$ in $(\\arccos{(\\frac{z}{c}}), \\pi)$ interval. This allowed me to adjust $\\alpha(\\beta)$ better so I can prevent it from being $(-)$. Since $\\arcsin{m}=\\theta\\pm k\\pi$, I can adjust the value of $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ in the $(0, \\arccos{(\\frac{z}{c}}))$ interval such that $\\alpha(\\beta)$ is not defined to be negative and that it manages to merge with the original function to make $\\alpha(\\beta)$ \"continuous\"(You could use calculus to figure out what value is $\\alpha(\\beta)$ when $\\beta=\\arccos{(\\frac{z}{c}})$). Here's the graph of it: Anyways, how is my explanation of the solution? Is this enough to justify that $\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right) & \\text{ if } \\arccos{(\\frac{z}{c})}<\\beta<\\pi \\end{cases}$ is indeed a solution to this problem? 14. Jun 6, 2017 ### McFluffy Oh no. I forgot that $\\alpha(\\beta)$ has to be defined for all values in $(0, \\pi)$. Which means the solution should be this:$\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\lim_{\\beta\\rightarrow \\arccos{(\\frac{z}{c})}}\\alpha(\\beta) & \\text{ if } \\beta=\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan", "arcsin \\ left (\\ cos \\ left ({\\ frac {\\ pi} {n}} \\ right) \\ cdot {\\ sqrt {\\ tan ^ {2} \\ left ({\\ frac {\\ pi} {n}} \\ right) - \\ tan ^ {2} \\ left ({\\ frac {\\ alpha} {2}} \\ right)}} \\ right)}$ With the number of edges / faces at a corner and the interior angle of the equilateral triangle, the following applies ${\\ displaystyle {n = 4}}$${\\ displaystyle {a = 60 ^ {\\ circ}}}$ {\\ displaystyle {\\ begin {aligned} {\\ mathsf {(1)}} \\; \\ Omega & = 2 \\ pi -2 \\ cdot 4 \\ cdot \\ arcsin \\ left (\\ cos \\ left ({\\ frac {\\ pi } {4}} \\ right) \\ cdot {\\ sqrt {\\ tan ^ {2} \\ left ({\\ frac {\\ pi} {4}} \\ right) - \\ tan ^ {2} \\ left ({\\ frac {60 ^ {\\ circ}} {2}} \\ right)}} \\ right) \\ Rightarrow \\\\ & = 2 \\ pi -8 \\ cdot \\ arcsin \\ left ({\\ frac {1} {\\ sqrt {3} }} \\ right) \\ approx 1 {,} 359347638 \\; \\ mathrm {sr} \\\\\\ end {aligned}}} because of it ${\\ displaystyle \\ cos (\\ arcsin x) = {\\ sqrt {1-x ^ {2}}}}$ {\\ displaystyle {\\ begin {aligned} {\\ mathsf {(2)}} \\; \\ cos (\\ arcsin", "$\\displaystyle r1^2\\arcsin \\frac {d^2+r1^2-r2^2}{2dr1}$. $$A=-\\frac {(d^2-r1^2+r2^2)}{2d}\\sqrt{r2^2-(\\frac {d^2-r1^2+r2^2}{2d}) ^2} - r2^2\\arcsin \\frac {d^2-r1^2+r2^2}{2dr2} +\\frac {\\pi}{2} r2^2 - \\frac {(d^2+r1^2-r2^2)}{2d} \\sqrt{r1^2-(\\frac {d^2+r1^2-r2^2}{2d})^2} - r1^2\\arcsin \\frac {d^2+r1^2-r2^2}{2dr1} + \\frac {\\pi}{2} r1^2$$ Further moving the terms containing $\\pi/2$ into those containing the $\\arcsin$ we finally get $$A=-\\frac {(d^2-r1^2+r2^2)}{2d}\\sqrt{r2^2-(\\frac {d^2-r1^2+r2^2}{2d}) ^2} + r2^2\\arccos \\frac {d^2-r1^2+r2^2}{2dr2} - \\frac {(d^2+r1^2- r2^2)}{2d} \\sqrt{r1^2-(\\frac {d^2+r1^2-r2^2}{2d})^2} + r1^2\\arccos \\frac {d^2+r1^2-r2^2}{2dr1}$$", "the arccos (or $\\cos^{-1}$... $3x = \\arccos{\\frac{1}{2}}$ $3x = \\left \\{ \\frac{\\pi}{3}, 2\\pi - \\frac{\\pi}{3} \\right \\} + 2\\pi n, n \\in \\mathbf{Z}$, since the period of cosine is $2\\pi$. $3x = \\left \\{ \\frac{\\pi}{3}, \\frac{5\\pi}{3} \\right \\} + 2\\pi n$ $x = \\left \\{ \\frac{\\pi}{9}, \\frac{5\\pi}{9} \\right \\} + \\frac{2\\pi n}{3}$. Case 2: $\\cos{(3x)} = -\\frac{1}{2}$ $3x = \\arccos{\\left(-\\frac{1}{2}\\right)}$ $3x = \\left \\{ \\pi - \\frac{\\pi}{3}, \\pi + \\frac{\\pi}{3} \\right \\} + 2\\pi n$, since cosine is negative in the second and third quadrants and the period is $2\\pi$ $3x = \\left \\{ \\frac{2\\pi}{3}, \\frac{4\\pi}{3} \\right \\} + 2\\pi n$ $x = \\left \\{ \\frac{2\\pi}{9}, \\frac{4\\pi}{9} \\right \\} + \\frac{2\\pi n}{3}$. So putting it all together, the possibilities are $\\left \\{ \\frac{\\pi}{9}, \\frac{2\\pi}{9}, \\frac{4\\pi}{9}, \\frac{5\\pi}{9} \\right \\} + \\frac{2\\pi n}{3}$.", "- \\arcsin \\left(- \\sin \\left(\\frac{\\pi}{4}\\right)\\right) - \\arccos \\left(\\cos \\left(\\frac{\\pi}{3}\\right)\\right) + \\arcsin \\left(\\sin \\left(\\frac{\\pi}{3}\\right)\\right)$ $= \\arccos \\left(\\cos \\left(\\pi - \\frac{\\pi}{6}\\right)\\right) - \\arcsin \\sin \\left(- \\frac{\\pi}{4}\\right) - \\arccos \\left(\\cos \\left(\\frac{\\pi}{3}\\right)\\right) + \\arcsin \\left(\\sin \\left(\\frac{\\pi}{3}\\right)\\right)$ $= \\frac{5 \\pi}{6} + \\frac{\\pi}{4} - \\frac{\\pi}{3} + \\frac{\\pi}{3}$ $= \\frac{10 \\pi + 3 \\pi}{12} = \\frac{13 \\pi}{12}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 2).$\\arcsin \\left(- \\frac{1}{2}\\right) + \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right)$ $= \\arcsin \\left(\\sin \\left(- \\frac{\\pi}{6}\\right)\\right) + \\arcsin \\left(\\sin \\left(- \\frac{\\pi}{3}\\right)\\right)$ $= - \\frac{\\pi}{6} - \\frac{\\pi}{3} = - \\frac{\\pi}{2}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 3).sin(arccos(-√3/2)) =sin(arccos(cos(pi-pi/6)) $= \\sin \\left(\\pi - \\frac{\\pi}{6}\\right) = \\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$", "${\\displaystyle \\tan(\\alpha \\pm \\beta )={\\frac {\\tan \\alpha \\pm \\tan \\beta }{1\\mp \\tan \\alpha \\tan \\beta }}}$[13][15] ${\\displaystyle \\csc(\\alpha \\pm \\beta )={\\frac {\\sec \\alpha \\sec \\beta \\csc \\alpha \\csc \\beta }{\\sec \\alpha \\csc \\beta \\pm \\csc \\alpha \\sec \\beta }}}$[16] ${\\displaystyle \\sec(\\alpha \\pm \\beta )={\\frac {\\sec \\alpha \\sec \\beta \\csc \\alpha \\csc \\beta }{\\csc \\alpha \\csc \\beta \\mp \\sec \\alpha \\sec \\beta }}}$[16] ${\\displaystyle \\cot(\\alpha \\pm \\beta )={\\frac {\\cot \\alpha \\cot \\beta \\mp 1}{\\cot \\beta \\pm \\cot \\alpha }}}$[13][17] ${\\displaystyle \\arcsin x\\pm \\arcsin y=\\arcsin \\left(x{\\sqrt {1-y^{2}}}\\pm y{\\sqrt {1-x^{2}}}\\right)}$[18] ${\\displaystyle \\arccos x\\pm \\arccos y=\\arccos \\left(xy\\mp {\\sqrt {\\left(1-x^{2}\\right)\\left(1-y^{2}\\right)}}\\right)}$[19] ${\\displaystyle \\arctan x\\pm \\arctan y=\\arctan \\left({\\frac {x\\pm y}{1\\mp xy}}\\right)}$[20] ${\\displaystyle \\operatorname {atan2} (y_{1},x_{1})\\pm \\operatorname {atan2} (y_{2},x_{2})=\\operatorname {atan2} (y_{1}x_{2}\\pm y_{2}x_{1},x_{1}x_{2}\\mp y_{1}y_{2})}$ ${\\displaystyle \\operatorname {arccot} x\\pm \\operatorname {arccot} y=\\operatorname {arccot} \\left({\\frac {xy\\mp 1}{y\\pm x}}\\right)}$ Matrix form The sum and difference formulae for sine and cosine follow from the fact that a rotation of the plane by angle α, following a rotation by β, is equal to a rotation by α+β. In terms of rotation matrices: {\\displaystyle {\\begin{aligned}&{}\\quad \\left({\\begin{array}{rr}\\cos \\alpha &-\\sin \\alpha \\\\\\sin \\alpha &\\cos \\alpha \\end{array}}\\right)\\left({\\begin{array}{rr}\\cos \\beta &-\\sin \\beta \\\\\\sin \\beta &\\cos \\beta \\end{array}}\\right)\\\\[12pt]&=\\left({\\begin{array}{rr}\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta &-\\cos \\alpha \\sin \\beta -\\sin \\alpha \\cos \\beta \\\\\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta &-\\sin \\alpha \\sin \\beta +\\cos \\alpha \\cos \\beta", "# arccos+arccot • October 2nd 2012, 07:56 AM Petrus arccos+arccot Calculate https://webwork.math.su.se/webwork2_...f7e026e1b1.png. The answer may not contain cyklometriska functions. got no ide to do this • October 2nd 2012, 09:14 AM Soroban Re: arccos+arccot Hello, Petrus! Quote: $\\text{Calculate: }\\:\\theta \\;=\\;\\underbrace{\\arccos\\left(\\tfrac{3}{5}\\right)} _{\\alpha} + \\underbrace{\\text{arccot}\\left(\\tfrac{1}{7}\\right) }_{\\beta}$ Let $\\alpha \\,=\\,\\arccos\\left(\\tfrac{3}{5}\\right) \\quad\\Rightarrow\\quad \\cos\\alpha \\,=\\,\\tfrac{3}{5} \\,=\\,\\tfrac{adj}{hyp}$ . . Then: . $opp \\,=\\,4\\quad\\Rightarrow\\quad \\tan\\alpha \\,=\\,\\tfrac{4}{3}$ Let $\\beta \\,=\\,\\text{arccot}\\left(\\tfrac{1}{7}\\right) \\quad\\Rightarrow\\quad \\cot\\beta \\,=\\,\\tfrac{1}{7}$ . . Then: . $\\tan\\beta \\,=\\,\\tfrac{7}{1} \\,=\\,7$ We have: . $\\theta \\;=\\;\\alpha + \\beta$ . . $\\tan\\theta \\;=\\;\\frac{\\tan\\alpha + \\tan\\beta}{1 - (\\tan\\alpha)(\\tan\\beta)} \\;=\\;\\frac{\\frac{4}{3}+7}{1-\\frac{4}{3}(7)} \\;=\\;\\frac{\\frac{25}{3}}{\\text{-}\\frac{25}{3}} \\;=\\;-1$ $\\text{Therefore: }\\:\\theta \\;=\\;\\frac{3\\pi}{4} + \\pi n\\;\\text{ for any integer }n.$ • October 2nd 2012, 01:18 PM Siron Re: arccos+arccot Let $\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)+\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)=x$ Now, let's take the $\\cos$ of both sides. Thus $\\cos\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)+\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]=\\cos(x)$ $\\Leftrightarrow \\cos\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)\\right]\\cos\\left[\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]-\\sin\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)\\right]\\sin\\left[\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]=\\cos(x)$ If we use the following identities: $\\cos\\left[\\mbox{arc}\\cos(x)\\right]=x$ $\\cos\\left[\\mbox{arc}\\cot(x)\\right]=\\frac{1}{\\sqrt{\\left(\\frac{1}{x}\\right)^2+1}}$ $\\sin\\left[\\mbox{arc}\\cos(x)\\right]=\\sqrt{1-x^2}$ $\\sin\\left[\\mbox{arc}\\cot(x)\\right] =\\frac{1}{\\sqrt{1+x^2}}$ then we can write the equation as $\\left(\\frac{3}{5}\\right)\\frac{1}{\\sqrt{50}}-\\sqrt{1-\\left(\\frac{3}{5}\\right)^2}\\frac{1}{\\sqrt{1+\\left( \\frac{1}{7}\\right)^2}}=\\cos(x)$ $\\Leftrightarrow \\frac{3}{5\\sqrt{50}} - \\frac{4}{5}\\left(\\frac{7}{\\sqrt{50}}\\right)=\\cos(x )$ $\\Leftrightarrow \\frac{3}{25\\sqrt{2}} - \\frac{28}{25\\sqrt{2}}=\\cos(x)$ $\\Leftrightarrow \\cos(x) = \\frac{-\\sqrt{2}}{2}$ $\\Leftrightarrow x = \\frac{3\\pi}{4}+2k\\pi$ • October 2nd 2012, 06:38 PM Nervous Re: arccos+arccot Quote: Originally Posted by Soroban Hello, Petrus! Let $\\alpha \\,=\\,\\arccos\\left(\\tfrac{3}{5}\\right) \\quad\\Rightarrow\\quad \\cos\\alpha \\,=\\,\\tfrac{3}{5} \\,=\\,\\tfrac{adj}{hyp}$ . . Then: . $opp \\,=\\,4\\quad\\Rightarrow\\quad \\tan\\alpha \\,=\\,\\tfrac{4}{3}$ Let $\\beta \\,=\\,\\text{arccot}\\left(\\tfrac{1}{7}\\right) \\quad\\Rightarrow\\quad \\cot\\beta \\,=\\,\\tfrac{1}{7}$ . . Then: . $\\tan\\beta \\,=\\,\\tfrac{7}{1} \\,=\\,7$ We have: . $\\theta \\;=\\;\\alpha + \\beta$ . . $\\tan\\theta", "# How do you solve sin(arcsin(3/5) + arccos(3/5))? May 15, 2015 Picture a right angled triangle with sides 3, 4 and 5 (since ${3}^{2} + {4}^{2} = {5}^{2}$). Call the smallest angle $\\alpha$ and the next smallest $\\beta$. $\\sin \\alpha = \\frac{3}{5}$, being the length of the opposite side divided by the length of the hypotenuse. $\\cos \\beta = \\frac{3}{5}$, being the length of the adjacent side divided by the length of the hypotenuse. So $\\arcsin \\left(\\frac{3}{5}\\right) + \\arccos \\left(\\frac{3}{5}\\right) = \\alpha + \\beta = \\frac{\\pi}{2}$ (or ${90}^{o}$) since together with the right angle, the internal angles of the triangle must add up to $\\pi$ (${180}^{o}$). $\\sin \\left(\\frac{\\pi}{2}\\right) = 1$. So $\\sin \\left(\\arcsin \\left(\\frac{3}{5}\\right) + \\arccos \\left(\\frac{3}{5}\\right)\\right) = 1$.", "some ideas on how to do it ....) 3. Use the identity: $\\arctan a+\\arctan b=\\arctan\\frac{a+b}{1-ab}$ a) $\\arctan\\frac{3x}{1-2x^2}=\\frac{\\pi}{2}\\Rightarrow 1-2x^2=0\\Rightarrow x=\\pm\\frac{\\sqrt{2}}{2}$ The correct solution is $x=\\frac{\\sqrt{2}}{2}$ b) $arctan\\frac{3}{8}+\\arctan\\frac{1}{3}+\\arctan\\frac{ 2}{19}=\\arctan\\frac{\\frac{3}{8}+\\frac{1}{3}}{1-\\frac{1}{8}}+\\arctan\\frac{2}{19}=$ $=\\arctan\\frac{17}{21}+\\arctan\\frac{2}{19}=\\arctan 1=\\frac{\\pi}{4}$ 4. Originally Posted by mr fantastic (a) Let $\\tan^{-1} (x) = \\alpha \\Rightarrow \\tan \\alpha = x$ and $\\tan^{-1} (2x) = \\beta \\Rightarrow \\tan \\beta = 2x$. Then you need the value(s) of x such that $\\alpha + \\beta = \\frac{\\pi}{2}$ $\\Rightarrow \\sin ( \\alpha + \\beta ) = 1 \\Rightarrow \\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha = 1$ $\\Rightarrow \\frac{x}{\\sqrt{x^2 + 1}} \\cdot \\frac{1}{\\sqrt{4x^2 + 1}} + \\frac{2x}{\\sqrt{4x^2 + 1}} \\cdot \\frac{1}{\\sqrt{x^2 + 1}} = 1$ $\\Rightarrow \\frac{3x}{\\sqrt{x^2 + 1} \\sqrt{4x^2 + 1}} = 1$ $\\Rightarrow 3x = \\sqrt{x^2 + 1} \\sqrt{4x^2 + 1} \\Rightarrow 9x^2 = 4x^4 + 5x^2 + 1$ $\\Rightarrow (2x^2 - 1)^2 = 0$. Solve this equation carefully - one of the solutions is extraneous (Google it). I might have time for (b) later (although I'd prefer to see if you now have some ideas on how to do it ....) thanks , i will give it a shot ! 5. ## arctan x + arctan 2x = pi/2 below is the graph of arctan x + arctan 2x = pi/2; (sqrt of 2)/2 is indeed correct 6. Originally Posted by", "Prove that $\\sin(2\\arcsin x + \\arccos x) = \\sqrt{1-x^2}$ I'm trying to prove that $$\\sin(2\\arcsin x + \\arccos x) = \\sqrt{1-x^2}.$$ Let $\\alpha = 2 \\arcsin x$ and $\\beta = \\arccos x$; meaning $\\sin\\alpha = \\frac{x}{2}, \\cos\\beta = x$. We know that: $$\\sin(\\alpha + \\beta) = \\sin\\alpha\\cos\\beta + \\cos\\alpha\\sin\\beta.$$ Finding $\\cos\\alpha$ and $\\sin\\beta$: \\begin{align} \\cos\\alpha & = \\frac{\\sqrt{4-x^2}}{2}, \\\\[0.1in] \\sin\\beta & = \\sqrt{1-x^2}. \\end{align} So plugging everything in: \\begin{align} \\sin(2\\arcsin x + \\arccos x) & = \\frac{x}{2} \\cdot x + \\frac{\\sqrt{4-x^2}}{2} \\cdot \\sqrt{1-x^2}\\\\[0.1in] & = \\frac{x^2 + \\sqrt{4-x^2}\\sqrt{1-x^2}}{2}. \\end{align} But this doesn't seem to lead to the right side. Is my method incorrect? • $\\alpha = 2\\arcsin x$ means that $x = \\sin\\frac{\\alpha}{2}$, not that $\\frac{x}{2} = \\sin\\alpha$. – rogerl Oct 10 '14 at 18:51 $$\\arcsin x+\\arccos x=\\frac\\pi2$$ So, we have $$\\sin\\left(\\frac\\pi2+\\arcsin x\\right)=\\cos(\\arcsin x)$$ • Why does $\\sin(\\frac{\\pi}{2}+\\arcsin x)=\\cos(\\arcsin x)$? – hohner Oct 10 '14 at 18:58 • @hohner, $\\sin(\\pi/2+y)=\\cos y$ as sine ratio is positive in the second quadrant – lab bhattacharjee Oct 10 '14 at 18:59 • Thanks. Is there any other way of proving this? I'm studying $R\\cos(x\\pm\\alpha)$ and inverse trig functions at the moment :) – hohner Oct 10 '14 at 19:06 • Also, why does $\\arcsin x=\\arccos\\sqrt{1-x^2}$? – hohner Oct 10 '14 at 19:10 • @hohner, Have not you noticed the proof? –" ]

[ "$P$ can be written as: ${x}_{P}=cos\\left(\\alpha \\right)$ and ${y}_{P}=sin\\left(\\alpha \\right)$ . Using Pythagoras' theorem you can check that: ${\\left(sin\\left(\\alpha \\right)\\right)}^{2}+{\\left(cos\\left(\\alpha \\right)\\right)}^{2}=1$ To decrease the number of brackets you can use: ${sin}^{2}\\left(\\alpha \\right)+{cos}^{2}\\left(\\alpha \\right)=1$ There are some values that are easy to use: • $cos\\left(0\\right)=1$ • $cos\\left(\\frac{1}{6}\\pi \\right)=\\frac{1}{2}\\sqrt{3}$ • $cos\\left(\\frac{1}{4}\\pi \\right)=\\frac{1}{2}\\sqrt{2}$ • $cos\\left(\\frac{1}{3}\\pi \\right)=\\frac{1}{2}$ • $cos\\left(\\frac{1}{2}\\pi \\right)=0$ and vice versa: • $arccos\\left(0\\right)=\\frac{1}{2}\\pi$ • $arccos\\left(\\frac{1}{2}\\right)=\\frac{1}{3}\\pi$ • $arccos\\left(\\frac{1}{2}\\sqrt{2}\\right)=\\frac{1}{4}\\pi$ • $arccos\\left(\\frac{1}{2}\\sqrt{3}\\right)=\\frac{1}{6}\\pi$ • $arccos\\left(1\\right)=1$ Use these values, when exact answers are required.", "# What does arccos(cos ((-2pi)/3)) equal? Dec 24, 2015 $\\frac{2 \\pi}{3}$ #### Explanation: It would look strange how is that possible! A question usually which pops up isn't $\\arccos \\left(\\cos \\left(A\\right)\\right) = A$. To understand this we can use $\\cos \\left(- \\theta\\right) = \\cos \\left(\\theta\\right)$ Therefore $\\cos \\left(- \\frac{2 \\pi}{3}\\right) = \\cos \\left(\\frac{2 \\pi}{3}\\right)$ Following it up with $\\arccos \\left(\\cos \\left(- \\frac{2 \\pi}{3}\\right)\\right) = \\arccos \\left(\\cos \\left(\\frac{2 \\pi}{3}\\right)\\right)$ That leads us to our answer $\\frac{2 \\pi}{3}$. Let us understand the same in a different manner. The range of $\\arccos \\left(x\\right)$ is $\\left[0 , \\pi\\right]$. $\\cos \\left(\\frac{- 2 \\pi}{3}\\right) = - \\frac{1}{2}$ The angle $\\frac{- 2 \\pi}{3}$ is not in the range of the function. So we select the angle in the range $\\left[0 , \\pi\\right]$ which gives cos(x) = -1/2 that works out to $\\frac{2 \\pi}{3}$.", "edited the post above you by the way. 11. Jun 6, 2017 ### McFluffy My bad. I had edited the post above you by the way. 12. Jun 6, 2017 ### ehild The angle which tangent is given as m, can have multiple values, so it is θ=arctan(m) ± k pi, where k is any integer. Choose k so as 0<alpha<pi. 13. Jun 6, 2017 ### McFluffy Alright! It seems that I've figured it out thanks to you. So the solution should be a piecewise function: $\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right) & \\text{ if } \\arccos{(\\frac{z}{c})}<\\beta<\\pi \\end{cases}$ The reason why it's defined as a piecewise function is because of the original solution which is: $\\alpha(\\beta)=\\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ is defined to be from $(-)$ to $(+)$ during the $0<\\beta<\\frac{\\pi}{2}$ interval. I tried finding at what value of $\\beta$ when $\\alpha(\\beta)$ starts to change from being $(-)$ to $(+)$. I noticed that the term: $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ can be undefined for a certain value of $\\beta$ that makes the denominator equal to 0. Thus, setting ${c\\cos \\beta-z}=0$ to had enabled me to conclude that $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ is undefined when $\\beta=\\arccos{(\\frac{z}{c})}$. What's useful from this is that I", "visualize the angle $$\\theta = \\mbox{arccot}(-2)$$ radians and note that it is a Quadrant II angle with $$\\tan(\\theta) = -\\frac{1}{2}$$. This means it is exactly $$\\pi$$ units away from $$\\beta$$, and we get $$\\theta = \\pi + \\beta = \\pi + \\arctan\\left(-\\frac{1}{2}\\right) \\approx 2.6779$$ radians. Hence, as before, $$\\mbox{arccot}(-2) \\approx 2.6779$$. \\item If the range of arccosecant is taken to be $$\\left[-\\frac{\\pi}{2}, 0\\right) \\cup \\left(0, \\frac{\\pi}{2}\\right]$$, we can use Theorem \\ref{arcsecantcosecantfunctionprops1} to get $$\\mbox{arccsc}\\left(-\\frac{3}{2}\\right) = \\arcsin\\left(-\\frac{2}{3}\\right) \\approx -0.7297$$. If, on the other hand, the range of arccosecant is taken to be $$\\left(0, \\frac{\\pi}{2}\\right] \\cup \\left(\\pi, \\frac{3\\pi}{2}\\right]$$, then we proceed as in the previous problem by letting $$t = \\mbox{arccsc}\\left(-\\frac{3}{2}\\right)$$. Then $$t$$ is a real number with $$\\csc(t) = -\\frac{3}{2}$$. Since $$\\csc(t) < 0$$, we have that $$\\pi < \\theta \\leq \\frac{3\\pi}{2}$$, so $$t$$ corresponds to a Quadrant III angle, $$\\theta$$. As above, we let $$\\alpha$$ be the reference angle for $$\\theta$$. Then $$0 < \\alpha < \\frac{\\pi}{2}$$ and $$\\csc(\\alpha) =\\frac{3}{2}$$, which means $$\\alpha = \\mbox{arccsc}\\left(\\frac{3}{2}\\right)$$ radians. Since the argument of arccosecant is now positive, we may use Theorem \\ref{arcsecantcosecantfunctionprops2} to get $$\\alpha = \\mbox{arccsc}\\left(\\frac{3}{2}\\right) = \\arcsin\\left(\\frac{2}{3}\\right)$$ radians. Since $$\\theta = \\pi + \\alpha = \\pi + \\arcsin\\left(\\frac{2}{3}\\right) \\approx 3.8713$$ radians, $$\\mbox{arccsc}\\left(-\\frac{3}{2}\\right) \\approx 3.8713$$. \\item \\begin{enumerate} \\item Since the domain of $$F(x) = \\arccos(x)$$ is $$-1 \\leq x \\leq", "(by congruence) the distance to the farther position on the right, and the two nearer distances are equal as well for the same reason. #### MarkFL ##### Super Moderator Staff member I would convert angles to radians, and write (using the Law of Sines): $$\\displaystyle \\frac{3}{\\sin\\left(\\dfrac{\\pi}{15}\\right)}=\\frac{5}{\\sin(\\alpha)}$$ And this implies: $$\\displaystyle \\alpha=\\frac{\\pi}{2}\\pm\\arccos\\left(\\frac{5}{3}\\sin\\left(\\frac{\\pi}{15}\\right)\\right)$$ Now, we know, since the sum of the 3 interior angles is $$\\pi$$, that we must have: $$\\displaystyle \\frac{\\pi}{15}+\\alpha+\\beta=\\pi$$ Hence: $$\\displaystyle \\beta=\\frac{14\\pi}{15}-\\left(\\frac{\\pi}{2}\\pm\\arccos\\left(\\frac{5}{3}\\sin\\left(\\frac{\\pi}{15}\\right)\\right)\\right)$$ Or: $$\\displaystyle \\beta=\\frac{13\\pi}{30}\\pm\\arccos\\left(\\frac{5}{3}\\sin\\left(\\frac{\\pi}{15}\\right)\\right)$$ Using the Law of Sines again, we may state: $$\\displaystyle b=\\frac{3\\sin(\\beta)}{\\sin\\left(\\dfrac{\\pi}{15}\\right)}$$ And so the two possible values of $$b$$ in km are: $$\\displaystyle b_1\\approx2.07660968027428$$ $$\\displaystyle b_2\\approx7.70486632706377$$", "can now determine at which interval does $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ starts to change its sign. I managed to construct a sign diagram and conclude that $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)>0$ in $(0, \\arccos{(\\frac{z}{c}}))$ interval and $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)<0$ in $(\\arccos{(\\frac{z}{c}}), \\pi)$ interval. This allowed me to adjust $\\alpha(\\beta)$ better so I can prevent it from being $(-)$. Since $\\arcsin{m}=\\theta\\pm k\\pi$, I can adjust the value of $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ in the $(0, \\arccos{(\\frac{z}{c}}))$ interval such that $\\alpha(\\beta)$ is not defined to be negative and that it manages to merge with the original function to make $\\alpha(\\beta)$ \"continuous\"(You could use calculus to figure out what value is $\\alpha(\\beta)$ when $\\beta=\\arccos{(\\frac{z}{c}})$). Here's the graph of it: Anyways, how is my explanation of the solution? Is this enough to justify that $\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right) & \\text{ if } \\arccos{(\\frac{z}{c})}<\\beta<\\pi \\end{cases}$ is indeed a solution to this problem? 14. Jun 6, 2017 ### McFluffy Oh no. I forgot that $\\alpha(\\beta)$ has to be defined for all values in $(0, \\pi)$. Which means the solution should be this:$\\alpha(\\beta)=\\begin{cases} \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)+\\pi & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\lim_{\\beta\\rightarrow \\arccos{(\\frac{z}{c})}}\\alpha(\\beta) & \\text{ if } \\beta=\\arccos{(\\frac{z}{c})} \\\\ \\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan", "Алгебра # Вычислить arccos(-√2/2)-arcsin(1) Решить уравнения: а) cos x=-√2/2 б) cos(2x-π/4)=√2/2 c)3ctgx+2=0 $1)$ $arccos(- \\frac{ \\sqrt{2} }{2})-arcsin1= \\frac{3 \\pi }{4} - \\frac{ \\pi }{2} = \\frac{3 \\pi }{4} - \\frac{2 \\pi }{4} =\\frac{ \\pi }{4}$ P. S. $arccos(- \\frac{ \\sqrt{2} }{2})= \\pi - arccos \\frac{ \\sqrt{2} }{2}= \\pi - \\frac{ \\pi }{4} = \\frac{3 \\pi }{4}$ $2)$ $a)$ $cosx= - \\frac{ \\sqrt{2} }{2}$ $x=бarccos(- \\frac{ \\sqrt{2} }{2} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б( \\pi -arccos\\frac{ \\sqrt{2} }{2} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б( \\pi - \\frac{ \\pi }{4} )+2 \\pi n,$ $n$ ∈ $Z$ $x=б \\frac{3 \\pi }{4} +2 \\pi n,$ $n$ ∈ $Z$ $b)$ $cos(2x- \\frac{ \\pi }{4} )= \\frac{ \\sqrt{2} }{2}$ $2x- \\frac{ \\pi }{4} =бarccos \\frac{ \\sqrt{2} }{2} +2 \\pi k,$ $k$ ∈ $Z$ $2x- \\frac{ \\pi }{4} =б \\frac{ \\pi }{4} +2 \\pi k,$ $k$ ∈ $Z$ $2x=б \\frac{ \\pi }{4} + \\frac{ \\pi }{4} +2 \\pi k,$ $k$ ∈ $Z$ $x=б \\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi k,$ $k$ ∈ $Z$ $x= \\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi k,$ $k$ ∈ $Z$ или $x=-\\frac{ \\pi }{8} + \\frac{ \\pi }{8} + \\pi n,$ $n$ ∈ $Z$ $x=\\frac{ \\pi }{4} + \\pi k,$ $k$ ∈ $Z$ или $x= \\pi n,$ $n$ ∈ $Z$ $c)$ $3ctgx+2=0$ $3ctgx=-2$", "# Solve algebraically and graphically: $\\arcsin(x) + \\arccos\\left(\\frac{x}{2}\\right) = \\frac{5\\pi}{6}$ Solve: $$\\arcsin(x) + \\arccos\\left(\\frac{x}{2}\\right) = \\frac{5\\pi}{6}$$ I think the algebraic solution should start like: $$\\arcsin(x) = \\frac{5\\pi}{6} - \\arccos\\left(\\frac{x}{2}\\right)$$ $$x = \\sin\\left(\\frac{5\\pi}{6} - \\arccos\\left(\\frac{x}{2}\\right)\\right)$$ at that stage probably I should use some trigonometric relation or property of $\\arccos(x)$ to convert it to $\\arcsin(x)$, however I can't figure it out. As for the solution by graph I can't even think what steps should I follow to build it. $$\\arcsin x + \\arccos\\frac{x}{2} = \\frac{5\\pi}{6}$$ $$\\arcsin x = \\frac{5\\pi}{6} - \\arccos\\frac{x}{2}$$ $$x = \\sin{\\left(\\frac{5\\pi}{6} - \\arccos\\frac{x}{2}\\right)}$$ $$x=\\sin\\frac{5\\pi}6\\cos \\left(\\arccos\\frac{x}{2}\\right)-\\cos\\frac{5\\pi}6 \\sin\\left(\\arccos\\frac{x}{2}\\right)$$ $$x=\\frac12\\cdot\\frac x2+\\frac{\\sqrt3}{2} \\cdot\\sqrt{1-\\left(\\frac x2\\right)^2}$$ Answer: $x=1$ First you have the sine of one thing minus another. $$\\sin(\\alpha-\\beta) = \\sin\\alpha\\cos\\beta - \\cos\\alpha\\sin\\beta.$$ After applying that, you will have $$\\sin\\arccos w = \\sqrt{1-w^2} \\tag 1$$ $$\\cos\\arccos w = w$$ To get the identity $(1)$, just draw a triangle. If $\\cos\\alpha =w$ then you have $\\text{adjacent} = w$ and $\\text{hypotenuse} = 1$, so by the Pythagorean theorem you have $\\text{opposite} = \\sqrt{1-w^2}.$ Then recall that $\\sin=\\dfrac{\\text{opposite}}{\\text{hypotenuse}}.$ HINT: $$\\arccos\\dfrac x2-\\dfrac\\pi3=\\dfrac\\pi2-\\arcsin x=\\arccos x$$ Applying cosine $$\\dfrac x2\\cdot\\cos\\dfrac\\pi3+\\sin\\dfrac\\pi3\\sqrt{1-\\left(\\dfrac x2\\right)^2}=x$$ $$\\implies x=1$$", "θ= pi - arcsin(m). Last edited: Jun 6, 2017 8. Jun 6, 2017 ### McFluffy Why did you define $\\theta=\\frac{a}{b}$? Are you talking about the $\\arctan$ function? Because I think if the $\\frac{a}{b}$ you are talking about are the triangle side ratios then it should be $\\arctan$ because it maps side ratios to the appropriate angle. After taking some time thinking about the terms in the equation, I noticed that $\\arcsin \\left(\\frac{z\\sin {\\beta}}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)$ will always be $(+)$ for $0<\\beta<\\pi$ but for $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ however, it can be $(+)$ or $(-)$ for $0<\\beta<\\pi$ which I'm having problem interpreting it because of the $-z$ term in the denominator of the argument. Because of that $-z$ term, when $0<\\beta<\\frac{\\pi}{2}$, $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ can be $(+)$ or $(-)$ because as $\\beta$ → $\\frac{\\pi}{2}$, $c\\cos \\beta$ becomes smaller and smaller and it'll eventually get overpowered by the $-z$ term. But for $\\frac{\\pi}{2}<\\beta<\\pi$ it's less complicated because it will always of the form of $\\arctan{(\\frac{a}{-b})}$. So my problem is how do I adjust $\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)$ for $0<\\beta<\\pi$ such that it doesn't make $\\alpha(\\beta)$ become negative? Last edited: Jun 6, 2017 9. Jun 6, 2017 ### ehild I did not. Read my previous post. It was m=a/b = tan(θ). 10. Jun 6, 2017 ### McFluffy My bad. I had", "# Find the value of? ## (1) $\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right) - \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right) - \\arccos \\left(\\frac{1}{2}\\right) + \\arcsin \\left(\\frac{\\sqrt{3}}{2}\\right)$ (2) $\\arcsin \\left(- \\frac{1}{2}\\right) + \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right)$ (3) $\\sin \\left(\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right)\\right) = \\sin {150}^{o}$ Dec 15, 2016 (1) ${210}^{o} = \\frac{7 \\pi}{12}$ (2) $- {90}^{o} = - \\frac{\\pi}{2}$ (3) $\\sin \\left(\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right)\\right) = \\frac{1}{2}$ #### Explanation: The trigonometric ratios of standard angles are given in However, before we use this let us remember that range for inverse trigonometric functions are - $\\left[- \\frac{\\pi}{2.} \\frac{\\pi}{2}\\right]$ for arcsin, arccsc, arctan and arccot, while for arccos and arcsec tange is $\\left[0 , p\\right]$. Considering this we solve above as follows: (1) $\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right) - \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right) - \\arccos \\left(\\frac{1}{2}\\right) + \\arcsin \\left(\\frac{\\sqrt{3}}{2}\\right)$ = ${150}^{o} - \\left(- {60}^{o}\\right) - {60}^{o} + {60}^{o}$ = ${150}^{o} + {60}^{o} - {60}^{o} + {60}^{o} = {210}^{o}$ or $\\frac{7 \\pi}{12}$ (2) $\\arcsin \\left(- \\frac{1}{2}\\right) + \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right)$ = $- {30}^{o} - {60}^{o} = - {90}^{o} = - \\frac{\\pi}{2}$ (3) $\\sin \\left(\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right)\\right) = \\sin {150}^{o} = \\frac{1}{2}$ Dec 15, 2016 To solve this type of problem we are to remember the range of inverse trigonometric function as shown in above figure 1) $\\arccos \\left(- \\frac{\\sqrt{3}}{2}\\right) - \\arcsin \\left(- \\frac{\\sqrt{2}}{2}\\right) - \\arccos \\left(\\frac{1}{2}\\right) + \\arcsin \\left(\\frac{\\sqrt{3}}{2}\\right)$ $= \\arccos \\left(- \\cos \\left(\\frac{\\pi}{6}\\right)\\right)", "Re^{i[0,\\pi]}$$ $\\hspace{3.5cm}$ to compute \\begin{align} &PV\\int_{-\\infty}^\\infty\\frac{\\cos(\\alpha x)}{\\beta^2-x^2}\\mathrm{d}x\\\\ &=PV\\int_{-\\infty}^\\infty\\frac{e^{i\\alpha x}}{\\beta^2-x^2}\\mathrm{d}x\\\\ &=\\int_\\gamma\\frac{e^{i\\alpha z}}{\\beta^2-z^2}\\mathrm{d}z+\\pi i\\operatorname*{Res}_{z=-\\beta}\\left(\\frac{e^{i\\alpha z}}{\\beta^2-z^2}\\right)+\\pi i\\operatorname*{Res}_{z=\\beta}\\left(\\frac{e^{i\\alpha z}}{\\beta^2-z^2}\\right)\\\\ &=0+\\pi i\\frac{e^{-i\\alpha\\beta}}{2\\beta}-\\pi i\\frac{e^{i\\alpha\\beta}}{2\\beta}\\\\[6pt] &=\\frac\\pi\\beta\\sin(\\alpha\\beta) \\end{align} where $\\alpha\\ge0$ (needed so that the integral vanishes along the large arc); however, the principal value integral is even in $\\alpha$. Setting $\\alpha=0$ gives $$PV\\int_{-\\infty}^\\infty\\frac1{\\beta^2-x^2}\\mathrm{d}x=0$$ Now we can use that $$\\log\\left(\\cos^2(\\alpha x)\\right)=-2\\log(2)+2\\sum_{k=1}^\\infty\\frac{(-1)^{k-1}}{k}\\cos(2k\\alpha x)$$ to get, for $\\alpha\\ge0$, \\begin{align} \\int_0^\\infty\\frac{\\log\\left(\\cos^2(\\alpha x)\\right)}{\\beta^2-x^2}\\mathrm{d}x &=\\frac12\\int_{-\\infty}^\\infty\\frac{\\log\\left(\\cos^2(\\alpha x)\\right)}{\\beta^2-x^2}\\mathrm{d}x\\\\ &=\\frac\\pi\\beta\\sum_{k=1}^\\infty\\frac{(-1)^{k-1}}{k}\\sin(2k\\alpha\\beta)\\\\ &=\\frac\\pi\\beta\\frac1{2i}\\sum_{k=1}^\\infty\\frac{(-1)^{k-1}}{k}\\left(e^{2ik\\alpha\\beta}-e^{-2ik\\alpha\\beta}\\right)\\\\ &=\\frac\\pi\\beta\\frac1{2i}\\left(\\log\\left(1+e^{2i\\alpha\\beta}\\right)-\\log\\left(1+e^{-2i\\alpha\\beta}\\right)\\right) \\end{align} Note that $$\\theta(\\alpha,\\beta)=\\frac\\pi\\beta\\frac1{2i}\\left(\\log\\left(1+e^{2i\\alpha\\beta}\\right)-\\log\\left(1+e^{-2i\\alpha\\beta}\\right)\\right)$$ is an odd function of $\\alpha$ with a period of $\\frac\\pi{|\\beta|}$ and equals $\\pi\\alpha$ for $|\\alpha|\\lt\\frac\\pi{2|\\beta|}$. Furthermore, in general, $$\\int_0^\\infty\\frac{\\log\\left(\\cos^2(\\alpha x)\\right)}{\\beta^2-x^2}\\mathrm{d}x=\\theta(|\\alpha|,\\beta)$$ and $\\theta(|\\alpha|,\\beta)=\\pi|\\alpha|$ for $|\\alpha|\\le\\frac\\pi{2|\\beta|}$. Thus, the integral is not totally independent of $\\beta$. • Sir Robjohn. Thank you for another method to solving this problem! – Jeff Faraci Jun 1 '14 at 17:20", "# arccos+arccot • October 2nd 2012, 07:56 AM Petrus arccos+arccot Calculate https://webwork.math.su.se/webwork2_...f7e026e1b1.png. The answer may not contain cyklometriska functions. got no ide to do this • October 2nd 2012, 09:14 AM Soroban Re: arccos+arccot Hello, Petrus! Quote: $\\text{Calculate: }\\:\\theta \\;=\\;\\underbrace{\\arccos\\left(\\tfrac{3}{5}\\right)} _{\\alpha} + \\underbrace{\\text{arccot}\\left(\\tfrac{1}{7}\\right) }_{\\beta}$ Let $\\alpha \\,=\\,\\arccos\\left(\\tfrac{3}{5}\\right) \\quad\\Rightarrow\\quad \\cos\\alpha \\,=\\,\\tfrac{3}{5} \\,=\\,\\tfrac{adj}{hyp}$ . . Then: . $opp \\,=\\,4\\quad\\Rightarrow\\quad \\tan\\alpha \\,=\\,\\tfrac{4}{3}$ Let $\\beta \\,=\\,\\text{arccot}\\left(\\tfrac{1}{7}\\right) \\quad\\Rightarrow\\quad \\cot\\beta \\,=\\,\\tfrac{1}{7}$ . . Then: . $\\tan\\beta \\,=\\,\\tfrac{7}{1} \\,=\\,7$ We have: . $\\theta \\;=\\;\\alpha + \\beta$ . . $\\tan\\theta \\;=\\;\\frac{\\tan\\alpha + \\tan\\beta}{1 - (\\tan\\alpha)(\\tan\\beta)} \\;=\\;\\frac{\\frac{4}{3}+7}{1-\\frac{4}{3}(7)} \\;=\\;\\frac{\\frac{25}{3}}{\\text{-}\\frac{25}{3}} \\;=\\;-1$ $\\text{Therefore: }\\:\\theta \\;=\\;\\frac{3\\pi}{4} + \\pi n\\;\\text{ for any integer }n.$ • October 2nd 2012, 01:18 PM Siron Re: arccos+arccot Let $\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)+\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)=x$ Now, let's take the $\\cos$ of both sides. Thus $\\cos\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)+\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]=\\cos(x)$ $\\Leftrightarrow \\cos\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)\\right]\\cos\\left[\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]-\\sin\\left[\\mbox{arc}\\cos\\left(\\frac{3}{5}\\right)\\right]\\sin\\left[\\mbox{arc} \\cot\\left(\\frac{1}{7}\\right)\\right]=\\cos(x)$ If we use the following identities: $\\cos\\left[\\mbox{arc}\\cos(x)\\right]=x$ $\\cos\\left[\\mbox{arc}\\cot(x)\\right]=\\frac{1}{\\sqrt{\\left(\\frac{1}{x}\\right)^2+1}}$ $\\sin\\left[\\mbox{arc}\\cos(x)\\right]=\\sqrt{1-x^2}$ $\\sin\\left[\\mbox{arc}\\cot(x)\\right] =\\frac{1}{\\sqrt{1+x^2}}$ then we can write the equation as $\\left(\\frac{3}{5}\\right)\\frac{1}{\\sqrt{50}}-\\sqrt{1-\\left(\\frac{3}{5}\\right)^2}\\frac{1}{\\sqrt{1+\\left( \\frac{1}{7}\\right)^2}}=\\cos(x)$ $\\Leftrightarrow \\frac{3}{5\\sqrt{50}} - \\frac{4}{5}\\left(\\frac{7}{\\sqrt{50}}\\right)=\\cos(x )$ $\\Leftrightarrow \\frac{3}{25\\sqrt{2}} - \\frac{28}{25\\sqrt{2}}=\\cos(x)$ $\\Leftrightarrow \\cos(x) = \\frac{-\\sqrt{2}}{2}$ $\\Leftrightarrow x = \\frac{3\\pi}{4}+2k\\pi$ • October 2nd 2012, 06:38 PM Nervous Re: arccos+arccot Quote: Originally Posted by Soroban Hello, Petrus! Let $\\alpha \\,=\\,\\arccos\\left(\\tfrac{3}{5}\\right) \\quad\\Rightarrow\\quad \\cos\\alpha \\,=\\,\\tfrac{3}{5} \\,=\\,\\tfrac{adj}{hyp}$ . . Then: . $opp \\,=\\,4\\quad\\Rightarrow\\quad \\tan\\alpha \\,=\\,\\tfrac{4}{3}$ Let $\\beta \\,=\\,\\text{arccot}\\left(\\tfrac{1}{7}\\right) \\quad\\Rightarrow\\quad \\cot\\beta \\,=\\,\\tfrac{1}{7}$ . . Then: . $\\tan\\beta \\,=\\,\\tfrac{7}{1} \\,=\\,7$ We have: . $\\theta \\;=\\;\\alpha + \\beta$ . . $\\tan\\theta", "# How do you evaluate arc cos(-1/2)? Nov 26, 2016 $\\arccos \\left(- \\frac{1}{2}\\right) = \\frac{2 \\pi}{3}$ #### Explanation: $\\arccos x$ is an angle whose cosine ratio is $- \\frac{1}{2}$. Now as $\\cos \\left(\\frac{\\pi}{3}\\right) = \\frac{1}{2}$ and $\\cos \\left(\\pi - A\\right) = - \\cos A$ $\\cos \\left(\\pi - \\frac{\\pi}{3}\\right)$ i.e. $\\cos \\left(\\frac{2 \\pi}{3}\\right) = - c o e \\left(\\frac{\\pi}{3}\\right) = - \\frac{1}{2}$ As $\\cos \\left(\\frac{2 \\pi}{3}\\right) = - \\frac{1}{2}$, we have $\\arccos \\left(- \\frac{1}{2}\\right) = \\frac{2 \\pi}{3}$", "that $$\\sin^{-1}(\\sin \\theta) =\\theta \\iff \\theta\\in \\left[-\\frac{\\pi}2,\\frac{\\pi}2\\right]$$ otherwise we need to adjust the result adding a suitable $$k\\pi$$ term. We can also use $$\\sin^{-1}(\\cos x)=\\sin^{-1}\\left(\\sin \\left(\\frac{\\pi}2+x\\right)\\right)$$ and in that case we obtain different range for $$x$$. Hints: You have to play with these relations/definitions: • $$y=\\arcsin x\\iff \\sin y=x\\;\\text{ AND }\\;-\\frac\\pi2\\le y\\le \\frac\\pi2$$, • $$y=\\arccos x\\iff \\cos y=x\\;\\text{ AND }\\;0\\le y\\le \\pi$$, • $$\\arcsin x+\\arccos x=\\frac\\pi2$$, • $$\\sin(\\arcsin x)=x$$ BUT only $$\\;\\arcsin (\\sin x)\\equiv \\begin{cases}x\\\\[0.5ex]\\frac\\pi 2-x\\end{cases}\\mod 2\\pi$$. Let $$x=\\pm\\frac{\\pi}{2}\\mp(y+2\\pi n),y\\in [-\\frac{\\pi}{2},\\frac{\\pi}{2}]$$. Then: \\begin{align}\\arcsin(\\cos x)&=\\arcsin(\\cos [\\pm\\frac{\\pi}{2}\\mp(y+2\\pi n)])=\\\\ &=\\arcsin(\\sin (y+2\\pi n))=\\\\ &=\\arcsin(\\sin y)=\\\\ &=y.\\end{align} Example 1: If $$x=300^\\circ$$, then $$300^\\circ=-90^\\circ+(30^\\circ+360^\\circ)$$ and: $$\\arcsin(\\cos 300^\\circ)=30^\\circ.$$ Example 2: If $$x=-300^\\circ$$, then $$-300^\\circ=90^\\circ-(30^\\circ+360^\\circ)$$ and: $$\\arcsin(\\cos (-300^\\circ))=30^\\circ.$$ Example 3: If $$x=180^\\circ$$, then $$180^\\circ=90^\\circ-(-90^\\circ+0^\\circ)$$ and: $$\\arcsin(\\cos (180^\\circ))=-90^\\circ.$$", "of Ray, I had some general idea to adjust the functions so it will conform to.the correct solution. Interestingly enough, I found that the other two solution can only be defined as a piecewise function. Here they are: With $\\arcsin$: $\\alpha(\\beta)=\\begin{cases} \\pi -\\arcsin \\left(\\frac{z\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arcsin \\left(\\frac{c\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right) & \\text{ if } 0<\\beta<\\arccos{(\\frac{z}{c})} \\\\ \\lim_{\\beta\\to\\arccos{(\\frac{z}{c})}}\\alpha(\\beta) & \\text{ if } \\beta=\\arccos{(\\frac{z}{c})} \\\\ \\pi -\\arcsin \\left(\\frac{z\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\left(\\pi -\\arcsin \\left(\\frac{c\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)\\right) & \\text{ if } \\arccos{(\\frac{z}{c})}<\\beta<\\pi \\end{cases}$ It's graph: Which is the same as ehild and Ray's solution except that I left one point not included because: The values of these two separate functions are not equal to each other when I set $\\beta=\\arccos{(\\frac{z}{c})}$ but from the graph it seems to approach to some finite value so I'd defined it as the limit of $\\alpha(\\beta)$ as $\\beta \\to \\arccos{(\\frac{z}{c})}$. I think it's approaching $\\arccos{(\\frac{z}{c})}$. The last one would be: $\\alpha(\\beta)= \\begin{cases} \\pi -\\arcsin \\left(\\frac{z\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)-\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right) & \\text{ if } 0<\\beta<\\arccos \\left(\\frac{z}{c}\\right) \\\\ \\lim_{\\beta\\to\\arccos \\left(\\frac{z}{c}\\right)}\\alpha(\\beta) & \\text{ if } \\beta=\\arccos \\left(\\frac{z}{c}\\right) \\\\ -\\left(\\arcsin \\left(\\frac{z\\sin \\beta}{\\sqrt{c^2-2cz\\cos \\beta+z^2}}\\right)+\\arctan \\left(\\frac{c\\sin \\beta}{c\\cos \\beta-z}\\right)\\right) & \\text{ if } \\arccos \\left(\\frac{z}{c}\\right)<\\beta<\\pi \\end{cases}$ The graph being identical as the last one. Overall, I think ehild and Ray's solution is better because $\\alpha(\\beta)$ wasn't defined to be a piecewise function for them. Thank", "}y\\leq \\frac{5}{4}\\pi^2 \\\\ f(y+2\\pi) = f(y)\\end{array} \\right.$ . . . .[*] Now if $y\\leq \\tfrac{5}{4}\\pi^2$ then, $(\\arcsin x)^2 + (\\arccos x)^2 = (\\arcsin x)^2 + (\\tfrac{\\pi}{2} - \\arcsin x)^2$ would give, $\\left( \\frac{\\pi+\\sqrt{8y - \\pi^2}}{4} \\right)^2 + \\left( \\frac{\\pi - \\sqrt{8y - \\pi^2}}{4} \\right)^2 = y$ What happens if $y > \\tfrac{5}{4}\\pi^2$? Therefore the range is $\\left[ \\frac{\\pi^2}{8}, \\frac{5\\pi^2}{4} \\right]$. $(\\arcsin x )^2 + (\\arccos x)^2 = (\\arcsin x + \\arccos x)^2 - 2 \\arcsin x + \\arccos x$ $= \\left( \\frac{\\pi}{2}\\right)^2 - 2 ( \\arcsin x ) ( \\arccos x )$. So all you have to do is get the range of $( \\arcsin x ) ( \\arccos x )$.", "everything back to radians: $$\\alpha = 30^{\\circ} = \\frac{\\pi}{6}$$ radians, $$\\gamma = \\arcsin\\left(\\frac{2}{3}\\right)$$ radians and $$180^{\\circ} = \\pi$$ radians. Hence, $$\\beta = \\pi - \\frac{\\pi}{6} - \\arcsin\\left(\\frac{2}{3}\\right) = \\frac{5\\pi}{6} - \\arcsin\\left(\\frac{2}{3}\\right)$$ radians $$\\approx 108.19^{\\circ}$$.} $$\\beta \\approx 180^{\\circ} - 30^{\\circ} - 41.81^{\\circ} = 108.19^{\\circ}$$. Using the Law of Sines with the angle-side opposite pair $$(\\alpha, a)$$ and $$\\beta$$, we find $$b \\approx \\frac{3 \\sin\\left(108.19^{\\circ}\\right)}{\\sin\\left(30^{\\circ}\\right)} \\approx 5.70$$ units. In the case $$\\gamma = \\pi - \\arcsin\\left(\\frac{2}{3}\\right)$$ radians $$\\approx 138.19^{\\circ}$$, we repeat the exact same steps and find $$\\beta \\approx 11.81^{\\circ}$$ and $$b \\approx 1.23$$ units.\\footnote{An exact answer for $$\\beta$$ in this case is $$\\beta = \\arcsin\\left(\\frac{2}{3}\\right) - \\frac{\\pi}{6}$$ radians $$\\approx 11.81^{\\circ}$$.} Both triangles are drawn below. 1. For this last problem, we repeat the usual Law of Sines routine to find that $$\\frac{\\sin(\\gamma)}{4} = \\frac{\\sin\\left(30^{\\circ}\\right)}{4}$$ so that $$\\sin(\\gamma) = \\frac{1}{2}$$. Since $$\\gamma$$ must inhabit a triangle with $$\\alpha = 30^{\\circ}$$, we must have $$0^{\\circ} < \\gamma < 150^{\\circ}$$. Since the measure of $$\\gamma$$ must be \\textit{strictly} less than $$150^{\\circ}$$, there is just one angle which satisfies both required conditions, namely $$\\gamma = 30^{\\circ}$$. So $$\\beta = 180^{\\circ} - 30^{\\circ} - 30^{\\circ} = 120^{\\circ}$$ and, using the Law of Sines one last time, $$b = \\frac{4\\sin\\left(120^{\\circ}\\right)}{\\sin\\left(30^{\\circ}\\right)} = 4\\sqrt{3} \\approx 6.93$$ units. Some remarks about Example \\ref{losex} are in order. We first", "- \\arcsin \\left(- \\sin \\left(\\frac{\\pi}{4}\\right)\\right) - \\arccos \\left(\\cos \\left(\\frac{\\pi}{3}\\right)\\right) + \\arcsin \\left(\\sin \\left(\\frac{\\pi}{3}\\right)\\right)$ $= \\arccos \\left(\\cos \\left(\\pi - \\frac{\\pi}{6}\\right)\\right) - \\arcsin \\sin \\left(- \\frac{\\pi}{4}\\right) - \\arccos \\left(\\cos \\left(\\frac{\\pi}{3}\\right)\\right) + \\arcsin \\left(\\sin \\left(\\frac{\\pi}{3}\\right)\\right)$ $= \\frac{5 \\pi}{6} + \\frac{\\pi}{4} - \\frac{\\pi}{3} + \\frac{\\pi}{3}$ $= \\frac{10 \\pi + 3 \\pi}{12} = \\frac{13 \\pi}{12}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 2).$\\arcsin \\left(- \\frac{1}{2}\\right) + \\arcsin \\left(- \\frac{\\sqrt{3}}{2}\\right)$ $= \\arcsin \\left(\\sin \\left(- \\frac{\\pi}{6}\\right)\\right) + \\arcsin \\left(\\sin \\left(- \\frac{\\pi}{3}\\right)\\right)$ $= - \\frac{\\pi}{6} - \\frac{\\pi}{3} = - \\frac{\\pi}{2}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 3).sin(arccos(-√3/2)) =sin(arccos(cos(pi-pi/6)) $= \\sin \\left(\\pi - \\frac{\\pi}{6}\\right) = \\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$", "# If $f(x)=\\frac{1}{\\pi}\\left(\\arcsin x+\\arccos x+\\arctan x\\right)+\\frac{x+1}{x^2+2x+10}\\;,$ Then $\\max$ value of $f(x)$ If $\\displaystyle f(x)=\\frac{1}{\\pi}\\left(\\arcsin x+\\arccos x+\\arctan x\\right)+\\frac{x+1}{x^2+2x+10}\\;,$ Then $\\max$ value of $f(x)$ $\\bf{My\\; Try::}$ Here Domain of $\\arcsin x\\;,\\arccos x$ is $\\displaystyle x\\in \\left[-1,1\\right]$ and for $\\arctan x$ is $(-\\infty,\\infty)$ So overall domain of function $f(x)$ is $\\displaystyle x\\in \\left[-1,1\\right]$ Now Using $\\displaystyle \\arcsin x+\\arccos x= \\frac{\\pi}{2}\\;,\\forall x \\in \\left[-1,1\\right]$ So we get $\\displaystyle f(x) =\\frac1\\pi\\left(\\frac{\\pi}{2}+\\arctan x\\right)+\\frac{1}{(x+1)+\\frac{9}{(x+1)}}$ How should I proceed? • Have you tried taking the derivative and setting this equal to zero to find the extremal points in the interior, and then comparing these to the values at the boudary? I mean... that is how I would proceed in any extremal value problem – b00n heT May 31 '16 at 11:28 • Domain of $\\arccos$ is $[-\\pi/2,\\pi/2]$? Do you mean image? – Emre May 31 '16 at 11:41 • As user E.Girgin suggests, the domain of $\\arcsin$ and $\\arccos$ is $\\left[-1,1\\right]$. Since the roots of $x^2+2x+10$ are not real, you need not add existence conditions for $\\frac{x+1}{x^2+2x+10}+\\arctan x$. If I were you, I would not divide numerator and denominator by $x+1$, because it unnecessarily alters the domain and it may hinder your possibility of evaluating the derivative of $f$. – user228113 May 31 '16 at 12:01 • Actually it was all by mistake, – juantheron May", "write $$x=\\cos\\alpha$$ and $$y=\\sin\\alpha$$ for $$\\alpha=\\arctan(y/x)=\\arccos(x)$$. So to get the $$\\theta$$ for our eigenvectors, as an expression with an $$\\arccos$$, it is enough to look at its first component. You also want a final expression containing half the angle, and thus we have $$\\theta/2=\\arccos(b/\\|v_\\pm\\|)$$. 3. We thus reduced the problem to finding a nice expression for $$\\theta = 2\\arccos\\left(\\frac{|b|}{\\|v_\\pm\\|}\\right). %= \\arccos\\left(\\frac{a-d}{\\lambda_+-\\lambda_-}\\right) %= \\arccos\\left(\\frac{a-d}{2S}\\right).$$ Remembering that $$\\cos(2\\arccos(A))=2A^2-1$$, we get $$2\\arccos\\left(\\frac{|b|}{\\|v_\\pm\\|}\\right) = \\arccos\\left(2\\frac{|b|^2}{\\|v_\\pm\\|^2}-1\\right) = \\arccos\\left(\\frac{2|b|^2-\\|v_\\pm\\|^2}{\\|v_\\pm\\|^2}\\right).\\tag A$$ We are pretty much there now. Observe that $$2|b|^2-\\|v_\\pm\\|^2 = -2(\\Delta^2 \\mp \\Delta S) = 2\\Delta(-\\Delta\\pm S),$$ and thus (A) becomes $$\\theta = \\arccos\\left(\\frac{2\\Delta(-\\Delta\\pm S)}{\\pm2S (-\\Delta\\pm S)}\\right) = \\arccos\\left(\\pm\\frac{\\Delta}{S}\\right).$$ But also, $$\\lambda_+-\\lambda_-=2S$$, and thus you get the final expression you were looking for. 4. To be closer to the way the problem was stated: you can think of the above procedure as applied only to the eigenvalue $$\\lambda_+$$, and thus obtaining your expression for $$|m_1\\rangle$$. Once you have that, the expression for $$|m_2\\rangle$$ follows immediately, because given any vector $$(a,b)$$ you can write its orthogonal as $$(\\bar b,-\\bar a)$$. You then just need to remember that you can always multiply by a global phase factor without problems." ]

[ "D \\sim \\triangle EHA$, so that $\\angle HEA \\cong \\angle HA^\\prime D$ (a right angle). To unify the trig, let $k$ be the circumdiameter of $\\triangle ABC$. Thus, $$|BC| = k \\sin A = k\\sin(B+C) \\qquad |CA| = k \\sin B \\qquad |AB| = \\sin C$$ (Throughout, \"$\\angle A$\", \"$\\angle B$\", and \"$\\angle C$\" refer to the angles in the corners of $\\triangle ABC$.) We'll show $\\triangle MA^\\prime A \\sim \\triangle H A^\\prime D$ by showing $$\\frac{|MA^\\prime|}{|AA^\\prime|} = \\frac{|HA^\\prime|}{|DA^\\prime|}$$ Let's take each element of the proportion in turn. Two are easy: \\begin{align} |MA^\\prime| &= |MC| - |A^\\prime C| \\\\ &=\\frac{1}{2}|BC| - |CA|\\cos C \\\\ &=\\frac{k}{2}\\left( \\sin(B+C) - 2 \\sin B \\cos C \\right)\\\\ &=\\frac{k}{2}\\sin(C-B) \\\\[8pt] |AA^\\prime| &= |AB| \\sin B \\\\ &= k \\sin B \\sin C \\end{align} For $|HA^\\prime|$, first note that $\\triangle A^\\prime HB \\sim \\triangle B^\\prime CB$, as these are two right triangles that share an acute angle (at $B$). Therefore, $\\angle A^\\prime H B \\cong \\angle C$, and we have $$|HA^\\prime| = |A^\\prime B| \\cot C = |AB| \\cos B \\cot C = k \\sin C \\cos B \\cot C = k \\cos B \\cos C$$ Segment $|DA^\\prime|$ is the interesting one. First, we write $$|DA^\\prime| = |CA^\\prime| + |DC| = |AC|\\cos C + |DC| = k \\sin B \\cos C + |DC|$$ Then,", "$c=5$ and from the formula $s=c h/2$, $h=4$. Obviously, all third points $C_k$ of triangles with the base $AB$, the height $h$ and the area $s$ will be located on the two parallel lines below and above $AB$, $h$ units apart, we just need to sort out all the points that make any two sides of $\\triangle ABC$ the same. 1) The first obvious choice is $a=b=\\sqrt{89}/2\\approx 4.717$. Two other cases are symmetric: 2) $a=c=5$, $b$ is to find out, 3) $b=c=5$, $a$ is to find out. To deal with cases 2) and 3), let's find possible length $a,b$ using a formula for the area in terms of the sides of triangle: \\begin{align} s&=1/4\\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}. \\end{align} For $a=c$, $b^4-4 c^2 b^2+16 s^2=0$ and we have two positive roots $b_0=2\\sqrt{5}$, $b_1=4\\sqrt{5}$. This results in three different shapes of isosceles triangles and ten points: $C_{00},C_{01},C_{02},C_{11},C_{12}$, $C^{\\prime}_{00},C^{\\prime}_{01},C^{\\prime}_{02},C^{\\prime}_{11},C^{\\prime}_{12}$. Five of them $(C_{00},C_{01},C_{02},C_{11},C^{\\prime}_{02})$ have $y>0$.", "# 2013 AIME II Problems/Problem 15 ## Problem 15 Let $A,B,C$ be angles of an acute triangle with \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ and} \\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which $$\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. ## Solutions ### Solution 1 Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &= 1-\\sin^2A + 1-\\sin^2B + 2\\sin A\\sin B\\cos C \\\\ &= 2-(\\sin^2A + \\sin^2B - 2\\sin A\\sin B\\cos C) \\end{align*} Now we can use the Law of Cosines to simplify this: $$= 2-\\sin^2C$$ Therefore: $$\\sin C = \\sqrt{\\dfrac{1}{8}},\\cos C = \\sqrt{\\dfrac{7}{8}}.$$ Similarly, $$\\sin A = \\sqrt{\\dfrac{4}{9}},\\cos A = \\sqrt{\\dfrac{5}{9}}.$$ Note that the desired value is", "# 2013 AIME II Problems/Problem 15 ## Problem 15 Let $A,B,C$ be angles of an acute triangle with \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ and} \\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which $$\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. ## Solutions ### Solution 1 Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &= 1-\\sin^2A + 1-\\sin^2B + 2\\sin A\\sin B\\cos C \\\\ &= 2-(\\sin^2A + \\sin^2B - 2\\sin A\\sin B\\cos C) \\end{align*} Now we can use the Law of Cosines to simplify this: $$= 2-\\sin^2C$$ Therefore: $$\\sin C = \\sqrt{\\dfrac{1}{8}},\\cos C = \\sqrt{\\dfrac{7}{8}}.$$ Similarly, $$\\sin A = \\sqrt{\\dfrac{4}{9}},\\cos A = \\sqrt{\\dfrac{5}{9}}.$$ Note that the desired value is", "- c^\\prime$ Now, suppose $$f(A^\\prime, B^\\prime, C^\\prime, a^\\prime, b^\\prime, c^\\prime) = 0$$ is any relation between the sides and angles of the triangle $$\\text{A}^\\prime \\text{B}^\\prime \\text{C}^\\prime$$. We may replace $$a^\\prime$$ by $$180^\\circ − A, \\ b^\\prime$$ by $$180^\\circ − B$$, and so on, and this will result in a relation between $$A, \\ B, \\ C, \\ a, \\ b$$ and $$c$$ ; that is, it will result in a relation between the sides and angles of the triangle $$\\text{ABC}$$. For example, the equation $\\cos a^\\prime = \\cos b^\\prime \\cos c^\\prime + \\sin b^\\prime \\sin c^\\prime \\cos A^\\prime \\label{3.5.14} \\tag{3.5.14}$ is valid for the triangle $$\\text{A}^\\prime \\text{B}^\\prime \\text{C}^\\prime$$. By making these substitutions, we find the following formula valid for triangle $$\\text{ABC}$$: $- \\cos A = \\cos B \\cos C - \\sin B \\sin C \\cos a, \\label{3.5.15} \\tag{3.5.15}$ which is the polar cosine formula. The reader will doubtless like to try starting from the sine and cotangent formulas for the triangle $$\\text{A}^\\prime \\text{B}^\\prime \\text{C}^\\prime$$ and deduce corresponding polar formulas for the triangle $$\\text{ABC}$$, though this, unfortunately, may give rise to some anticlimactic disappointment. I know of no particularly interesting derivation of the cotangent formula, and I leave it to the reader to work through the rather pedestrian algebra. Start from $\\cos a = \\cos b \\cos c +", "TrigInTriangle AIME Difficult 2013 Problem - 209 Let $A,B,C$ be angles of an acute triangle with $$\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$$ and $$\\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A = \\frac{14}{9}$$ There are positive integers $p$, $q$, $r$, and $s$ for which $\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "###### back to index | new Define the sequence $a_1, a_2, a_3, \\ldots$ by $a_n = \\sum\\limits_{k=1}^n \\sin{k}$, where $k$ represents radian measure. Find the index of the $100^{th}$ term for which $a_n < 0$. Suppose that the angles of $\\triangle ABC$ satisfy $cos(3A)+cos(3B)+cos(3C)=1.$ Two sides of the triangle have lengths $10$ and $13$. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\\triangle ABC$ is $\\sqrt{m}$. Find $m$. For $\\pi \\le \\theta < 2\\pi$, let\\begin{align*} P &= \\frac12\\cos\\theta - \\frac14\\sin 2\\theta - \\frac18\\cos 3\\theta + \\frac{1}{16}\\sin 4\\theta + \\frac{1}{32} \\cos 5\\theta - \\frac{1}{64} \\sin 6\\theta - \\frac{1}{128} \\cos 7\\theta + \\cdots \\end{align*} and \\begin{align*} Q &= 1 - \\frac12\\sin\\theta -\\frac14\\cos 2\\theta + \\frac18 \\sin 3\\theta + \\frac{1}{16}\\cos 4\\theta - \\frac{1}{32}\\sin 5\\theta - \\frac{1}{64}\\cos 6\\theta +\\frac{1}{128}\\sin 7\\theta + \\cdots \\end{align*} so that $\\frac{P}{Q} = \\frac{2\\sqrt2}{7}$. Then $\\sin\\theta = -\\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Let $A,B,C$ be angles of an acute triangle with $$\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$$ and $$\\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A = \\frac{14}{9}$$ There are positive integers $p$, $q$, $r$, and $s$ for which $\\cos^2 C + \\cos^2 A + 2 \\sin C", "angles are integers.[23] ### Integer triangles with one angle with a given rational cosine Some integer triangles with one angle at vertex A having given rational cosine h/k (h<0 or >0; k>0) are given by[24] $a=p^2-2pqh+q^2k^2,$ $b=p^2-q^2k^2,$ $c=2qk(p-qh),$ where p and q are any coprime positive integers such that p>qk. #### Integer triangles with a 60° angle (angles in arithmetic progression) All integer triangles with a 60° angle have their angles in an arithmetic progression. All such triangles are proportional to:[6] $a=4mn \\,$ $b=3m^2+n^2\\,$ $c=2mn+|3m^2-n^2| \\,$ with coprime integers m, n and 1 ≤ n ≤ m or 3m ≤ n. From here, all primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor. Integer triangles with a 60° angle can also be generated by[25] $a=m^2-mn+n^2 \\,$ $b=2mn - n^2\\,$ $c=m^2-n^2 \\,$ with coprime integers m, n with 0 < n < m (the angle of 60° is opposite to the side of length a). From here, all primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor (e.g. an equilateral triangle solution is obtained by taking m = 2 and n = 1, but this produces a = b = c = 3, which is not a primitive solution). See also [26][27] #### Integer triangles", "## Content description Use similarity to investigate the constancy of the sine, cosine and tangent ratios for a given angle in right-angled triangles (ACMMG223) Source: Australian Curriculum, Assessment and Reporting Authority (ACARA) ### Similarity of right-angled triangles The triangles pictured are similar (SSS), that is the corresponding sides are in the same ratio. Hence the matching angles are the same. For any right-angled triangle similar to triangle ABC, the ratio of the matching sides will be the same. \\begin{align}\\dfrac{BC}{B^\\prime C^\\prime}=\\dfrac{CA}{C^\\prime A^\\prime}=\\dfrac{AB}{A^\\prime B^\\prime}=\\dfrac{2}{1}\\end{align} Notice that we can also write \\begin{align}\\dfrac{BC}{CA}=\\dfrac{B^\\prime C^\\prime}{C^\\prime A^\\prime}=\\dfrac{4}{5}, \\ \\dfrac{BC}{AB}=\\dfrac{B^\\prime C^\\prime}{A^\\prime B^\\prime}=\\dfrac{4}{3} \\ \\text{and} \\ \\dfrac{CA}{AB}=\\dfrac{C^\\prime A^\\prime}{A^\\prime B^\\prime}=\\dfrac{5}{3}\\end{align} That is, once the angles of a triangle are fixed, the ratios of the sides of the triangle are constant. In a right-angled triangle, we only need to know one other angle and then the angle sum of a triangle gives us the third angle. Hence, in a right-angled triangle, if we know one other angle, then the ratios of the sides of the triangle are constant. This is the basis of trigonometry.", "then \\begin{align*} 4(\\cos^2 A \\cos^2 B + &\\sin^2 A \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &= 4\\left(\\frac{1}{4} \\times \\frac{3}{4} + \\frac{3}{4} \\times \\frac{1}{4}\\right) - 2 \\times \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2}\\\\ &= \\frac{3}{2} - \\frac{3}{2} \\\\ &= 0. \\end{align*} So we have found a triangle for which the equation does not hold, and we know that this is an equation that we should try to solve, rather than an identity that we need to prove. We have little choice but to think about the left-hand side. We have \\begin{align*} &4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &\\ = 4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 8 \\sin A \\cos A \\sin B \\cos B. \\end{align*} We might think of reordering the terms in the final product: we could think of it as $\\sin A \\sin B \\cos A \\cos B$, and it all starts to look a bit as though it involves things like $(\\cos A \\cos B - \\sin A \\sin B)^2 = \\cos^2(A + B)$. We have \\begin{align*} &\\cos^2 (A + B) \\\\ &\\qquad = (\\cos A \\cos B - \\sin A \\sin B)^2 \\\\ &\\qquad = \\cos^2 A \\cos^2 B + \\sin^2 A \\sin^2 B - 2\\sin A \\sin B \\cos A \\cos B, \\end{align*} and", "by the Law of Sines, $$\\frac{|DC|}{|CB^\\prime|} = \\frac{\\sin \\angle DB^\\prime C}{\\sin \\angle B^\\prime DC} \\implies |DC| = |BC| \\cos C \\cdot \\frac{\\sin \\angle DB^\\prime C}{\\sin \\angle B^\\prime DC} = k \\sin A \\cos C \\cdot \\frac{\\sin \\angle DB^\\prime C}{\\sin \\angle B^\\prime DC}$$ To get a handle on those final angles, it helps to have a picture. We'll use the fact that $AB^\\prime H C^\\prime$ lie on a circle (with diameter $AH$, which is subtended by right angles at $B^\\prime$ and $C^\\prime$). Note that $\\angle BAA^\\prime$ is the complement of $\\angle B$ in $\\triangle AA^\\prime B$. In the circle, $\\angle BAA^\\prime$ and $C^\\prime B^\\prime H$ subtend the same chord, $C^\\prime H$, and are therefore congruent. Since $\\angle C^\\prime B^\\prime H$ and $\\angle A B^\\prime C^\\prime$ form a right angle at $B^\\prime$, we have that the latter must be congruent to $\\angle B$; further $\\angle DB^\\prime C \\cong \\angle A B^\\prime C^\\prime$ as vertical angles, so that $\\angle DB^\\prime C \\cong \\angle B$. Moreover, $\\angle DCB^\\prime$ is the supplement of $\\angle C$, and we have $\\angle CDB^\\prime = \\angle C - \\angle B$, whence $$|DC| = \\frac{k \\sin(B+C) \\sin B \\cos C}{\\sin(C-B)}$$ and \\begin{align} |DA^\\prime| &= k \\sin B \\cos C + \\frac{k\\sin(B+C)\\sin B\\cos C}{\\sin(C-B)} \\\\ &= \\frac{k \\sin B \\cos C \\; \\left(\\sin(C-B)+\\sin(B+C)\\right)}{\\sin(C-B)} \\\\ &= \\frac{2 k \\cos B", "###### back to index | new Suppose that the angles of $\\triangle ABC$ satisfy $cos(3A)+cos(3B)+cos(3C)=1.$ Two sides of the triangle have lengths $10$ and $13$. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\\triangle ABC$ is $\\sqrt{m}$. Find $m$. Let $A,B,C$ be angles of an acute triangle with $$\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$$ and $$\\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A = \\frac{14}{9}$$ There are positive integers $p$, $q$, $r$, and $s$ for which $\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,5,x$. The sum of the possible values of x equals $a+\\sqrt{b}+\\sqrt{c}$ where $a, b$, and $c$ are positive integers. What is $a+b+c$? In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? An infinite number of equilateral triangles are constructed as shown on the right. Each inner triangle is inscribed in its immediate outsider and is shifted by a constant", "either (or both) of the denominators is zero. But that has nothing specifically to do with trig. – Ilmari Karonen May 14 '15 at 23:08 We have \\begin{align}\\frac{\\cos x}{1-\\sin x}-\\tan x & = \\frac{\\cos x}{1-\\sin x}-\\frac{\\sin x}{\\cos x} \\\\[1.5ex] & =\\frac{\\cos^2 x-\\sin x(1-\\sin x)}{\\cos x(1-\\sin x)} \\\\[1.5ex] & =\\frac{\\cos^2 x+\\sin^2 x-\\sin x}{\\cos x(1-\\sin x)} \\\\[1.5ex] & =\\frac{1-\\sin x}{\\cos x(1-\\sin x)} \\\\[1.5ex] & =\\frac{1}{\\cos x} \\\\[2.8ex] & =\\sec x\\end{align} For fun, I found a \"trigonograph\" of this identity (for acute $\\theta$). In the diagram, $\\overline{AB}$ is tangent to the unit circle at $P$. The \"trig lengths\" (except for $|\\overline{AQ}|$) should be clear. We note that $\\angle BPR \\cong \\angle RPP^\\prime$, since these inscribed angles subtend congruent arcs $\\stackrel{\\frown}{PR}$ and $\\stackrel{\\frown}{RP^\\prime}$. Very little angle chasing gives that $\\triangle APQ$ is isosceles, with $\\overline{AP} \\cong \\overline{AQ}$ (justifying that last trig length). Then, $$\\triangle SPR \\sim \\triangle OQR \\implies \\frac{|\\overline{SP}|}{|\\overline{SR}|} = \\frac{|\\overline{OQ}|}{|\\overline{OR}|} \\implies \\frac{\\cos\\theta}{1-\\sin\\theta} = \\frac{\\sec\\theta+\\tan\\theta}{1}$$ Hint $$\\frac{\\cos(x)}{1-\\sin(x)}-\\tan(x)=\\frac{\\cos(x)}{1-\\sin(x)}-\\frac{\\sin(x)}{\\cos(x)}=\\frac{\\cos^2(x)-\\sin(x)+\\sin^2(x)}{\\cos(x)(1-\\sin(x))}$$ Now what is $\\cos^2(x)+\\sin^2(x)$? • Thank you for the hint. This really help me. – user203103 May 14 '15 at 1:36 $$1 = \\sec^2 x - \\tan^2 x = (\\sec x + \\tan x )(\\sec x - \\tan x)$$ dividing by the second factor on the RHS: $$\\frac1{\\sec x - \\tan x} = \\sec x + \\tan x$$ multiplying LHS numerator and denominator by", "\\parallel FE & (\\text{given})\\\\ \\therefore PQ & \\parallel FE & \\end{array}$ OR \\begin{align*} \\frac{AF}{AP} & = \\frac{1}{6}\\\\ \\frac{AE}{AQ} & = \\frac{1}{6}\\\\ \\frac{AF}{AP} & = \\frac{AE}{AQ}\\\\ \\therefore PQ & \\parallel FE \\quad(\\text{converse prop. theorem}) \\end{align*} If $$PQ$$ is $$\\text{60}$$ $$\\text{cm}$$, calculate the length of $$FE$$. In $$\\triangle AEF$$ and $$\\triangle APQ$$: 1. $$\\hat{A} \\quad (\\text{ is common})$$ 2. $$A\\hat{E}F = A\\hat{Q}P \\quad (\\text{corresp. } \\angle \\text{s, } FE \\parallel PQ)$$ 3. $$A\\hat{F}E = A\\hat{P}Q \\quad (\\text{corresp. } \\angle \\text{s, } FE \\parallel PQ)$$ $$\\therefore \\triangle BHD \\enspace ||| \\enspace \\triangle FED \\quad (\\angle \\angle \\angle)$$ \\begin{align*} \\frac{FE}{PQ} &= \\frac{AF}{AP} \\quad (||| \\enspace \\triangle\\text{s}) \\\\ \\frac{FE}{60} &= \\frac{1}{6} \\\\ FE &= \\text{10}\\text{ cm} \\end{align*} OR In $$\\triangle ADC$$ and $$\\triangle APQ$$: 1. $$\\hat{A} \\quad (\\text{ is common})$$ 2. $$A\\hat{D}C = A\\hat{Q}P \\quad (\\text{corresp. } \\angle \\text{s, } CD \\parallel PQ)$$ 3. $$A\\hat{C}D = A\\hat{P}Q \\quad (\\text{corresp. } \\angle \\text{s, } CD \\parallel PQ)$$ $$\\therefore \\triangle ADC \\enspace ||| \\enspace \\triangle APQ \\quad (\\angle \\angle \\angle)$$ $\\begin{array}{[email protected]{\\;}[email protected]{\\quad}l} \\frac{AC}{AP} &= \\frac{AD}{AQ} = \\frac{1}{3} & (||| \\triangle\\text{s}) \\\\ CD & = \\frac{1}{3}PQ & \\\\ CD & = \\text{20}\\text{ cm} & \\\\ \\text{But } AF & = FC &\\\\ AE & = ED & (\\text{mid-point th.}) \\\\ FE & = \\frac{1}{2}CD & \\\\ FE & = \\text{20}\\text{ cm} & \\end{array}$", "# Problem 6 : Strange Fractions Geometry Level 5 In a non-isoceles triangle $$ABC$$ let $$O$$ and $$I$$ be the circumcenter and the incenter respectively. Let $$D, E, F$$ be the midpoints of the sides $$BC, AC, AB$$ respectively. Let $$T$$ be the foot of perpendicular from $$I$$ to $$AB$$, $$P$$ be the circumcenter of triangle $$DEF$$, and $$Q$$ be the midpoint of $$OI$$. If $$A, P, Q$$ are collinear, the maximum possible value of $$|\\frac{AO}{OD} - \\frac{BC}{AT}|$$ is $$\\frac{a\\sqrt{b}}{c}$$, where $$b$$ is not divisible by the square of any prime and $$a$$ and $$c$$ are relatively prime positive integers. Determine the value of $$a + b + c$$. ×", "\\frac{k-1}{2}}(-1)^{i}\\binom{k}{2i+1}p^{k-2i-1}(q^2-p^2)^i,$ $c = q^{h+k-1} \\frac{\\sin (h+k) \\alpha}{\\sin \\alpha} = \\sum_{0 \\leq i \\leq \\frac{h+k-1}{2}}(-1)^{i}\\binom{h+k}{2i+1}p^{h+k-2i-1}(q^2-p^2)^i,$ where $\\alpha = \\cos^{-1} \\frac{p}{q}$ and p and q are any relatively prime integers such that $\\cos \\frac{\\pi}{h+k} < \\frac{p}{q} < 1$. #### Integer triangles with one angle equal to twice another With angle A opposite side $a$ and angle B opposite side $b$, some triangles with B=2A are generated by[30] $a=n^2, \\,$ $b = mn \\,$ $c=m^2 - n^2, \\,$ with integers m, n such that 0 < n < m < 2n. Note that all triangles with B = 2A (whether integer or not) have[31] $a(a+c)=b^2$. #### Integer triangles with one angle equal to 3/2 times another The equivalence class of similar triangles with $\\ B=\\tfrac{3}{2}A$ are generated by[30] $a=mn^3, \\,$ $b=n^2(m^2-n^2), \\,$ $c=(m^2 - n^2)^2 - m^2 n^2, \\,$ with integers $\\ m, n$ such that $\\ 0<\\varphi n, where $\\ \\varphi$ is the golden ratio $\\varphi = \\frac{1+\\sqrt{5}}{2}\\approx 1.61803$. Note that all triangles with $\\ B=\\tfrac{3}{2}A$ (whether with integer sides or not) satisfy $\\ (b^{2}-a^{2})(b^{2}-a^{2}+bc) = a^{2}c^{2}$. #### Integer triangles with one angle three times another We can generate the full equivalence class of similar triangles that satisfy B=3A by using the formulas [32] $a=n^{3}, \\,$ $b=n(m^{2}-n^{2}), \\,$ $c=m(m^{2}-2n^{2}), \\,$ where $m$ and $n$ are integers such that $\\sqrt{2}n <", "Let $$O$$ be a midpoint of a segment $$AC$$. Rotate $$F$$ for $$90^{\\circ}$$ around $$E$$ to a new point $$H$$. This rotation takes $$O$$ to $$C$$. Then $$EF=EH$$ and $$\\angle FEH = 90^{\\circ}$$ so $$EG$$ is angle bisector for $$\\angle FEH$$ so $$GH = GF = x$$. It is easy to see that $$CH = FO = {x+1\\over 2}$$ Now $$FHCE$$ is cyclic quadrilateral since $$\\angle CFE = \\angle CHE$$ so $$\\angle FCH = \\angle FEH =90^{\\circ}$$ Finaly we use Pythagoras theorem in triangle $$\\triangle CGH$$: $$x^2 = 4^2+\\Big({x+1\\over 2}\\Big)^2 \\implies x=5$$ • Your solution encourages me to think that there may be even more geometric solutions. – blackened Feb 27 '19 at 18:30 With the figure as labeled below, define $$a := |AA^\\prime| \\qquad b := |BB^\\prime| \\qquad c := |A^\\prime B^\\prime| \\qquad d := |A^\\prime M|$$ Note that the perpendicular from midpoint $$M$$ to the diagonal necessarily quadrisects that diagonal; moreover, $$\\overline{MM^\\prime}\\cong\\overline{BM^\\prime}$$. \\begin{align} \\triangle A^\\prime M^\\prime M \\text{ is a right triangle} &\\quad\\to\\quad d^2 = \\left(c+b-\\frac14(a+b+c)\\right)^2+\\left(\\frac14(a+b+c)\\right)^2 \\tag{1}\\\\[4pt] \\triangle A^\\prime B^\\prime M \\sim \\triangle A^\\prime M B \\text{ (Angle-Angle)} &\\quad\\to\\quad \\frac{|A^\\prime M|}{|A^\\prime B^\\prime|} = \\frac{|A^\\prime B|}{|A^\\prime M|} \\quad\\to\\quad d^2 = c(b+c) \\tag{2} \\end{align} Equating $$d^2$$ with $$d^2$$ yields $$3c^2 + 2c(a-b) - a^2 + 2 a b - 5 b^2 = 0 \\tag{3}$$ which suggests that there's", "$CF$: Further, we see that quadrilaterals $BPCD$, $APCE$, and $FAPB$ are all cyclic. As $\\angle BDC = 60^\\circ$, $\\angle BPC = 120^\\circ$, and similarly $\\angle APC = 120^\\circ$ and $\\angle APB = 120^\\circ$. ### Designing an algorithm We have enough information now to start working on an algorithm. Let us come back to the previous diagram: So knowing that the central angles are all $120^\\circ$, we can apply the cosine law ($\\cos 120^\\circ = -\\frac{1}{2}$): $a^2 = q^2 + r^2 - 2qr \\cos 120$ $a^2 = q^2 + r^2 +qr$ A similar formula applies to sides $b$ and $c$. We call $(x,y)$ a pair if $x^2 + xy + y^2$ is a perfect square. We have found a Torricelli triangle if for three integers $p$, $q$, $r$, all of $(p,q)$, $(p,r)$, $(q,r)$ are all pairs. This leaves us with an outline of an algorithm: 1. Generate all pairs $(x,y)$ under the limit (with $x+y < 120000$ and $x^2 + xy + y^2$ being a square) 2. Sort the list of pairs and index them (to be explained in step 3) 3. For each pair $(a,b)$ in the list, search through the list to check if there exists some $c$ where $(a,c)$ is a pair and $(b,c)$ is a pair. We index the list to drastically improve searching time.", "$CF$: Further, we see that quadrilaterals $BPCD$, $APCE$, and $FAPB$ are all cyclic. As $\\angle BDC = 60^\\circ$, $\\angle BPC = 120^\\circ$, and similarly $\\angle APC = 120^\\circ$ and $\\angle APB = 120^\\circ$. ### Designing an algorithm We have enough information now to start working on an algorithm. Let us come back to the previous diagram: So knowing that the central angles are all $120^\\circ$, we can apply the cosine law ($\\cos 120^\\circ = -\\frac{1}{2}$): $a^2 = q^2 + r^2 - 2qr \\cos 120$ $a^2 = q^2 + r^2 +qr$ A similar formula applies to sides $b$ and $c$. We call $(x,y)$ a pair if $x^2 + xy + y^2$ is a perfect square. We have found a Torricelli triangle if for three integers $p$, $q$, $r$, all of $(p,q)$, $(p,r)$, $(q,r)$ are all pairs. This leaves us with an outline of an algorithm: 1. Generate all pairs $(x,y)$ under the limit (with $x+y < 120000$ and $x^2 + xy + y^2$ being a square) 2. Sort the list of pairs and index them (to be explained in step 3) 3. For each pair $(a,b)$ in the list, search through the list to check if there exists some $c$ where $(a,c)$ is a pair and $(b,c)$ is a pair. We index the list to drastically improve searching time.", "angle are given • if three sides are given ## The cosine rule Exercise 6.12 Solve the following triangles (that is, find all unknown sides and angles): $$\\triangle ABC$$ in which $$\\hat{A}=\\text{70}\\text{°}$$; $$b=4$$ and $$c=9$$ \\begin{align*} {a}^{2} &= {b}^{2}+{c}^{2}-2bc \\cos \\hat{A} \\\\ {a}^{2} &= {4}^{2}+{9}^{2}-2(9)(4) \\cos \\hat{\\text{70}\\text{°}} \\\\ &= \\text{72,374} \\ldots \\\\ \\therefore a &= \\text{8,5} \\end{align*} \\begin{align*} {c}^{2} &= {b}^{2}+{a}^{2}-2ba \\cos \\hat{C} \\\\ {9}^{2} &= {4}^{2}+{\\text{8,5}}^{2}-2(4)(\\text{8,5}) \\cos \\hat{C} \\\\ \\therefore \\hat{C} &= \\text{83,9}\\text{°} \\end{align*} \\begin{align*} \\hat{B} &= \\text{180}\\text{°} - \\text{70}\\text{°} -\\text{83,9}\\text{°} \\\\ \\therefore \\hat{B} &= \\text{26,1}\\text{°} \\end{align*} $$\\triangle RST$$ in which $$RS=14$$; $$ST=26$$ and $$RT=16$$ \\begin{align*} {r}^{2} &= {s}^{2}+{t}^{2}-2st \\cos \\hat{R} \\\\ {26}^{2} &= {16}^{2}+{\\text{14}}^{2}-2(16)(\\text{14}) \\cos \\hat{R} \\\\ \\therefore \\hat{R} &= \\text{120}\\text{°} \\end{align*} \\begin{align*} {s}^{2} &= {r}^{2}+{t}^{2}-2rt \\cos \\hat{S} \\\\ {16}^{2} &= {26}^{2}+{\\text{14}}^{2}-2(26)(\\text{14}) \\cos \\hat{S} \\\\ \\therefore \\hat{S} &= \\text{32,2}\\text{°} \\end{align*} \\begin{align*} \\hat{T} &= \\text{180}\\text{°} - \\text{120}\\text{°} -\\text{32,2}\\text{°} \\\\ \\therefore \\hat{T} &= \\text{27,8}\\text{°} \\end{align*} $$\\triangle KLM$$ in which $$KL=5$$; $$LM=10$$ and $$KM=7$$ \\begin{align*} {m}^{2} &= {k}^{2}+{l}^{2}-2kl \\cos \\hat{M} \\\\ {5}^{2} &= {10}^{2}+{\\text{7}}^{2}-2(10)(\\text{7}) \\cos \\hat{M} \\\\ \\therefore \\hat{M} &= \\text{27,7}\\text{°} \\end{align*} \\begin{align*} {l}^{2} &= {k}^{2}+{m}^{2}-2km \\cos \\hat{L} \\\\ {7}^{2} &= {10}^{2}+{\\text{5}}^{2}-2(10)(\\text{5}) \\cos \\hat{L} \\\\ \\therefore \\hat{L} &= \\text{40,5}\\text{°} \\end{align*} \\begin{align*} \\hat{K} &= \\text{180}\\text{°} - \\text{27,7}\\text{°} -\\text{40,5}\\text{°} \\\\ \\therefore \\hat{K} &= \\text{111,8}\\text{°} \\end{align*} $$\\triangle JHK$$ in which $$\\hat{H}=\\text{130}\\text{°}$$; $$JH=13$$ and $$HK=8$$ \\begin{align*} {h}^{2} &= {j}^{2}+{k}^{2}-2jk \\cos \\hat{H} \\\\ {h}^{2}" ]

[ "(\\frac{1 + \\cos b}{\\cos b} \\right ) \\\\ & = \\frac{1}{\\cos b} \\div \\frac{1 + \\cos b}{\\sin b \\cos b} \\\\ & = \\frac{1}{\\cos b} \\cdot \\frac{\\sin b \\cos b}{1 + \\cos b} \\\\ & = \\frac{\\sin b}{1 + \\cos b}\\end{align*} Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula. \\begin{align*}\\sqrt{\\frac{1 - \\cos b}{1 + \\cos b}} & = \\frac{\\sin b}{1 + \\cos b} \\\\ \\frac{1 - \\cos b}{1 + \\cos b} & = \\frac{\\sin^2 b}{(1 + \\cos b)^2} \\\\ (1 - \\cos b)(1 + \\cos b)^2 & = \\sin^2 b (1 + \\cos b) \\\\ (1 - \\cos b)(1 + \\cos b) & = \\sin^2 b \\\\ 1 - \\cos^2 b & = \\sin^2 b\\end{align*} 2. Step 1: change cotangent to cosine over sine, then cross-multiply. \\begin{align*}\\cot \\frac{c}{2} & = \\frac{\\sin c}{1 - \\cos c} \\\\ & = \\frac{\\cos \\frac{c}{2}}{\\sin \\frac{c}{2}} = \\sqrt{\\frac{1 + \\cos c}{1 - \\cos c}} \\\\ \\sqrt{\\frac{1 + \\cos c}{1 - \\cos c}} & = \\frac{\\sin c}{1 - \\cos c} \\\\ \\frac{1 + \\cos c}{1 - \\cos c} & = \\frac{\\sin^2 c}{(1 - \\cos c)^2} \\\\ (1 + \\cos c)(1 - \\cos c)^2 & = \\sin^2 c (1 - \\cos c) \\\\ (1 +", "\\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$. Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\\cos C = \\cos(A + B)$, gives us: $\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos (A+B) = \\frac{15}{8}$ Expanding out gives us $\\cos^2 A + \\cos^2 B + 2 \\sin^2 A \\sin^2 B - 2 \\sin A \\sin B \\cos A \\cos B = \\frac{15}{8}$. Using the double angle formula $\\cos^2 k = \\frac{\\cos (2k) + 1}{2}$, we can substitute for each of the squares $\\cos^2 A$ and $\\cos^2 B$. Next we can use the Pythagorean identity on the $\\sin^2 A$ and $\\sin^2 B$ terms. Lastly we can use the sine double angle to simplify. $\\cos^2 A + \\cos^2 B + 2(1 - \\cos^2 A)(1 - \\cos^2 B) - \\frac{1}{2} \\cdot \\sin 2A \\sin 2B = \\frac{15}{8}$. Expanding and canceling yields, and again using double angle substitution, $1 + 2 \\cdot \\frac{\\cos (2A) + 1}{2} \\cdot \\frac{\\cos (2B) + 1}{2} - \\frac{1}{2} \\cdot \\sin 2A \\sin 2B = \\frac{15}{8}$. Further simplifying yields: $\\frac{3}{2} + \\frac{\\cos 2A \\cos 2B - \\sin 2A \\sin 2B}{2} = \\frac{15}{8}$. Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields:", "have $$B=C < A$$. If the angles summed to $$2\\pi$$, we would have a nice solution $$\\frac A2=\\frac{\\pi}{2}, \\frac B2=\\frac C2 = \\frac{\\pi}{4}$$ - but no, we're not that lucky. Instead, we apply $$\\frac A2 = \\frac{\\pi}{2}-B$$ to eliminate $$A$$: $$1-\\cos B=2\\sin^2 \\frac B2 = \\sin^2 \\frac A2 = \\cos^2 B$$ so $$\\cos^2 B + \\cos B - 1 = 0$$ and $$\\sin \\frac A2 = \\cos B = \\dfrac{\\sqrt{5}-1}{2}$$. Then \\begin{align*}\\cos\\frac A2 &= \\sqrt{1-\\left(\\frac{\\sqrt{5}-1}{2}\\right)^2} = \\sqrt{\\frac{\\sqrt{5}-1}{2}}\\\\ \\sin\\frac B2 &= \\sqrt{\\frac{3-\\sqrt{5}}{4}} = \\frac1{\\sqrt{2}}\\cdot \\frac{\\sqrt{5}-1}{2}\\\\ \\cos\\frac B2 &= \\sqrt{\\frac{1+\\sqrt{5}}{4}} = \\frac1{\\sqrt{2}}\\cdot\\sqrt{\\frac{\\sqrt{5}+1}{2}}\\\\ \\cot\\frac A2 &= \\sqrt{\\frac{2}{\\sqrt{5}-1}} = \\sqrt{\\frac{\\sqrt{5}+1}{2}}\\\\ \\cot\\frac B2 &= \\sqrt{\\frac{\\sqrt{5}+1}{3-\\sqrt{5}}} = \\sqrt{\\sqrt{5}+2}\\\\ p &= 2\\cot\\frac A2 + 2\\cot\\frac B2 = 2\\sqrt{\\frac{\\sqrt{5}+1}{2}}+2\\sqrt{\\sqrt{5}+2}\\approx 6.66\\end{align*} A truly fiendish problem. Numerically, the minimizing angles are about $$76^\\circ$$ at $$A$$ and $$52^\\circ$$ at $$B$$ and $$C$$.", "\\,(1+ \\cos\\;A)\\,(1- \\cos\\;A) ~.$ By the Law of Cosines we know that \\nonumber \\begin{align*} 1 + \\cos\\;A ~&=~ 1 + \\frac{b^2 + c^2 - a^2}{2bc} ~=~ \\frac{2bc + b^2 + c^2 - a^2}{2bc} ~=~ \\frac{(b+c)^2 - a^2}{2bc} ~=~ \\frac{((b+c) + a)\\,((b+c) - a)}{2bc}\\\\ \\nonumber &=~ \\frac{(a + b + c)\\,(b + c - a)}{2bc} ~, \\end{align*} and similarly \\nonumber \\begin{align*} 1 - \\cos\\;A ~&=~ 1 - \\frac{b^2 + c^2 - a^2}{2bc} ~=~ \\frac{2bc - b^2 - c^2 + a^2}{2bc} ~=~ \\frac{a^2 - (b-c)^2}{2bc} ~=~ \\frac{(a - (b-c))\\,(a + (b-c))}{2bc}\\\\ \\nonumber &=~ \\frac{(a - b + c)\\,(a + b - c)}{2bc} ~. \\end{align*} Thus, substituting these expressions into Equation \\ref{2.32}, we have \\nonumber \\begin{align} K^2 ~&=~ \\tfrac{1}{4}\\,b^2 c^2 \\;\\frac{(a + b + c)\\,(b + c - a)}{2bc} \\;\\cdot\\; \\frac{(a - b + c)\\,(a + b - c)}{2bc}\\\\ \\nonumber &=~ \\frac{a + b + c}{2} \\;\\cdot\\; \\frac{b + c - a}{2} \\;\\cdot\\; \\frac{a - b + c}{2} \\;\\cdot\\; \\frac{a + b - c}{2} ~,\\\\ \\end{align} and since we defined $$s = \\frac{1}{2}\\,(a+b+c)$$, we see that $\\nonumber K^2 ~=~ s\\,(s-a)\\,(s-b)\\,(s-c) ~,$ so upon taking square roots we get $\\nonumber K ~=~ \\sqrt{s\\,(s-a)\\,(s-b)\\,(s-c)} ~~.\\quad \\textbf{QED}$ Example 2.15 Find the area of the triangle $$\\triangle\\,ABC$$ given $$a=5$$, $$b=4$$, and $$c = 7$$. Solution: Using Heron's formula with $$s = \\frac{1}{2}\\,(a+b+c) = \\frac{1}{2}\\,(5+4+7) =", "\\\\ \\cos^2B & \\sin^2B & 0 \\\\ \\cos^2C & 0 & \\sin^2C \\\\ \\end{vmatrix}=0.$$ Expanding, we must prove $$\\sin(2A)\\sin^2B\\sin^2C=\\cos^2C\\sin^2B\\sin(2C)+\\sin^2C\\cos^2B\\sin(2B)$$ $$\\frac{\\sin(2A)}{2}=\\sin^2B(1-\\sin^2C)\\sin(2C)+\\sin^2C(1-\\sin^2B)\\sin(2B)$$ \\begin{align*} \\frac{\\sin(2A)+\\sin(2B)+\\sin(2C)}{2}&=\\sin^2B\\sin(2C)+\\sin^2C\\sin(2B)\\\\ &=2\\sin B\\sin C(\\sin B\\cos C+\\cos B\\sin C) \\\\ &=2\\sin B\\sin C\\sin A.\\end{align*} Let $x=e^{iA}, y=e^{iB}, z=e^{iC},$ such that $xyz=-1.$ The left side is equal to $$\\frac{x^2+y^2+z^2-\\frac{1}{x^2}-\\frac{1}{y^2}-\\frac{1}{z^2}}{4i}.$$ The right side is equal to \\begin{align*} 2\\cdot \\frac{x-\\frac{1}{x}}{2i}\\cdot \\frac{y-\\frac{1}{y}}{2i}\\cdot \\frac{z-\\frac{1}{z}}{2i}&=\\frac{xyz-\\frac{1}{xyz}-\\frac{xy}{z}-\\frac{yz}{x}-\\frac{xz}{y}+\\frac{x}{yz}+\\frac{y}{xz}+\\frac{z}{xy}}{-4i}\\\\ &=\\frac{\\frac{1}{x^2}+\\frac{1}{y^2}+\\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\\end{align*} which is equivalent to the left hand side. Therefore, the determinant is $0,$ and $O,P,Q$ are collinear. $\\blacksquare$ ## Solution 3 For convenience, let $a, b, c$ denote the lengths of segments $BC, CA, AB,$ respectively, and let $\\alpha, \\beta, \\gamma$ denote the measures of $\\angle CAB, \\angle ABC, \\angle BCA,$ respectively. Let $R$ denote the circumradius of $\\triangle ABC.$ Since the central angle $\\angle AOB$ subtends the same arc as the inscribed angle $\\angle ACB$ on the circumcircle of $\\triangle ABC,$ we have $\\angle AOB = 2\\gamma.$ Note that $OA = OB,$ so $\\angle OAB = \\angle OBA.$ Thus, $\\angle OAB = \\frac{\\pi}{2} - \\gamma.$ Similarly, one can show that $\\angle OAC = \\frac{\\pi}{2} - \\beta.$ (One could probably cite this as well-known, but I have proved it here just in case.) Clearly, $AO = R.$ Since $AH^2 = 2\\cdot AO^2,$ we have $AH = \\sqrt{2}R.$ Thus, $AH\\cdot AO = \\sqrt{2}R^2.$ Note that $AH = b\\sin\\gamma =", "# Tag Info 14 This is what I'm trying to show on a diagram indicating explicitly all quantities and the approximation is quite rough. Here is a circle with center $O$. Quantities $OA=OB=OC=OD=AB=1$ and $OC\\perp OA, OD\\parallel OA$. Line segments $AC$ and $BD$ have an intersection $E$. We can easily deduce the following quantity. $$AC^2=OA^2+OC^2=2\\qquad ... 8 Riffing on @Shuchang's answer ... Starting with unit circle \\bigcirc O, one easily constructs A, B, C, D with |\\overline{AB}| = \\sqrt{2} and |\\overline{CD}| = \\sqrt{3}. Quadrisecting \\overline{AB} and \\overline{CD} one draws \\bigcirc A and \\bigcirc{B} to provide chords of length \\sqrt{2}/4 and \\sqrt{3}/4. Chains of congruent ... 6 Here is the integral with the Weierstrass substitution:$$ \\begin{align} \\int\\frac{\\mathrm{d}x}{1+\\sin(x)} &=\\int\\frac1{1+\\frac{2z}{1+z^2}}\\frac{2\\,\\mathrm{d}z}{1+z^2}\\\\ &=\\int\\frac{2\\,\\mathrm{d}z}{1+2z+z^2}\\\\ &=\\frac{-2}{1+z}+C\\\\ &=\\frac{-2}{1+\\tan(x/2)}+C \\end{align} $$so your answer is correct. Now consider$$ \\begin{align} ... 5 Note that $\\sqrt 2 \\sin^2 x = \\sqrt 2(1 - \\cos^2 x)$. Make that substitution, \\begin{align} \\sqrt 2 \\sin^2 x + \\cos x = 0 & \\iff \\sqrt 2(1-\\cos^2 x) + \\cos x = 0\\\\ \\\\ &\\iff 1 - \\cos^2 x + \\frac 1{\\sqrt 2}\\cos x = 0\\\\ \\\\ &\\iff \\cos^2 x - \\frac 1{\\sqrt 2}\\cos x - 1 = 0\\end{align} and put $y = \\cos x$...You'll have a quadratic equation in ... 5 There is no need for complicated Cauchy products. We will simply use $\\Big(f^2(x)\\Big)'=2f(x)f'(x)$.", "cosines ($C^2 = A^2 + B^2 - 2AB\\cos \\theta$), we get that $(\\sqrt{33})^2 = 2^2 + 5^2 - 2 \\cdot 2 \\cdot 5 \\cos \\theta$. Simplifying this, we get $33 = 29 - 20 \\cos \\theta$, which means that $\\displaystyle \\cos \\theta = -\\frac{1}{5}$ Because $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we get that $\\displaystyle \\sin^2 \\theta = \\frac{24}{25}$. This means that $\\displaystyle \\sin \\theta = \\frac{2\\sqrt{6}}{5}$ (note that, because $0 \\le \\theta \\le \\pi$, $\\sin \\theta \\ge 0$). The area of the triangle is $\\displaystyle \\frac{1}{2} A B \\sin \\theta = \\frac{1}{2} \\cdot 2 \\cdot 5 \\cdot \\frac{2\\sqrt{6}}{5} = 2\\sqrt{6}$. - Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. – Asimov Apr 28 '14 at 23:48 Use the $\\frac{1}{4}\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula. \\begin{align} & \\frac{1}{4}\\sqrt{4\\cdot4\\cdot25-(4+25-33)^2} \\\\ = & \\frac{1}{4}\\sqrt{4^2\\cdot25-4^2} \\\\ = & \\sqrt{25-1} \\\\ = & 2\\sqrt{6} \\end{align} - This is the way I would have done it, nice. – Sawarnik Apr 29 '14 at 6:40 You could've used Heron's formula straight away, actually. \\begin{align} T & = \\tfrac{1}{4} \\sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\\\ & = \\tfrac{1}{4} \\sqrt{(2+5-\\sqrt{33})(2-5+\\sqrt{33})(-2+5+\\sqrt{33})(2+5+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7-\\sqrt{33})(-3+\\sqrt{33})(3+\\sqrt{33})(7+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7+\\sqrt{33})(7-\\sqrt{33})(\\sqrt{33}+3)(\\sqrt{33}-3)} \\\\ & = \\tfrac{1}{4} \\sqrt{(49-33)(33-9)} \\\\ & = \\tfrac{1}{4} \\sqrt{16", "cos^2 B}$ = $\\sqrt {1 - (\\frac 5{13})^2}$ = $\\sqrt {1 - \\frac {25}{169}}$ = $\\sqrt {\\frac {169 - 25}{169}}$ = $\\sqrt {\\frac {144}{169}}$ = $\\frac {12}{13}$ Now, cos (A + B) = cos A cos B - sin A sin B = $\\frac 45$⋅ $\\frac 5{13}$ - $\\frac 35$⋅ $\\frac {12}{13}$ = $\\frac {20}{65}$ - $\\frac {36}{65}$ = $\\frac {20 - 36}{65}$ = -$\\frac {16}{65}$ Ans Here, sin A = $\\frac 1{\\sqrt {10}}$, sin B = $\\frac 1{\\sqrt 5}$ cos A = $\\sqrt {1 - sin^2 A}$ = $\\sqrt {1 - (\\frac 1{\\sqrt {10}})^2}$ = $\\sqrt {1 - \\frac 1{10}}$ = $\\sqrt {\\frac {10 - 1}{10}}$ = $\\sqrt {\\frac {9}{10}}$ = $\\frac 3{\\sqrt {10}}$ cos B = $\\sqrt {1 - sin^2 B}$ = $\\sqrt {1 - (\\frac 1{\\sqrt 5})^2}$ = $\\sqrt {1 - \\frac 15}$ = $\\sqrt {\\frac {5 - 1}{5}}$ = $\\sqrt {\\frac {4}{5}}$ = $\\frac 2{\\sqrt 5}$ sin (A + B) = sin A cos B + cos A sin B or, sin (A + B) = $\\frac 1{\\sqrt {10}}$× $\\frac 2{\\sqrt 5}$ + $\\frac 3{\\sqrt {10}}$× $\\frac 1{\\sqrt 5}$ or, sin (A + B) = $\\frac 2{\\sqrt {50}}$ + $\\frac 3{\\sqrt {50}}$ or, sin (A + B) = $\\frac {2 + 3}{\\sqrt {50}}$ or, sin (A + B) = $\\frac 5{\\sqrt {50}}$ or, sin", "\\sin^2 B + \\sin^2 C) &=6 ,\\\\ \\sin^2 A + \\sin^2 B + \\sin^2 C &=2 ,\\\\ 4R^2\\sin^2 A + 4R^2\\sin^2 B + 4R^2\\sin^2 C &=2\\cdot4R^2 ,\\\\ a^2+b^2+c^2&=8R^2 ,\\\\ 2\\rho^2-8r\\,R-2r^2&=8R^2 ,\\\\ \\rho^2-(r+2R)^2&=0 ,\\\\ \\frac{\\rho^2-(r+2R)^2}{4R^2} =0 . \\end{align} And since \\begin{align} \\cos A \\cos B \\cos C &=\\frac{\\rho^2-(r+2R)^2}{4R^2} , \\end{align} one of $\\cos A,\\cos B,\\cos C$ must be 0, hence, one of the angles must be $\\tfrac\\pi2$. COMMENT.- First note that if $C=90^{\\circ}$ then you have $\\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\\circ}$ you can consider that $\\sin C=\\sin (180^{\\circ}-(A+B))=\\sin(A+B)$ and $\\cos C= \\cos (180^{\\circ}-(A+B))=-\\cos(A+B)$ so you get $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B)=2(\\cos^2A+\\cos ^2 B+\\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish. EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B) = 2\\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\\dfrac{\\pi}{2}$ (the \"if\" is obvious and we need the \"only if\"). So put $A + B = \\dfrac{\\pi}{2}\\pm h$ with $h\\ne 0$ and look at the functions $$\\sin^2(x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h-x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h),\\quad0\\lt x\\lt\\dfrac{\\pi}{2}$$ which becomes for $h$ positive and $h$", "\\sin(r))\\frac{d}{dr}[4^r - 2\\cos(r)] \\\\ =\\mathstrut \\amp [15r^2 + \\cos(r)](4^r - 2\\cos(r))+(5r^3 + \\sin(r))[4^r \\ln(4) + 2\\sin(r)]\\text{.} \\end{align*} 2. We use the quotient rule on $p\\text{,}$ followed by the product rule to differentiate the denominator, finding that \\begin{align*} p'(t) =\\mathstrut \\amp \\frac{\\frac{d}{dt}[\\cos(t)]t^6 \\cdot 6^t - \\cos(t) \\frac{d}{dt}[t^6 \\cdot 6^t]}{(t^6 \\cdot 6^t)^2}\\\\ =\\mathstrut \\amp \\frac{[-\\sin(t)]t^6 \\cdot 6^t - \\cos(t) [6t^5 \\cdot 6^t + t^6\\cdot 6^t \\ln(6)]}{(t^6 \\cdot 6^t)^2}\\text{.} \\end{align*} 3. Using the sum and constant multiple rules, it follows first that \\begin{equation*} g'(z) = 3 \\frac{d}{dz}[z^7 e^z] - 2\\frac{d}{dz}[z^2 \\sin(z)] + \\frac{d}{dz}\\left[ \\frac{z}{z^2 + 1} \\right]\\text{.} \\end{equation*} Applying the product rule in the first two terms and the quotient rule in the third, we find that \\begin{equation*} g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\\sin(z)+z^2 \\cos(z)] + \\frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\\text{.} \\end{equation*} 4. The particle's instantaneous velocity at the moment $t = 1$ is given by $s'(1)\\text{.}$ We use the quotient rule to find $s'(t)\\text{,}$ and then simplify by removing a common factor of $e^t$ to get \\begin{align*} s'(t) =\\mathstrut \\amp \\frac{(-3\\sin(t) - \\cos(t))e^t-(3\\cos(t) - \\sin(t))e^t}{(e^t)^2}\\\\ =\\mathstrut \\amp \\frac{-2\\sin(t)-4\\cos(t)}{e^t}\\text{.} \\end{align*} Thus, $s'(1) = \\frac{-2\\sin(1)-4\\cos(1)}{e^1} \\approx -1.414\\text{,}$ which is the particle's instantaneous velocity in feet per second at the moment $t = 1\\text{.}$ 5. Since $p(x) = f(x) \\cdot g(x)\\text{,}$ the product rule tells us \\begin{equation*} p'(x) = f'(x)g(x)+f(x)g'(x)\\text{,} \\end{equation*} and", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "# Thread: A few HW questions I dont get 1. ## A few HW questions I dont get Using the conditions below sinA=3/4, A lies in quadrant 2 CosB=12/13, B lies in quadrant 1 Find the following: Cos (A+B) Tan(A-B) Cos B/2 Solve the equation in th interval [0,2pie) 2cos^2x-3cosx+1 Use Half angle forumlas to find the exact value of the expression cos 22.5 cos^2 75 - sin^2 75 2. Originally Posted by slaying_ice Using the conditions below sinA=3/4, A lies in quadrant 2 CosB=12/13, B lies in quadrant 1 Find the following: Cos (A+B) Tan(A-B) Cos B/2 First: $cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$ Second: $tan(A - B) = \\frac{sin(A - B)}{cos(A - B)}$ $tan(A - B) = \\frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$ Now: $sin(A) = \\frac{3}{4}$ and is in the 2nd quadrant. Thus: $cos(A) = -\\sqrt{1 - sin^2(A)} = -\\frac{1}{2}$ Again: $cos(B) = \\frac{12}{13}$ and is in the first quadrant. Thus: $sin(B) = +\\sqrt{1 - cos^2(B)} = \\frac{5}{13}$ To the problems: $cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$ $cos(A + B) = -\\frac{1}{2} \\cdot \\frac{12}{13} - \\frac{3}{4} \\cdot \\frac{5}{13} = -\\frac{3}{4}$ $tan(A - B) = \\frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$ $tan(A - B) = \\frac{\\frac{3}{4} \\cdot \\frac{12}{13} - \\frac{5}{13} \\cdot -\\frac{1}{2}}{-\\frac{1}{2} \\cdot \\frac{12}{13} + \\frac{3}{4} \\cdot \\frac{5}{13}} = - \\frac{46}{9}$ So", "the given equation \\begin{align}\\sqrt 2 |OI|\\; = \\;|AB| - |AC|\\end{align} gives \\begin{align}|O\\,I{|^2} = \\frac{{{{(c - b)}^2}}}{2} = 2{R^2}{(\\sin C - \\sin B)^2}.\\end{align} Therefore \\begin{align}2{(\\sin C - \\sin B)^2} = 1 - 2(\\sin A + \\sin C - \\sin B)\\left( {\\sqrt 2 - 1} \\right)\\end{align} or \\begin{align}1 - 2{\\left( {\\sin C - \\frac{{\\sqrt 2 }}{2}} \\right)^2} = 2\\left( {\\sin A + \\sin C - \\frac{{\\sqrt 2 }}{2}} \\right)\\left( {\\sqrt 2 - 1} \\right) (*)\\end{align}. The addition and subtraction formulas give \\begin{align} \\sin C & =\\sin (180{}^\\text{o}-B-A)=\\sin (135{}^\\text{o}-A) \\\\ & =\\sin 135{}^\\text{o}\\cos A-\\cos 135{}^\\text{o}\\sin A=\\frac{\\sqrt{2}(\\sin A+\\cos A)}{2} \\\\ \\end{align} and so \\begin{align}\\sin C-\\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{2}(\\sin A+\\cos A-1).\\end{align} Plugging the last equation into equation (*) yields \\begin{align}1-{{(\\sin A+\\cos A-1)}^{2}}=2\\left( \\sqrt{2}-1 \\right)\\left[ \\sin A+\\frac{\\sqrt{2}}{2}(\\sin A+\\cos A-1) \\right]\\end{align}, Expanding both sides of the last equation gives \\begin{align} &1-{{(\\sin A+\\cos A)}^{2}}+2(\\sin A+\\cos A)-1 \\\\ & =\\left( \\sqrt{2}-1 \\right)\\left( 2+\\sqrt{2} \\right)\\sin A+\\left( 2-\\sqrt{2} \\right)(\\cos A-1) \\\\ \\end{align} or \\begin{align}{{\\sin }^{2}}A+{{\\cos }^{2}}A+2\\sin A\\cos A=\\left( 2-\\sqrt{2} \\right)\\sin A+\\sqrt{2}\\cos A+\\left( 2-\\sqrt{2} \\right).\\end{align} Consequently, we have \\begin{align}2\\sin A\\cos A-\\left( 2-\\sqrt{2} \\right)\\sin A-\\sqrt{2}\\cos A+\\left( \\sqrt{2}-1 \\right)=0\\ ;\\end{align} that is, \\begin{align}\\left( \\sqrt{2}\\sin A-1 \\right)\\left( \\sqrt{2}\\cos A-\\sqrt{2}+1 \\right)=0.\\end{align} This implies that \\begin{align}\\sin A=\\frac{\\sqrt{2}}{2}\\ \\text{or}\\ \\cos A=1-\\frac{\\sqrt{2}}{2}\\end{align}. Therefore, the answer to the problem is \\begin{align}\\sin A=\\frac{\\sqrt{2}}{2}\\ \\text{or}\\ \\sin A=\\sqrt{1-{{\\cos }^{2}}A}=\\frac{\\sqrt{4\\sqrt{2}-2}}{2}.\\end{align} Example - 20 A circle \\begin{align}\\omega \\end{align} is inscribed in a quadrilateral $$ABCD.$$ Let I", "– nyxlotus Feb 4 at 16:32 I got misled a bit in my initial reasoning. So, back to it. If we have Equation $$(1)$$ here below \\begin{align} H\\left(e^{j0.5\\pi}\\right) &= \\frac{1 - e^{-j0.5\\pi}}{1 - 0.25e^{-j\\pi}}\\\\ &= \\frac{1 - e^{-j\\frac{\\pi}{2}}}{1 - \\frac 14e^{-j\\pi}}\\tag{1}\\\\ \\end{align} And we know that $$e^{j\\theta} = \\cos(\\theta) + j\\sin(\\theta)$$. With this we have that \\begin{align} e^{-j\\frac{\\pi}{2}} &= -j\\\\ e^{-j\\pi} &= -1\\\\ e^{j\\frac{\\pi}{4}} &= \\frac{\\sqrt2}{2} + j\\frac{\\sqrt2}{2}\\\\ \\end{align} Equation $$(1)$$ then becomes \\begin{align} H\\left(e^{j\\frac{\\pi}{2}}\\right) &= \\frac 45\\cdot \\left(1 + j\\right)\\\\ &=\\frac 45\\cdot \\left(\\frac{\\sqrt{2}\\sqrt{2}}{2} + j\\frac{\\sqrt{2}\\sqrt{2}}{2}\\right)\\\\ &= \\underbrace{\\frac 45\\cdot \\sqrt{2}}_{\\approx 1.13}\\underbrace{\\left(\\frac{\\sqrt{2}}{2} + j\\frac{\\sqrt{2}}{2}\\right)}_{e^{j\\frac{\\pi}{4}}} \\equiv 1.13\\cdot e^{j\\frac{\\pi}{4}}\\\\ \\end{align} • $e^{-j\\pi}=1$${}$? – Matt L. Feb 4 at 12:25 • Well spotted, my bad, let me fix that. – Gilles Feb 4 at 12:26 • But then it turns out that the original solution is correct, i.e., the magnitude is indeed $1.13$. – Matt L. Feb 4 at 12:31 • Back online, thanks for the sharp eye on it. – Gilles Feb 4 at 15:01 • Hey, thanks Gilles. Your and ZR Han's answer is helping me understand this a bit more. My class is challenging and I'm mainly using the textbook to learn the material. – nyxlotus Feb 4 at 16:34 By plotting the unit circle you will find that $$1-e^{-j\\pi/2} = \\sqrt{2}e^{j\\pi/4}$$. • I'm sorry I'm still having some", "30^\\circ+r\\sec 30^\\circ}{2r-r\\tan 30^\\circ}$$ So far as the conjecture goes, we can stop here, since there's clearly no chance of introducing$\\sqrt 5, and thus no appearance of the golden ratio. For completeness, though, we can evaluate the ratio and get ... $$\\frac{4}{11}\\;\\left(\\;1 + 2 \\sqrt{3}\\;\\right) = ... 1 We know that ab = cd by the power of the point P, and also that a + b = c + d because the lengths of the chords AB and CD are equal. So$$ b = \\frac{cd}{a} \\\\ \\implies a + \\frac{cd}{a} = c + d \\\\ \\implies a^2 + cd = ac + ad \\\\ \\implies a^2 - ac + cd - ad = 0 \\\\ \\implies (a-c)(a-d) = 0 \\\\ \\implies a = c\\quad \\text{OR}\\quad a = d $$We obtain that ... 1 So applying cosine rule I got$$\\small PI^2=4R^2+16R^2\\sin^2\\frac{B}{2}\\sin^2\\frac{C}{2} -16R^2\\cos A\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\Bigg(\\cos\\frac{B}{2}\\cos\\frac{C}{2}+\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\Bigg)$$I think that you have a typo (the red part) :$$\\begin{align}&\\small PI^2=4R^2\\color{red}{\\cos^2A}+16R^2\\sin^2\\frac{B}{2}\\sin^2\\frac{C}{2} ... 1 Suppose by contradiction there exists a linea$passing through a single point$A$. By axiom 3 there exist two other points$B$and$C$and by axiom 1 they belong to a line$r$parallel to$a$. By axiom 1 there exists a line$s$passing through$A$and$B$and by axiom 2 there exists a line$t$passing through$C$and parallel to$s$. Line$t$does not ... 1 That's from Kedlaya I believe? First of all, let's mention a useful theorem:", "at 13:13 I'll assume that $a$ is the side opposite to $\\alpha$ and similarly for $b$ and $c$, opposite to $\\beta$ and $\\gamma$. The cosine law tells you that $$a^2=b^2+c^2-2bc\\cos\\alpha$$ so $$2bc+2bc\\cos\\alpha=b^2+2bc+c^2-a^2=(b+c)^2-a^2=(a+b+c)(b+c-a)$$ which means $$\\frac{1}{15}bc=b+c-a$$ Similarly, $$2ac(1+\\cos\\beta)=(a+b+c)(a+c-b)$$ that is $$\\frac{8}{45}ac=a+c-b$$ Summing the two relations we get $$\\frac{c}{45}(8a+3b)=2c$$ that implies $8a+3b=90$ (because we assume $c\\ne0$ in a triangle). The sine law tells you that $$\\frac{a}{\\sin\\alpha}=\\frac{b}{\\sin\\beta}$$ Since $$\\sin\\alpha=\\sqrt{1-\\frac{1}{9}}=\\frac{2}{3}\\sqrt{2}$$ and $$\\sin\\beta=\\sqrt{1-\\frac{49}{81}}=\\frac{4}{9}\\sqrt{2}$$ we have $$\\frac{3a}{2\\sqrt{2}}=\\frac{9b}{4\\sqrt{2}}$$ that simplifies to $2a=3b$. Now we have \\begin{cases} 8a+3b=90\\\\[3px] 2a=3b\\\\[3px] a+b+c=20 \\end{cases} Ok. I'll take $\\alpha$ corresponds to vertice $A$ and $\\beta$ to vertice $B$. Also, i´ll denonte by $a,b,c$ the oposite sides to vertices $A,B,C$ respectively. By Cosine's Law, and using hypothesis about $P$, we have the system: $$\\left\\{ \\begin{eqnarray} 20&=&a+b+c&\\\\ a^2&=&b^2+c^2+2bc\\frac{1}{3}\\\\ b^2&=&c^2+a^2-2ac\\frac{7}{9}\\\\ \\end{eqnarray}\\right.$$ Solving, we have three tripletes of solutions: $(a,b,c)=(9,6,5)$ or $(a,b,c)=(10,10,0)$ or $(a,b,c)=(15,-10,15)$. But the last two tripletes can't be solution due to the fact that 0, in the second triplete, and -10, in the third triplete, can't be the measure of the sides of a triangle. Then, the solution is $(a,b,c)=(9,6,5)$", "$b$ and equal length $l$ is $\\frac{b\\sqrt{4l^2-b^2}}{4}$. Plugging in $b = 2\\sqrt2$ and $l = 2\\sqrt5$, $$\\frac{2\\sqrt2 \\cdot \\sqrt{80-8}}{4} = \\frac{\\sqrt{576}}{4} = \\frac{24}{4} = \\boxed{\\textbf{(C) }6}$$ ## Solution 10 (Trig) Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$, $B (4, 6)$, and $C (6, 4)$. By the Pythagorean Theorem, $AB = AC = 2\\sqrt5$ and $BC = 2\\sqrt2$. By the Law of Cosines, $$\\cos A = \\frac{(2\\sqrt5)^2 + (2\\sqrt5)^2 - (2\\sqrt2)^2}{2 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5} = \\frac{20 + 20 - 8}{40} = \\frac{32}{40} = \\frac45$$ Therefore, $\\sin A = \\sqrt{1 - \\cos^2 A} = \\frac35$. By the extended Law of Sines, $$2R = \\frac{a}{\\sin A} = \\frac{2\\sqrt2}{\\frac35} = \\frac{10\\sqrt2}{3}$$ $$R = \\frac{5\\sqrt2}{3}$$ Then the area is $\\frac{abc}{4R} = \\frac{2\\sqrt2 \\cdot 2\\sqrt5^2}{4 \\cdot \\frac{5\\sqrt2}{3}} = \\boxed{\\textbf{(C) }6}$. ## Solution 11 The area of a triangle formed by three lines, $$a_1x + a_2y + a_3 = 0$$ $$b_1x + b_2y + b_3 = 0$$ $$c_1x + c_2y + c_3 = 0$$ is the absolute value of $$\\frac12 \\cdot \\frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \\cdot \\begin{vmatrix} a_1&a_2&a_3\\\\ b_1&b_2&b_3\\\\ c_1&c_2&c_3\\\\ \\end{vmatrix}^2$$ Plugging in the three lines, $$-x + 2y - 2 = 0$$ $$-2x + y + 2 = 0$$ $$x + y - 10 = 0$$ the area is the absolute value of $$\\frac12 \\cdot \\frac{1}{(-2-1)(-1-2)(-1+4)} \\cdot", "# Prove $\\sin(A+B)\\sin(A-B)=\\sin^2A-\\sin^2B$ How would I verify the following double angle identity. $$\\sin(A+B)\\sin(A-B)=\\sin^2A-\\sin^2B$$ So far I have done this. $$(\\sin A\\cos B+\\cos A\\sin B)(\\sin A\\cos B-\\cos A\\sin B)$$But I am not sure how to proceed. - use this formula $$2 \\sin(A+B)\\sin(A-B)=\\cos2B-\\cos2A$$ it will like this $$\\dfrac12 \\cdot (\\cos2B-\\cos2A)$$ $$\\dfrac {(1-2\\sin^2B)-(1-2\\sin^2A)}{2}$$ - \\begin{eqnarray} \\sin(A+B)\\sin(A-B) &=& (\\sin A\\cos B+\\cos A\\sin B)(\\sin A\\cos B-\\cos A\\sin B)\\\\\\\\ &=& \\sin^2 A \\cos^2 B -\\sin^2 B \\cos^2 A\\\\\\\\ &=& \\sin^2 A \\cos^2 B -\\sin^2 B (1-\\sin^2 A)\\\\\\\\ &=& \\sin^2 A (\\cos^2 B + \\sin^2 B) - \\sin^2 B\\\\\\\\ &=& \\sin^2 A - \\sin^2 B \\end{eqnarray} - It makes sense.Thank you. – El Bananero Jul 25 '12 at 17:47 \\begin{align*}\\sin (A+B)\\cdot\\sin (A-B)&=\\frac{\\cos(2B)-\\cos (2A)}{2}\\\\&=\\frac{(1-2\\sin^2B)-(1-2\\sin^2A)}{2}\\\\&=\\sin^2A-\\sin^2B\\end{align*} Here i have used $$\\sin x\\cdot\\sin y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$ - Your question involves the basic algebra identity which says, $(a + b)(a - b) = a^2 - b^2$. For targeting your question, it is easy to assume $a = \\sin A\\cos B$ and $b = \\cos A \\sin B$. The process becomes easy now. \\begin{align}(a + b)(a - b)& =& a^2 - b^2\\\\ & = & (\\sin A \\cos B)^2 - (\\cos A \\sin B)^2\\\\ & = & \\sin^2A\\cos^2 B - \\cos^2A\\sin^2B \\\\ & = & \\sin^2A(1 - \\sin^2 B) - \\cos^2 A\\sin^2B \\end{align}Proceed. - Open the bracket of what you got and substitute", "### Problem asked by Nitesh Kumar Singh If $A$, $B$ and $C$ are the angles of a triangle, such that $\\sin^2 A + \\sin^2 B + \\sin^2 C$ is a constant, then show that $\\frac{dA}{dB} = \\frac{\\tan B - \\tan C}{\\tan C - \\tan A}$. Proof: $pi = A + B + C$, so $C = pi - (A + B)$, and $\\frac{dC}{dB} = -(1 + \\frac{dA}{dB})$ We have $\\sin^2 A + \\sin^2 B + \\sin^2 C = k$. Differentiating with respect to $B$ we get $\\sin 2A \\frac{dA}{dB} + \\sin 2B + \\sin 2C \\frac{dC}{dB} = 0$ or, $\\sin 2A \\frac{dA}{dB} + \\sin 2B - \\sin 2C (1 + \\frac{dA}{dB}) = 0$ or, $\\frac{dA}{dB} = \\frac{\\sin 2C - \\sin 2B}{\\sin 2A - \\sin 2C}$ $= \\frac{\\cos(B + C) \\cdot \\sin (B - C)}{\\cos (C + A) \\cdot \\sin (C - A)}$ $= \\frac{\\cos A}{\\cos B} \\cdot \\frac{\\sin (B - C)}{\\sin (C - A)}$ $\\frac{\\tan B - \\tan C}{\\tan C - \\tan A}$ $= \\frac{\\sin B \\cos C - \\sin C \\cos B}{\\sin C \\cos A - \\sin A \\cos C} \\cdot \\frac{\\cos C \\cos A}{\\cos B \\cos C}$ $= \\frac{\\sin (B - C)}{\\sin (C - A)} \\cdot \\frac{\\cos A}{\\cos B}$ thus $\\frac{dA}{dB} = \\frac{\\tan B - \\tan C}{\\tan C - \\tan A}$", "\\triangle ABC & \\\\ &= \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} + \\frac{1}{2} AB \\cdot BC \\sin \\hat{B} & \\\\ &= \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} + \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} & \\\\ &= DC \\cdot DA \\sin \\hat{D} & \\end{array}$ Write down two expressions for $$CA^{2}$$: one in terms of $$\\cos\\hat{D}$$ and one in terms of $$\\cos\\hat{B}$$. \\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \\cos \\hat{D}\\\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \\cos \\hat{B} \\end{align*} Show that $$2C{A}^{2}=C{D}^{2}+D{A}^{2}+A{B}^{2}+B{C}^{2}$$. \\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \\cos \\hat{D}\\\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \\cos \\hat{B}\\\\ &= AB^{2} + BC^{2} - 2(DC)(DA) \\cos \\hat{B} \\quad (DC \\cdot DA = AB \\cdot BC) \\\\ 2CA^{2}& = CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA) \\cos \\hat{D} \\\\ & - 2(DC)(DA) \\cos ( \\text{180} ° -\\hat{D})\\\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA)\\cos \\hat{D} \\\\ & + 2(DC)(DA) \\cos \\hat{D}\\\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} \\end{align*} Suppose that $$BC=10$$ units, $$CD=15$$ units, $$AD=4$$ units and $$AB=6$$ units. Calculate $$C{A}^{2}$$ (correct to one decimal place). \\begin{align*} 2CA^{2} &= (15)^{2} + (4)^{2} + (6)^{2} + (10)^{2} \\\\ &= 377\\\\ \\therefore CA^{2} &= \\text{188,5}\\text{ units$^{2}$} \\end{align*} Find the angle $$\\hat{B}$$. Hence, calculate the area of $$ABCD$$ (correct to one" ]

[ "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "then \\begin{align*} 4(\\cos^2 A \\cos^2 B + &\\sin^2 A \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &= 4\\left(\\frac{1}{4} \\times \\frac{3}{4} + \\frac{3}{4} \\times \\frac{1}{4}\\right) - 2 \\times \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2}\\\\ &= \\frac{3}{2} - \\frac{3}{2} \\\\ &= 0. \\end{align*} So we have found a triangle for which the equation does not hold, and we know that this is an equation that we should try to solve, rather than an identity that we need to prove. We have little choice but to think about the left-hand side. We have \\begin{align*} &4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &\\ = 4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 8 \\sin A \\cos A \\sin B \\cos B. \\end{align*} We might think of reordering the terms in the final product: we could think of it as $\\sin A \\sin B \\cos A \\cos B$, and it all starts to look a bit as though it involves things like $(\\cos A \\cos B - \\sin A \\sin B)^2 = \\cos^2(A + B)$. We have \\begin{align*} &\\cos^2 (A + B) \\\\ &\\qquad = (\\cos A \\cos B - \\sin A \\sin B)^2 \\\\ &\\qquad = \\cos^2 A \\cos^2 B + \\sin^2 A \\sin^2 B - 2\\sin A \\sin B \\cos A \\cos B, \\end{align*} and", "# 2013 AIME II Problems/Problem 15 ## Problem 15 Let $A,B,C$ be angles of an acute triangle with \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ and} \\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which $$\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. ## Solutions ### Solution 1 Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &= 1-\\sin^2A + 1-\\sin^2B + 2\\sin A\\sin B\\cos C \\\\ &= 2-(\\sin^2A + \\sin^2B - 2\\sin A\\sin B\\cos C) \\end{align*} Now we can use the Law of Cosines to simplify this: $$= 2-\\sin^2C$$ Therefore: $$\\sin C = \\sqrt{\\dfrac{1}{8}},\\cos C = \\sqrt{\\dfrac{7}{8}}.$$ Similarly, $$\\sin A = \\sqrt{\\dfrac{4}{9}},\\cos A = \\sqrt{\\dfrac{5}{9}}.$$ Note that the desired value is", "# 2013 AIME II Problems/Problem 15 ## Problem 15 Let $A,B,C$ be angles of an acute triangle with \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ and} \\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which $$\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. ## Solutions ### Solution 1 Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &= 1-\\sin^2A + 1-\\sin^2B + 2\\sin A\\sin B\\cos C \\\\ &= 2-(\\sin^2A + \\sin^2B - 2\\sin A\\sin B\\cos C) \\end{align*} Now we can use the Law of Cosines to simplify this: $$= 2-\\sin^2C$$ Therefore: $$\\sin C = \\sqrt{\\dfrac{1}{8}},\\cos C = \\sqrt{\\dfrac{7}{8}}.$$ Similarly, $$\\sin A = \\sqrt{\\dfrac{4}{9}},\\cos A = \\sqrt{\\dfrac{5}{9}}.$$ Note that the desired value is", "C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which $$\\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B = \\frac{p-q\\sqrt{r}}{s},$$ where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$. 2013 AIME II (Problems • Answer Key • Resources) Preceded by2013 AIME I Problems Followed by2014 AIME I Problems 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "$PM$ and $PN$ perpendicular to $OM$ and $ON$ respectively. From $N$ we draw $NQ$ and $NR$ perpendicular to $OQ$ and $MP$ respectively. $\\angle RPN = 90^{\\circ} - \\angle PNR = \\angle QON = A$ Thus, we can write $\\sin(A - B) = \\sin MOP = \\frac{MP}{OP} = \\frac{MR - PR}{OP} = \\frac{QN}{ON}\\frac{ON}{OP} - \\frac{PR}{PN}\\frac{PN}{OP}$ Thus, $\\sin(A - B) = \\sin A\\cos B - \\cos A\\sin B$ Also, $\\cos(A - B) = \\frac{OM}{OP} = \\frac{OQ + QM}{OP} = \\frac{OQ}{ON}\\frac{ON}{OP} + \\frac{RN}{NP}\\frac{NP}{OP}$ $= \\cos A\\cos B + \\sin A\\sin B$ We have shown that subtraction formula is true when angles involved are acute angles. The same proof can be applied to prove the results for all values of $A$ and $B$. From the results obtained we find upon division that $\\tan(A - B) = \\frac{\\tan A - \\tan B}{1 + \\tan A\\tan B}$ ## 8.3. Important Deductions¶ 1. $\\sin(A + B)\\sin(A - B) = \\sin^2A - \\sin^2B = \\cos^2B - \\cos^2A$ Proof: L.H.S. $= (\\sin A\\cos B + \\sin B\\cos A)(\\sin A\\cos B - \\sin B\\cos A)$ $= \\sin^2A\\cos^2B - \\sin^2B\\cos^2A = \\sin^2A(1 - \\sin^2B) - \\sin^2B(1 - \\sin^2A)$ $= \\sin^2A - \\sin^2A\\sin^2B - \\sin^2B + \\sin^2B\\sin^2A$ $= =\\sin^2A - \\sin^2B = (1 - \\cos^2A) - (1 - \\cos^2B)$ $= \\cos^2B - \\cos^2A$ 2. $\\cos(A + B)\\cos(A - B) =", "the left-hand side. We now want to consider $1 + \\cos B$, in the hope that it may turn out to be $\\cos^2(\\frac{B}{2})$. Now $\\cos B = \\cos\\left(2 \\times \\frac{B}{2}\\right) = 2\\cos^2\\left(\\frac{B}{2}\\right) - 1,$ and similarly for $A$, and so \\begin{align*} &\\sin(A + B) + \\sin(B + C) + \\sin(C + A) \\\\ & \\qquad = \\sin A \\left(2 \\cos^2 \\left(\\frac{B}{2}\\right)\\right) + \\sin B \\left(2 \\cos^2 \\left(\\frac{A}{2}\\right)\\right), \\end{align*} and so we have proved the identity. 1. $\\sin(A + B) = \\cos C$. We might notice that $A + B = \\pi - C$, so this statement starts to look like a more familiar statement. Let’s dive in and try to prove it (and if it turns out that we’re wrong then that will become clear because our proof won’t work). Since $A$, $B$ and $C$ are the angles of a triangle, we see that \\begin{align*} \\sin(A + B) &= \\sin(\\pi - C) \\\\ &= \\sin \\pi \\cos C - \\cos \\pi \\sin C \\\\ &= \\sin C. \\end{align*} Oh dear, that’s not quite what we wanted. But this effort isn’t wasted, we can just try to solve the equation using this extra information. So the triangle satisfies the equation if and only if $\\sin C = \\cos C$. We cannot have $\\cos C = 0$ (because we cannot", "$\\sin A \\sin (B-C) = \\sin^2 B - \\sin^2 C$, therefore, $2 \\sin A \\sin B \\cos C = \\sin^2 A + \\sin^2 B - \\sin^2 C$. - Hmm which two would I cancel out. – Fernando Martinez Jul 31 '12 at 15:45 No, I meant to cancel out the $2$ in $2 \\sin A \\sin (B-C) = 2 ( \\sin^2 B - \\sin^2 C)$, giving us $\\sin A \\sin (B-C) = \\sin^2 B - \\sin^2 C$ – Rijul Saini Jul 31 '12 at 16:02 I see my now so technically would I have now sin^2B-sin^2C=cos(a-b+c)-cos(a+b-c)=cos2c-cos2b=2sin^2B-2sin^2C – Fernando Martinez Jul 31 '12 at 16:17 What would I do next I am not sure what to do..... – Fernando Martinez Jul 31 '12 at 16:20 I edited my post. – Rijul Saini Jul 31 '12 at 17:10 One can deduce this identity from the conjunction of the law of sines and the law of cosines. The law of sines says that for the three angles $A$, $B$, $C$ of a triangle, with opposite sides $a$, $b$, $c$, we have $$\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = d.$$ The last equality merely defines $d$, and one can omit it and still have a statement of the law of sines. The common value $d$ is actually the diameter of", "so that $$\\cos{A} = \\frac{b^2 + c^2 – a^2}{2bc}.$$ The double-angle formulas are $$\\sin(2x) = 2\\sin(x)\\cos(x)\\\\ \\cos(2x) = \\cos^2(x) – \\sin^2(x) = 2\\cos^2(x) – 1 = 1 – \\sin^2(x).$$ (He uses a period for multiplication.) We use the formula area = (1/2)bc.sin(A) And sin(A) = 2 sin(A/2).cos(A/2) so area = bc.sin(A/2).cos(A/2) Next we find expressions for cos(A/2) and sin(A/2) in terms of s, a, b, c, where a, b, c are the sides of the triangle and s is the semi-perimeter. So s = (a+b+c)/2 b^2 + c^2 - a^2 cos(A) = 2cos^2(A/2) - 1 = --------------- 2bc b^2 + c^2 - a^2 2cos^2(A/2) = 1 + --------------- 2bc (b+c)^2 - a^2 (b+c-a)(b+c+a) = ------------- = -------------- 2bc 2bc (2s-2a)2s = --------- 2bc s(s-a) cos^2(A/2) = ------ bc Next cos(A) = 1 - 2 sin^2(A/2) b^2 + c^2 - a^2 = --------------- 2bc b^2 + c^2 - a^2 2sin^2(A/2) = 1 - ---------------- 2bc a^2 - (b-c)^2 (a-b+c)(a+b-c) = ------------- = -------------- 2bc 2bc (2s-2b)(2s-2c) = -------------- 2bc (s-b)(s-c) sin^2(A/2) = ---------- bc We can now return to the formula for the area of the triangle: (1/2)bc.sin(A) = bc.sin(A/2).cos(A/2) bc.sqrt[s(s-a)(s-b(s-c)] = ------------------------ bc = sqrt[s(s-a)(s-b)(s-c)] It isn’t obvious how you would know to do this (that’s true of many proofs in their final form), but it’s beautiful how", "$3-2k = \\cos 2A + \\cos 2B + \\cos (2A+2B)$ i.e $3-2k = 2\\cos(A+B)\\cos(A-B) + 2\\cos^2(A+B) -1$ i.e $4-2k = 4\\cos(A+B)\\cos A\\cos B$ So if $k > 2$, then $\\cos(A+B) \\lt 0$ hence acute. $k = 2$, then $\\cos(A+B) = 0$ hence right triangled. $k < 2$, then $\\cos(A+B) \\gt 0$, hence obtuse. In fact, we can go further and show that the maximum possible value of $k$ is $k = \\frac{9}{4}$ which corresponds to $\\triangle{ABC}$ being equilateral, as follows: $4\\sin^2 A + 4\\sin^2 B + 4\\sin^2 C = 9 + \\delta$ i.e. $(2 - 2\\cos2A) + (2-2\\cos 2B) + 4(1- \\cos^2 (A+B)) = 9 + \\delta$ i.e. $1 + 2\\cos2A + 2\\cos 2B + 4\\cos^2(A+B) = -\\delta$ i.e. $1 + 4\\cos(A+B)\\cos(A-B) + 4\\cos^2(A+B) = -\\delta$ i.e. $\\sin^2(A-B) + \\cos^2(A-B) + 4\\cos(A+B)\\cos(A-B) + 4\\cos^2(A+B) = -\\delta$ i.e. $\\sin^2(A-B) + (\\cos (A-B) + 2\\cos(A+B))^2 = -\\delta$. Hence $\\delta \\le 0$ and so $\\sin^2 A + \\sin^2 B + \\sin^2 C \\le \\frac{9}{4}$ The case $\\delta = 0$ gives us $\\sin(A-B) = 0$ and $\\cos(A+B) = \\frac{-1}{2}$. Hence $A=B=C$. Thus the max value of $\\sin^2 A + \\sin^2 B + \\sin^2 C$ is $\\frac{9}{4}$ and is achieved when $A=B=C$. - $$\\sin^2A+\\sin^2B+\\sin^2C$$ $$=1-(\\cos^2A-\\sin^2B)+1-\\cos^2C$$ $$=2-\\cos(A+B)\\cos(A-B)-\\cos C\\cdot\\cos C$$ $$=2-\\cos(\\pi-C)\\cos(A-B)-\\cos\\{\\pi-(A+B)\\}\\cdot\\cos C$$ $$=2+\\cos C\\cos(A-B)+\\cos(A+B)\\cdot\\cos C\\text{ as }\\cos(\\pi-x)=-\\cos x$$ $$=2+\\cos C\\{\\cos(A-B)+\\cos(A+B)\\}$$ $$=2+2\\cos A\\cos B\\cos C$$ $(1)$", "finally, a complete description of primitive pythagorean triples is well known. You can deduce your formula from them rather easily. Explicitly, suppose that $(a,b,c)$ is a primitive Pythagorean triple, and let us assume that $a$ is even; we have $$(c+b)(c-b) = c^2-b^2 = a^2;$$ since $a,b,c$ are pairwise coprime, and $a$ is even, then $b$ and $c$ are both odd, so $c+b$ and $c-b$ are both even. Any common divisor of $c+b$ and $c-b$ will divide $(c-b)+(c+b) = 2c$, and also $(c+b)-(c-b) = 2b$; since $\\gcd(2c,2b) = 2\\gcd(b,c) = 2$, the gcd of $c-b$ and $c+b$ is $2$. Dividing through by $4$ (which we can do since $a$ is even, so $a^2$ is a multiple of $4$, we get $$\\left(\\frac{c+b}{2}\\right)\\left(\\frac{c-b}{2}\\right) = \\left(\\frac{a}{2}\\right)^2.$$ Since $\\frac{c+b}{2}$ and $\\frac{c-b}{2}$ are relatively prime, and their product is a square, they are each a square. So we can write $\\frac{c+b}{2} = s^2$, $\\frac{c-b}{2}=r^2$, $\\frac{a}{2} = rs$, with $r$ and $s$ positive, $s\\gt r$, relatively prime, and of opposite parity (since $c$ is odd). This gives \\begin{align*} c &= \\frac{c+b}{2} + \\frac{c-b}{2} = s^2+r^2.\\\\ b &= \\frac{c+b}{2} - \\frac{c-b}{2} = s^2 - r^2.\\\\ a &= 2rs. \\end{align*} In your case, you want $a=2^m$; since $s$ and $r$ are of opposite parity, one must be equal to $1$; since $s\\gt r\\gt 0$, we will have", "in the given inequality – Dr. Sonnhard Graubner May 25 at 5:55 A purely \"trig-bashing\" algebraic way: We have \\begin{align*} &\\sin\\left(B-\\frac\\pi4\\right)+\\sin\\left(C-\\frac\\pi4\\right)\\\\ &=2\\sin\\left(\\frac{B+C}2-\\frac\\pi4\\right)\\cos\\frac{B-C}2\\\\ &=2\\sin\\left(\\frac\\pi4-\\frac{A}2\\right)\\cos\\frac{B-C}2. \\end{align*} So we want to prove $$2\\cos\\frac{B-C}2-1>0$$ when $$A<\\pi/4$$, $$B,C$$ acute. Note that we have $$\\lvert B-C\\rvert$$ is at most $$A$$ (only in the degenerate case when $$B$$ or $$C$$ is $$\\pi/2$$), so $$2\\cos\\frac{B-C}2-1>2\\cos\\frac{A}2-1>2\\cos\\frac\\pi8-1>0$$ as desired. Note that A,B,C are are positive acute angles, so both LHS and RHS are positive. Square both LHS and RHS, Then do LHS - RHS we get. $$\\cos2A + \\cos2B + \\cos2C + 2(\\cos(A+B)+\\cos(A+C)+\\cos(B+C))$$ note that $$2(\\cos(A+B)+\\cos(A+C)+\\cos(B+C))$$ is always -ve for acute triangle (all these angles fall in second quadrant). Check $$\\cos2A + \\cos2B + \\cos2C$$, If all are greater than 45 deg. then this becomes negative. Suppose $$A= 45 -\\theta$$ (in deg.), writing $$\\cos2(45 -\\theta) + \\cos2(45 +\\theta + a) + \\cos2(45 +\\theta + b)$$ where $$0\\leq \\theta < 45$$ , $$0< a + \\theta < 45$$, $$0< b + \\theta < 45$$ This part becomes $$\\sin2\\theta - \\sin(2a+2\\theta) - \\sin(2b+2\\theta)$$, all these angles fall in first quadrant, so this is also negative. So $$LHS < RHS$$", "but I wouldn't have the term in $x$ less than the coefficient. – Sebastiano Jun 25 '20 at 22:53 $$I=\\int\\frac{2\\cos x-\\sin x}{3\\sin x+5\\cos x }\\,dx$$ $$2\\cos x-\\sin x=\\sqrt5 \\cos(x+a)$$, where $$\\tan a=1/2$$ and $$3\\sin x+5\\cos x=\\sqrt{34}\\cos(x-b)$$, where $$\\tan b=3/5$$ $$I=\\int \\frac{\\sqrt 5 \\cos(x+a) } {\\sqrt{34}\\cos(x-b)} \\, dx$$ Substitute $$t=x-b$$ so that $$I=\\sqrt{5/34}\\int \\frac{\\cos(t+a+b) } {\\cos t}\\, dt$$ The integrand is now: $$\\cos(a+b) - \\tan t\\sin(a+b)$$ Can you take it from here? • My $\\infty$ thank you. I think that this way is more complicated. – Sebastiano Jun 27 '20 at 9:01 • @Sebastiano: You are welcome. This way isn't complicated. It only involves calculating $\\cos (a+b)$. Given $\\tan a, \\tan b$, we can find $\\sin, \\cos$ and after finding $\\sin (a+b)$, we can find $\\cos (a+b)$. Why do you think that it's complicated? – Koro Jun 27 '20 at 9:05 • @Sebastiano: What I would do is imagine a rt. angled triangle (for angle $a$), then its perp. is $1$,base is $2$ and hence hypot. is $\\sqrt 5$. Hence we can find $cos a$, $sin a$ etc. Similarly for angle $b$. You don't have to do $\\sin(arctan (1/2)+arc tan (3/5))$ – Koro Jun 27 '20 at 9:10 • Why there are roots square $\\cos \\theta=\\pm 1/\\sqrt{1+\\tan^2\\theta}$, $\\sin \\theta=\\pm \\tan \\theta/\\sqrt{1+\\tan^2\\theta}$ and the choice of angle depends on the", "3 Why are you not comfortable about the structure of your first proof? With the little claim about the case when $\\sin b=0$ mentioned in the comments under OP, it's OK. What you did is that you managed to reduce the problem to showing that the equality $$\\frac{\\cos^2a}{\\cos^2b}=\\frac 12$$ is true. Then you used the given hypothesis to prove this (more ... 0 If a triangle has the same proportions as one that fits a unit circle, it is called similar. Under these conditions, if you take the ratios of any $2$ sides, you will come up with the same $number(s)$ which mean the same angle(s) as though it fit the unit circle. 1 Starting from this line continue as follows (instead of dividing by $\\cos ^2b$, express it as $1-\\sin^2b$): $$(\\sin^2a - \\cos^2a) \\cos^2b - 2\\sin^2b\\cos^2a = 0 \\\\ (\\sin^2a - \\cos^2a) (1-\\sin^2b) - 2\\sin^2b\\cos^2a = 0\\\\ -\\cos 2a-\\sin^2a\\sin^2b+\\sin^2b\\cos^2a-2\\sin^2b\\cos^2a=0 \\\\ -\\cos2a-\\sin^2a\\sin^2b-\\sin^2b\\cos^2a=0\\\\ \\cos2a+\\sin^2b=0.$$ 2 $\\cos x$ is strictly decreasing and positive for $x\\in [0,1].$ So $\\cos a +\\cos b$ cannot be a minimum if $a<c-b$ because then $\\cos a +\\cos b >\\cos (c-b)+\\cos b.$ For $a=c-b$ let $M=(a+b)/2=c/2$ and $D=(a-b)/2.$ We have $$\\cos a=\\cos (M+D)=\\cos M \\cos D-\\sin M\\sin D$$ and we have $$\\cos b=\\cos (M-D)=\\cos M \\cos D+\\sin M \\sin D.$$ Adding, we have (... 0 $\\cos(a)", "done it 8 years ago: We can WLOG assume that $A$, $B$, $C$ are the angles of a triangle (this means that $A$, $B$, $C$ lie in the interval $\\left[0,\\pi\\right]$, additionally to satisfying $A+B+C=\\pi$), because everything is analytic (or by casebash). We denote the sides of this triangle by $a$, $b$, $c$ (so we forget about the old $a$, $b$, $c$), its semiperimeter $\\dfrac{a+b+c}{2}$ by $s$, its area by $\\Delta$ and its circumradius by $R$. Then, $\\sin A=\\dfrac{a}{2R}$ (by the Extended Law of Sines) and similarly $\\sin B=\\dfrac{b}{2R}$ and $\\sin C=\\dfrac{c}{2R}$, so that $\\sin A+\\sin B+\\sin C=\\dfrac{a}{2R}+\\dfrac{b}{2R}+\\dfrac{c}{2R}=\\frac{a+b+c}{2R}=\\frac{s}{R}$. On the other hand, one of the half-angle formulas shows that $\\cos\\dfrac{A}2=\\sqrt{\\dfrac{s\\left(s-a\\right)}{bc}}$, and similar formulas hold for $\\cos\\dfrac{B}2$ and $\\cos\\dfrac{C}2$, so that $4\\cdot\\cos\\dfrac{A}2\\cdot\\cos\\dfrac{B}2\\cdot\\cos\\dfrac{C}{2}$ $=4\\cdot\\sqrt{\\dfrac{s\\left(s-a\\right)}{bc}}\\cdot\\sqrt{\\dfrac{s\\left(s-b\\right)}{ca}}\\cdot\\sqrt{\\dfrac{s\\left(s-c\\right)}{ab}}$ $=\\frac{4s}{abc}\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}$. Now, $\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}=\\Delta$ (by Heron's formula) and $\\Delta=\\frac{abc}{4R}$ (by another formula for the area of the triangle), so tis becomes $4\\cdot\\cos\\dfrac{A}2\\cdot\\cos\\dfrac{B}2\\cdot\\cos\\dfrac{C}{2}=\\dfrac{4s}{abc}\\cdot\\dfrac{abc}{4R}=\\dfrac{s}{R}$. This is exactly what we got for the left hand side, qed.", "# Mock AIME 1 Pre 2005 Problems/Problem 15 ## Problem Triangle $ABC$ has an inradius of $5$ and a circumradius of $16$. If $2\\cos{B} = \\cos{A} + \\cos{C}$, then the area of triangle $ABC$ can be expressed as $\\frac{a\\sqrt{b}}{c}$, where $a, b,$ and $c$ are positive integers such that $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Compute $a+b+c$. ## Solution Using the identity $\\cos A + \\cos B + \\cos C = 1+\\frac{r}{R}$, we have that $\\cos A + \\cos B + \\cos C = \\frac{21}{16}$. From here, combining this with $2\\cos B = \\cos A + \\cos C$, we have that $\\cos B = \\frac{7}{16}$ and $\\sin B = \\frac{3\\sqrt{23}}{16}$. Since $\\sin B = \\frac{b}{2R}$, we have that $b = 6\\sqrt{23}$. By the Law of Cosines, we have that: $$b^2 = a^2 + c^2-2ac\\cdot \\cos B \\implies a^2+c^2-\\frac{7ac}{8} = 36 \\cdot 23.$$But one more thing: noting that $\\cos A = \\frac{b^2+c^2-a^2}{2cb}$. and $\\cos C = \\frac{a^2+b^2-c^2}{2ab}$, we know that $\\frac{36 \\cdot 23 + b^2+c^2-a^2}{bc} + \\frac{36 \\cdot 23+a^2+b^2-c^2}{ab} = \\frac{7}{4} \\implies$ $\\frac{36 \\cdot 23 + c^2-a^2}{c} + \\frac{36 \\cdot 23 + a^2-c^2}{a} = \\frac{21\\sqrt{23}}{2} \\implies$ $\\frac{(a+c)(36 \\cdot 23 + 2ac-c^2-a^2)}{ac} = \\frac{21\\sqrt{23}}{2}$. Combining this with the fact that $a^2+c^2 - \\frac{7ac}{8} = 36 \\cdot 23$, we have", ". $\\cos A \\:=\\:\\cos2B$ . . Therefore: . $A \\,=\\,2B$ Very nice. In my example I confused between $a, A\\,\\,and\\,\\,b, B$ , since clearly $A=\\frac{\\pi}{2}=2\\frac{\\pi}{4}=2B$...*sigh* Of course, this would have hardly happend had the OP written $\\angle A=2\\angle B$ ... Tonio 6. Originally Posted by razemsoft21 ABC is a triangle with sides a,b and c. show that : if $a^2-b^2=bc$ then A=2B I tried to use the cosine law, but no result. Another way.... $a^2-b^2=\\left[b^2+c^2-2bcCosA\\right]-\\left[a^2+c^2-2acCosB\\right]=b^2-a^2-2bcCosA+2acCosB$ $\\Rightarrow\\ 2\\left(a^2-b^2\\right)=2bc=2\\left[acCosB-bcCosA\\right]\\Rightarrow\\ b=aCosB-bCosA$ $\\displaystyle\\Rightarrow\\ b(1+CosA)=aCosB\\Rightarrow\\frac{b}{a}=\\frac{CosB}{ 1+CosA}$ We have an alternative expression for this fraction using the Sine Law $\\displaystyle\\frac{a}{SinA}=\\frac{b}{SinB}\\Rightar row\\frac{b}{a}=\\frac{SinB}{SinA}$ $\\displaystyle\\Rightarrow\\frac{CosB}{1+CosA}=\\frac{ SinB}{SinA}\\Rightarrow\\ SinACosB=SinB+SinBCosA$ $\\Rightarrow\\ SinACosB-CosASinB=SinB$ Using the identity $Sin(A-B)=SinACosB-CosASinB$ $\\Rightarrow\\ Sin(A-B)=SinB\\Rightarrow\\ A=2B$ 7. Originally Posted by Archie Meade Another way.... $a^2-b^2=\\left[b^2+c^2-2bcCosA\\right]-\\left[a^2+c^2-2acCosB\\right]=b^2-a^2-2bcCosA+2acCosB$ $\\Rightarrow\\ 2\\left(a^2-b^2\\right)=2bc=2\\left[acCosB-bcCosA\\right]\\Rightarrow\\ b=aCosB-bCosA$ $\\displaystyle\\Rightarrow\\ b(1+CosA)=aCosB\\Rightarrow\\frac{b}{a}=\\frac{CosB}{ 1+CosA}$ We have an alternative expression for this fraction using the Sine Law $\\displaystyle\\frac{a}{SinA}=\\frac{b}{SinB}\\Rightar row\\frac{b}{a}=\\frac{SinB}{SinA}$ $\\displaystyle\\Rightarrow\\frac{CosB}{1+CosA}=\\frac{ SinB}{SinA}\\Rightarrow\\ SinACosB=SinB+SinBCosA$ $\\Rightarrow\\ SinACosB-CosASinB=SinB$ Using the identity $Sin(A-B)=SinACosB-CosASinB$ $\\Rightarrow\\ Sin(A-B)=SinB\\Rightarrow\\ A=2B$ GOOD job .... under stood .... THANKS", "the cosine of the angle between them, namely $$\\cos a$$, from which we obtain immediately $\\cos a = \\cos b \\cos c + \\sin b \\sin c \\cos A. \\label{3.5.9} \\tag{3.5.9}$ To obtain the sine formula, we isolate $$\\cos A$$ from this equation, square both sides, and write $$1 − \\sin^2 A$$ for $$\\cos^2 A$$. Thus, $(\\sin b \\sin c \\cos A )^2 = (\\cos a - \\cos b \\cos c )^2 , \\label{3.5.10} \\tag{3.5.10}$ and when we have carried out these operations we obtain $\\sin^2 A = \\frac{\\sin^2 b \\sin^2 c - \\cos^2 a - \\cos^2 b \\cos^2 c + 2\\cos a \\cos b \\cos c}{\\sin^2 b \\sin^2 c} . \\label{3.5.11} \\tag{3.5.11}$ In the numerator, write $$1 − \\cos^2 b$$ for $$\\sin^2 b$$ and $$1 − \\cos^2 c$$ for $$\\sin^2 c$$, and divide both sides by $$\\sin^2 a$$. This results in $\\frac{\\sin^2 A}{\\sin^2 a} = \\frac{1-\\cos^2 a - \\cos^2 b - \\cos^2 c + 2\\cos a \\cos b \\cos c}{\\sin^2 a \\sin^2 b \\sin^2 c}. \\label{3.5.12} \\tag{3.5.12}$ At this stage the reader may feel that we are becoming bogged down in heavier and heavier algebra and getting nowhere. But, after a careful look at equation $$\\ref{3.5.12}$$, it may be noted with some delight that the next line is: Therefore $\\frac{\\sin A}{\\sin a} = \\frac{\\sin B}{\\sin b} =", "###### back to index | new Define the sequence $a_1, a_2, a_3, \\ldots$ by $a_n = \\sum\\limits_{k=1}^n \\sin{k}$, where $k$ represents radian measure. Find the index of the $100^{th}$ term for which $a_n < 0$. Suppose that the angles of $\\triangle ABC$ satisfy $cos(3A)+cos(3B)+cos(3C)=1.$ Two sides of the triangle have lengths $10$ and $13$. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\\triangle ABC$ is $\\sqrt{m}$. Find $m$. For $\\pi \\le \\theta < 2\\pi$, let\\begin{align*} P &= \\frac12\\cos\\theta - \\frac14\\sin 2\\theta - \\frac18\\cos 3\\theta + \\frac{1}{16}\\sin 4\\theta + \\frac{1}{32} \\cos 5\\theta - \\frac{1}{64} \\sin 6\\theta - \\frac{1}{128} \\cos 7\\theta + \\cdots \\end{align*} and \\begin{align*} Q &= 1 - \\frac12\\sin\\theta -\\frac14\\cos 2\\theta + \\frac18 \\sin 3\\theta + \\frac{1}{16}\\cos 4\\theta - \\frac{1}{32}\\sin 5\\theta - \\frac{1}{64}\\cos 6\\theta +\\frac{1}{128}\\sin 7\\theta + \\cdots \\end{align*} so that $\\frac{P}{Q} = \\frac{2\\sqrt2}{7}$. Then $\\sin\\theta = -\\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Let $A,B,C$ be angles of an acute triangle with $$\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$$ and $$\\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A = \\frac{14}{9}$$ There are positive integers $p$, $q$, $r$, and $s$ for which $\\cos^2 C + \\cos^2 A + 2 \\sin C", "$$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$ Sort by: let PQ=a;QR=b;SR=c and PS=d then a=c and $d=(\\sqrt{3}+1)b$ we have $PR^{2}=c^{2}+d^{2}-2cdcos\\angle S=a^{2}+b^{2}-2abcos\\angle Q$ FROM THIS $b^{2}-d^{2}=2abcos\\angle Q-2cdcos\\angle S$ similarly $b^{2}-d^{2}=2bccos\\angle R-2adcos\\angle P$ comparing we get $2abcos\\angle Q-2cdcos\\angle S=2bccos\\angle R-2adcos\\angle P$ or simflifying it we get [putting the values a=c and $d=(\\sqrt{3}+1)b$]=== $cosQ-(\\sqrt{3}+1)cosS=cosR-(\\sqrt{3}+1)cosP$ OR $cosQ-cosR=(\\sqrt{3}+1)(cosS-cosP)$ OR $sin\\frac{Q+R}{2}sin\\frac{Q-R}{2}=(\\sqrt{3}+1)sin\\frac{S+P}{2}sin\\frac{S-P}{2}$ we see $sin\\frac{Q+R}{2}=sin\\frac{S+P}{2}$ putting this value in the above equation we get $sin\\frac{Q-R}{2}=(\\sqrt{3}+1)sin\\frac{30}{2}$[as S-P=30]...... putting the value of sin15 we get $sin\\frac{Q-R}{2}=\\frac{1}{\\sqrt{2}}$ so $\\frac{Q-R}{2}=45$ or Q-R=90. so we have done now. now you give me your solutin;kishan. i know this is not the shortest solution;nor the best!!! - 4 years, 1 month ago My solution is slightly shorter than yours.I would like you to do it.I would just give you a path to my solution.Try to get the area of the quadrilateral in 2 different ways.$$[PQS]+[QRS]=[PQR]+[PSR]$$ and try to use the sine rule area formula.After few simplifications,you will get your answer. - 4 years, 1 month ago thanks!! i will be waiting for your next beautiful problems;kishan - 4 years, 1 month ago" ]

[ "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution" ]

[ "so $c = \\frac{7}{9}d$. You can see $b + 2c = d$ $b + 2\\left(\\frac{7}{9}d\\right) = d$ $b + \\frac{14}{9}d = d$ $b = -\\frac{5}{9}d$. You can see $a - b = d$ $a - \\left(-\\frac{5}{9}d\\right) = d$ $a + \\frac{5}{9}d = d$ $a = \\frac{4}{9}d$. $d$ is a free variable, you can let it be whatever you like to get a consistent solution set. So why not let $d = 9$. Therefore a correct equation of the plane is $4x - 5y + 7z = 9$. The vectors normal to the plane has the same coefficients as the coefficients of the plane itself. Therefore a vector normal to the plane is $\\mathbf{n} = 4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}$. To find the unit vector, divide by its length. $\\frac{\\mathbf{n}}{|\\mathbf{n}|} = \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{4^2 + (-5)^2 + 7^2}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{16 + 25 + 49}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{90}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{3\\sqrt{10}}$ $= \\frac{4\\sqrt{10}}{30}\\mathbf{i} - \\frac{5\\sqrt{10}}{30}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$ $= \\frac{2\\sqrt{10}}{15}\\mathbf{i} - \\frac{\\sqrt{10}}{6}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$. • May 15th 2010, 01:00 AM Sudharaka Quote: Originally Posted by kandyfloss I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to. Here is the question :- A,B and C are the point (0,1,2),(3,2,1) and", "$b + 2\\left(\\frac{7}{9}d\\right) = d$ $b + \\frac{14}{9}d = d$ $b = -\\frac{5}{9}d$. You can see $a - b = d$ $a - \\left(-\\frac{5}{9}d\\right) = d$ $a + \\frac{5}{9}d = d$ $a = \\frac{4}{9}d$. $d$ is a free variable, you can let it be whatever you like to get a consistent solution set. So why not let $d = 9$. Therefore a correct equation of the plane is $4x - 5y + 7z = 9$. The vectors normal to the plane has the same coefficients as the coefficients of the plane itself. Therefore a vector normal to the plane is $\\mathbf{n} = 4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}$. To find the unit vector, divide by its length. $\\frac{\\mathbf{n}}{|\\mathbf{n}|} = \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{4^2 + (-5)^2 + 7^2}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{16 + 25 + 49}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{90}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{3\\sqrt{10}}$ $= \\frac{4\\sqrt{10}}{30}\\mathbf{i} - \\frac{5\\sqrt{10}}{30}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$ $= \\frac{2\\sqrt{10}}{15}\\mathbf{i} - \\frac{\\sqrt{10}}{6}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$. 3. Originally Posted by kandyfloss I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to. Here is the question :- A,B and C are the point (0,1,2),(3,2,1) and (1,-1,0) respectively.Find the unit vector perpendicular to the plane ABC. I've just started learning this topic,and i'm not very", "the numbers are 8 and 16-8 = 8 Question 31. Find ∫$$\\frac{x}{(x+1)(x+2)} d x$$ Let I = $$\\frac{x}{(x+1)(x+2)} d x$$ Resolve into partial fractions, Question 32. Evaluate ∫$$\\frac{1}{1+\\tan x} d x$$ Question 33. Find the area of the region bounded by the curve y = x2 and the line y=4 Given y = x2 and y = 4 To find A and B: put y = 4 in y = x2 ⇒ x2 = 4 ⇒ x =±2 put x = 2, y = 4 ⇒(2,4) x = -2, y = 4 ⇒ (-2,4) Question 34. Prove that the equation x2$$\\frac{d y}{d x}$$ = x2 – 2y2 + xy is a homogeneous differential equation. x2$$\\frac{d y}{d x}$$ = x2 – 2y2 + xy :. (1) is homogeneous DiffentiaI Equation. Question 35. Find a vector perpendicular to each of the vectors $$\\overrightarrow{\\mathbf{a}}=2$$ $$\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$$ $$\\overrightarrow{\\mathbf{b}}=3 \\hat{\\mathbf{i}}+5 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}}$$ which has magnitude 10 units. Let $$\\overrightarrow{\\mathbf{a}}=2$$ $$\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$$ and $$\\overrightarrow{\\mathbf{b}}=3 \\hat{\\mathbf{i}}+5 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}}$$ = î[- 2 – 15] – ĵ[-4 – 9] + k̂[10 – 3] = î[-17] – ĵ [-13] + k̂[7] = -17î + 13 ĵ + 7k̂ Question 36. Show that the points and A (-1, 4, -3), B(3,2,-5) C (-3,8,-5) D(-3,2,1)are coplanar. $$\\overrightarrow{\\mathrm{AB}}=4 \\hat{\\mathrm{i}}+(-2) \\hat{\\mathrm{j}}+(-2) \\hat{\\mathrm{k}}$$ $$\\overrightarrow{\\mathrm{BC}}=-6 \\hat{\\mathrm{i}}+6 \\hat{\\mathrm{j}}+0 \\hat{\\mathrm{k}}$$ $$\\overrightarrow{\\mathrm{CD}}=0 \\hat{\\mathrm{i}}-6 \\hat{\\mathrm{j}}+6 \\hat{\\mathrm{k}}$$ = 4[36", "## Elementary Linear Algebra 7th Edition $\\mathbf{z}=(\\frac{7}{2},3,\\frac{5}{2})$ $\\mathbf{u}=(1,2,3)\\quad\\mathbf{v}=(2,2-1)\\quad\\mathbf{w}=(4,0,-4)$ Find $\\mathbf{z}$ if $2\\mathbf{z}-3\\mathbf{u}=\\mathbf{w}$ We can solve for $\\mathbf{z}$: $2\\mathbf{z}=\\mathbf{w}+3\\mathbf{u}$ $\\mathbf{z}=\\frac{1}{2}\\mathbf{w}+\\frac{3}{2}\\mathbf{u}$. All we need to do now is plug in the vectors $\\mathbf{w}$ and $\\mathbf{u}$: $\\mathbf{z}=\\frac{1}{2}(4,0,-4)+\\frac{3}{2}(1,2,3)$ $\\mathbf{z}=(2,0,-2)+(\\frac{3}{2},3,\\frac{9}{2})=(\\frac{7}{2},3,\\frac{5}{2})$", "$dy = a\\,dx$ $\\frac{dy}{dx} = a$ $dy = \\mathbf{a}\\,d\\mathbf{x}$ $\\frac{dy}{d\\mathbf{x}} = \\mathbf{a}$ $dy = {\\rm tr}(\\mathbf{A}\\,d\\mathbf{X})$ $\\frac{dy}{d\\mathbf{X}} = \\mathbf{A}$ $d\\mathbf{y} = \\mathbf{a}\\,dx$ $\\frac{d\\mathbf{y}}{dx} = \\mathbf{a}$ $d\\mathbf{y} = \\mathbf{A}\\,d\\mathbf{x}$ $\\frac{d\\mathbf{y}}{d\\mathbf{x}} = \\mathbf{A}$ $d\\mathbf{Y} = \\mathbf{A}\\,dx$ $\\frac{d\\mathbf{Y}}{dx} = \\mathbf{A}$ 4. ^ a b c Here, $\\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 5. ^ a b Here, $\\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 6. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\\frac{\\partial \\mathbf{Y}}{\\partial x},$ by X in $\\frac{\\partial y}{\\partial \\mathbf{X}}.$ $\\frac{d\\,\\ln au}{dx} = \\frac{1}{au}\\frac{d(au)}{dx} = \\frac{1}{au}a\\frac{du}{dx} = \\frac{1}{u}\\frac{du}{dx} = \\frac{d\\,\\ln u}{dx}.$", "14 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{728}$. This slightly improves upon the lower bound of 1/2124 coming from Corollary 10. Proof Combine Corollary 4 and Lemma 13. $\\Box$ Lemma 15 One has ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{2}{3}$ and $2{\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq 1$. Proof As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length $\\sqrt{3}$ and three of length $2$. If we let a denote the number of monochromatic $\\sqrt{3}$ edges and b the number of monochromatic $2$-edges, we see from inspection of all four colorings that $(a,b)$ is either $(6, 0), (4,0), (2, 1)$, or $(0,3)$. In particular, one always has $\\frac{a}{6} + \\frac{b}{3} \\geq \\frac{2}{3}$ and $2\\frac{a}{6} + \\frac{b}{3} \\geq 1$. Taking expectations, we obtain the claim. $\\Box$ Corollary 16 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{1}{6}$, ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3} ) ) \\geq \\frac{1}{4}$ and ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{28}$. Proof Combine Lemma 2, Lemma 15, and Lemma 13. $\\Box$ Lemma 17 If $a,b,c \\gt 0$ are the lengths", "### Home > APCALC > Chapter 9 > Lesson 9.3.2 > Problem9-102 9-102. Locate vectors $\\mathbf{f}$ and $\\mathbf{g}$ on the Section 9.3 Resource Page. 1. Write $\\mathbf{f}$ and $\\mathbf{g}$ in $\\mathbf{i}, \\mathbf{j}$ form. Then write the $\\mathbf{i}, \\mathbf{j}$ form for $2\\mathbf{f}-\\frac { 1 } { 3 }\\mathbf{g}$. $2\\mathbf{f}-\\frac{1}{3}\\mathbf{g}=2\\Big<4,2\\Big>-\\frac{1}{3}\\Big<3,-6\\Big>$ 2. Draw and label $2\\mathbf{f}-\\frac { 1 } { 3 }\\mathbf{g}$. Does your diagram confirm your answer from part (a)?", "v and w. Sci Advisor PF Gold P: 1,352 Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. P: 101 Quote by DrClaude Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? Sci Advisor PF Gold P: 1,352 I'm", "this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt},\\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. 8. Feb 14, 2013 ### strokebow Thanks for your help. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? 9. Feb 14, 2013 ### Staff: Mentor I'm not sure I follow you. The \"first step\" is the calculation of $\\mathbf{k}_1$. The first element of $\\mathbf{f}(\\mathbf{a}_n,t_n)$ is $$\\left.\\frac{du}{dt}\\right|_{t_n} = K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n )$$ so the first element of $\\mathbf{k}_1$ is $$h \\times \\left[ K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n ) \\right]$$ and so on. For instance, the 4th element of $\\mathbf{k}_1$ is $$u_n$$ Once you have calculated all six elements of $\\mathbf{k}_1$, the \"second step\" is do", "# Finding $P(D)$ given we know $P(C)$, $P(C|D')$ and $P(C|D)$ As the title states we need to find $P(D)$ given that $P(C)={9\\over20}$, $P(C|D')={3\\over7}$ and $P(C|D)={6\\over13}$. I have been hitting my head against a brick wall for the past half an hour or so, unable to make much progress. So far all that I have managed to do is some algebraic manipulation with ${6\\over13}={P(C\\cap D)\\over P(D)}$ and ${3\\over7}={P(C\\cap D')\\over P(D')}$. However I don't seem to be making any progress. Does anyone have any suggestions on a continuation? - I'll assume that $D'$ is the complement of the event $D$, since if not we don't have enough information to solve this problem. You have $$\\mathbb P(C) = \\mathbb P(C \\, | \\, D) \\mathbb P(D) + \\mathbb P(C \\, | \\, D') \\mathbb P(D') = \\mathbb P(C \\, | \\, D) \\mathbb P(D) + \\mathbb P(C \\, | \\, D') (1-\\mathbb P(D)).$$ Now you have a linear equation in $\\mathbb P(D)$, so it's pretty easy to find it. It gives you $$\\frac 9{20} = \\frac 6{13} \\times \\mathbb P(D) + \\frac 37 (1 - \\mathbb P(D)) = \\frac 37 + \\left( \\frac 6{13} - \\frac 37 \\right) \\mathbb P(D)$$ which leads to $\\mathbb P(D) = \\frac {13}{20}$.", "On comparing the ratios $\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}, \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ and $\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{\\mathbf{z}}}$ Question. On comparing the ratios $\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}, \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ and $\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$, find out whether the following pairs of linear equations are consistent, or inconsistent. (i) $3 x+2 y=5 ; 2 x-3 y=7$ (ii) $2 x-3 y=8 ; 4 x-6 y=9$ (iii) $\\frac{\\mathbf{3}}{\\mathbf{z}} \\mathrm{x}+\\frac{\\mathbf{5}}{\\mathbf{3}} \\mathrm{y}=7 ; 9 \\mathrm{x}-10 \\mathrm{y}=14$ (iv) $5 x-3 y=11 ;-10 x+6 y=-22$ (v) $\\frac{\\mathbf{4}}{\\mathbf{3}} x+2 y=8 ; 2 x+3 y=12$ Solution: (i) $3 x+2 y-5=0$ …(i) $2 x-3 y-7=0$ …(ii) $\\frac{a_{1}}{a_{2}}=\\frac{3}{2} ; \\frac{h_{1}}{b_{2}}=\\frac{2}{-3}=-\\frac{2}{3}$ $\\Rightarrow \\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}$ The equations have a unique solution. Hence, consistent. (ii) $2 x-3 y=8$ …(i) $4 x-6 y=9$ …(ii) $\\frac{a_{1}}{a_{2}}=\\frac{2}{4}, \\frac{b_{1}}{b_{2}}=\\frac{-3}{-6}, \\frac{c_{1}}{c_{2}}=\\frac{8}{9}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}=\\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}} \\neq \\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$  The equations have no solution. ​Hence inconsistent. (iii) $\\frac{\\mathbf{3}}{\\mathbf{2}} \\mathbf{x}+\\frac{\\mathbf{5}}{\\mathbf{3}} \\mathbf{y}=7$ …..(i) $9 x-10 y=14$ …..(ii) $\\frac{a_{1}}{a_{2}}=\\frac{3 / 2}{9}=\\frac{1}{6}, \\frac{b_{1}}{b_{2}}=\\frac{5 / 3}{-10}=\\frac{-1}{6}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}} \\neq \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}$ The equations have infinite solutions. Hence, consistent. (iv) $5 x-3 y=11$ ….(i) $-10 x+6 y=-22$ ….(ii) $\\frac{a_{1}}{a_{2}}=\\frac{5}{-10}=\\frac{-1}{2}, \\frac{b_{1}}{b_{2}}=\\frac{-3}{6}=\\frac{-1}{2}$ $\\frac{c_{1}}{c_{2}}=\\frac{11}{-22}=\\frac{-1}{2}$ $\\Rightarrow \\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}=\\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}=\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}$ The equations have infinite solutions. Hence, consistent. (v) $\\frac{4}{3} \\mathbf{x}+\\mathbf{2} \\mathbf{y}=8$ …..(i) $2 x+3 y=12$ ….(ii) $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$ The equations have infinite solutions. Hence, consistent. Editor", "ago, # ^ | +11 A more intuitive way (for me) is to just write the dp as and move to the left hand side. • » » » » » » 3 years ago, # ^ | +10 Why extra \"1\" is added?? • » » » » » » » 3 years ago, # ^ | ← Rev. 4 → +10 let $k = a + b + c$$\\mathbf{E}(a,b,c) := Expected no. of trials required to reach (0,0,0) from (a,b,c)$$\\mathbf{E}(a,b,c) = \\frac{n-k}{n} * (1 + \\mathbf{E}(a,b,c)) + \\frac{a}{n} * (1 + \\mathbf{E}(a-1,b,c)) + \\frac{b}{n} * (1 + \\mathbf{E}(a+1,b-1,c)) + \\frac{c}{n} * (1 + \\mathbf{E}(a,b+1,c-1))$ or $\\mathbf{E}(a,b,c) = 1 + \\frac{n-k}{n} * \\mathbf{E}(a,b,c) + \\frac{a}{n} * \\mathbf{E}(a-1,b,c) + \\frac{b}{n} * \\mathbf{E}(a+1,b-1,c) + \\frac{c}{n} * \\mathbf{E}(a,b+1,c-1)$hence, $\\mathbf{E}(a,b,c) = \\frac{n}{k} + \\frac{a}{k} * \\mathbf{E}(a-1,b,c) + \\frac{b}{k} * \\mathbf{E}(a+1,b-1,c) + \\frac{c}{k} * \\mathbf{E}(a,b+1,c-1)$ • » » » » » » » 2 years ago, # ^ | 0 My understanding for the reason why we have to add 1: To reach state $(cnt_1, cnt_2, cnt_3)$ we can go from 4 states $(cnt_1, cnt_2, cnt_3)$, $(cnt_1-1, cnt_2, cnt_3)$, $(cnt_1+1, cnt_2-1, cnt_3)$ or $(cnt_1, cnt_2+1, cnt_3-1)$ with probability as above. For each \"from\" state, we must take the last fixed turn to reach the required state, for example if we go from", "## Parallel Lines Vector Question In the diagram P splits AB in the ratio 1:2. BQ=k(BC) and PQD is a straight line. If $\\mathbf{AB}=\\mathbf{a}, \\; \\mathbf{AC}=\\mathbf{c}, \\; \\mathbf{CD}=3\\mathbf{a}$ find the value of $k$ . Since P splits AB in the ratio 1:2, $\\mathbf{AP}=\\frac{1}{3} \\mathbf{a}, \\; \\mathbf{PB}=\\frac{2}{3} \\mathbf{a}$ . $\\mathbf{BC}=- \\mathbf{a}+ \\mathbf{b}$ so $\\mathbf{BQ}=k(- \\mathbf{a}+ \\mathbf{b})$ . Then $\\mathbf{PQ}= \\mathbf{PB}+ \\mathbf{BQ} =\\frac{2}{3} \\mathbf{a}+k(- \\mathbf{a}+ \\mathbf{b})=(\\frac{2}{3}-k) \\mathbf{a} + k \\mathbf{b}$ $\\mathbf{PD}= - \\frac{1}{3} \\mathbf{a} ++4 \\mathbf{b}$ . $PQD$ is a straight line so $\\mathbf{PQ}, \\; \\mathbf{PD}$ are parallel and the coefficients of $\\mathbf{a}, \\; \\mathbf{b}$ are in the same ratio. Hence $\\frac{2/3-k}{-1/3}= \\frac{k}{4} \\rightarrow\\frac{8}{3}-4k=- \\frac{k}{3} \\rightarrow \\frac{8}{3}= \\frac{11k}{3} \\rightarrow k =\\frac{\\frac{8}{3}}{\\frac{11}{3}}=\\frac{8}{11}$", "$$Y_2=\\left\\{ \\left(\\frac{2}{7} , 0 , \\frac{3}{7} , \\frac{-6}{7} , 0 \\right)^T, \\left(\\frac{-3}{7} , 0 , \\frac{6}{7} , \\frac{2}{7} , 0 \\right)^T, \\left(0 , \\frac{4}{5} , 0 , 0 , \\frac{3}{5} \\right)^T, \\left(0 , \\frac{-3}{5} , 0 , 0 , \\frac{4}{5}\\right)^T \\right\\}$$. To extend it to an orthonormal basis of $$\\mathbb R^5$$ it is enough to take any vector linearly independent of the vectors of $$Y_2$$ e.g. the vector $$\\mathbf e^5$$ and To obtain the Gram-Schmidt orthonormalization from it the extension $$X_2=\\left\\{\\left(\\frac{6}{7},0,\\frac{2}{7},\\frac{3}{7},0\\right)^T\\right\\}$$ Since $$\\ker (\\mathbf A)$$ is the orthogonal complement of $$\\mathcal R (\\mathbf A)$$, $$Y_2$$ is also a basis of $$\\mathcal R (\\mathbf A)$$ as well as $$X_2$$ is a basis of $$\\ker (\\mathbf A)$$ (in our case indeed $$X_1 = X_2$$). In fact, it would be possible to take the basis $$\\mathcal R (\\mathbf A)$$ as $$Y_3=\\left\\{ \\left(\\frac{2}{7} , 0 , \\frac{3}{7} , \\frac{-6}{7} , 0 \\right)^T, \\left(\\frac{-3}{7} , 0 , \\frac{6}{7} , \\frac{2}{7} , 0 \\right)^T, \\mathbf e^2, \\mathbf e^5\\right\\}$$.", "## Educators Problem 1 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=0.3 \\mathbf{i}-0.4 \\mathbf{j}$$ Check back soon! Problem 2 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=\\frac{1}{2} x \\mathbf{i}+y \\mathbf{j}$$ Check back soon! Problem 3 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=-\\frac{1}{2} \\mathbf{i}+(y-x) \\mathbf{j}$$ Check back soon! Problem 4 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=y \\mathbf{i}+(x+y) \\mathbf{j}$$ Check back soon! Problem 5 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=\\frac{y \\mathbf{i}+x \\mathbf{j}}{\\sqrt{x^{2}+y^{2}}}$$ Check back soon! Problem 6 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y)=\\frac{y \\mathbf{i}-x \\mathbf{j}}{\\sqrt{x^{2}+y^{2}}}$$ Check back soon! Problem 7 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y, z)=\\mathbf{k}$$ Check back soon! Problem 8 $1-10$ \" Sketch the vector ficld $\\mathbf{F}$ by drawing a diagram like Figure 4 or Figure $8 .$ $$\\mathbf{F}(x, y, z)=-y \\mathbf{k}$$ Check back soon!", "and point $$O$$ is the intersection point of the diagonals. Taking $$\\mathbf{AB} = \\mathbf{a}$$ and $$\\mathbf{AD} = \\mathbf{b}$$ as the basis vectors, find in this basis the coordinates of the vectors $$\\mathbf{BD},$$ $$\\mathbf{CO}$$ and $$\\mathbf{MD}.$$ Solution. Express the vector $$\\mathbf{BD}$$ in terms of vectors $$\\mathbf{AB} = \\mathbf{a}$$ and $$\\mathbf{AD} = \\mathbf{b}.$$ By the vector subtraction rule, $\\mathbf{BD} = \\mathbf{AD} - \\mathbf{AB} = -\\mathbf{a} + \\mathbf{b}, \\Rightarrow \\mathbf{BD} = \\left({-1, 1}\\right).$ Find now the vector $$\\mathbf{CO}.$$ Notice that $\\mathbf{AC} = \\mathbf{AB} + \\mathbf{AD} = \\mathbf{a} + \\mathbf{b}.$ Since $$\\mathbf{CA} = - \\mathbf{AC}$$ and $$\\mathbf{CO} = \\frac{1}{2}\\mathbf{CA},$$ we get $\\mathbf{CO} = \\frac{1}{2}\\mathbf{CA} = -\\frac{1}{2}\\mathbf{AC} = -\\frac{1}{2}\\left({\\mathbf{a} + \\mathbf{b}}\\right) = -\\frac{1}{2}\\mathbf{a} - \\frac{1}{2}\\mathbf{b},$ or $\\mathbf{CO} = \\left({-\\frac{1}{2}, -\\frac{1}{2}}\\right).$ Calculate the vector $$\\mathbf{MD}:$$ $\\mathbf{MD} = \\mathbf{AD} - \\mathbf{AM},$ where $\\mathbf{AM} = \\mathbf{AB} + \\mathbf{BM} = \\mathbf{AB} + \\frac{1}{2}\\mathbf{AD} = \\mathbf{a} + \\frac{1}{2}\\mathbf{b}.$ Hence $\\mathbf{MD} = \\mathbf{AD} - \\mathbf{AM} = \\mathbf{b} - \\left({\\mathbf{a} + \\frac{1}{2}\\mathbf{b}}\\right) = -\\mathbf{a} - \\frac{1}{2}\\mathbf{b}, \\Rightarrow \\mathbf{MD} = \\left({-1, -\\frac{1}{2}}\\right).$ ### Example 4. Find a basis for the set of $$2 \\times 2$$ matrices and determine the coordinates of the matrix $\\left[ \\begin{array}{*{20}{r}} -1 & 5\\\\ 3 & 0 \\end{array} \\right]$ in this basis. Solution. As a basis we can take the following matrices $\\left[ \\begin{array}{*{20}{r}} 1 & 0\\\\ 0 & 0 \\end{array} \\right],\\left[ \\begin{array}{*{20}{r}} 0 & 1\\\\ 0", "is, $A^{\\trans}=-A$. Then prove the following statements. (a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number. (b) The rank of $A$ is even. Proof. (a) Each […] • Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let $\\mathbf{a}$ and $\\mathbf{b}$ be vectors in $\\R^n$ such that their length are $\\|\\mathbf{a}\\|=\\|\\mathbf{b}\\|=1$ and the inner product $\\mathbf{a}\\cdot \\mathbf{b}=\\mathbf{a}^{\\trans}\\mathbf{b}=-\\frac{1}{2}.$ Then determine the length $\\|\\mathbf{a}-\\mathbf{b}\\|$. (Note […] • Inner Product, Norm, and Orthogonal Vectors Let $\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3$ are vectors in $\\R^n$. Suppose that vectors $\\mathbf{u}_1$, $\\mathbf{u}_2$ are orthogonal and the norm of $\\mathbf{u}_2$ is $4$ and $\\mathbf{u}_2^{\\trans}\\mathbf{u}_3=7$. Find the value of the real number $a$ in […] • Find the Inverse Matrix of a Matrix With Fractions Find the inverse matrix of the matrix $A=\\begin{bmatrix} \\frac{2}{7} & \\frac{3}{7} & \\frac{6}{7} \\\\[6 pt] \\frac{6}{7} &\\frac{2}{7} &-\\frac{3}{7} \\\\[6pt] -\\frac{3}{7} & \\frac{6}{7} & -\\frac{2}{7} \\end{bmatrix}.$ Hint. You may use the augmented matrix […] • Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is […]", "Find the components of Let $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components ofLet $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components of (a) $\\mathbf{u}+\\mathbf{w}$ (b) $\\mathbf{v}-3 \\mathbf{u}$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vectorShow that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector ... Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components ofLet $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of (a) $\\mathbf{v}-\\mathbf{w}$ (b) $6 \\mathrm{u}+2 \\ma ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$0 answers 58 views Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$0 answers 88 views Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)\\$ ...", "15. (a) $\\overrightarrow{A B}=\\vec{b}-\\vec{a}$ (b) Show (c) Show (d) $\\lambda=\\frac{2}{3}$ 16. (a)(i) Prove (ii) Show (b) $\\frac{4}{5}$ 17. (a)(i) $\\frac{2}{3} \\vec{b}-\\vec{a}$ (ii) $\\frac{3}{5} \\vec{a}+\\frac{4}{15} \\vec{b}$ (iii) $\\frac{3}{5} \\vec{a}-\\frac{11}{15} \\vec{b}$ (b) $\\lambda=\\frac{9}{11}$ (c) $\\frac{110}{27}$ 18. $(a) \\overrightarrow{C D}=\\frac{2}{3} \\vec{b}-\\frac{1}{2} \\vec{a}$ (b) $\\overrightarrow{\\mathrm{CE}}=2 \\vec{b}-\\frac{3}{2} \\vec{a}$ (c) Show 19. \\begin{aligned}&\\overrightarrow{A B}=-\\mathbf{a}+\\mathbf{b} \\\\&\\overrightarrow{P Q}=\\frac{\\mathbf{a}}{4}+\\frac{1}{2}(-\\mathbf{a}+\\mathbf{b}),=\\frac{1}{4}(2 \\mathbf{b}-\\mathbf{a}) \\text { oe } \\\\&\\overrightarrow{Q R}=\\frac{\\mu}{4}(2 \\mathbf{b}-\\mathbf{a}) \\\\&\\overrightarrow{Q R}=\\frac{1}{2}(\\mathbf{b}-\\mathbf{a})+\\lambda \\mathbf{b} \\\\&\\frac{2 \\mu}{4} \\mathbf{b}-\\frac{\\mu}{4} \\mathbf{a}=\\frac{1}{2} \\mathbf{b}-\\frac{1}{2} \\mathbf{a}+\\lambda \\mathbf{b} \\Rightarrow-\\frac{\\mu}{4}=-\\frac{1}{2} \\Rightarrow \\mu=2 \\\\&\\frac{2 \\mu}{4}=\\frac{1}{2}+\\lambda \\Rightarrow \\lambda=\\frac{1}{2} \\end{aligned}", "and simplify: $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} – \\mathbf{V}_{IN} (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})}{\\mathbf{R}_{1}} \\tag{3}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{4}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – (1 + \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}})]}{\\mathbf{R}_{1}} \\tag{5}\\\\$$ $$\\mathbf{I}_{1} = \\frac{\\mathbf{V}_{IN} [1 – 1 – \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}]}{\\mathbf{R}_{1}} \\tag{6}\\\\$$ $$\\mathbf{I}_{1} = \\frac{-\\mathbf{V}_{IN} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{7}\\\\$$ We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation: $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\frac{- \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{3}}}{\\mathbf{R}_{1}} \\tag{8}\\\\$$ $$\\frac{\\mathbf{I}_{1}}{\\mathbf{V}_{IN}} = \\mathbf{-} \\frac{\\mathbf{R}_{2}}{\\mathbf{R}_{1}\\mathbf{R}_{3}} \\tag{9}\\\\$$ Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at: $$\\mathbf{R}_{IN} = \\frac{\\mathbf{V}_{IN}}{\\mathbf{I}_{1}} = \\mathbf{-} \\frac{\\mathbf{R}_{1}\\mathbf{R}_{3}}{\\mathbf{R}_{2}} \\tag{10}\\\\$$ This is the input resistance seen looking into the input of the NIC circuit. If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1. If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3." ]

[ "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); [/asy]$ Let $\\text{E}$ be the origin. Then, $\\text{D}=(1, 0)$ $\\text{A}=(0, 2)$ $\\text{B}=(1, 2)$ $\\text{F}=(2, 1)$ ${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\\frac{1}{2}x+2$ Subtracting the second equation from the first gives us $\\frac{5}{2}x-2=0$. Thus, $x=\\frac{4}{5}$. Plugging this into the first equation gives us $y=\\frac{8}{5}$. Since $\\text{C} (0.8, 1.6)$, $G$ is $(0.8, 2)$, ${AB}=1$ and ${CG}=0.4$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot 0.4=0.2=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\ \\frac15}$. ## Solution 3 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); label(\"H\",H,W); label(\"I\",I,E); [/asy]$ Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$ Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \\frac{3}{2} = 2: 3$. Based on similarity the length of the height of $GC$ is thus $\\frac{2}{5}\\cdot1 = \\frac{2}{5}$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{5}=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "# Difference between revisions of \"2004 AMC 10B Problems/Problem 20\" ## Problem In $\\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $\\frac{AT}{DT}=3$ and $\\frac{BT}{ET}=4$, what is $\\frac{CD}{BD}$? $\\mathrm{(A) \\ } \\frac{1}{8} \\qquad \\mathrm{(B) \\ } \\frac{2}{9} \\qquad \\mathrm{(C) \\ } \\frac{3}{10} \\qquad \\mathrm{(D) \\ } \\frac{4}{11} \\qquad \\mathrm{(E) \\ } \\frac{5}{12}$ $[asy] unitsize(1cm); defaultpen(0.8); pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,2*WNW); [/asy]$ ## Solution 1 (Triangle Areas) We use the square bracket notation $[\\cdot]$ to denote area. WLOG, we can assume $[\\triangle BTD] = 1$. Then $[\\triangle BTA] = 3$, and $[\\triangle ATE] = 3/4$. We have $CD/BD = [\\triangle ACD]/[\\triangle ABD]$, so we need to find the area of quadrilateral $TDCE$. Draw the line segment $TC$ to form the two triangles $\\triangle TDC$ and $\\triangle TEC$. Let $x = [\\triangle TDC]$, and $y = [\\triangle TEC]$. By considering triangles $\\triangle BTC$ and $\\triangle ETC$, we obtain $(1+x)/y=4$, and by considering triangles $\\triangle ATC$ and $\\triangle DTC$, we obtain $(3/4+y)/x=3$. Solving, we get $x=4/11$, $y=15/44$, so the area of quadrilateral $TDEC$ is $x+y=31/44$. Therefore $\\frac{CD}{BD}=\\frac{\\frac{3}{4}+\\frac{31}{44}}{3+1}=\\boxed{\\textbf{(D)} \\frac{4}{11}}$ ## Solution 2 (Mass points) The presence of only ratios in the problem", "3); C = (4, 0); H = foot(C, A, B); draw(A--B--C--cycle); draw(C--H); draw(rightanglemark(A, C, B)); draw(rightanglemark(C, H, B)); label(\"A\", A, SSW); label(\"B\", B, ENE); label(\"C\", C, SE); label(\"H\", H, NNW); [/asy]$ Since $ABC, CBH, ACH$ are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G," ]

[ "PtolemyTheorem Circle AMC10/12 2014 Problem - 475 Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?", "# ABCD Geometry Level 5 $$ABCD$$ is a quadrilateral inscribed in a circle with $$AB = 1$$, $$BC = 3$$, $$CD = 4$$ and $$DA = 6$$. What is the value of $$\\sec^2 \\angle BAD$$? Details and assumptions A quadrilateral is inscribed in a circle if all four of its vertices lie on the circle. ×", "A quadrilateral $$ABCD$$ is inscribed in a circle with $$AD$$ as diameter. If $$AD=4$$ and $$AB=BC=1,$$ find the length of $$CD.$$", "# Difference between revisions of \"Mock AIME 3 Pre 2005 Problems/Problem 7\" ## Problem $ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$. If $AB = 1, CD = 4,$ and $BP : DP = 3 : 8,$ then the area of the inscribed circle of $ABCD$ can be expressed as $\\frac{p\\pi}{q}$, where $p$ and $q$ are relatively prime positive integers. Determine $p + q$.", "A quadrilateral $$ABCD$$ in which $$AB=a, BC=b, CD=c, DA=d$$ such that one circle can be inscribed in it and another circle can be circumscribed about it. Then compute $$\\cos A$$.", "$$ABCD$$ is a cyclic quadrilateral inscribed in a circle with radius $$20\\sqrt{2}$$. Given that $$AB=BC=CD=20$$, what is the length of $$AD$$?", "back to index | new Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? A quadrilateral is inscribed in a circle of radius $200\\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side? As shown, $\\angle{ACB} = 90^\\circ$, $AD=DB$, $DE=DC$, $EM\\perp AB$, and $EN\\perp CD$. Prove $$MN\\cdot AB = AC\\cdot CB$$ What is the Ptolemy Theorem?", "• Mar 13th 2010, 01:53 PM disclaimer Quadrilateral $ABCD$ is inscribed in a circle, where the tangents to this circle at points $B$ and $D$ intersect on the straight line joining $A$ and $C$. Show that: $AB\\cdot{CD}=AD\\cdot{BC}$ • Mar 14th 2010, 09:34 PM sa-ri-ga-ma Quote: Originally Posted by disclaimer Quadrilateral $ABCD$ is inscribed in a circle, where the tangents to this circle at points $B$ and $D$ intersect on the straight line joining $A$ and $C$. Show that: $AB\\cdot{CD}=AD\\cdot{BC}$", "# Circle Inscribed in a Trapezium Geometry Level 3 $$ABCD$$ is a trapezium with $$AB < CD$$ and $$AB$$ parallel to $$CD$$. $$\\Gamma$$ is a circle inscribed in $$ABCD$$, such that $$\\Gamma$$ is tangent to all four sides. If $$AD = BC = 25$$ and the area of $$ABCD$$ is $$600$$, what is the radius of $$\\Gamma$$? ×", "# Quadrilateral Geometry Level 3 A quadrilateral $$ABCD$$ is inscribed in a circle with $$AD$$ as diameter. If $$AD=4$$ and $$AB=BC=1,$$ find the length of $$CD.$$ × Problem Loading... Note Loading... Set Loading...", "$$ABCD$$ is a square inscribed in a circle of diameter $$3\\sqrt{2}$$. $$E$$ and $$F$$ are the midpoints of $$AB$$ and $$BC$$, respectively. What is the area of the shaded region?", "Half Inscribed Geometry Level 1 Given that $AB=12, CD=1, \\angle BAC = 90^\\circ,$ and the semicircle is tangent to $BC,$ find the radius of the semicircle. ×", "# I've seen it before - 3 Geometry Level 3 If $$ABCD$$ is an isosceles trapezium inscribed in a semi-circle with diameter $$AD$$ and $$AB=CD=2\\text{ cm}$$ and the radius of the semi-circle is $$4 \\text{ cm}$$, then the length of $$BC$$ (in cm) is: ×", "Semi-circles are drawn on $AB$, $AC$, $BD$ and $CD$ where $AB = BC = CD$ Show that the large circle is divided into $3$ portions of equal area.", "# Another Cyclic Trapezoid Geometry Level 4 A trapezoid $$ABCD$$ is circumscribed around a circle with radius $$r$$ and inscribed in a circle with radius $$R$$, with bases $$AB$$ and $$CD$$, such that $$AB>CD$$ and $$BC=AD=R$$. If $$R=k\\cdot AB$$, find $$k$$ to the nearest 3 decimal places. ×", "Accounting (Commercial Banking) Followers: 2 Kudos [?]: 97 [0], given: 16 Re: Help! [#permalink] 15 Dec 2012, 09:47 Bunuel wrote: Attachment: Circle.PNG Quadrilateral ABCD is inscribed in circle K. The diameter of K is 20. AC is perpendicular to BD. What is the area of ABCD? First note that as AC is perpendicular to BD and CA is a diameter then triangles CDA and CBA are congruent (they are mirror images of each other). Thus the area of quadrilateral ABCD is twice the area of triangle CDA, which equals to 1/2*DE*CA=1/2*DE*20=10*DE (AC is perpendicular to BD means DE is the height of CDA), so area of ABCD=2*CDA=20*DE. All we need to find is the lengths of the line segment DE. Alternately as ABCD is a kite (triangles CDA and CBA are congruent then CB=CD and AB = AD --> ABCD is a kite) then its area equals to $$\\frac{d_1*d_2}{2}$$ and as one diagonal is a diameter CA, which equals to 20, then we need the second diagonal BD which equals to 2*DE. Next, you should know the following properties to solve this question: A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that", "Image via Wikipedia Quadrilateral ABCD is inscribed in a circle. Let AB = a, BC = b, CD = c, DA = d, AC = p and BD = q. Prove Ptolemy’s second theorem that $p \\cdot q = \\frac{ad+bc}{ab+cd}$ . This question is also taken from Bertley Math Circle . It is easy question, just use $R = \\frac{abc}{4S}$ a, b, c are sides of triangle and S is area of triangle. If you couldn’t prove send me mail, and i will give few tips.", "Find the two numbers. (c) In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC//AE. If ∠BAC = 50°, find giving reasons: (i) ∠ACB (ii) ∠EDC (iii) ∠BEC Hence prove that BE is also a diameter VIEW SOLUTION What are you looking for? Syllabus", "# Let $ABCD$ be a quadrilateral with area 18 with side AB parallel to the side CD and AB=2CD.Let AD be $\\perp$ to AB and CD.If a circle is drawn inside the quadrilateral ABCD touching all the sides then its radius is $(a)\\;3\\qquad(b)\\;2\\qquad(c)\\;\\large\\frac{3}{2}$$\\qquad(d)\\;1$", "# Inscribed circle and trapezoid Author(s): Circle O(r) is inscribed in an isosceles trapezoid ABCD, where AB is parallel to CD and AD = BC. Show that 4r2 = (AB)(CD). (Japan, Edo Period, 1603-1867)" ]

[ "as well, and $\\triangle BCE$ is equilateral, so $BC=CE=BE=7$. $[asy] size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot(\"W\",circumcenter(A,B,C),dir(180)); label(\"\\gamma\",gamma,dir(180)); [/asy]$ In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \\cdot \\cos \\angle AEF$ and $XF=AF \\cdot \\cos \\angle AFE$, we get \\begin{align}\\tag{1} EF = AE \\cdot \\cos \\angle AEF + AF \\cdot \\cos \\angle AFE. \\end{align} Note $\\angle AFE = \\angle ACE$, and the Law of Cosines in $\\triangle ACE$ gives $\\cos \\angle ACE = -\\tfrac 17$. Also, $\\angle AEF = \\angle DEF$, and $\\angle DFE = \\tfrac{\\pi}{2}$ ($DE$ is a diameter), so $\\cos \\angle AEF = \\tfrac{EF}{DE} = \\tfrac{8}{49}\\cdot EF$. Plugging in all our values into equation $(1)$, we get: $$EF = \\tfrac{64}{49} EF -\\tfrac{1}{7} AF \\quad \\Longrightarrow \\quad EF = \\tfrac{7}{15} AF.$$ The Law of Cosines in $\\triangle AEF$, with $EF=\\tfrac 7{15}AF$ and $\\cos\\angle AFE = -\\tfrac 17$ gives $$8^2 = AF^2 + \\tfrac{49}{225} AF^2 + \\tfrac 2{15} AF^2 = \\tfrac{225+49+30}{225}\\cdot AF^2$$ Thus $AF^2 = \\frac{900}{19}$. The answer is $\\boxed{919}$. ## Solution 2 Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\\angle MAD=\\angle DAF$.", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "= Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"\\angle C obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"\\angle B obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: 1. The region in which $\\angle B$ is obtuse is determined by construction. 2. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"A\", A, 1.5*S, linewidth(4.5)); dot(\"B\", B, 1.5*S, linewidth(4.5)); dot(\"D\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"C_2\", C2, 1.5*N, linewidth(4.5)); dot(\"C_1\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: • If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$,", "(30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly. Notice that because $BC = AD, 20+x = 6+DT_2 \\implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_2CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \\implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\\triangle ACF$; we have $26^2 +", "angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); [/asy]$ Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives$$DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.$$By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields $$[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.$$Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,$$\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.$$Therefore $$[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.$$The requested sum is $15+7+14 = 36$. ## Solution 10 (Official MAA 2) Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "\\begin{asydef} void ABCD(guide AB, guide BC, guide CD, guide DA ,pen linepen=deepblue+1.2bp ,pen areapen=red+1.4bp ,pen fillpen=palegreen){ real tA[], tB[], tC[], tD[]; pair A,B,C,D; tA=intersect(AB,DA); // tA[0] - AB time of intersection, t[1] - DA time of intersection tB=intersect(BC,AB); // tB[0] - BC time of intersection, t[1] - AB time of intersection tC=intersect(CD,BC); // tC[0] - CD time of intersection, t[1] - BC time of intersection tD=intersect(DA,CD); // tD[0] - DA time of intersection, t[1] - CD time of intersection A=point(AB,tA[0]); B=point(BC,tB[0]); C=point(CD,tC[0]); D=point(DA,tD[0]); guide area=buildcycle(AB,BC,CD,DA); draw(AB^^BC^^CD^^DA,linepen); filldraw(area,fillpen,areapen); label(\"$A$\",A,2N); label(\"$B$\",B,2W); label(\"$C$\",C,2S); label(\"$D$\",D,2E); dot(new pair[]{A,B,C,D},UnFill); } \\end{asydef} % \\begin{document} % \\begin{figure} \\captionsetup[subfigure]{justification=centering} \\centering \\begin{subfigure}{0.49\\textwidth} \\begin{asy} size(200); ABCD( arc(( 1,-1),2, 90,180) ,arc(( 1, 1),2,180,270) ,arc((-1, 1),2,-90, 0) ,arc((-1,-1),2, 0, 90) ); \\end{asy} % \\caption{} \\label{fig:1a} \\end{subfigure} % \\begin{subfigure}{0.49\\textwidth} \\begin{asy} size(200); ABCD( (1,2)..(-1.2,0.1)..(-1.7,-1.8) ,(-2,1)..(-1,1)..(-0.5,0)..(0.5,-0.8) ,arc((-1, 1),2,-90, 0) ,(2,-1)..(2,0)..(0.5,1)..(0,1.7) ,olive+0.4bp ,orange+1.3bp ,lightyellow ); \\end{asy} % \\caption{} \\label{fig:1b} \\end{subfigure} \\caption{} \\label{fig:1} \\end{figure} % \\end{document} % % Process : % % pdflatex xsect.tex % asy xsect-*.asy % pdflatex xsect.tex With PSTricks. Intersection is not necessary in this case as a simple logic can help you to find the angles at which the arcs start and stop. See the following explanation how to calculate the angles. \\documentclass[pstricks,border=12pt]{standalone} \\degrees[12] \\psset{dimen=monkey} \\begin{document} \\begin{pspicture}(-3,-3)(3,3) \\psframe(-3,-3)(3,3) \\pscustom*[linecolor=yellow] { \\foreach \\x/\\y/\\a in {-3/-3/1, 3/-3/4, 3/3/7, -3/3/10} } \\foreach \\x/\\y/\\a in {-3/-3/0,", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "# 2014 USAJMO Problems/Problem 6 ## Problem Let $ABC$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$. Let $M,N,P$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ and let $E,F$ be the tangency points of $\\gamma$ with $\\overline{CA}$ and $\\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\\Gamma$. (a) Prove that $I$ lies on ray $CV$. (b) Prove that line $XI$ bisects $\\overline{UV}$. ## Solution $[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label(\"A\", A, W); label(\"B\", B, dir(90)); label(\"C\", C, dir(0)); dot(I); label(\"I\", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label(\"\\Gamma\", dir(35), dir(35)); label(\"\\gamma\", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label(\"M\", M, NE); label(\"N\", N, SE); label(\"P\", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label(\"E\", E, SW); label(\"F\", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label(\"U\", U, S); label(\"V\",", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let" ]

[ "\\begin{asydef} void ABCD(guide AB, guide BC, guide CD, guide DA ,pen linepen=deepblue+1.2bp ,pen areapen=red+1.4bp ,pen fillpen=palegreen){ real tA[], tB[], tC[], tD[]; pair A,B,C,D; tA=intersect(AB,DA); // tA[0] - AB time of intersection, t[1] - DA time of intersection tB=intersect(BC,AB); // tB[0] - BC time of intersection, t[1] - AB time of intersection tC=intersect(CD,BC); // tC[0] - CD time of intersection, t[1] - BC time of intersection tD=intersect(DA,CD); // tD[0] - DA time of intersection, t[1] - CD time of intersection A=point(AB,tA[0]); B=point(BC,tB[0]); C=point(CD,tC[0]); D=point(DA,tD[0]); guide area=buildcycle(AB,BC,CD,DA); draw(AB^^BC^^CD^^DA,linepen); filldraw(area,fillpen,areapen); label(\"$A$\",A,2N); label(\"$B$\",B,2W); label(\"$C$\",C,2S); label(\"$D$\",D,2E); dot(new pair[]{A,B,C,D},UnFill); } \\end{asydef} % \\begin{document} % \\begin{figure} \\captionsetup[subfigure]{justification=centering} \\centering \\begin{subfigure}{0.49\\textwidth} \\begin{asy} size(200); ABCD( arc(( 1,-1),2, 90,180) ,arc(( 1, 1),2,180,270) ,arc((-1, 1),2,-90, 0) ,arc((-1,-1),2, 0, 90) ); \\end{asy} % \\caption{} \\label{fig:1a} \\end{subfigure} % \\begin{subfigure}{0.49\\textwidth} \\begin{asy} size(200); ABCD( (1,2)..(-1.2,0.1)..(-1.7,-1.8) ,(-2,1)..(-1,1)..(-0.5,0)..(0.5,-0.8) ,arc((-1, 1),2,-90, 0) ,(2,-1)..(2,0)..(0.5,1)..(0,1.7) ,olive+0.4bp ,orange+1.3bp ,lightyellow ); \\end{asy} % \\caption{} \\label{fig:1b} \\end{subfigure} \\caption{} \\label{fig:1} \\end{figure} % \\end{document} % % Process : % % pdflatex xsect.tex % asy xsect-*.asy % pdflatex xsect.tex With PSTricks. Intersection is not necessary in this case as a simple logic can help you to find the angles at which the arcs start and stop. See the following explanation how to calculate the angles. \\documentclass[pstricks,border=12pt]{standalone} \\degrees[12] \\psset{dimen=monkey} \\begin{document} \\begin{pspicture}(-3,-3)(3,3) \\psframe(-3,-3)(3,3) \\pscustom*[linecolor=yellow] { \\foreach \\x/\\y/\\a in {-3/-3/1, 3/-3/4, 3/3/7, -3/3/10} } \\foreach \\x/\\y/\\a in {-3/-3/0,", "= Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"\\angle C obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"\\angle B obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: 1. The region in which $\\angle B$ is obtuse is determined by construction. 2. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"A\", A, 1.5*S, linewidth(4.5)); dot(\"B\", B, 1.5*S, linewidth(4.5)); dot(\"D\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"C_2\", C2, 1.5*N, linewidth(4.5)); dot(\"C_1\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: • If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "dots and labels */ dot((-1,4),dotstyle); label(\"A\", (-0.92,4.2), NE * labelscalefactor); dot((-4.08,3.78),dotstyle); label(\"B\", (-4.42,4), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label(\"C\", (-3.4,-3.94), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label(\"D\", (1.8,-0.4), NE * labelscalefactor); dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle); label(\"O\", (-1.34,1.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ (Diagram not to scale) Since $AO \\cdot OC = BO \\cdot OD$, $ABCD$ is cyclic through power of a point. From the given information, we see that $\\triangle{AOB}\\sim \\triangle{DOC}$ and $\\triangle{BOC} \\sim \\triangle{AOD}$. Hence, we can find $CD=\\frac{9}{2}$ and $AD=2 \\cdot BC$. Letting $BC$ be $x$, we can use Ptolemy's to get $$6 \\cdot \\frac{9}{2} + 2x^2=10 \\cdot 11 \\implies x=\\sqrt{\\frac{83}{2}}$$ Since we are solving for $AD=2x=2\\cdot\\sqrt{\\frac{83}{2}}=\\sqrt{4\\cdot\\frac{83}{2}} = \\boxed{\\textbf{(E)}~\\sqrt{166}}$ - PhunsukhWangdu", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "$AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry. We notice that $\\triangle ACD \\cong \\triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\\triangle AED \\sim \\triangle ADC,$ with $\\angle EAD \\cong \\angle DAC$ and $\\angle CDA \\cong \\angle DEA.$ This similarity means that we can create a proportion: $\\frac{AD}{AB}=\\frac{DE}{CD}.$ We plug in $AD=\\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\\frac{8}{17} \\cdot 15= \\boxed{\\textbf{(B) }\\frac{120}{17}}.$ (By the way, we could also use $\\triangle DEC \\sim \\triangle ADC.$) ## Solution 4: Inscribed Circle $[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(\"B\",B,SW); label(\"D\",D,SE); label(\"A\",A,N); label(\"M\",M,S); label(\"C\",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$ Call this triangle $ABD$ and let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of triangle $ABD$ is $\\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=\\boxed{288}$. ### Solution 2 $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy]$ Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$. ### Solution 3 We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "# 2005 AIME II Problems/Problem 14 ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of", "# 2014 USAJMO Problems/Problem 6 ## Problem Let $ABC$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$. Let $M,N,P$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ and let $E,F$ be the tangency points of $\\gamma$ with $\\overline{CA}$ and $\\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\\Gamma$. (a) Prove that $I$ lies on ray $CV$. (b) Prove that line $XI$ bisects $\\overline{UV}$. ## Solution $[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label(\"A\", A, W); label(\"B\", B, dir(90)); label(\"C\", C, dir(0)); dot(I); label(\"I\", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label(\"\\Gamma\", dir(35), dir(35)); label(\"\\gamma\", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label(\"M\", M, NE); label(\"N\", N, SE); label(\"P\", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label(\"E\", E, SW); label(\"F\", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label(\"U\", U, S); label(\"V\",", "# Problem #2091 2091 Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way? $[asy] size(200); defaultpen(0.9); real r = 5/dir(54).x, h = 5 tan(54*pi/180); pair A = (5,0), B = A+10*dir(72), C = (0,r+h), E = (-5,0), D = E+10*dir(108); draw(A--B--C--D--E--cycle); label(\"$A$\",A+(0,-0.5),SSE); label(\"$B$\",B+(0.5,0),ENE); label(\"$C$\",C+(0,0.5),N); label(\"$D$\",D+(-0.5,0),WNW); label(\"$E$\",E+(0,-0.5),SW); // real l = 5*sqrt(3); pair ab = (h+l)*dir(72), bc = (h+l)*dir(54); pair AB = (ab.y, h-ab.x), BC = (bc.x,h+bc.y), CD = (-bc.x,h+bc.y), DE = (-ab.y, h-ab.x), EA = (0,-l); draw(A--AB--B^^B--BC--C^^C--CD--D^^D--DE--E^^E--EA--A, dashed); // dot(A); dot(B); dot(C); dot(D); dot(E); dot(AB); dot(BC); dot(CD); dot(DE); dot(EA); [/asy]$ $\\text {(A) } 1 \\qquad \\text {(B) } 2 \\qquad \\text {(C) } 3 \\qquad \\text {(D) } 4 \\qquad \\text {(E) } 5$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved." ]

[ "comes out to be $$e^{j 2\\pi k/100}$$", "\\prod_{i=1}^{k_j} h_{t_{i,j}}=\\prod_{j=1}^m x_j^{-1} g^{k_j} x_j. \\ \\Box$", "$\\prod_q S^{2k_q-1}$ where the numbers $2k_q-1$ are the dimensions of the generators.", "= k; //since par[k][j][0] = k } } }", "\\prod_{k=1}^n k^{n+1} = (n!)^{n+1}$ Thus $(\\prod_{k=1}^n k!)^2 \\sim (2\\pi)^n \\sqrt{n!} \\frac{(n!)^{n+1}}{e^{n(n+1)/2}}$ Then we get, applying Stirling's approximation to $\\sqrt{n!}$ and $(n!)^{n+1}$: $(\\prod_{k=1}^n k!)^2 \\sim (2\\pi)^{2n+3/2}n^{3/4 + n/2}e^{-n(n+1)/2-n/2-n(n+1)} n^{n(n+1)}$ Thus $\\prod_{k=1}^n k! \\sim (2\\pi)^{n+3/4} e^{-3n^2/4 - n} n^{n^2/2 + 3n/4 + 3/8}$", "| \\; \\pi(k+1) = k+1 \\} \\cong \\mathcal{Sym}(k) \\subset \\mathcal{Sym}(k+1)## Last edited:", "{j} -12 \\hat {k}) + 13 = 0, \\$ then find the value of p. To see more, click for all the questions in this category.", "j }$$ + 15$$\\hat { k }$$", "the equation is defined as pi,j,k=12fj+12fk+94gi−J=(0,0,0,…,j12,…,k12,…,0n,−1,…,i54,…", "4 \\[Pi]}], {k, 1, 5}] - This is exactly I was looking for, Thank you very much! – Joseph Dec 21 '12 at 20:10", "to obtain those for (12^), 2 and 3 in the results for (12^1-) to obtain those for (126), 3 and k in the results for (126) to obtain those for (12??). The above discussion aiid Theorem 11 prove the following Theorems 12 and I3. Theorem 12 . Let D = P-j^pgP^Pli whe.^e p-j_ < P2 < P^ < P2|. ^-^^ distinct odd primes. Let t-j_ + u-j^V^ tie the fundamental solu tion of (1), t^ Du^ = 1, Inhere = 2\\j^ + 1. If ^ y^, (-P^iPj) = ^-PilPk^ ^ (-P^iPj) = +1, (21g) Pi s 3 (mod ^) , and either at least t\\io of (p^|pj^), (p^ipj), ^Pj^lPj) = -1, (219) and at least one of Pj , r>^, pj = 1 (mod H-) , PAGE 127 121 or (220) or (221) or (222) or (223) = s 3Pj = 1 (mod H-) , (p^lPj) =+1. and (p^ll^) = -(PjlPj) 3Pj S Pi^ S 3Pj =1 (mod ^) , (PjIPj) = +1, and (PjIPj^) = -(p^Ipj) 3Pj = 3Pk 5 Pj =1 (mod , (Pjlpi^) = +1, e-nd (PjIpj) = -(p^iPj). P, = P, = P. 3 3 (mod k) and (Pjlp^) = -(PjlPj) = ^Pk'Pj^' then either equation (122), {12k), (126), or (12S) has a solution in Integers x, y according as i =1, 2, 3,", "p\\mid (10^b-10^a)$ This can be established using Fermat's Little Theorem,too. -", ", we obtain P(Wjsϕ(Nj−1(x))+bjs≤0|~Acj−1,x)=12. Therefore,", "\\pi + 21 {\\pi}^{3}\\right)$ $= 45177.2 J$", "1 * = SUk Sj Ujk, : p :n), :n)= Figure 2.3. The direction , in which the elements of G( U) are computed, when the number of processors are greater than half of the total blocks. Suppose p Then 2NB .2 we use Proc(1) to compute GNBj, - 1, , - 1 in the increasing order of the column index j, and the other p 1 processors to compute the remaining elements of G(S, U) in the decreasing order of the row index i as we explain below . We use Proc(2) to compute Gn, Proc(3) to compute Gi2, NB 1,...,3. Continuing the assignment in this way, we use Proc(p) to compute Gip-1, -NB - 1, ..., p. NB - then will continue this ascending order of assignments to those processors except Proc(1). That means , Proc(2) to compute Gip, = NB ,..,p+ 1, Proc(3) to compute = NB- and in 1, ... this order if necessary Proc(p) to compute Gi2p-2, NB-1,..., p- 1. Pictorially it can be presented by arrow heads, where ,...,2, Gip+l, ,p+ -1p Figure 2.4. The direction, in which the elements of G(S, U) are computed, when the number of processors are less than half of the total blocks. Next, suppose p [NBt + 1. 2 JI This time we will assign each", "(20), we see E(k)+k∑i=12s[(r−3)(i+r−2)+1]r−1", "compute: p_1^0 p_2^0=1 p_1^0 p_2^1=3 p_1^1 p_2^0=2 p_1^1 p_2^1=6 p_1^2 p_2^0=4 p_1^2 p_2^1=12 p_1^3 p_2^0=8 p_1^3 p_2^1=24 i.e. all (k_1 + 1) \\times (k_2 + 1) possible divisors. 2 Likes Amazing, thanks a lot man… U save a day… 1 Like", "\\int_0^5 k6\\pi t \\,dt\\\\ &= \\left.k3 \\pi t^2 \\right|_0^5 = 3\\pi (25)k = 75\\pi k \\end{align*} #### Notation systems More information on notation systems", "bad, I worked out that $1_{E}(ax)=1_{a^{-1}E}(x)$ for $\\mathbb{R}$ and then we extend this to $\\mathbb{R}^d$ by using the product measure, since if $E=E_1\\times\\cdots\\times E_d$, then $m(a^{-1}E)=\\prod^{d}_{j=1}m(a_{j}^{-1}E_j)=\\prod^{d}_{j=1}a_{j}^{-1}m(E)$. Is that right? – G the Stackman Mar 16 at 21:50", "what does m equal to find k(sub)e?" ]

[ "the polynomial by $(x-x^*_j)$ and restart Newton. Therefore, this methods calculates the root $x_{k+1}$ if you already got the roots $x_1...x_k$. So I dont think the sum in your formula iterates over all roots, but rather goes from $i=1..k$ if we are calculating the root $x_{k+1}$. Thats why I would also suggest to add another index $k$ for the number of the root. So in my notation, $i$ describes the iteration of Newton and $k$ the root we want to find. Assuming we already know the first $k$ roots. We now want to find the roots of the polynomial \\begin{align} P^{k+1}(x) = P(x) \\cdot \\frac{1}{\\prod_{j=1}^{k} (x-x^*_j)} \\end{align} So we apply Newton to this polynomial: \\begin{align} x^{k+1}_{i+1} = x^{k+1}_{i} - \\frac{P^{k+1}(x^{k+1}_{i})}{(P^{k+1}(x^{k+1}_{i}))'} \\end{align} We obtain $P^{k+1}(x)'$ by chainrule: \\begin{align} P^{k+1}(x)' &=(P(x) \\cdot \\frac{1}{\\prod_{j=1}^{k} (x-x^*_j)})' \\\\ &=P'(x)\\cdot \\frac{1}{\\prod_{j=1}^{k} (x-x^*_j)} +(P(x)\\sum_{m=1}^k \\{ \\frac{-1}{(x-x^*_m)}\\prod_{j=1}^k\\frac{1}{ (x-x^*_j)} \\}) \\\\ &= \\prod_{j=1}^k\\frac{1}{ (x-x^*_j)}\\{P'(x) -P(x) \\sum_{m=1}^k\\frac{1}{ (x-x^*_m)}\\} \\end{align} Using this and the definition of $P^{k+1}$ we get \\begin{align} \\frac{P^{k+1}(x)}{(P^{k+1}(x))'} &= \\frac{P(x)}{P'(x)-P(x)\\sum_{m=1}^k \\frac{1}{x-x^*_m}} \\\\ &= \\frac{\\frac{P(x)}{P'(x)}}{1-\\frac{P(x)}{P'(x)}\\sum_{m=1}^k \\frac{1}{x-x^*_m}} \\end{align} So if we finally write the complete Newton iteration we get: \\begin{align} x^{k+1}_{i+1} = x^{k+1}_i -{\\frac{P(x^{k+1}_i)}{P'(x^{k+1}_i)}}\\frac{1}{1-\\frac{P(x^{k+1}_i)}{P'(x^{k+1}_i)}\\sum_{m=1}^k \\frac{1}{x^{k+1}_i-x^*_m}} \\end{align} Which differs from your formula. I am not sure whether or where I made a mistake, but I think this gives an overview over the background of the Newton Horners Method. -", "roots in $(0,1)$. If not, then there is a polynomial $Q(x)$ of degree $< n$ that has the same sign as $P(x)$ for all $x\\in[0,1]$. Then for some $\\epsilon>0$, $\\int_0^1|P(x)-\\epsilon Q(x)|\\mathrm{d}x<\\int_0^1|P(x)|\\mathrm{d}x$. Let $0< x_1< x_2<\\dots< x_n<1$ be the roots of $P(x)$. By $(2)$ we have that $$\\frac{1}{k+1}\\left[(x_1^{k+1}-0)-(x_2^{k+1}-x_1^{k+1})+(x_3^{k+1}-x_2^{k+1})-\\dots+(-1)^{n-1}(1-x_n^{k+1})\\right]=0$$ for each $k=0\\dots n-1$, which is equivalent to $$x_n^k-x_{n-1}^k+x_{n-2}^k-\\dots+(-1)^{n-1}x_1^k=\\tfrac12\\tag{3}$$ for each $k=1\\dots n$. Equation $(3)$ has roots $$x_j=\\sin^2\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\tag{4}$$ for $j=1\\dots n$. That is, since $\\displaystyle \\operatorname{Re}\\left(\\frac{1}{e^{ix}+1}\\right)=\\frac12$, \\begin{align} &\\sum_{j=1}^n(-1)^{n-j}\\sin^{2k}\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\\\ &=(-\\tfrac14)^k\\sum_{j=1}^n(-1)^{n-j}\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^m\\exp\\left(i\\pi\\frac{(k-m)j}{n+1}\\right)\\\\ &=(-\\tfrac14)^k(-1)^{n}\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^m\\left(\\frac{(-1)^{k-m+n}-1}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}-1\\right)\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^{k-m+n}\\frac{(-1)^{k-m+n}-1}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac{1-(-1)^{k-m+n}}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac12\\left(1-(-1)^{k-m+n}\\right)\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac12\\\\ &=\\frac12\\tag{5} \\end{align} Thus, the minimal monic polynomial is $$P(x)=\\prod_{j=1}^n\\left(x-\\sin^2\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\right)\\tag{6}$$ -", "-3 (with multiplicity 3), \\begin{align*}-1, \\sqrt{3}i, -\\sqrt{3}i\\end{align*} 3. 5 (with multiplicity 2), -1 (with multiplicity 2), \\begin{align*}2i, -2i\\end{align*} 4. \\begin{align*}i, -i, \\sqrt{2}i, -\\sqrt{2}i\\end{align*} For each polynomial, factor into its linear factorization and state all of its roots. 5. \\begin{align*}f(x)=4x^2+x-3\\end{align*} 6. \\begin{align*}g(x)=x^4-1\\end{align*} 7. \\begin{align*}h(x)=x^3+5x^2\\end{align*} 8. \\begin{align*}j(x)=x^4-7x^3+17x^2-17x+6\\end{align*} 9. \\begin{align*}k(x)=x^3-64\\end{align*} 10. \\begin{align*}m(x)=x^6-2x^4-9x^2+18\\end{align*} 11. \\begin{align*}n(x)=2x^3-8x^2-3x-12\\end{align*} 12. \\begin{align*}p(x)=-x^3+2x^2-25x+50\\end{align*} 13. How can you tell the number of roots that a polynomial has from its equation? 14. Explain the meaning of the term “multiplicity”. 15. A polynomial with real coefficients has one root that is \\begin{align*}\\sqrt{3}i \\end{align*}. What other root(s) must the polynomial have? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes", "out, though, this firs step is all one needs to do: primitive roots mod $p^2$ are always primitive roots mod $p^m$. Proposition: Any primitive root mod $p^2$ is a primitive root mod $p^m$, where p is an odd prime. Proof: Let r be a primitive root mod $p^2$, and write n for $\\mbox{ord}_{p^m}(r)$. As before, we know that (10) \\begin{align} r^{\\phi(p^m)} \\equiv r^{p^{m-1}(p-1}} \\equiv 1 \\mod{p^m} \\LongRightarrow n \\mid p^{m-1}(p-1). \\end{align} Moreover we have that (11) \\begin{align} r^{n} \\equiv 1 \\mod{p^m} \\Rightarrow r^n \\equiv 1 \\mod{p^2} \\Rightarrow \\mbox{ord}_{p^2}(r) \\mid n. \\end{align} Since we know that $\\mbox{ord}_{p^2}(r) = p(p-1)$ (since r is a primitive root mod $p^2$), Equations (10) and (11) together imply (12) \\begin{align} n = \\mbox{ord}_{p^m}(r) = p^k(p-1) \\quad \\mbox{for some }1 \\leq k \\leq m-1. \\end{align} Now we would like to show that $n = p^{m-1}(p-1)$, so we need to show that no choice of $k<m-1$ is possible in the above equation. To do this, we'll show that (13) \\begin{align} r^{p^{m-2}(p-1)} \\not\\equiv 1 \\mod{p^m} \\end{align} in the next lemma. In particular, this equation will force $\\mbox{ord}_{p^m}(r) \\neq p^k(p-1)$ for $k<m-1$, and hence we will be forced into the condition $\\mbox{ord}_{p^m}(r) = p^{m-1}(p-1)$; i.e., r will be a primitive root. $\\square$ Lemma: Suppose that r is a primitive root mod $p^2$. Then $r^{p^{m-2}(p-1)} \\not\\equiv 1 \\mod{p^m}$. Proof:", ".\\end{align*} ## 4.6 Cyclotomic Polynomials $$\\deg \\Phi_n(x) = \\phi(n)$$ for $$\\phi$$ the totient function. (Click to expand) $$\\deg \\Phi_n(x)$$ is the number of $$n$$th primitive roots, which is the number of numbers less than and coprime to $$n$$. Computing $$\\Phi_n$$: 1. \\begin{align*} \\Phi_{n}(z)=\\prod_{d | n, d>0}\\left(z^{d}-1\\right)^{\\mu\\left(\\frac{n}{d}\\right)} \\end{align*} where \\begin{align*} \\mu(n) \\equiv\\left\\{ \\begin{array}{ll}{0} & {\\text { if } n \\text { has one or more repeated prime factors }} \\\\ {1} & {\\text { if } n=1} \\\\ {(-1)^{k}} & {\\text { if } n \\text { is a product of } k \\text { distinct primes, }}\\end{array}\\right. \\end{align*} 2. \\begin{align*} x^{n}-1=\\prod_{d | n} \\Phi_{d}(x) \\implies \\Phi_n(x) = \\frac{x^n-1}{\\prod_{d | n \\atop d < n} \\Phi_{d}(x)}, \\end{align*} so just use polynomial long division. \\begin{align*} \\Phi_{p}(x) &= x^{p-1}+x^{p-2}+\\cdots+x+1 \\\\ \\Phi_{2 p}(x) &= x^{p-1}-x^{p-2}+\\cdots-x+1 \\end{align*} \\begin{align*} k\\divides n \\implies \\Phi_{n k}(x)=\\Phi_{n}\\left(x^{k}\\right) \\end{align*} Every finite separable extension is simple. $${\\mathbb{GF}}(p^n)$$ is a simple extension over $${\\mathbb{F}}_p$$. # 5 Modules ## 5.1 General Modules A module is simple iff it has no nontrivial proper submodules. A free module is a module with a basis (i.e. a spanning, linearly independent set). A module $$M$$ is torsion-free if and only if for every $$x\\in M$$, $$mx = 0_M \\implies m=0_M$$. I.e. the only torsion element of $$M$$ is zero itself. Free $$\\implies$$ torsionfree. $${\\mathbb{Z}}/(6)$$ is", "$0,1,2,\\dots$ and $(2, -n\\pi)$ where $n=1,3,5,\\dots$. 37 ##### Web Bonus Problems / Re: Phaseportrait « on: March 28, 2018, 01:50:57 PM » $\\text{My attempt:}$ \\begin{align}x'&=\\sin(x)\\cos(y)=F(x,y)\\\\y'&=-\\cos(x)\\sin(y)=G(x,y)\\end{align} Let \\begin{align}\\cases{F(x,y)=\\sin(x)\\cos(y)=0\\\\G(x,y)=-\\cos(x)\\sin(y)=0}\\end{align} $$\\implies\\cases{x=k\\pi,y=k\\pi\\\\x={\\pi\\over2}+k\\pi,y={\\pi\\over2}+k\\pi}$$ where $k$ can only be $-1,0,1$, since we have $x\\in(-4,4)$ and $y\\in(-4,4)$, $\\pi\\cong 3.14159265359$. $\\\\$ Thus we have the following critical points:$\\\\$ Case#1: $(0,0);(\\pi,0);(-\\pi,0);(0,\\pi);(0,-\\pi);(\\pi,\\pi);(-\\pi,\\pi);(-\\pi,-\\pi);(\\pi,-\\pi)$. $\\\\$ Case#2: $({\\pi\\over2},{\\pi\\over2});(-{\\pi\\over2},-{\\pi\\over2});({\\pi\\over2},-{\\pi\\over2});(-{\\pi\\over2},{\\pi\\over2})$. $$J=\\begin{pmatrix}F_x(x,y)&F_y(x,y)\\\\G_x(x,y)&G_x(x,y)\\end{pmatrix}=\\begin{pmatrix}\\cos(x)\\cos(y)&-\\sin(x)\\sin(y)\\\\\\sin(x)\\sin(y)&-\\cos(x)\\cos(y)\\end{pmatrix}$$ When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}$ or $\\begin{pmatrix}-1&0\\\\0&0\\end{pmatrix}$, wtih eigenvalues $\\lambda=\\pm1$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\0\\end{pmatrix}$ or $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#1 are all $\\text{Saddle Points}$. $\\\\$ When $(x,y)=$ critical points in Case#2: we get matrices $\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}$ or $\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}$ with eigenvalues $\\lambda=\\pm i$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\i\\end{pmatrix}$ or $\\begin{pmatrix}1\\\\-i\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#2 are all $\\text{Center}$. $\\\\$ Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that. 38 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018, 09:11:15 AM » Thanks for the correction. I made a mistake copying the result integral. I got the same result. I think there is a calculation error when computing the initial condition, hence the $c_1,c_2$ values. (if I am not mistaken ) $$c_1+c_2=-2-15\\ln2$$ and $$-2c_2=3-15\\ln2$$ Therefore, $$c_1=-{1\\over2}-{45\\over2}\\ln2$$ $$c_2=-{3\\over2}+{15\\over2}\\ln2$$ 39 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018,", "four solutions of the given equation are ($$0,7$$), ($$1,5$$), ($$2, 3$$) and ($$3,1$$). (ii) $$\\quad \\pi x+y=9$$ Given, Linear Equation, $$\\pi x+y=9$$ $$y=9-\\pi x \\qquad \\dots \\dots \\text{Equation (1)}$$ Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For \\begin{align},x = 0,y = 9 - \\pi \\left( 0 \\right) \\Rightarrow y=9.\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (0,9 )\\end{align} For \\begin{align},x = 1,y = 9 - \\pi \\left( 1 \\right) \\Rightarrow 9 - \\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (1,9 - \\pi )\\end{align} For \\begin{align},x = 2,y = 9 - \\pi \\left( 2 \\right) \\Rightarrow 9 - 2\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (2,9 - 2\\pi )\\end{align} For \\begin{align},x = 3,y = 9 - \\pi \\left( 3 \\right) \\Rightarrow 9 - 3\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (3,9 - 3\\pi )\\end{align} Therefore, the four solutions of the given equation are $$(0,9),(1,9-\\pi),(2,9-2 \\pi),(3,9-3 \\pi)$$. (iii) $$x = 4y$$ Given, Linear Equation, $$x = 4y$$ \\begin{align}\\therefore y = \\frac{x}{4} \\qquad \\dots \\dots \\text{Equation (1)}\\end{align} Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For $$x = 0$$, \\begin{align}y =\\frac{0}{4} = 0\\end{align}. Hence, we get $$(x, y)$$ $$= (0, 0)$$ For $$x = 1$$, \\begin{align}y =\\frac{1}{4}\\end{align} Hence, we get", "and $$\\ P\\left(\\left.\\xi_j=0\\,\\right|N=n\\right)=1-p\\$$—that is, if $$\\ i_1,i_2, \\dots, i_n\\$$ is a sequence of ones and zeros containing $$\\ m\\$$ ones and $$\\ n-m\\$$ zeros, then \\begin{align} P\\left(\\xi_1=i_1, \\xi_2=i_2, \\dots, \\xi_n=i_n |\\,N=n\\right)&=\\prod_{j=1}^nP\\left(\\xi_j=i_j|\\,N=n\\right)\\\\ &=p^m(1-p)^{n-m} \\end{align} Therefore \\begin{align} P \\left(\\xi_1+ \\xi_2+ \\dots +\\xi_n=m \\,|\\,N=n\\right)&=\\sum_{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n:\\\\ i_1+i_2+ \\dots+i_n=m} p^m(1-p)^{n-m}\\\\ &={n\\choose m} p^m(1-p)^{n-m}\\ , \\end{align} because $$\\ \\left|\\left\\{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n\\,|\\, i_1+i_2+ \\dots+i_n=m\\right\\}\\right|= {n\\choose m}\\$$.", "= (3, 1)$$ Therefore, the four solutions of the given equation are ($$0,7$$), ($$1,5$$), ($$2, 3$$) and ($$3,1$$). (ii) $$\\quad \\pi x+y=9$$ Given, Linear Equation, $$\\pi x+y=9$$ $$y=9-\\pi x \\qquad \\dots \\dots \\text{Equation (1)}$$ Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For \\begin{align},x = 0,y = 9 - \\pi \\left( 0 \\right) \\Rightarrow y=9.\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (0,9 )\\end{align} For \\begin{align},x = 1,y = 9 - \\pi \\left( 1 \\right) \\Rightarrow 9 - \\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (1,9 - \\pi )\\end{align} For \\begin{align},x = 2,y = 9 - \\pi \\left( 2 \\right) \\Rightarrow 9 - 2\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (2,9 - 2\\pi )\\end{align} For \\begin{align},x = 3,y = 9 - \\pi \\left( 3 \\right) \\Rightarrow 9 - 3\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (3,9 - 3\\pi )\\end{align} Therefore, the four solutions of the given equation are $$(0,9),(1,9-\\pi),(2,9-2 \\pi),(3,9-3 \\pi)$$. (iii) $$x = 4y$$ Given, Linear Equation, $$x = 4y$$ \\begin{align}\\therefore y = \\frac{x}{4} \\qquad \\dots \\dots \\text{Equation (1)}\\end{align} Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For $$x = 0$$, \\begin{align}y =\\frac{0}{4} = 0\\end{align}. Hence, we get $$(x, y)$$ $$= (0, 0)$$ For $$x = 1$$,", "\\pi=\\left\\{\\pi_{t}\\right\\}_{t=1}^{T^{\\prime}} , where label π t ∈ V \\pi_{t} \\in \\mathcal{V} . The probability of alignment path π \\pi given the input sequence is defined as follows, p ( π ∣ o ) = ∏ t = 1 T ′ p ( π t ∣ o ) = ∏ t = 1 T ′ y t , π t p(\\pi \\mid \\mathbf{o})=\\prod_{t=1}^{T^{\\prime}} p\\left(\\pi_{t} \\mid \\mathbf{o}\\right)=\\prod_{t=1}^{T^{\\prime}} y_{t, \\pi_{t}} . (10) CTC 创建带有空白标签 “ − − ”的扩展词汇表 V \\mathcal{V} 。这里的 V = V origin ∪ { − } \\mathcal{V}=\\mathcal{V}_{\\text {origin }} \\cup\\{-\\} 。空白的标签代表 stillness and transitions，没有确切的意义。将输入序列的 alignment path 表示为 π = { π t } t = 1 T ′ \\pi=\\left\\{\\pi_{t}\\right\\}_{t=1}^{T^{\\prime}} ，其中标签 π t ∈ V \\pi_{t} \\in \\mathcal{V} 。给定输入序列， alignment path π \\pi 的概率定义为: p ( π ∣ o ) = ∏ t = 1 T ′ p ( π t ∣ o ) = ∏ t = 1 T ′ y t , π t p(\\pi \\mid \\mathbf{o})=\\prod_{t=1}^{T^{\\prime}} p\\left(\\pi_{t} \\mid \\mathbf{o}\\right)=\\prod_{t=1}^{T^{\\prime}} y_{t, \\pi_{t}} . (10) Define a many-to-one mapping operation B \\mathcal{B} which removes all blanks and repeated words in the alignment path (e.g., B ( I I − m i s s − − y o u ) = I , \\mathcal{B}(I I-m i s s--y o u)=I, miss, y o u y o u ). In", "show. This might help but I have no idea. – Per Alexandersson Jun 18 '15 at 1:11 • also note that the final coeficients (after multiplications) should be divisable by $p$ or be $0 \\mod p$ (except few cases, which should be the diagonals, why? because they would be multiplied by similar coeficients) – Nikos M. Jun 24 '15 at 5:06 • \"is a polynomial in $p$ with integer coefficients\": rather, with coefficients whose denominators are coprime to $p$. – darij grinberg Aug 16 '15 at 19:15 Notice that ${k+j\\choose j} \\equiv (-1)^j{p-k-1 \\choose j} \\pmod p$ for $0\\le k+j \\le p-1$ and ${k+j\\choose j} \\equiv 0 \\pmod p$ for $k+j\\ge p$, $j\\le p-1$ (since $j!$ is coprime to $p$). Therefore, \\begin{align} \\sum_{j = 0}^{p-1} {k+j\\choose j} z^j = \\sum_{j = 0}^{p-k-1} (-1)^j{p-k-1 \\choose j} z^j = (1-z)^{p-k-1} \\tag{1} \\end{align} in $\\mathbb F_p(z)$. Then for $x\\in \\mathbb{F}_p$, $x\\neq 1$, using that $(1-x)^{p-1} = 1$ in $\\mathbb F_p$, $$\\sum_{j=0}^{p-1} {k+j\\choose j} x^j = \\frac{1}{(1-x)^{k}},\\ k = 0,\\dots,p-1.\\tag{2}$$ Consider now the vectors $e(x) = (1,x,x^2,\\dots,x^{p-1})^{\\top}$, $x\\in \\mathbb{F}_p$. For $x\\notin \\{0,1\\}$, from (2) we have $$e(x) \\overset{A}{\\longrightarrow} e((1-x)^{-1}) \\overset{A}{\\longrightarrow} e(-(1-x) x^{-1})\\overset{A}{\\longrightarrow} e(x).\\tag{3}$$ Further, $$e(0) \\overset{A}{\\longrightarrow} e(1) \\overset{A}{\\longrightarrow} (0,\\dots,0,1)^\\top \\overset{A}{\\longrightarrow} e(0),$$ where the middle arrow follows from plugging $z=1$ to (1). The vectors $e(x)$, $x\\in \\mathbb F_p$, are linearly independent. Therefore, $A^3", "$$m =\\prod_{p \\in P} p^{m_i} =\\prod_{p \\in P} p^{2(u_i-v_i)} =\\left(\\prod_{p \\in P} p^{u_i-v_i}\\right)^2$$ so $$m$$ is a perfect square. Therefore the square root of an integer is rational only if it is a square. I'll now try to generalize this to $$k$$-th roots, with as much cut-and-paste as possible. Suppose $$\\sqrt[k]{m} =\\dfrac{u}{v}$$ where $$m = \\prod_{p \\in P} p^{m_i}, u = \\prod_{p \\in P} p^{u_i}, v = \\prod_{p \\in P} p^{v_i}$$. Then $$\\prod_{p \\in P} p^{m_i} =\\dfrac{u^k}{v^k} =\\dfrac{ \\prod_{p \\in P} p^{ku_i}}{\\prod_{p \\in P} p^{kv_i}}$$ so $$\\prod_{p \\in P} p^{m_i}\\prod_{p \\in P} p^{kv_i} =\\prod_{p \\in P} p^{ku_i}$$ or $$\\prod_{p \\in P} p^{m_i+kv_i} =\\prod_{p \\in P} p^{ku_i}$$. By unique factorization, $$m_i+kv_i =ku_i$$, so $$m_i =ku_i-kv_i =k(u_i-v_i)$$. Therefore $$m =\\prod_{p \\in P} p^{m_i} =\\prod_{p \\in P} p^{k(u_i-v_i)} =\\left(\\prod_{p \\in P} p^{u_i-v_i}\\right)^k$$ so $$m$$ is a perfect $$k$$-th power. Therefore the $$k$$-th root of an integer is rational only if it is a $$k$$-th power. When you get $$2p^2 = 3q^2$$ that means that $$2|q^2$$ so $$2|q$$ and that $$3|p^2$$ so $$3|p$$. Then if you replace $$p = 3p'$$ for some integer $$p'$$ and $$q=2q'$$ for some integer $$q'$$ you get $$2(3p')^2 = 3(2q')^2$$ $$18p'^2 = 12q'^2$$ $$3p'^3 = 2q'^2$$. Can you finish from there? This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $$p$$ is", "a_3 = ka_4, a_0 = ka_1,\\,$ then −k is a root and we can factor out $x+k\\,$, \\begin{align} Q(x) &=a_4 x^4 + k a_4 x^3 + a_1 x + ka_1 \\&=(x + k)a_4x^3 + (x + k)a_1 \\&=(x + k)(a_4x^3 + a_1). \\end{align} And if $a_0=0, a_3=ka_4, a_1 = ka_2,\\,$ then both $0\\,$ and $-k\\,$ are roots Now we can factor out $x(x + k)\\,$ and get \\begin{align} Q(x) &=x(a_4x^3+ka_4x^2+a_2x+ka_2) \\&=x(x+k)(a_4x^2+a_2). \\end{align} To get Q 's other roots, we simply solve the quadratic factor. Biquadratic equations If $a_3=a_1=0,\\,$ then $Q(x) = a_4x^4+a_2x^2+a_0.\\,\\!$ We call such a polynomial a biquadratic, which is easy to solve. Let $z=x^2.\\,$ Then Q becomes a quadratic q in z, $q(z) = a_4z^2+a_2z+a_0.\\,\\!$ Let $z_+\\,$ and $z_-\\,$ be the roots of q. Then the roots of our quartic Q are \\begin{align} x_1&=+\\sqrt{z_+}, \\x_2&=-\\sqrt{z_+}, \\x_3&=+\\sqrt{z_-}, \\x_4&=-\\sqrt{z_-}. \\end{align} Quasi-symmetric equations $a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2=0 \\,$ Steps: 1) Divide by x 2. 2) Use variable change z = x + m/x. The general case, along Ferrari's lines To begin, the quartic must first be converted to a depressed quartic. Converting to a depressed quartic Let $A x^4 + B x^3 + C x^2 + D x + E = 0 \\qquad\\qquad(1')$ be the general quartic equation which it is desired to solve. Divide both sides by A,", "of $$1/k^n$$. Question: is the above sequence of polynomials well known? Or is it related to a well-known sequence of polynomials via a scaling and/or shifting? (Such as: making the leading term monic, or clearing denominators, or zeroing the constant term, or whatever.) EDIT: We can get some nicer lists if we clear denominators in the above. Doing so, we get \\begin{align} P_1(x) &= \\pi (x) \\\\ P_2(x) &= \\pi^2(12x^2 - 1) \\\\ P_3(x) &= \\pi^3(8x^3 - 2x) \\\\ P_4(x) &= \\pi^4(240x^4 - 120x^2 + 7) \\\\ P_5(x) &= \\pi^5(96x^5 - 80x^3 + 14x) \\\\ P_6(x) &= \\pi^6(1344x^6 - 1680x^4 + 588x^2 - 31) \\\\ P_7(x) &= \\pi^7(384x^7 - 672x^5 + 392x^3 - 62x) \\\\ P_8(x) &= \\pi^8(3840x^8 - 8960x^6 + 7840x^4 - 2480x^2 + 127) \\\\ \\end{align} It also so happens that if $$n$$ is odd, the GCD of the coefficients of $$P(n)$$ is 2, except for $$P(n)$$. If we divide by the GCD, we get \\begin{align} P_1(x) &= \\pi(x) \\\\ P_2(x) &= \\pi^2(12x^2 - 1) \\\\ P_3(x) &= \\pi^3(4x^3 - x) \\\\ P_4(x) &= \\pi^4(240x^4 - 120x^2 + 7) \\\\ P_5(x) &= \\pi^5(48x^5 - 40x^3 + 7x) \\\\ P_6(x) &= \\pi^6(1344x^6 - 1680x^4 + 588x^2 - 31) \\\\ P_7(x) &= \\pi^7(192x^7 - 336x^5 + 196x^3 - 31x) \\\\ P_8(x) &= \\pi^8(3840x^8 - 8960x^6 + 7840x^4 -", "\\right)$ will be a factor of the polynomial. We can rearrange the polynomial as: \\begin{align} & {{x}^{3}}-6{{x}^{2}}+11x-6 \\\\ & ={{x}^{2}}\\left( x-1 \\right)-5x\\left( x-1 \\right)+6\\left( x-1 \\right) \\\\ \\end{align} We have $\\left( x-1 \\right)$ with each of the terms. Therefore we can take it out as a common factor. $=\\left( x-1 \\right)\\left( {{x}^{2}}-5x+6 \\right)$ \\begin{align} & =\\left( x-1 \\right)\\left( {{x}^{2}}-2x-3x+6 \\right) \\\\ & =\\left( x-1 \\right)\\left( x\\left( x-2 \\right)-3\\left( x-2 \\right) \\right) \\\\ & =\\left( x-1 \\right)\\left( x-2 \\right)\\left( x-3 \\right) \\\\ \\end{align} Therefore, ${{x}^{3}}-6{{x}^{2}}+11x-6=\\left( x-1 \\right)\\left( x-2 \\right)\\left( x-3 \\right)$ Now we have, \\begin{align} & {{x}^{3}}-6{{x}^{2}}+11x-6=0 \\\\ & \\Rightarrow \\left( x-1 \\right)\\left( x-2 \\right)\\left( x-3 \\right)=0 \\\\ \\end{align} Hence the roots of the equation are, \\begin{align} & x-1=0\\Rightarrow x=1 \\\\ & x-2=0\\Rightarrow x=2 \\\\ & x-3=0\\Rightarrow x=3 \\\\ \\end{align} The three roots are 1, 2 and 3. Therefore, the length, breadth and height are 1, 2 and 3 respectively. Note: Alternatively, we can solve the equation by using the relation between roots and coefficients of the polynomial equation. We know that one root is 1. Sum of the three roots = 6 Therefore sum of the other two roots $=6-1=5$. And the product of the three roots is also 6. One of them is 1. Therefore the product of two roots is 6. Hence the other two roots", "we get that b is a root of $q(x)$ mod p. So we see that any root of f is either a root of $x-a$ or a root of $q(x)$, so that (10) \\begin{align} \\#\\mbox{roots of }f(x) \\leq \\#\\mbox{roots of }(x-a) + \\#\\mbox{roots of }q(x) \\end{align} Since $q(x)$ is a polynomial of degree n which doesn't have all its coefficients divisible by p, there are at most n roots of $q(x)$ mod p. Equation (10) then says that there are at most n+1 solutions to $f(x)$ mod p. $\\square$ Though this result gives an upper bound on the number of solutions a particular polynomial can have mod p, for certain special polynomials this is enough to count solutions exactly. Lemma: For $d \\mid p-1$, the polynomial $x^d-1$ has exactly d roots mod p. Proof: Fermat's Little Theorem says that $a^{p-1} \\equiv 1 \\mod{p}$ whenever $(a,p) = 1$. This means that the polynomial $x^{p-1} - 1$ has $p-1$ distinct solutions mod p. Now if $d \\mid p-1$ then we get $de \\ p-1$ for some integer e. Hence we can factor the polynomial $x^{p-1}-1$ has (11) \\begin{align} x^{p-1}-1 = (x^d-1)(x^{p-1-d}+x^{p-1-2d}+\\cdots+x^{p-1-(e-1)d}+x^{p-1-ed} = (x^d-1)(x^{d(e-1)}+x^{d(e-2)}+\\cdots+x^{d(1)}+1). \\end{align} Now the polynomial on the left hand side has exactly p-1 roots. By Lagrange's theorem, the first polynomial in the factorization on the right hand side", "vertex $i$ with parent $j$, we include the variable $X(i,j)$ in the monomial. In total $X(T)$ is a product of $n-1$ monomials, one for every edge in the tree. Let $M$ be the $(n,n)$ minor of $L$. We will show that $\\det M = \\sum_T X(T)$. You can derive the matrix-tree theorem from this statement by substituting the actual graph for the indeterminates $X(i,j)$. If you wish, you can run the entire proof after doing the substitution, and then you don't have to talk about indeterminates. Recall that $$\\det M = \\sum_{\\pi \\in S_{n-1}} (-1)^\\pi \\prod_{i=1}^{n-1} M_{i\\pi(i)}.$$ We can write every permutation $\\pi$ as a product of cycles $\\pi_1,\\ldots,\\pi_{\\ell(\\pi)}$ and of fixed points $F(\\pi)$, and then we get \\begin{align*} \\det M &= \\sum_\\pi (-1)^\\pi \\prod_{t=1}^{\\ell(\\pi)} \\prod_{i \\in \\pi_t} (-X(i,\\pi(i))) \\prod_{i \\in F(\\pi)} \\left(\\sum_{j \\neq i} X(i,j)\\right) \\\\ &= \\sum_\\pi (-1)^{\\ell(\\pi)} \\prod_{t=1}^{\\ell(\\pi)} \\prod_{i \\in \\pi_t} X(i,\\pi(i)) \\prod_{i \\in F(\\pi)} \\left(\\sum_{j \\neq i} X(i,j)\\right), \\end{align*} using the formula $(-1)^\\pi = \\prod_{t=1}^{\\ell(\\pi)} (-1)^{|\\pi_t|+1}$. We can rewrite this as $$\\det M = \\sum_\\pi D(\\pi),$$ where $D(\\pi)$ is the expression appearing above. If we open up all the sums, we find that $\\det M$ is a (weighted) sum of monomials of the form $X(\\alpha) = \\prod_{i=1}^{n-1} X(i,\\alpha(i))$. For each $\\alpha$, we can look at the directed graph in which $i$ points at $\\alpha(i)$.", "\\sqrt{\\prod_i x_i^{-2x_i}}$. Which is enough for the lower bound, because $\\prod_i x_i^{x_i}<1$. – martin cripps Apr 12 '16 at 9:27 • Sorry too many square roots in the last inequality - it should read $\\sum_i\\sqrt{x_i}\\geq \\prod_ix_i^{-2x_i}$! – martin cripps Apr 12 '16 at 9:46 The upper bound is not correct. E.g., let $n=101$, $x_1=\\dots=x_{100}=1/1000$, $x_{101}=9/10$. Then $$\\left(\\prod_{i=1}^{n}x_{i}^{x_{i}}\\right)\\left(\\sum_{i=1}^{n}x_{i}^{1/2}\\right)^{2}>7>2.$$ More generally, take any $a\\in(0,1)$ and any natural $k$, and then let $n=k+1$, $x_1=\\dots=x_k=a/k$, $x_{k+1}=1-a$. Then $$\\ln\\left[\\left(\\prod_{i=1}^{n}x_{i}^{x_{i}}\\right)\\left(\\sum_{i=1}^{n}x_{i}^{1/2}\\right)^{2}\\right]=a\\ln\\frac ak+(1-a)\\ln(1-a)+2\\ln(\\sqrt{ak}+\\sqrt{1-a})\\to\\infty$$ as $k\\to\\infty$.", "+ 5) 8 6 4 2 (x - 8 x + 19 x - 12 x + 1) 3 2 3 2 20 (x - 1) (x + 1) (x - x - 2 x + 1) (x + x - 2 x - 1) 6 5 4 3 2 (x - x - 6 x + 6 x + 8 x - 8 x + 1) 6 5 4 3 2 (x + x - 6 x - 6 x + 8 x + 8 x + 1) Quote by Dodo I think this may be the case. Since\\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) \\end{align*}if 1 (or -1, for that matter) is a root of P(n-3) then it is a root of P(n). And since P(2) = x^2 - 1 and has roots 1 and -1, then this is true for n=2,5,8,... For similar reasons, 0 is a root (as mentioned) for n=1,3,5,7,... to extend further, \\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) &= (x^3-x-1)P(n-3)-(x^2-1)P(n-2) \\end{align*} and we could continue to extend this procedure to obtain as many repeated roots as we like. Not yet there, but here is another", "on why you are looking for the root. For most applications coming within 0.01 of the root is a reasonable approximation, but for some applications (such as building a bridge or launching a rocket) you need much more accuracy. #### Example A Show that f(x)=3x3+5x\\begin{align*}f(x)=-3x^{3}+5x\\end{align*} has at least one root in the interval [1, 2] Solution Since f(x)\\begin{align*}f(x)\\end{align*} is a polynomial we know that it is continuous. f(1)=2\\begin{align*}f(1)=2\\end{align*} and f(2)=14\\begin{align*}f(2)=-14\\end{align*}. Let n=0[14,2]\\begin{align*}n=0\\in[-14,2]\\end{align*}. Applying the Intermediate Value Theorem, there must exist some point c[1,2]\\begin{align*}c\\in[1,2]\\end{align*} such that f(c)=0\\begin{align*}f(c)=0\\end{align*}. This proves that f(x)\\begin{align*}f(x)\\end{align*} has a root in [1, 2]. #### Example B The table below shows several sample values of a polynomial p(x)\\begin{align*}p(x)\\end{align*}. xp(x)444.1526.6204.1214.0941.166088.741024.071549.89183.41 Based on the information in the table (a) What is the minimum number of roots of p(x)\\begin{align*}p(x)\\end{align*}? (b) What are bounds on the roots of p(x)\\begin{align*}p(x)\\end{align*} that you identified in (a)? Solution Since p(x)\\begin{align*}p(x)\\end{align*} is a polynomial we already know that it is continuous. We can use the Intermediate Value Theorem to identify roots by looking at when p(x)\\begin{align*}p(x)\\end{align*} changes from negative to positive, or from positive to negative. (a) There are four sign changes of p(x)\\begin{align*}p(x)\\end{align*} in the table, so at minimum, p(x)\\begin{align*}p(x)\\end{align*} has four roots. (b) The roots are in the following intervals x[2,0],x[1,4],x[15,18],\\begin{align*}x\\in[-2,0], x\\in[1,4], x\\in[15,18],\\end{align*} and the table also tells us that one root" ]

[ "roots in $(0,1)$. If not, then there is a polynomial $Q(x)$ of degree $< n$ that has the same sign as $P(x)$ for all $x\\in[0,1]$. Then for some $\\epsilon>0$, $\\int_0^1|P(x)-\\epsilon Q(x)|\\mathrm{d}x<\\int_0^1|P(x)|\\mathrm{d}x$. Let $0< x_1< x_2<\\dots< x_n<1$ be the roots of $P(x)$. By $(2)$ we have that $$\\frac{1}{k+1}\\left[(x_1^{k+1}-0)-(x_2^{k+1}-x_1^{k+1})+(x_3^{k+1}-x_2^{k+1})-\\dots+(-1)^{n-1}(1-x_n^{k+1})\\right]=0$$ for each $k=0\\dots n-1$, which is equivalent to $$x_n^k-x_{n-1}^k+x_{n-2}^k-\\dots+(-1)^{n-1}x_1^k=\\tfrac12\\tag{3}$$ for each $k=1\\dots n$. Equation $(3)$ has roots $$x_j=\\sin^2\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\tag{4}$$ for $j=1\\dots n$. That is, since $\\displaystyle \\operatorname{Re}\\left(\\frac{1}{e^{ix}+1}\\right)=\\frac12$, \\begin{align} &\\sum_{j=1}^n(-1)^{n-j}\\sin^{2k}\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\\\ &=(-\\tfrac14)^k\\sum_{j=1}^n(-1)^{n-j}\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^m\\exp\\left(i\\pi\\frac{(k-m)j}{n+1}\\right)\\\\ &=(-\\tfrac14)^k(-1)^{n}\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^m\\left(\\frac{(-1)^{k-m+n}-1}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}-1\\right)\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}(-1)^{k-m+n}\\frac{(-1)^{k-m+n}-1}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac{1-(-1)^{k-m+n}}{\\exp\\left(i\\pi\\frac{k-m}{n+1}\\right)+1}\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac12\\left(1-(-1)^{k-m+n}\\right)\\\\ &=(\\tfrac14)^k\\sum_{m=0}^{2k}\\binom{2k}{m}\\frac12\\\\ &=\\frac12\\tag{5} \\end{align} Thus, the minimal monic polynomial is $$P(x)=\\prod_{j=1}^n\\left(x-\\sin^2\\left(\\frac{\\pi}{2}\\frac{j}{n+1}\\right)\\right)\\tag{6}$$ -", "i/n} \\right) \\prod_{k=1}^{n-1} \\left(1-e^{-2k\\pi i/n} \\right). \\end{align*} The first product simplifies to $$e^{\\sum_{k=1}^{n-1} k\\pi i/n} = e^{(n-1)\\pi i/2} = i^{n-1}$$ which cancels out with the $i^{n-1}$ in the numerator. The second product can be recognized as the polynomial $f(X) = \\prod_{k=1}^{n-1} (X-e^{-2k\\pi i/n})$ evaluated at $X = 1$. The roots of this polynomial are the non-trivial $n$-th roots of unity, so $f(X) = \\frac{X^n-1}{X-1} = 1+X+X^2+\\ldots+X^{n-1}$. Plugging in $1$ for $X$ yields $$\\prod_{k=1}^{n-1} \\left(1-e^{-2k\\pi i/n} \\right) = f(1) = n.$$ Altogether, we have $$\\prod_{k=1}^{n-1} \\sin(k\\pi/n) = \\frac{n}{2^{n-1}}.$$ -", "&= \\frac{1}{i^{n/2}} e^{in\\pi/4} \\prod_{k=1}^{n-1} 1 - e^{-i2k\\pi / n} \\big / \\prod_{k=1}^{\\frac{n}{2}-1} 1 - e^{-i2\\pi k / \\frac{n}{2}} \\\\ \\end{aligned} Let $f(x) = \\prod_{k=1}^{n-1} (x - e^{-i2\\pi k/n})$. The roots of this polynomial are the non-trivial $n$-th roots of unity, so $$f(x)=\\frac{x^n−1}{x−1} =1+x+x^2+…+x^{n−1}$$ Plugging in $1$ for $x$ yields $f(1) = \\prod_{k=1}^{n-1} (1 - e^{-i2\\pi k/n}) = n$. Using the above result, we have \\begin{aligned} \\prod_{k=1}^{n-1} 1 - e^{-i2k\\pi/n} &= n\\\\ \\prod_{k=1}^{\\frac{n}{2}-1} 1 - e^{-i2\\pi k/ \\frac{n}{2}} &= n/2 \\end{aligned} Therefore, when $n=1000$, the required product $P = 2e^{in\\pi}/i^{n/2}$ has the value $\\textbf{2}$. ### Extra Credit For the case when $n$ is odd, we need to calculate the product $$P = \\prod_{k=1}^{\\frac{n-1}{2}} V_k = \\prod_{k=1}^{\\frac{n-1}{2}} 2 \\sin\\left((2k - 1) \\frac{π}{n}\\right)$$ We make use of the result below. You can find the proof here. $$\\prod_{k=1}^{n-1} 2 \\sin \\frac{k \\pi}{n} = n$$ We have, \\begin{aligned} \\prod_{k=1}^{n-1} 2 \\sin \\frac{k \\pi}{n} &= n \\\\ \\implies \\prod_{k=1}^{\\frac{n-1}{2}} 2 \\sin \\frac{(2k-1) \\pi}{n} \\cdot \\prod_{k=1}^{\\frac{n-1}{2}} 2 \\sin \\frac{2k \\pi}{n} &= n \\\\ \\implies P \\cdot \\prod_{k=1}^{\\frac{n-1}{2}} 2 \\sin \\left(\\pi - \\frac{2k \\pi}{n} \\right) &= n \\\\ \\implies P \\cdot P &= n \\\\ \\implies P &= \\sqrt{n} \\end{aligned} When $n=1001$, the required product has the value $\\sqrt{1001} \\approx \\textbf{31.63854}$.", ".\\end{align*} ## 4.6 Cyclotomic Polynomials $$\\deg \\Phi_n(x) = \\phi(n)$$ for $$\\phi$$ the totient function. (Click to expand) $$\\deg \\Phi_n(x)$$ is the number of $$n$$th primitive roots, which is the number of numbers less than and coprime to $$n$$. Computing $$\\Phi_n$$: 1. \\begin{align*} \\Phi_{n}(z)=\\prod_{d | n, d>0}\\left(z^{d}-1\\right)^{\\mu\\left(\\frac{n}{d}\\right)} \\end{align*} where \\begin{align*} \\mu(n) \\equiv\\left\\{ \\begin{array}{ll}{0} & {\\text { if } n \\text { has one or more repeated prime factors }} \\\\ {1} & {\\text { if } n=1} \\\\ {(-1)^{k}} & {\\text { if } n \\text { is a product of } k \\text { distinct primes, }}\\end{array}\\right. \\end{align*} 2. \\begin{align*} x^{n}-1=\\prod_{d | n} \\Phi_{d}(x) \\implies \\Phi_n(x) = \\frac{x^n-1}{\\prod_{d | n \\atop d < n} \\Phi_{d}(x)}, \\end{align*} so just use polynomial long division. \\begin{align*} \\Phi_{p}(x) &= x^{p-1}+x^{p-2}+\\cdots+x+1 \\\\ \\Phi_{2 p}(x) &= x^{p-1}-x^{p-2}+\\cdots-x+1 \\end{align*} \\begin{align*} k\\divides n \\implies \\Phi_{n k}(x)=\\Phi_{n}\\left(x^{k}\\right) \\end{align*} Every finite separable extension is simple. $${\\mathbb{GF}}(p^n)$$ is a simple extension over $${\\mathbb{F}}_p$$. # 5 Modules ## 5.1 General Modules A module is simple iff it has no nontrivial proper submodules. A free module is a module with a basis (i.e. a spanning, linearly independent set). A module $$M$$ is torsion-free if and only if for every $$x\\in M$$, $$mx = 0_M \\implies m=0_M$$. I.e. the only torsion element of $$M$$ is zero itself. Free $$\\implies$$ torsionfree. $${\\mathbb{Z}}/(6)$$ is", "$$m =\\prod_{p \\in P} p^{m_i} =\\prod_{p \\in P} p^{2(u_i-v_i)} =\\left(\\prod_{p \\in P} p^{u_i-v_i}\\right)^2$$ so $$m$$ is a perfect square. Therefore the square root of an integer is rational only if it is a square. I'll now try to generalize this to $$k$$-th roots, with as much cut-and-paste as possible. Suppose $$\\sqrt[k]{m} =\\dfrac{u}{v}$$ where $$m = \\prod_{p \\in P} p^{m_i}, u = \\prod_{p \\in P} p^{u_i}, v = \\prod_{p \\in P} p^{v_i}$$. Then $$\\prod_{p \\in P} p^{m_i} =\\dfrac{u^k}{v^k} =\\dfrac{ \\prod_{p \\in P} p^{ku_i}}{\\prod_{p \\in P} p^{kv_i}}$$ so $$\\prod_{p \\in P} p^{m_i}\\prod_{p \\in P} p^{kv_i} =\\prod_{p \\in P} p^{ku_i}$$ or $$\\prod_{p \\in P} p^{m_i+kv_i} =\\prod_{p \\in P} p^{ku_i}$$. By unique factorization, $$m_i+kv_i =ku_i$$, so $$m_i =ku_i-kv_i =k(u_i-v_i)$$. Therefore $$m =\\prod_{p \\in P} p^{m_i} =\\prod_{p \\in P} p^{k(u_i-v_i)} =\\left(\\prod_{p \\in P} p^{u_i-v_i}\\right)^k$$ so $$m$$ is a perfect $$k$$-th power. Therefore the $$k$$-th root of an integer is rational only if it is a $$k$$-th power. When you get $$2p^2 = 3q^2$$ that means that $$2|q^2$$ so $$2|q$$ and that $$3|p^2$$ so $$3|p$$. Then if you replace $$p = 3p'$$ for some integer $$p'$$ and $$q=2q'$$ for some integer $$q'$$ you get $$2(3p')^2 = 3(2q')^2$$ $$18p'^2 = 12q'^2$$ $$3p'^3 = 2q'^2$$. Can you finish from there? This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $$p$$ is", "\\sqrt{\\prod_i x_i^{-2x_i}}$. Which is enough for the lower bound, because $\\prod_i x_i^{x_i}<1$. – martin cripps Apr 12 '16 at 9:27 • Sorry too many square roots in the last inequality - it should read $\\sum_i\\sqrt{x_i}\\geq \\prod_ix_i^{-2x_i}$! – martin cripps Apr 12 '16 at 9:46 The upper bound is not correct. E.g., let $n=101$, $x_1=\\dots=x_{100}=1/1000$, $x_{101}=9/10$. Then $$\\left(\\prod_{i=1}^{n}x_{i}^{x_{i}}\\right)\\left(\\sum_{i=1}^{n}x_{i}^{1/2}\\right)^{2}>7>2.$$ More generally, take any $a\\in(0,1)$ and any natural $k$, and then let $n=k+1$, $x_1=\\dots=x_k=a/k$, $x_{k+1}=1-a$. Then $$\\ln\\left[\\left(\\prod_{i=1}^{n}x_{i}^{x_{i}}\\right)\\left(\\sum_{i=1}^{n}x_{i}^{1/2}\\right)^{2}\\right]=a\\ln\\frac ak+(1-a)\\ln(1-a)+2\\ln(\\sqrt{ak}+\\sqrt{1-a})\\to\\infty$$ as $k\\to\\infty$.", "use a hint in Marsden's book. 1st, $\\cos(A-B)-\\cos(A+B)=2\\sin A \\sin B$ (1), which follows by angle summation formulas. Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy $$(1-z)^n=1 \\leftrightarrow (1-z) \\in \\left\\{\\cos \\frac{2 \\pi k}{n}+i \\sin \\frac{2\\pi k}{n}:k=0,...,n-1 \\right\\}$$ (the set of nth roots of 1) $$\\leftrightarrow z \\in \\left\\{z_k= 1-\\cos \\frac{2 \\pi k}{n}-i \\sin \\frac{2\\pi k}{n}:k=0,...,n-1\\right\\}\\;\\;\\; (2)$$. Since $z_0,....,z_{n-1}$ are the roots of $(1-z)^n-1$, we have by factorization that $$(1-z)^n-1=\\prod_{k=0}^{n-1}(z_k-z)=-z \\prod_{k=1}^{n-1}(z_k-z) \\;\\;(3)$$ (since, by (2), $z_0=0$) In (3), the LHS and RHS are polynomials in z. Equating the coeffs in front of z, we get $$-n=-\\prod_{k=1}^{n-1}z_k \\leftrightarrow n=\\prod_{k=1}^{n-1}z_k\\,.$$ Note $$\\prod_{k=1}^{n-1} \\bar{z}_k=\\overline{\\prod_{k=1}^{n-1}z_k}=n$$ (since $n\\in \\mathbb{R}$), so $$\\prod_{k=1}^{n-1}|z_k|^2=\\prod_{k=1}^{n-1} z_k \\bar{z}_k=\\prod_{k=1}^{n-1} z_k \\prod_{k=1}^{n-1} \\bar{z}_k=n^2\\;\\; (4).$$ Next, $$|z_k|^2=(1-\\cos \\frac{2 \\pi k}{n})^2+ \\sin^2 \\frac{2\\pi k}{n}=2(1-\\cos \\frac{2 \\pi k}{n})$$ using this in (4) gives $$2^{n-1} \\prod_{k=1}^{n-1}(1-\\cos \\frac{2 \\pi k}{n})=n^2\\;\\;(5)$$. Next, $$(\\prod_{k=1}^{n-1} \\sin \\frac{k \\pi}{n})^2=\\prod_{k=1}^{n-1} \\sin \\frac{k \\pi}{n} \\prod_{k=1}^{n-1} \\sin \\frac{(n-k) \\pi}{n}=\\prod_{k=1}^{n-1} \\sin \\frac{k \\pi}{n} \\sin \\frac{(n-k) \\pi}{n}=$$ (where in the last 2 steps, we exploit that the order of taking a product doesn't matter) $$=\\frac{1}{2^{n-1}} \\prod_{k=1}^{n-1} (\\cos \\frac{(n-2k) \\pi}{n}-\\cos \\pi)=$$ (by (1)) $$=\\frac{1}{2^{n-1}} \\prod_{k=1}^{n-1} (1-\\cos \\frac{2k \\pi}{n})=$$ (using $\\cos (\\pi -x)=-\\cos x$) $$=n^2 /2^{2(n-1)}\\;.$$ Applying a sqrt to everything gives the desired result. Consider $z^n=1$, each root is $$\\xi_k = \\cos\\frac{2k\\pi}{n} + i\\sin\\frac{2k\\pi}{n} = e^{i\\frac{2k\\pi}{n}}, k=0,1,2,...,n-1$$ So, we have $$z^n", "# $\\sum_{k=1}^{n-1}\\left(1-e^{\\frac{2\\pi ki}{n}}\\right)^{-1}$ How can I go about computing $$\\sum_{k=1}^{n-1}\\left(1-e^{\\frac{2\\pi ki}{n}}\\right)^{-1}?$$ I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $$\\frac{1}{n}$$, but now I realized that it is the sum of the reciprocals. I've tried using $$e^{ix}=\\cos x+i\\sin x,$$ but I didn't get anywhere with that. • do $n=2$ and $n=3$ explicitly. – Will Jagy Feb 10 at 0:33 • @WillJagy, okay, I'm getting that it comes out to $\\frac{n-1}{2}$. Thanks. – Jake Feb 10 at 0:39 The answer is $$\\frac{n-1}{2}$$. There are multiple ways to approach it, here is one way; Let $$\\displaystyle \\alpha_k = e^{i \\frac{2\\pi k}{n}}$$ for $$k = 1, \\dots, n-1$$. We seek to evaluate, $$\\sum_{k=1}^{n-1} \\frac{1}{1-\\alpha_k}, \\ \\ (*)$$ We know that $$\\alpha_1, \\dots , \\alpha_{n-1}$$ are roots of the polynomial, $$P(z) = 1 + z+ \\dots + z^{n-1}$$ This is because $$1, \\alpha, \\dots, \\alpha_{n-1}$$ are the roots of the polynomial $$1 - z^n$$ (roots of unity), so when we take away $$1$$ as a root we obtain $$P$$. But $$\\alpha_k$$ is a root of $$P$$ simply means that $$P(\\alpha_k) = 0$$ for each $$k$$. Our goal now is to find a polynomial that has roots, $$\\frac{1}{1-\\alpha_k} = f(\\alpha_k)$$ as roots instead of $$\\alpha_k$$, with this polynomial, we find the sum of", "coauthors) that uses a generalisation of this. Thanks. – Brendan McKay Mar 4 '19 at 23:54 Choose $$n$$ numbers $$x_1,\\dots,x_n$$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $$f_n$$. We should get zero value for odd $$n$$. Well, what are these numbers? The roots of $$x^{n}-x^{n-1}+x^{n-2}-\\ldots-1=(x^{n+1}-1)/(x+1)$$. This polynomial indeed has two roots with sum equal to 0 when $$n$$ is odd. If $$n=2k$$ is even, we substitute the roots $$w_1,\\dots,w_n$$ of the polynomial $$f(x)=x^{2k}-x^{2k-1}+\\ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)\\dots (x-w_n)$$. Then your claim reads as $$A:=\\prod_{1\\leqslant i This is done by the standard trick (and is well known itself). At first, $$|A|^2=\\prod_{i=1}^n \\prod_{j\\ne i,1\\leqslant j\\leqslant n}|w_i+w_j|=2^{-n}\\prod_{i=1}^n \\prod_{j=1}^n|(-w_i)-w_j|=2^{-n}\\prod_{i=1}^n |f(w_i)|=\\\\=2^{-n}\\prod_{i=1}^n\\left|\\frac{(-w_i)^{2k+1}+1}{-w_i+1}\\right|=1,$$ since $$1+(-w_i)^{2k+1}=2$$ for all $$i=1,2,\\dots,n$$ and $$\\prod_{i=1}^n (1-w_i)=f(1)=1$$. At second, we need to find the argument of the complex number $$A$$. This may be done for example as follows: all pairs $$w_i+w_j$$ for which $$w_i$$ and $$w_j$$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $$w_i$$ and $$w_j$$ are complex conjugate, the sum $$w_i+w_j$$ is a real number whose sign is the sign of the real part of $$w_i$$. Therefore $$A$$ is the real number whose sign equals $$(1)^{m/2}$$, where $$m$$ is the number of $$w$$'s in the left half-plane. It is easy to see that $$m/2=[k/2]$$", "$0,1,2,\\dots$ and $(2, -n\\pi)$ where $n=1,3,5,\\dots$. 37 ##### Web Bonus Problems / Re: Phaseportrait « on: March 28, 2018, 01:50:57 PM » $\\text{My attempt:}$ \\begin{align}x'&=\\sin(x)\\cos(y)=F(x,y)\\\\y'&=-\\cos(x)\\sin(y)=G(x,y)\\end{align} Let \\begin{align}\\cases{F(x,y)=\\sin(x)\\cos(y)=0\\\\G(x,y)=-\\cos(x)\\sin(y)=0}\\end{align} $$\\implies\\cases{x=k\\pi,y=k\\pi\\\\x={\\pi\\over2}+k\\pi,y={\\pi\\over2}+k\\pi}$$ where $k$ can only be $-1,0,1$, since we have $x\\in(-4,4)$ and $y\\in(-4,4)$, $\\pi\\cong 3.14159265359$. $\\\\$ Thus we have the following critical points:$\\\\$ Case#1: $(0,0);(\\pi,0);(-\\pi,0);(0,\\pi);(0,-\\pi);(\\pi,\\pi);(-\\pi,\\pi);(-\\pi,-\\pi);(\\pi,-\\pi)$. $\\\\$ Case#2: $({\\pi\\over2},{\\pi\\over2});(-{\\pi\\over2},-{\\pi\\over2});({\\pi\\over2},-{\\pi\\over2});(-{\\pi\\over2},{\\pi\\over2})$. $$J=\\begin{pmatrix}F_x(x,y)&F_y(x,y)\\\\G_x(x,y)&G_x(x,y)\\end{pmatrix}=\\begin{pmatrix}\\cos(x)\\cos(y)&-\\sin(x)\\sin(y)\\\\\\sin(x)\\sin(y)&-\\cos(x)\\cos(y)\\end{pmatrix}$$ When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}$ or $\\begin{pmatrix}-1&0\\\\0&0\\end{pmatrix}$, wtih eigenvalues $\\lambda=\\pm1$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\0\\end{pmatrix}$ or $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#1 are all $\\text{Saddle Points}$. $\\\\$ When $(x,y)=$ critical points in Case#2: we get matrices $\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}$ or $\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}$ with eigenvalues $\\lambda=\\pm i$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\i\\end{pmatrix}$ or $\\begin{pmatrix}1\\\\-i\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#2 are all $\\text{Center}$. $\\\\$ Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that. 38 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018, 09:11:15 AM » Thanks for the correction. I made a mistake copying the result integral. I got the same result. I think there is a calculation error when computing the initial condition, hence the $c_1,c_2$ values. (if I am not mistaken ) $$c_1+c_2=-2-15\\ln2$$ and $$-2c_2=3-15\\ln2$$ Therefore, $$c_1=-{1\\over2}-{45\\over2}\\ln2$$ $$c_2=-{3\\over2}+{15\\over2}\\ln2$$ 39 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018,", "\\pi=\\left\\{\\pi_{t}\\right\\}_{t=1}^{T^{\\prime}} , where label π t ∈ V \\pi_{t} \\in \\mathcal{V} . The probability of alignment path π \\pi given the input sequence is defined as follows, p ( π ∣ o ) = ∏ t = 1 T ′ p ( π t ∣ o ) = ∏ t = 1 T ′ y t , π t p(\\pi \\mid \\mathbf{o})=\\prod_{t=1}^{T^{\\prime}} p\\left(\\pi_{t} \\mid \\mathbf{o}\\right)=\\prod_{t=1}^{T^{\\prime}} y_{t, \\pi_{t}} . (10) CTC 创建带有空白标签 “ − − ”的扩展词汇表 V \\mathcal{V} 。这里的 V = V origin ∪ { − } \\mathcal{V}=\\mathcal{V}_{\\text {origin }} \\cup\\{-\\} 。空白的标签代表 stillness and transitions，没有确切的意义。将输入序列的 alignment path 表示为 π = { π t } t = 1 T ′ \\pi=\\left\\{\\pi_{t}\\right\\}_{t=1}^{T^{\\prime}} ，其中标签 π t ∈ V \\pi_{t} \\in \\mathcal{V} 。给定输入序列， alignment path π \\pi 的概率定义为: p ( π ∣ o ) = ∏ t = 1 T ′ p ( π t ∣ o ) = ∏ t = 1 T ′ y t , π t p(\\pi \\mid \\mathbf{o})=\\prod_{t=1}^{T^{\\prime}} p\\left(\\pi_{t} \\mid \\mathbf{o}\\right)=\\prod_{t=1}^{T^{\\prime}} y_{t, \\pi_{t}} . (10) Define a many-to-one mapping operation B \\mathcal{B} which removes all blanks and repeated words in the alignment path (e.g., B ( I I − m i s s − − y o u ) = I , \\mathcal{B}(I I-m i s s--y o u)=I, miss, y o u y o u ). In", "plus z, close paren 10 $2\\left(y+1\\right)$ 2 times, open paren, y plus 1, close paren 11 $\\left(2-1\\right)x$ open paren, 2 minus 1, close paren, times x 12 ${p}_{1}\\left(3+7\\right)$ p sub 1, times, open paren, 3 plus 7, close paren 13 ${p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}$ p sub 1, raised to the, a sub 1, power, p sub 2, raised to the, a sub 2, power 14 ${\\left(x+y\\right)}^{-4}{\\left(x-y\\right)}^{-4}$ open paren, x plus y, close paren, to the negative 4 power, times, open paren, x minus y, close paren, to the negative 4 power 15 ${2}^{4\\left(x+y\\right)}$ 2 raised to the 4 times, open paren, x plus y, close paren, power 16 $xy$ x y 17 ${x}^{2}{y}^{3}$ x squared, y cubed 18 ${x}^{y+1}{x}^{y+2}$ x raised to the y plus 1 power, x raised to the y plus 2 power 19 $\\sqrt{a}\\sqrt{b}=\\sqrt{ab}$ the square root of a, the square root of b, equals the square root of a b 20 $\\sqrt{3}\\sqrt{10}=\\sqrt{30}$ the square root of 3, the square root of 10, equals the square root of 30 21 $2\\sqrt{3}$ 2 the square root of 3 22 $1+2\\sqrt{3}$ 1 plus 2 the square root of 3 23 $f\\left(x\\right)={x}^{2}\\left(x+1\\right)$ f of x, equals x squared times, open paren, x plus 1, close paren 24 $\\mathrm{sin}x\\mathrm{cos}y+\\mathrm{cos}x\\mathrm{sin}y$ sine x cosine y, plus, cosine x sine y 25 $\\mathrm{sin}\\left(x+y\\right)\\mathrm{cos}\\left(x+y\\right)$ the sine", "\\choose k-1}+\\cdots \\pm {n \\choose 0} \\end{align*} ### Solution We have \\begin{align*} (1+x)^n-x(1+x)^n + x^2(1+x)^n - \\dots x^k(1+x)^n \\\\ = (1+x)^n(1-x+x^2-\\dots\\pm x^k) \\\\ = (1+x)^n \\frac{(1 - (-x)^{-k+1})}{1-(-x)} \\\\ = (1+x)^{n-1} (1-(-x)^{k+1}) \\end{align*} The coefficient of $x^k$ on the left hand side is \\begin{align*} {n \\choose k}-{n \\choose k-1}+\\cdots \\pm {n \\choose 0} \\end{align*} as we only need the term $x^{k-i}$ from $(1+x)^n$ from the $i^{th}$ term. The coefficient of $x^k$ on the right hand side is \\begin{align*} {n-1 \\choose k} \\end{align*} ## Problem 4.3.19 a. Solve the equation $x^3-3x^2+4=0$, given that two of its roots are equal. b. Solve the equation $x^3-9x^2+23x-15=0$, given that its roots are in arithmetical progression. ### Solution a. Let $f(x)= x^3-3x^2+4$. As $f(x)=0$ has two equal roots, $f(x)=0$ and $f'(x)=0$ share that root. The roots of $f'(x)=3x^2-6x=0$ are $x=0$ and $x=2$. As $x=0$ is not a root of $f(x)=0$, $x=2$ is a common root of $f(x)$ and $f'(x)$.As the product of the roots of $f(x)=0$ is $-4$, the roots of $f(x)=0$ are $-1,2$ and $2$. b. Let $a-d,a,a+d$ be the roots of $x^3-9x^2+23x-15=0$. We have sum of the roots is $3a=9$ and the product of the roots is $3(9-d^2)=15$. Therefore, the roots of the equation are $1,3$ and $5$. ## Problem 4.3.20 Given $r,s,t$ are the roots of $x^3+ax^2+bx+c=0$, a. Evaluate $1/r^2+1/s^2+1/t^2$,", "s\\left(F^i\\left(u\\right)\\right) . \\end{align} Comparing this with \\eqref{darij1.pf.t1.Fi}, we obtain $$a^{q^i} = s\\left(F^i\\left(u\\right)\\right)$$. The same argument (applied to $$j$$ instead of $$i$$) yields $$a^{q^j} = s\\left(F^j\\left(u\\right)\\right)$$. Hence, \\begin{align} a^{q^i} = s\\left(\\underbrace{F^i\\left(u\\right)}_{= F^j\\left(u\\right)}\\right) = s\\left(F^j\\left(u\\right)\\right) = a^{q^j} . \\end{align} In other words, $$a^{q^j} - a^{q^i} = 0$$. Now, forget that we fixed $$a$$. We thus have shown that $$a^{q^j} - a^{q^i} = 0$$ for each $$a \\in L$$. In other words, each $$a \\in L$$ is a root of the polynomial $$x^{q^j} - x^{q^i} \\in K\\left[x\\right]$$. This polynomial is nonzero (since $$q^i < q^j$$, so that the terms $$x^{q^j}$$ and $$x^{q^i}$$ do not cancel) and has degree $$q^j$$ (since $$q^i < q^j$$), and thus has at most $$q^j$$ many roots in $$L$$ (since any nonzero polynomial over $$L$$ of degree $$d$$ has at most $$d$$ many roots in $$L$$). Since each $$a \\in L$$ is a root of this polynomial, we thus conclude that there are at most $$q^j$$ many $$a \\in L$$. In other words, $$\\left|L\\right| \\leq q^j$$. But $$L = K\\left[t\\right] / \\left(f\\left(t\\right)\\right)$$ is a $$K$$-vector space of dimension $$n$$ (since the polynomial $$f$$ has degree $$n$$), and thus has size $$\\left|K\\right|^n = q^n$$ (since $$\\left|K\\right| = q$$). Hence, $$\\left|L\\right| = q^n$$, so that $$q^n = \\left|L\\right| \\leq q^j$$ and therefore $$n \\leq j$$. This contradicts $$j \\in", "&= c' + 2bd, \\end{align*} so $$c' = c^2 - 2bd$$. Thus, our transformation which squares roots takes a monic polynomial $$t^3 + bt^2 + ct + d$$ and transforms it into the monic polynomial $$t^3 + (2c - b^2)t^2 + (c^2 - 2bd)t - d^2.$$ Let's apply it to $$t^3 - t + 1$$ to get a polynomial whose roots are $$x^2, y^2, z^2$$. We obtain the polynomial: $$t^3 + (2(-1) - 0^2)t^2 + ((-1)^2 - 2(0)(1))t - 1^2 = t^3 - 2t^2 + t - 1.$$ Apply again to get a polynomial whose roots are $$x^4, y^4, z^4$$: $$t^3 + (2(1) - (-2)^2)t^2 + (1^2 - 2(-2)(-1))t - (-1)^2 = t^3 - 2t^2 - 3t - 1.$$ We could apply this transformation one final time to get the polynomial whose roots are $$x^8, y^8, z^8$$, but really we just need the coefficient of $$t^2$$. So, partially applying the above formula, we get: $$t^3 + (2(-3) - (-2)^2)t^2 + \\underline{\\hspace{12pt}} t + \\underline{\\hspace{12pt}} = t^3 - 10t^2 + \\underline{\\hspace{12pt}} t + \\underline{\\hspace{12pt}}.$$ So, by Vieta, $$x^8 + y^8 + z^8 = -(-10) = 10.$$ You got already that $$x^2+y^2+z^2=2.$$ Also, we have: $$\\sum_{cyc}x^2y^2=\\left(\\sum_{cyc}xy\\right)^2-2xyz(x+y+z)=1$$ and $$x^2y^2z^2=1.$$ Thus, $$\\sum_{cyc}x^8=\\left(\\sum_{cyc}x^4\\right)^2-2\\sum_{cyc}x^4y^4=$$ $$=\\left(\\left(\\sum_{cyc}x^2\\right)^2-2\\sum_{cyc}x^2y^2\\right)^2-2\\left(\\left(\\sum_{cyc}x^2y^2\\right)^2-2x^2y^2z^2\\sum_{cyc}x^2\\right)=$$ $$=(4-2)^2-2(1-2\\cdot2)=10.$$ Question: \"@hm2020 ,hi , I can't relate the equation you are proposing to the problem, could you clarify?", "we have at $(n-1)$ roots of $P'(x)$ that are different from the roots of $P(x)$. Hence $H(x)$ has $(n-1)$ real roots and thus has all its roots real . • Neat. I have a vested interest that my answer is accepted ;) but I really like this one! – user491874 Dec 12, 2017 at 18:05 • how do we prove that each of these roots is unique so that the total of the roots is exactly k - 1? Aug 27, 2019 at 18:27 • @DimitrisPapadimitriou the roots of $P'(x)$ lie between the roots of $P(x)$. So in each interval formed by consecutive roots of $P(x)$ you have a root of $P'(x)$ thus netting us $n-1$ distinct roots of $P(x)$. Aug 29, 2019 at 5:11 Multiply with all denomiators. You get a polynomial \\begin{align*} p(x) = \\sum_{i=1}^k b_i\\prod_{j\\neq i} (x-a_j) \\end{align*} of degree $k-1$ on the left-hand side. None of the $a_i$ is a zero of that polynomial because for $x=a_i$ the left-hand side simplifies to $b_i \\prod_{j\\neq i} (x-a_j)$ and $a_i$ is not a zero of that simplified term. Therefore, the zeros of the polynomial are the zeros of the original left-hand side. Because of $a_i < a_{i+1}$ and $b_i>0$ you have \\begin{align*} (-1)^{k-i} p(a_i) > 0. \\end{align*} for $k=1,\\ldots,k$ and therefore $k$ sign-changes for the continuous", "four solutions of the given equation are ($$0,7$$), ($$1,5$$), ($$2, 3$$) and ($$3,1$$). (ii) $$\\quad \\pi x+y=9$$ Given, Linear Equation, $$\\pi x+y=9$$ $$y=9-\\pi x \\qquad \\dots \\dots \\text{Equation (1)}$$ Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For \\begin{align},x = 0,y = 9 - \\pi \\left( 0 \\right) \\Rightarrow y=9.\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (0,9 )\\end{align} For \\begin{align},x = 1,y = 9 - \\pi \\left( 1 \\right) \\Rightarrow 9 - \\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (1,9 - \\pi )\\end{align} For \\begin{align},x = 2,y = 9 - \\pi \\left( 2 \\right) \\Rightarrow 9 - 2\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (2,9 - 2\\pi )\\end{align} For \\begin{align},x = 3,y = 9 - \\pi \\left( 3 \\right) \\Rightarrow 9 - 3\\pi .\\end{align} Hence, we get \\begin{align}\\left( {x,y} \\right) = (3,9 - 3\\pi )\\end{align} Therefore, the four solutions of the given equation are $$(0,9),(1,9-\\pi),(2,9-2 \\pi),(3,9-3 \\pi)$$. (iii) $$x = 4y$$ Given, Linear Equation, $$x = 4y$$ \\begin{align}\\therefore y = \\frac{x}{4} \\qquad \\dots \\dots \\text{Equation (1)}\\end{align} Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get For $$x = 0$$, \\begin{align}y =\\frac{0}{4} = 0\\end{align}. Hence, we get $$(x, y)$$ $$= (0, 0)$$ For $$x = 1$$, \\begin{align}y =\\frac{1}{4}\\end{align} Hence, we get", "Let $$p \\in \\mathbb{R}[x]$$. Let $$\\alpha$$ be a root of $$p$$. Then so is $$\\bar{\\alpha}$$. ProOF. Let $$p(x)=\\sum_{k=0}^{N} a_{k} x^{k}$$. Then $p(\\alpha)=\\sum_{k=0}^{N} a_{k} \\alpha^{k}=0,$ SO $p(\\bar{\\alpha})=\\sum_{k=0}^{N} a_{k} \\bar{\\alpha}^{k}=\\overline{p(\\alpha)}=0 .$ Let $$\\alpha=a+i b$$. Then \\begin{aligned} (x-\\alpha)(x-\\bar{\\alpha}) &=(x-(a+i b))(x-(a-i b)) \\\\ &=x^{2}-2 a x+a^{2}+b^{2} \\\\ &=(x-a)^{2}+b^{2} . \\end{aligned} So applying the Fundamental Theorem of Algebra to the real polynomial $$p$$, we first factor out the real roots, and for each pair of complex conjugate roots we get a factor as in (9.47). Thus we get: THEOREM 9.48. Let $$p \\in \\mathbb{R}[x]$$ be a polynomial of degree $$N$$. Then $$p$$ can be factored into a product of linear factors $$\\left(x-c_{k}\\right)$$ and quadratic factors $$\\left(\\left(x-a_{k}\\right)^{2}+b_{k}^{2}\\right)$$ : $p(x)=c\\left(\\prod_{k=1}^{N_{1}}\\left(x-c_{k}\\right)\\right)\\left(\\prod_{j=1}^{N_{2}}\\left(\\left(x-a_{j}\\right)^{2}+b_{j}^{2}\\right)\\right)$ for some (not necessarily distinct) real numbers $$c, c_{j}, a_{j}, b_{j}$$. We have $$N_{1}+2 N_{2}=N$$, and the factoring is unique, up to ordering and replacing any $$b_{j} b y-b_{j}$$. ## Further remarks In Chapter 5 we defined cosine and sine in terms of power series. In Section 9.2, we interpreted them geometrically and used trigonometric identities. Showing that the power series and the trigonometric interpretation are really describing the same function is part of a course in Complex Analysis. There are two main ingredients to a first course in Complex Analysis. The first is to show that if a function $$f$$ has a derivative everywhere", "z^{-3} \\\\ \\prod\\nolimits_{i=1}^{4} (1-q_i z_{-1}) &= 1 \\\\ &- (q_1 + q_2 + q_3 + q_4) z^{-1} \\\\ &+ (q_1 q_2 + q_1 q_3 + q_1 q_4 + q_2 q_3 + q_2 q_4 + q_3 q_4) z^{-2} \\\\ &- (q_1 q_2 q_3 + q_1 q_2 q_4 + q_1 q_3 q_4 + q_2 q_3 q_4) z^{-3} \\\\ &+ (q_1 q_2 q_3 q_4) z^{-4} \\end{aligned} It looks like there might be straightforward closed form for any of these coefficients (pardon the awkward notation): \\begin{align} b_0 &= 1 \\\\ b_i &= (-1)^i \\sum_{ Q \\in \\text{combs}(i) } \\bigg[ \\prod_{ j \\in Q } q_j \\bigg] \\end{align} Where $$\\text{combs}(i): \\mathbb{Z} \\mapsto \\mathbb{Z}^i$$ is the set of all length-$$i$$ combinations of coefficients. Here's a possibly even harder to understand version in Julia/SymPy: z = symbols(\"z\") function build_expr(n_roots) qs = symbols(\"q1:$(n_roots+1)\") # generates q1,q2,... poly = 1 for i in 1:n_roots prods = [reduce(*, c) for c in combinations(qs, i)] poly += ((-1)^i * reduce(+, prods))/z^i end return poly end Again, using SymPy, it's trivial to test if the code works for a given number of roots: function test_expr(n_roots) mine = build_expr(n) # first generate the actual solution qs = symbols(\"q1:$(n_roots+1)\") # generates q1,q2,... actual = reduce(*, [(1-qs[i]/z) for i in 1:n]) # if these exprs cancel out, then the exprs are equivalent return", "initiated with $U_1=1$ and $U_3=3-T$ (I'd even take $U_{-1}=-1$), so that $U_q$ has integer coefficients, degree $(q-1)/2$ and its leading coefficient is $(-1)^{(q-1)/2}$. Since $\\sin(qx)$ vanishes for $x=k\\pi/q\\pmod{2\\pi}$, the roots of $U_q$ are the $3\\sin^2(k\\pi/q)$, for $1\\leq k\\leq (q-1)/2$, and we have the factorization $\\sin (qx) = (-1)^{(q-1)/2} \\sin(x) \\prod_{k=1}^{(q-1)/2} \\big( 4\\sin^2(x) - 4\\sin^2(k\\pi/q ) \\big) .$ Consequently, $\\big( \\frac qp\\big) = (-4)^{(q-1)(p-1)/4} \\prod_{a=1}^{(p-1)/2} \\prod_{k=1}^{(q-1)/2} \\left( \\sin ^2(2\\pi a/p)- \\sin^2(2\\pi k/q)\\right).$ By symmetry, $\\big( \\frac pq\\big) = (-4)^{(q-1)(p-1)/4} \\prod_{a=1}^{(q-1)/2} \\prod_{k=1}^{(p-1)/2} \\left( \\sin ^2(2\\pi a/q)- \\sin^2(2\\pi k/p)\\right).$ The big product in the two expressions just differ by the sign of each factor. Consequently, $\\big( \\frac pq\\big) = (-1)^{(p-1)(q-1)/4} \\big( \\frac qp\\big),$ which is the quadratic reciprocity law. #### First variation (Kronecker) The analytic property of the sine function is not really used. In fact, one can deduce from Gauss's lemma the formula $\\big(\\frac qp\\big) = \\mathop{\\rm sign} \\left( \\prod_{a=1}^{(p-1)/2} \\prod_{k=1}^{(q-1)/2} \\big( \\frac qp-\\frac kq\\big)\\right) .$ This implies the quadratic reciprocity law in a similar way. #### Second variation (learnt from Jean-François Mestre) The products appearing in Eisenstein's proof are reminiscent of resultants. In fact, one has the following formula: $\\big(\\frac qp\\big) = \\mathop{\\rm Res}(U_q,U_p),$ unless I fucked up the signs. This equality follows from the analysis above, since the zeroes of $U_q$ are $4\\sin^2(k\\pi/q)$, and those of $U_p$ are" ]

[ "\\circ$.", ")$ and $(5, 160^\\circ )$ .", "is $55^\\circ$.", "\\circ h' =g$.", "{45}^{\\circ}$.", "6^\\circ$ -", "How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "$^\\circ$C.", "larger than $180^\\circ$.", "is equal to ${1260}^{\\circ}$. Hope this helps!", "# How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "= 50^{\\circ}$?", "^{\\circ}$", "^ {\\ circ}}$", "to this problem is $$\\{114.3^\\circ,335.7^\\circ\\}$$", "\\circ m(f)$", "you get the $45^\\circ$ line.", "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "70.85^\\circ)$$", "H - N - H = {109.5}^{\\circ}$." ]

[ "How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "# How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "$$C = (180-2x)^\\circ$$ and we have $$a=b$$ and $$c = 2a\\cos x^\\circ$$. Then $\\begin{array}{rcl} \\frac{aA+bB+cC}{a+b+c} & = & \\frac{ax + ax + 2a(180-2x)\\cos x^\\circ}{a + a + 2a\\cos x^\\circ} \\\\ & = & \\frac{2a[x + (180-2x)\\cos x^\\circ]}{2a(1 + \\cos x^\\circ)} \\; = \\; \\frac{x + (180-2x)\\cos x^\\circ}{1 + \\cos x^\\circ} \\end{array}$ and this tends to $$90$$ as $$x \\to 0$$. · 3 years, 11 months ago Thank you for noticing. I will use $$\\frac{\\pi}{3}$$ and $$\\frac{\\pi}{2}$$ instead, on the next time if i meet those numbers again. · 3 years, 11 months ago", "## Algebra 2 (1st Edition) Since $\\cos(180-theta)=-\\cos \\theta$ we obtain $\\cos(180-135)=-\\cos 135\\\\\\cos 45=-\\cos 135\\\\-\\cos 45=\\cos 135\\\\\\cos 135=-\\frac{\\sqrt 2}{2}$ Now find $\\sec135=\\frac{1}{\\cos 135}=\\frac{1}{-\\frac{\\sqrt 2}{2}}=-\\frac{2}{\\sqrt 2}=-\\sqrt 2$", "$$x = \\frac{-14 + \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – \\frac{1}{2}$$ $$x = \\frac{-14 – \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – 3$$ 8- $$x^{\\frac{ 15}{4}}$$ $$(x^6)^{\\frac{5}{8}} = x^{6 × \\frac{5}{8}} = x^{\\frac{ 30}{8}} =x^{\\frac{ 15}{4}}$$ 9- 32 $$32 + 35 + 29 = 96$$ Average = $$\\frac{96}{3} = 32$$ 10- $$45^\\circ$$ $$90^\\circ + 45^\\circ = 135^\\circ$$ $$180^\\circ- 135^\\circ = 45^\\circ$$ 11- $$(x – 1)(x + 6)$$ $$x^2 + 5x – 6 = (x – 1)(x + 6)$$ 12- The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: 13- 5 $$C = \\sqrt{(x_A- x_B)^2 + (y- y_B)^2 ) }$$ $$C = \\sqrt{(1-(-2))^2 + (3- 7)^2}$$ $$C = \\sqrt{(3)^2 + (- 4)^2 )}$$ $$C = \\sqrt{9 + 16}$$ $$C = \\sqrt{25} = 5$$ 14- $$0.02857143 \\cong 0.0286$$ $$\\frac{4}{140} = \\frac{1}{35} = 0.02857143 \\cong 0.0286$$ 15- $$\\frac{20}{3}$$ $$f(x) = x – \\frac{5}{3} ⇒( y = x) (– \\frac{5}{3} ) ⇒y+ \\frac{5}{3} = x$$ $$f^{-1} = x+ \\frac{5}{3}$$ $$f ^{–1}(5) = 5 + \\frac{5}{3} = \\frac{20}{3}$$ 16- $$\\frac{ 1}{4}$$ Probability $$=\\frac{number of desired outcomes}{number of total outcomes} = \\frac{18}{12+18+18+24} = \\frac{18}{72} =\\frac{ 1}{4}$$ 17- $$\\frac{\\sqrt{3}}{2}$$ $$cos 30^\\circ = \\frac{\\sqrt{3}}{2}$$ 18- cosθ=$$\\frac{4}{5}$$ sinθ$$=\\frac{3}{5}$$⇒ we have following triangle, then $$c=\\sqrt{5^2-3^2 }=\\sqrt{25-9}=\\sqrt{16}=4)$$ cosθ$$=\\frac{4}{5}$$ 19- 5 $$|9 – (12 ÷ | 2 –", "= 0$$ ⇒then: a = 4, b = 14 and c = 6 $$x = \\frac{-14 + \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – \\frac{1}{2}$$ $$x = \\frac{-14 – \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – 3$$ 8- $$x^{\\frac{ 15}{4}}$$ $$(x^6)^{\\frac{5}{8}} = x^{6 × \\frac{5}{8}} = x^{\\frac{ 30}{8}} =x^{\\frac{ 15}{4}}$$ 9- 32 $$32 + 35 + 29 = 96$$ Average = $$\\frac{96}{3} = 32$$ 10- $$45^\\circ$$ $$90^\\circ + 45^\\circ = 135^\\circ$$ $$180^\\circ- 135^\\circ = 45^\\circ$$ 11- $$(x – 1)(x + 6)$$ $$x^2 + 5x – 6 = (x – 1)(x + 6)$$ 12- The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: 13- 5 $$C = \\sqrt{(x_A- x_B)^2 + (y- y_B)^2 ) }$$ $$C = \\sqrt{(1-(-2))^2 + (3- 7)^2}$$ $$C = \\sqrt{(3)^2 + (- 4)^2 )}$$ $$C = \\sqrt{9 + 16}$$ $$C = \\sqrt{25} = 5$$ 14- $$0.02857143 \\cong 0.0286$$ $$\\frac{4}{140} = \\frac{1}{35} = 0.02857143 \\cong 0.0286$$ 15- $$\\frac{20}{3}$$ $$f(x) = x – \\frac{5}{3} ⇒( y = x) (– \\frac{5}{3} ) ⇒y+ \\frac{5}{3} = x$$ $$f^{-1} = x+ \\frac{5}{3}$$ $$f ^{–1}(5) = 5 + \\frac{5}{3} = \\frac{20}{3}$$ 16- $$\\frac{ 1}{4}$$ Probability $$=\\frac{number of desired outcomes}{number of total outcomes} = \\frac{18}{12+18+18+24} = \\frac{18}{72} =\\frac{ 1}{4}$$ 17- $$\\frac{\\sqrt{3}}{2}$$ $$cos 30^\\circ = \\frac{\\sqrt{3}}{2}$$ 18- cosθ=$$\\frac{4}{5}$$ sinθ$$=\\frac{3}{5}$$⇒ we have following triangle,", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "• 33 + $$\\frac{1}{3}$$x = 135.2 D. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = 135.2 B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, Explanation: Given to show the correct use of the Distributive Property when solving (33 – x) = 135.2? as the correct answer is B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, if we multiply both sides with $$\\frac{1}{3}$$ we get the correct equation as (33 – x) = 135.2 because $$\\frac{1}{3}$$(33 – x) = $$\\frac{1}{3}$$ X135.2 means, $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, therefore the correct answer is B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2. Practice & Problem Solving Leveled Practice For 7-10, use the Distributive Property to solve the equations. Question 7. -2(x + 5) = 4 Explanation: Given equation as -2(x + 5) = 4, using Distributive Property we get -2 X x + (-2 X 5) = 4, -2x – 10 = 4, -2x = 14, So x = -14/2 = -7, therefore x = -7. Question 8. 3.2 = $$\\frac{4}{5}$$(b – 5) Explanation: Given equation as 3.2 = $$\\frac{4}{5}$$(b – 5) using Distributive Property, we get 3.2 = $$\\frac{4}{5}$$ X b + $$\\frac{4}{5}$$(-5), 3.2 = $$\\frac{4}{5}$$b – 4, 3.2 + 4 = $$\\frac{4}{5}$$b, 7.2 = $$\\frac{4}{5}$$b, b = $$\\frac{5}{4}$$ X 7.2, b", "## Trigonometry (11th Edition) Clone $$2\\cos^267\\frac{1^\\circ}{2}-1=-\\frac{\\sqrt2}{2}$$ $$2\\cos^267\\frac{1^\\circ}{2}-1$$ Recall the Double-Angle Identity for cosine: $$2\\cos^2A-1=\\cos2A$$ Therefore, replace $A=67\\frac{1^\\circ}{2}$, we can apply the above identity to $2\\cos^267\\frac{1^\\circ}{2}-1$. $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos\\Big(2\\times67\\frac{1^\\circ}{2}\\Big)$$ $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos135^\\circ$$ As $135^\\circ+45^\\circ=180^\\circ$, the absolute value of $\\cos135^\\circ$ would equal the (already positive) value of $\\cos45^\\circ$. $$|\\cos135^\\circ|=\\cos45^\\circ$$ However, $135^\\circ$ lies in quadrant II, where $\\cos\\theta\\lt0$, so $\\cos135^\\circ\\lt0$. Thus, $$\\cos135^\\circ=-\\cos45^\\circ=-\\frac{\\sqrt2}{2}$$ Now bring the value of $\\cos135^\\circ$ back to the calculation of $2\\cos^267\\frac{1^\\circ}{2}-1$ $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos135^\\circ=-\\frac{\\sqrt2}{2}$$", "## A community for students. Sign up today! Here's the question you clicked on: ## haleyking345 one year ago Solve the equation: a.) 2 cos^2x-cos x=1 • This Question is Closed 1. Marx let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$ #### Ask your own question Ask a Question Find more explanations on OpenStudy", "2 x = \\frac{4}{20} = \\frac{1}{5}$ a. $\\cos 2 x = \\frac{1}{2}$ --> $2 x = \\pm \\frac{\\pi}{3}$, $x = \\pm \\frac{\\pi}{6}$, or $x = \\pm {30}^{\\circ}$ b. $\\cos 2 x = \\frac{1}{5}$ --> $2 x = \\pm {78}^{\\circ} 46$ --> $x = \\pm {39}^{\\circ} 23$ $x = {30}^{\\circ}$ and $x = {39}^{\\circ} 23$", "cos 138 b² = 400 + 144 – 480 (-0.74) b² = 899.2 b = 29.9 $$\\frac { sin B }{ b }$$ = $$\\frac { sin A }{ a }$$ $$\\frac { sin 138 }{ 29.9 }$$ = $$\\frac { sin A }{ 20 }$$ $$\\frac { 0.66 }{ 29.9 }$$ = $$\\frac { sin A }{ 20 }$$ sin A = 0.44 ∠A = 26.1 $$\\frac { sin B }{ b }$$ = $$\\frac { sin C }{ c }$$ $$\\frac { sin 138 }{ 29.9 }$$ = $$\\frac { sin C }{ 12 }$$ sin C = 0.26 ∠C = 15.07 Question 21. Question 22. ∠A = 107.3, ∠B = 51.6, ∠C = 21.1 Explanation: b² = a² + c² – 2ac cos B 28² = 18² + 13² – 2(18)(13) cos B 784 = 324 + 169 – 468 cos B 291 = 468 cos B cos B = 0.62 ∠B = 51.6 $$\\frac { sin C }{ c }$$ = $$\\frac { sin B }{ b }$$ $$\\frac { sin C }{ 13 }$$ = $$\\frac { sin 51.6 }{ 28 }$$ sin C = 0.36 ∠C = 21.1 ∠A + ∠B + ∠C = 180 51.6 + 21.1 + ∠A = 180 ∠A = 107.3 Question 23. Question 24. ∠A = 23,", "= \\cos 45^\\circ.$$ Therefore, we can express the desired value as $$1 + \\frac{\\sin 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}{\\cos 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}.$$ Because $\\cos 45^\\circ = \\sin 45^\\circ$, we see that the fractional part is $1$, and so the sum is $1 + 1 = 2$, which brings us to the answer $\\boxed{B}$.", "Hence, $r^2=x(x-5)$. Similarly, $\\cot\\frac{B}{2}=\\frac{14-x}{r}$, and $\\cot\\frac{D}{2}=\\frac{12-x}{r}=\\tan\\frac{B}{2}=\\frac{r}{14-x}$. Therefore, $r^2=\\frac{12-x}{14-x}$. Putting this together with the above equation yields $$(12-x)(14-x)=x(x-5)$$$$\\Longrightarrow x^2-26x+168=x^2-5x$$$$\\Longrightarrow 21x=168$$$$\\Longrightarrow x=8.$$ Thus, $r^2=8(8-5)=8(3)=24$, and $r=\\boxed{\\textbf{(C)}\\ 2\\sqrt{6}}$.", "1 - \\frac{\\sqrt{3}}{2} = 0.134$ $\\sin x + \\cos x = \\sqrt{2} \\cos \\left(x - \\frac{\\pi}{4}\\right) = 0.134$ $\\cos \\left(x - \\frac{\\pi}{4}\\right) = \\frac{0.134}{\\sqrt{2}} = 0.0947$ $x - \\frac{\\pi}{4} = \\pm {84.564}^{\\circ}$ $x = 84.564 + 45 = {129.564}^{\\circ}$ and $x = 360 - 84.564 + 45 = {321.564}^{\\circ}$", "{\\sqrt {3}}{2}$,$\\cos (120) = \\cos (180 -60) =- \\cos 60= – \\frac {1}{2}$,$\\tan 120 = \\frac {\\sin 120}{\\cos 120} = -\\sqrt {3}$,$\\sin (135) = \\sin (180 -45) = \\sin 45= \\frac {1}{\\sqrt {2}}$,$\\cos (135) = \\cos (180 -45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$,$\\tan 135 = \\frac {\\sin 135}{ \\cos 135} = -1$,$\\tan 180 = \\frac {\\sin 180}{\\cos 180} = 0$,$\\csc 120 = \\frac {1}{\\sin 120} = \\frac {2}{\\sqrt 3}$,$\\cot 120 = \\frac {1}{\\tan 120} = – \\frac {1}{\\sqrt 3}$,$\\csc 135 = \\frac {1}{\\sin 135} = \\sqrt 2$,$\\sec 135 = \\frac {1}{\\cos 135} = -\\sqrt 2$, Now Trigonometric table for 210 to 270 is given by,$\\sin (210) = \\sin (180 +30) =- \\sin 30= -\\frac {1}{2}$,$\\cos (210) = \\cos (180 +30) =- \\cos 30=-\\frac {\\sqrt {3}}{2}$,$\\tan (210) = \\frac {\\sin 210}{ \\cos 210} = \\frac {1}{\\sqrt {3}}$,$\\sin (225) = \\sin (180 +45) =- \\sin 45= -\\frac {1}{\\sqrt {2}}$,$\\cos (225) = \\cos (180+45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$,$\\tan 225 = \\frac {\\sin 225}{\\cos 225} = 1$,$\\sin (270) = \\sin (180 +90) =- \\sin 90= -1$,$\\cos (270) = \\cos (180+90) =- \\cos 90= 0$,$\\tan 270 = \\frac {\\sin 270}{\\cos 270} = -\\frac {1}{0}$Undefined value,$\\csc (210) = \\frac {1}{\\sin (210)} = -2$,$\\sec (210) = \\frac {1}{\\cos (210)}=-\\frac {2}{\\sqrt 3}$,$\\cot (210) = \\frac {1}{\\tan (210)} = \\sqrt {3}$,$\\csc (225) = \\frac", "4A - 1} = \\frac{\\tan 8A}{\\tan 2A}$$ L.H.S. $$= \\frac{\\sec 8A - 1}{\\sec 4A - 1} = \\frac{1 - \\cos 8A}{1 - \\cos 4A}.\\frac{\\cos4A}{\\cos8A}$$ $$= \\frac{2\\sin^24A}{2\\sin^22A}.\\frac{\\cos 4A}{\\cos8A} = \\frac{(2\\sin4A\\cos4A).\\sin4A}{2\\sin^22A.\\cos8A}$$ $$= \\frac{\\sin8A}{\\cos8A}.\\frac{\\sin4A}{2\\sin^22A} = \\frac{\\tan8A. 2\\sin2A\\cos2A}{2\\sin^22A} = \\frac{\\tan8A}{\\tan2A} =$$ R.H.S. 13. We have to prove that $$\\frac{1 + \\tan^2(45^\\circ - A)}{1 - \\tan^2(45^\\circ - A)} = \\cosec 2A$$ L.H.S. $$= \\frac{1 + \\tan^2(45^\\circ - A)}{1 - \\tan^2(45^\\circ - A)}$$ $$= \\frac{\\cos^2(45^\\circ - A) + \\sin^2(45^\\circ - A)}{\\cos^2(45^\\circ - A) - \\sin^2(45^\\circ - A)}$$ $$= \\frac{1}{\\cos(90^\\circ - 2A)} = \\frac{1}{\\sin2A} = \\cosec2A =$$ R.H.S. 14. We have to prove that $$\\frac{\\sin A + \\sin B}{\\sin A - \\sin B} = \\frac{\\tan \\frac{A + B}{2}}{\\tan \\frac{A - B}{2}}$$ L.H.S. $$= \\frac{\\sin A + \\sin B}{\\sin A - \\sin B} = \\frac{2\\sin\\frac{A + B}{2}\\cos\\frac{A - B}{2}}{2\\cos\\frac{A + B}{2}\\sin\\frac{A - B}{2}}$$ $$= \\frac{\\tan \\frac{A + B}{2}}{\\tan \\frac{A - B}{2}} =$$ R.H.S. 15. We have to prove that $$\\frac{\\sin^2A - \\sin^2B}{\\sin A\\cos A - \\sin B\\cos B} = \\tan(A + B)$$ L.H.S. $$= \\frac{2(\\cos^2B - \\cos^2A)}{\\sin2A - \\sin2B} = \\frac{\\cos2B - \\cos2A}{\\sin2A - \\sin2B}$$ $$= \\frac{\\sin(A + B)\\sin(A - B)}{\\cos(A + B)\\sin(A - B)} = \\tan(A + B) =$$ R.H.S. 16. We have to prove that $$\\tan\\left(\\frac{\\pi}{4} + A\\right) - \\tan\\left(\\frac{\\pi}{4} - A\\right) = 2\\tan 2A$$ L.H.S. $$= \\frac{1 + \\tan A}{1 - \\tan A}", "+ $$\\sqrt 3$$ = 0 or, 2 cos$$\\theta$$ = - $$\\sqrt 3$$ or, cos$$\\theta$$ = - $$\\frac {\\sqrt 3}2$$ or, cos$$\\theta$$ = cos (180° - 30°) ∴ $$\\theta$$ = 90° and 150° Ans Here, cos2$$\\frac {\\theta}2$$ - cos$$\\frac {\\theta}2$$ + $$\\frac 14$$ = 0 or, (cos$$\\frac {\\theta}2$$)2 - 2 . cos$$\\frac {\\theta}2$$ . ($$\\frac 12$$)2 = 0 or, (cos$$\\frac {\\theta}2$$ - $$\\frac 12$$)2 = 0 or, cos$$\\frac {\\theta}2$$ - $$\\frac 12$$ = 0 or, cos$$\\frac {\\theta}2$$ = $$\\frac 12$$ or, cos$$\\frac {\\theta}2$$ = cos 60° or, $$\\frac {\\theta}2$$ = 60° or, $$\\theta$$ = 2× 60° ∴ $$\\theta$$ = 120° Ans 0% 40(^o) 30(^o) 50(^o) 60(^o) 40(^o) 60(^o) 30(^o) 50(^o) • ### 2cos2θ - 1 = 0. 50(^o) , -135(^o) 45(^o) , 135(^o) -45(^o) , -135(^o) 35(^o) , 35(^o) 0o , 180o 100o , 100o 0o , -180o 360o , 80o 155(^o) 135(^o) -35(^o) 360(^o) -60(^o) 60(^o) 10(^o) 30(^o) 90(^0) 30(^0) -30(^0) 60(^0) - 45(^0) 45(^0) 35(^0) 55(^0) 30(^0) 60(^0) 90(^0) 120(^0) 90(^0) 60(^0) 260(^0) 180(^0) • ### Sollve: 2 cos2 ( heta) = 3 sin ( heta) ( 00 ≤ ( heta) ≤ 1800 ) 50(^0) , 190(^0) 40(^0) , 180(^0) 30(^0) , 150(^0) -80(^0) , 90(^0) • ### Solve:2cos2 ( heta) - cos ( heta) = 2 [00≤ ( heta)≤ 1800] 10(^o) ,39(^o) ,120(^o) , 180(^o) 0(^o)", "by dividing $\\sin^2 x + \\cos^2 x = 1$ by $\\cos^2 x$). Define $v = \\tan x$, so we need to solve $3 \\sec^2 x = \\tan x + 5, \\quad \\text{ or} \\quad 3(1+v^2)-v-5=0, \\quad \\text{ or} \\quad 3v^2 -v -2=0.$ This is $(3v+2)(v-1)=0$, so $v = -\\frac {2}{3}, 1$, which (since $\\tan x$ has period $\\pi$) gives the four solutions $x = 45^\\circ, 225^\\circ, 146.3^\\circ, 326.3^\\circ.$", "--> $\\cos 3 t = \\cos {216}^{\\circ} = - 0.809$. OK $t = 2 \\pi$ --> $\\cos 2 t = \\cos 4 \\pi = 1$ and $\\cos 3 t = \\cos 6 \\pi = 1$. OK" ]

[ "How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "# How do you evaluate sec 137? $\\sec {137}^{\\circ} = - 1.3672$ $\\sec {137}^{\\circ} = \\frac{1}{\\cos} \\left({137}^{\\circ}\\right) = \\frac{1}{\\cos} \\left(180 - {137}^{\\circ}\\right)$ $\\implies - \\frac{1}{\\cos} \\left({43}^{\\circ}\\right) = - \\left(\\frac{1}{0.7314}\\right) = - 1.3672$", "$$x = \\frac{-14 + \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – \\frac{1}{2}$$ $$x = \\frac{-14 – \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – 3$$ 8- $$x^{\\frac{ 15}{4}}$$ $$(x^6)^{\\frac{5}{8}} = x^{6 × \\frac{5}{8}} = x^{\\frac{ 30}{8}} =x^{\\frac{ 15}{4}}$$ 9- 32 $$32 + 35 + 29 = 96$$ Average = $$\\frac{96}{3} = 32$$ 10- $$45^\\circ$$ $$90^\\circ + 45^\\circ = 135^\\circ$$ $$180^\\circ- 135^\\circ = 45^\\circ$$ 11- $$(x – 1)(x + 6)$$ $$x^2 + 5x – 6 = (x – 1)(x + 6)$$ 12- The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: 13- 5 $$C = \\sqrt{(x_A- x_B)^2 + (y- y_B)^2 ) }$$ $$C = \\sqrt{(1-(-2))^2 + (3- 7)^2}$$ $$C = \\sqrt{(3)^2 + (- 4)^2 )}$$ $$C = \\sqrt{9 + 16}$$ $$C = \\sqrt{25} = 5$$ 14- $$0.02857143 \\cong 0.0286$$ $$\\frac{4}{140} = \\frac{1}{35} = 0.02857143 \\cong 0.0286$$ 15- $$\\frac{20}{3}$$ $$f(x) = x – \\frac{5}{3} ⇒( y = x) (– \\frac{5}{3} ) ⇒y+ \\frac{5}{3} = x$$ $$f^{-1} = x+ \\frac{5}{3}$$ $$f ^{–1}(5) = 5 + \\frac{5}{3} = \\frac{20}{3}$$ 16- $$\\frac{ 1}{4}$$ Probability $$=\\frac{number of desired outcomes}{number of total outcomes} = \\frac{18}{12+18+18+24} = \\frac{18}{72} =\\frac{ 1}{4}$$ 17- $$\\frac{\\sqrt{3}}{2}$$ $$cos 30^\\circ = \\frac{\\sqrt{3}}{2}$$ 18- cosθ=$$\\frac{4}{5}$$ sinθ$$=\\frac{3}{5}$$⇒ we have following triangle, then $$c=\\sqrt{5^2-3^2 }=\\sqrt{25-9}=\\sqrt{16}=4)$$ cosθ$$=\\frac{4}{5}$$ 19- 5 $$|9 – (12 ÷ | 2 –", "$$C = (180-2x)^\\circ$$ and we have $$a=b$$ and $$c = 2a\\cos x^\\circ$$. Then $\\begin{array}{rcl} \\frac{aA+bB+cC}{a+b+c} & = & \\frac{ax + ax + 2a(180-2x)\\cos x^\\circ}{a + a + 2a\\cos x^\\circ} \\\\ & = & \\frac{2a[x + (180-2x)\\cos x^\\circ]}{2a(1 + \\cos x^\\circ)} \\; = \\; \\frac{x + (180-2x)\\cos x^\\circ}{1 + \\cos x^\\circ} \\end{array}$ and this tends to $$90$$ as $$x \\to 0$$. · 3 years, 11 months ago Thank you for noticing. I will use $$\\frac{\\pi}{3}$$ and $$\\frac{\\pi}{2}$$ instead, on the next time if i meet those numbers again. · 3 years, 11 months ago", "## Algebra 2 (1st Edition) Since $\\cos(180-theta)=-\\cos \\theta$ we obtain $\\cos(180-135)=-\\cos 135\\\\\\cos 45=-\\cos 135\\\\-\\cos 45=\\cos 135\\\\\\cos 135=-\\frac{\\sqrt 2}{2}$ Now find $\\sec135=\\frac{1}{\\cos 135}=\\frac{1}{-\\frac{\\sqrt 2}{2}}=-\\frac{2}{\\sqrt 2}=-\\sqrt 2$", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "= 0$$ ⇒then: a = 4, b = 14 and c = 6 $$x = \\frac{-14 + \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – \\frac{1}{2}$$ $$x = \\frac{-14 – \\sqrt{14^2 – 4 ,4 ,6} }{2 ,4} = – 3$$ 8- $$x^{\\frac{ 15}{4}}$$ $$(x^6)^{\\frac{5}{8}} = x^{6 × \\frac{5}{8}} = x^{\\frac{ 30}{8}} =x^{\\frac{ 15}{4}}$$ 9- 32 $$32 + 35 + 29 = 96$$ Average = $$\\frac{96}{3} = 32$$ 10- $$45^\\circ$$ $$90^\\circ + 45^\\circ = 135^\\circ$$ $$180^\\circ- 135^\\circ = 45^\\circ$$ 11- $$(x – 1)(x + 6)$$ $$x^2 + 5x – 6 = (x – 1)(x + 6)$$ 12- The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: 13- 5 $$C = \\sqrt{(x_A- x_B)^2 + (y- y_B)^2 ) }$$ $$C = \\sqrt{(1-(-2))^2 + (3- 7)^2}$$ $$C = \\sqrt{(3)^2 + (- 4)^2 )}$$ $$C = \\sqrt{9 + 16}$$ $$C = \\sqrt{25} = 5$$ 14- $$0.02857143 \\cong 0.0286$$ $$\\frac{4}{140} = \\frac{1}{35} = 0.02857143 \\cong 0.0286$$ 15- $$\\frac{20}{3}$$ $$f(x) = x – \\frac{5}{3} ⇒( y = x) (– \\frac{5}{3} ) ⇒y+ \\frac{5}{3} = x$$ $$f^{-1} = x+ \\frac{5}{3}$$ $$f ^{–1}(5) = 5 + \\frac{5}{3} = \\frac{20}{3}$$ 16- $$\\frac{ 1}{4}$$ Probability $$=\\frac{number of desired outcomes}{number of total outcomes} = \\frac{18}{12+18+18+24} = \\frac{18}{72} =\\frac{ 1}{4}$$ 17- $$\\frac{\\sqrt{3}}{2}$$ $$cos 30^\\circ = \\frac{\\sqrt{3}}{2}$$ 18- cosθ=$$\\frac{4}{5}$$ sinθ$$=\\frac{3}{5}$$⇒ we have following triangle,", "5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d = \\sqrt{34 - 30\\cos 57^\\circ}$ 4. $d \\approx 4.202$ 5. $\\displaystyle \\frac{\\sin 57^\\circ}{d} = \\frac{\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} = \\frac{\\sin E}{3}$ 6. $\\displaystyle\\sin E = \\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} \\approx 0.5987$ 7. $\\displaystyle E = \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\approx 36.78^\\circ$ 8. $\\displaystyle\\,F = 180^\\circ - 57^\\circ - E \\ =\\ 180^\\circ - 57^\\circ - \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\ \\approx\\ 86.22^\\circ$ It's just scary looking. (But, every ‘$\\,=\\,$’ is strictly correct.) Potential Remedy #3 with Comments Another potential remedy is to use ‘$\\,\\approx\\,$’ for everything that is truly an approximation. In this case, pretty much everything after the first occurrence of an approximation gets ‘$\\,\\approx\\,$’: 1. $d^2 = 3^2 + 5^2 - 2\\cdot 3\\cdot 5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d \\approx 4.202$ 4. $\\displaystyle\\frac{\\sin 57^\\circ}{4.202} \\approx \\frac{\\sin E}{3}$ 5. $\\displaystyle\\sin E \\approx \\frac{3\\sin 57^\\circ}{4.202} \\approx 0.5988$ 6. $E \\approx \\arcsin(0.5988) \\approx 36.78^\\circ$ 7. $\\,F \\approx 180^\\circ - 57^\\circ - 36.78^\\circ = 86.22^\\circ$ I don't like this at all. It makes it ‘sound like’ (say) the Law of Sines only sort of works. Potential Remedy #4 with Comments One final potential remedy is to use an ellipsis ‘$\\,\\ldots\\,$’ to indicate missing digits of a number. In this case,", "## A community for students. Sign up today! Here's the question you clicked on: ## haleyking345 one year ago Solve the equation: a.) 2 cos^2x-cos x=1 • This Question is Closed 1. Marx let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$ #### Ask your own question Ask a Question Find more explanations on OpenStudy", "by dividing $\\sin^2 x + \\cos^2 x = 1$ by $\\cos^2 x$). Define $v = \\tan x$, so we need to solve $3 \\sec^2 x = \\tan x + 5, \\quad \\text{ or} \\quad 3(1+v^2)-v-5=0, \\quad \\text{ or} \\quad 3v^2 -v -2=0.$ This is $(3v+2)(v-1)=0$, so $v = -\\frac {2}{3}, 1$, which (since $\\tan x$ has period $\\pi$) gives the four solutions $x = 45^\\circ, 225^\\circ, 146.3^\\circ, 326.3^\\circ.$", "## Trigonometry (11th Edition) Clone $$2\\cos^267\\frac{1^\\circ}{2}-1=-\\frac{\\sqrt2}{2}$$ $$2\\cos^267\\frac{1^\\circ}{2}-1$$ Recall the Double-Angle Identity for cosine: $$2\\cos^2A-1=\\cos2A$$ Therefore, replace $A=67\\frac{1^\\circ}{2}$, we can apply the above identity to $2\\cos^267\\frac{1^\\circ}{2}-1$. $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos\\Big(2\\times67\\frac{1^\\circ}{2}\\Big)$$ $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos135^\\circ$$ As $135^\\circ+45^\\circ=180^\\circ$, the absolute value of $\\cos135^\\circ$ would equal the (already positive) value of $\\cos45^\\circ$. $$|\\cos135^\\circ|=\\cos45^\\circ$$ However, $135^\\circ$ lies in quadrant II, where $\\cos\\theta\\lt0$, so $\\cos135^\\circ\\lt0$. Thus, $$\\cos135^\\circ=-\\cos45^\\circ=-\\frac{\\sqrt2}{2}$$ Now bring the value of $\\cos135^\\circ$ back to the calculation of $2\\cos^267\\frac{1^\\circ}{2}-1$ $$2\\cos^267\\frac{1^\\circ}{2}-1=\\cos135^\\circ=-\\frac{\\sqrt2}{2}$$", "• 33 + $$\\frac{1}{3}$$x = 135.2 D. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = 135.2 B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, Explanation: Given to show the correct use of the Distributive Property when solving (33 – x) = 135.2? as the correct answer is B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, if we multiply both sides with $$\\frac{1}{3}$$ we get the correct equation as (33 – x) = 135.2 because $$\\frac{1}{3}$$(33 – x) = $$\\frac{1}{3}$$ X135.2 means, $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2, therefore the correct answer is B. $$\\frac{1}{3}$$ • 33 – $$\\frac{1}{3}$$x = $$\\frac{1}{3}$$ • 135.2. Practice & Problem Solving Leveled Practice For 7-10, use the Distributive Property to solve the equations. Question 7. -2(x + 5) = 4 Explanation: Given equation as -2(x + 5) = 4, using Distributive Property we get -2 X x + (-2 X 5) = 4, -2x – 10 = 4, -2x = 14, So x = -14/2 = -7, therefore x = -7. Question 8. 3.2 = $$\\frac{4}{5}$$(b – 5) Explanation: Given equation as 3.2 = $$\\frac{4}{5}$$(b – 5) using Distributive Property, we get 3.2 = $$\\frac{4}{5}$$ X b + $$\\frac{4}{5}$$(-5), 3.2 = $$\\frac{4}{5}$$b – 4, 3.2 + 4 = $$\\frac{4}{5}$$b, 7.2 = $$\\frac{4}{5}$$b, b = $$\\frac{5}{4}$$ X 7.2, b", "Browse Questions # The value of $\\cos 1^{\\large\\circ}\\cos 2^{\\large\\circ}\\cos 3^{\\large\\circ}.......\\cos 179^{\\large\\circ}$ is $\\begin{array}{1 1}(A)\\;\\large\\frac{1}{\\sqrt 2}&(B)\\;0\\\\(C)\\;1&(D)\\;-1\\end{array}$ $\\cos 1^{\\large\\circ}\\cos 2^{\\large\\circ}\\cos 3^{\\large\\circ}$.......$\\cos 90^{\\large\\circ}$.......$\\cos 91^{\\large\\circ}$........$\\cos 179^{\\large\\circ}$ $\\therefore \\cos 90^{\\large\\circ}=0$ So the product is 0 Hence (B) is the correct answer.", "{}^{\\circ}$ and $a_2=\\tan ^{-1}\\left(\\frac{3}{4}\\right)$ are known and $c_1$ and $c_2$ need to be determined. $$750 \\cos \\left(a_1\\right)-c_2 \\cos \\left(a_2\\right)=c_1$$ $$c_2 \\sin \\left(a_2\\right)+750 \\sin \\left(a_1\\right)=0$$ This will give $$c_1=875 \\sqrt{2}=1237.44$$ and $$c_2=-625 \\sqrt{2}=-883.883$$", "1 - \\frac{\\sqrt{3}}{2} = 0.134$ $\\sin x + \\cos x = \\sqrt{2} \\cos \\left(x - \\frac{\\pi}{4}\\right) = 0.134$ $\\cos \\left(x - \\frac{\\pi}{4}\\right) = \\frac{0.134}{\\sqrt{2}} = 0.0947$ $x - \\frac{\\pi}{4} = \\pm {84.564}^{\\circ}$ $x = 84.564 + 45 = {129.564}^{\\circ}$ and $x = 360 - 84.564 + 45 = {321.564}^{\\circ}$", "= \\cos 45^\\circ.$$ Therefore, we can express the desired value as $$1 + \\frac{\\sin 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}{\\cos 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}.$$ Because $\\cos 45^\\circ = \\sin 45^\\circ$, we see that the fractional part is $1$, and so the sum is $1 + 1 = 2$, which brings us to the answer $\\boxed{B}$.", "2 x = \\frac{4}{20} = \\frac{1}{5}$ a. $\\cos 2 x = \\frac{1}{2}$ --> $2 x = \\pm \\frac{\\pi}{3}$, $x = \\pm \\frac{\\pi}{6}$, or $x = \\pm {30}^{\\circ}$ b. $\\cos 2 x = \\frac{1}{5}$ --> $2 x = \\pm {78}^{\\circ} 46$ --> $x = \\pm {39}^{\\circ} 23$ $x = {30}^{\\circ}$ and $x = {39}^{\\circ} 23$", "## Trigonometry (11th Edition) Clone $$\\cos67.5^\\circ=\\frac{\\sqrt{2-\\sqrt2}}{2}$$ 10 goes with B. $$\\cos67.5^\\circ$$ $67.5^\\circ$ is the half angle of $135^\\circ$ $$67.5^\\circ=\\frac{135^\\circ}{2}$$ Therefore, $$\\cos67.5^\\circ=\\cos\\Big(\\frac{135^\\circ}{2}\\Big)$$ - Recall the half-angle identities: $$\\cos\\Big(\\frac{A}{2}\\Big)=\\pm\\sqrt{\\frac{1+\\cos A}{2}}$$ Apply the identity to $\\cos67.5^\\circ$ with $A=135^\\circ$, we have $$\\cos67.5^\\circ=\\pm\\sqrt{\\frac{1+\\cos135^\\circ}{2}}$$ The sign of $\\cos67.5^\\circ$ would determine whether positive or negative square root should be selected. $67.5^\\circ$ is in quadrant I, where $\\cos\\theta\\gt0$. Thus, $\\cos67.5^\\circ\\gt0$. We need to pick the positive square root as a result. $$\\cos67.5^\\circ=\\sqrt{\\frac{1+\\cos135^\\circ}{2}}$$ As $135^\\circ+45^\\circ=180^\\circ$, $|\\cos135^\\circ|=\\cos45^\\circ$ $135^\\circ$ lies in quadrant II, where $\\cos\\theta\\lt0$, meaning $\\cos135^\\circ\\lt0$ Therefore, $\\cos135^\\circ=-\\cos45^\\circ=-\\frac{\\sqrt2}{2}$ So, $$\\cos67.5^\\circ=\\sqrt{\\frac{1-\\frac{\\sqrt2}{2}}{2}}$$ $$\\cos67.5^\\circ=\\sqrt{\\frac{\\frac{2-\\sqrt2}{2}}{2}}$$ $$\\cos67.5^\\circ=\\sqrt{\\frac{2-\\sqrt2}{4}}$$ $$\\cos67.5^\\circ=\\frac{\\sqrt{2-\\sqrt2}}{2}$$ 10 goes with B.", "##### Problem 2 $x -2 = 133 \\\\ x = 133 + 2 \\\\ x = \\boxed{ 135}$ ##### Problem 3 $2x + 5 = 105 \\\\ 2x = 100 \\\\ \\frac 1 2 (2x) = \\frac 1 2 (100) \\\\ x = \\boxed{ 50}$ ##### Problem 4 $4x + 7 = 131 \\\\ 4x = 124 \\\\ \\frac 1 4 (4x) = \\frac 1 4 (124) \\\\ x = \\boxed{ 31}$", "## Algebra 2 (1st Edition) $C=180-50-74=56^\\circ$. Using the law of sines, $\\frac{a}{\\sin50^\\circ}=\\frac{12}{\\sin56^\\circ}\\\\a=\\frac{12\\sin50^\\circ}{\\sin56^\\circ}\\approx11.09$ $\\frac{b}{\\sin74^\\circ}=\\frac{12}{\\sin56^\\circ}\\\\b=\\frac{12\\sin74^\\circ}{\\sin56^\\circ}\\approx14.0$" ]

[ "\\frac{\\sqrt{1}}{2} = \\frac{1}{2}, & \\cos 90^\\circ &= \\frac{\\sqrt{0}}{2} = 0 , \\end{aligned} using the usual triangles.", "## Algebra: A Combined Approach (4th Edition) x = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$ Given, that $\\frac{80}{x}$ and $\\frac{100}{x}$ are complementary angles. Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = 90$^{\\circ}$ Since denominators are same (x), thus we can directly add the numerators ( 80 and 100) Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = $\\frac{180}{x}$ Since $\\frac{180}{x}$ = 90$^{\\circ}$ thus x = $\\frac{180}{90}$$^{\\circ}$ = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$", "$\\frac{1}{x}=\\frac{1}{60} \\quad$ and $\\quad \\frac{1}{y}=\\frac{1}{80}$ $\\therefore \\quad x=60 \\mathrm{~km} / \\mathrm{hr}$ and $y=80 \\mathrm{~km} / \\mathrm{hr}$", "# The number of roots of the equation, Question: The number of roots of the equation, $(81)^{\\sin ^{2} x}+(81)^{\\cos ^{2} x}=30$ in the interval $[0, \\pi]$ is equal to: 1. (1) 3 2. (2) 4 3. (3) 8 4. (4) 2 Correct Option: , 2 Solution: $(81)^{\\sin ^{2} x}+(81)^{\\cos ^{2} x}=30$ $(81)^{\\sin ^{2} x}+\\frac{(81)^{1}}{(18)^{\\sin ^{2} x}}=30$ $(81)^{\\sin ^{2} x}=t$ $\\mathrm{t}+\\frac{81}{t}=30$ $(t-3)(t-27)=0$ $(81)^{\\sin ^{2} x}=3^{1} \\quad$ or $\\quad(81)^{\\sin ^{2} x}=3^{3}$ $3^{4 \\sin ^{2} x}=3^{1} \\quad$ or $\\quad 3^{4 \\sin ^{2} x}=3^{3}$ $\\sin ^{2} x=\\frac{1}{4} \\quad$ or $\\quad \\sin ^{2} x=\\frac{3}{4}$", "\\sin^{100} x}{100}\\Big) = \\frac{\\sin (100x) \\sin^{100} x}{100}$ so $\\int \\sin (101x) \\sin^{99}x \\ dx = \\frac{\\sin (100x) \\sin^{100} x}{100} + C$", "have already pointed out. So $$cos30^\\circ =\\frac{\\sqrt3}{2}\\\\ so\\\\ \\frac{5}{AD}=\\frac{\\sqrt3}{2}\\\\ \\frac{AD}{5}=\\frac{2}{\\sqrt3}\\\\ AD=\\frac{10}{\\sqrt3}\\\\ AD=\\frac{10\\sqrt3}{3}$$", "$\\sin(32^{\\circ})=\\frac{x} {2}$ $(\\theta)=\\cos^{-1}(\\frac{3} {5})$", "\\sin 40^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos (40^o-20^o) - \\cos (40^o+20^o) \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o - \\cos 60^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o -$ $\\frac{1}{2}$ $\\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ 2 \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (80^o+20^o) + \\sin (80^o-20^o) \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin 100^o + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (180^o - 80^o) + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\sin 80^o +$ $\\frac{\\sqrt{3}}{8}$ $-$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{\\sqrt{3}}{8}$ $=$ RHS. Hence proved. iv) LHS $= \\cos 20^o \\cos 40^o \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ 2 \\cos 20^o \\cos 40^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos ( 40^o+ 20^o) + \\cos ( 40^o - 20^o) \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos 60^o + \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[$ $\\frac{1}{2}$ $+ \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + 2\\cos 20^o \\cos 80^o \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos ( 80^o + 20^o) + \\cos ( 80^o - 20^o) \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos 100^o + \\cos 60^o \\Big]$ $=", "}}{4}} \\right) = \\frac{1}{4}\\sin {40^\\circ} - \\frac{1}{2}\\left( {\\frac{1}{2}\\sin {{40}^\\circ} - \\frac{{\\sqrt 3 }}{4}} \\right) = \\cancel{\\frac{1}{4}\\sin {40^\\circ}} - \\cancel{\\frac{1}{4}\\sin {40^\\circ}} + \\frac{{\\sqrt 3 }}{8} = \\frac{{\\sqrt 3 }}{8}.$", "## Basic College Mathematics (10th Edition) $\\frac{80\\ cal}{\\frac{1}{3}\\ cup} = \\frac{x\\ cal}{\\frac{1}{2}\\ cup}$ $\\frac{1}{3}x = 80\\times\\frac{1}{2}$ $3(\\frac{1}{3}x)= 3(40)$ $x = 120$ cal $\\frac{8\\ gm}{\\frac{1}{3}\\ cup} = \\frac{x\\ gm}{\\frac{1}{2}\\ cup}$ $\\frac{1}{3}x = 8\\times\\frac{1}{2}$ $3(\\frac{1}{3}x)= 3(4)$ $x = 12$ gm", "Evaluate each of the following Question: Evaluate each of the following $\\sin 60 \\cos 30^{\\circ}+\\cos 60^{\\circ} \\sin 30^{\\circ}$ Solution: We have to find the value of the expression $\\sin 60^{\\circ} \\cos 30^{\\circ}+\\cos 60^{\\circ} \\sin 30^{\\circ}$....(1) Now $\\sin 60^{\\circ}=\\cos 30^{\\circ}=\\frac{\\sqrt{3}}{2}, \\sin 30^{\\circ}=\\cos 60^{\\circ}=\\frac{1}{2}$ So by substituting above values in equation (1) We get, $\\sin 60^{\\circ} \\cos 30^{\\circ}+\\cos 60^{\\circ} \\sin 30^{\\circ}$ $=\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\times \\frac{1}{2}$ $=\\frac{3}{4}+\\frac{1}{4}$ $=\\frac{3+1}{4}$ $=\\frac{4}{4}$ $=1$ Therefore, $\\sin 60^{\\circ} \\cos 30^{\\circ}+\\cos 60^{\\circ} \\sin 30^{\\circ}=1$", "$\\sin 90^\\circ= 1$, $c \\left( \\frac{1}{2} \\right) = (10) (1)$ $c = 20$ If the above problem asked to find the radius of the circumcircle of $\\vartriangle ABC$, the law of sines could help to find the diameter. $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2r$ $\\frac{b}{\\sin B} = 2r$ Substituting the values for $b, B$, $\\frac{10}{\\sin 30^\\circ} = 2r$ Since $\\sin 30^\\circ= \\frac{1}{2}$, $\\cfrac{10}{\\left( \\frac{1}{2} \\right)} = 2r$ $20 = 2r$ $10 = r$ # References All images were made by the page's author using Adobe Photoshop and Cinderella2.", "$\\sin 60^{\\circ}=\\frac{\\sqrt{3}}{2}$ Therefore by substituting above values in equation (4) We get, $\\sin B=\\sin 60^{\\circ}$ $\\Rightarrow \\frac{A C}{A B}=\\sin 60^{\\circ}$ $\\Rightarrow \\frac{A C}{15}=\\frac{\\sqrt{3}}{2}$ Now by cross multiplying we get, $\\frac{A C}{15}=\\frac{\\sqrt{3}}{2}$ $\\Rightarrow 2 \\times A C=\\sqrt{3} \\times 15$ $\\Rightarrow \\quad A C=\\frac{\\sqrt{3} \\times 15}{2}$ $\\Rightarrow \\quad A C=\\frac{15}{2} \\sqrt{3}$ Therefore, $A C=\\frac{15}{2} \\sqrt{3}$ units Hence, $\\angle A=30^{\\circ}$ $B C=7.5$ units $A C=\\frac{15}{2} \\sqrt{3}$ units", "\\cos {50^\\circ}.$ Therefore, $A = \\frac{1}{4}\\cos {50^\\circ} + \\frac{1}{2}\\cos {50^\\circ}\\cos {80^\\circ} = \\frac{1}{4}\\cos {50^\\circ} + \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} + \\frac{1}{2}\\cos {{130}^\\circ}} \\right) = \\frac{1}{4}\\cos {50^\\circ} + \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} - \\frac{1}{2}\\cos {{50}^\\circ}} \\right) = \\cancel{\\frac{1}{4}\\cos {50^\\circ}} + \\frac{{\\sqrt 3 }}{8} - \\cancel{\\frac{1}{4}\\cos {50^\\circ}} = \\frac{{\\sqrt 3 }}{8}.$ ### Example 8. Calculate $\\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ}.$ Solution. We denote this expression by $$B.$$ Transform the product $$\\sin {20^\\circ}\\sin {80^\\circ}:$$ $\\sin {20^\\circ}\\sin {80^\\circ} = \\frac{1}{2}\\left[ {\\cos \\left( {{{20}^\\circ} - {{80}^\\circ}} \\right) - \\cos \\left( {{{20}^\\circ} + {{80}^\\circ}} \\right)} \\right] = \\frac{1}{2}\\left[ {\\cos \\left( { - {{60}^\\circ}} \\right) - \\cos {{100}^\\circ}} \\right] = \\frac{1}{2}\\cos {60^\\circ} - \\frac{1}{2}\\cos {100^\\circ} = \\frac{1}{4} - \\frac{1}{2}\\cos {100^\\circ}.$ Substitute this into the original triple product: $B = \\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ} = \\sin {40^\\circ}\\left( {\\frac{1}{4} - \\frac{1}{2}\\cos {{100}^\\circ}} \\right) = \\frac{1}{4}\\sin {40^\\circ} - \\frac{1}{2}\\sin {40^\\circ}\\cos {100^\\circ}.$ Now we convert the product $$\\sin {40^\\circ}\\cos {100^\\circ}:$$ $\\sin {40^\\circ}\\cos {100^\\circ} = \\frac{1}{2}\\left[ {\\sin \\left( {{{40}^\\circ} - {{100}^\\circ}} \\right) + \\sin \\left( {{{40}^\\circ} + {{100}^\\circ}} \\right)} \\right] = \\frac{1}{2}\\left[ {\\sin \\left( { - {{60}^\\circ}} \\right) + \\sin {{140}^\\circ}} \\right] = \\frac{1}{2}\\sin {140^\\circ} - \\frac{1}{2}\\sin {60^\\circ} = \\frac{1}{2}\\sin {140^\\circ} - \\frac{{\\sqrt 3 }}{4}.$ Note that $\\sin {140^\\circ} = \\sin \\left( {{{180}^\\circ} - {{40}^\\circ}} \\right) = \\sin {40^\\circ}.$ Hence, $B = \\frac{1}{4}\\sin {40^\\circ} - \\frac{1}{2}\\left( {\\frac{1}{2}\\sin {{140}^\\circ} - \\frac{{\\sqrt 3", "\\sin 60=\\left(-\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{3}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\ right)=.5$", "-1 \\end{smallmatrix}\\right] \\left(\\frac{-9}{80} \\sin(2 t)+ \\frac{1}{30} \\cos(2 t)+\\frac{9 t}{40}-\\frac{\\cos(t)}{30}\\right)$", "= $\\frac{1}{\\sqrt{3}}$ => $\\theta$ = 30° In $\\triangle$ABC => $tan\\theta = \\frac{AB}{BC}$ => $tan30^{\\circ} = \\frac{AB}{100}$ => $\\frac{1}{\\sqrt{3}} = \\frac{AB}{100}$ => $AB = \\frac{100}{\\sqrt{3}}$", "\\frac{20}{\\sin} \\frac{3 \\pi}{4} \\cdot \\sin \\frac{2 \\pi}{3} \\sin \\frac{\\pi}{12}$ ${S}_{\\Delta} = 10 h = 200 \\cdot \\frac{2}{\\sqrt{2}} \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\sin \\frac{\\pi}{12}$", "\\frac{ 2 \\cos 20^{\\circ} \\sin 80^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + 2 \\cos 20^{\\circ} \\cos 80^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (80+20) + \\sin (80^{\\circ}-20^{\\circ}) - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (80^{\\circ}+20^{\\circ}) + \\cos (80^{\\circ}-20^{\\circ}) } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 100^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos 100^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (180^{\\circ}-80^{\\circ}) + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (180^{\\circ}-80^{\\circ}) + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 80^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} - \\cos 80^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\tan 60^{\\circ}$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\times \\sqrt{3} = 1 = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{vii) } \\text{LHS } = \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin (90^{\\circ} - 80^{\\circ}) \\sin (90^{\\circ} - 40^{\\circ}) \\sin (90^{\\circ} - 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\cos 80^{\\circ} \\cos 40^{\\circ} \\cos 20^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos 80^{\\circ} (2 \\cos 40^{\\circ} \\cos 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos (40^{\\circ} + 20^{\\circ}) + \\cos (40^{\\circ} - 20^{\\circ}) \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{4}", "=-cos 60 = -\\frac {1}{2}$ $sin (-120)= -sin(120) = -sin 60 = – \\frac {\\sqrt {3}}{2}$ This site uses Akismet to reduce spam. Learn how your comment data is processed. Subscribe Notify of" ]

[ "\\frac{ 2 \\cos 20^{\\circ} \\sin 80^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + 2 \\cos 20^{\\circ} \\cos 80^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (80+20) + \\sin (80^{\\circ}-20^{\\circ}) - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (80^{\\circ}+20^{\\circ}) + \\cos (80^{\\circ}-20^{\\circ}) } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 100^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos 100^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (180^{\\circ}-80^{\\circ}) + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (180^{\\circ}-80^{\\circ}) + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 80^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} - \\cos 80^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\tan 60^{\\circ}$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\times \\sqrt{3} = 1 = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{vii) } \\text{LHS } = \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin (90^{\\circ} - 80^{\\circ}) \\sin (90^{\\circ} - 40^{\\circ}) \\sin (90^{\\circ} - 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\cos 80^{\\circ} \\cos 40^{\\circ} \\cos 20^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos 80^{\\circ} (2 \\cos 40^{\\circ} \\cos 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos (40^{\\circ} + 20^{\\circ}) + \\cos (40^{\\circ} - 20^{\\circ}) \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{4}", "4. . \\scriptsize \\begin{align*}\\cos {{50}^\\circ} & =\\cos ({{25}^\\circ}+{{25}^\\circ})\\\\=2{{\\cos }^{2}}{{25}^\\circ}-1\\\\\\therefore 2{{\\cos }^{2}}{{25}^\\circ} & =\\cos {{50}^\\circ}+1\\\\\\therefore {{\\cos }^{2}}{{25}^\\circ} & =\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}\\\\\\therefore \\cos {{25}^\\circ} & =\\sqrt{{\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}}}\\quad \\quad \\cos {{25}^\\circ} \\gt 0\\therefore {{\\cos }^{2}}{{25}^\\circ} \\gt 0\\\\&=\\sqrt{{\\displaystyle \\frac{{\\cos ({{{90}}^\\circ}-{{{40}}^\\circ})+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{\\sin {{{40}}^\\circ}+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{b+1}}{2}}}\\end{align*} 5. . \\scriptsize \\begin{align*}\\cos {{80}^\\circ}&=\\cos ({{40}^\\circ}+{{40}^\\circ})\\\\&=1-2{{\\sin }^{2}}{{40}^\\circ}\\\\&=1-2{{b}^{2}}\\end{align*} 6. . \\scriptsize \\begin{align*}\\cos (-{{800}^\\circ})&=\\cos (-{{80}^\\circ}-2\\cdot {{360}^\\circ})\\\\&=\\cos (-{{80}^\\circ})\\\\&=\\cos {{80}^\\circ}\\\\&=1-2{{b}^{2}}\\quad \\quad \\text{From e}\\text{.}\\end{align*} 2. . \\scriptsize \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{1}{{\\tan 2A}}\\\\&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{{\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-(1-2{{{\\sin }}^{2}}A)}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{2{{{\\sin }}^{2}}A}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{\\sin A}}{{\\cos A}}\\\\&=\\tan A=\\text{RHS}\\end{align*} 3. $\\scriptsize 13\\cos \\theta =-5$ and $\\scriptsize {{180}^\\circ}\\le \\theta \\le {{360}^\\circ}$ 1. . \\scriptsize \\begin{align*}\\cos \\theta & =-\\displaystyle \\frac{5}{{13}}\\\\\\cos 2\\theta & =2{{\\cos }^{2}}\\theta -1\\\\&=2{{\\left( {-\\displaystyle \\frac{5}{{13}}} \\right)}^{2}}-1\\\\&=2\\left( {\\displaystyle \\frac{{25}}{{169}}} \\right)-1\\\\&=\\displaystyle \\frac{{50}}{{169}}-1\\\\&=\\displaystyle \\frac{{50-169}}{{169}}\\\\&=-\\displaystyle \\frac{{119}}{{169}}\\end{align*} 2. . \\scriptsize \\begin{align*}\\sin 2\\theta &=2\\sin \\theta \\cos \\theta \\\\&=2\\times \\left( {-\\displaystyle \\frac{{12}}{{13}}\\times -\\displaystyle \\frac{5}{{13}}} \\right)\\\\&=2\\times \\left( {\\displaystyle \\frac{{60}}{{169}}} \\right)\\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} 3. . \\scriptsize \\begin{align*}\\tan 2\\theta &=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\cos 2\\theta }}\\\\&=\\displaystyle \\frac{{\\displaystyle \\frac{{120}}{{169}}}}{{-\\displaystyle \\frac{{119}}{{169}}}}\\\\&=-\\displaystyle \\frac{{120}}{{169}}\\times \\displaystyle \\frac{{169}}{{119}}\\\\&=-\\displaystyle \\frac{{120}}{{119}}\\end{align*} 4. . \\scriptsize \\begin{align*}\\sin ({{180}^\\circ}-2\\theta )&=\\sin 2\\theta \\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} Back to Exercise 1.2 ### Unit 1: Assessment 1. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\sin {{75}^\\circ}+\\sin {{15}^\\circ}\\\\&=\\sin ({{45}^\\circ}+{{30}^\\circ})+\\sin ({{45}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{45}^\\circ}\\cos {{30}^\\circ}+\\cos {{45}^\\circ}\\sin {{30}^\\circ}+\\sin {{45}^\\circ}\\cos {{30}^\\circ}-\\cos {{45}^\\circ}\\sin {{30}^\\circ}\\\\&=2\\sin {{45}^\\circ}\\cos {{30}^\\circ}\\\\&=2\\times \\displaystyle \\frac{1}{{\\sqrt{2}}}\\times \\displaystyle \\frac{{\\sqrt{3}}}{2}\\\\&=\\displaystyle \\frac{{\\sqrt{3}}}{{\\sqrt{2}}}\\\\&=\\displaystyle \\frac{{\\sqrt{6}}}{2}=\\text{RHS}\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\sin \\theta }}-\\displaystyle \\frac{{\\cos 2\\theta }}{{\\cos \\theta }}\\\\&=\\displaystyle \\frac{{2\\sin \\theta", "# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 11 $A=80{}^\\circ,a\\approx 39.5$ and $c\\approx 10.4$. #### Work Step by Step First we will find the value of A Properties of the triangle: Sum of three angles is $A+B+C=180{}^\\circ$ \\begin{align} & A+B+C=180{}^\\circ \\\\ & A+85{}^\\circ +15{}^\\circ =180{}^\\circ \\\\ & A=180{}^\\circ -100{}^\\circ \\\\ & A=80{}^\\circ \\end{align} Now, we will find the remaining sides using the ratio $\\frac{b}{\\sin B}$,or $\\frac{40}{\\sin 85{}^\\circ }$, Now, we will use the law of sines to find a. \\begin{align} & \\frac{a}{\\sin A}=\\frac{b}{\\sin B} \\\\ & \\frac{a}{\\sin 80{}^\\circ }=\\frac{40}{\\sin 85{}^\\circ } \\\\ & a=\\frac{40\\sin 80{}^\\circ }{\\sin 85{}^\\circ } \\\\ & a\\approx 39.5 \\end{align} Using the law of sines again, we will find c. \\begin{align} & \\frac{c}{\\sin C}=\\frac{b}{\\sin B} \\\\ & \\frac{c}{\\sin 15{}^\\circ }=\\frac{40}{\\sin 85{}^\\circ } \\\\ & c=\\frac{40\\sin 15{}^\\circ }{\\sin 85{}^\\circ } \\\\ & c\\approx 10.4 \\end{align} After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "F \\\\ T_0 &= 185^{\\circ}\\ F \\\\ T(30) &= 150^{\\circ}\\ F \\end{align*} {/eq} {eq}\\displaystyle \\begin{align*} T(t) &= T_s + (T_0 - T_s) e^{-kt} \\\\ T(30) &= 75^{\\circ}\\ F + (185^{\\circ}\\ F - 75^{\\circ}\\ F) e^{-k(30\\ min)} \\\\ 150^{\\circ}\\ F &= 75^{\\circ}\\ F + (185^{\\circ}\\ F - 75^{\\circ}\\ F) e^{-k(30\\ min)} \\\\ 150^{\\circ}\\ F - 75^{\\circ}\\ F &= 110^{\\circ}\\ Fe^{-k(30\\ min)} \\\\ \\frac{75^{\\circ}\\ F }{110^{\\circ}\\ F} &= e^{-k(30\\ min)} \\\\ \\frac{15 }{22} &= e^{-k(30\\ min)} \\end{align*} {/eq} Taking the natural logarithm of both sides to solve for k. {eq}\\displaystyle \\begin{align*} \\ln \\bigg( \\frac{15 }{22} &= e^{-k(30\\ min)} \\bigg) \\\\ \\ln \\bigg( \\frac{15 }{22} \\bigg) &= -k(30\\ min) \\\\ k&= -\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1} \\\\ k &\\approx 0.012766 \\end{align*} {/eq} Now we can write the explicit form of Newton's cooling law as {eq}\\displaystyle \\boxed{ T(t) = 75^{\\circ}\\ F + (110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)t}} {/eq}. a.) To solve for the temperature after 40 minutes we set t = 40 mins and use the function we derived above. {eq}\\displaystyle \\begin{align*} T(t) &= 75^{\\circ}\\ F + 110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)t} \\\\ &= 75^{\\circ}\\ F + 110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)(40\\ min)} \\\\ T(t) &= \\boxed{ 141.01 ^{\\circ}\\ F} \\end{align*} {/eq} The temperature of the turkey after 40 minutes is {eq}\\boxed{", "\\sin 40^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos (40^o-20^o) - \\cos (40^o+20^o) \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o - \\cos 60^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o -$ $\\frac{1}{2}$ $\\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ 2 \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (80^o+20^o) + \\sin (80^o-20^o) \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin 100^o + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (180^o - 80^o) + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\sin 80^o +$ $\\frac{\\sqrt{3}}{8}$ $-$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{\\sqrt{3}}{8}$ $=$ RHS. Hence proved. iv) LHS $= \\cos 20^o \\cos 40^o \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ 2 \\cos 20^o \\cos 40^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos ( 40^o+ 20^o) + \\cos ( 40^o - 20^o) \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos 60^o + \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[$ $\\frac{1}{2}$ $+ \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + 2\\cos 20^o \\cos 80^o \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos ( 80^o + 20^o) + \\cos ( 80^o - 20^o) \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos 100^o + \\cos 60^o \\Big]$ $=", "## Algebra: A Combined Approach (4th Edition) x = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$ Given, that $\\frac{80}{x}$ and $\\frac{100}{x}$ are complementary angles. Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = 90$^{\\circ}$ Since denominators are same (x), thus we can directly add the numerators ( 80 and 100) Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = $\\frac{180}{x}$ Since $\\frac{180}{x}$ = 90$^{\\circ}$ thus x = $\\frac{180}{90}$$^{\\circ}$ = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$", "50^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ (\\cos 10^{\\circ} \\cos 50^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ (2 \\cos 50^{\\circ} \\cos 10^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\Big( \\cos (50^{\\circ}+10^{\\circ}) + \\cos (50^{\\circ}-10^{\\circ}) \\Big) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\cos 60^{\\circ} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\frac{1}{2} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{4} \\cos 40^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} (2\\cos 40^{\\circ} \\cos 70^{\\circ})$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} ( \\cos (70^{\\circ}+40^{\\circ}) + \\cos (70^{\\circ}-40^{\\circ}) )$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} \\cos 110 + \\frac{\\sqrt{3}}{8} \\cos 30^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos 110 + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos (180 - 70) + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} - \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{3}{16} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{ii) } \\text{LHS } = \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} ( 2 \\cos 160^{\\circ} \\cos 40^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} [ \\cos ( 160^{\\circ}+40^{\\circ})", "\\Big[ \\cos \\frac{2\\pi}{3} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ \\cos \\Big( \\frac{\\pi}{2} + \\frac{\\pi}{6} \\Big) + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\sin \\frac{\\pi}{6} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\frac{1}{2} + \\cos 2x \\Big]$ $\\displaystyle = - \\cos x + 2 \\cos x \\cos 2x$ $\\displaystyle = - \\cos x + [ \\cos (2x + x) + \\cos ( 2x - x) ]$ $\\displaystyle = - \\cos x + \\cos 3x + \\cos x$ $\\displaystyle = \\cos 3x =$ RHS $\\displaystyle \\\\$ Question 5: Prove that: $\\displaystyle \\text{ i) } \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos 50^{\\circ} \\cos 70^{\\circ} = \\frac{3}{16} \\hspace{1.0cm} \\text{ ii) } \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} = - \\frac{1}{8}$ $\\displaystyle \\text{ iii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 80^{\\circ} = \\frac{\\sqrt{3}}{8} \\hspace{1.0cm} \\text{ iv) } \\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ} = \\frac{1}{8}$ $\\displaystyle \\text{ v) } \\tan 20^{\\circ} \\tan 40^{\\circ} \\tan 60^{\\circ} \\tan 80^{\\circ} = 3 \\hspace{1.0cm} \\text{ vi) } \\tan 20^{\\circ} \\tan 30^{\\circ} \\tan 40^{\\circ} \\tan 80^{\\circ} = 1$ $\\displaystyle \\text{ vii) } \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ} = \\frac{\\sqrt{3}}{16} \\hspace{1.0cm} \\text{ viii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 60^{\\circ} \\sin 80^{\\circ} = \\frac{3}{16}$ $\\displaystyle \\text{i) } \\text{LHS } = \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos", "into a sum of trig functions: ${\\cos {50^\\circ}\\cos {80^\\circ} }={ \\frac{1}{2}\\left[ {\\cos \\left( {{{50}^\\circ} – {{80}^\\circ}} \\right) + \\cos \\left( {{{50}^\\circ} + {{80}^\\circ}} \\right)} \\right] }={ \\frac{1}{2}\\left[ {\\cos \\left( { – {{30}^\\circ}} \\right) + \\cos {{130}^\\circ}} \\right] }={ \\frac{1}{2}\\cos {30^\\circ} + \\frac{1}{2}\\cos {130^\\circ} }={ \\frac{{\\sqrt 3 }}{4} + \\frac{1}{2}\\cos {130^\\circ}.}$ By the reduction formula, ${\\cos {130^\\circ} = \\cos \\left( {{{180}^\\circ} – {{50}^\\circ}} \\right) }={ – \\cos {50^\\circ}.}$ Therefore, ${A = \\frac{1}{4}\\cos {50^\\circ} + \\frac{1}{2}\\cos {50^\\circ}\\cos {80^\\circ} }={ \\frac{1}{4}\\cos {50^\\circ} }+{ \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} + \\frac{1}{2}\\cos {{130}^\\circ}} \\right) }={ \\frac{1}{4}\\cos {50^\\circ} }+{ \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} – \\frac{1}{2}\\cos {{50}^\\circ}} \\right) }={ \\cancel{\\frac{1}{4}\\cos {50^\\circ}} + \\frac{{\\sqrt 3 }}{8} – \\cancel{\\frac{1}{4}\\cos {50^\\circ}} }={ \\frac{{\\sqrt 3 }}{8}.}$ ### Example 8. Calculate $\\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ}.$ Solution. We denote this expression by $$B.$$ Transform the product $$\\sin {20^\\circ}\\sin {80^\\circ}:$$ ${\\sin {20^\\circ}\\sin {80^\\circ} }={ \\frac{1}{2}\\left[ {\\cos \\left( {{{20}^\\circ} – {{80}^\\circ}} \\right) – \\cos \\left( {{{20}^\\circ} + {{80}^\\circ}} \\right)} \\right] }={ \\frac{1}{2}\\left[ {\\cos \\left( { – {{60}^\\circ}} \\right) – \\cos {{100}^\\circ}} \\right] }={ \\frac{1}{2}\\cos {60^\\circ} – \\frac{1}{2}\\cos {100^\\circ} }={ \\frac{1}{4} – \\frac{1}{2}\\cos {100^\\circ}.}$ Substitute this into the original triple product: ${B = \\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ} }={ \\sin {40^\\circ}\\left( {\\frac{1}{4} – \\frac{1}{2}\\cos {{100}^\\circ}} \\right) }={ \\frac{1}{4}\\sin {40^\\circ} – \\frac{1}{2}\\sin {40^\\circ}\\cos {100^\\circ}.}$ Now we convert the product $$\\sin {40^\\circ}\\cos {100^\\circ}:$$ ${\\sin {40^\\circ}\\cos {100^\\circ} }={ \\frac{1}{2}\\left[", "0 \\\\ 0 & 0 & 1 } \\right] \\\\ \\\\ \\\\ & = & \\left[ \\matrix { \\;\\;\\; \\cos \\psi \\cos \\theta \\cos \\phi - \\sin \\psi \\sin \\phi & \\;\\;\\; \\sin \\psi \\cos \\theta \\cos \\phi + \\cos \\psi \\sin \\phi & -\\sin \\theta \\cos \\phi \\\\ -\\cos \\psi \\cos \\theta \\sin \\phi - \\sin \\psi \\cos \\phi & -\\sin \\psi \\cos \\theta \\sin \\phi + \\cos \\psi \\cos \\phi & \\;\\;\\; \\sin \\theta \\sin \\phi \\\\ \\cos \\psi \\sin \\theta & \\sin \\psi \\sin \\theta & \\cos \\theta } \\right] \\end{eqnarray}$ Even though the rotation angles are applied in $$\\psi, \\theta, \\phi$$ order, the matrices are indeed written in the opposite order: $$\\phi, \\theta, \\psi$$. ### Roe Angles Example In the figure above, $$\\psi = 60^\\circ, \\theta = 30^\\circ,$$ and $$\\phi = 45^\\circ$$. These Roe angles give a coordinate transformation matrix equal to $\\begin{eqnarray} {\\bf Q} & = & \\left[ \\matrix { \\;\\;\\; \\cos 60^\\circ \\cos 30^\\circ \\cos 45^\\circ - \\sin 60^\\circ \\sin 45^\\circ & \\;\\;\\; \\sin 60^\\circ \\cos 30^\\circ \\cos 45^\\circ + \\cos 60^\\circ \\sin 45^\\circ & -\\sin 30^\\circ \\cos 45^\\circ \\\\ -\\cos 60^\\circ \\cos 30^\\circ \\sin 45^\\circ - \\sin 60^\\circ \\cos 45^\\circ & -\\sin 60^\\circ \\cos 30^\\circ \\sin 45^\\circ + \\cos 60^\\circ \\cos 45^\\circ & \\;\\;\\; \\sin 30^\\circ \\sin 45^\\circ \\\\ \\cos 60^\\circ", "; 30^{\\circ} \\rightarrow 1 ; 45^{\\circ} \\rightarrow 2 ; 60^{\\circ} \\rightarrow 3 ; 90^{\\circ} \\rightarrow 4$$. Then divide the values by $$4$$ and square root the entire value. $$0^{\\circ} \\rightarrow \\sqrt{\\frac{0}{4}}=0 ; 30^{\\circ} \\rightarrow \\sqrt{\\frac{1}{4}}=\\frac{1}{2} ; 45^{\\circ} \\rightarrow \\sqrt{\\frac{2}{4}}=\\frac{1}{\\sqrt{2}} ; 60^{\\circ} \\rightarrow \\sqrt{\\frac{3}{4}}=\\frac{\\sqrt{3}}{2} ; 90^{\\circ} \\rightarrow \\sqrt{\\frac{4}{4}}=1$$. This gives the values of sine for these five angles. Now for the remaining three, use:$$\\sin \\left(180^{\\circ}-x\\right)= \\sin x \\quad \\sin \\left(180^{\\circ}+x\\right)=\\,- \\sin x \\sin \\left(360^{\\circ}-x\\right)=\\,- \\sin x$$This means, $$\\sin \\left(180^{\\circ}-0^{\\circ}\\right)=\\sin 0^{\\circ} \\quad \\sin \\left(180^{\\circ}+90^{\\circ}\\right)=\\,- \\sin 90^{\\circ} \\sin \\left(360^{\\circ}-0^{\\circ}\\right)=\\,- \\sin 0^{\\circ}$$ Step 3: Determine the value of $$\\cos$$.$$\\sin \\left(90^{\\circ}-x\\right)=\\cos x$$To find values for $$\\cos x$$, use this formula.For example, equals $$\\left(90^{\\circ}-45^{\\circ}\\right)= \\sin 45^{\\circ},\\left(90^{\\circ}-30^{\\circ}\\right)=\\sin 60^{\\circ}$$ and vice versa.You can quickly determine the value of the $$\\cos$$ function by using this method: Step 4: Determine the value of $$\\tan$$. We know that $$\\sin$$ divided by $$\\cos$$ equals the $$\\tan$$.$$\\frac{{\\sin }}{{\\cos }} = \\tan$$Divide the value of $$\\sin$$ at $$0^{\\circ}$$ by the value of $$\\cos$$ at $$0^{\\circ}$$ to get the value of $$\\tan$$ at $$0^{\\circ}$$. Take a look at the sample below.$$\\tan 0^{\\circ}=\\frac{0}{1}=0$$In the same way, the table would be as follows. Step 5: Determine the value of $$\\cot$$.The reciprocal of $$\\tan$$ is equal to the value of $$\\cot$$. Divide $$1$$ by the value of $$\\tan$$ at $$0^{\\circ}$$ to get the value of $$\\cot$$", "and \\begin{align*}b = 30^\\circ\\end{align*}. \\begin{align*}\\cos (45^\\circ - 30^\\circ) & = \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{6} + \\sqrt{2}}{4}\\end{align*} #### Example C Find the exact value of \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*}, in radians. Solution: \\begin{align*}\\cos \\frac{5 \\pi}{12} = \\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right )\\end{align*}, notice that \\begin{align*}\\frac{\\pi}{4} = \\frac{3 \\pi}{12}\\end{align*} and \\begin{align*}\\frac{\\pi}{6} = \\frac{2 \\pi}{12}\\end{align*} \\begin{align*}\\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right ) & = \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} \\\\ \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ & = \\frac{\\sqrt{6} - \\sqrt{2}}{4}\\end{align*} ### Vocabulary Cosine Sum Formula: The cosine sum formula relates the cosine of a sum of two arguments to a set of sine and cosines functions, each containing one argument. Cosine Difference Formula: The cosine difference formula relates the cosine of a difference of two arguments to a set of sine and cosines functions, each containing one argument. ### Guided Practice 1. Find the exact value for \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*} 2. Find the exact value for \\begin{align*}\\cos \\frac{7 \\pi}{12}\\end{align*} 3. Find the exact value for \\begin{align*}\\cos 345^\\circ\\end{align*} Solutions: 1. \\begin{align*}\\cos \\frac{5 \\pi}{12} & = \\cos \\left (\\frac{2 \\pi}{12} + \\frac{3", "\\Big[ \\cos 20^{\\circ} \\sin 80^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ 2 \\cos 20^{\\circ} \\sin 80^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin (80^{\\circ}+20^{\\circ}) + \\sin (80^{\\circ}-20^{\\circ}) \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin 100^{\\circ} + \\sin 60^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin (180^{\\circ} - 80^{\\circ}) + \\sin 60^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\sin 80^{\\circ} + \\frac{\\sqrt{3}}{8} - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{iv) } \\text{LHS } = \\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 20^{\\circ} \\cos 40^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos ( 40^{\\circ}+ 20^{\\circ}) + \\cos ( 40^{\\circ} - 20^{\\circ}) \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 60^{\\circ} + \\cos 20^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\frac{1}{2} + \\cos 20^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + 2\\cos 20^{\\circ} \\cos 80^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos ( 80^{\\circ} + 20^{\\circ}) + \\cos ( 80^{\\circ} - 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos 100^{\\circ} + \\cos 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos (180^{\\circ} - 80^{\\circ}) + \\cos 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} -", "30^{\\circ}$$ $$a = \\displaystyle \\frac {b\\sin(130^{\\circ})}{\\sin(20^{\\circ})} = 22.398$$ $$c = \\displaystyle \\frac {b\\sin(30^{\\circ})}{\\sin(20^{\\circ})} = 14.619$$ 2. $$A = 180 ^{\\circ} - 120 ^{\\circ} - 30 ^{\\circ} = 30 ^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(120^{\\circ})}{\\sin30^{\\circ})} = 27.713$$ $$c = \\displaystyle \\frac {a\\sin(30^{\\circ})}{sin(30^{\\circ})} = 16$$ 3. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(30^{\\circ})} = 17.851$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 107.76^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 42.24^{\\circ}$$ 4. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(108^{\\circ})} = 20.438^{\\circ}$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 27.73^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 44.27^{\\circ}$$ 5. $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 38.44^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 61.73^{\\circ}$$ $$C = \\arccos\\left(\\displaystyle \\frac {a^2 + b^2 - c^2}{2ab}\\right) = 79.84^{\\circ}$$ 6. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 66.67^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 66.67^{\\circ} = 73.33^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(73.33^{\\circ})}{\\sin(40^{\\circ})} = 10.432$$ 7. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 32.39^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 32.39^{\\circ} = 107.61^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(107.61^{\\circ})}{\\sin(40^{\\circ})} = 17.794$$ ### Exercise $$\\PageIndex{15}$$ 1. A ladder $$10$$ meters long is leaning against the side of a building. It makes an angle of $$75^{\\circ}$$ with the ground. How far up", "\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} + 1}{2\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} + 1}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)}{4} \\end{align*} $$\\cos \\text{15} °$$ \\begin{align*} \\cos \\text{15} ° &= \\cos ( \\text{60} ° - \\text{45} °) \\\\ &= \\cos \\text{60} ° \\cos \\text{45} ° + \\sin \\text{60} ° \\sin \\text{45} ° \\\\ &= \\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{1 + \\sqrt{3}}{2\\sqrt{2}} \\\\ &= \\frac{1 + \\sqrt{3}}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( 1 + \\sqrt{3} \\right)}{4} \\end{align*} $$\\sin \\text{15} °$$ \\begin{align*} \\sin \\text{15} ° &= \\sin ( \\text{60} ° - \\text{45} °) \\\\ &= \\sin \\text{60} ° \\cos \\text{45} ° - \\cos \\text{60} ° \\sin \\text{45} ° \\\\ &= \\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}} - \\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} - 1}{2\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} - 1}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\end{align*} $$\\tan \\text{15} °$$ \\begin{align*} \\tan \\text{15} ° &= \\frac{\\sin \\text{15} °}{\\cos \\text{15} °} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\div \\frac{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)}{4} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\times \\frac{4}{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)} \\\\ &= \\frac{ \\sqrt{3} - 1 }{ \\sqrt{3} + 1 } \\\\ &= \\frac{ \\sqrt{3} - 1 }{ \\sqrt{3} + 1 } \\times \\frac{\\sqrt{3} - 1}{\\sqrt{3} - 1} \\\\ &= \\frac{", "20o + i sin 20o) c. Multiply: \\begin{align*}2 \\left (\\mbox{cos}\\ \\frac{\\pi} {8} + i\\ \\mbox{sin}\\ \\frac{\\pi} {8}\\right ) \\times 2 \\left (\\mbox{cos}\\ \\frac{\\pi} {10} + i\\ \\mbox{sin}\\ \\frac{\\pi} {10}\\right )\\end{align*} d. Divide: 2(cos 80o + i sin 80o) ÷ 6(cos 200o + i sin 200o) e. Divide: 3cis(130o) ÷ 4cis(270o) 1. a. \\begin{align*}4\\sqrt{2} (\\mbox{cos}\\ 15^\\circ + i\\ \\mbox{sin}\\ 15^\\circ)\\end{align*} b. 8 cis (60o) c. \\begin{align*}4\\ \\mbox{cis}\\ \\left (\\frac{9 \\pi} {40}\\right )\\end{align*} d. \\begin{align*}\\frac{1} {3} \\mbox{cis}\\ (240^\\circ)\\end{align*} e. \\begin{align*}\\frac{3} {4} \\mbox{cis}\\ (220^\\circ)\\end{align*} Show Hide Details Description Tags: Subjects: Date Created: Feb 23, 2012", "$\\frac{BM}{CM}=\\frac{sin(10^\\circ)sin(30^\\circ)}{sin(20^\\circ)sin(40^\\circ)}$. Let $\\angle MBC=x$. Then, $\\angle MCB=80^\\circ-x$. By the law of sines, $\\frac{BM}{CM}=\\frac{sin(80^\\circ-x)}{sin(x)}$. Combining, we have $\\frac{sin(80^\\circ-x)}{sin(x)}=\\frac{sin(10^\\circ)sin(30^\\circ)}{sin(20^\\circ)sin(40^\\circ)}$. From here, we can use the given trigonometric identities at each step: \\begin{equations*}[t]{llr} \\frac{sin(80^\\circ-x)}{sin(x)}&=\\frac{sin(10^\\circ)sin(30^\\circ)}{sin(20^\\circ)sin(40^\\circ)}\\$10] sin(80^\\circ-x)sin(20^\\circ)sin(40^\\circ)&=sin(10^\\circ)sin(30^\\circ)sin(x)\\\\[10] sin(80^\\circ-x)sin(20^\\circ)sin(40^\\circ)&=\\frac{1}{2}sin(10^\\circ)sin(x)&[sin(30^\\circ)=1/2]\\\\[10] sin(80^\\circ-x)sin(30^\\circ-10^\\circ)sin(30^\\circ+10^\\circ)&=\\frac{1}{2}sin(10^\\circ)sin(x)\\\\[10] sin(80^\\circ-x)(cos^2(10^\\circ)-cos^2(30^\\circ))&=\\frac{1}{2}sin(10^\\circ)sin(x)&[sin(A-B)sin(A+B)=cos^2 B-cos^2 A]\\\\[10] sin(80^\\circ-x)(cos^2(10^\\circ)-\\frac{3}{4})&=\\frac{1}{2}sin(10^\\circ)sin(x)&[cos(30^\\circ)=\\frac{\\sqrt{3}}{2}]\\\\[10] sin(80^\\circ-x) \\frac{4cos^3(10^\\circ)-3cos(10^\\circ)}{4cos(10^\\circ)}&=\\frac{1}{2}sin(10^\\circ)sin(x)\\\\[10] sin(80^\\circ-x) \\frac{cos(30^\\circ)}{4cos(10^\\circ)}&=\\frac{1}{2}sin(10^\\circ)sin(x)&[cos(3A)=4cos^3 A-3cos A]\\\\[10] sin(80^\\circ-x)cos(30^\\circ)&=2sin(10^\\circ)cos(10^\\circ)sin(x)\\\\[10] sin(80^\\circ-x)cos(30^\\circ)&=sin(20^\\circ)sin(x)&[sin(2A)=2sin A cos A ]\\\\[10] sin(80^\\circ-x)sin(60^\\circ)&=sin(20^\\circ)sin(x)&[cos(30^\\circ)=sin(60^\\circ)]\\\\[10] \\frac{1}{2}(cos(20^\\circ-x)-cos(140^\\circ-x))&=\\frac{1}{2}(cos(20^\\circ-x)-cos(20^\\circ+x))&[sin A sin B=\\frac{1}{2}(cos(A-B)-cos(A+B))]\\\\[10] cos(140^\\circ-x)&=cos(20^\\circ+x) \\end{equations*} (Error compiling LaTeX. ! LaTeX Error: Environment equations* undefined.) The only acute angle satisfying this equality is $x=60^\\circ$. Therefore, $\\angle ACB=80^\\circ-x+30^\\circ=50^\\circ$ and $\\angle BAC=10^\\circ+40^\\circ=50^\\circ$. Thus, $\\triangle ABC$ is isosceles. ### Solution 3 If $\\angle{MBC} = x$ then by Angle Sum in a Triangle we have $\\angle{MCB} = 80^\\circ - x$. By Trig Ceva we have $\\[\\sin 10^\\circ \\sin x \\sin 30^\\circ = \\sin (80^\\circ - x) \\sin 40^\\circ \\sin 20^\\circ.$$ Because $\\dfrac{\\sin x}{\\sin (80^\\circ - x)}$ is monotonic increasing over $(0, \\dfrac{\\pi}{2})$, there is only one solution $0 \\le x \\le \\dfrac{\\pi}{2}$ to the equation. We claim it is $x = 60^\\circ$, which will make $ABC$ isosceles with $\\angle{A} = \\angle{C}$. Notice that $$\\sin 20^\\circ \\sin 20^\\circ \\sin 40^\\circ = 2 \\sin 10^\\circ \\cos 10^\\circ \\sin 20^\\circ \\sin 40^\\circ$$ $$= \\sin 10^\\circ (\\sin 10^\\circ + \\frac{1}{2}) \\sin 40^\\circ$$ $$= \\sin 10^\\circ (\\frac{1}{2} \\sin 40^\\circ + \\frac{1}{2} (\\cos 30^\\circ - \\cos 50^\\circ))$$ $$= \\sin 10^\\circ \\frac{1}{2} \\cos 30^\\circ$$ $$= \\sin 10^\\circ \\sin", "from Appendix C for our superposition: \\begin{aligned} x &= (20t - 30x) \\\\ y &= (20t + 30x)\\end{aligned} \\begin{aligned} y_s &= 4 \\cos{x} + (-4 \\cos{y}) \\\\ &= 4 \\left[ \\cos{x} - \\cos{y} \\right] \\\\ &= 4 \\left[ -2 \\sin{\\frac{x+y}{2}} \\sin{\\frac{x-y}{2}} \\right] \\\\ &= 4 \\left[ -2 \\left( \\sin{\\frac{(20t - 30x)+(20t + 30x)}{2}} \\right) \\left( \\sin{\\frac{(20t - 30x)-(20t + 30x)}{2}} \\right) \\right] \\\\ &= 4 \\left[ -2 \\left( \\sin{\\frac{20t - 30x+20t + 30x}{2}} \\right) \\left( \\sin{\\frac{20t - 30x-20t - 30x)}{2}} \\right) \\right] \\\\ &= 4 \\left[ -2 \\left( \\sin{\\frac{40t}{2}} \\right) \\left( \\sin{\\frac{-60x)}{2}} \\right) \\right] \\\\ &= 4 \\left[ -2 \\left( \\sin{20t} \\right) \\left( \\sin{-30x} \\right) \\right] \\\\ &= -8 \\left( \\sin{20t} \\right) \\left( \\sin{-30x} \\right) \\\\ &= -8 \\left( \\sin{20t} \\right) \\left( -\\sin{30x} \\right) \\\\ &= 8 \\left( \\sin{20t} \\right) \\left( \\sin{30x} \\right) \\\\\\end{aligned} Now substitute $$t=\\frac{\\pi}{50}$$: \\begin{aligned} y_s &= 8 \\left( \\sin{ \\left( \\frac{ 20 \\pi}{50} \\right)} \\right) \\left( \\sin{30x} \\right) \\\\ &= 7.60845 \\sin{(30x)}\\end{aligned} A sine function reaches its maximum with its phase angle, $$\\phi(x,t)$$, is equal to $$n \\pi + 90^{\\circ}$$, and reaches its minimum when $$\\phi(x,t)$$ is equal to $$n \\pi$$. ## Solution to Part (a) • Since $$x$$ and $$t$$ in wave function $$y_1$$ have different signs, we conclude that $$y_1$$ is traveling in the positive $$x$$-direction. • Since $$x$$ and $$t$$ in", "= \\frac{\\sin B}{6.1}, B & = 84.8^\\circ \\\\ 180^\\circ - 32.6^\\circ - 84.8^\\circ, C & = 62.6^\\circ\\end{align*} (d) No solution (e) Both cases need angle \\begin{align*}C\\end{align*} and side \\begin{align*}c\\end{align*}. \\begin{align*}\\text{Case}\\ 1: C & = 180^\\circ - 47^\\circ - 77.2^\\circ = 55.8^\\circ, \\frac{\\sin 55.8^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 50.9 \\\\ \\text{Case}\\ 2: C & = 180^\\circ - 47^\\circ - 102.8^\\circ = 30.2^\\circ,\\frac{\\sin 30.2^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 30.95\\end{align*} 2. To find the area of the rhombus, use the formula \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} and then multiply that by 2. We first need to find one of the angles that are opposite the given diagonal (they are both the same measurement). We will call it angle \\begin{align*}A\\end{align*}. \\begin{align*}21.5^2 = 12^2 + 12^2 - 2(12)(12) \\sin A, A = 127.2^\\circ\\end{align*}, which means the other two angles are both \\begin{align*}52.8^\\circ (360^\\circ - 127.2^\\circ - 127.2^\\circ\\end{align*} and then divide by 2). \\begin{align*}K = 2 \\left (\\frac{1}{2}(12)(12) \\sin 127.2^\\circ \\right ) = 114.7\\end{align*} 3. Divide the pentagon into three triangles, drawing segments from \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{5}\\end{align*}, called \\begin{align*}x\\end{align*} below, and \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{4}\\end{align*}, called \\begin{align*}y\\end{align*} below. With these three triangles, only the middle triangle needs us to find two sides and the angle between them (called \\begin{align*}\\angle{Z}\\end{align*} below) to use \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} (the outer" ]

[ "\\frac{ 2 \\cos 20^{\\circ} \\sin 80^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + 2 \\cos 20^{\\circ} \\cos 80^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (80+20) + \\sin (80^{\\circ}-20^{\\circ}) - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (80^{\\circ}+20^{\\circ}) + \\cos (80^{\\circ}-20^{\\circ}) } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 100^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos 100^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin (180^{\\circ}-80^{\\circ}) + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} + \\cos (180^{\\circ}-80^{\\circ}) + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\Big[ \\frac{ \\sin 80^{\\circ} + \\sin 60^{\\circ} - \\sin 80^{\\circ} }{ \\cos 80^{\\circ} - \\cos 80^{\\circ} + \\cos 60^{\\circ} } \\Big]$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\tan 60^{\\circ}$ $\\displaystyle = \\frac{1}{\\sqrt{3}} \\times \\sqrt{3} = 1 = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{vii) } \\text{LHS } = \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\sin (90^{\\circ} - 80^{\\circ}) \\sin (90^{\\circ} - 40^{\\circ}) \\sin (90^{\\circ} - 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ \\cos 80^{\\circ} \\cos 40^{\\circ} \\cos 20^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos 80^{\\circ} (2 \\cos 40^{\\circ} \\cos 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\cos (40^{\\circ} + 20^{\\circ}) + \\cos (40^{\\circ} - 20^{\\circ}) \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{4}", "4. . \\scriptsize \\begin{align*}\\cos {{50}^\\circ} & =\\cos ({{25}^\\circ}+{{25}^\\circ})\\\\=2{{\\cos }^{2}}{{25}^\\circ}-1\\\\\\therefore 2{{\\cos }^{2}}{{25}^\\circ} & =\\cos {{50}^\\circ}+1\\\\\\therefore {{\\cos }^{2}}{{25}^\\circ} & =\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}\\\\\\therefore \\cos {{25}^\\circ} & =\\sqrt{{\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}}}\\quad \\quad \\cos {{25}^\\circ} \\gt 0\\therefore {{\\cos }^{2}}{{25}^\\circ} \\gt 0\\\\&=\\sqrt{{\\displaystyle \\frac{{\\cos ({{{90}}^\\circ}-{{{40}}^\\circ})+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{\\sin {{{40}}^\\circ}+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{b+1}}{2}}}\\end{align*} 5. . \\scriptsize \\begin{align*}\\cos {{80}^\\circ}&=\\cos ({{40}^\\circ}+{{40}^\\circ})\\\\&=1-2{{\\sin }^{2}}{{40}^\\circ}\\\\&=1-2{{b}^{2}}\\end{align*} 6. . \\scriptsize \\begin{align*}\\cos (-{{800}^\\circ})&=\\cos (-{{80}^\\circ}-2\\cdot {{360}^\\circ})\\\\&=\\cos (-{{80}^\\circ})\\\\&=\\cos {{80}^\\circ}\\\\&=1-2{{b}^{2}}\\quad \\quad \\text{From e}\\text{.}\\end{align*} 2. . \\scriptsize \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{1}{{\\tan 2A}}\\\\&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{{\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-(1-2{{{\\sin }}^{2}}A)}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{2{{{\\sin }}^{2}}A}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{\\sin A}}{{\\cos A}}\\\\&=\\tan A=\\text{RHS}\\end{align*} 3. $\\scriptsize 13\\cos \\theta =-5$ and $\\scriptsize {{180}^\\circ}\\le \\theta \\le {{360}^\\circ}$ 1. . \\scriptsize \\begin{align*}\\cos \\theta & =-\\displaystyle \\frac{5}{{13}}\\\\\\cos 2\\theta & =2{{\\cos }^{2}}\\theta -1\\\\&=2{{\\left( {-\\displaystyle \\frac{5}{{13}}} \\right)}^{2}}-1\\\\&=2\\left( {\\displaystyle \\frac{{25}}{{169}}} \\right)-1\\\\&=\\displaystyle \\frac{{50}}{{169}}-1\\\\&=\\displaystyle \\frac{{50-169}}{{169}}\\\\&=-\\displaystyle \\frac{{119}}{{169}}\\end{align*} 2. . \\scriptsize \\begin{align*}\\sin 2\\theta &=2\\sin \\theta \\cos \\theta \\\\&=2\\times \\left( {-\\displaystyle \\frac{{12}}{{13}}\\times -\\displaystyle \\frac{5}{{13}}} \\right)\\\\&=2\\times \\left( {\\displaystyle \\frac{{60}}{{169}}} \\right)\\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} 3. . \\scriptsize \\begin{align*}\\tan 2\\theta &=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\cos 2\\theta }}\\\\&=\\displaystyle \\frac{{\\displaystyle \\frac{{120}}{{169}}}}{{-\\displaystyle \\frac{{119}}{{169}}}}\\\\&=-\\displaystyle \\frac{{120}}{{169}}\\times \\displaystyle \\frac{{169}}{{119}}\\\\&=-\\displaystyle \\frac{{120}}{{119}}\\end{align*} 4. . \\scriptsize \\begin{align*}\\sin ({{180}^\\circ}-2\\theta )&=\\sin 2\\theta \\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} Back to Exercise 1.2 ### Unit 1: Assessment 1. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\sin {{75}^\\circ}+\\sin {{15}^\\circ}\\\\&=\\sin ({{45}^\\circ}+{{30}^\\circ})+\\sin ({{45}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{45}^\\circ}\\cos {{30}^\\circ}+\\cos {{45}^\\circ}\\sin {{30}^\\circ}+\\sin {{45}^\\circ}\\cos {{30}^\\circ}-\\cos {{45}^\\circ}\\sin {{30}^\\circ}\\\\&=2\\sin {{45}^\\circ}\\cos {{30}^\\circ}\\\\&=2\\times \\displaystyle \\frac{1}{{\\sqrt{2}}}\\times \\displaystyle \\frac{{\\sqrt{3}}}{2}\\\\&=\\displaystyle \\frac{{\\sqrt{3}}}{{\\sqrt{2}}}\\\\&=\\displaystyle \\frac{{\\sqrt{6}}}{2}=\\text{RHS}\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\sin \\theta }}-\\displaystyle \\frac{{\\cos 2\\theta }}{{\\cos \\theta }}\\\\&=\\displaystyle \\frac{{2\\sin \\theta", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "the problem gives us: We don’t seem to know much about the side $PR$, whereas parts of $PQ$ and $QR$ are known. Hence, in calculating the area of $\\Delta PQR$, it seems like using the formula $\\text{Area of } \\Delta PQR = \\displaystyle\\frac{1}{2} |PQ||QR|\\sin PQR = \\frac{1}{2} |PQ||QR| \\sin 60^\\circ$ could work. Let’s label the diagram further to give more info on $PQ$ and $QR$: We have 2 unknown quantities we would like to find: $|PQ|$ and $|QR|$. We can use the sine rule to reduce that to just finding $|PQ|$: \\begin{aligned} \\displaystyle\\frac{|PQ|}{\\sin 45^\\circ} &= \\frac{|QR|}{\\sin 75^\\circ}, \\\\ |QR| &= \\frac{|PQ|}{\\sin 45^\\circ} \\sin 75^\\circ = \\sqrt{2}|PQ| \\sin 75^\\circ. \\end{aligned} From the diagram, we know that $|PQ| = 2 + |AP|$. We can use the sine rule again to find the value of $|AP|$: \\begin{aligned} \\displaystyle\\frac{|AP|}{\\sin 45^\\circ} &= \\frac{|AF|}{\\sin 75^\\circ}, \\\\ |AP| &= \\frac{1}{\\sin 75^\\circ} \\sin 45^\\circ = \\frac{1}{\\sqrt{2}\\sin 75^\\circ}. \\end{aligned} Hence, we have \\begin{aligned} \\text{Area} &= \\displaystyle\\frac{1}{2}|PQ| \\cdot \\sqrt{2} |PQ| \\sin 75^\\circ \\cdot \\sin 60^\\circ \\\\ &= \\frac{\\sqrt{2}}{2}\\frac{\\sqrt{3}}{2} \\sin 75^\\circ |PQ|^2 \\\\ &= \\frac{\\sqrt{6}}{4} \\sin 75^\\circ \\left( 2 + \\frac{1}{\\sqrt{2}\\sin 75^\\circ} \\right)^2 \\\\ &= \\frac{\\sqrt{6}}{4}\\sin 75^\\circ \\left( 4 + \\frac{4}{\\sqrt{2}\\sin 75^\\circ} + \\frac{1}{2\\sin^2 75^\\circ} \\right) \\\\ &= \\sqrt{6}\\sin 75^\\circ + \\sqrt{3} + \\frac{\\sqrt{6}}{8 \\sin 75^\\circ}. \\end{aligned} It remains to get rid of the $\\sin 75^\\circ$ term. $\\sin 75^\\circ", "F \\\\ T_0 &= 185^{\\circ}\\ F \\\\ T(30) &= 150^{\\circ}\\ F \\end{align*} {/eq} {eq}\\displaystyle \\begin{align*} T(t) &= T_s + (T_0 - T_s) e^{-kt} \\\\ T(30) &= 75^{\\circ}\\ F + (185^{\\circ}\\ F - 75^{\\circ}\\ F) e^{-k(30\\ min)} \\\\ 150^{\\circ}\\ F &= 75^{\\circ}\\ F + (185^{\\circ}\\ F - 75^{\\circ}\\ F) e^{-k(30\\ min)} \\\\ 150^{\\circ}\\ F - 75^{\\circ}\\ F &= 110^{\\circ}\\ Fe^{-k(30\\ min)} \\\\ \\frac{75^{\\circ}\\ F }{110^{\\circ}\\ F} &= e^{-k(30\\ min)} \\\\ \\frac{15 }{22} &= e^{-k(30\\ min)} \\end{align*} {/eq} Taking the natural logarithm of both sides to solve for k. {eq}\\displaystyle \\begin{align*} \\ln \\bigg( \\frac{15 }{22} &= e^{-k(30\\ min)} \\bigg) \\\\ \\ln \\bigg( \\frac{15 }{22} \\bigg) &= -k(30\\ min) \\\\ k&= -\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1} \\\\ k &\\approx 0.012766 \\end{align*} {/eq} Now we can write the explicit form of Newton's cooling law as {eq}\\displaystyle \\boxed{ T(t) = 75^{\\circ}\\ F + (110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)t}} {/eq}. a.) To solve for the temperature after 40 minutes we set t = 40 mins and use the function we derived above. {eq}\\displaystyle \\begin{align*} T(t) &= 75^{\\circ}\\ F + 110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)t} \\\\ &= 75^{\\circ}\\ F + 110^{\\circ}\\ F e^{\\bigg(\\frac{1}{30}\\ln \\bigg( \\frac{15 }{22} \\bigg)\\ min^{-1}\\bigg)(40\\ min)} \\\\ T(t) &= \\boxed{ 141.01 ^{\\circ}\\ F} \\end{align*} {/eq} The temperature of the turkey after 40 minutes is {eq}\\boxed{", "50^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ (\\cos 10^{\\circ} \\cos 50^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ (2 \\cos 50^{\\circ} \\cos 10^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\Big( \\cos (50^{\\circ}+10^{\\circ}) + \\cos (50^{\\circ}-10^{\\circ}) \\Big) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\cos 60^{\\circ} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\frac{1}{2} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{4} \\cos 40^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} (2\\cos 40^{\\circ} \\cos 70^{\\circ})$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} ( \\cos (70^{\\circ}+40^{\\circ}) + \\cos (70^{\\circ}-40^{\\circ}) )$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} \\cos 110 + \\frac{\\sqrt{3}}{8} \\cos 30^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos 110 + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos (180 - 70) + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} - \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{3}{16} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{ii) } \\text{LHS } = \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} ( 2 \\cos 160^{\\circ} \\cos 40^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} [ \\cos ( 160^{\\circ}+40^{\\circ})", "\\sin 40^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos (40^o-20^o) - \\cos (40^o+20^o) \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o - \\cos 60^o \\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o -$ $\\frac{1}{2}$ $\\Big] \\sin 80^o$ $=$ $\\frac{1}{2}$ $\\Big[ \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ 2 \\cos 20^o \\sin 80^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (80^o+20^o) + \\sin (80^o-20^o) \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin 100^o + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\Big[ \\sin (180^o - 80^o) + \\sin 60^o \\Big] -$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{1}{4}$ $\\sin 80^o +$ $\\frac{\\sqrt{3}}{8}$ $-$ $\\frac{1}{4}$ $\\sin 80^o$ $=$ $\\frac{\\sqrt{3}}{8}$ $=$ RHS. Hence proved. iv) LHS $= \\cos 20^o \\cos 40^o \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ 2 \\cos 20^o \\cos 40^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos ( 40^o+ 20^o) + \\cos ( 40^o - 20^o) \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[ \\cos 60^o + \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{2}$ $\\Big[$ $\\frac{1}{2}$ $+ \\cos 20^o \\Big] \\cos 80^o$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + 2\\cos 20^o \\cos 80^o \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos ( 80^o + 20^o) + \\cos ( 80^o - 20^o) \\Big]$ $= \\frac{1}{4}$ $\\Big[ \\cos 80^o + \\cos 100^o + \\cos 60^o \\Big]$ $=", "\\sin 80^{\\circ} + \\sin 100^{\\circ} + \\sin 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ - \\sin 80^{\\circ} + \\sin (180^{\\circ} - 80^{\\circ}) + \\sin 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ - \\sin 80^{\\circ} + \\sin 80^{\\circ} + \\sin 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\sin 60^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{16} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\\\$ Question 6: Show that: $\\displaystyle \\text{ i) } \\sin A \\sin (B-C) + \\sin B \\sin (C-A) + \\sin C \\sin (A-B) = 0$ $\\displaystyle \\text{ ii) } \\sin (B-C) \\cos (A - D) + \\sin (C-A) \\cos (B-D) + \\sin (A-B) \\cos (C-D) = 0$ $\\displaystyle \\text{i) } \\text{LHS } = \\sin A \\sin (B-C) + \\sin B \\sin (C-A) + \\sin C \\sin (A-B)$ $\\displaystyle = \\frac{1}{2} \\Big[ 2\\sin A \\sin (B-C) + 2\\sin B \\sin (C-A) + 2\\sin C \\sin (A-B) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos ( A - B + C) - \\cos ( A + B - C) + \\cos ( B - C + A ) - \\cos ( B + C - A ) + \\cos ( C - A + B ) - \\cos ( C + A - B) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos ( A - B + C) - \\cos ( A - B + C) - \\cos ( A", "\\Big[ \\cos \\frac{2\\pi}{3} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ \\cos \\Big( \\frac{\\pi}{2} + \\frac{\\pi}{6} \\Big) + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\sin \\frac{\\pi}{6} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\frac{1}{2} + \\cos 2x \\Big]$ $\\displaystyle = - \\cos x + 2 \\cos x \\cos 2x$ $\\displaystyle = - \\cos x + [ \\cos (2x + x) + \\cos ( 2x - x) ]$ $\\displaystyle = - \\cos x + \\cos 3x + \\cos x$ $\\displaystyle = \\cos 3x =$ RHS $\\displaystyle \\\\$ Question 5: Prove that: $\\displaystyle \\text{ i) } \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos 50^{\\circ} \\cos 70^{\\circ} = \\frac{3}{16} \\hspace{1.0cm} \\text{ ii) } \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} = - \\frac{1}{8}$ $\\displaystyle \\text{ iii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 80^{\\circ} = \\frac{\\sqrt{3}}{8} \\hspace{1.0cm} \\text{ iv) } \\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ} = \\frac{1}{8}$ $\\displaystyle \\text{ v) } \\tan 20^{\\circ} \\tan 40^{\\circ} \\tan 60^{\\circ} \\tan 80^{\\circ} = 3 \\hspace{1.0cm} \\text{ vi) } \\tan 20^{\\circ} \\tan 30^{\\circ} \\tan 40^{\\circ} \\tan 80^{\\circ} = 1$ $\\displaystyle \\text{ vii) } \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ} = \\frac{\\sqrt{3}}{16} \\hspace{1.0cm} \\text{ viii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 60^{\\circ} \\sin 80^{\\circ} = \\frac{3}{16}$ $\\displaystyle \\text{i) } \\text{LHS } = \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos", "# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 11 $A=80{}^\\circ,a\\approx 39.5$ and $c\\approx 10.4$. #### Work Step by Step First we will find the value of A Properties of the triangle: Sum of three angles is $A+B+C=180{}^\\circ$ \\begin{align} & A+B+C=180{}^\\circ \\\\ & A+85{}^\\circ +15{}^\\circ =180{}^\\circ \\\\ & A=180{}^\\circ -100{}^\\circ \\\\ & A=80{}^\\circ \\end{align} Now, we will find the remaining sides using the ratio $\\frac{b}{\\sin B}$,or $\\frac{40}{\\sin 85{}^\\circ }$, Now, we will use the law of sines to find a. \\begin{align} & \\frac{a}{\\sin A}=\\frac{b}{\\sin B} \\\\ & \\frac{a}{\\sin 80{}^\\circ }=\\frac{40}{\\sin 85{}^\\circ } \\\\ & a=\\frac{40\\sin 80{}^\\circ }{\\sin 85{}^\\circ } \\\\ & a\\approx 39.5 \\end{align} Using the law of sines again, we will find c. \\begin{align} & \\frac{c}{\\sin C}=\\frac{b}{\\sin B} \\\\ & \\frac{c}{\\sin 15{}^\\circ }=\\frac{40}{\\sin 85{}^\\circ } \\\\ & c=\\frac{40\\sin 15{}^\\circ }{\\sin 85{}^\\circ } \\\\ & c\\approx 10.4 \\end{align} After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "\\cot 30^{\\circ } ={\\frac {BD}{AD}}}$ substituting $BD = \\sqrt{3}$ and $AD = 1$ ${\\displaystyle \\cot 30^{\\circ } ={\\frac {\\sqrt{3}}{1}}}$ $\\cot 30^{\\circ } = \\sqrt{3}$ When m A = 60o: The following Figure 7-6 represents the right-angled triangle from the perspective of the special angle $A = 60^{\\circ }$. Now, we can easily determine the values of trigonometric ratio for $A = 60^{\\circ }$. Looking at the diagram 7-6 from the perspective of m A = 60o Sine function ${\\displaystyle \\sin 60^{\\circ } ={\\frac {\\mathrm {opposite} }{\\mathrm {hypotenuse} }}}$ ${\\displaystyle \\sin 60^{\\circ } ={\\frac {BD}{AB}}}$ substituting $BD = \\sqrt{3}$ and $AB = 2$ ${\\displaystyle \\sin 60^{\\circ } ={\\frac {\\sqrt{3}}{2}}}$ Cosine function ${\\displaystyle \\cos 60^{\\circ } ={\\frac {\\mathrm {adjacent} }{\\mathrm {hypotenuse} }}}$ ${\\displaystyle \\cos 60^{\\circ } ={\\frac {AD}{AB}}}$ substituting $AD = 1$ and $AB = 2$ ${\\displaystyle \\cos 60^{\\circ } ={\\frac {1}{2}}}$ Tangent function ${\\displaystyle \\tan 60^{\\circ } ={\\frac {\\mathrm {opposite} }{\\mathrm {adjacent} }}}$ ${\\displaystyle \\tan 60^{\\circ } ={\\frac {BD}{AD}}}$ substituting $BD = \\sqrt{3}$ and $AD = 1$ ${\\displaystyle \\tan 60^{\\circ } ={\\frac {\\sqrt{3}}{1}}}$ $\\tan 60^{\\circ } = \\sqrt{3}$ Cosecant function ${\\displaystyle \\csc 60^{\\circ } ={\\frac {\\mathrm {hypotenuse} }{\\mathrm {opposite} }}}$ ${\\displaystyle \\csc 60^{\\circ } ={\\frac {AB}{BD}}}$ substituting and $AB = 2$ and $BD = \\sqrt{3}$ ${\\displaystyle \\csc 60^{\\circ } ={\\frac {2}{\\sqrt{3}}}}$ Secant function ${\\displaystyle \\sec 60^{\\circ } ={\\frac {\\mathrm", "## Algebra: A Combined Approach (4th Edition) x = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$ Given, that $\\frac{80}{x}$ and $\\frac{100}{x}$ are complementary angles. Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = 90$^{\\circ}$ Since denominators are same (x), thus we can directly add the numerators ( 80 and 100) Thus, $\\frac{80}{x}$ + $\\frac{100}{x}$ = $\\frac{180}{x}$ Since $\\frac{180}{x}$ = 90$^{\\circ}$ thus x = $\\frac{180}{90}$$^{\\circ}$ = 2$^{\\circ}$ So angles are: $\\frac{80}{x}$ = $\\frac{80}{2}$ = 40$^{\\circ}$ $\\frac{100}{x}$ = $\\frac{100}{2}$ = 50$^{\\circ}$", "into a sum of trig functions: ${\\cos {50^\\circ}\\cos {80^\\circ} }={ \\frac{1}{2}\\left[ {\\cos \\left( {{{50}^\\circ} – {{80}^\\circ}} \\right) + \\cos \\left( {{{50}^\\circ} + {{80}^\\circ}} \\right)} \\right] }={ \\frac{1}{2}\\left[ {\\cos \\left( { – {{30}^\\circ}} \\right) + \\cos {{130}^\\circ}} \\right] }={ \\frac{1}{2}\\cos {30^\\circ} + \\frac{1}{2}\\cos {130^\\circ} }={ \\frac{{\\sqrt 3 }}{4} + \\frac{1}{2}\\cos {130^\\circ}.}$ By the reduction formula, ${\\cos {130^\\circ} = \\cos \\left( {{{180}^\\circ} – {{50}^\\circ}} \\right) }={ – \\cos {50^\\circ}.}$ Therefore, ${A = \\frac{1}{4}\\cos {50^\\circ} + \\frac{1}{2}\\cos {50^\\circ}\\cos {80^\\circ} }={ \\frac{1}{4}\\cos {50^\\circ} }+{ \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} + \\frac{1}{2}\\cos {{130}^\\circ}} \\right) }={ \\frac{1}{4}\\cos {50^\\circ} }+{ \\frac{1}{2}\\left( {\\frac{{\\sqrt 3 }}{4} – \\frac{1}{2}\\cos {{50}^\\circ}} \\right) }={ \\cancel{\\frac{1}{4}\\cos {50^\\circ}} + \\frac{{\\sqrt 3 }}{8} – \\cancel{\\frac{1}{4}\\cos {50^\\circ}} }={ \\frac{{\\sqrt 3 }}{8}.}$ ### Example 8. Calculate $\\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ}.$ Solution. We denote this expression by $$B.$$ Transform the product $$\\sin {20^\\circ}\\sin {80^\\circ}:$$ ${\\sin {20^\\circ}\\sin {80^\\circ} }={ \\frac{1}{2}\\left[ {\\cos \\left( {{{20}^\\circ} – {{80}^\\circ}} \\right) – \\cos \\left( {{{20}^\\circ} + {{80}^\\circ}} \\right)} \\right] }={ \\frac{1}{2}\\left[ {\\cos \\left( { – {{60}^\\circ}} \\right) – \\cos {{100}^\\circ}} \\right] }={ \\frac{1}{2}\\cos {60^\\circ} – \\frac{1}{2}\\cos {100^\\circ} }={ \\frac{1}{4} – \\frac{1}{2}\\cos {100^\\circ}.}$ Substitute this into the original triple product: ${B = \\sin {20^\\circ}\\sin {40^\\circ}\\sin {80^\\circ} }={ \\sin {40^\\circ}\\left( {\\frac{1}{4} – \\frac{1}{2}\\cos {{100}^\\circ}} \\right) }={ \\frac{1}{4}\\sin {40^\\circ} – \\frac{1}{2}\\sin {40^\\circ}\\cos {100^\\circ}.}$ Now we convert the product $$\\sin {40^\\circ}\\cos {100^\\circ}:$$ ${\\sin {40^\\circ}\\cos {100^\\circ} }={ \\frac{1}{2}\\left[", "; 30^{\\circ} \\rightarrow 1 ; 45^{\\circ} \\rightarrow 2 ; 60^{\\circ} \\rightarrow 3 ; 90^{\\circ} \\rightarrow 4$$. Then divide the values by $$4$$ and square root the entire value. $$0^{\\circ} \\rightarrow \\sqrt{\\frac{0}{4}}=0 ; 30^{\\circ} \\rightarrow \\sqrt{\\frac{1}{4}}=\\frac{1}{2} ; 45^{\\circ} \\rightarrow \\sqrt{\\frac{2}{4}}=\\frac{1}{\\sqrt{2}} ; 60^{\\circ} \\rightarrow \\sqrt{\\frac{3}{4}}=\\frac{\\sqrt{3}}{2} ; 90^{\\circ} \\rightarrow \\sqrt{\\frac{4}{4}}=1$$. This gives the values of sine for these five angles. Now for the remaining three, use:$$\\sin \\left(180^{\\circ}-x\\right)= \\sin x \\quad \\sin \\left(180^{\\circ}+x\\right)=\\,- \\sin x \\sin \\left(360^{\\circ}-x\\right)=\\,- \\sin x$$This means, $$\\sin \\left(180^{\\circ}-0^{\\circ}\\right)=\\sin 0^{\\circ} \\quad \\sin \\left(180^{\\circ}+90^{\\circ}\\right)=\\,- \\sin 90^{\\circ} \\sin \\left(360^{\\circ}-0^{\\circ}\\right)=\\,- \\sin 0^{\\circ}$$ Step 3: Determine the value of $$\\cos$$.$$\\sin \\left(90^{\\circ}-x\\right)=\\cos x$$To find values for $$\\cos x$$, use this formula.For example, equals $$\\left(90^{\\circ}-45^{\\circ}\\right)= \\sin 45^{\\circ},\\left(90^{\\circ}-30^{\\circ}\\right)=\\sin 60^{\\circ}$$ and vice versa.You can quickly determine the value of the $$\\cos$$ function by using this method: Step 4: Determine the value of $$\\tan$$. We know that $$\\sin$$ divided by $$\\cos$$ equals the $$\\tan$$.$$\\frac{{\\sin }}{{\\cos }} = \\tan$$Divide the value of $$\\sin$$ at $$0^{\\circ}$$ by the value of $$\\cos$$ at $$0^{\\circ}$$ to get the value of $$\\tan$$ at $$0^{\\circ}$$. Take a look at the sample below.$$\\tan 0^{\\circ}=\\frac{0}{1}=0$$In the same way, the table would be as follows. Step 5: Determine the value of $$\\cot$$.The reciprocal of $$\\tan$$ is equal to the value of $$\\cot$$. Divide $$1$$ by the value of $$\\tan$$ at $$0^{\\circ}$$ to get the value of $$\\cot$$", "and \\begin{align*}b = 30^\\circ\\end{align*}. \\begin{align*}\\cos (45^\\circ - 30^\\circ) & = \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{6} + \\sqrt{2}}{4}\\end{align*} #### Example C Find the exact value of \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*}, in radians. Solution: \\begin{align*}\\cos \\frac{5 \\pi}{12} = \\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right )\\end{align*}, notice that \\begin{align*}\\frac{\\pi}{4} = \\frac{3 \\pi}{12}\\end{align*} and \\begin{align*}\\frac{\\pi}{6} = \\frac{2 \\pi}{12}\\end{align*} \\begin{align*}\\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right ) & = \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} \\\\ \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ & = \\frac{\\sqrt{6} - \\sqrt{2}}{4}\\end{align*} ### Vocabulary Cosine Sum Formula: The cosine sum formula relates the cosine of a sum of two arguments to a set of sine and cosines functions, each containing one argument. Cosine Difference Formula: The cosine difference formula relates the cosine of a difference of two arguments to a set of sine and cosines functions, each containing one argument. ### Guided Practice 1. Find the exact value for \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*} 2. Find the exact value for \\begin{align*}\\cos \\frac{7 \\pi}{12}\\end{align*} 3. Find the exact value for \\begin{align*}\\cos 345^\\circ\\end{align*} Solutions: 1. \\begin{align*}\\cos \\frac{5 \\pi}{12} & = \\cos \\left (\\frac{2 \\pi}{12} + \\frac{3", "\\Big[ \\cos 20^{\\circ} \\sin 80^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ 2 \\cos 20^{\\circ} \\sin 80^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin (80^{\\circ}+20^{\\circ}) + \\sin (80^{\\circ}-20^{\\circ}) \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin 100^{\\circ} + \\sin 60^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\sin (180^{\\circ} - 80^{\\circ}) + \\sin 60^{\\circ} \\Big] - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\sin 80^{\\circ} + \\frac{\\sqrt{3}}{8} - \\frac{1}{4} \\sin 80^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{iv) } \\text{LHS } = \\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 20^{\\circ} \\cos 40^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos ( 40^{\\circ}+ 20^{\\circ}) + \\cos ( 40^{\\circ} - 20^{\\circ}) \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 60^{\\circ} + \\cos 20^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ \\frac{1}{2} + \\cos 20^{\\circ} \\Big] \\cos 80^{\\circ}$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + 2\\cos 20^{\\circ} \\cos 80^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos ( 80^{\\circ} + 20^{\\circ}) + \\cos ( 80^{\\circ} - 20^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos 100^{\\circ} + \\cos 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} + \\cos (180^{\\circ} - 80^{\\circ}) + \\cos 60^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{4} \\Big[ \\cos 80^{\\circ} -", "& = \\frac{1}{\\sqrt{2}} \\div \\frac{1}{2} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{2}} \\times \\frac{2}{1}\\right) \\div \\frac{1}{\\sqrt{2}}\\\\ & = \\frac{2}{\\sqrt{2}} \\times \\frac{\\sqrt{2}}{1} \\\\ & = 2 \\end{align*} $$\\tan{60^\\circ} - \\tan{60^\\circ} - \\sin{60^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{1}{\\sqrt{3}} &-\\frac{1}{2} &-\\frac{1}{\\sqrt{2}} &-\\frac{1}{1} &\\frac{-\\sqrt{3}}{2} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{60^\\circ} - \\tan{60^\\circ} - \\sin{60^\\circ} & = \\frac{\\sqrt{3}}{1} - \\frac{\\sqrt{3}}{1} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{(2)\\left(\\sqrt{3}\\right) - (2)\\left(\\sqrt{3}\\right) - \\sqrt{3}}{2} \\\\ & = \\frac{-\\sqrt{3}}{2} \\end{align*} $$\\cos{45^\\circ} - \\sin{60^\\circ} - \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{1}{2} &-\\frac{1}{\\sqrt{2}} &-\\frac{7}{2\\sqrt{3}} &\\frac{-\\sqrt{3}}{2} &-\\frac{1}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\cos{45^\\circ} - \\sin{60^\\circ} - \\sin{45^\\circ} & = \\frac{1}{\\sqrt{2}} - \\frac{\\sqrt{3}}{2} - \\frac{1}{\\sqrt{2}} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{2}\\right) - 2}{(2)\\left(\\sqrt{2}\\right)} \\\\ & = \\frac{-\\sqrt{3} \\sqrt{2}}{2\\sqrt{2}} \\\\ & = \\frac{-\\sqrt{3}}{2} \\end{align*} Use special angles to show that: $$\\dfrac{\\sin 60°}{\\cos 60°} = \\tan 60°$$ We are told to use special angles, so we first write each ratio using special angles and then simplify each side of the equation. LHS: \\begin{align*} \\frac{\\sin 60°}{\\cos 60°} & = \\frac{\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}} \\\\ & = \\frac{\\sqrt{3}}{2} \\times \\frac{2}{1} \\\\ & = \\sqrt{3}", "\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} + 1}{2\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} + 1}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)}{4} \\end{align*} $$\\cos \\text{15} °$$ \\begin{align*} \\cos \\text{15} ° &= \\cos ( \\text{60} ° - \\text{45} °) \\\\ &= \\cos \\text{60} ° \\cos \\text{45} ° + \\sin \\text{60} ° \\sin \\text{45} ° \\\\ &= \\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{1 + \\sqrt{3}}{2\\sqrt{2}} \\\\ &= \\frac{1 + \\sqrt{3}}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( 1 + \\sqrt{3} \\right)}{4} \\end{align*} $$\\sin \\text{15} °$$ \\begin{align*} \\sin \\text{15} ° &= \\sin ( \\text{60} ° - \\text{45} °) \\\\ &= \\sin \\text{60} ° \\cos \\text{45} ° - \\cos \\text{60} ° \\sin \\text{45} ° \\\\ &= \\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}} - \\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} - 1}{2\\sqrt{2}} \\\\ &= \\frac{\\sqrt{3} - 1}{2\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\end{align*} $$\\tan \\text{15} °$$ \\begin{align*} \\tan \\text{15} ° &= \\frac{\\sin \\text{15} °}{\\cos \\text{15} °} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\div \\frac{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)}{4} \\\\ &= \\frac{\\sqrt{2} \\left( \\sqrt{3} - 1 \\right)}{4} \\times \\frac{4}{\\sqrt{2} \\left( \\sqrt{3} + 1 \\right)} \\\\ &= \\frac{ \\sqrt{3} - 1 }{ \\sqrt{3} + 1 } \\\\ &= \\frac{ \\sqrt{3} - 1 }{ \\sqrt{3} + 1 } \\times \\frac{\\sqrt{3} - 1}{\\sqrt{3} - 1} \\\\ &= \\frac{", "20o + i sin 20o) c. Multiply: \\begin{align*}2 \\left (\\mbox{cos}\\ \\frac{\\pi} {8} + i\\ \\mbox{sin}\\ \\frac{\\pi} {8}\\right ) \\times 2 \\left (\\mbox{cos}\\ \\frac{\\pi} {10} + i\\ \\mbox{sin}\\ \\frac{\\pi} {10}\\right )\\end{align*} d. Divide: 2(cos 80o + i sin 80o) ÷ 6(cos 200o + i sin 200o) e. Divide: 3cis(130o) ÷ 4cis(270o) 1. a. \\begin{align*}4\\sqrt{2} (\\mbox{cos}\\ 15^\\circ + i\\ \\mbox{sin}\\ 15^\\circ)\\end{align*} b. 8 cis (60o) c. \\begin{align*}4\\ \\mbox{cis}\\ \\left (\\frac{9 \\pi} {40}\\right )\\end{align*} d. \\begin{align*}\\frac{1} {3} \\mbox{cis}\\ (240^\\circ)\\end{align*} e. \\begin{align*}\\frac{3} {4} \\mbox{cis}\\ (220^\\circ)\\end{align*} Show Hide Details Description Tags: Subjects: Date Created: Feb 23, 2012", "= \\frac{\\pi}{6}\\end{align*} (or \\begin{align*}30^\\circ\\end{align*}). So we replace \\begin{align*}x\\end{align*} with \\begin{align*}x' \\cos(30) - y' \\sin (30) = \\frac{\\sqrt{3}}{2}x' - \\frac{1}{2}y'\\end{align*}, and \\begin{align*}y\\end{align*} with \\begin{align*}x' \\sin(30)+y'\\cos(30) = \\frac{1}{2}x' + \\frac{\\sqrt{3}}{2}y'\\end{align*}, giving us: \\begin{align*}& 79 \\left (\\frac{\\sqrt{3}}{2}x' - \\frac{1}{2}y' \\right )^2 + 37 \\left (\\frac{1}{2}x' + \\frac{\\sqrt{3}}{2}y' \\right )^2 + 42 \\sqrt{3} \\left (\\frac{\\sqrt{3}}{2}x' - \\frac{1}{2}y' \\right ) \\left (\\frac{1}{2}x' + \\frac{\\sqrt{3}}{2}y' \\right ) + \\\\ & \\left (-200 \\sqrt{3} -8 \\right ) \\left (\\frac{\\sqrt{3}}{2}x' - \\frac{1}{2}y' \\right ) + \\left (8 \\sqrt{3} - 200 \\right ) \\left (\\frac{1}{2}x' + \\frac{\\sqrt{3}}{2}y' \\right ) + 4 = 0\\end{align*} Multiplying through by 4 we have: \\begin{align*}& 79 \\left (\\sqrt{3}x' - y' \\right )^2 + 37 \\left (x' + \\sqrt{3}y' \\right )^2 + 42 \\sqrt{3} \\left (\\sqrt{3}x' - y' \\right ) \\left (x' + \\sqrt{3}y' \\right ) + \\\\ & \\left (-200 \\sqrt{3} - 8 \\right ) \\left (2 \\sqrt{3}x' - 2y' \\right ) + \\left (8 \\sqrt{3} - 200 \\right ) \\left (2x' + 2\\sqrt{3}y' \\right ) + 16 = 0\\end{align*} Multiplying out we have: \\begin{align*}& 237(x')^2 - 158 \\sqrt{3}x'y' + 79(y')^2 + 37(x')^2 + 74 \\sqrt{3}x'y' + 111(y')^2 + 126(x')^2 + 84 \\sqrt{3}x'y' - 126(y')^2 \\\\ & - 1200x' - 16\\sqrt{3} x' + 400 \\sqrt{3}y' + 16y' + 16\\sqrt{3}x' - 400 x' + 48y' -400 \\sqrt{3}y' + 16 = 0\\end{align*} Which simplifies" ]

[ "+0 # Convert $2 e^{\\pi i/6}$ to rectangular form. 0 132 1 +474 Convert $$2 e^{\\pi i/6}$$ to rectangular form. Jan 4, 2019 #1 +99333 +2 This is probably the easiest question you have ever asked. But why would I answer it? https://web2.0calc.com/questions/let-a-and-b-be-real-numbers-the-complex-number-4 Jan 4, 2019", "length $1$ and argument $x$. Converting $e^{ix}$ to rectangular coordinates, we get $$e^{ix}=\\cos(x)+i\\sin(x)\\tag{7}$$ Comparing the real and imaginary parts of $(2)$ and $(7)$, we get the series for $\\sin(x)$ and $\\cos(x)$. -", "here $$e(z) = e^{2 \\pi i z}$$", "of the required form. $$\\square$$", "rectangular form back them up with references or personal experience MATH 1113 at of.", "be written in the form of $$e^{i \\pi}+1=0$$.", "# How do you convert e^3e^{-4i} into cartesian form? ${e}^{3} {e}^{- 4 i} = {e}^{3} \\cos \\left(- 4\\right) + {e}^{3} \\sin \\left(- 4\\right) i$", "\\pi i \\mathrm{e}^{-1}$$", "e^{-i \\pi /2}$ Since |z_4-z_1| = |z_2-z_1| (side of square are equal ) Now you will get z_4 from above Use the same technique and find z_3", "i {\\rm e}^\\pi$$", "Form Math Calculus Precalculus 11/12/15 #### Convert from Rectangular to Polar Form Convert the following from Rectangular to Polar Form: 2x-9y+3=0 Rectangular Form 05/10/15 #### approximate [100, angle 2.95 rad] Approximate in rectangular form to the nearest 0.01 Rectangular Form Parametric Equations 12/16/13 #### How do I convert from parametric equations to rectangular form? Write the rectangular form for the parametric equation x=cosθ ; y=4sinθ ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "outer shape should be $(x-a)^2+(y-b)^2=(r+1)^2$. How about if the inner shape is an ellipsis? Or a square? -", "= \\sqrt{e^{j \\pi /2}}$$", "e n}{11 - 2 \\sqrt{30}}$", "the form $1+2\\pi i n /\\ln 2, n\\in\\mathbb{Z}-\\{0\\}$. – anon Aug 11 '11 at 4:01", "i \\pi}}}$. Let $z=1/2$, then so $0.065988 \\approx e^{-e} = \\LARGE e^{e^{i \\pi-e^{i \\pi}}}$ The complex function $\\LARGE e^{e^{2 z i \\pi-e^{2 z i \\pi}}}$", "2.4.4.11. $\\square$", "# What is the trigonometric form of 8 e^( ( 7 pi)/12 i ) ? $8 {e}^{\\frac{7 \\pi}{12} i} = 8 \\left(\\cos \\left(7 \\frac{\\pi}{12}\\right) + i \\sin \\left(7 \\frac{\\pi}{12}\\right)\\right)$", "2 \\pi i (-1) = -2\\pi i$", "= E(13)^2+E(13)^11" ]

[ "= \\boxed{\\frac{\\pi}{4}.}$$ $$\\int_{0}^{\\frac{\\pi}{2}}\\frac{{cos}^{n}\\left(x\\right)\\:dx}{{cos}^{n}\\left(x\\right)+ {sin}^{n}\\left(x\\right)} .$$", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "i} = \\frac{\\pi}{4}$ So the given series $11 - \\frac{11}{3} + \\frac{11}{5} - \\frac{11}{7} + \\ldots = \\frac{11}{4} \\pi$ (b) sum_(n=2)^oo ((-1)^n(8^n))/(n!) = sum_(n=0)^oo (-8)^n/(n!)-(-8)^0/(0!)-(-8)^1/(1!) $q \\quad = {e}^{-} 8 - 1 + 8 = 7 + {e}^{-} 8$", "\\pi i \\mathrm{e}^{-1}$$", "\\boxed{\\frac{\\pi}{2^{n+1}}}$$", "\\pi} \\right)^{\\frac{1}{3}} \\approx 0.894$ sphere $\\frac{4}{3} \\pi r^3$ $4 \\pi\\,r^2$ $1\\,$", "Subtracting $\\frac 12$ from $h$ gives us $$x = h-\\frac 12 = \\frac 1{16},$$ so our answer is $\\boxed{\\text{E}}$.", "$$P(\\pi/4) = \\frac{16}{2\\sqrt2\\pi^2}\\\\ \\boxed{P(\\pi/4) = \\frac{4\\sqrt2}{\\pi^2}}$$", "{120 \\pi}} = {E^2 r^2 \\over {30 G}} \\space [W]\\$$", "$$x = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ So, from this, we can conclude that $$\\cos(5\\pi t)\\cos(6 \\pi t) = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ Other excursions into complex numbers can be explained similarly. We simply use this embedding of $$\\mathbb{R}$$ into $$\\mathbb{C}$$, and use the properties proven for all complex numbers to help us with real numbers.", "are $\\frac{\\sqrt2}}{2}$ and $\\frac{\\sqrt{3}}{2}$, respectively. Therefore: $cos(\\frac{5\\pi}{12}) + cos(\\frac{\\pi}{12}) = 2(\\frac{ \\sqrt{2}}{2}) (\\frac{\\sqrt{3} }{2})$. Finally, this becomes: $\\frac{ \\sqrt{2}\\sqrt{3} }{2} = \\frac{ \\sqrt{6}}{2}$.", "${\\{{\\mathbb R} e^{\\frac{i\\pi}{3}}, {\\mathbb R} e^{\\frac{i\\pi}{3}}, {\\mathbb R} e^{\\frac{i\\pi}{3}}}$, which clearly have successive pairwise cone–angle ${\\frac{\\pi}{3}}$. It follows that the cusp cross section in ${\\mathop{\\mathbb H}^3/\\Gamma_d}$ is topologically ${S^2}$ with three singular points of order ${3}$, i.e ${S^2(3,3,3)}$. $\\Box$", "# How do you evaluate e^( ( pi)/12 i) - e^( ( 11 pi)/8 i) using trigonometric functions? Apr 19, 2016 $1.139 + 1.183 i$ #### Explanation: Euler's formula says that ${e}^{i x} = \\cos x + i \\sin x$. Using values from the question of $x = \\frac{\\pi}{12}$ and $x = \\frac{11 \\pi}{8}$, ${e}^{\\frac{\\pi}{12} i} = \\cos \\left(\\frac{\\pi}{12}\\right) + i \\sin \\left(\\frac{\\pi}{12}\\right)$ $= \\cos 15 + i \\sin 15$ $= 0.966 + 0.259 i$ ${e}^{\\frac{11 \\pi}{8} i} = \\cos \\left(\\frac{11 \\pi}{8}\\right) + i \\sin \\left(\\frac{11 \\pi}{8}\\right)$ $= \\cos 247.5 + i \\sin 247.5$ $= - 0.383 - 0.924 i$ Substituting these values into the question, $\\left(0.966 + 0.259 i\\right) - \\left(- 0.383 - 0.924 i\\right)$ $= 0.966 + 0.383 + 0.259 i + 0.924 i$ $= 1.349 + 1.183 i$", "\\sin 5 \\approx \\boxed{10.46\\, m} {/eq}", "### Home > PC > Chapter 11 > Lesson 11.2.2 > Problem11-87 11-87. Show that $i^{n}$ can be written as $\\cos(\\frac { n \\pi } { 2 }) + i \\sin (\\frac { n \\pi } { 2 })$. Recall from problem 11-85 that $i$ can be written as: Now use DeMoivre's theorem to write $i^{n}$. $\\textit{i}=\\textit{z}=1\\left( \\cos\\frac{\\pi}{2}+\\textit{i}\\sin\\frac{\\pi}{2} \\right)$", "$$\\boxed{x(t) = \\ell - \\frac{m g}{k} + \\frac{a k}{k-m \\omega^2} \\sin(\\omega t) }$$ • This looks correct to me, although you're using a different definition of $x$ from the one the OP defines. – Brionius Jan 21 '15 at 4:17", "# What is the trigonometric form of 13 e^( ( pi)/12 i ) ? $13 {e}^{\\frac{\\pi}{12} i} = 13 \\left(\\cos \\left(\\frac{\\pi}{12}\\right) + i \\sin \\left(\\frac{\\pi}{12}\\right)\\right)$ ${e}^{i x} = \\cos x + i \\sin x$ $13 {e}^{\\frac{\\pi}{12} i} = 13 \\left(\\cos \\left(\\frac{\\pi}{12}\\right) + i \\sin \\left(\\frac{\\pi}{12}\\right)\\right)$", "\\frac{\\pi}{3} + e$ $-\\frac{\\pi}{6} = e$. Therefore $y = 2\\sin{\\left(\\frac{\\pi t}{3} - \\frac{\\pi}{6}\\right)}$.", "$\\frac{53}{49} I = - \\frac{1}{7} {e}^{2 x} \\cos 7 x + \\frac{2}{49} {e}^{2 x} \\sin 7 x$ $I = - \\frac{7}{53} {e}^{2 x} \\cos 7 x + \\frac{2}{53} {e}^{2 x} \\sin 7 x + C$ $I = \\frac{1}{53} {e}^{2 x} \\left(2 \\sin 7 x - 7 \\cos 7 x\\right) + C$", "proceed as follows: $$\\sum_{k=0}^{m-1} \\sin \\frac{2 \\pi k}{n} = \\sum_{k=0}^{m-1} \\Im(e^{2 \\pi ik/n}) = \\Im \\frac{e^{2 \\pi im/n} -1}{e^{2 \\pi i/n}-1}$$ When you plug in $m = n$, it turns out that: $$\\sum_{k=0}^{m-1} \\sin \\frac{2 \\pi k}{n} = \\Im \\frac{e^{2 \\pi i} -1}{e^{2 \\pi i/n}-1} = \\Im \\frac{1 -1}{e^{2 \\pi i/n}-1} = 0$$" ]

[ "## anonymous 4 years ago how to convert z = 2(cos60 + isin60) to rectangular form 1. anonymous for this you only need $\\cos(60)=\\frac{1}{2}$ adn $\\sin(60)=\\frac{\\sqrt{3}}{2}$ plug em in 2. anonymous then multiply by 2 and get $1+\\sqrt{3}i$ 3. anonymous going the other way is harder, but changing from trig form to rectangular means evaluate the trig functions", "and trials, now in $$sin x = \\sqrt{1-e^{i\\pi/3}} \\\\$$ express ##1-e^{i\\pi/3}## in ##x+iy## form. Not that complicated if you have put some more creativity and trials, now in $$sin x = \\sqrt{1-e^{i\\pi/3}} \\\\$$ express ##1-e^{i\\pi/3}## in ##x+iy## form. Hmmm... I'm new to complex number... and, I don't know how to convert ##1-e^{i\\pi/3}## to rectangular form since there is a constant (in this case, 1)... blue_leaf77 Homework Helper Hmmm... I'm new to complex number... and, I don't know how to convert ##1-e^{i\\pi/3}## to rectangular form since there is a constant (in this case, 1)... In rectangular form of a complex number ##x+iy##, both ##x## and ##y## are real and every complex number can be transformed to this form. To do what you need to do, transform ##e^{i\\pi/3}## into rectangular form using Euler formula. In rectangular form of a complex number ##x+iy##, both ##x## and ##y## are real and every complex number can be transformed to this form. To do what you need to do, transform ##e^{i\\pi/3}## into rectangular form using Euler formula. Alright, so it becomes like this ##1-e^{i\\pi/3} = 1 - cos \\frac{\\pi}{3} + i\\ sin \\frac{\\pi}{3} = \\frac{1}{2} + \\frac{1}{2}\\sqrt{3}i = \\frac{1}{2} (1+\\sqrt{3}i) \\\\ sin\\ x = \\sqrt{\\frac{1}{2} (1+\\sqrt{3}i)} ## and I convert cos x to rectangular form, ## cos\\ x = \\frac{1}{2}\\sqrt{3}+\\frac{1}{2}i## ##sin\\ 2x = 2\\", "Home > PC > Chapter 11 > Lesson 11.2.1 > Problem11-71 11-71. Find the rectangular form of each complex number. Use exact values. 1. $z = 6(\\cos(\\frac { 5 \\pi } { 6 })+ i \\sin (\\frac { 5 \\pi } { 6 }))$ Substitute the exact values for sine and cosine of $5π/6$. $z=6\\left( -\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i \\right)$ Distribute. 1. $z = 7(\\cos(\\frac { 3 \\pi } { 2 })+ i \\sin(\\frac { 3 \\pi } { 2 }))$", "## Trigonometry (11th Edition) Clone The rectangular form is $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$ $\\sqrt{3}$ $cis$ $315^\\circ$ = $\\sqrt{3}$ $(cos 315^\\circ$ + $isin 315^\\circ)$ = $\\sqrt{3} \\cdot \\frac{\\sqrt{2}}{2}$ + $\\sqrt{3} \\cdot i \\cdot (- \\frac{\\sqrt{2}}{2})$ = $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$ The rectangular form is $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$", "My Math Forum Rectangular Form Complex Analysis Complex Analysis Math Forum September 24th, 2013, 04:41 AM #1 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Rectangular Form Hi, does anyone know how to write $e^{\\frac{2-i \\pi}{4}}$ in rectagular form? September 24th, 2013, 04:53 AM #2 Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Rectangular Form You need to use Euler's formula: $e^{ix}=cos(x)+isin(x)$ September 24th, 2013, 05:08 AM #3 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Rectangular Form so if I write it $e^{\\frac{2-i\\pi}{4}}= e^{\\frac{2}{4}-{\\frac{i\\pi}{4}}} = e^{\\frac{2}{4}} e^{{\\frac{-i\\pi}{4}}$ I have $e^{\\frac{1}{2}} cos(\\frac{\\pi}{4}) - i e^{\\frac{1}{2}} sin(\\frac{\\pi}{4})$ but then I am lost there is that already rectangular form? I think it is bc the real part (cosine) is not in terms of z. September 24th, 2013, 07:23 AM #4 Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Rectangular Form Do you not know what \"rectangular form\" means? You have a+ bi with $a= e^{1/2}cos(\\pi/4)$ and $b= -e^{1/2}sin(\\pi/4)$. I don't know what you mean by \"the real part (cosine) is not in terms of z\". There is no \"z\" in anything you write. September 24th, 2013, 07:40 AM #5 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0", "square is $r\\sqrt{2}$ and $p=\\frac{S_{square}}{S_{circle}}=\\frac{2 r^2}{\\pi r^2}=\\frac{2}{\\pi}$", "+0 # Convert $2 e^{\\pi i/6}$ to rectangular form. 0 132 1 +474 Convert $$2 e^{\\pi i/6}$$ to rectangular form. Jan 4, 2019 #1 +99333 +2 This is probably the easiest question you have ever asked. But why would I answer it? https://web2.0calc.com/questions/let-a-and-b-be-real-numbers-the-complex-number-4 Jan 4, 2019", "# How do you find the corresponding rectangular coordinates for the point (-1, (-3pi)/4 )? Jan 23, 2018 The rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$. #### Explanation: We can use $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = - 1 \\cos \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ $y = - 1 \\sin \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ So the rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$.", "# What is the trigonometric form of 13 e^( ( pi)/12 i ) ? $13 {e}^{\\frac{\\pi}{12} i} = 13 \\left(\\cos \\left(\\frac{\\pi}{12}\\right) + i \\sin \\left(\\frac{\\pi}{12}\\right)\\right)$ ${e}^{i x} = \\cos x + i \\sin x$ $13 {e}^{\\frac{\\pi}{12} i} = 13 \\left(\\cos \\left(\\frac{\\pi}{12}\\right) + i \\sin \\left(\\frac{\\pi}{12}\\right)\\right)$", "# Math Help - Parametric->Rectangular 1. ## Parametric->Rectangular Find the equivalent rectangular equation: $x=sec(t)$ $y=cos(t)$ $\\frac{-\\pi}{2}\\le t \\le\\frac{\\pi}{2}$ Also, do circle/elliptical parameters always remain between $0\\le y \\le 2\\pi$??? It seems like they do but I feel I am not seeing the whole picture in gestalt. 2. ## Re: Parametric->Rectangular Originally Posted by Greymalkin Find the equivalent rectangular equation: $x=sec(t)$ $y=cos(t)$ $\\frac{-\\pi}{2}\\le t \\le\\frac{\\pi}{2}$ Is it true that $xy=1~?$", "### Home > PC3 > Chapter 11 > Lesson 11.2.3 > Problem11-115 11-115. Write the rectangular form of each complex number. Use exact values. 1. $z = 6(\\cos(\\frac { 5 \\pi } { 6 })+ i \\sin (\\frac { 5 \\pi } { 6 }))$ Substitute the exact values for sine and cosine of $5π/6$. $z=6\\left( -\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i \\right)$ Distribute. 1. $z = 7(\\cos(\\frac { 3 \\pi } { 2 })+ i \\sin(\\frac { 3 \\pi } { 2 }))$", "# What is the trigonometric form of 8 e^( (3pi)/2 i ) ? $8 {e}^{\\frac{3 \\pi}{2} i} = - 8 i$ Trigonometric form of $8 {e}^{\\frac{3 \\pi}{2} i}$ is $8 \\left(\\cos \\left(\\frac{3 \\pi}{2}\\right) + i \\sin \\left(\\frac{3 \\pi}{2}\\right)\\right)$ = 8(0+ixx(-1)=-8i#", "$\\frac{\\sqrt 3} 2 + \\frac 1 2 i, -1 + 0i$, and $\\frac{\\sqrt 3}2 - i\\frac 1 2$. In polar form, these are $e^{i\\pi/3}, e^{i\\pi}$, and $e^{i 5\\pi/3}$.", "the square: $7t - 11 - 2t^2 = -2(t-\\frac{7}4)^2-\\ldots$ -", "= \\boxed{\\frac{\\pi}{4}.}$$ $$\\int_{0}^{\\frac{\\pi}{2}}\\frac{{cos}^{n}\\left(x\\right)\\:dx}{{cos}^{n}\\left(x\\right)+ {sin}^{n}\\left(x\\right)} .$$", "# How do you convert (4, 4pi/3) into rectangular forms? Dec 6, 2015 $x = - 2$ $y = - 2 \\sqrt{3}$ #### Explanation: If you know the poalr coordinates $\\left(\\setminus \\rho , \\setminus \\theta\\right)$, then the transformation to have the rectangular ones is $x = r \\cos \\left(\\theta\\right)$ $y = r \\sin \\left(\\theta\\right)$ $x = 4 \\cos \\left(\\frac{4 \\pi}{3}\\right)$ $y = 4 \\sin \\left(\\frac{4 \\pi}{3}\\right)$ since $\\cos \\left(\\frac{4 \\pi}{3}\\right) = - \\frac{1}{2}$, and $\\sin \\left(\\frac{4 \\pi}{3}\\right) = - \\frac{\\sqrt{3}}{2}$, we have $x = 4 \\cdot \\left(- \\frac{1}{2}\\right) = - 2$ $y = 4 \\cdot \\left(- \\frac{\\sqrt{3}}{2}\\right) = - 2 \\sqrt{3}$", "# What is the trigonometric form of 8 e^( ( 7 pi)/12 i ) ? $8 {e}^{\\frac{7 \\pi}{12} i} = 8 \\left(\\cos \\left(7 \\frac{\\pi}{12}\\right) + i \\sin \\left(7 \\frac{\\pi}{12}\\right)\\right)$", "# How do you convert (cos120+isin120) to its rectangular form? Feb 7, 2015 You evaluate the $\\cos$ and $\\sin$ and multiply the modulus (in this case $1$): You have: 1[cos(120°)+isin(120°)]= =cos(120°)+isin(120°)=# $= - \\frac{1}{2} + i \\frac{\\sqrt{3}}{2}$ which is in rectangular form: $a + i b$ Hope it helps", "# How do you convert e^3e^{-4i} into cartesian form? ${e}^{3} {e}^{- 4 i} = {e}^{3} \\cos \\left(- 4\\right) + {e}^{3} \\sin \\left(- 4\\right) i$", "Free Version Difficult # Product Complex Numbers, Euler’s Formula, Rectangular Form TRIG-P5BVQO What is product of $4{e^{i\\frac{\\pi }{4}}}\\,\\cdot\\,9{e^{i\\frac{{5\\pi }}{6}}}$ written in rectangular form? (Round your answer to the nearest thousandth) A $34.773 + 9.317i$ B $- 9.317 – 34.773i$ C $36i$ D $- 34.773 - 9.317i$" ]

[ "Find the positive value of $x$ that satisfies $x = 1.900 \\sin(x)$. Give the answer to four places of accuracy. $x\\approx$ Remember to calculate the trig functions in radian mode.", "0}^n {k^{\\left\\lfloor {\\cos k} \\right\\rfloor } } \\le \\sum\\limits_{k = 1}^n {k^0 } = n.$$ @Shai: What if $\\lfloor \\cos k \\rfloor$ is replaced by $\\lceil \\cos k \\rceil$? – JavaMan May 31 '11 at 5:15 It is not hard to see that $\\cos k$ is often positive, since we are advancing by about $60$ degrees each time. That forces the altered version of $f$ to lie between $an^2$ and $bn^2$ for positive constants $a$ and $b$. – André Nicolas May 31 '11 at 5:47 @Shai: Thank you. @user6312: $1$ radian is about $57^{\\circ}$, but I would think that $\\{\\cos k\\}_{k \\in \\Bbb{N}}$ would be pretty equidistributed. I may think about this more, and possibly start a new thread for this new question. – JavaMan May 31 '11 at 5:56", "Geometry # Trigonometric Graphs - Amplitude and Periodicity Consider two functions $f(x)=25-\\sin 6x, g(x)=\\left\\lfloor \\frac{x}{2} \\right\\rfloor.$ Find the sum of all the elements in the range of $g(f(x))$. Details and assumptions The function $\\lfloor x \\rfloor: \\mathbb{R} \\rightarrow \\mathbb{Z}$ refers to the greatest integer smaller than or equal to $x$. For example, $\\lfloor 2.3 \\rfloor = 2$ and $\\lfloor -5 \\rfloor = -5$. If the period of the function $f(x)=\\sqrt{224+\\cos x}+\\sqrt{226-\\cos x}$ is $a$, what is the value of $f(a)$? Consider two functions $f(x)=7\\sin x+12, g(x)=\\pi\\cos \\pi x.$ If $a$ and $b$ are the maximum and minimum values, respectively, of $f(g(x))$, what is $a+b$? The graph of $y=7 \\tan (2\\pi x-\\pi)+18$ has period $\\frac{1}{a}$. What is the value of $a$? What is the sum of the maximum and minimum values of the function $y=|\\sin x-7|+20?$ ×", "$x_{n+1}=x_n -\\frac{f(x_n)}{f'(x_n)}$. We have that $f(1)$ is positive, so if $x_0=1$, $x_1$ will be less than $1$. Is your calculator set to radian mode? If you are in degree mode, of course there will be trouble! – André Nicolas Sep 5 '15 at 0:30 • You're right @AndréNicolas. Calculator was set to degree mode. Ops! – Fabiano Araujo Sep 5 '15 at 0:37", "# Inverse Sine calculator ### Inverse Sine Calculator Note • Enter the value of x and unit in order to calculate inverse sine values • Click on the calculate button. ### Inverse Sine Calculator Example of Few questions where you can use this formula • Find the value of $sin^{-1} 1$ in radian • Find the value of $sin^{-1} .5$ in radian • Find the value of $sin^{-1} .25$ in °", "# Thread: Trig Question - Urgent 1. ## Trig Question - Urgent Sorry for bothering you again but this is rather (very?) urgent. I have one question that I want to ask and if you could help me, it would be greatly appreciated. Also, would you be able to do them in radian measure and give the exact answers (eg. without decimals).: ----- Given f(x) = 2sinx - cosx Write in the form of Asin(x + a) I've managed to do this one with the answer being: square root 5 (x + arctan -0.5) Find the maximum value of 2sinx - cosx and the corresponding values of x Okay, I've found that the maximum value is square root 5 (using my trusty calculator) but I'm quite stuck on finding the corresponding values of x. Could you perhaps provide a starting point of some sort or guide me through this one? Find the least positive value of p such that f(x) = f(x + p) Now this is the one I need urgent help on. I have absolutely no idea how to go about this one. I assume the p would mean the period. Do I simply find the period of square root 5 (x + arctan -0.5)? I tried some working out but am uncertain about it: 2sinx", "# Math Help - smallest \"p\" value for sin 1. ## smallest \"p\" value for sin If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees? How can I show that it really is 360? 2. Hello donnagirl Originally Posted by donnagirl If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees? How can I show that it really is 360? Divide by $2$; then expand $\\sin (x+p)$, and say: $\\sin x \\equiv \\sin(x+p)$ $\\Rightarrow \\sin x \\equiv \\sin x \\cos p + \\cos x \\sin p$ Then compare the coefficients of $\\sin x$ and $\\cos x$: $\\Rightarrow \\left\\{ \\begin{array}{l} \\cos p =1\\\\ \\sin p = 0\\end{array}\\right .$ And the smallest positive value of $p$ that satisfies these equations is $p = 360^o$ 3. Hmmm, I still don't see it Grandad--how did you obtain the coefficients values? 4. Hello donnagirl Originally Posted by donnagirl Hmmm, I still don't see it Grandad--how did you obtain the coefficients values? You'll see that I've used a $\\equiv$ sign, rather than a $=$ sign, to show that the", "# New Post When $$a \\ne 0$$, there are two solutions $$ax^2 + bx + c = 0$$ $$\\lim_{x \\rightarrow 0} \\frac{Sin x}{x} = 1$$ [where x is measured in radian]", "\\\\ {2} & {0} & {-1} \\\\ {b} & {1} & {0}\\end{array}\\right]$is skew-symmetric, find the values of’a’ and ‘b’. [1] Solution: Question 3. Find the magnitude of each of the two vectors $\\vec{a} \\text { and } \\vec{b}$, having the same magnitude such that the angle between them is 60° and their scalar product is $\\frac{9}{2}$. [1] Solution: Let $\\vec{a} \\text { and } \\vec{b}$be two such vectors. Question 4. If a * b denotes the larger of ‘a’ and ‘b’ and if a o b = (a * b) + 3, then write the value of (5) o (10), where * and o are binary operations.** [1] Section – B Question 5. Prove that: [2] Solution: R.H.S = sin-1(3x – 4x3) Putting x = sin θ in R.H.S, we get R.H.S = sin-1(3 sin θ – 4 sin3 θ) = sin-1(sin 3θ) Question 6. Given A = $\\left[\\begin{array}{rr}{2} & {-3} \\\\ {-4} & {7}\\end{array}\\right]$, compute A-1 and show that 2A-1 = 9I – A. [2] Solution: Question 7. Differentiate $\\tan ^{-1}\\left(\\frac{1+\\cos x}{\\sin x}\\right)$with respect to x. [2] Solution: Question 8. The total cost C(x) associated with the pro-duction of x units of an item is given by C(x) = 0.005x3 – 0.02x2 + 30s + 5000. Find the marginal cost when 3 units are produced, where by", "= \\pi/6$, for example. The general formula is $r(x) = \\frac{\\pi}{180}x$. So using the chain rule, $$\\tan'(r(x)) = (1 + \\tan^2(r(x))) r'(x) = \\frac{\\pi}{180} (1 + \\tan^2(r(x))).$$ So, you're missing the extra factor of $\\pi/180$ in the derivative. -- The reason why this is true goes back to the definition of the derivative. Take $\\sin x$. Calculating its derivative at $x = 0$ with the difference quotient formula gives $$\\lim_{h \\to 0} \\frac{\\sin h - \\sin 0}{h} = \\lim_{h \\to 0} \\frac{\\sin h}{h} = 1 .$$ The way this limit is computed is geometrically, it'll be on the internet somewhere. The takeaway is this limit is only equal to $1$ when $h$ is measured in radians. That is why we pick radians as the \"natural\" way to measure angles in math. If you pick degrees instead, there will be an extra factor of $\\frac{\\pi}{180}$ floating around, as we saw above. • +1 Very nice and clear explanation – Ovi Aug 17 '18 at 4:19 A possible way to understand what is going on is to think a little about what the quantities represent. One nice description of an angle is that it corresponds to an arc length on the unit circle. That is, the measure of an angle $\\theta$ is exactly the length of the arc on the", "# Math Help - sin a sinx+ b cosx 1. ## sin a sinx+ b cosx Hey. The question is find the value of R and the value of A. sinx+ 2cosx = Rcos(x-A) i can find R which is square root of 5, but i keep getting 1.11 for A and the answer is 0,46 (radians) can anyone help? 2. Hello Oasis1993 Originally Posted by Oasis1993 Hey. The question is find the value of R and the value of A. sinx+ 2cosx = Rcos(x-A) i can find R which is square root of 5, but i keep getting 1.11 for A and the answer is 0,46 (radians) can anyone help? $R\\cos(x-A) = R\\cos x \\cos A + R \\sin x \\sin A=\\sin x + 2\\cos x$ $\\Rightarrow R\\sin A = 1,\\;R\\cos A = 2$ $\\Rightarrow \\tan A =\\tfrac12$ $\\Rightarrow A = 0.46...$", "is $$S$$, find $$\\lfloor 100 S \\rfloor$$. If the polynomial $$f(x)=4x^4-a.x^3+b.x^2-c.x+5$$ where $$a,b,c \\in \\mathbb{R}$$ has $$4$$ positive real zeroes(roots) say $$r_1,r_2,r_3, \\ and \\ r_4$$, such that $\\frac{r_1}{2}+\\frac{r_2}{4}+\\frac{r_3}{5}+\\frac{r_4}{8}=1$ Find the value of $$a$$. ×", "and store it in a variable. gvn_expression = sin(45)*cos(45) # Apply evalf() function on the above given mathematical expression to # evaluate the given mathematical expression. # Here we multiply the values of sin(45) and cos(45). # Store it in another variable. rslt = gvn_expression.evalf() # Print the result of the evaluated mathematical expression. print(\"The result of the evaluated mathematical expression:\") print(rslt) Output: The result of the evaluated mathematical expression: 0.446998331800279", "# Math Help - cos x 1. ## cos x I would appreciate if someone helps with this problem. Thanks. Tom and Jerry use their calculators to calculate cos x. Tom enters x as x degrees, while Jerry enters x radians. Tom gets an answer y, while Jerry gets the value -y. Let x be the smallest positive value for which this happens. Find x. 2. Originally Posted by chil2e I would appreciate if someone helps with this problem. Thanks. Tom and Jerry use their calculators to calculate cos x. Tom enters x as x degrees, while Jerry enters x radians. Tom gets an answer y, while Jerry gets the value -y. Let x be the smallest positive value for which this happens. Find x. 1. Let r denote the value of an angle in radian and d the value of the same angle in degree. Then the conversion of degree into radian is calculated by: $r = \\dfrac d{180} \\cdot \\pi$ 2. If the angle x is measured in degree the cos-function has the property $\\cos(x) =- \\cos(180^\\circ - x)$ 3. According to the text of the question you have to solve the equation for x: $x = \\dfrac{180-x}{180} \\cdot \\pi$ 4. I've got $x \\approx 3.087702086...$ 3. That is a better and more correct approach. Touche.", "have maximum value over all x of 1, as you say. Of course, the derivative of sin(x) is cos(x) and the derivative or cos(x) is -sin(x) only when x is measured in radians so $x_0= 0$ and $x= \\frac{3}{180}\\pi= 0.03491$, approximately. So the errror is less than $\\frac{1}{(N+1)!}(.03491)^{N+1}$ and you want to make that .00001. I would recommend just evaluating that at N= 3, 4, 5, ... until it gets small enough.", "# If f(x)=cos(x) - sin(x) then what is the value of f(1)? Dev kumar basant 38 Points 5 years ago anwer will be negative and in fraction you can see easily in the graph of sinx and cosx they intersect at 45 and after that sinx is above then cosx. Arun 25757 Points 5 years ago here in f(1), 1 is angle in radian. so f(1)= cos1−sin1 this will be a negative answer as in [0,$\\pi$/4) cos will be greater than sin. and 1 will be less than $\\pi$/4.", "Find the positive value of $x$ that satisfies $x = 4.7 \\cos(x)$. Give the answer to six places of accuracy. $x\\approx$ Remember to calculate the trig functions in radian mode. Your overall score for this problem is", "SWR on the other hand uses just the magnitude of the reflection coefficient: $$VSWR\\;=\\;\\frac{1+|\\Gamma|}{1-| \\Gamma |}$$ VSWR is always a real number because we have taken the magnitude of a complex number and it is always positive.", "Find the least positive integer $$k$$ for which the equation $$\\left\\lfloor\\frac{2002}{n}\\right\\rfloor=k$$ has no integer solutions for $$n$$. (The notation $$\\lfloor x\\rfloor$$ means the greatest integer less than or equal to $$x$$.) (第二十届AIME2 2002 第8题)", "$$f(x,y)=(2x^2+y^2-1)(r^2-1)+1,$$ which much easier. So, apply Lagrange multiplies to get maximum/minimum values. In your special case the minima and maxima can be computed without using Lagrange multipliers or partial derivatives. For all points $(x, y)$ on the circle with center $(0, 0)$ and radius $r$ you have $x^2 + y^2 = r^2$ and therefore $$f(x, y) = (x^2 + r^2 - 1)(r^2 - 1) + 1 \\, .$$ Every term apart is constant apart from $x \\in [-r, r]$, and the factor$r^2 -1$ is positive. Therefore $f$ is minimal for $x = 0$ (which implies $y = \\pm r$) and maximal for $x = \\pm r$ (which implies $y = 0$). The same result can be obtained by using polar coordinates: $$g(\\theta) = f(r \\cos \\theta, r \\sin \\theta) = (r^2 \\cos^2 \\theta + r^2 - 1)(r^2 -1)+1$$ is minimal/maximal exactly where $\\cos^2 \\theta$ is." ]

[ "^{3} x \\sin x$ $\\therefore f^{\\prime}(x)=0$ $\\Rightarrow 4 \\sin ^{3} x \\cos x-4 \\cos ^{3} x \\sin x=0$ $\\Rightarrow \\sin ^{3} x \\cos x-\\cos ^{3} x \\sin x=0$ $\\Rightarrow \\tan ^{3} x-\\tan x=0$ $\\Rightarrow \\tan x\\left(\\tan ^{2} x-1\\right)=0$ $\\Rightarrow \\tan x=0, \\tan ^{2} x=1$ $\\Rightarrow \\tan x=0, \\tan x=\\pm 1$ $\\Rightarrow x=0, x=\\frac{\\pi}{4}, \\frac{3 \\pi}{4}$ Thus, $c=\\frac{\\pi}{4} \\in\\left(0, \\frac{\\pi}{2}\\right)$ such that $f^{\\prime}(c)=0$. ​Hence, Rolle’s theorem is verified. (xviii) The given function is $f(x)=\\sin x-\\sin 2 x$. Since $\\sin x$ and $\\sin 2 x$ are everywhere continuous and differentiable, $f(x)$ is continuous on $[0, \\pi]$ and differentiable on $(0, \\pi)$. Also, $f(\\pi)=f(0)=0$ Thus, $f(x)$ satisfies all the conditions of Rolle’s theorem. Now, we have to show that there exists $c \\in(0, \\pi)$ such that $f^{\\prime}(c)=0$. We have $f(x)=\\sin x-\\sin 2 x$ $\\Rightarrow f^{\\prime}(x)=\\cos x-2 \\cos 2 x$ $\\therefore f^{\\prime}(x)=0$ $\\Rightarrow \\cos x-2 \\cos 2 x=0$ $\\Rightarrow \\cos x-2 \\cos ^{2} x+1=0$ $\\Rightarrow 2 \\cos ^{2} x-\\cos x-1=0$ $\\Rightarrow(\\cos x-1)(2 \\cos x+1)=0$ $\\Rightarrow \\cos x=1, \\cos x=\\frac{-1}{2}$ $\\Rightarrow \\cos x=\\cos \\frac{\\pi}{2}, \\cos x=\\frac{2 \\pi}{3}$ $\\Rightarrow x=\\frac{\\pi}{2}, \\frac{2 \\pi}{3}$ Thus, $c=\\frac{\\pi}{2}, \\frac{2 \\pi}{3} \\in(0, \\pi)$ such that $f^{\\prime}(c)=0$. Hence, Rolle’s theorem is verified.​", "$[-1,1]$, because of $\\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\\tan\\arcsin x$. (Note: your $x\\in[-\\pi/4,\\pi/4]$ is wrong and the probable cause for the low grade.) The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\\in(0,1)$. For every root we find, also its negative will be a root. If $y=\\sin\\arctan x$, then $\\arctan x=\\arcsin y$ and so $x=\\tan\\arcsin y$ so $$x^2=\\frac{\\sin^2\\arcsin y}{1-\\sin^2\\arcsin y}=\\frac{y^2}{1-y^2}$$ whence $$y^2=\\frac{x^2}{1+x^2}$$ If $z=\\tan\\arcsin x$, then $$z^2=\\frac{\\sin^2\\arcsin x}{1-\\sin^2\\arcsin x}=\\frac{x^2}{1-x^2}$$ So your equation $\\frac{1}{y^2}-\\frac{1}{z^2}=4x^2$ becomes $$\\frac{1+x^2}{x^2}-\\frac{1-x^2}{x^2}=4x^2$$ that is, $$2x^2=4x^4$$ and, since $x\\ne0$, $x^2=\\frac{1}{2}$. The only positive root is $$x=\\frac{1}{\\sqrt{2}}$$ which indeed belongs to $(0,1)$. Thus the solutions are $$\\frac{1}{\\sqrt{2}}\\qquad\\text{and}\\qquad{-}\\frac{1}{\\sqrt{2}}$$ • Yes, ofcourse! arctan is undefined at $-\\pi/2$, $0$ and $\\pi/2$. So the domain is $(-1,0)\\cup(0,1)$, if I'm not mistaken? Which in turn means that the solutions to the equation $x\\in[\\pi/4,\\pi/4] / {0}$? Sep 18, 2017 at 20:53 • @Parseval No: $x\\in(-1,0)\\cup(0,1)$. Why should you take the arctangent? Sep 18, 2017 at 20:54 • Because it's 22:57 and I've been staring at math for 18 hours. This is the kind of blind mistakes you do in such a condition. Thank you very much for clearing up everything, I really appreciate it mate! Sep 18, 2017 at 20:57 Taking derivative of", "ns was watching this video clip and at 5:30 $\\cos \\frac\\pi2$ is substituted with $0$. Utilizing my calculator the result I acquired was $-0.5$. I m sorry one is correct? $\\cos\\left(\\frac\\pi2\\right)=0$. The factor you are gaining $-0.5$ is since you are not putting brackets approximately $\\pi/2$. Thus, you room obtaining the value of $(\\cos\\pi)/2=(-1)/2=-0.5$. You need to use the brackets together follows: $\\cos\\left(\\frac\\pi2\\right)$. There room a pair ways to present that $\\cos (\\pi/2)=0$ 1) The value $\\tan(\\pi/2)$ is undefined. Because $\\tan(x)=\\frac\\sin(x)\\cos(x)$, and both $\\sin(x)$ and also $\\cos(x)$ are continuous throughout the reals, this suggests that it have the right to only occur if $\\cos (\\pi/2)=0$. You are watching: Cos (-pi/2) 2) The co-function identification $\\cos(\\pi/2-x)=\\sin(x)$ means that $\\cos (\\pi/2)=\\cos(\\pi/2-0)=\\sin(0)=0$ 3) Looking at the unit circle where $y=\\sin(\\theta)$ and also $x=\\cos(\\theta)$. When $\\theta= 90^0=\\pi/2$, $\\sin (\\theta)=1$ and also $\\cos(\\theta)=0$. 4) the derivative of $\\sin(x)$ is $0$ only as soon as $\\sin(x)=1$, which only occurs once $x= \\fracn\\pi2$ for any type of $n\\in \\sirhenryjones-museums.orgbb N$. Because the derivative the $\\sin(x)$ is $\\cos(x)$ this suggests that $\\cos(\\pi/2)=\\cos((1)\\pi/2)=0$. If girlfriend have any qustions let me recognize below. share point out follow reply Mar 3 \"17 at 23:21 Sentinel135Sentinel135 $\\endgroup$ Thanks for contributing response to sirhenryjones-museums.orgematics stack Exchange! Please be certain to answer the question. Provide details and share her research! But avoid", "I think, assuming that $$\\frac{-1}{\\phi} = \\frac{1-\\sqrt{5}}{2}$$ Which I seem to remember a similar property about phi... 12. Sep 13, 2010 Helios note that tan( x ) = cos( x ) = 1 / sqr( phi ) so the solution can be made in terms of arctan or arccos. 13. Sep 15, 2010 willem2 Since tan(x) increases monotonically from 0 to infinity in the domain (0, pi/1) and cos (x) decreases from 1 to 0 in the same domain, there should be a solution x between 0 and pi/2 and 0< sin(x) < 1. There's a solution in the second quadrant as well. cos(x) and tan(x) have different signs in the 3rd and fourth quadrant, so a solution where sin(x) < 0 is not possible Once you get all the signs right in the quadratic formula, you'll get $$sin(x) = \\frac {1} {\\phi} = \\phi - 1$$ from one of the solutions. The other solution has sin(x) >1 so that isn't a valid solution. }", "the other limit, the situation is different: for $0<x<\\pi/2$ we have $\\tan x>x$ (also this can be verified by using the derivative). Since $$\\lim_{x\\to0^+}\\frac{\\tan x}{x}=1$$ there is $\\delta>0$ such that, for $0<x<\\delta$, $1<\\frac{\\tan x}{x}<2$ and so, for $0<x<\\delta$, $$\\left\\lfloor\\frac{\\tan x}{x}\\right\\rfloor=1$$ which gives, together with the fact that $\\frac{\\tan x}{x}$ is even, $$\\lim_{x\\to 0}\\left\\lfloor\\frac{\\tan x}{x}\\right\\rfloor=1$$ In both cases you are actually doing the limit of a constant function (in a punctured neighborhood of $0$, for the second one). • this can't possibly be true, you can see that from using l'hopital's rule or expanding sinx (tanx ) by McLauren expansion, you see that sinx/x -> 1 – user247608 Jun 19 '15 at 20:49 • @david Yes, I know that $\\lim_{x\\to0}\\frac{\\sin x}{x}=1$, but this has little to do with the problem at hand. The “floor” function is not continuous at $1$, so you can't “put the limit inside”. – egreg Jun 19 '15 at 20:51 Intuitively: Note that on $\\lfloor \\sin(x)/x \\rfloor$, the graph approaches $1$ from below, while on $\\lfloor \\tan(x)/x \\rfloor$, the graph approaches $1$ from above. Therefore, while the graph of $\\sin(x)/x$ approaches 1, it never reaches there: therefore, for small values of $x$ around 0, the value of $\\lfloor \\sin(x)/x \\rfloor$ will be 0. However, even for small values of $x$ around 0, the value of", "}$ is (a) 2 cosθ (b) 2i sin nθ (c) 2i sin nθ (d) 2i cos nθ 7. If x is real and $\\frac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 }$ then (a) $\\frac{1}{3}$ ≤k≤ (b) k≥5 (c) k≤0 (d) none 8. lf the root of the equation x3 +bx2+cx-1=0 form an lncreasing G.P, then (a) one of the roots is 2 (b) one of the rots is 1 (c) one of the rots is -1 (d) one of the rots is -2 9. If ax2 + bx + c = 0, a, b, c E R has no real zeros, and if a + b + c < 0, then __________ (a) c>0 (b) c<0 (c) c=0 (d) c≥0 10. ·If $\\alpha ={ tan }^{ -1 }\\left( \\cfrac { \\sqrt { 3 } }{ 2y-x } \\right) ,\\beta ={ tan }^{ -1 }\\left( \\cfrac { 2x-y }{ \\sqrt { 3y } } \\right)$ then $\\alpha -\\beta$ (a) $\\cfrac { \\pi }{ 6 }$ (b) $\\cfrac { \\pi }{ 3 }$ (c) $\\cfrac { \\pi }{ 2 }$ (d) $\\cfrac { -\\pi }{ 3 }$ 11. If tan-1(3)+tan-1(x)=tan-1(8)then x= (a) 5 (b) $\\cfrac { 1 }{ 5 }$ (c) $\\cfrac { 5 }{ 14 }$ (d) $\\cfrac { 14 }{ 5 }$", "are going to need $$\\arctan\\left(\\frac{x-\\pi}{x-4}\\right)\\gt 0.$$ When is the arctangent positive? When the argument is positive. So we need $$\\frac{x-\\pi}{x-4}\\gt 0.$$ When is a fraction positive? When both numerator and denominator are positive, or when they are both negative. So we need either $x-\\pi\\gt 0$ and $x-4\\gt 0$ (this happens when $x\\gt 4$); or $x-\\pi\\lt 0$ and $x-4\\lt 0$ (this happens when $x\\lt \\pi$). So we now need to restrict our $x$s to $(-\\infty,\\pi)\\cup(4,\\infty)$. (Note that this also maintains the exclusion of $4$). 5. Finally, we need to take the square root of the answer. That means that the logarithm must be nonnegative. When is $\\log_{\\frac{1}{2}}(a)\\geq 0$? When $0\\lt a \\leq 1$ (taking exponentials with base $\\frac{1}{2}$ flips the inequality, because $(\\frac{1}{2})^x$ is decreasing). So we actually need $$0\\lt \\arctan\\left(\\frac{x-\\pi}{x-4}\\right)\\leq 1.$$ When is $0\\lt \\arctan(a)\\leq 1$? When $0\\lt a \\lt \\frac{\\pi}{4}$ (thanks to Jonas Meyer for the heads up!). When $\\tan(0)\\lt a \\leq \\tan(1)$. So we need $$0 \\lt \\frac{x-\\pi}{x-4}\\leq \\tan(1).$$ Since $\\tan(1)\\gt 0$ This happens if either $$0\\lt (x-\\pi) \\lt \\tan(1)(x-4)$$ or $$\\tan(1)(x-4)\\lt (x-\\pi)\\lt 0.$$ So check the inequalities; then remember that $x$ must be greater than $4$ for both $x-\\pi$ and $x-4$ to be positive; or less than $\\pi$ for both to be negative. - I think we made the same mistake of saying $\\arctan(x)<1\\Leftrightarrow x<\\frac{\\pi}{4}$.", "to added integer multiples of $$2\\pi$$). Namely, $$X = \\frac{2\\pi}{3} + 2\\pi k,$$ for some integer $$k$$. Since $$\\sin$$ and $$\\cos$$ are $$2\\pi$$-periodic, we get $$\\sin(X) = \\sin\\left(\\frac{2\\pi}{3}\\right) = \\frac{\\sqrt 3}{2} \\qquad \\text{ and } \\qquad \\cos(X) = \\cos\\left(\\frac{2\\pi}{3}\\right) = -\\frac{1}{2}.$$ You have definitely seen trig identities such as $$\\sin ^2x +\\cos^2 x =1$$ And $$\\sec ^2x=1+\\tan ^2 x$$ Well you need to apply the identities to get the unknown parts from the given information. In this case, your professor is using $$A = \\theta$$ and $$B = X$$. You are given $$\\cos(A) = \\frac{3}{5}$$. From this, you know that for a right triangle, the adjacent side is 3 and the hypotenuse is 5. By the Pythagorean Theorem, this means that the opposite side is 4 since $$3^{2} + 4^{2} = 5^{2}$$. Also, since you are in the 4th quadrant, your opposite side must be negative ($$y$$ values are negative in the 4th quadrant). Thus, $$\\sin(A) = \\frac{-4}{5}$$. For $$\\tan(B) = -\\sqrt{3}$$, you know that $$\\tan(B) = \\frac{\\frac{\\sqrt{3}}{2}}{\\frac{-1}{2}} = \\frac{\\sin(B)}{\\cos(B)}$$ (you are in the second quadrant, so your cosine is negative and your sine is positive).", "# Finding smallest positive root $\\sqrt{\\sin(1-x)}=\\sqrt{\\cos x}$ Finding smallest positive root of the equation $\\sqrt{\\sin(1-x)}=\\sqrt{\\cos x}$ Try: $\\sin(1-x)=\\cos x= \\sin\\bigg(\\frac{\\pi}{2}-x\\bigg)$ $1-x=n\\pi+(-1)^n\\bigg(\\frac{\\pi}{2}-x\\bigg)$ Could some help me to solve it, Thanks If $n$ is even, then the equation becomes $$1-x=n\\pi +\\frac\\pi2 -x$$ which simplifies to $$1=n\\pi + \\frac\\pi2$$ which is clearly not true for any value of $n\\in\\mathbb N$. For odd values of $n$, the equation is $$1-x=n\\pi -\\frac\\pi2 + x$$ simplifying to $$1-n\\pi + \\frac\\pi2 = 2x$$ and is solvable for $x$. This gives you a set of candidate values for $x$ which you then still need to check, because if $\\sin(1-x)<0$, then $x$ cannot be a solution! • Neither can there be a solution if $\\cos x\\lt 0$ – Rohan Shinde Feb 14 '18 at 14:39 If you square both sides, and apply sum of angles formula, you get $$\\sin 1\\cos x - \\sin x \\cos 1 = \\cos x.$$ Divide by $\\cos x \\cos 1$ to get $$\\tan 1 -\\tan x = \\sec 1$$ so that $x = \\arctan(\\tan 1 - \\sec 1).$ This is not as satisfying as @5xum 's answer, but I thought it was interesting. • Just a small doubt. Shouldn't that be $\\sin 1\\cos x \\color{red}{-} \\sin x \\cos 1$ – Rohan Shinde Feb 14 '18 at 14:33 • @Manthanein Yep. Thanks. –", "really confused about that and entering things like $\\tan(30\\sin30)$ instead of $\\tan30\\cdot\\sin30$, etc. Maybe some explicit report on the sequence of key strokes could shed some light on what's going on. – Michael Hardy Apr 27 '15 at 11:40 By definition, $$\\tan x=\\frac{\\sin x}{\\cos x}$$ but in order to give sense to this definition, the $\\cos$ must be nonzero, i.e. $x\\neq\\frac{\\pi}2+k\\pi$. Hence $\\tan$ is defined only for $x\\in\\Bbb R\\setminus \\{\\frac{\\pi}{2}+k\\pi\\}$. For these values you can multiply both sides for $\\cos$, obtaining $\\tan x\\cos x=\\sin x$. But pay attention: this holds only for the values I wrote! Otherwise $\\tan$ is not even defined, so it wouldn't make sense to consider the relation. Now in your case, $x=30°$ is the same as $x=\\pi/6$ in radiant and in this value the $\\tan$ is defined. So the relation holds. You must had a bad work with your calculator, that's all. As tan(x)≡ Sin(x)/Cos(x), you are right in that Tan(x) * cos(x) ≡ Sin(x). This is by definition of the Tan function, which is defined as Sin(x) / Cos(x). If it helps consider the right angle triangle from the unit circle, where cos(x) = Hypotenuse / adjacent and Sin(x) = opposite / hypotenuse, so Tan(x) as equalling opposite / adjacent is easy to see by dividing one by the other since the hypotenuse", "2\\pi$. $\\tan x=1$ if and only if $x=\\tan^{-1} 1=\\frac{\\pi}{4}$ or $x=\\pi+\\tan^{-1} 1=\\frac{5\\pi}{4}$. ($\\frac{\\pi}{4}$ is the principal value and the tangent is positive in the first and third quadrants.) - (1) $f$ is increasing or decreasing accordingly as $f'(x)>0$ or $<0$ Now, $f'(x)=\\cos x-\\sin x=\\sqrt2\\cos(x+\\frac \\pi 4)$ $f'(x)>0$ if $\\cos(x+\\frac \\pi 4)>0$ if $x+\\frac \\pi 4$ lies in the 1st or 4th quadrant. As,$0\\le x\\le 2\\pi,$ for $f'(x)>0, 0\\le x<\\frac \\pi 4$ or $\\frac{3\\pi}2-\\frac \\pi4<x\\le 2\\pi$ Similarly for $f'(x)<0$ (2)For the maxima/minima, $f'(x)=0\\implies \\tan x=1=\\tan \\frac \\pi 4 \\implies x=m\\pi+\\frac \\pi 4$ where $m$ is any integer. $f''(x)=-(\\sin x+\\cos x)=-\\sqrt2\\cos(x-\\frac \\pi 4)$ So, $f''(m\\pi+\\frac \\pi 4)= -\\sqrt2\\cos m\\pi$ which is $<0$ if $m$ is even$=2r$(say) where $r$ is any integer So,the local maximum of $f(x)$ is at $x=2r\\pi+\\frac \\pi 4$ As $0\\le x\\le 2\\pi,$ for local maximum of $f(x),x=\\frac \\pi 4$ $f_{max}=f(\\frac \\pi 4)=\\sqrt 2$ Similarly, for local minimum. (3) Using this, $f(x)$ is concave up if $f''(x) > 0$ So, we need $\\cos(x-\\frac \\pi 4)<0$ i.e., $x-\\frac \\pi 4$ will lie in the 2nd or in the 3rd quadrant. $\\frac \\pi2<x-\\frac \\pi 4< \\frac {3\\pi}2\\implies \\frac \\pi2+\\frac \\pi 4<x< \\frac {3\\pi}2+\\frac \\pi 4$ as $0\\le x\\le 2\\pi$ Similar for the concave down if $f''(x) < 0$ For the point of inflexion, $f''(x)=0$ or $f''(x)$ does not exist. Here", "and I think expanding the $\\sin(2x)$ is a good idea: $\\int_{\\piby 6}^{\\piby 3} \\frac{3 \\cos^2(x)}{4 + 4\\sin(x)\\cos(x)} \\times \\sec^2(\\theta)\\d \\theta$. If we divide top and bottom by $\\cos^2(x)$, we get something nicer still: $\\int_{\\piby 6}^{\\piby 3} \\frac{3 }{4\\sec^2(x) + 4\\tan(x)} \\times \\sec^2(\\theta)\\d \\theta$. Aha! There’s our tan! Also, $\\sec^2(x) \\equiv 1 + \\tan^2(x)$, so the whole of the denominator can be written in terms of $\\tan(x)$. ### Step 4: tidy up any remaining $x$s Let’s define $t = \\tan(x)$ and $T = \\tan(\\theta)$; our substitution is $2t + 1 = \\sqrt{3}T$ and the bottom of the fraction is $4t^2 + 4t + 4$. By squaring, we have $4t^2 + 4t + 1 = 3T^2$, so the bottom is clearly $3T^2 + 3$. Now we have $\\int_{\\piby 6}^{\\piby 3} \\frac{3 }{3\\tan^2(\\theta)+ 3} \\times \\sec^2(\\theta)\\d \\theta$. ### Step 5: Do the actual integration That integrand, once you cancel the 3s and remember your relationship between $\\tan^2(\\theta)$ and $\\sec^2(\\theta)$, turns out to be … 1. And $\\int_{\\piby 6}^{\\piby 3} \\d \\theta$ is trivially $\\piby 6$. Phew! I’m glad I got to use ‘trivially’ at the end there. Hope that helps! - Uncle Colin", "# Thread: curious about this 1. ## curious about this Find x , where 0<x<360 , which satisfy $\\sec x +\\tan x=4$ $\\frac{1+\\sin x}{\\cos x}=4\\Rightarrow 1+\\sin x=4\\cos x$ putting $4\\cos x-\\sin x=1$ and continuing , i found that x=61.9 using $r\\cos (x+\\alpha)=1$ where $r=\\sqrt{17}$ and $\\alpha=\\tan^{-1}\\frac{1}{4}$ But how come it doesn't work when i put $\\sin x-4\\cos x=$ -1 , doing the same thing .. $r\\sin (x-\\alpha)=-1$ and i don get the same result , x=61.9 ? is it because -1 is not positive ? Is it one of the conditions ? 2. Originally Posted by thereddevils ... $\\frac{1+\\sin x}{\\cos x}=4\\Rightarrow 1+\\sin x=4\\cos x$ $4\\cos x-\\sin x=1$ and continuing , i found that x=61.9 But how come it doesn't work when i put $\\sin x-4\\cos x=$ -1 I do not understand what you are saying; if you multiply both sides of this $4\\cos x-\\sin x=1.0019...$ by -1, after rearranging, you will get this: $\\sin x-4\\cos x=$ -1.0019... 3. Originally Posted by aidan I do not understand what you are saying; if you multiply both sides of this $4\\cos x-\\sin x=1.0019...$ by -1, after rearranging, you will get this: $\\sin x-4\\cos x=$ -1.0019... sorry bout that Aidan , edited .", "Dear Uncle Colin, In an answer sheet, they’ve made a leap from $\\arctan\\left(\\frac{\\cos(x)+\\sin(x)}{\\cos(x)-\\sin(x)}\\right)$ to $x + \\frac{\\pi}{4}$ and I don’t understand where it’s come from. Can you help? -- Awful Ratio Converted To A Number Hello, ARCTAN, and thank you for your message! There’s a principle I want to introduce here that’s not an obvious one, and possibly not a hard-and-fast rule: when you’re working with the arctangent of something, it often helps if the something is expressed in tangents. In this case, the argument of the arctangent is $\\frac{\\cos(x)+\\sin(x)}{\\cos(x)-\\sin(x)}$, which - if you divide top and bottom by $\\cos(x)$ - is $\\frac{1 + \\tan(x)}{1 - \\tan(x)}$. If we notice that $\\tan(A+B) \\equiv \\frac{\\tan(A)+\\tan(B)}{1-\\tan(A)\\tan(B)}$, then we can see that if $\\tan(A)=1$ (so $A=\\frac{\\pi}{4}$) and $B=x$, we recover exactly what we have above. Your expression is $\\arctan\\left(\\tan\\left(x + \\frac{\\pi}{4}\\right)\\right)$, which is clearly $x + \\frac{\\pi}{4}$. Hope that helps, -- Uncle Colin", "• Apr 5th 2009, 10:53 PM Hello goldenroll Quote: Originally Posted by goldenroll Can someone help with these problems I got a test on Thursday and I don't know how to do these. I can't use a cal by the way. 1. What is tan(2π/3)? I know that tan is y/x so it is -1/2, sqroot 3/2 so how would I put this in an equation? Use $\\tan \\pi/3 = \\sqrt3$, and then $\\tan 2\\pi /3 = -\\tan(\\pi - \\pi /3)$ Quote: 2. If t^(2) = ln(1 + x) − ln(1 − x), what is x in terms of t? $\\ln(1+x) -\\ln(1-x) = \\ln\\left(\\frac{1+x}{1-x}\\right) = t^2$ Raise $e$ to the power of both sides: $\\frac{1+x}{1-x} = e^{t^2}$ Solve for $x$ . Quote: 3. What are all solutions of the equation: sin^(2) x + 2 sin x = 5/4? Write the equation as $4\\sin^2x+8\\sin x - 5 =0$ This is a quadratic in $\\sin x$. Factorise and solve. Quote: 4. If f (x) = √4x^(2) + 8, x ≥ 0, what is f ^(− 1) (x)? I'm not quite sure what the square root sign covers here. I assume you mean: $f(x) = \\sqrt{4x^2 + 8}$ So, to find $f^{-1}$, let $f(x)=y$. Then $y = \\sqrt{4x^2 + 8}$ Now make $x$ the subject of this equation, to give $x", "to be positive. I'm pretty sure that's the reason why I wasn't getting the right values. Thank you for your answer though. That's probably a better way to anaylize the problem. – Modesto Rosado Sep 19 '20 at 6:56 • OK, I had added some more stuff to my answer. – Jean Marie Sep 19 '20 at 6:59 We have that for $$z\\neq 0 \\implies \\Re(z)\\neq 0$$ $$z^{n}\\Re(z)=z^{-n}\\Im(z) \\iff z^{2n}=\\frac{\\Im (z)}{\\Re (z)}\\in \\mathbb R$$ $$\\rho^{2n}\\left(\\cos(2n\\theta)+i\\sin(2n\\theta)\\right)=\\frac{\\sin \\theta}{\\cos \\theta}$$ therefore we have $$\\sin(2n\\theta)=0 \\iff 2n\\theta=k\\pi$$ and then $$\\rho^{2n}\\cos(k\\pi)=\\tan\\left(\\frac{k\\pi}{2n}\\right)$$ that is for $$k=2h$$ even $$\\rho^{2n}=\\tan\\left(\\frac{h\\pi}{n}\\right)$$ for $$k=2h+1$$ odd $$\\rho^{2n}=-\\tan\\left(\\frac{(2h+1)\\pi}{n}\\right)$$", "# How do you simplify sin[Tan^-1 (-2/5) + Cos^-1 (3/5)]? Jul 2, 2016 $\\frac{14}{5 \\sqrt{29}}$, against principal values of inverse functions, in the expression. General values are $\\pm \\frac{14}{5 \\sqrt{29}} \\mathmr{and} \\pm \\frac{26}{5 \\sqrt{29}}$. #### Explanation: Let $a = {\\tan}^{- 1} \\left(- \\frac{2}{5}\\right)$. Then, $\\tan a = - \\frac{2}{5} < 0$. So, principal a is in the 4th quadrant. And so, sin is negative and cos is positive. Thus, in this case, $\\cos a = \\frac{5}{\\sqrt{29}} \\mathmr{and} \\sin a = - \\frac{2}{\\sqrt{29}}$. Let $b = {\\cos}^{- 1} \\left(\\frac{3}{5}\\right)$. Then, #cos b =3/5 > 0. So, principal b is in the 1st quadrant. And so, sin is positive.. Thus, in this case, sin b = 4/5. The given expression = sin (a + b)=sin a cos b+cos a sin b $= \\left(- \\frac{2}{\\sqrt{29}}\\right) \\left(\\frac{3}{5}\\right) + \\left(\\frac{5}{\\sqrt{29}}\\right) \\left(\\frac{4}{5}\\right)$ $= \\frac{14}{5 \\sqrt{29}}$. In general, a is in either 4th quadrant or in the 2nd and b is in either 1st or the 2nd. So, sin a and cos a can take both signs. ln general, b is in either 1st or 4th. So, sin b can take both signs. Accordingly, the general values are $\\pm \\frac{14}{5 \\sqrt{29}} \\mathmr{and} \\pm \\frac{26}{5 \\sqrt{20}}$.", "How do I solve 3 sin x+2 tan x=0 for 0<=x<=360 ? Sep 28, 2015 $x = 0$, $x = 180$, $x = 360$, $x = \\arccos \\left(- \\frac{2}{3}\\right)$, $x = - \\arccos \\left(- \\frac{2}{3}\\right)$ Explanation: The first step is to determine the domain of the functions involved. With the restriction $0 \\le x \\le 360$ (degrees, I assume) we have exclude the points where $\\tan \\left(x\\right)$ is not defined. By definition, $\\tan \\left(x\\right) = \\sin \\frac{x}{\\cos} \\left(x\\right)$ Therefore, we have to exclude points where $\\cos \\left(x\\right) = 0$, that is the domain is described as $x \\ne 90$ and $x \\ne 270$ With the above restrictions in mind we can transform the equation as follows: $3 \\sin \\left(x\\right) + 2 \\sin \\frac{x}{\\cos} \\left(x\\right) = 0$ The immediate temptation to reduce the above equation by $\\sin \\left(x\\right)$ should be accompanied by checking if it can be equal to zero to avoid losing solutions. Indeed, $\\sin \\left(x\\right) = 0$ can occur within a domain defined above. It happens when $x = 0$, $x = 180$ and $x = 360$. So the three values above are solutions. After specifying these three solutions we can reduce the equation by $\\sin \\left(x\\right)$ getting $3 + \\frac{2}{\\cos} \\left(x\\right) = 0$, from which follows $\\cos \\left(x\\right) = - \\frac{2}{3}$, solutions of this are $x", "$|\\cos(x)| < \\frac{|\\sin(x)|}{y} < \\frac{1}{|y|}$ and since both functions $\\sin$ and $\\cos$ are positive for these values of $x$ we have $\\frac{\\sin(x)}{\\cos(x)} > y.$ You can prove the $-\\infty$ part similarly. - Are you claiming that if $\\lim_{x \\to a^-} f(x) = 1$ and $\\lim_{x \\to a^-} g(x) = 0$, then $\\lim_{x \\to a^-} f(x)/g(x) = +\\infty$? That ain't right. – TonyK Nov 27 '11 at 17:22 @TonyK, that is proved here in the case of $a = \\pi/2$, $f = \\sin$ and $g = \\cos$, do you have an objection to this proof? – user16697 Nov 27 '11 at 17:47 Your \"therefore\" is premature. It looks like you are claiming that the third equation is a consequence of the first two. It looked like that to me, anyway. – TonyK Nov 27 '11 at 18:19 @TonyK: Yes I state two hypothesis (which are direct consequences of continuity) and then claim \"therefore X\" then unfold all the definitions and show how to prove X given the hypothesis. – user16697 Nov 27 '11 at 18:23 The simple facts are that $\\lim \\limits_{x \\to -\\frac{\\pi}{2}} \\tan(x)$ is not equal to $-\\infty$, and $\\lim \\limits_{x \\to \\frac{\\pi}{2}} \\tan(x)$ is not equal to $+\\infty$. This is because in every neighbourhood of either $\\pi/2$ or $-\\pi/2$, $\\tan(x)$ takes on arbitrarily large positive and", "# Math Help - Not sure about the reasoning for this answer? 1. ## Not sure about the reasoning for this answer? the problem is Given cos2x = 7/32, tan2x<0 and -\\pi/2<x<\\pi/2, find sin x $1-2sin^2x=7/32$ $-2sin^2x=7/32-1$ $-2sin^2x= -25/32$ $sin^2x=-25/32 * -1/2$ $sin^2x=25/64$ $sin x = -5/8 but not 5/8$ because -\\pi/2<x<\\pi/2 means the answer would be either in the first or forth quadrant and tan2x has to be smaller than 0 so if the cos is positive, then the sin has to be negative? Is this why there is only -5/8 as the answer instead of + or - 5/8 because of the square root when I took out the $sin^2$? 2. Originally Posted by dorkymichelle the problem is Given cos2x = 7/32, tan2x<0 and -\\pi/2<x<\\pi/2, find sin x $1-2sin^2x=7/32$ $-2sin^2x=7/32-1$ $-2sin^2x= -25/32$ $sin^2x=-25/32 * -1/2$ $sin^2x=25/64$ $sin x = -5/8 but not 5/8$ because -\\pi/2<x<\\pi/2 means the answer would be either in the first or forth quadrant and tan2x has to be smaller than 0 so if the cos is positive, then the sin has to be negative? Is this why there is only -5/8 as the answer instead of + or - 5/8 because of the square root when I took out the $sin^2$? Look at the last line again. Should be $\\sin{x}=\\frac{+5}{8}$ 3. The" ]

[ "What is the value of $\\lim _{x\\to 0}\\left\\lfloor\\frac{\\tan x \\sin x}{x^2}\\right\\rfloor$ How do I evaluate the limit $$\\lim _{x\\to 0}\\left\\lfloor\\frac{\\tan x \\sin x}{x^2}\\right\\rfloor$$ where $\\lfloor\\cdot\\rfloor$ denotes greatest integer function. I know that $x>\\sin x$ and $x < \\tan x$ but how do I use these results here? $\\tan x \\over x$ tends to $1+$ whereas $\\sin x \\over x$ tends to $1-$ • To dheeraj is your question is $\\displaystyle \\lim_{x\\rightarrow 0}\\lfloor \\frac{\\tan x\\sin x}{x^2} \\rfloor$, Where $\\lfloor x \\rfloor$ represent floor function of $x$ – juantheron Feb 18 '16 at 17:05 • Yah There is a greatest integer function. – Dhiraj Barnwal Feb 18 '16 at 17:06 • Then answer is $=1$ – juantheron Feb 18 '16 at 17:06 • How do we determine that ? As one quantity tries to reduce the value and the other tries to increase it. – Dhiraj Barnwal Feb 18 '16 at 17:10 • Using $f(x) = \\tan x\\sin x-x^2$ – juantheron Feb 18 '16 at 17:11 Using Taylor–Young expansions: we know that $$\\tan(x)=x+\\frac{x^3}3+o(x^4),\\qquad\\sin(x)=x-\\frac{x^3}6+o(x^4),$$ hence $$\\tan(x)\\sin(x)=x^2+\\frac{x^4}6+o(x^5),$$ and hence $$\\frac{\\tan(x)\\sin(x)}{x^2}=1+\\frac{x^2}6+o(x^3).$$ From here, we conclude that there exists a punctured neighborhood $V$ of $0$ such that $$\\forall x\\in V,\\ \\frac{\\tan(x)\\sin(x)}{x^2}>1.$$ Since $$\\lim_{x\\to0}\\frac{\\tan(x)\\sin(x)}{x^2}=1$$ we can assume that $V$ has been chosen such that $$\\forall x\\in V,\\ 1<\\frac{\\tan(x)\\sin(x)}{x^2}<2.$$ Hence $$\\forall x\\in V,\\ \\left\\lfloor\\frac{\\tan(x)\\sin(x)}{x^2}\\right\\rfloor=1,$$ from which", "the other limit, the situation is different: for $0<x<\\pi/2$ we have $\\tan x>x$ (also this can be verified by using the derivative). Since $$\\lim_{x\\to0^+}\\frac{\\tan x}{x}=1$$ there is $\\delta>0$ such that, for $0<x<\\delta$, $1<\\frac{\\tan x}{x}<2$ and so, for $0<x<\\delta$, $$\\left\\lfloor\\frac{\\tan x}{x}\\right\\rfloor=1$$ which gives, together with the fact that $\\frac{\\tan x}{x}$ is even, $$\\lim_{x\\to 0}\\left\\lfloor\\frac{\\tan x}{x}\\right\\rfloor=1$$ In both cases you are actually doing the limit of a constant function (in a punctured neighborhood of $0$, for the second one). • this can't possibly be true, you can see that from using l'hopital's rule or expanding sinx (tanx ) by McLauren expansion, you see that sinx/x -> 1 – user247608 Jun 19 '15 at 20:49 • @david Yes, I know that $\\lim_{x\\to0}\\frac{\\sin x}{x}=1$, but this has little to do with the problem at hand. The “floor” function is not continuous at $1$, so you can't “put the limit inside”. – egreg Jun 19 '15 at 20:51 Intuitively: Note that on $\\lfloor \\sin(x)/x \\rfloor$, the graph approaches $1$ from below, while on $\\lfloor \\tan(x)/x \\rfloor$, the graph approaches $1$ from above. Therefore, while the graph of $\\sin(x)/x$ approaches 1, it never reaches there: therefore, for small values of $x$ around 0, the value of $\\lfloor \\sin(x)/x \\rfloor$ will be 0. However, even for small values of $x$ around 0, the value of", "^{3} x \\sin x$ $\\therefore f^{\\prime}(x)=0$ $\\Rightarrow 4 \\sin ^{3} x \\cos x-4 \\cos ^{3} x \\sin x=0$ $\\Rightarrow \\sin ^{3} x \\cos x-\\cos ^{3} x \\sin x=0$ $\\Rightarrow \\tan ^{3} x-\\tan x=0$ $\\Rightarrow \\tan x\\left(\\tan ^{2} x-1\\right)=0$ $\\Rightarrow \\tan x=0, \\tan ^{2} x=1$ $\\Rightarrow \\tan x=0, \\tan x=\\pm 1$ $\\Rightarrow x=0, x=\\frac{\\pi}{4}, \\frac{3 \\pi}{4}$ Thus, $c=\\frac{\\pi}{4} \\in\\left(0, \\frac{\\pi}{2}\\right)$ such that $f^{\\prime}(c)=0$. ​Hence, Rolle’s theorem is verified. (xviii) The given function is $f(x)=\\sin x-\\sin 2 x$. Since $\\sin x$ and $\\sin 2 x$ are everywhere continuous and differentiable, $f(x)$ is continuous on $[0, \\pi]$ and differentiable on $(0, \\pi)$. Also, $f(\\pi)=f(0)=0$ Thus, $f(x)$ satisfies all the conditions of Rolle’s theorem. Now, we have to show that there exists $c \\in(0, \\pi)$ such that $f^{\\prime}(c)=0$. We have $f(x)=\\sin x-\\sin 2 x$ $\\Rightarrow f^{\\prime}(x)=\\cos x-2 \\cos 2 x$ $\\therefore f^{\\prime}(x)=0$ $\\Rightarrow \\cos x-2 \\cos 2 x=0$ $\\Rightarrow \\cos x-2 \\cos ^{2} x+1=0$ $\\Rightarrow 2 \\cos ^{2} x-\\cos x-1=0$ $\\Rightarrow(\\cos x-1)(2 \\cos x+1)=0$ $\\Rightarrow \\cos x=1, \\cos x=\\frac{-1}{2}$ $\\Rightarrow \\cos x=\\cos \\frac{\\pi}{2}, \\cos x=\\frac{2 \\pi}{3}$ $\\Rightarrow x=\\frac{\\pi}{2}, \\frac{2 \\pi}{3}$ Thus, $c=\\frac{\\pi}{2}, \\frac{2 \\pi}{3} \\in(0, \\pi)$ such that $f^{\\prime}(c)=0$. Hence, Rolle’s theorem is verified.​", "# Math Help - Solving 13sinx-84sinx = 17 1. ## Solving 13sinx-84cosx = 17 Find all $x$ such that $13\\sin{x}-84\\cos{x} = 17$. EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17. 2. Originally Posted by Resilient Find all $x$ such that $13\\sin{x}-84\\cos{x} = 17$. EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17. $A\\sin{x} - B\\cos{x} = C$ $A\\sin{x} - B\\cos{x} = R\\sin(x - \\alpha)$ $R = \\sqrt{A^2+B^2}$ $\\alpha = \\arctan\\left(\\dfrac{B}{A}\\right) = \\arctan\\left(\\dfrac{84}{13}\\right)$ so, assuming $0 \\le x < 2\\pi$ ... $13\\sin{x} - 84\\cos{x} = 85\\sin(x - \\alpha)$ $85\\sin(x - \\alpha) = 17 $ $\\sin(x - \\alpha) = \\dfrac{1}{5}$ $x - \\alpha = \\arcsin\\left(\\dfrac{1}{5}\\right)$ $x - \\alpha = \\pi - \\arcsin\\left(\\dfrac{1}{5}\\right)$ 3. Originally Posted by skeeter ... Thanks. Can we find $\\tan^{-1}\\frac{84}{13}$ and $\\sin^{-1}\\frac{1}{5}$ in terms of $\\pi$? 4. Originally Posted by Resilient Thanks. Can we find $\\tan^{-1}\\frac{84}{13}$ and $\\sin^{-1}\\frac{1}{5}$ in terms of $\\pi$? would be really difficult ... use the calculator approximation. 5. @Resilient I had one just like this and I found a way to get the correct solution with pencil but... your's is a bit to nasty to get a clean radian solution as skeeter says. you can check my thread how I did almost the same problem. http://www.mathhelpforum.com/math-he...on-172605.html I got yours into quadratic form in sin function but", "# Solving 13sinx-84sinx = 17 • Feb 24th 2011, 02:19 PM Resilient Solving 13sinx-84cosx = 17 Find all $x$ such that $13\\sin{x}-84\\cos{x} = 17$. EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17. • Feb 24th 2011, 04:26 PM skeeter Quote: Originally Posted by Resilient Find all $x$ such that $13\\sin{x}-84\\cos{x} = 17$. EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17. $A\\sin{x} - B\\cos{x} = C$ $A\\sin{x} - B\\cos{x} = R\\sin(x - \\alpha)$ $R = \\sqrt{A^2+B^2}$ $\\alpha = \\arctan\\left(\\dfrac{B}{A}\\right) = \\arctan\\left(\\dfrac{84}{13}\\right)$ so, assuming $0 \\le x < 2\\pi$ ... $13\\sin{x} - 84\\cos{x} = 85\\sin(x - \\alpha)$ $85\\sin(x - \\alpha) = 17 $ $\\sin(x - \\alpha) = \\dfrac{1}{5}$ $x - \\alpha = \\arcsin\\left(\\dfrac{1}{5}\\right)$ $x - \\alpha = \\pi - \\arcsin\\left(\\dfrac{1}{5}\\right)$ • Feb 24th 2011, 05:14 PM Resilient Quote: Originally Posted by skeeter ... Thanks. Can we find $\\tan^{-1}\\frac{84}{13}$ and $\\sin^{-1}\\frac{1}{5}$ in terms of $\\pi$? • Feb 25th 2011, 12:35 PM skeeter Quote: Originally Posted by Resilient Thanks. Can we find $\\tan^{-1}\\frac{84}{13}$ and $\\sin^{-1}\\frac{1}{5}$ in terms of $\\pi$? would be really difficult ... use the calculator approximation. • Feb 25th 2011, 06:04 PM skoker @Resilient I had one just like this and I found a way to get the correct solution with pencil but... your's is a bit to nasty to get a", "# Finding smallest positive root $\\sqrt{\\sin(1-x)}=\\sqrt{\\cos x}$ Finding smallest positive root of the equation $\\sqrt{\\sin(1-x)}=\\sqrt{\\cos x}$ Try: $\\sin(1-x)=\\cos x= \\sin\\bigg(\\frac{\\pi}{2}-x\\bigg)$ $1-x=n\\pi+(-1)^n\\bigg(\\frac{\\pi}{2}-x\\bigg)$ Could some help me to solve it, Thanks If $n$ is even, then the equation becomes $$1-x=n\\pi +\\frac\\pi2 -x$$ which simplifies to $$1=n\\pi + \\frac\\pi2$$ which is clearly not true for any value of $n\\in\\mathbb N$. For odd values of $n$, the equation is $$1-x=n\\pi -\\frac\\pi2 + x$$ simplifying to $$1-n\\pi + \\frac\\pi2 = 2x$$ and is solvable for $x$. This gives you a set of candidate values for $x$ which you then still need to check, because if $\\sin(1-x)<0$, then $x$ cannot be a solution! • Neither can there be a solution if $\\cos x\\lt 0$ – Rohan Shinde Feb 14 '18 at 14:39 If you square both sides, and apply sum of angles formula, you get $$\\sin 1\\cos x - \\sin x \\cos 1 = \\cos x.$$ Divide by $\\cos x \\cos 1$ to get $$\\tan 1 -\\tan x = \\sec 1$$ so that $x = \\arctan(\\tan 1 - \\sec 1).$ This is not as satisfying as @5xum 's answer, but I thought it was interesting. • Just a small doubt. Shouldn't that be $\\sin 1\\cos x \\color{red}{-} \\sin x \\cos 1$ – Rohan Shinde Feb 14 '18 at 14:33 • @Manthanein Yep. Thanks. –", "ns was watching this video clip and at 5:30 $\\cos \\frac\\pi2$ is substituted with $0$. Utilizing my calculator the result I acquired was $-0.5$. I m sorry one is correct? $\\cos\\left(\\frac\\pi2\\right)=0$. The factor you are gaining $-0.5$ is since you are not putting brackets approximately $\\pi/2$. Thus, you room obtaining the value of $(\\cos\\pi)/2=(-1)/2=-0.5$. You need to use the brackets together follows: $\\cos\\left(\\frac\\pi2\\right)$. There room a pair ways to present that $\\cos (\\pi/2)=0$ 1) The value $\\tan(\\pi/2)$ is undefined. Because $\\tan(x)=\\frac\\sin(x)\\cos(x)$, and both $\\sin(x)$ and also $\\cos(x)$ are continuous throughout the reals, this suggests that it have the right to only occur if $\\cos (\\pi/2)=0$. You are watching: Cos (-pi/2) 2) The co-function identification $\\cos(\\pi/2-x)=\\sin(x)$ means that $\\cos (\\pi/2)=\\cos(\\pi/2-0)=\\sin(0)=0$ 3) Looking at the unit circle where $y=\\sin(\\theta)$ and also $x=\\cos(\\theta)$. When $\\theta= 90^0=\\pi/2$, $\\sin (\\theta)=1$ and also $\\cos(\\theta)=0$. 4) the derivative of $\\sin(x)$ is $0$ only as soon as $\\sin(x)=1$, which only occurs once $x= \\fracn\\pi2$ for any type of $n\\in \\sirhenryjones-museums.orgbb N$. Because the derivative the $\\sin(x)$ is $\\cos(x)$ this suggests that $\\cos(\\pi/2)=\\cos((1)\\pi/2)=0$. If girlfriend have any qustions let me recognize below. share point out follow reply Mar 3 \"17 at 23:21 Sentinel135Sentinel135 $\\endgroup$ Thanks for contributing response to sirhenryjones-museums.orgematics stack Exchange! Please be certain to answer the question. Provide details and share her research! But avoid", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "Browse Questions # Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $f (x) = \\sin x+\\cos x, x \\in [0,\\pi]$ This is second part of multipart Q5, which appeared in model paper 2012. Toolbox: • $\\large\\frac{d}{dx}$$(\\sin x)=\\cos x • \\large\\frac{d}{dx}$$(\\cos x)=-\\sin x$ Step 1: $f(x)=\\sin x+\\cos x$ Differentiate with respect to x, $f'(x)=\\cos x-\\sin x$ Given interval $[0,\\pi]$ For extreme values $f'(x)=0$ $\\cos x-\\sin x=0$ On simplifying we get $\\Rightarrow \\cos x\\big[1-\\large\\frac{\\sin x}{\\cos x}\\big]$$=0 But \\sin x/\\cos x =\\ tan x \\Rightarrow \\cos x[1-\\tan x]=0 1-\\tan x=0 \\tan x=1 x=\\tan^{-1}1 x=\\large\\frac{\\pi}{4} Step 2: Now we find f(x) at x=0,\\large\\frac{\\pi}{4}$$,\\pi$ $f(0)=\\sin 0+\\cos 0$ But $\\sin 0= 0$ and $\\cos 0 = 1$ $\\qquad=0+1$ $\\qquad=1$ Step 3: $f\\big(\\large\\frac{\\pi}{4}\\big)$$=\\sin\\large\\frac{\\pi}{4}+$$\\cos\\large\\frac{\\pi}{4}$ But $\\sin\\large\\frac{\\pi}{4}=\\cos\\large\\frac{\\pi}{4}= \\frac{1}{\\sqrt2}$ $\\qquad\\;\\;=\\large\\frac{1}{\\sqrt 2}+\\frac{1}{\\sqrt 2}$ $\\qquad\\;\\;=\\large\\frac{2}{\\sqrt 2}$ $\\qquad\\;\\;=\\sqrt 2$ Step 4: $f(\\pi)=\\sin \\pi+\\cos\\pi$ $\\qquad=0-1=-1$ Step 5: Absolute maximum value $=\\sqrt 2$ at $x=\\large\\frac{\\pi}{4}$ and absolute minimum value =-1 at $x=\\pi$", "# Show that $f(x)=\\tan^{-1}(\\sin x+\\cos x)$ is an increasing function in$\\bigg(0,\\Large \\frac{1}{4}\\normalsize \\bigg).$ Toolbox: • Let $f(x)$ be a function defined on $(a,b)$. If $f'(x) > 0$ for all $x \\in (a,b)$ except for a finite number of points, where $f'(x)>0,$ then $f(x)$ is increasing on $(a,b)$ Step 1 $f(x)= \\tan^{-1} ( \\sin x+ \\cos x)$ Differentiating w.r.t $x$ we get, $f'(x)=\\large\\frac{1}{1+ ( \\sin x+ \\cos x)^2} \\times ( \\cos x- \\sin x)$ $= \\large\\frac{\\cos x- \\sin x}{1+( \\sin x+ \\cos x)^2}= \\large\\frac{\\cos x(1- \\tan x)}{1+ ( \\sin x+ \\cos x)^2}$ Now let us consider the interval $0 < x < \\large\\frac{\\pi}{4}$ $\\Rightarrow \\cos x > 0, \\large\\frac{1}{1+(\\sin x+ \\cos x)^2}\\: and \\: (1- \\tan x) > 0$ $( \\because \\tan x < 1$ in the interval $0 < x < \\large\\frac{\\pi}{4})$ $\\Rightarrow \\large\\frac{\\cos x(1-\\tan x)}{1+ ( \\sin x+ \\cos x)^2} > 0$ $\\Rightarrow f'(x) > 0$ Hence $f'(x)>0$ for all $x \\in \\bigg( 0, \\large\\frac{\\pi}{4} \\bigg)$ Hence $f(x)$ is increasing on $\\bigg( 0, \\large\\frac{\\pi}{4} \\bigg)$ answered Aug 5, 2013", "I think, assuming that $$\\frac{-1}{\\phi} = \\frac{1-\\sqrt{5}}{2}$$ Which I seem to remember a similar property about phi... 12. Sep 13, 2010 Helios note that tan( x ) = cos( x ) = 1 / sqr( phi ) so the solution can be made in terms of arctan or arccos. 13. Sep 15, 2010 willem2 Since tan(x) increases monotonically from 0 to infinity in the domain (0, pi/1) and cos (x) decreases from 1 to 0 in the same domain, there should be a solution x between 0 and pi/2 and 0< sin(x) < 1. There's a solution in the second quadrant as well. cos(x) and tan(x) have different signs in the 3rd and fourth quadrant, so a solution where sin(x) < 0 is not possible Once you get all the signs right in the quadratic formula, you'll get $$sin(x) = \\frac {1} {\\phi} = \\phi - 1$$ from one of the solutions. The other solution has sin(x) >1 so that isn't a valid solution. }", "# Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $f (x) =\\sin x-\\cos x,0 < x <2\\pi$ $\\begin{array}{1 1} Maxima =\\sqrt 3\\; Minima = -\\sqrt 2 \\\\ Maxima =\\sqrt 2\\; Minima = -\\sqrt 2 \\\\ Maxima =\\sqrt 2 Minima = \\sqrt 2 \\\\ Maxima =\\sqrt 2 Minima = -\\sqrt 3 \\end{array}$ Toolbox: • $\\large\\frac{d}{dx}$$(\\sin x)=\\cos x • \\large\\frac{d}{dx}$$(\\cos x)=-\\sin x$ • For Maxima and Minima $f'(x)=0$ Step 1: $f(x)=\\sin x-\\cos x$ Differentiating with respect to x $f'(x)=\\cos x+\\sin x$ $\\qquad=\\cos x[1+\\large\\frac{\\sin x}{\\cos x}]$ $\\qquad=\\cos x[1+\\tan x]$ For Maxima and Minima $f'(x)=0$ $\\Rightarrow \\cos x+\\sin x=0$ $\\cos x[1+\\tan x]=0$ $1+\\tan x=0$ $\\tan x=-1$ $x=\\large\\frac{3\\pi}{4},\\frac{7\\pi}{4}$ Step 2: At $x=\\large\\frac{3\\pi}{4}$ When $x$ is slightly $<\\large\\frac{3\\pi}{4}$ $\\cos x=-ve$ $\\tan x=-(1+h)$ $\\qquad=-1-h$ $1+\\tan x=1+(-1-h)=-h$=-ve $\\cos x(1+\\tan x)=(-)(-)=+ve$ Step 3: when $x$ is slightly >$\\large\\frac{3\\pi}{4}$ $\\cos x=-ve$ $\\tan x=-(1-h)$ $\\qquad=-1+h$ $1+\\tan x=-1-1+h=+h$=+ve $f'(x)=\\cos x(1+\\tan x)$ $\\qquad=(-)(+)=-ve$ $f'(x)$ changes sign from +ve to -ve. Therefore At $x=\\large\\frac{3\\pi}{4}$,there is a point of local maxima. Step 4: Local maximum value=$f(\\large\\frac{3\\pi}{4})=$$\\sin\\large\\frac{3\\pi}{4}$$-\\cos\\large\\frac{3\\pi}{4}$ $\\Rightarrow \\large\\frac{1}{\\sqrt 2}+\\frac{1}{\\sqrt 2}$ $\\Rightarrow \\large\\frac{2}{\\sqrt 2}$ $\\Rightarrow \\sqrt 2$ Step 5: At $x=\\large\\frac{7\\pi}{4}$ Let $x$ is slightly < $\\large\\frac{7\\pi}{4}$ $\\cos x=+ve$ $\\tan x=-(1+h)$ $1+\\tan x=1-(1+h)=-h$=-ve $f'(x)=\\cos x(1+\\tan x)$ $\\qquad=(+)(-)=-ve$ When $x$ is slightly >$\\large\\frac{7\\pi}{4}$ $\\cos x=+ve$", "in the form $M \\cos 2 x+N \\sin 2 x$, where $M$ and $N$ are constants, giving the exact value of $M$ and the exact value of $N$ (b) Solve, for $0^{\\circ} \\leqslant x \\leqslant 180^{\\circ}$, the equation $\\cos 2 x-\\sin 2 x=1$ The maximum value of $\\cos 2 x-\\sin 2 x$ is $k$. (c) Find the exact value of $k$ (d) Find the smallest positive value of $x$ for which a maximum occurs. 5. $\\left(2013 / \\mathrm{jan} / \\mathrm{Paper01} / \\mathrm{q8}\\right)$ Solve, for $0 \\leqslant \\theta \\leqslant \\pi$, giving each solution to 3 significant figures, (a) $5 \\sin \\theta-1=0$ (3) (b) $\\tan \\left(2 \\theta+\\frac{\\pi}{3}\\right)=0.4$ (4) (c) $4 \\sin ^{2} \\theta-7 \\cos \\theta=2$ (4) 6. (2013/june/paper01/q4) Solve, for $-90<x \\leqslant 90$, the equation $$6 \\sin ^{2} x^{\\circ}-\\cos x^{\\circ}-4=0$$ 7. (2014/jan/paper01/q8) $$\\begin{gathered}\\sin (A+B)=\\sin A \\cos B+\\cos A \\sin B \\\\\\tan A=\\frac{\\sin A}{\\cos A}\\end{gathered}$$ (a) Show that the equation $$4 \\sin (x+\\alpha)=7 \\sin (x-\\alpha)$$ can be written in the form $$3 \\tan x=11 \\tan \\alpha$$ (b) Hence solve, to 1 decimal place, $$4 \\sin (3 y+45)^{\\circ}=7 \\sin (3 y-45)^{\\circ} \\text { for } 0 \\leqslant y \\leqslant 180$$ 8. (2014/june/paper01/q2) Solve, in degrees to 1 decimal place, for $0 \\leqslant \\theta<180^{\\circ}$ (a) $\\tan 2 \\theta=1.5$ (b) $(3 \\cos \\theta+1)(2 \\cos \\theta+3)=-2$ 9. (2014/june/paper02/q10) Using the identities $\\quad \\cos (A+B)=\\cos A", "week, 3 days ago Problem 6 An odd function satisfies $$f(-x)=-f(x)$$, so let's plug in $$-x$$ and simplify. Note that $$\\sin x$$, $$\\tan x$$ and $$x$$ are all odd functions. \\begin{align} f(-x) &= \\frac {2(-x)(\\sin(-x)+\\tan(-x))} {2\\left\\lfloor \\frac {-x+2\\pi}{\\pi} \\right\\rfloor -3} \\\\ &= \\frac {(-1)2x(-\\sin x -\\tan x)} {2\\left\\lfloor -\\frac {x}{\\pi}+2 \\right\\rfloor -3} \\\\ &= \\frac {2x(\\sin x +\\tan x)} {2\\left\\lfloor -\\frac {x}{\\pi}+2 \\right\\rfloor -3} \\\\ &= \\frac {2x(\\sin x +\\tan x)} {2\\left\\lfloor -\\frac {x}{\\pi}\\right\\rfloor +2-3} \\\\ &= \\frac {2x(\\sin x +\\tan x)} {2\\left\\lfloor -\\frac {x}{\\pi}\\right\\rfloor -1} \\\\ &{\\color{red}\\stackrel{?}{=}} {\\color{red}\\frac {2x(\\sin x +\\tan x)} {2\\left(\\left\\lfloor \\frac {x}{\\pi}\\right\\rfloor-1\\right)-1}} \\\\ &= \\frac {2x(\\sin x +\\tan x)} {2\\left\\lfloor \\frac {x}{\\pi}\\right\\rfloor-2-1} \\\\ &= \\frac {2x(\\sin x +\\tan x)} {2\\left\\lfloor \\frac {x}{\\pi}\\right\\rfloor-3} \\\\ &= -\\frac {2x(\\sin x +\\tan x)} {3-2\\left(\\left\\lfloor\\frac {x}{\\pi}+\\frac{2\\pi}{\\pi}-2\\right\\rfloor\\right)} \\\\ &= -\\frac {2x(\\sin x +\\tan x)} {3-2\\left(\\left\\lfloor\\frac {x+2\\pi}{\\pi}\\right\\rfloor\\right)+4} \\\\ &= -\\frac {2x(\\sin x +\\tan x)} {7-2\\left\\lfloor\\frac {x+2\\pi}{\\pi}\\right\\rfloor} \\\\ \\end{align} Here, I am kind of stuck, because it doesn't look like this function is really odd, maybe someone can go on or correct my steps. I'm not quite sure if the step marked with $${\\color{red}\\stackrel{?}{=}}$$ is valid. - 1 week, 4 days ago Your technique is good but I too think that there is a mistake. - 1 week, 3 days ago I'm still not sure about the marked step, but at least", "point. Step 1: $h(x)=\\sin x+\\cos x$ Differentiating with respect to x $h'(x)=\\cos x-\\sin x$ $\\qquad=\\cos x[1-\\large\\frac{\\sin x}{\\cos x}]$ $\\qquad=\\cos x[1-\\tan x]$ For maxima and minima $h'(x)=0$ $\\Rightarrow \\cos x-\\sin x=0$ (or)$\\tan x=1$ $x=\\tan^{-1}1$ $x=\\large\\frac{\\pi}{4}$ $x=\\large\\frac{\\pi}{4}$$\\in [0,\\large\\frac{\\pi}{2}] Step 2: At x=\\large\\frac{\\pi}{4} h'(x)=\\cos x[1-\\tan x] When x is slightly < \\large\\frac{\\pi}{4} \\cos x=+ve \\tan x=1-h Where h is small. 1-\\tan x=1-(1-h)=+ve h'(x)=\\cos x(1-\\tan x) \\qquad=(+)(+)=+ve Step 3: When x is slightly > \\large\\frac{\\pi}{4} \\cos x=+ve \\tan x=1+h Where h is small. 1-\\tan x=1-(1-h) \\qquad\\;\\;\\;\\;\\;\\;=-h=-ve Step 4: Hence there is a local maxima at x=\\large\\frac{\\pi}{4} Local maximum value =h\\big(\\large\\frac{\\pi}{4}\\big)$$=\\sin\\large\\frac{\\pi}{4}+$$\\cos \\large\\frac{\\pi}{4} \\qquad\\qquad\\qquad\\qquad\\qquad\\quad=\\large\\frac{1}{\\sqrt 2}+\\frac{1}{\\sqrt 2} \\qquad\\qquad\\qquad\\qquad\\qquad\\quad=\\large\\frac{2}{\\sqrt 2}$$=\\sqrt 2$", "to added integer multiples of $$2\\pi$$). Namely, $$X = \\frac{2\\pi}{3} + 2\\pi k,$$ for some integer $$k$$. Since $$\\sin$$ and $$\\cos$$ are $$2\\pi$$-periodic, we get $$\\sin(X) = \\sin\\left(\\frac{2\\pi}{3}\\right) = \\frac{\\sqrt 3}{2} \\qquad \\text{ and } \\qquad \\cos(X) = \\cos\\left(\\frac{2\\pi}{3}\\right) = -\\frac{1}{2}.$$ You have definitely seen trig identities such as $$\\sin ^2x +\\cos^2 x =1$$ And $$\\sec ^2x=1+\\tan ^2 x$$ Well you need to apply the identities to get the unknown parts from the given information. In this case, your professor is using $$A = \\theta$$ and $$B = X$$. You are given $$\\cos(A) = \\frac{3}{5}$$. From this, you know that for a right triangle, the adjacent side is 3 and the hypotenuse is 5. By the Pythagorean Theorem, this means that the opposite side is 4 since $$3^{2} + 4^{2} = 5^{2}$$. Also, since you are in the 4th quadrant, your opposite side must be negative ($$y$$ values are negative in the 4th quadrant). Thus, $$\\sin(A) = \\frac{-4}{5}$$. For $$\\tan(B) = -\\sqrt{3}$$, you know that $$\\tan(B) = \\frac{\\frac{\\sqrt{3}}{2}}{\\frac{-1}{2}} = \\frac{\\sin(B)}{\\cos(B)}$$ (you are in the second quadrant, so your cosine is negative and your sine is positive).", "# Maxima and minima of $f(x)=\\sin x+\\cos x+\\tan x+\\arcsin x+\\arccos x+\\arctan x$ Let $f(x)=\\sin x+\\cos x+\\tan x+\\arcsin x+\\arccos x+\\arctan x$ . If $M$ and $m$ are maximum and minimum values of $f(x)$ then their arithmetic mean is equal to? My Approach: $\\arcsin x + \\arccos x =\\frac{\\pi}{2}$, so the equation becomes $\\sin x + \\cos x + \\tan x + \\frac{\\pi}{2} + \\arctan x$ . After this I differentiated this equation but did not came to any conclusion . Also I tried to solve the equation to make something relevant but the equation did not simplified . Please help... • Maximum and minimum over what set? All of the functions listed are bounded except for $\\tan x$, which is unbounded over $(-\\pi/2,\\pi/2)$. – Joey Zou Sep 1 '16 at 18:07 • @JoeyZou Obviously, on a subset of $[-1,1]$. – Stop hurting Monica Sep 1 '16 at 18:08 • Since there is a inverse function so domain can be [-1,1]. – saladi Sep 1 '16 at 18:14 Hint: We have $f(x) = \\sin x + \\cos x + \\tan x + \\frac{\\pi}{2} + \\arctan x$ on $[-1,1]$. Now $$f'(x) = \\cos x - \\sin x + \\sec^2 x + \\frac{1}{1+x^2}.$$ On $[-1,1]$, we have $\\cos x > 0$, $-\\sin x > -1$, $\\sec^2x\\ge 1$, and $\\frac{1}{1+x^2}>0$, so for all", "The second formula is a fancy way to write $$\\lfloor x \\rfloor = x - (x),$$ where $(x)$ is the fractional part. To find the fractional part, we take some monotone increasing periodic function, apply it and \"reverse\" it, to find the location within the period. If $f$ is a monotone increasing periodic function such that $f^{-1}$ maps to the period $(a,b)$, then $$(x) = \\frac{f^{-1}(f(a + (b-a)x))-a}{b-a}.$$ For example, if $f = \\tan$ then $[a,b] = [-\\pi/2,\\pi/2]$ (given the usual choice of period for $\\arctan$), and so $$(x) = \\frac{\\arctan(\\tan(\\pi(x-1/2))+\\pi/2}{\\pi} = \\frac{1}{\\pi} \\arctan(\\tan(\\pi(x-1/2)) + \\frac{1}{2}.$$ But why would you define $\\lfloor x \\rfloor$ this way? Both \"definitions\" have the drawback that they don't work at the discontinuities (note that a monotone periodic function must have discontinuities for obvious reasons). More importantly, they are far more obscure than the usual, textual definition, which is also easier to work with. Gone are the days when \"function\" implicitly meant \"formula\". -", "- 4\\tan^3\\!x + 6\\tan^2\\!x - 4\\tan x + 1 \\;=\\;\\tan^4\\!x + \\tan^2\\!x$ And we have: . . . . $4\\tan^3\\!x - 5\\tan^2\\!x + 4\\tan x - 1 \\:=\\;0$ . . This cubic equation has no rational roots. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ I realize that the roots need not be rational. For example, if $x = \\tfrac{\\pi}{6}\\;(30^o)$, then $\\tan\\tfrac{\\pi}{6} = \\tfrac{1}{\\sqrt{3}}$ could be a root . . . (It isn't.) And if $x = \\tfrac{\\pi}{12}\\;(15^o)$, then $\\tan\\tfrac{\\pi}{12} = 2-\\sqrt{3}$ could be a root . . . (It isn't.) At this point, I surrendered . . . 3. Hello, Let's multiply by $\\cos(x)$. This gives : $\\sin^2(x)+\\sin(x)-2\\sin(x)\\cos(x)+\\cos^2(x)=0$ but we know that $\\cos^2(x)+\\sin^2(x)=1$ and $\\sin(2x)=2\\sin(x)\\cos(x)$ so the equation is now $\\sin(x)-\\sin(2x)+1=0$ Perhaps it has a better form this way... 4. For what it is worth, pursuing Moo's initial stages, by iteration the solution is 1.1131Pi or 3.49683 radians. There appears to be only the one real solution.", "$[-1,1]$, because of $\\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\\tan\\arcsin x$. (Note: your $x\\in[-\\pi/4,\\pi/4]$ is wrong and the probable cause for the low grade.) The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\\in(0,1)$. For every root we find, also its negative will be a root. If $y=\\sin\\arctan x$, then $\\arctan x=\\arcsin y$ and so $x=\\tan\\arcsin y$ so $$x^2=\\frac{\\sin^2\\arcsin y}{1-\\sin^2\\arcsin y}=\\frac{y^2}{1-y^2}$$ whence $$y^2=\\frac{x^2}{1+x^2}$$ If $z=\\tan\\arcsin x$, then $$z^2=\\frac{\\sin^2\\arcsin x}{1-\\sin^2\\arcsin x}=\\frac{x^2}{1-x^2}$$ So your equation $\\frac{1}{y^2}-\\frac{1}{z^2}=4x^2$ becomes $$\\frac{1+x^2}{x^2}-\\frac{1-x^2}{x^2}=4x^2$$ that is, $$2x^2=4x^4$$ and, since $x\\ne0$, $x^2=\\frac{1}{2}$. The only positive root is $$x=\\frac{1}{\\sqrt{2}}$$ which indeed belongs to $(0,1)$. Thus the solutions are $$\\frac{1}{\\sqrt{2}}\\qquad\\text{and}\\qquad{-}\\frac{1}{\\sqrt{2}}$$ • Yes, ofcourse! arctan is undefined at $-\\pi/2$, $0$ and $\\pi/2$. So the domain is $(-1,0)\\cup(0,1)$, if I'm not mistaken? Which in turn means that the solutions to the equation $x\\in[\\pi/4,\\pi/4] / {0}$? Sep 18, 2017 at 20:53 • @Parseval No: $x\\in(-1,0)\\cup(0,1)$. Why should you take the arctangent? Sep 18, 2017 at 20:54 • Because it's 22:57 and I've been staring at math for 18 hours. This is the kind of blind mistakes you do in such a condition. Thank you very much for clearing up everything, I really appreciate it mate! Sep 18, 2017 at 20:57 Taking derivative of" ]

[ "period for Y is (n+1 year)", "Period of sin^6 3x+ cos^6 3x", "− cos2 A. ## What is the period of y sin 3x? 1 Answer. Nghi N. So, the period of sin 3x is 2π3.", "# What is the period of sin (pix)? May 5, 2016 2. #### Explanation: In general, the period of both sin kx and cos kx is $\\frac{2 \\pi}{k}$. Here, k = $\\pi$.", "# What is the period of y = –5 sin (5x)? Period: $\\frac{2 \\pi}{5}$ Period of sin x is $2 \\pi .$ Period of sin 5x is $\\frac{2 \\pi}{5}$", "# What is the period of y=cos(6x)? $\\frac{\\pi}{3}$ Period of y --> $\\frac{2 \\pi}{6} = \\frac{\\pi}{3}$", "# What is the period of y=sin4theta? $\\frac{\\pi}{2}$ The period of y = sin 4t --> $\\frac{2 \\pi}{4} = \\frac{\\pi}{2}$", "Period of sin^6 3x+ cos^6 3x Period of sin^6 3x+ cos^6 3x", "cos^6 3x Period of sin^6 3x+ cos^6 3x", "# What is the period of y = 5 sin(4x−2)−3? $y = 5 \\sin \\left(4 x - 2\\right) - 3$ Period: $\\frac{2 \\pi}{4} = \\frac{\\pi}{2}$", "i t e}{\\text{XXXXXXX}}$ That is, it has a period of $\\pi$", "I am sorry, I am not a native speaker. I mean period just as $2\\pi$ for $\\sin(x)$. – Fejwin Aug 20 '12 at 23:12 There are many cases where the period is $y$ (and, as Andre points out, many cases where it is not). If $y$ is squarefree (that is, if there is no integer $d\\gt1$ such that $y$ is a multiple of $d^2$), then $x^n$ is zero if and only if $x$ is 0 (modulo $y$), so the period has to be $y$. The period need not be $y$. For example, let $y=9$, and consider the function $x^3$. This has period $3$. One can build many similar examples.", "# How do you determine the period for y=2 sin x? Period of sin x is $2 \\pi$. Hence the period of 2 sin x would also be the same.", "## anonymous one year ago What is the period of y = cot x? A. 2pi B. pi/2 C. pi D.1/pi 1. ganeshie8 |dw:1433391386504:dw| 2. ganeshie8 Notice that the pattern repeats every $$\\pi$$ units, so the period of cotx is $$\\pi$$", "How do you find the period of y = cot(2x/3)? Jul 30, 2016 (3pi)/2 Explanation: Period of cot y --> $\\pi$ Period of $\\cot \\left(\\frac{2 \\pi}{3}\\right)$ --> $\\frac{3 \\left(\\pi\\right)}{2}$", "# How do you find the period of y=sin(x/3)? ##### 1 Answer Nov 17, 2017 We have that the period of $\\sin \\left(x\\right)$ is $2 \\pi$. Thus, with a horizontal stretch of a factor of $3$, as given in the above equation, the period will become $2 \\pi \\cdot 3 = 6 \\pi$. Graphical confirmation: graph{sin(x/3) [-10, 10, -5, 5]} Hopefully this helps!", "## Calculus with Applications (10th Edition) The period of both $\\sin x$ and $\\cos x$ is $2\\pi$. So, it is not right to say $f(x)=\\cos x$ is periodic with a period of $\\pi$", "Precalculus (10th Edition) We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the funcion. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$ We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the function. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$", "you understand how the period of $\\sin nx$ is related to that of $\\sin x$, you should be able to see how the period of $[nx]$ is related to that of $[x]$. – Brian M. Scott Jun 18 '12 at 8:05", "# What is the period of y=sin(2/3x)? $3 \\pi$ Period of $\\sin \\left(\\frac{2 x}{3}\\right)$ --> $\\frac{\\left(3\\right) \\left(2 \\pi\\right)}{2} = 3 \\pi$" ]

[ "So for $x$ in the interval $[-\\pi/2, \\pi/2]$, $\\arcsin(\\sin(x)) = x$ satisfies that definition. For $x$ in the interval $[\\pi/2, 3 \\pi/2]$, $\\arcsin(\\sin(x))$ can't be $x$, but it can be $\\pi - x$ which is in the interval $[-\\pi/2, \\pi/2]$ (notice that $\\sin(\\pi - x) = \\sin(x)$). So that takes care of this part of the graph: Now notice that $\\sin(x)$ is periodic with period $2\\pi$, so the graph looks the same if you shift it left or right by $2\\pi$. That takes care of the rest of it. - So I understand the jist of it; but, I don't get why $sin(π−x)=sin(x)$. Did you mean that this only works for the interval $[−π/2,π/2]$? – evamvid Feb 19 at 3:19 To understand why sin(π−x)=sin(x), we need to start from the extended definition of sine for angles greater than π/2. sin(x) is defined as y-ordinate to the radius of the circle in question. That circle is centered at O(0, 0) and radius = $sqrt(x^2 + y^2)$. – Mick Feb 19 at 4:43 So let $f(x)=\\sin(x)$ and let $g(x)=\\arcsin(x)$. Let $\\operatorname{Dom}(f)$ be the domain of $f$ and let $\\operatorname{Rng}(f)$ be the range of $f$. Thus we have \\begin{align} \\operatorname{Dom}(f)&=\\mathbb{R}\\\\ \\operatorname{Rng}(f)&=[-1,1]\\\\ \\operatorname{Dom}(g)&=[-1,1]\\\\ \\operatorname{Rng}(g)&=\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right] \\end{align} If we restrict the $\\operatorname{Dom}(f)$ to $\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$, then $f$ and $g$ are inverses of one another and", "calculate that part and end up with $35.75$. We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$. The answer is then $\\boxed{(D) 36.8}$. ## Solution 13 1、First, move terms to get $1+5cos3x=3sinx$. After graphing, we find that there are $\\boxed{6}$ solutions (two in each period of $5cos3x$). -dstanz5 2、We can graph two functions in this case: $5\\cos{3x}$ and $3\\sin{x} -1$.$$\\newline$$Using transformation of functions, we know that $5\\cos{3x}$ is just a cos function with amplitude 5 and period $\\frac{2\\pi}{3}$. Similarly, $3\\sin{x} -1$ is just a sin function with amplitude 3 and shifted 1 unit downwards. So:$[asy] import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,\"3\\sin{x} -1 \"); draw(graph(Cos,0, 2pi),blue,\"5\\cos{3x}\"); xaxis(\"x\",BottomTop,LeftTicks); yaxis(\"y\",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); [/asy]$We have $\\boxed{(A) 6}$ ## Solution 14 1、This question is just about pythagorean theorem$$a^2+(a+2)^2-b^2 = (a+4)^2$$$$2a^2+4a+4-b^2 = a^2+8a+16$$$$a^2-4a+4-b^2 = 16$$$$(a-2+b)(a-2-b) = 16$$$$a=3, b=7$$With these calculation, we find out answer to be $\\boxed{\\textbf{(A) }24\\sqrt5}$ ~Lopkiloinm 2、Let $\\overline{AD}$ be $b$$\\overline{CD}$ be $a$$\\overline{MD}$ be $x$$\\overline{MC}$$\\overline{MA}$$\\overline{MB}$ be $t$$t-2$$t+2$ respectively. We have three equations:$$a^2 + x^2 = t^2$$$$a^2 + b^2 + x^2 = t^2 + 4t + 4$$$$b^2 + x^2 = t^2 - 4t + 4$$ Subbing in the first and third equation into the second equation, we get:$$t^2", "addition formulas). • Using that $$\\cos$$ is even and that $$\\sin$$ is odd, the addition formulas give the subtraction formulas: \\begin{align*} \\cos(x-y)=&\\cos(x)\\cos(y)+\\sin(x)\\sin(y) \\\\ \\sin(x-y)=&\\sin(x)\\cos(y)-\\cos(x)\\sin(y). \\end{align*} • Dividing the formulas we get: $\\tan(x\\pm y)=\\frac{\\sin(x\\pm y)}{\\cos(x\\pm y)}=\\frac{\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y)}=\\frac{\\tan(x)\\pm\\tan(y)}{1\\mp\\tan(x)\\tan(y)}.$ • Putting $$x=y$$ in the addition formulas, we get the double-angle formulas: \\begin{align*} \\cos(2x)=&(\\cos x)^2-(\\sin x)^2 \\\\ \\sin(2x)=&2\\sin(x)\\cos(x) \\end{align*} • Using the identity $$(\\cos x)^2+(\\sin x)^2=1$$ in the double-angle formula for $$\\cos$$, we get the following alternative forms. \\begin{align*} \\cos(2x)=&(\\cos x)^2-(1-(\\cos x)^2)=2(\\cos x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2 \\end{align*} # Trigonometric identities -- from the addition formulas 2 • Once again, the addition and subtraction formulas are \\begin{align*} \\cos(x\\pm y)=&\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y) \\\\ \\sin(x\\pm y)=&\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y), \\end{align*} and the alternative double-angle formulas for $$\\cos$$ are \\begin{align*} \\cos(2x)=&\\cos(x)^2-(1-\\cos(x)^2)=2\\cos(x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2. \\end{align*} • Rearranging the latter, we get the following half-angle formulas. \\begin{align*} (\\cos x)^2=&\\frac{\\cos(2x)+1}{2} \\\\ (\\sin x)^2=&\\frac{1-\\cos(2x)}{2} \\end{align*} • Finally, rearranging the addition formulas, we get the following product formulas. \\begin{align*} \\cos(x)\\cos(y)=&\\frac{\\cos(x+y)+\\cos(x-y)}{2} \\\\ \\sin(x)\\cos(y)=&\\frac{\\sin(x+y)+\\sin(x-y)}{2} \\\\ \\sin(x)\\sin(y)=&\\frac{\\cos(x-y)-\\cos(x+y)}{2}. \\end{align*} • The half-angle and product formulas will be used frequently in integral calculus, since they are linearization formulas. • Exercises: D.45, 47, 67 # Exponentiation by rational numbers • An exponential function is a function of the form $$f(x)=b^x$$ for some fixed positive number $$b>0$$. Let's recall what the expression $$b^x$$ means. • If $$n$$ is a", "# Graphical integration of $$\\sin^2(x)$$ and $$\\cos^2(x)$$ ## Visualisation; beats using double angle formulas! Here's a nice way of visualising the integration of $$\\sin^2(x)$$ and $$\\cos^2(x)$$ between $$\\pi$$ and $$2\\pi$$. Below the graphic is the algebraic version, which involves the use of double angle formulas. I don't like to condone \"tricks\" over understanding, but this is quite a nice demonstration I saw not long ago for speed of remembering! ## ...and now the slightly more gritty written solution First use the double angle formula for $$\\cos(2x)$$ to express $$\\sin^2(x)$$ differently: \\begin{align} \\cos(2x)&=\\cos^2(x)-\\sin^2(x) \\\\[8pt] &=(1-\\sin^2(x))-\\sin^2(x) \\\\[8pt] &=1-2\\sin^2(x) \\\\[8pt] 2\\sin^2(x)&=1-\\cos(2x) \\\\[8pt] \\sin^2(x)&=\\tfrac{1}{2}-\\tfrac{1}{2}\\cos(2x) \\end{align} Then integrate using the substitution from above: \\begin{align} &\\int^{2\\pi}_0 \\sin^2(x)\\,dx \\\\[12pt] &= \\int^{2\\pi}_0 \\tfrac{1}{2}-\\tfrac{1}{2}\\cos(2x) \\,dx \\\\[12pt] &= \\left[\\tfrac{1}{2}x-\\tfrac{1}{4}\\sin(2x)\\right]^{2\\pi}_0 \\\\[12pt] &= \\left(\\tfrac{1}{2}(2\\pi)-\\tfrac{1}{4}\\sin(4\\pi)\\right)-\\left(\\tfrac{1}{2}(0)-\\tfrac{1}{4}\\sin(0)\\right) \\\\[12pt] &= \\pi \\end{align} Intergrating $$\\cos^2(x)$$ is equivalent, but you will need to substitute out the $$\\sin^2(x)$$ term instead on the second line. In case you didn't realise, this uses $$\\sin^2(x)+\\cos^2(x)=1$$.", "Review question # Can we sketch $y=3 \\cos x - \\cos 3x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5251 ## Solution Prove that $\\begin{equation*} \\frac{d}{dx} (3 \\cos x - \\cos 3x)= 6 \\sin x \\cos 2x. \\end{equation*}$ Using the chain rule or by thinking about horizontal scaling, we have that $\\begin{equation*} \\frac{d}{dx} \\cos 3x = - 3 \\sin 3x. \\end{equation*}$ By the compound angle formula, \\begin{align*} \\sin 3x = \\sin(x + 2x) &= \\sin x \\cos 2x + \\sin 2x \\cos x \\\\ &= \\sin x \\cos 2x + 2 \\sin x \\cos^2 x \\\\ &= \\sin x \\cos 2x + \\sin x (1 + \\cos 2x) \\\\ &= 2 \\sin x \\cos 2x + \\sin x \\end{align*} so that \\begin{align*} \\frac{d}{dx} \\left( 3 \\cos x - \\cos 3x \\right) &= - 3 \\sin x + 3 \\sin 3x \\\\ &= - 3 \\sin x + 6 \\sin x \\cos 2x + 3 \\sin x \\\\ &= 6 \\sin x \\cos 2x. \\end{align*} Alternatively, we can use the standard formula, $\\sin P-\\sin Q = 2\\cos\\left(\\dfrac{P+Q}{2}\\right)\\sin\\left(\\dfrac{P-Q}{2}\\right)$. For what two ranges of values of $x$ between $0$ and $\\pi$ is $\\begin{equation*} 3 \\cos x - \\cos 3x \\end{equation*}$ an increasing function of $x$? A", "# Question #114e8 Feb 7, 2017 #### Explanation: (f+g)((x) is performing the algebraic addition of f and g, for $x \\in \\left[0 , 2 \\pi\\right]$. graph{2sinx+sin(2x) [0, 6.2832, -5, 5]} graph{(y-2sinx)(y-sin(2x))(2sinx+sin(2x)-y)=0 [0, 6.2832, -4, 4]} Demonstrative addition is depicted by the line x = 1, for x = 1 radian See the last local graph, for clarity, in this addition. $\\left(f + g\\right) \\left(1\\right) = f \\left(1\\right) + g \\left(1\\right) = 2 \\sin \\left({57.3}^{o}\\right) + \\sin \\left({114.8}^{o}\\right) = 1.7 + 0.9 = 2.6 ,$ nearly graph{(y-2sinx)(y-sin(2x))(2sinx+sin(2x)-y)=0 [0.95 1.05, 0, 3]}", "curve backward, then the equation would be \\begin{align*}y=\\cos \\left(x+\\frac{5 \\pi}{4}\\right)+3\\end{align*}. ### Review For questions 1-4, match the equation with its graph. 1. \\begin{align*}y=\\sin \\left(x-\\frac{\\pi}{2}\\right)\\end{align*} 2. \\begin{align*}y=\\cos \\left(x-\\frac{\\pi}{4}\\right)+3\\end{align*} 3. \\begin{align*}y=\\cos \\left(x+\\frac{\\pi}{4}\\right)-2\\end{align*} 4. \\begin{align*}y=\\sin \\left(x-\\frac{\\pi}{4}\\right)+2\\end{align*} Which graph above also represents these equations? 1. \\begin{align*}y=\\cos(x-\\pi)\\end{align*} 2. \\begin{align*}y=\\sin \\left(x+\\frac{3 \\pi}{4}\\right)-2\\end{align*} 3. Write another sine equation for graph A. 4. Writing How many sine (or cosine) equations can be generated for one curve? Why? 5. Fill in the blanks below. 1. \\begin{align*}\\sin x=\\cos(x-\\underline{\\;\\;\\;\\;\\;})\\end{align*} 2. \\begin{align*}\\cos x=\\sin(x-\\underline{\\;\\;\\;\\;\\;})\\end{align*} For questions 10-15, graph the following equations from \\begin{align*}[-2\\pi, 2\\pi]\\end{align*}. 1. \\begin{align*}y=\\sin \\left(x+\\frac{\\pi}{4}\\right)\\end{align*} 2. \\begin{align*}y=1+\\cos x\\end{align*} 3. \\begin{align*}y=\\cos(x+\\pi)-2\\end{align*} 4. \\begin{align*}y=\\sin(x+3)-4\\end{align*} 5. \\begin{align*}y=\\sin \\left(x-\\frac{\\pi}{6}\\right)\\end{align*} 6. \\begin{align*}y=\\cos(x-1)-3\\end{align*} 7. Critical Thinking Is there a difference between \\begin{align*}y=\\sin x +1\\end{align*} and \\begin{align*}y=\\sin(x+1)\\end{align*}? Explain your answer. ### Answers for Review Problems To see the Review answers, open this PDF file and look for section 14.2. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes", "{x+\\frac{\\pi }{{24}}} \\right)$$. To get $$b$$, we first find the period of the graph by seeing how long it goes before repeating itself (we can subtract the two $$x$$ values to get this new period). We see that the graph repeats itself from when $$\\displaystyle x=-\\frac{\\pi }{{24}}$$to when $$\\displaystyle x=\\frac{\\pi }{8}$$, so the new period is $$\\displaystyle \\frac{\\pi }{8}-\\left( {-\\frac{\\pi }{{24}}} \\right)=\\frac{{3\\pi }}{{24}}+\\frac{\\pi }{{24}}=\\frac{\\pi }{6}$$. Then we use the equation $$\\displaystyle b=\\frac{\\pi }{{\\text{new period}}}=\\frac{\\pi }{{\\left( {\\frac{\\pi }{6}} \\right)}}=\\frac{\\pi }{1}\\times \\frac{6}{\\pi }=6$$. So now we have $$\\displaystyle y=-2\\tan 6\\left( {x+\\frac{\\pi }{{24}}} \\right)$$. Try these on the graphing calculator! # Transformation of all Trig Functions without T-Charts Let’s go over once again how to transform trig functions without t-charts. Note that in order to perform the transformations accurately and quickly, you must know your 6 trig functions graphs inside out! Here are the steps for sin and cos graphs: 1. Put the trig function in the $$y=a\\sin b\\left( {x-c} \\right)+d\\,\\,\\,\\text{or}\\,\\,\\,y=a\\cos b\\left( {x-c} \\right)+d$$ format. 2. The new domain will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 3. The new range will be $$\\left[ {\\text{vertical shift (}d\\text{)}-\\left| a \\right|\\,\\text{,}\\,\\,\\text{vertical shift (}d\\text{)}+\\left| a \\right|} \\right]$$. ($$\\left| a \\right|$$ is amplitude) 4. New period will be $$\\displaystyle \\frac{{2\\pi }}{b}$$. 5. Phase shift will be $$c$$; start graph at this point, and if $$a$$", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> 2.13: Order of Real Numbers Difficulty Level: Basic Created by: CK-12 Estimated9 minsto complete % Progress Practice Order of Real Numbers Progress Estimated9 minsto complete % Suppose that you and three friends were playing a game where you each drew a number from a hat and the person with the highest number won. Let's say that you drew the number 32\\begin{align*} \\frac{3}{2}\\end{align*}, while your friends drew the numbers 3\\begin{align*}\\sqrt{3}\\end{align*}, 1.7, and π3\\begin{align*} \\frac{\\pi}{3}\\end{align*}, respectively. Could you figure out who won the game? Ordering Real Numbers Classifying Real Numbers The chart above shows the relationships between different real numbers. If a number is in one category, it is also in all the categories connected above it. Let's use the chart to categorize the following numbers: 1. 0 Zero is a whole number, an integer, a rational number, and a real number. 1. –1 –1 is an integer, a rational number, and a real number. 1. π3\\begin{align*}\\frac{\\pi}{3}\\end{align*} π3\\begin{align*}\\frac{\\pi}{3}\\end{align*} is an irrational number and a real number. 1. 369\\begin{align*}\\frac{\\sqrt{36}}{9}\\end{align*} 369=69=23\\begin{align*}\\frac{\\sqrt{36}}{9} = \\frac{6}{9} = \\frac{2}{3}\\end{align*}. This is a rational number and a real number. Graphing and Ordering Real Numbers Every real number can be positioned between two integers. Many times you will need to organize real numbers to determine the least value, greatest value, or", "this expansion from $$\\cos^6x$$ in the first place. 11. anonymous Actually, all the terms are easy with that trig identity... 12. anonymous The way I understand your question is that you want to compute the integral without having to expand anything of the form $$(a+b)^n$$ for $$n\\ge2$$. Am I right? 13. dinamix yup 14. dinamix this what i mean dude 15. dinamix i think no away without use binomial theorem 16. anonymous Right, so the expansion above doesn't abide by your rules. Continuing where I left off: \\begin{align*} \\cos^6x+\\sin^6x&=\\cos^4x-\\cos^2x\\sin^2x+\\sin^4x\\\\[2ex] &=\\cos^4x-\\frac{4}{4}\\cos^2x\\sin^2x+\\sin^4x\\\\[2ex] &=\\cos^4x-\\left(\\frac{1}{2}\\sin x\\cos x\\right)^2+\\sin^4x\\\\[2ex] &=\\cos^4x-\\left(\\frac{1}{4}\\sin 2x\\right)^2+\\sin^4x\\\\[2ex] &=\\cos^4x-\\frac{1}{16}\\sin^2 2x+\\sin^4x \\end{align*} The middle term can be handled with another half-angle identity, $$\\sin^2x=\\dfrac{1-\\cos2x}{2}$$. So we have \\begin{align*}2\\int_0^{\\pi/2}\\cos^6x\\,dx&=\\int_0^{\\pi/2}(\\cos^6x+\\sin^6x)\\,dx\\\\[2ex] &=\\int_0^{\\pi/2}\\left(\\cos^4x-\\frac{1}{32}(1-\\cos4x)+\\sin^4x\\right)\\,dx \\end{align*} Using the same reasoning as before, you have \\begin{align*}\\int_0^{\\pi/2}\\cos^4x\\,dx&=\\int_0^{\\pi/2}\\sin^4x\\,dx\\\\[2ex] 0&=\\int_0^{\\pi/2}(\\cos^4x-\\sin^4x)\\,dx\\\\[2ex] &=\\int_0^{\\pi/2}(\\cos^2x-\\sin^2x)(\\cos^2x+\\sin^2x)\\,dx\\\\[2ex] &=\\int_0^{\\pi/2}\\cos2x\\,dx\\end{align*} leaving you with $\\int_0^{\\pi/2}\\cos^6x\\,dx=\\frac{1}{64}\\int_0^{\\pi/2}(\\cos4x-1)\\,dx$ 17. anonymous No wait, there's a mistake somewhere up there... 18. anonymous Still looking for the error, but I hope you see the general idea? 19. freckles did you change 4/4 to 1/4? 20. freckles \\begin{align*} \\cos^6x+\\sin^6x&=\\cos^4x-\\cos^2x\\sin^2x+\\sin^4x\\\\[2ex] &=\\cos^4x-\\color{red}{\\frac{4}{4}}\\cos^2x\\sin^2x+\\sin^4x\\\\[2ex] &=\\cos^4x-\\left(\\color{red}{\\frac{1}{2}}\\sin x\\cos x\\right)^\\color{red}{2}+\\sin^4x\\\\[2ex] &=\\cos^4x-\\left(\\frac{1}{4}\\sin 2x\\right)^2+\\sin^4x\\\\[2ex] &=\\cos^4x-\\frac{1}{16}\\sin^2 2x+\\sin^4x \\end{align*} 21. anonymous There it is! 22. anonymous Oh actually a much bigger mistake: the integral of $$\\cos^4x+\\sin^4x$$ is not zero. 23. anonymous So in fact, we're back to this stage: \\begin{align*}2\\int_0^{\\pi/2}\\cos^6x\\,dx&=\\int_0^{\\pi/2}(\\cos^6x+\\sin^6x)\\,dx\\\\[2ex] &=\\int_0^{\\pi/2}\\left(\\cos^4x-\\frac{1}{32}(1-\\cos4x)+\\sin^4x\\right)\\,dx\\\\[2ex] &=\\int_0^{\\pi/2}\\left(2\\cos^4x-\\frac{1}{32}(1-\\cos4x)\\right)\\,dx\\end{align*} ...which WA is also telling me is not", "# A Trigonometric Inequality from the RMM ### Proof Due to the addition formula for sine, $\\sin (x+y)=\\sin x\\cos y+\\cos x\\sin y,\\;$ to prove the required inequality suffice it to establish that, for $\\displaystyle z\\in\\left(0,\\frac{\\pi}{2}\\right),\\;$ $\\displaystyle\\cos z\\lt\\left(\\frac{\\sin z}{z}\\right)^3.$ This is the same as $f(z)=\\sin^2z\\tan z - z^3 \\gt 0.\\;$ Note that $f(0)=0.$ We shall differentiate repeatedly. \\displaystyle\\begin{align} f'(z) &= \\sin^2z\\,\\sec^2z+\\tan z(2\\sin z\\cos z)-3z^2\\\\ &=\\tan^2 z+2\\sin^2 z-3z^2. \\end{align} Introduce $g(z)=f'(z)\\;$ and note that $g(0)=0.$ $\\displaystyle g'(z)= 2\\tan z\\,\\sec^2z+4\\sin z\\cos z-6z.$ Introduce $h(z)=\\frac{1}{2}g'(z)\\;$ and note that $h(0)=0.$ \\displaystyle\\begin{align} h'(z) &= \\left(\\sec^2z\\right)^2+(\\tan z)(2\\sec z)(\\sec z\\,\\tan z)+2(\\cos^2z-\\sin^2z)-3\\\\ &=\\left(1+\\tan^2z\\right)^2+2\\tan^2z(1+\\tan^2z)+2(2\\cos^2z-1)-3\\\\ &=(1+t^2)^2+2t(1+t^2)+\\frac{4}{1+t}-5\\\\ &=\\frac{3t^3+7t^2}{1+t}, \\end{align} $t=\\tan z\\;\\gt 0$ and so $h'(z)\\gt 0.\\;$ Hence, $h(z)\\gt h(0)=0,\\;$ so that $g'(z)\\gt 0,\\;$ and $g(z)\\gt g(0)=0,\\;$ meaning $f'(z)\\gt 0\\;$ and $f(z)\\gt f(0)=0.$ This completes the proof. ### Acknowledgment Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. The proof is by Soumava Chakraborty.", "obvious that: \\begin{align} P(X\\le1)&=1 \\\\ P(X\\le0)&=0.5 \\\\ P(X\\le-1)&=0 \\\\ \\end{align} It's also clear that the CDF will be the same for both sin(Y) and cos(Y) since it does not depend on the phase, only on the range of Y. I have no idea how to calculate, for example, P(X ≤ 0.5). I feel like I'm missing something really obvious. 10. Mar 26, 2014 ### vela Staff Emeritus Take a look at this plot (which is of sine, not cosine, but it's the same idea). Does this help? The red portion of the domain corresponds to the green part of the graph. I plotted it in units of $\\pi$ on the horizontal axis. #### Attached Files: • ###### plot.png File size: 4 KB Views: 145 11. Mar 26, 2014 ### thapyhap I'm not sure- I still feel like I'm missing something simple. Looking at the sin plot I can see a geometric solution as follows: $$A_1=y(\\pi-2\\sin^{-1}y)$$ \\begin{align} A_2&=2\\int^{\\sin^{-1}y}_0 \\sin x \\,dx \\\\ &=2\\left[ \\cos 0 - \\cos(\\sin^{-1} y) \\right] \\\\ &=2\\left( 1 - \\sqrt{1-y^2} \\right) \\end{align} $$A_3 = 2$$ Now I have that $P(Y \\le y) = \\frac{A_1 + A_2 + A_3}{4}$, but this doesn't seem to simplify to anything and I have the feeling there is a much simpler approach. I'm also not sure how to put it", "the equation would be \\begin{align*}y=\\sin (x-3.28)+1\\end{align*}. To determine the value of the horizontal shift, you might have to estimate. For example, we estimated that the negative shift was -3 because the maximum value of the parent graph is at \\begin{align*}x=\\frac{\\pi}{2}\\end{align*} and the maximum to the left of it didn’t quite make it to \\begin{align*}x=- \\frac{\\pi}{2}\\end{align*} (the distance between \\begin{align*}\\frac{\\pi}{2}\\end{align*} and \\begin{align*}-\\frac{\\pi}{2}\\end{align*} is \\begin{align*}\\pi\\end{align*}). Then, to determine the positive shift equation, recall that a period is \\begin{align*}2 \\pi\\end{align*}, which is \\begin{align*}6.28 \\ldots\\end{align*} So, the positive shift would be \\begin{align*}2 \\pi - 3\\end{align*} or \\begin{align*}6.28 - 3 = 3.28.\\end{align*}. Guided Practice Graph the following functions from \\begin{align*}[\\pi, 3 \\pi]\\end{align*}. 1. \\begin{align*}y=-1+\\sin x\\end{align*} 2. \\begin{align*}y=\\cos\\left(x+\\frac{\\pi}{3}\\right) - 2\\end{align*} 3. Find the equation of the cosine curve below. 1. Shift the parent graph down one unit. 2. Shift the parent graph to the left \\begin{align*}\\frac{\\pi}{3}\\end{align*} units and down 2 units. 3. The parent graph is in green. It moves up 3 units and to the right \\begin{align*}\\frac{3 \\pi}{4}\\end{align*} units. Therefore, the equation is \\begin{align*}y=\\cos \\left(x-\\frac{3 \\pi}{4}\\right)+3\\end{align*}. If you moved the cosine curve backward, then the equation would be \\begin{align*}y=\\cos \\left(x+\\frac{5 \\pi}{4}\\right)+3\\end{align*}. Vocabulary Phase Shift The horizontal shift of a trigonometric function. Problem Set For questions 1-4, match the equation with its graph. 1. \\begin{align*}y=\\sin \\left(x-\\frac{\\pi}{2}\\right)\\end{align*} 2. \\begin{align*}y=\\cos \\left(x-\\frac{\\pi}{4}\\right)+3\\end{align*} 3. \\begin{align*}y=\\cos \\left(x+\\frac{\\pi}{4}\\right)-2\\end{align*} 4. \\begin{align*}y=\\sin", "10:12 • math.stackexchange.com/questions/61605/… If you want further discussion ping me in chat. – N3buchadnezzar Oct 8 '14 at 10:32 Split the integral into two terms with limit $\\left[0,\\frac{\\pi}{2}\\right]$ and $\\left[\\frac{\\pi}{2},\\pi\\right]$ \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi/2} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x +\\int_{\\pi/2}^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\\\ \\end{align} Using identity \\begin{align} \\int_a^{b} f(x) \\,\\mathrm{d}x = \\int_a^{b} f(a+b-x) \\,\\mathrm{d}x \\end{align} Also the facts that $\\cos\\left(\\frac{\\pi}{2}-x\\right)=\\sin x$, $\\sin\\left(\\frac{\\pi}{2}-x\\right)=\\cos x$, $\\cos\\left(\\frac{3\\pi}{2}-x\\right)=-\\sin x$, and $\\sin\\left(\\frac{3\\pi}{2}-x\\right)=-\\cos x$, we get \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x &=\\int_0^{\\pi/2} \\frac{(\\sin x)^2}{1 + \\sin x \\cos x} \\,\\mathrm{d}x +\\int_{\\pi/2}^{\\pi} \\frac{(-\\sin x)^2}{1 + \\sin x \\cos x} \\,\\mathrm{d}x \\\\ &=\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\tag{1} \\end{align} Let \\begin{align} I=\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\end{align} Since \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\end{align} then \\begin{align} 2I&=\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x +\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\\\ &=\\int_0^{\\pi} \\frac{(\\cos x)^2 + (\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\\\ I&=\\frac{1}{2}\\int_0^{\\pi} \\frac{1}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\tag{2} \\end{align} Using $(2)$, we get \\begin{align} I&=\\int_0^{\\pi} \\frac{1}{2 + 2\\cos x \\sin x} \\,\\mathrm{d}x\\\\ &=\\int_0^{\\pi} \\frac{1}{2 + \\sin (2x)} \\,\\mathrm{d}x\\qquad\\Rightarrow\\qquad", "\\cos (x) + i \\sin(x)$. We have that \\begin{align*} \\cos (x+y) + i \\sin (x+y) &= e^{i(x+y)} \\\\ &= e^{ix} \\cdot e^{iy} \\\\ &= (\\cos (x) + i \\sin (x))(\\cos (y) + i \\sin (y)) \\\\ &= (\\cos (x) \\cos (y) - \\sin (x) \\sin(y)) + i(\\sin (x) \\cos(y) + \\cos(x) \\sin(y)) \\end{align*} By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by $\\cos (x) \\cos (y)$, and simplify. \\begin{align*} \\tan (x+y) &= \\frac{\\sin (x+y)}{\\cos (x+y)} \\\\ &= \\frac{\\sin (x) \\cos(y) + \\cos(x) \\sin(y)}{\\cos (x) \\cos (y) - \\sin (x) \\sin(y)} \\\\ &= \\frac{\\frac{\\sin(x)}{\\cos(x)} + \\frac{\\sin(y)}{\\cos(x)}}{1 - \\frac{\\sin (x) \\sin(y)}{\\cos (x) \\cos(y)}} \\\\ &= \\frac{\\tan (x) + \\tan (y)}{1 - \\tan (x) \\tan(y)} \\end{align*} as desired. ## Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$. Doing so yields: • $\\sin (2x) = 2\\sin (x) \\cos (x)$ • $\\cos (2x) = \\cos^2 (x) - \\sin^2 (x)$ • $\\tan (2x) = \\frac{2\\tan (x)}{1-\\tan^2 (x)}$ ### Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle", "2r^2 = 7r^2 \\\\ \\end{align} This shows that $$r = \\frac{1}{\\sqrt 7}$$. Now by the Law of Sines, \\begin{align} \\frac{{\\sin A}}{{\\sin {{120}^ \\circ }}} = \\frac{{r}}{{1}} = \\frac{1}{{\\sqrt 7 }} \\quad &\\Rightarrow \\sin A = \\frac{1}{{\\sqrt 7 }}\\sin {120^ \\circ } = \\frac{{\\sqrt 3 }}{{2\\sqrt 7 }} \\\\ \\\\ &\\Rightarrow A = \\arcsin \\left( {\\frac{{\\sqrt 3 }}{{2\\sqrt 7 }}} \\right) \\approx {19.1066^ \\circ } \\\\ \\end{align} The red hexagon does not need to be translated, but each of the other 6 scaled and rotated hexagons needs to be translated so that the appropriate vertex is translated back to the corresponding vertex in the original hexagon, as shown in the figure below. The matrices in the IFS involve the sine and cosine values for the rotation angle A. Since $$A = \\arcsin \\left( {\\frac{{\\sqrt 3 }}{{2\\sqrt 7 }}} \\right)$$, we have $$\\sin (A) = \\dfrac{{\\sqrt 3 }}{{2\\sqrt 7 }} = \\dfrac{{\\sqrt 3 \\cdot \\sqrt 7 }}{{14}}$$ $$\\cos (A) = \\sqrt {1 - {{\\sin }^2}(A)} = \\sqrt {1 - \\dfrac{{21}}{{{{14}^2}}}} = \\dfrac{{5\\sqrt 7 }}{{14}}$$ The scaling/rotation matrix is therefore $M = r\\left[ {\\begin{array}{*{20}{c}} {\\cos A} & { - \\sin A} \\\\ {\\sin A} & {\\cos A} \\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} {5/14} & { - \\sqrt 3 /14} \\\\ {\\sqrt 3 /14} & {5/14} \\\\ \\end{array}} \\right] = \\left[", "+ 4 y^2 - 2 z^2 - 5 y == 0 /. First[ans] // Simplify Which yielded: 4 + 97 y^2 == 110 y + 36 z^2. Then I completed the square and got: $$\\frac{(y-55/97)^2}{2637/9409}-\\frac{z^2}{293/388}=1$$ Using the fact that $\\cosh^2t-\\sinh^2t=1$, I parametrized this in the yz-plane: \\begin{align*} x&=0\\\\ y&=\\frac{55}{97}+\\sqrt{\\frac{2637}{9409}}\\cosh t\\\\ z&=\\sqrt{\\frac{293}{388}}\\sinh t \\end{align*} Pushing this toward the plane $6x+5y=2$, the y- and z-values do not change. Solving for x: Solve[6 x + 5 y == 2 /. y -> 55/97 + Sqrt[2637/9409] Cosh[t], x] Which yielded: \\begin{align*} x&=-\\frac{27}{194}-\\frac{5\\sqrt{293}}{194}\\cosh t\\\\ y&=\\frac{55}{97}+\\sqrt{\\frac{2637}{9409}}\\cosh t\\\\ z&=\\sqrt{\\frac{293}{388}}\\sinh t \\end{align*} My final code: Show[ContourPlot3D[{x^2 + 2 y^2 - z^2 + 3 x == 1, 2 x^2 + 4 y^2 - 2 z^2 - 5 y == 0}, {x, -5, 3.2}, {y, -3, 3}, {z, -3, 3}, ContourStyle -> Opacity[0.5], Mesh -> None, AxesLabel -> Automatic], ParametricPlot3D[{-27/194 - (5 Sqrt[293])/194 Cosh[t], 55/97 + Sqrt[2637/9409] Cosh[t], Sqrt[293/388] Sinh[t]}, {t, -3, 3}, PlotStyle -> {Thick, Red}], ViewPoint -> {-0.5, 3, 1.5} ] Which yielded this image. Similar work yield the curve of intersection on the other side. Show[ContourPlot3D[{x^2 + 2 y^2 - z^2 + 3 x == 1, 2 x^2 + 4 y^2 - 2 z^2 - 5 y == 0}, {x, -5, 3.2}, {y, -3, 3}, {z, -3, 3}, ContourStyle -> Opacity[0.5], Mesh -> None, AxesLabel", "# Precalculus Let $\\mathcal{G}$ be the graph of the parametric equations \\begin{align*} x &= \\cos(4t),\\\\ y &= \\sin(6t). \\end{align*}What is the length of the smallest interval $I$ such that the graph of these equations for all $t\\in I$ produces the entire graph $\\mathcal{G}$? 1. 👍 0 2. 👎 0 3. 👁 1,008 1. since cos(4t) has period pi/2 and sin(6t) has period pi/3 the LCM is pi 1. 👍 0 2. 👎 0 2. That answer keeps on coming as wrong, I had gotten that answer earlier. I am not sure what is wrong :( 1. 👍 0 2. 👎 0 3. its pi/2 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Trig Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) - 2. ### geometry Find the measure of the acute angle x, if : sin(x)=0.0175; sin(x)=0.5015; cos(x)=0.06814; cos(x)=0.0670. I know that Sin(x)=opp./hyp. and that cos(x)=adj./hyp. but i have no clue about how to find the xs in these equations 3. ### Pre-Cal (Trig) Help? The following relationship is known to be true for two angles A and B: cos(A)cos(B)-sin(A)sin(B)=0.957269 Express A in terms of the angle B. Work in degrees", "$\\frac{RN}{NP}$ $\\frac{NP}{OP}$ = cosA cosB - sinA sinB Hence, sin formula of compound angle (A + B) is sin (A + B) = sinA cosB + cosA sinB and consine formula of compound angle (A + B) and cos(A + B) = cosA cosB - sinA sinB ### Trigonometric Ratios of A - B(Subtraction formula) Let a revolving line start from OX and trace out an angle XOY = A and then revolve back through an angle YOZ = B ∴ ∠XOZ = A - B Let P be any point in the Line OZ. Draw PM perpendicular to OX and PN perpendicular to OY. From N Draw NQ perpendicular to OX and perpendicular to MP produced. Here, ∠RPN = 900 - ∠PNR = ∠PNY = ∠XOY = A Again QMRN is a rectangle. So, QN = MR and QM = NR Now sin(A-B) =$\\frac{PM}{OP}$ = $\\frac{MR - PR}{OP}$ =$\\frac{QN - PR}{OP}$ = $\\frac{QN}{OP}$ - $\\frac{PR}{OP}$ =$\\frac{QN}{ON}$ $\\frac{ON}{OP}$ - $\\frac{PR}{NP}$ $\\frac{NP}{OP}$ = sinA cosB - cosA sinB cos(A - B) = $\\frac{OM}{OP}$ =$\\frac{OQ + QM}{OP}$ =$\\frac{OQ + NR}{OP}$ =$\\frac{OQ}{OP}$ + $\\frac{NR}{OP}$ =$\\frac{OQ}{ON}$ $\\frac{ON}{OP}$ - $\\frac{NR}{NP}$ $\\frac{NP}{OP}$ = cosA cosB + sinA sinB Hence, sine formula of compound angle (A - B) is sin (A - B) = sinA cosB - cosA sinB and cosine formula of compound angle", "dimensional rescaling as givens. (If you do not, see below for my comments on that.) Start be decomposing a $1\\times1\\times1$ cube into six equal square-based pyramids, using a vertex at the center of the cube. So the volume of these pyramids are each ... 1 The distance from the axis of rotation to the curve is $r = 2\\sqrt a - \\sqrt x$. So a vertical slice of the solid of revolution, of thickness $dx$, will have area $\\pi r^2dx = \\pi(4a + x - 4\\sqrt{ax})dx$. You just have to integrate this expression (over the interval $[0,4a]$, if that is what you want). 1 You did $dx/dt = 2t - 3t^2$ and $dy/dt = 1 + 4t^3$ which leads us to have $$\\frac{dy}{dx}=\\frac{1 + 4t^3}{2t - 3t^2}$$ Now you were using this formula: $$P=\\int_a^b2\\pi y \\sqrt{1+y'^2}dy$$ But what is $\\sqrt{y'^2+1}$. $$\\sqrt{y'^2+1}=\\sqrt{1+\\left(\\frac{1 + 4t^3}{2t - 3t^2}\\right)^2}=\\frac{\\sqrt{(2t-3t^2)^2+(1+4t^3)^2}}{\\color{red}{|2t-3t^2|}}$$ Now you ... 1 Let $\\displaystyle\\arcsin x=\\phi\\implies\\sin\\phi=x$ and $-\\frac\\pi2\\le \\phi\\le\\frac\\pi2$ based the principal value of inverse sine ratio So, $\\displaystyle\\cos(\\arcsin x)=\\cos\\phi=+\\sqrt{1-\\sin^2\\phi}=+\\sqrt{1-x^2}$ 1 $$\\int\\limits_{-\\pi}^0e^{-x}\\cos(nx)\\,\\mathrm{d}x=-\\int\\limits_{-\\pi}^0(e^{-x})'\\cos(nx)\\,\\mathrm{d}x =-e^{-0}\\cos(n0)+e^{-\\pi}\\cos(n\\pi)-n\\int\\limits_{-\\pi}^0e^{-x}\\sin(nx)\\,\\mathrm{d}x= -1+(-1)^ne^{-\\pi}+n\\int\\limits_{-\\pi}^0(e^{-x})'\\sin(nx)\\,\\mathrm{d}x=$$ ... 1 Only top voted, non community-wiki answers of a minimum length are eligible" ]

[ "\\; Sin({\\pi})=0,\\; \\; Sin\\left(\\frac{3{\\pi}}{2}\\right)=-1,\\; \\; Sin(2{\\pi})=0$ $Sinx$ goes through a full period in $2{\\pi}$ radians. Finding 3 consecutive zeros will discover the period, as the centreline of the graph is the x-axis. Simplest is to start at zero. $\\displaystyle\\ Sin(2x)=0$ for $\\displaystyle\\ x=0$ and $\\displaystyle\\ 2x={\\pi}\\Rightarrow\\ x=\\frac{{\\pi}}{2}$ and $2x=2{\\pi}\\Rightarrow\\ x={\\pi}$ therefore $Sin(2x)$ has period ${\\pi}$ radians. $\\displaystyle\\ Sin\\left(\\frac{x}{2}\\right)=0\\Rightarrow\\ x=0, \\;\\; x=2{\\pi}, \\;\\; x=4{\\pi}$ so the period of $\\displaystyle\\ Sin\\left(\\frac{x}{2}\\right)$ is $4{\\pi}$ radians. Alternatively...... $SinA=Sin(A+2{\\pi})$ $Sinx=Sin(x+2{\\pi})$ has a period of $2{\\pi}$ radians $Sin(2x)=Sin(2x+2{\\pi})=Sin[2(x+{\\pi})]$ which therefore has a period of ${\\pi}$ radians $\\displaystyle\\ Sin\\left(\\frac{x}{2}\\right)=Sin\\left(\\frac{x}{2}+2 {\\pi}\\right)=Sin\\left[\\frac{1}{2}(x+4{\\pi})\\right]$ has a period of $4{\\pi}$ radians • Sep 6th 2010, 04:42 AM stupidguy Quote: Originally Posted by Educated A period is how long it takes for the wave to complete one full cycle. Sin(x) has a period of 2 $\\pi$ in radians, or $360^{\\circ}$ in degrees. This means that if you draw the graph of sin(x), the graph would take 2 $\\pi$ or $360^{\\circ}$ to complete one full cycle. Which is a period. Sin(2x) has a period of half of sin(x) Sin(0.5x) has a period of double of sin(x) The 2 in front of the x doubles the speed of one wave, which means that it halves the time, and since a period is the length of time for one full cycle, the period is", "{x+\\frac{\\pi }{{24}}} \\right)$$. To get $$b$$, we first find the period of the graph by seeing how long it goes before repeating itself (we can subtract the two $$x$$ values to get this new period). We see that the graph repeats itself from when $$\\displaystyle x=-\\frac{\\pi }{{24}}$$to when $$\\displaystyle x=\\frac{\\pi }{8}$$, so the new period is $$\\displaystyle \\frac{\\pi }{8}-\\left( {-\\frac{\\pi }{{24}}} \\right)=\\frac{{3\\pi }}{{24}}+\\frac{\\pi }{{24}}=\\frac{\\pi }{6}$$. Then we use the equation $$\\displaystyle b=\\frac{\\pi }{{\\text{new period}}}=\\frac{\\pi }{{\\left( {\\frac{\\pi }{6}} \\right)}}=\\frac{\\pi }{1}\\times \\frac{6}{\\pi }=6$$. So now we have $$\\displaystyle y=-2\\tan 6\\left( {x+\\frac{\\pi }{{24}}} \\right)$$. Try these on the graphing calculator! # Transformation of all Trig Functions without T-Charts Let’s go over once again how to transform trig functions without t-charts. Note that in order to perform the transformations accurately and quickly, you must know your 6 trig functions graphs inside out! Here are the steps for sin and cos graphs: 1. Put the trig function in the $$y=a\\sin b\\left( {x-c} \\right)+d\\,\\,\\,\\text{or}\\,\\,\\,y=a\\cos b\\left( {x-c} \\right)+d$$ format. 2. The new domain will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 3. The new range will be $$\\left[ {\\text{vertical shift (}d\\text{)}-\\left| a \\right|\\,\\text{,}\\,\\,\\text{vertical shift (}d\\text{)}+\\left| a \\right|} \\right]$$. ($$\\left| a \\right|$$ is amplitude) 4. New period will be $$\\displaystyle \\frac{{2\\pi }}{b}$$. 5. Phase shift will be $$c$$; start graph at this point, and if $$a$$", "# How do you graph y=-2csc(2x)+1? Nov 18, 2016 Using the Socratic graphic facility, I have inserted a graph. #### Explanation: The period of y is $\\frac{2 \\pi}{2} = \\pi$. It is convenient to choose one period as #x in (-pi/2, pi/2). Note that y has infinite discontinuity at the ends and the middle x = 0. A Table for making the graph: $\\left(x , y\\right) : \\left(3.31 , - \\frac{\\pi}{3}\\right) \\left(3 , - \\frac{\\pi}{4}\\right) \\left(3.1 , - \\frac{\\pi}{6}\\right) \\left(\\infty , 0\\right)$ $\\left(- 1.3 , \\frac{\\pi}{6}\\right) \\left(- 1 , \\frac{\\pi}{4}\\right) \\left(- 1.3 , \\frac{\\pi}{3}\\right)$ The part of the inserted graph that braces the y-axis is the graph for this period. ygraph{(y-1) sin (2x)+2=0 [-9.91, 10.09, -3.68, 6.32]}", "So for $x$ in the interval $[-\\pi/2, \\pi/2]$, $\\arcsin(\\sin(x)) = x$ satisfies that definition. For $x$ in the interval $[\\pi/2, 3 \\pi/2]$, $\\arcsin(\\sin(x))$ can't be $x$, but it can be $\\pi - x$ which is in the interval $[-\\pi/2, \\pi/2]$ (notice that $\\sin(\\pi - x) = \\sin(x)$). So that takes care of this part of the graph: Now notice that $\\sin(x)$ is periodic with period $2\\pi$, so the graph looks the same if you shift it left or right by $2\\pi$. That takes care of the rest of it. - So I understand the jist of it; but, I don't get why $sin(π−x)=sin(x)$. Did you mean that this only works for the interval $[−π/2,π/2]$? – evamvid Feb 19 at 3:19 To understand why sin(π−x)=sin(x), we need to start from the extended definition of sine for angles greater than π/2. sin(x) is defined as y-ordinate to the radius of the circle in question. That circle is centered at O(0, 0) and radius = $sqrt(x^2 + y^2)$. – Mick Feb 19 at 4:43 So let $f(x)=\\sin(x)$ and let $g(x)=\\arcsin(x)$. Let $\\operatorname{Dom}(f)$ be the domain of $f$ and let $\\operatorname{Rng}(f)$ be the range of $f$. Thus we have \\begin{align} \\operatorname{Dom}(f)&=\\mathbb{R}\\\\ \\operatorname{Rng}(f)&=[-1,1]\\\\ \\operatorname{Dom}(g)&=[-1,1]\\\\ \\operatorname{Rng}(g)&=\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right] \\end{align} If we restrict the $\\operatorname{Dom}(f)$ to $\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$, then $f$ and $g$ are inverses of one another and", "# How do you sketch one cycle of y=-3sin4theta? Jul 22, 2018 One petal graph only, for one cyle $\\theta \\in \\left(0 , \\frac{\\pi}{2}\\right)$. #### Explanation: $r = - 3 \\sin 4 \\theta \\ge 0 \\Rightarrow \\sin 4 \\theta < 0$ $\\Rightarrow 3 \\theta \\in {Q}_{1}$or ${Q}_{2} \\Rightarrow \\theta \\in {Q}_{1}$ If cycle means period, here, the period = (2pi)/4 = pi/2. Considering one cycle $\\theta \\in \\left[0 , \\frac{\\pi}{2}\\right]$, $r \\ge 0 \\in \\left(0 , \\frac{\\pi}{4}\\right) \\mathmr{and} r < 0 \\in \\left(\\frac{\\pi}{4} , \\frac{\\pi}{2}\\right)$. For creating a graph, convert to Cartesian form, usimg $r = \\sqrt{{x}^{2} + {y}^{2}} \\ge 0 , r \\left(\\cos \\theta , \\sin \\theta\\right)$ and $\\sin 4 \\theta = 4 \\left({\\cos}^{3} \\theta \\sin \\theta - \\cos \\theta {\\sin}^{3} \\theta\\right)$ as ( x^2 + y^2 )^2.5 + 12 xy( ( x^2 - y^2 ) = 0#. Now, the Socratic graph is ready. graph{( x^2 + y^2 )^2.5 + 12 xy ( x^2 - y^2 ) = 0[-0.1 8 -0.1 4] } Slide the graph $\\rightarrow \\uparrow$, to view the other three petals.", "# Verifying the period of $f(x)=\\sin(x)+\\cos(x/2)$ It seems clear from the graph of $$f(x)=\\sin(x)+\\cos(x/2)$$ that the period $$p$$ of the function is equal to $$4\\pi$$. To verify that $$4\\pi$$ is a period of $$f(x)$$, note that \\begin{align} \\sin(x + 4\\pi) + \\cos\\left(\\frac{x + 4\\pi}{2}\\right) & =\\sin(x)\\cos(4\\pi)+\\cos(x)\\sin(4\\pi)+\\cos(x/2)\\cos(4\\pi/2)-\\sin(x/2)\\sin(4\\pi/2) \\\\ & =\\sin(x)+\\cos(x/2) \\end{align} Thus $$4\\pi$$ is indeed a period of $$f$$. My question is, how would one go about trying to prove that $$4\\pi$$ is the smallest $$p>0$$ such that $$f(x+p)=f(x)$$? • Why not find it as $lcm(2\\pi,4\\pi)=4\\pi$? Jul 8, 2020 at 10:53 • Maybe solve $f(x)=0$, noting that if $f(x_0)=0$ and if $p$ is the period in question, then $f(x_0+p)=0$. – lulu Jul 8, 2020 at 11:06 • The $lcm$ will not always work. Examples: $\\sin{(\\omega x)}(\\sin^2{x}+\\cos^2{x}), \\sin(x)\\cos(x)$ Jul 8, 2020 at 12:00 If $$f$$ is periodic with period $$T$$ then so is $$f'$$. This means that $$f'(x) = c \\implies f'(x+T) = c.$$ When looking at the graph of $$f$$ it looks like the solutions of the equation $$f'(x) = f'(\\pi)$$ are exactly the points $$S = \\{ \\pi + 4k \\pi : k \\in \\mathbb Z\\}$$ . If we manage to prove this then we are finished since for a smaller period $$\\tilde T,$$ $$\\pi + \\tilde T$$ would not be a solution of the equation which contradicts", "# What is the period of $f(x)=\\sin x\\cos x$? Problem We need to find the period of the following: $f(x)=(\\sin(x))(\\cos(x))$ using basic trigonometric identities which is as follows: My steps disclaimer! I know the steps but I will pin point where I am confused and please explain so Steps: • 1) $f(x)=\\sin(x)\\cos(x)$ • 2) $f(x)=\\frac{1}{2}\\sin(2x)$ <-- I do not understand the transition from line 1 to line 2 • 3) therefore period is $\\pi$ • Are you familiar with trig identities such as $\\sin(x+y)=\\sin(x)\\cos(y)+\\cos(x)\\sin(y)$? – stewbasic Jul 11 '16 at 6:18 • yes its just the basic trig identities – John Rawls Jul 11 '16 at 6:22 • @JohnRawls: Then put $y=x$ and you get $\\sin(2x)=2\\sin(x)\\cos(x)$. – André Nicolas Jul 11 '16 at 6:26 Use 1) $2\\sin \\alpha \\cdot \\cos \\alpha = \\sin 2 \\alpha$ 2) If $y=a\\sin(kx+b)+c$, then period is $T=\\frac{2\\pi}{k}$. Hence, $$f(x)=\\sin x \\cos x= \\frac12 \\cdot 2\\sin x \\cos x=\\frac12 \\sin 2x$$ $$T=\\frac{2\\pi}{2}=\\pi$$ One of the trig identities is the following: $$\\sin(2\\theta) = 2\\sin(\\theta)\\cos(\\theta)$$ from which the result follows immediately. Various proofs for this identity exist, but I prefer using complex analysis to prove the result: Given that $\\sin(z) = \\frac{e^{iz}-e^{-iz}}{2i}$ and $\\cos(z)=\\frac{e^{iz}+e^{-iz}}{2}$ one has: $$2\\sin(z)\\cos(z) = 2\\cdot \\frac{e^{iz}-e^{-iz}}{2i}\\cdot \\frac{e^{iz}+e^{-iz}}{2} = \\frac{e^{2iz}+1-1-e^{-2iz}}{2i}=\\frac{e^{i(2z)}-e^{-i(2z)}}{2i}=\\sin(2z)$$", "why. Visualise the graph) $$z{\\rm{ }}\\left( x \\right) = \\left| {{\\rm{ }}sin{\\rm{ }}x{\\rm{ }}} \\right| + \\left| {{\\rm{ }}cos{\\rm{ }}x{\\rm{ }}} \\right|{\\rm{ }}$$ is, although periodic with period $$\\pi$$ {LCM ( $$\\pi$$,$$\\pi$$ )}, but the fundamental period is\\begin{align}\\frac{\\pi }{2}\\end{align}. (Verify). Hence, the LCM rule is not applicable everywhere. Consider $$y(x) = a|\\sin x| + b|\\cos x|.$$ Whatis the fundamental period? Think carefully and you will realise that the answer depends upon whether $$a=b$$ or $$a\\ne \\, b.$$ If $$a=b,$$ then $$y(x) = a(|\\sin x| + |\\cos x|)$$ and the fundamental period is \\begin{align}\\frac{\\pi }{2},\\end{align} as you just saw. If $$a \\, \\ne \\, b,$$ then the fundamental period is $$\\pi.$$ Similarly consider $$v(x) = |\\sin nx| + |\\cos mx|.$$ Since $$|\\text{sin}x |$$ is periodic with $$\\pi$$, $$|\\text{sin}nx |$$ is periodic with \\begin{align}\\frac{\\pi }{n}\\end{align}, $$|\\text{cos}mx |$$ is periodic with \\begin{align}\\frac{\\pi }{m}\\end{align}. Hence, $$v(x)$$ is periodic with \\begin{align}{\\rm{LCM\\;}}\\left( {\\frac{\\pi }{n},\\frac{\\pi }{m}} \\right)\\end{align}. (Evaluate this LCM for a few examples, like $$m = 2, n = 4 \\text{ or } m =3, n = 7$$). ## TRY YOURSELF – V Q. 1 What are the periods for the followingfunctions? (a) $$f(x) = |\\sin 2x| + |\\sin 3x|$$ (b) $$f(x) = {\\sin ^4}x + {\\cos^4}x$$ (c) \\begin{align}f(x) = 2\\cos \\frac{{x - \\pi }}{3}\\end{align} (d) $$f(x) = {\\sin ^2}\\pi x$$ Q. 2 If", "# Precalculus Let $\\mathcal{G}$ be the graph of the parametric equations \\begin{align*} x &= \\cos(4t),\\\\ y &= \\sin(6t). \\end{align*}What is the length of the smallest interval $I$ such that the graph of these equations for all $t\\in I$ produces the entire graph $\\mathcal{G}$? 1. 👍 0 2. 👎 0 3. 👁 1,008 1. since cos(4t) has period pi/2 and sin(6t) has period pi/3 the LCM is pi 1. 👍 0 2. 👎 0 2. That answer keeps on coming as wrong, I had gotten that answer earlier. I am not sure what is wrong :( 1. 👍 0 2. 👎 0 3. its pi/2 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Trig Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) - 2. ### geometry Find the measure of the acute angle x, if : sin(x)=0.0175; sin(x)=0.5015; cos(x)=0.06814; cos(x)=0.0670. I know that Sin(x)=opp./hyp. and that cos(x)=adj./hyp. but i have no clue about how to find the xs in these equations 3. ### Pre-Cal (Trig) Help? The following relationship is known to be true for two angles A and B: cos(A)cos(B)-sin(A)sin(B)=0.957269 Express A in terms of the angle B. Work in degrees", "Example $$\\PageIndex{8}$$ Find the period of $$y=\\cos\\;6x + \\sin\\;4x$$. Solution: The period of $$\\cos\\;6x$$ is $$\\frac{2\\pi}{6} = \\frac{\\pi}{3}$$, and the period of $$\\sin\\;4x$$ is $$\\frac{2\\pi}{4} = \\frac{\\pi}{2}$$. The lowest common multiple of $$\\frac{\\pi}{3}$$ and $$\\frac{\\pi}{2}$$ is $$\\pi$$: \\nonumber \\begin{alignat*}{4} 1 \\;\\cdot\\; \\tfrac{\\pi}{3} ~&=~ \\tfrac{\\pi}{3} \\quad\\quad\\quad &1 \\;&\\cdot\\; \\tfrac{\\pi}{2} ~&=~ \\tfrac{\\pi}{2}\\\\ \\nonumber 2 \\;\\cdot\\; \\tfrac{\\pi}{3} ~&=~ \\tfrac{2\\pi}{3} \\quad\\quad\\quad &2 \\;&\\cdot\\; \\tfrac{\\pi}{2} ~&=~ \\pi\\\\ \\nonumber 3 \\;\\cdot\\; \\tfrac{\\pi}{3} ~&=~ \\pi \\quad\\quad\\quad &{} &{}\\\\ \\nonumber \\end{alignat*} Thus, the period of $$y=\\cos\\;6x + \\sin\\;4x$$ is $$\\pi$$. We can see this from its graph in Figure 5.2.9: Figure 5.2.9 $$y = \\cos 6x+\\sin 4x$$ What about the amplitude? Unfortunately we can not use the technique from Example 5.9, since we are not taking the cosine and sine of the same angle; we are taking the cosine of $$6x$$ but the sine of $$4x$$. In this case, it appears from the graph that the maximum is close to $$2$$ and the minimum is close to $$-2$$. In Chapter 6, we will describe how to use a numerical computation program to show that the maximum and minimum are $$\\pm\\,1.90596111871578$$, respectively (accurate to within $$\\approx 2.2204 \\times 10^{-16}$$). Hence, the amplitude is $$1.90596111871578$$. Generalizing Example 5.9, an expression of the form $$a\\,\\sin\\;\\omega x \\;+\\; b\\,\\cos\\;\\omega x$$ is equivalent to $$\\sqrt{a^2 + b^2}\\;\\sin\\;(x+\\theta)$$, where $$\\theta$$ is an", "same transformations as they did before. One important difference: the period of sine and cosine is defined as \\begin{align*}\\frac{2\\pi}{B}\\end{align*}. The period of tangent and cotangent is only \\begin{align*}\\pi\\end{align*}, so the period would be \\begin{align*}\\frac{\\pi}{B}\\end{align*}. Example 1: Sketch the graph of \\begin{align*}g(x)=-2+ \\cot \\frac{1}{3}x\\end{align*} over the interval \\begin{align*}[0, 6\\pi]\\end{align*}. Solution: Starting with \\begin{align*}y=\\cot x\\end{align*}, \\begin{align*}g(x)\\end{align*} would be shifted down two and frequency is \\begin{align*}\\frac{1}{3}\\end{align*}, which means the period would be \\begin{align*}3\\pi\\end{align*}, instead of \\begin{align*}\\pi\\end{align*}. So, in our interval of \\begin{align*}[0, 6\\pi]\\end{align*} there would be two complete repetitions. The red graph is \\begin{align*}y=\\cot x\\end{align*} . Example 2: Sketch the graph of \\begin{align*}y=-3 \\tan \\left(x-\\frac{\\pi}{4}\\right)\\end{align*} over the interval \\begin{align*}[-\\pi, 2\\pi]\\end{align*}. Solution: If you compare this graph to \\begin{align*}y=\\tan x\\end{align*}, it will be stretched and flipped. It will also have a phase shift of \\begin{align*}\\frac{\\pi}{4}\\end{align*} to the right. The red graph is \\begin{align*}y=\\tan x\\end{align*}. ## Secant and Cosecant Because secant is the reciprocal of cosine, it will have the same period, \\begin{align*}2\\pi\\end{align*}. Notice that an entire period encompasses an upward \\begin{align*}\\bigcup\\end{align*} and downward \\begin{align*}\\bigcap\\end{align*} and the asymptote between them. There are no \\begin{align*}x-\\end{align*}intercepts and only one \\begin{align*}y-\\end{align*}intercept at \\begin{align*}(0, 1)\\end{align*}. The vertical asymptotes are everywhere cosine is zero, so \\begin{align*}\\pi n \\pm \\frac{\\pi}{2}\\end{align*} and \\begin{align*}\\pi n \\pm \\frac{3\\pi}{2}\\end{align*}, where \\begin{align*}n\\end{align*} is any integer. The general equation of a secant function is just", "// Alternatively, a period of $t\\mapsto\\sin t$ is $2\\pi$ and a period of $t\\mapsto\\sin 3t$ is $2\\pi/3$. Hence a period of their linear combination $t\\mapsto(\\sin t)^3$ is $\\max(2\\pi,2\\pi/3)=2\\pi$. – Did Sep 9 '13 at 15:09 • Hm, how do we know that the maximum of one of the two periods actually are the linear combination's period? I've checked it in Wolfram Alpha, just want to know. – Curtain Sep 9 '13 at 16:51 • Maximum was stupid, I should have said: least common multiple. Thanks for the remark. – Did Sep 9 '13 at 17:42 I would write it in complex form because trying to guess your way through rearranging terms from known trigonometric identities is a little silly. Why spend time guessing when you can quickly get the answer? $$\\sin t = \\frac{1}{2i}(e^{it}-e^{-it})$$ Thus we have $\\sin^3t = \\frac{i}{8}(e^{it}-e^{-it})^3 = \\frac{i}{8}(e^{3it}-3e^{it}+3e^{-it}-e^{-3it})$. Can you take it from here? hint: use power reduction identities. Seeing as $\\sin 3x=3\\sin x-4\\sin^3x$ it follows $\\sin^3x=\\dfrac34\\sin x-\\dfrac14\\sin 3x$ hence $a_i=0$ and $b_1=3/4,b_3=-1/4$ and other $b_j=0$. The period is essentially the 'least common multiple' i.e. $\\operatorname{lcm}(1,3)\\cdot\\dfrac{2\\pi}3=2\\pi$. • This is what I was about to write (+1) – robjohn Sep 9 '13 at 15:55 • Thanks for the correction, @robjohn. I posted this from my phone and it's difficult to use the small touchscreen keyboard", "# How do you graph the function r^2 = 9cos(2θ)? Nov 17, 2016 I have inserted the graph of this Lemniscate, using the Socratic facility for the cartesian frame. #### Explanation: The period of $\\cos k \\theta$ is $\\frac{2}{k} \\pi$. Here, k = 2. So, the period is $\\pi$. ${r}^{2} = 9 \\cos 2 \\theta \\ge 0 \\to \\cos 2 \\theta \\ge 0$. So, conveniently, one period can be chosen as $\\theta \\in \\left[0. \\pi\\right]$. wherein $\\cos 2 \\theta \\ge 0$. As $\\cos \\left(2 \\left(- \\theta\\right)\\right) = \\cos 2 \\theta$, the graph is symmetrical about the initial line. Also, as cos (2(pi-theta) = cos 2theta, the graph is symmetrical about the vertical $\\theta = \\frac{\\pi}{2}$. A Table for half period $\\left[0 , \\frac{\\pi}{4}\\right]$ is sufficient for the part in ${Q}_{1}$.. Use symmetry. for the other three quarters.. $\\left(r , \\theta\\right) : \\left(0 , 3\\right) \\left(\\frac{3}{\\sqrt{\\sqrt{2}}} , \\frac{\\pi}{8}\\right) \\left(\\frac{3}{2} \\sqrt{2} , \\frac{\\pi}{6}\\right) \\left(0 , \\frac{\\pi}{4}\\right)$ graph{(x^2+y^2)^2=9(x^2-y^2) [-10, 10, -5, 5]}", "can obtain our graph by multiplying each y-coordinate from our basic sine graph by 1/2. This graph will be vertically compressed by a factor of 1/2 when compared to our basic sine graph. Our range is now from -1/2 to 1/2. The period however is still unchanged. Let's get our five key points and sketch the graph: $$(0, 0)$$ $$\\left(\\frac{π}{2}, \\frac{1}{2}\\right)$$ $$(π, 0)$$ $$\\left(\\frac{3π}{2}, -\\frac{1}{2}\\right)$$ $$(2π, 0)$$ Let's also consider the graph of: $$f(x)=-2 \\cdot \\text{sin}\\hspace{.1em}x$$ We can obtain our graph by multiplying each y-coordinate from our basic sine graph by -2. Our range is now from -2 to 2. The period however is still unchanged. Notice that this graph can be obtained by sketching the graph of f(x) = 2 sin x and then reflecting across the x-axis. Recall this is the case when we have -f(x), this is the graph of f(x) reflected across the x-axis. In other words, when compared to our basic sine graph we have reflected across the x-axis and vertically stretched by a factor of 2. Let's get our five key points and sketch the graph: $$(0, 0)$$ $$\\left(\\frac{π}{2}, -2\\right)$$ $$(π, 0)$$ $$\\left(\\frac{3π}{2}, 2\\right)$$ $$(2π, 0)$$ In general, the graph of f(x) = a sin x will have the following properties: $$\\text{range}: \\left[-|a|, |a|\\right]$$ If |a| > 1, the basic sine curve", "period $2l$ and the left has period $2\\pi$ – Graham Hesketh May 31 '14 at 19:13 • @DanZimm :Fourier series of $sinx$ has to be found on $(-l,l)$ – clarkson Jun 1 '14 at 5:51 • @GrahamHesketh :I wrote the formula in the OP – clarkson Jun 1 '14 at 5:52 Yes it is correct, although conventionally when you give the expression in trig form the index usually runs only over the positive integers while in the exponential form it is usually writtena s running from $-\\infty$ to $+\\infty$, hence, the following notations are equivalent: \\begin{aligned} \\sin \\left( x \\right) &=\\sin \\left( l \\right) \\sum _{n=-\\infty}^{\\infty } \\frac{\\left( -1 \\right) ^{n}}{ -\\pi \\,n+l}\\sin \\left( {\\frac {\\pi \\,nx}{l}} \\right) \\quad :-l<x<l\\\\ &=\\sin \\left( l \\right) \\sum _{n=-\\infty}^{-1} \\frac{\\left( -1 \\right) ^{n}}{ -\\pi \\,n+l}\\sin \\left( {\\frac {\\pi \\,nx}{l}} \\right)+\\sin \\left( l \\right) \\sum _{n=1}^{\\infty } \\frac{\\left( -1 \\right) ^{n}}{ -\\pi \\,n+l}\\sin \\left( {\\frac {\\pi \\,nx}{l}} \\right)\\\\ &=\\sin \\left( l \\right) \\sum _{n=1}^{\\infty} \\frac{\\left( -1 \\right) ^{n}}{ \\pi \\,n+l}\\sin \\left( -{\\frac {\\pi \\,nx}{l}} \\right)+\\sin \\left( l \\right) \\sum _{n=1}^{\\infty } \\frac{\\left( -1 \\right) ^{n}}{ -\\pi \\,n+l}\\sin \\left( {\\frac {\\pi \\,nx}{l}} \\right)\\\\ &=\\sin \\left( l \\right) \\sum _{n=1}^{\\infty } \\left(\\frac{1}{ -\\pi \\,n+l}+ \\frac{1}{ -\\pi \\,n-l} \\right) \\left( -1 \\right) ^{n}\\sin \\left( {\\frac {\\pi \\,nx}{l}} \\right) \\end{aligned} where the properties $\\sin(0)=0$ and $\\sin(-x)=-x$", "# Graph of $y=4\\sin\\left(\\frac{x}{3}\\right)\\cos\\left(\\frac{x}{3}\\right)$ What is the graph of the equation $$y=4\\sin\\left(\\frac{x}{3}\\right)\\cos\\left(\\frac{x}{3}\\right)$$ along with its period and amplitude? I got confused when the equation got cosine and sine in it. I only know how to handle graphs of an equation when only there is either cosine or sine, not altogether with sine and cosine in one equation. • Hint:$\\sin 2A=2\\sin A \\cos A$. – Anurag A Jun 12 '17 at 11:40 • $y=4 \\sin(x/3) \\cos(x/3) = 2 \\sin(2x/3)$ So the amplitude is $2$ and the period id $3\\pi$ ... now check it ... desmos.com/calculator/kcoysbbfip – Donald Splutterwit Jun 12 '17 at 11:46 • Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jun 12 '17 at 11:50 Using the following double-angle formula makes graphing your curve a lot easier: $$\\sin(2\\theta)\\equiv 2\\sin{\\theta}\\cos{\\theta}$$ Hence, the equation of your curve is equivalent to: $$y=4\\sin\\left(\\frac{x}{3}\\right)\\cos\\left(\\frac{x}{3}\\right)=2\\cdot \\color{green}{2\\sin\\left(\\frac{x}{3}\\right)\\cos\\left(\\frac{x}{3}\\right)}=2\\color{green}{\\sin\\left(\\frac{2x}{3}\\right)}$$ Given a sinusoid $y=A\\sin(\\omega x+\\varphi)$, the amplitude is $A$ and the period is $\\dfrac{2\\pi}{\\omega}$.", "$2{\\pi}$ is the period of the $\\sin$ and the $\\cos$. 2. Fix ${\\varphi} = 0$, find the image with one parameter. $$\\begin{array}{} f( {\\theta}, 0 ) &= ( ( a + b \\cos 0 ) \\cos {\\theta}, ( a + b \\cos 0 ) \\sin {\\theta}, b \\sin 0 ) \\\\ &= ( ( a + b ) \\cos {\\theta}, ( a + b ) \\sin {\\theta}, 0 ). \\end{array}$$ What is it? A circle in the $y_1y_2$-plane with center $( 0, 0, 0 )$ and radius $( a + b )$. To put this together:", "obvious that: \\begin{align} P(X\\le1)&=1 \\\\ P(X\\le0)&=0.5 \\\\ P(X\\le-1)&=0 \\\\ \\end{align} It's also clear that the CDF will be the same for both sin(Y) and cos(Y) since it does not depend on the phase, only on the range of Y. I have no idea how to calculate, for example, P(X ≤ 0.5). I feel like I'm missing something really obvious. 10. Mar 26, 2014 ### vela Staff Emeritus Take a look at this plot (which is of sine, not cosine, but it's the same idea). Does this help? The red portion of the domain corresponds to the green part of the graph. I plotted it in units of $\\pi$ on the horizontal axis. #### Attached Files: • ###### plot.png File size: 4 KB Views: 145 11. Mar 26, 2014 ### thapyhap I'm not sure- I still feel like I'm missing something simple. Looking at the sin plot I can see a geometric solution as follows: $$A_1=y(\\pi-2\\sin^{-1}y)$$ \\begin{align} A_2&=2\\int^{\\sin^{-1}y}_0 \\sin x \\,dx \\\\ &=2\\left[ \\cos 0 - \\cos(\\sin^{-1} y) \\right] \\\\ &=2\\left( 1 - \\sqrt{1-y^2} \\right) \\end{align} $$A_3 = 2$$ Now I have that $P(Y \\le y) = \\frac{A_1 + A_2 + A_3}{4}$, but this doesn't seem to simplify to anything and I have the feeling there is a much simpler approach. I'm also not sure how to put it", "this means $t>0$ and $f(y)=f(y+t)$ for all real $y$, and that no smaller positive $t$ has the same property. Suppose that there is an integer $k$ such that $$kt<8<(k+1)t\\ ;$$ this can be rewritten as $$8=kt+t'\\quad\\hbox{with}\\quad 0<t'<t\\ .$$ Then we have $$f(y)=f(y+8)=f(y+t'+kt)=f(y+t')\\quad\\hbox{for all y},$$ which contradicts the fact that $f$ has period $t$. Therefore we must have $8=kt$ for some positive integer $k$, that is, $t=8/k$. Now $k$ cannot be even, for if $k=2m$ then for all $y$ we have $f(y)=f(y+mt)=f(y+4)$; in conjunction with $(*)$ this shows that $f(y)=1$ for all $y$, which is not (strictly speaking) periodic. However it is possible that the period could be $t=8/k$ with $k$ odd. For example, take $$f(y)=1+\\sin\\Bigl(\\frac{2\\pi y}{t}\\Bigr)\\ .$$ Then, as is well known, $f$ has period $t$; also $$f(y)+f(y+4) =1+\\sin\\Bigl(\\frac{2\\pi y}{t}\\Bigr)+1+\\sin\\Bigl(\\frac{2\\pi y}{t}+k\\pi\\Bigr)=2$$ for all $y$, as required. - Here $t$ is the period. So $f(y)=f(y+t)=f(y+2t)=f(y+12345t)=f(y+nt)$, where $n$ is any integer you like. BTW I found a better example for the last part of the question and have edited it - please re-read. – David Apr 16 '14 at 7:58 Thanks.. So u have proven that the period is $8/k$ where $k$ is odd... How do u prove $k=1$?? – Apurv Apr 16 '14 at 8:13 It is certainly possible to have $k=1$, just take my final example. If you", "# How do you graph y=2cot(2x+pi/3)+1? Jan 2, 2017 It is periodic, with period $\\frac{\\pi}{2}$. The graph is for ${12}_{+}$ periods. I am satisfied that this edition is bug-free. #### Explanation: For $\\cot \\left(2 x + \\frac{\\pi}{3}\\right) + 1$, the period is $\\frac{\\pi}{2}$. The asymptotes are given by $2 x + \\frac{\\pi}{3}$= a multiple of $\\pi = k \\pi$, giving $x = \\frac{k \\pi - \\frac{\\pi}{3}}{2} = \\left(\\frac{3 k - 1}{6}\\right) \\pi , k = 0 , \\pm 1 , \\pm 2 , \\pm 3 , \\ldots$ For one period, $x \\in \\left(- \\frac{\\pi}{6} , \\frac{\\pi}{3}\\right)$, there are two terminal asymptotes $\\uparrow x = - \\frac{\\pi}{6} \\downarrow \\mathmr{and} \\uparrow x = \\frac{\\pi}{3} \\downarrow$. graph{(y-1)tan(2x+1.047)-1=0 [-10, 10, -5.21, 5.21]}" ]

[ "+0 # help 0 76 1 Let $$x,y,z$$ be positive real numbers. Find the set of all possible values of $$f(x,y,z) = \\frac{x}{x + y} + \\frac{y}{y + z} + \\frac{z}{z + x}.$$ Apr 10, 2019", "+0 # help 0 205 1 Let $$x,y,z$$ be positive real numbers. Find the set of all possible values of $$f(x,y,z) = \\frac{x}{x + y} + \\frac{y}{y + z} + \\frac{z}{z + x}.$$ Apr 10, 2019", "$$x = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ So, from this, we can conclude that $$\\cos(5\\pi t)\\cos(6 \\pi t) = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ Other excursions into complex numbers can be explained similarly. We simply use this embedding of $$\\mathbb{R}$$ into $$\\mathbb{C}$$, and use the properties proven for all complex numbers to help us with real numbers.", "TrigIdentity AIME Difficult 2012 Problem - 233 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$. The value of $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y}$ can be expressed in the form $\\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "Let $x$, $y$, and $z$ be real numbers such that $x+y+z=6$ and $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}=2$. Find the value of $\\dfrac{(x+y)}{z}+\\dfrac{(y+z)}{x}+\\dfrac{(x+z)}{y}$.", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "+0 # Help 0 40 1 Let x, y, and z be nonzero real numbers, such that no two are equal, and $$x + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x}.$$Find all possible numeric values of xyz. Aug 28, 2020", "+0 # Polynomial Question 0 122 1 Let $x,$ $y,$ and $z$ be nonzero real numbers, such that no two are equal, and $x + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x}.$ Find all possible numeric values of $xyz.$ Feb 26, 2021", "y$ Ex-2 To see that it exists, try x = 4 and y be arbitrary real number. LHS = xy + x - 4 = 4y + 4 - 4 = 4y = RHS To prove it is unique, let say there exists another real number z s.t. (zy + z - 4 = 4y) Then, z(y + 1) = 4y + 4 => z = $\\frac{4y + 4}{y + 1} = \\frac{4(y + 1)}{(y + 1)}$ which is clearly not defined if y+1 = 0 else z = 4 = x. Ex-3 To see that it exists, try y = $\\frac{x^2}{x - 1}$ , its defined since $x - 1 \\neq 0$ $\\frac{y}{x}$ = $\\frac{x^2}{x(x - 1)}$ = $\\frac{x}{x - 1}$ y - x = $\\frac{x^2}{x - 1} - x$ = $\\frac{x^2 - x^2 + x}{x - 1}$ = $\\frac{x}{x - 1}$ clearly $\\frac{y}{x} = y - x$ To see that its unique. Let say there exists another real number z s.t.$\\frac{z}{x} = z - x$ => $z = xz - x^2$ => $xz - z = x^2$ => z = $\\frac{x^2}{x - 1}$ = y Ex-4 To see that it exists, try y = $\\frac{1}{x}$ For any real number z, zy = $\\frac{z}{x}$ To see that it is unique, Let say there exists another real number", "\\sin x - 2a &= 0 \\\\ 4y^3 +\\frac{1}{2}\\sin 2y +a &=0 \\end{array} \\right.$$ Compute the value of $\\cos(x+2y)$ Let $f$ be a real-valued function such that $f(x) + 2f(\\frac{2002}{x}) = 3x$ for all $x > 0$. Find $f(2)$.", "can let $\\frac{Z_1}{Z_2} = x + yi$ where $x, y \\in \\mathbb R$ So you end up with $|x+(y+1)i| = |x+(y-1)i|$ which gives $x^2 + (y+1)^2 = x^2 + (y-1)^2 \\implies (y+1)^2 - (y-1)^2 = 0 \\implies 2y(2) = 0 \\implies y = 0$. Since its imaginary part is zero, $\\frac{Z_1}{Z_2}$ is real.", "in the positive direction. Step 3: Reach the point interpreted as the final answer on the number line. ### Examples on Adding Real Numbers Given below is an example based on adding real numbers Add $\\frac{8}{12}$ + $\\frac{6}{12}$ Solution: Given $\\frac{8}{12}$ + $\\frac{6}{12}$ Step 1: The denominators are the same go to step 2. Step 2: Add the numerators $\\frac{8}{12}$ + $\\frac{6}{12}$ = $\\frac{8+6}{12}$ = $\\frac{14}{12}$ Step 3: Simplify the fraction $\\frac{14}{12}$ = $\\frac{7}{6}$ ## Properties of Real Numbers(with respect to Addition) • Closure : If a and b are two Real Numbers, the a+b or b+a is also a Real Number. • Commutativity : For two Real Numbers a and b, a+b=b+a. • Associativity : For three Real Numbers a, b and c, a+(b+c) = (a+b)+c. • Additive Identity : Since a+0 = a = 0+a , therefore 0 is the Identity element for Addition. • Distributivity : For three Real Numbers a, b and c, a×(b+c) = (a×b)+(a×c). • Density : Between two Real numbers a and b, there always lies another Real number x. such that a $\\leq$ x $\\leq$ b ## Subtracting Real Numbers The following Steps for subtracting real numbers will help us in finding the difference between two real numbers. Let us consider the sum of two negative numbers like (-3 -", "# $$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$$ Algebra Level 2 $$a$$, $$b$$ and $$c$$ are positive real numbers that satisfy $a^x=b^y=c^z=125, \\log_{5} abc=3.$ What is the value of $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}?$ ×", "Love Invariants Sep 7 '18 at 16:09 $ab + bc + ac = a(2 - \\frac12a -\\frac12c) + c(2 - \\frac12a - \\frac12c) + ac\\\\= 2a - \\frac12a^2 + 2c - \\frac12c^2$ $(2a - \\frac12a^2)$ can be shown to have a maximum value of $2$ for real $a$. It follows that $(2a - \\frac12a^2) + (2c - \\frac12c^2)$ has a maximum value of $4$ for real $a$ and $c$. As observed in other answers, the value of $4$ is attained at $(a, b, c) = (2, 0, 2)$.", "# Let's compute this manually! Algebra Level 4 Let $$a$$ be a real number such that $a=\\frac{1}{1}+\\frac{1}{3}+\\frac{1}{6}+\\frac{1}{10}+\\cdots+\\frac{1}{5050}.$ If $$a$$ can be expressed in the form $$\\frac{x}{y}$$, where $$x,y$$ are coprime, positive integers, find $$x+y$$. ×", "+0 0 38 1 Let $$x,y,z$$ be positive real numbers such that $$x + y + z = 1$$ Find the minimum value of $$\\frac{1}{x + y} + \\frac{1}{x + z} + \\frac{1}{y + z}.$$ Nov 12, 2019", "any real number $$p,$$ there is at least one real number, usually two, that could be the value of $$y$$ in the true equation $$2 y^2 - (p+3)y + 2(p-1) = 0.$$ But now we have to remember that $$\\cos x$$ cannot be just any real number. It has to be in the interval $$[-1,1].$$ So instead of just being concerned about whether a real number exists that makes $$2 y^2 - (p+3)y + 2(p-1) = 0$$ true when we plug this number in for $$y,$$ we need to think about which real number we might have to plug in for $$y,$$ and make sure that that number is in the interval $$[-1,1].$$ The use of the variable $$p$$ makes it a little difficult to guess a factorization of the polynomial $$2 y^2 - (p+3)y + 2(p-1),$$ so let's just apply the standard formula to find solutions of any solvable quadratic equation in $$y$$: $$y = \\frac{-b \\pm \\sqrt{b^2 - 4ac} }{2a}.$$ Putting $$a = 2,$$ $$b = -(p+3),$$ $$c = 2(p-1),$$ we have $$y = \\frac{(p+3) \\pm \\sqrt{(p+3)^2 - 16(p-1)} }{4} = \\frac{(p+3) \\pm (p - 5)}{4}.$$ If we choose $$+$$ for $$\\pm$$ we get $$y = (p - 1)/2$$ and if we choose $$-$$ we get $$y = 2.$$ Obviously $$y = 2$$ is no good", "60 views If $c=\\frac{16 x}{y}+\\frac{49 y}{x}$ for some non-zero real numbers $x$ and $y$, then $c$ cannot take the value 1. $-60$ 2. $-50$ 3. $60$ 4. $-70$ Let z=(x/y). Now, c = 16z + 49/z => 16z$^{2}$ – cz + 49 = 0. Since x & y are non zero real number. So, z is also a non zero real number. So, the equation 16z$^{2}$ – cz + 49 = 0 satisfies for some real number z, if b$^{2}$ – 4ac >= 0 => c$^{2}$ – 56$^{2}$ >= 0 => |c| >= 56. Therefore the absolute value of c is always >= 56. So, value of c can’t be -50. Ans is (B). by 154 points 1 vote 1 1 vote", "+0 +1 157 4 The problem is: Find the set of values $$a$$ for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. I got $$[-1 , ∞)$$ by finding x in terms of y and a, then finding when that x value would be real. May 31, 2021 #1 +1 [-1.inf) is correct. Good job! May 31, 2021 #2 0 Oh, it was incorrect for some reason. Guest May 31, 2021 #3 0 I got it right :) Guest May 31, 2021 #4 +13045 +1 Find the set of values a for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. Hello Guest! $$a\\in \\mathbb R\\ |\\ a\\neq -1$$ ! May 31, 2021", "$$\\frac{x^2}{4}+\\frac{y^2}{49}=1$$ Algebra Level 2 If real numbers $$x$$ and $$y$$ satisfy $\\frac{x^2}{4}+\\frac{y^2}{49}=1,$ the maximum value of $$x+y$$ is $$\\sqrt{k}.$$ What is $$k?$$ ×" ]

[ "# Difference between revisions of \"2012 AIME II Problems/Problem 9\" ## Problem 9 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$. The value of $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y}$ can be expressed in the form $\\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. ### The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\\sin y \\cos y} = \\frac{\\sin x}{\\sin y}\\cdot\\frac{\\cos x}{\\cos y}=3\\cdot\\frac{1}{2} = \\frac{3}{2}$ ### The Second Term Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2. Hungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$, we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$. Equation 1 then becomes: $\\sin x = 3\\sin y$. Equation 2 then becomes: $\\cos x = \\frac{1}{2} \\cos y$ Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: $1 = 9\\sin^2 y + \\frac{1}{4}", "# 2012 AIME II Problems/Problem 9 ## Problem 9 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$. The value of $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y}$ can be expressed in the form $\\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. ### The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\\sin y \\cos y} = \\frac{\\sin x}{\\sin y}\\cdot\\frac{\\cos x}{\\cos y}=3\\cdot\\frac{1}{2} = \\frac{3}{2}$ ### The Second Term Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2. Hungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$, we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$. Equation 1 then becomes: $\\sin x = 3\\sin y$. Equation 2 then becomes: $\\cos x = \\frac{1}{2} \\cos y$ Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ Applying the", "\\cos^2 y$ Applying the identity $\\cos^2 y = 1 - \\sin^2 y$ (which is similar to $\\sin^2\\theta + \\cos^2\\theta = 1$ but a bit different), we can change $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ into: $1 = 9\\sin^2 y + \\frac{1}{4} - \\frac{1}{4} \\sin^2 y$ Rearranging, we get $\\frac{3}{4} = \\frac{35}{4} \\sin^2 y$. So, $\\sin^2 y = \\frac{3}{35}$. Squaring Equation 1 (leading to $\\sin^2 x = 9\\sin^2 y$), we can solve for $\\sin^2 x$: $\\sin^2 x = 9\\left(\\frac{3}{35}\\right) = \\frac{27}{35}$ Using the identity $\\cos 2\\theta = 1 - 2\\sin^2\\theta$, we can solve for $\\frac{\\cos 2x}{\\cos 2y}$. $\\cos 2x = 1 - 2\\sin^2 x = 1 - 2\\cdot\\frac{27}{35} = 1 - \\frac{54}{35} = -\\frac{19}{35}$ $\\cos 2y = 1 - 2\\sin^2 y = 1 - 2\\cdot\\frac{3}{35} = 1 - \\frac{6}{35} = \\frac{29}{35}$ Thus, $\\frac{\\cos 2x}{\\cos 2y} = \\frac{-19/35}{29/35} = -\\frac{19}{29}$. ### Now Back to the Solution! Finally, $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y} = \\frac32 + \\left(-\\frac{19}{29} \\right) = \\frac{49}{58}$. So, the answer is $49+58=\\boxed{107}$.", "identity $\\cos^2 y = 1 - \\sin^2 y$ (which is similar to $\\sin^2\\theta + \\cos^2\\theta = 1$ but a bit different), we can change $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ into: $1 = 9\\sin^2 y + \\frac{1}{4} - \\frac{1}{4} \\sin^2 y$ Rearranging, we get $\\frac{3}{4} = \\frac{35}{4} \\sin^2 y$. So, $\\sin^2 y = \\frac{3}{35}$. Squaring Equation 1 (leading to $\\sin^2 x = 9\\sin^2 y$), we can solve for $\\sin^2 x$: $\\sin^2 x = 9\\left(\\frac{3}{35}\\right) = \\frac{27}{35}$ Using the identity $\\cos 2\\theta = 1 - 2\\sin^2\\theta$, we can solve for $\\frac{\\cos 2x}{\\cos 2y}$. $\\cos 2x = 1 - 2\\sin^2 x = 1 - 2\\cdot\\frac{27}{35} = 1 - \\frac{54}{35} = -\\frac{19}{35}$ $\\cos 2y = 1 - 2\\sin^2 y = 1 - 2\\cdot\\frac{3}{35} = 1 - \\frac{6}{35} = \\frac{29}{35}$ Thus, $\\frac{\\cos 2x}{\\cos 2y} = \\frac{-19/35}{29/35} = -\\frac{19}{29}$. ### Now Back to the Solution! Finally, $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y} = \\frac32 + \\left(-\\frac{19}{29} \\right) = \\frac{49}{58}$. So, the answer is $49+58=\\boxed{107}$. ## Solution 2 As mentioned above, the first term is clearly $\\frac{3}{2}.$ For the second term, we first wish to find $\\frac{\\cos 2x}{\\cos 2y} =\\frac{2\\cos^2 x - 1}{2 \\cos^2y -1}.$ Now we first square the first equation getting $\\frac{\\sin^2x}{\\sin^2y} =\\frac{1-\\cos^2x}{1 - \\cos^2y} =9.$ Squaring the second equation yields $\\frac{\\cos^2x}{\\cos^2y} =\\frac{1}{4}.$ Let $\\cos^2x = a$ and $\\cos^2y", "left-hand side. Let’s combine the right-hand side by giving them same denominator. $\\begin{array}{|c} \\frac{1}{\\sin x}- \\frac{\\sin^2x}{\\sin x} \\\\\\frac{1- \\sin^2x}{\\sin x} \\\\\\frac{\\cos^2x}{\\sin x}\\end{array}$ The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, $\\sin^2 \\theta+\\cos^2 \\theta=1$ , and isolated the $\\cos^2x=1- \\sin^2x$ . There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to $\\frac{\\cos^2x}{\\sin^2x}$ , we could have also substituted the identity $\\cot^2x=\\csc^2x-1$ . #### Example B Verify the identity $\\frac{\\sin x}{1- \\cos x}=\\frac{1+ \\cos x}{\\sin x}$ . Solution: Multiply the left-hand side of the equation by $\\frac{1+ \\cos x}{1+ \\cos x}$ . $\\frac{\\sin x}{1- \\cos x}&= \\frac{1+ \\cos x}{\\sin x} \\\\\\frac{1+ \\cos x}{1+ \\cos x} \\cdot \\frac{\\sin x}{1- \\cos x}&= \\\\\\frac{\\sin \\left(1+\\cos x\\right)}{1- \\cos^2x}&= \\\\\\frac{\\sin \\left(1+\\cos x\\right)}{\\sin^2x}&= \\\\\\frac{1+\\cos x}{\\sin x}&=$ The two sides are the same, so we are done. #### Example C Verify the identity $\\sec(-x)=\\sec x$ . Solution: Change secant to cosine. $\\sec(-x)= \\frac{1}{\\cos \\left(-x\\right)}$ From the Negative Angle Identities, we know that $\\cos (-x)=\\cos x$ . $&=\\frac{1}{\\cos x} \\\\&=\\sec x$ Concept Problem Revisit Start by simplifying the lefthand side of the equation. $\\frac{\\sin^2x}{\\tan^2x}=\\frac{\\sin^2x}{\\frac{\\sin^2x}{\\cos^2x}}\\\\=\\cos^2x$ . Now simplify the righthand side of the equation. By", "# Convert the polar equation to rectangular form. ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. LCKurtz Homework Helper Gold Member ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. You are making it way too difficult. Put ##\\sin\\theta =\\frac y r## in your first equation then see if you can get an xy equation from that. Dick Homework Helper ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. Try putting sinθ=y/r. You are making it way too difficult. Put ##\\sin\\theta =\\frac y r## in your first equation then see if you can get an xy equation from that. Like so: r = 3sinθ becomes (r=3$\\frac{y}{r}$) = (r =$\\sqrt{3y}$) ? or solving for y, (y = $\\frac{r^2}{3}$) or do you mean something like this: ($\\frac{y}{r} = sinθ$) becomes ($\\frac{y}{3sinθ} = sinθ$) = (y = 3sin$^{2}θ$) Dick Homework Helper Like so: r = 3sinθ becomes (r=3$\\frac{y}{r}$) = (r =$\\sqrt{3y}$) ? or solving for y, (y = $\\frac{r^2}{3}$) or do you mean something like this: ($\\frac{y}{r} = sinθ$) becomes ($\\frac{y}{3sinθ} =", "# Proving $\\sin x + \\sin x \\cdot \\cot^2 x = \\csc x$ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process: \\begin{align*} \\tan \\theta &= \\frac{\\sin \\theta}{\\cos \\theta}\\\\ (\\sin \\theta)^2 + (\\cos \\theta)^2 &= 1\\\\ \\csc \\theta &= \\frac{1}{\\sin \\theta}\\\\ \\sec \\theta &= \\frac{1}{\\cos \\theta}\\\\ \\cot \\theta &= \\frac{1}{\\tan \\theta}. \\end{align*} The trig identity given is: $$\\sin x + \\sin x \\cdot \\cot^2 x = \\csc x$$ I simplified it to: $$\\sin x + \\sin x \\cdot \\left(\\frac{\\cos x}{\\sin x}\\right)^2 = \\frac{1}{\\sin x}$$ After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated. - Multiply both sides by $\\sin x$ and see what you get. – Karolis Juodelė Jun 1 '12 at 19:03 Arturo, you are simply too quick at editing. You finished mere seconds before I was done. – Joe Jun 1 '12 at 19:04 you are at last step: $\\frac {1}{sinx}$=Cosecx – iostream007 May 10 '13 at 4:41 Taking the left hand side, try cancelling the $\\sin^2{x}$ with the $\\sin{x}$ in the numerator (i.e. $\\sin{x} .(\\frac{\\cos^2{x}}{\\sin^2{x}}) = \\frac{\\cos^2{x}}{\\sin{x}}$) Then add the remaining two terms ($\\sin{x} + \\frac{\\cos^2{x}}{\\sin{x}}$) and use the", "### Home > CCA2 > Chapter 12 > Lesson 12.2.2 > Problem12-118 12-118. 1. Prove that each of the following equations is an identity. In other words, prove that they are each true for all values for which the functions are defined. Homework Help ✎ 1. (sin θ + cos θ)2 = 1 + 2 sin θ cos θ 2. tan θ + cot θ = sec θ csc θ 3. (tan θ cos θ)(sin2θ + ) = sin θ Start by expanding the left side of the equation. You might have to rearrange the equation to see ways to simplify it after you expand it. Change the trigonometric identities on the left side of the equation into the ratios they represent. $\\frac{\\sin\\theta}{\\cos\\theta}+\\frac{\\cos\\theta}{\\sin\\theta}=$ Find a common denominator and then add the two terms. $\\frac{\\sin^{2}\\theta}{\\cos\\theta\\sin\\theta}+\\frac{\\cos^{2}\\theta}{\\cos\\theta\\sin\\theta}=$ Remember that sin²(θ) + cos²(θ) = 1, and rewrite the equation. $\\frac{1}{\\cos\\theta\\sin\\theta}=$ Substitute in the identities for the two terms. What does 1/cosθ equal? 1/sinθ? Substitute the trig identities in first, then expand by multiplying the two terms together. Remember to use Giant Ones to simplify when you can. Use methods from both parts (a) and (b).", "equation with only one trigonometric function: $y=a(1)+(b-a)\\sin^2 x$ $y=a+(b-a)\\sin^2 x$ and now we can readily invert it: $\\sin^2 x=\\frac{y-a}{b-a}$ $x=\\arcsin\\sqrt{\\frac{y-a}{b-a}}$ Now it’s time for the second one: $y=a\\cos x+b\\sin^2 x$ We will again use the pythagorean identity, but in a different way. This time, we will use the fact that since $\\cos^2\\theta+\\sin^2\\theta=1$ then $\\sin^2\\theta=1-\\cos^2\\theta$ and so, by substitution, our original equation can be turned into $y=a\\cos x+b(1-\\cos^2x)$ $y=a\\cos x+b-b\\cos^2x$ $b\\cos^2 x-a\\cos x-b+y$ And we have just reduced it to a quadratic, which can be solved using the quadratic formula: $\\cos x=\\frac{a\\pm\\sqrt{a^2-4b(y-b)}}{2b}$ $x=\\arccos \\frac{a\\pm\\sqrt{a^2-4b(y-b)}}{2b}$ Now for the third equation: $y=a\\cos x+b\\sin x$ Though it seems much simpler than the other two, this one is actually much more difficult. It, too, involves a “trick” and a trigonometric identity, but it does not use the pythagorean identity. Before I give it away, you should try this problem - once you discover the trick, the result can be very satisfying. To solve this, you must convert the equation into this equivalent equation: $\\frac{y}{\\sqrt{a^2+b^2}}=\\frac{a}{\\sqrt{a^2+b^2}}\\cos x+\\frac{b}{\\sqrt{a^2+b^2}}\\sin x$ Now we will need to use another identity - one that I did not mention at the beginning of the post: $\\sin \\arccos \\theta=\\sqrt{1-\\theta^2}$ which means that $\\sin \\arccos \\frac{b}{\\sqrt{a^2+b^2}}=\\sqrt{1-\\frac{b^2}{a^2+b^2}}=\\frac{a}{\\sqrt{a^2+b^2}}$ And so we can again transform our equation into the equivalent equation $\\frac{y}{\\sqrt{a^2+b^2}}=\\sin\\bigg(\\arccos\\frac{b}{\\sqrt{a^2+b^2}}\\bigg)\\cos x+\\cos\\bigg(\\arccos\\frac{b}{\\sqrt{a^2+b^2}}\\bigg)\\sin x$ Now recall", "linear, so we may easily solve for the parameter $t$: $t=\\frac{y}{7}$ Now, if we substitute for $t$ into the equation for $x$ we get an equation that relates $x$ and $y$ where $t$ has been \"eliminated.\" We then have a Cartesian equation: $x=2\\left(\\frac{y}{7} \\right)^2+3\\left(\\frac{y}{7} \\right)+5$ $x=\\frac{2}{49}y^2+\\frac{3}{7}y+5$ 2.) Suppose you are given the parametric equations: $x(t)=a\\cdot\\cos(t)$ $y(t)=b\\cdot\\sin(t)$ Rather than solving one of the equations for $t$, we may rewrite the equations as: $\\frac{x}{a}=\\cos(t)$ $\\frac{y}{b}=\\sin(t)$ If we square both equations, we may then add and take advantage of the Pythagorean identity $\\sin^2(\\theta)+\\cos^2(\\theta)=1$ $\\left(\\frac{x}{a} \\right)^2=\\cos^2(t)$ $\\left(\\frac{y}{b} \\right)^2=\\sin^2(t)$ $\\left(\\frac{x}{a} \\right)^2+\\left(\\frac{y}{b} \\right)^2=\\cos^2(t)+\\sin^2(t)$ $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$", "the left-hand side. Let’s combine the right-hand side by giving them same denominator. 1sinxsin2xsinx1sin2xsinxcos2xsinx$\\begin{array}{|c} \\frac{1}{\\sin x}- \\frac{\\sin^2x}{\\sin x} \\\\ \\frac{1- \\sin^2x}{\\sin x} \\\\ \\frac{\\cos^2x}{\\sin x}\\end{array}$ The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, sin2θ+cos2θ=1$\\sin^2 \\theta+\\cos^2 \\theta=1$ , and isolated the cos2x=1sin2x$\\cos^2x=1- \\sin^2x$ . There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to cos2xsin2x$\\frac{\\cos^2x}{\\sin^2x}$ , we could have also substituted the identity cot2x=csc2x1$\\cot^2x=\\csc^2x-1$ . #### Example B Verify the identity sinx1cosx=1+cosxsinx$\\frac{\\sin x}{1- \\cos x}=\\frac{1+ \\cos x}{\\sin x}$ . Solution: Multiply the left-hand side of the equation by 1+cosx1+cosx$\\frac{1+ \\cos x}{1+ \\cos x}$ . \\frac{\\sin x}{1- \\cos x}&= \\frac{1+ \\cos x}{\\sin x} \\\\ \\frac{1+ \\cos x}{1+ \\cos x} \\cdot \\frac{\\sin x}{1- \\cos x}&= \\\\ \\frac{\\sin \\left(1+\\cos x\\right)}{1- \\cos^2x}&= \\\\ \\frac{\\sin \\left(1+\\cos x\\right)}{\\sin^2x}&= \\\\ \\frac{1+\\cos x}{\\sin x}&= The two sides are the same, so we are done. #### Example C Verify the identity sec(x)=secx$\\sec(-x)=\\sec x$ . Solution: Change secant to cosine. sec(x)=1cos(x) From the Negative Angle Identities, we know that cos(x)=cosx$\\cos (-x)=\\cos x$ . &=\\frac{1}{\\cos x} \\\\ &=\\sec x Concept Problem Revisit Start by simplifying the left-hand side of the equation. sin2xtan2x=sin2xsin2xcos2x=cos2x . Now simplify the", "\\end{equation*} Equating the real parts we see that \\begin{equation*} \\frac{1}{2}+\\cos(\\theta)+\\cos(2\\theta)+\\cdots =0. \\end{equation*} Replacing $\\theta$ with $\\theta+\\pi$ & using the trigonometric addition formulae we get \\begin{equation*} \\frac{1}{2}-\\cos(\\theta)+\\cos(2\\theta)-\\cdots =0. \\end{equation*} Integrate from $\\theta=0$ to $\\theta=\\phi$ & write $\\theta$ again for $\\phi$ we obtain \\begin{equation*} \\sin(\\theta)-\\frac{1}{2}\\sin(2\\theta)+\\frac{1}{3}\\sin(3\\theta)-\\cdots =\\frac{1}{2}\\theta,~-\\pi<\\theta<\\pi \\end{equation*} & this series is convergent. Integrating again gives \\begin{equation*} 1-\\cos(\\theta)-\\frac{1-\\cos(2\\theta)}{2^2}+\\frac{1-\\cos(3\\theta)}{3^2}-\\cdots =\\frac{1}{4}\\theta^2. \\end{equation*} Setting $\\theta=\\pi$ gives \\begin{equation*} 1+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots =\\frac{1}{8}\\pi^2. \\end{equation*} Since \\begin{equation*} 1+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots =1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots -\\frac{1}{2^2}-\\frac{1}{4^2}-\\cdots =(1-\\frac{1}{4})(1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots), \\end{equation*} we deduce that \\begin{equation*} 1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots = \\sum^{\\infty}_{n=1} \\frac{1}{n^2} =\\frac{1}{6}\\pi^2. \\end{equation*} Remark: We also get the formula \\begin{equation*} 1-\\frac{1}{2^2}+\\frac{1}{3^2}-\\cdots =\\frac{1}{12}\\pi^2 \\end{equation*}", "by dividing the other two to get $\\tan^2 \\theta = \\frac{1-\\cos(2\\theta)}{1+\\cos(2\\theta)}$ To get the half angle formulas, replace $$\\theta$$ with $$\\frac12\\theta$$ in each equation. ## Product-to-Sum Now we're going to swap back to $$u$$'s and $$v$$'s, which means we're going back the sum-difference formula again. There's some cute linear algebra you can use here, but you can get along fine without it. Our objective is to express a product of trig functions in terms of a sum of trig functions. So, take a look at the [sum and difference formulas for sine,][(the ones for $$\\sin(u+v)$$ and $$\\sin(u-v)$$)] and focus on the product term $$\\sin u\\cos v$$ that doesn't change sign. If you were to add the two equations together, it'd cancel the other term, leaving you with two of this one, easy to solve for and get $\\sin u\\cos v = \\frac12\\big(\\sin(u+v) + \\sin(u-v)\\big)$ If you subtract the equations instead, you get $\\cos u\\sin v = \\frac12\\big(\\sin(u+v) - \\sin(u-v)\\big)$ (You could also have gotten that one by swapping the places of $$u$$ and $$v$$, applying the odd rule for sine on the right.) Doing the same thing for the cosine sum and difference formulas gives you $\\cos u\\cos v = \\frac12\\big(\\cos(u+v) + \\cos(u-v)\\big) \\\\ \\sin u\\sin v = -\\frac12\\big(\\cos(u+v) - \\cos(u-v)\\big)$ ## Sum-to-Product You may notice that we", "y_2 = y$ ;) The vector form is more general (because it works on all inner product spaces, not only $\\mathbb R^n$) – AlexR Feb 5 '14 at 22:29 • @AlexR Yes, you are right $y1=x, y_2 = y$. I might consider getting more familiar with vector form, because as you've mentioned it has much more wider use. – Stefan4024 Feb 5 '14 at 22:40 $\\frac12 (x^2+y^2)-(\\frac12 (x+y))^2$ $=\\frac12 (x^2+y^2)-\\frac14 (x^2+y^2+2xy)$ $=\\frac14 (2x^2+2y^2-x^2-y^2-2xy)$ $=\\frac14 (x^2+y^2-2xy)$ $=\\frac14 (x-y)^2\\ge 0$ This substitution is a nice way to do this: $$x=r\\sin\\theta , y=r\\cos\\theta$$ So, our identity: $$\\left(\\frac12(x+y)\\right)^2 \\leq \\frac12(x^2 + y^2)$$ $$\\frac{1}{4} (x^2+y^2+2xy) \\leq \\frac12(x^2 + y^2)$$ Applying our substitutions: $$\\frac{1}{4} (r^2+2\\cdot r\\sin\\theta\\cdot r\\cos\\theta) \\leq \\frac12(r^2)$$ $$\\frac{r^2}{2}\\cdot \\frac{1+\\sin2\\theta}{2} \\leq \\frac{r^2}{2}$$ Now, if you note that the maximum value of $\\sin 2\\theta$ is $1$, then it becomes obvious. Added: It could have also been done it this way: $$\\left(\\frac{1}{2}(x+y)\\right)^2 \\leq \\frac12(x^2 + y^2)$$ $$\\frac{1}{2}\\cdot\\frac{1}{2}(r\\cos\\theta+r\\sin\\theta)^2 \\leq \\frac{r^2}{2}$$ $$\\frac{r^2}{2}\\cdot\\frac{1}{2}(\\cos\\theta+\\sin\\theta)^2 \\leq \\frac{r^2}{2}$$ Now, note that $-\\sqrt{2} \\leq \\cos\\theta+\\sin\\theta = \\sqrt{2}\\cos(x+45^{\\circ}) \\leq \\sqrt{2}$. Thus the maximum value of $(\\cos\\theta+\\sin\\theta)^2$ is $2$, and the result follow. $$(x-y)^2\\geq0$$ $$2xy\\leq x^2+y^2$$ $$x^2+y^2+2xy\\leq 2x^2+2y^2$$ $$(x+y)^2\\leq 2x^2+2y^2$$ let investigate $$\\frac{(x+y)^2}{4}-\\frac{x^2+y^2}{2}=\\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\\frac{(x+y)^2}{4}-\\frac{x^2+y^2}{2}\\leq 0.$$ So $$\\frac{(x+y)^2}{4}\\leq\\frac{x^2+y^2}{2}.$$ Notice that $$(\\frac{1}{2}(x+y))^2=\\frac{1}{4}(x^2+2xy+y^2)\\leq \\frac{1}{4}(x^2+[x^2+y^2]+y^2)$$ (why?) • because x^2 + y^2>=2xy ! thanks – Sua Feb 5 '14 at 21:54", "sin\\theta}{2cos\\theta sin\\theta \\pi}=\\frac{1}{2\\pi}$ $PDF(\\phi)=\\frac{1}{2\\phi}$ Now we want to calculate the CDF of each function, $CDF(\\theta)=\\int_0^\\theta PDF(\\theta) d\\theta$ $CDF(\\theta)=\\int_0^\\theta 2cos(\\theta)sin(\\theta) d\\theta$ $CDF(\\theta)=\\int_0^\\theta sin(2\\theta) d\\theta$ $CDF(\\theta)=\\frac{1}{2}-\\frac{cos(2\\theta)}{2}$ $CDF(\\phi)=\\int_0^\\phi PDF(\\phi) d\\phi$ $CDF(\\phi)=\\int_0^\\phi\\frac{1}{2\\pi} d\\phi$ $CDF(\\phi)=\\frac{\\phi}{2\\pi}$ Now we want to invert our CDF to sample it using our random variable $y$, $y=CDF(\\theta)$ $y=\\frac{1}{2}-\\frac{cos(2\\theta)}{2}$ $\\frac{1}{2}-y=\\frac{cos(2\\theta)}{2}$ $1-2y=cos(2\\theta)$ $\\frac{cos^{-1}(1-2y)}{2}=\\theta$ (7) For $\\phi$, $y=CDF(\\phi)$ $y=\\frac{\\phi}{2\\pi}$ $y*2\\pi = \\phi$ (8) Now we have our CDFs and our PDFs, we can finally calculate our direction. In pseudo code you can simply do: $\\theta=\\frac{cos^{-1}(1-2rand01())}{2}$ $\\phi=rand01()*2\\pi$ With these directions, you can now sample your scene: $\\frac{SampleScene(SphericalTo3D(\\theta, \\phi))}{PDF(\\omega)}$ Plug in (5) $\\frac{SampleScene(SphericalTo3D(\\theta, \\phi))\\pi}{cos\\theta}$ ## Conclusion That’s it! These formulas will sample the hemisphere where it is receiving more light defined by the cosine lobe. The results are pretty awesome. ## Introduction Recently, I’ve been wrestling with attempting to define the mathematics behind the illumination of a triangular area light. Although I have not succeeded yet, I figured I would chronicle my recent attempts. Here goes! ## Premise I have recently been attempting to improve my understanding of perspective projection. This included a variety of topics such as deriving a perspective projection matrix and understanding interpolation of vertex shader outputs in perspective. However, one topic that evaded me, was the surprising result that depth is interpolated as 1/z instead of z.", "# How do you convert r = 2 sin theta into cartesian form? Jul 14, 2016 Make use of a few formulas and do some simplification. See below. #### Explanation: When dealing with transformations between polar and Cartesian coordinates, always remember these formulas: • $x = r \\cos \\theta$ • $y = r \\sin \\theta$ • ${r}^{2} = {x}^{2} + {y}^{2}$ From $y = r \\sin \\theta$, we can see that dividing both sides by $r$ gives us $\\frac{y}{r} = \\sin \\theta$. We can therefore replace $\\sin \\theta$ in $r = 2 \\sin \\theta$ with $\\frac{y}{r}$: $r = 2 \\sin \\theta$ $\\to r = 2 \\left(\\frac{y}{r}\\right)$ $\\to {r}^{2} = 2 y$ We can also replace ${r}^{2}$ with ${x}^{2} + {y}^{2}$, because ${r}^{2} = {x}^{2} + {y}^{2}$: ${r}^{2} = 2 y$ $\\to {x}^{2} + {y}^{2} = 2 y$ We could leave it at that, but if you're interested... Further Simplification If we subtract $2 y$ from both sides we end up with this: ${x}^{2} + {y}^{2} - 2 y = 0$ Note that we can complete the square on ${y}^{2} - 2 y$: ${x}^{2} + \\left({y}^{2} - 2 y\\right) = 0$ $\\to {x}^{2} + \\left({y}^{2} - 2 y + 1\\right) = 0 + 1$ $\\to {x}^{2} + {\\left(y - 1\\right)}^{2} = 1$ And how about that! We end up", "# Prove $\\cos^2 x \\,\\sin^3 x=\\frac{1}{16}(2 \\sin x + \\sin 3x - \\sin 5x)$ How would I prove the following? $$\\cos^2 x \\,\\sin^3 x=\\frac{1}{16}(2 \\sin x + \\sin 3x - \\sin 5x)$$ I do not know how to do do the problem I do know $\\sin(3x)$ can be $\\sin(2x+x)$ and such yet I am not sure how to commence. - Using $\\sin 2\\theta=2\\sin\\theta\\cos\\theta$ and $\\sin^2\\theta=(1-\\cos2\\theta)/2$ we get $$\\cos^2x ~\\sin^3x=(\\cos x\\sin x)^2\\sin x=\\left(\\frac{\\sin 2x}{2}\\right)^2\\sin x=\\frac{1}{4}\\frac{1-\\cos 4x}{2}\\sin x \\tag{\\circ}$$ With the sum rule we have $\\sin(\\alpha\\pm\\beta)=\\sin\\alpha\\cos\\beta\\pm\\sin\\beta\\cos\\alpha.$ Therefore we have the identity $$\\frac{\\sin(\\alpha+\\beta)-\\sin(\\alpha-\\beta)}{2}=\\sin\\beta\\cos\\alpha.$$ Apply with $\\alpha=4x$ and $\\beta=x$ to $(\\circ)$. - If you know that $$\\sin(x+y)=\\sin x\\cos y+\\cos x\\sin y\\tag{1}$$ you can apply that to get $\\sin (2x)\\cos y + \\cos(2x)\\sin y$. Then use $(1)$ again to get $\\sin(2x)=\\sin(x+x)=\\cdots$ (I'll let you fill in the blanks. Then recall that $\\cos(x+y)=\\cos x\\cos y-\\sin x\\sin y$, and apply that to get $\\cos(2x)=\\cos(x+x)=\\cdots$ (fill in the blanks here as well). Then $\\sin(5x) = \\sin(3x+2x)=\\cdots$ (and apply $(1)$ again here), and then you have to work with $\\sin(3x)$ and $\\sin(2x)$ again, and with $\\cos(3x)$ and $\\cos(2x)$. Use the identities for sine of a sum and cosine of a sum to reduce those further. Eventually the only thing you'll be taking sines and cosines of is $x$, and then you should be done. - This is", "Step 3: Simplify the equation. Let's look at some examples. Rewrite the equation $$r = 3\\sin\\theta$$ in rectangular form. Step 1: The first step is to rewrite it. Using the identity $$x = r\\sin\\theta$$, which is equivalent to the expression $\\sin\\theta = \\frac{x}{r},$ you can write $r = 3\\frac{x}{r} .$ Step 2: Multiplying both sides by $$r$$ gives you $r^2 = 3x.$ Then you can use the identity $$x^2 + y^2 = r^2$$ to write $x^2 + y^2 = 3x.$ Step 3: This is already simplified, so you are done! Let's look at an example where you actually need to simplify a little. Rewrite the equation $r = \\frac{5}{2+3\\sin\\theta }$ in rectangular form. Step 1: Rewrite all the $$\\theta$$ terms. In this case, you only need to use $\\sin\\theta = \\frac{y}{r}$ to get $r = \\frac{5}{2+3\\frac{y}{r} }.$ Step 2: Next, rearrange the equation, multiplying both sides by the denominator to get the $$r$$ terms on the same side of the equation $r\\left(2+3\\frac{y}{r} \\right) = 5,$ or $2r + 3y = 5.$ You can solve this for $$r$$ to get $2r = 5-3y.$ You really wish that was an $$r^2$$ on the left side because then you would be almost done. But it isn't, so you need to square both sides of the equation to get \\begin{align} 4r^2 &=", "$x_\\theta = -y , y_\\theta = x$ $\\frac{df}{d\\theta} = f_x (-y) + f_y (x)$ simplify it and find the sum you will get the solution $\\sin ^2 x + \\cos ^2 x =1$ use this 7. $x= e^r \\cos \\theta$ $y=e^r \\sin \\theta$ $x_\\theta = \\frac{\\partial x}{\\partial \\theta } = e^r (-\\sin \\theta) = -e^r\\sin \\theta = -y$ $y_\\theta = \\frac{\\partial y}{\\partial \\theta } = e^r \\cos \\theta = x$ 8. er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ 9. Originally Posted by billym er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ $f_r = f_x \\cdot x_r + f_y \\cdot y_r$ $f_{rr} = (f_{xx}\\cdot x_r)x_r + f_x\\cdot x_{rr} + (f_{yy} \\cdot y_r)y_r + f_y \\cdot y_{rr}$ now $x_{rr} = x_r=x , y_{rr} = y_r=y$ $f_{rr} = f_{xx}\\cdot x^2 + f_x \\cdot x + f_{yy}\\cdot y^2 + f_y \\cdot y$ .................(1) other one I will symbol theta with t ok $f_{t} = f_x \\cdot x_t + f_y \\cdot y_t$ $f_{tt}= (f_{xx} x_t)x_t + f_x \\cdot x_{tt} + (f_{yy} \\cdot y_t)y_t + f_y \\cdot y_{tt}$ $x_t = -y , y_t = x$ and $x_{tt} = -x , y_{tt} = -y$ $f_{tt} = f_{xx} \\cdot y^2 + f_x \\cdot -x + f_{yy} \\cdot x^2 + f_y \\cdot -y$.........(2) sub x value and y value in 1,2 and", "$x = -\\frac{1}{8}$. Putting the equations as $y$ in terms of $x$ gives: $y = \\pm \\sqrt{1 - x^2}$ and $y = \\pm \\sqrt{\\frac{9}{4} - (x - 1)^2}$. Since these are symmetrical about the $x$ axis, we can work out the areas enclosed by the above semicircles, and then double it. So we can finally say: $A = 2\\left(\\int_{-\\frac{1}{2}}^{-\\frac{1}{8}}{\\sqrt{\\frac{9}{4} - (x - 1)^2}\\,dx} + \\int_{-\\frac{1}{2}}^{1}{\\sqrt{1 - x^2}\\,dx}\\right)$ You will need to use trigonometric substitution to solve these. Try $x - 1 = \\frac{3}{2}\\sin{\\theta}$ for the first and $x = \\sin{\\theta}$ for the second. 3. Originally Posted by Prove It It always helps to draw a picture. You can have the second circle be centred anywhere on the first, I've just chosen the point $(x, y) = (1, 0)$. It's important to note that the equations of the circles are $x^2 + y^2 = 1$ and $(x - 1)^2 + y^2 = \\frac{9}{4}$. By subtracting the first equation from the second we find $(x - 1)^2 + y^2 - (x^2 + y^2) = \\frac{9}{4} - 1$ $x^2 - 2x + 1 + y^2 - x^2 - y^2 = \\frac{5}{4}$ $-2x + 1 = \\frac{5}{4}$ $-2x = \\frac{1}{4}$ $x = -\\frac{1}{8}$. So the equations intersect at $x = -\\frac{1}{8}$. Putting the equations as $y$ in terms of $x$ gives: $y" ]

[ "# 2012 AIME II Problems/Problem 9 ## Problem 9 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$. The value of $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y}$ can be expressed in the form $\\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. ### The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\\sin y \\cos y} = \\frac{\\sin x}{\\sin y}\\cdot\\frac{\\cos x}{\\cos y}=3\\cdot\\frac{1}{2} = \\frac{3}{2}$ ### The Second Term Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2. Hungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$, we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$. Equation 1 then becomes: $\\sin x = 3\\sin y$. Equation 2 then becomes: $\\cos x = \\frac{1}{2} \\cos y$ Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ Applying the", "# Difference between revisions of \"2012 AIME II Problems/Problem 9\" ## Problem 9 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$. The value of $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y}$ can be expressed in the form $\\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. ### The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\\sin y \\cos y} = \\frac{\\sin x}{\\sin y}\\cdot\\frac{\\cos x}{\\cos y}=3\\cdot\\frac{1}{2} = \\frac{3}{2}$ ### The Second Term Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2. Hungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$, we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$. Equation 1 then becomes: $\\sin x = 3\\sin y$. Equation 2 then becomes: $\\cos x = \\frac{1}{2} \\cos y$ Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: $1 = 9\\sin^2 y + \\frac{1}{4}", "# Convert the polar equation to rectangular form. ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. LCKurtz Homework Helper Gold Member ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. You are making it way too difficult. Put ##\\sin\\theta =\\frac y r## in your first equation then see if you can get an xy equation from that. Dick Homework Helper ## Homework Statement r = 3sin$\\theta$ since x= rcos$\\theta$ x = 3sin$\\theta$cos$\\theta$ and since: y = rsin$\\theta$ y = 3$sin^2\\theta$ Then I'm sort of stuck.. Try putting sinθ=y/r. You are making it way too difficult. Put ##\\sin\\theta =\\frac y r## in your first equation then see if you can get an xy equation from that. Like so: r = 3sinθ becomes (r=3$\\frac{y}{r}$) = (r =$\\sqrt{3y}$) ? or solving for y, (y = $\\frac{r^2}{3}$) or do you mean something like this: ($\\frac{y}{r} = sinθ$) becomes ($\\frac{y}{3sinθ} = sinθ$) = (y = 3sin$^{2}θ$) Dick Homework Helper Like so: r = 3sinθ becomes (r=3$\\frac{y}{r}$) = (r =$\\sqrt{3y}$) ? or solving for y, (y = $\\frac{r^2}{3}$) or do you mean something like this: ($\\frac{y}{r} = sinθ$) becomes ($\\frac{y}{3sinθ} =", "left-hand side. Let’s combine the right-hand side by giving them same denominator. $\\begin{array}{|c} \\frac{1}{\\sin x}- \\frac{\\sin^2x}{\\sin x} \\\\\\frac{1- \\sin^2x}{\\sin x} \\\\\\frac{\\cos^2x}{\\sin x}\\end{array}$ The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, $\\sin^2 \\theta+\\cos^2 \\theta=1$ , and isolated the $\\cos^2x=1- \\sin^2x$ . There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to $\\frac{\\cos^2x}{\\sin^2x}$ , we could have also substituted the identity $\\cot^2x=\\csc^2x-1$ . #### Example B Verify the identity $\\frac{\\sin x}{1- \\cos x}=\\frac{1+ \\cos x}{\\sin x}$ . Solution: Multiply the left-hand side of the equation by $\\frac{1+ \\cos x}{1+ \\cos x}$ . $\\frac{\\sin x}{1- \\cos x}&= \\frac{1+ \\cos x}{\\sin x} \\\\\\frac{1+ \\cos x}{1+ \\cos x} \\cdot \\frac{\\sin x}{1- \\cos x}&= \\\\\\frac{\\sin \\left(1+\\cos x\\right)}{1- \\cos^2x}&= \\\\\\frac{\\sin \\left(1+\\cos x\\right)}{\\sin^2x}&= \\\\\\frac{1+\\cos x}{\\sin x}&=$ The two sides are the same, so we are done. #### Example C Verify the identity $\\sec(-x)=\\sec x$ . Solution: Change secant to cosine. $\\sec(-x)= \\frac{1}{\\cos \\left(-x\\right)}$ From the Negative Angle Identities, we know that $\\cos (-x)=\\cos x$ . $&=\\frac{1}{\\cos x} \\\\&=\\sec x$ Concept Problem Revisit Start by simplifying the lefthand side of the equation. $\\frac{\\sin^2x}{\\tan^2x}=\\frac{\\sin^2x}{\\frac{\\sin^2x}{\\cos^2x}}\\\\=\\cos^2x$ . Now simplify the righthand side of the equation. By", "# Proving $\\sin x + \\sin x \\cdot \\cot^2 x = \\csc x$ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process: \\begin{align*} \\tan \\theta &= \\frac{\\sin \\theta}{\\cos \\theta}\\\\ (\\sin \\theta)^2 + (\\cos \\theta)^2 &= 1\\\\ \\csc \\theta &= \\frac{1}{\\sin \\theta}\\\\ \\sec \\theta &= \\frac{1}{\\cos \\theta}\\\\ \\cot \\theta &= \\frac{1}{\\tan \\theta}. \\end{align*} The trig identity given is: $$\\sin x + \\sin x \\cdot \\cot^2 x = \\csc x$$ I simplified it to: $$\\sin x + \\sin x \\cdot \\left(\\frac{\\cos x}{\\sin x}\\right)^2 = \\frac{1}{\\sin x}$$ After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated. - Multiply both sides by $\\sin x$ and see what you get. – Karolis Juodelė Jun 1 '12 at 19:03 Arturo, you are simply too quick at editing. You finished mere seconds before I was done. – Joe Jun 1 '12 at 19:04 you are at last step: $\\frac {1}{sinx}$=Cosecx – iostream007 May 10 '13 at 4:41 Taking the left hand side, try cancelling the $\\sin^2{x}$ with the $\\sin{x}$ in the numerator (i.e. $\\sin{x} .(\\frac{\\cos^2{x}}{\\sin^2{x}}) = \\frac{\\cos^2{x}}{\\sin{x}}$) Then add the remaining two terms ($\\sin{x} + \\frac{\\cos^2{x}}{\\sin{x}}$) and use the", "\\end{equation*} Equating the real parts we see that \\begin{equation*} \\frac{1}{2}+\\cos(\\theta)+\\cos(2\\theta)+\\cdots =0. \\end{equation*} Replacing $\\theta$ with $\\theta+\\pi$ & using the trigonometric addition formulae we get \\begin{equation*} \\frac{1}{2}-\\cos(\\theta)+\\cos(2\\theta)-\\cdots =0. \\end{equation*} Integrate from $\\theta=0$ to $\\theta=\\phi$ & write $\\theta$ again for $\\phi$ we obtain \\begin{equation*} \\sin(\\theta)-\\frac{1}{2}\\sin(2\\theta)+\\frac{1}{3}\\sin(3\\theta)-\\cdots =\\frac{1}{2}\\theta,~-\\pi<\\theta<\\pi \\end{equation*} & this series is convergent. Integrating again gives \\begin{equation*} 1-\\cos(\\theta)-\\frac{1-\\cos(2\\theta)}{2^2}+\\frac{1-\\cos(3\\theta)}{3^2}-\\cdots =\\frac{1}{4}\\theta^2. \\end{equation*} Setting $\\theta=\\pi$ gives \\begin{equation*} 1+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots =\\frac{1}{8}\\pi^2. \\end{equation*} Since \\begin{equation*} 1+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots =1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots -\\frac{1}{2^2}-\\frac{1}{4^2}-\\cdots =(1-\\frac{1}{4})(1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots), \\end{equation*} we deduce that \\begin{equation*} 1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots = \\sum^{\\infty}_{n=1} \\frac{1}{n^2} =\\frac{1}{6}\\pi^2. \\end{equation*} Remark: We also get the formula \\begin{equation*} 1-\\frac{1}{2^2}+\\frac{1}{3^2}-\\cdots =\\frac{1}{12}\\pi^2 \\end{equation*}", "linear, so we may easily solve for the parameter $t$: $t=\\frac{y}{7}$ Now, if we substitute for $t$ into the equation for $x$ we get an equation that relates $x$ and $y$ where $t$ has been \"eliminated.\" We then have a Cartesian equation: $x=2\\left(\\frac{y}{7} \\right)^2+3\\left(\\frac{y}{7} \\right)+5$ $x=\\frac{2}{49}y^2+\\frac{3}{7}y+5$ 2.) Suppose you are given the parametric equations: $x(t)=a\\cdot\\cos(t)$ $y(t)=b\\cdot\\sin(t)$ Rather than solving one of the equations for $t$, we may rewrite the equations as: $\\frac{x}{a}=\\cos(t)$ $\\frac{y}{b}=\\sin(t)$ If we square both equations, we may then add and take advantage of the Pythagorean identity $\\sin^2(\\theta)+\\cos^2(\\theta)=1$ $\\left(\\frac{x}{a} \\right)^2=\\cos^2(t)$ $\\left(\\frac{y}{b} \\right)^2=\\sin^2(t)$ $\\left(\\frac{x}{a} \\right)^2+\\left(\\frac{y}{b} \\right)^2=\\cos^2(t)+\\sin^2(t)$ $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$", "equation with only one trigonometric function: $y=a(1)+(b-a)\\sin^2 x$ $y=a+(b-a)\\sin^2 x$ and now we can readily invert it: $\\sin^2 x=\\frac{y-a}{b-a}$ $x=\\arcsin\\sqrt{\\frac{y-a}{b-a}}$ Now it’s time for the second one: $y=a\\cos x+b\\sin^2 x$ We will again use the pythagorean identity, but in a different way. This time, we will use the fact that since $\\cos^2\\theta+\\sin^2\\theta=1$ then $\\sin^2\\theta=1-\\cos^2\\theta$ and so, by substitution, our original equation can be turned into $y=a\\cos x+b(1-\\cos^2x)$ $y=a\\cos x+b-b\\cos^2x$ $b\\cos^2 x-a\\cos x-b+y$ And we have just reduced it to a quadratic, which can be solved using the quadratic formula: $\\cos x=\\frac{a\\pm\\sqrt{a^2-4b(y-b)}}{2b}$ $x=\\arccos \\frac{a\\pm\\sqrt{a^2-4b(y-b)}}{2b}$ Now for the third equation: $y=a\\cos x+b\\sin x$ Though it seems much simpler than the other two, this one is actually much more difficult. It, too, involves a “trick” and a trigonometric identity, but it does not use the pythagorean identity. Before I give it away, you should try this problem - once you discover the trick, the result can be very satisfying. To solve this, you must convert the equation into this equivalent equation: $\\frac{y}{\\sqrt{a^2+b^2}}=\\frac{a}{\\sqrt{a^2+b^2}}\\cos x+\\frac{b}{\\sqrt{a^2+b^2}}\\sin x$ Now we will need to use another identity - one that I did not mention at the beginning of the post: $\\sin \\arccos \\theta=\\sqrt{1-\\theta^2}$ which means that $\\sin \\arccos \\frac{b}{\\sqrt{a^2+b^2}}=\\sqrt{1-\\frac{b^2}{a^2+b^2}}=\\frac{a}{\\sqrt{a^2+b^2}}$ And so we can again transform our equation into the equivalent equation $\\frac{y}{\\sqrt{a^2+b^2}}=\\sin\\bigg(\\arccos\\frac{b}{\\sqrt{a^2+b^2}}\\bigg)\\cos x+\\cos\\bigg(\\arccos\\frac{b}{\\sqrt{a^2+b^2}}\\bigg)\\sin x$ Now recall", "y_2 = y$ ;) The vector form is more general (because it works on all inner product spaces, not only $\\mathbb R^n$) – AlexR Feb 5 '14 at 22:29 • @AlexR Yes, you are right $y1=x, y_2 = y$. I might consider getting more familiar with vector form, because as you've mentioned it has much more wider use. – Stefan4024 Feb 5 '14 at 22:40 $\\frac12 (x^2+y^2)-(\\frac12 (x+y))^2$ $=\\frac12 (x^2+y^2)-\\frac14 (x^2+y^2+2xy)$ $=\\frac14 (2x^2+2y^2-x^2-y^2-2xy)$ $=\\frac14 (x^2+y^2-2xy)$ $=\\frac14 (x-y)^2\\ge 0$ This substitution is a nice way to do this: $$x=r\\sin\\theta , y=r\\cos\\theta$$ So, our identity: $$\\left(\\frac12(x+y)\\right)^2 \\leq \\frac12(x^2 + y^2)$$ $$\\frac{1}{4} (x^2+y^2+2xy) \\leq \\frac12(x^2 + y^2)$$ Applying our substitutions: $$\\frac{1}{4} (r^2+2\\cdot r\\sin\\theta\\cdot r\\cos\\theta) \\leq \\frac12(r^2)$$ $$\\frac{r^2}{2}\\cdot \\frac{1+\\sin2\\theta}{2} \\leq \\frac{r^2}{2}$$ Now, if you note that the maximum value of $\\sin 2\\theta$ is $1$, then it becomes obvious. Added: It could have also been done it this way: $$\\left(\\frac{1}{2}(x+y)\\right)^2 \\leq \\frac12(x^2 + y^2)$$ $$\\frac{1}{2}\\cdot\\frac{1}{2}(r\\cos\\theta+r\\sin\\theta)^2 \\leq \\frac{r^2}{2}$$ $$\\frac{r^2}{2}\\cdot\\frac{1}{2}(\\cos\\theta+\\sin\\theta)^2 \\leq \\frac{r^2}{2}$$ Now, note that $-\\sqrt{2} \\leq \\cos\\theta+\\sin\\theta = \\sqrt{2}\\cos(x+45^{\\circ}) \\leq \\sqrt{2}$. Thus the maximum value of $(\\cos\\theta+\\sin\\theta)^2$ is $2$, and the result follow. $$(x-y)^2\\geq0$$ $$2xy\\leq x^2+y^2$$ $$x^2+y^2+2xy\\leq 2x^2+2y^2$$ $$(x+y)^2\\leq 2x^2+2y^2$$ let investigate $$\\frac{(x+y)^2}{4}-\\frac{x^2+y^2}{2}=\\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\\frac{(x+y)^2}{4}-\\frac{x^2+y^2}{2}\\leq 0.$$ So $$\\frac{(x+y)^2}{4}\\leq\\frac{x^2+y^2}{2}.$$ Notice that $$(\\frac{1}{2}(x+y))^2=\\frac{1}{4}(x^2+2xy+y^2)\\leq \\frac{1}{4}(x^2+[x^2+y^2]+y^2)$$ (why?) • because x^2 + y^2>=2xy ! thanks – Sua Feb 5 '14 at 21:54", "### Home > CCA2 > Chapter 12 > Lesson 12.2.2 > Problem12-118 12-118. 1. Prove that each of the following equations is an identity. In other words, prove that they are each true for all values for which the functions are defined. Homework Help ✎ 1. (sin θ + cos θ)2 = 1 + 2 sin θ cos θ 2. tan θ + cot θ = sec θ csc θ 3. (tan θ cos θ)(sin2θ + ) = sin θ Start by expanding the left side of the equation. You might have to rearrange the equation to see ways to simplify it after you expand it. Change the trigonometric identities on the left side of the equation into the ratios they represent. $\\frac{\\sin\\theta}{\\cos\\theta}+\\frac{\\cos\\theta}{\\sin\\theta}=$ Find a common denominator and then add the two terms. $\\frac{\\sin^{2}\\theta}{\\cos\\theta\\sin\\theta}+\\frac{\\cos^{2}\\theta}{\\cos\\theta\\sin\\theta}=$ Remember that sin²(θ) + cos²(θ) = 1, and rewrite the equation. $\\frac{1}{\\cos\\theta\\sin\\theta}=$ Substitute in the identities for the two terms. What does 1/cosθ equal? 1/sinθ? Substitute the trig identities in first, then expand by multiplying the two terms together. Remember to use Giant Ones to simplify when you can. Use methods from both parts (a) and (b).", "# Prove $\\cos^2 x \\,\\sin^3 x=\\frac{1}{16}(2 \\sin x + \\sin 3x - \\sin 5x)$ How would I prove the following? $$\\cos^2 x \\,\\sin^3 x=\\frac{1}{16}(2 \\sin x + \\sin 3x - \\sin 5x)$$ I do not know how to do do the problem I do know $\\sin(3x)$ can be $\\sin(2x+x)$ and such yet I am not sure how to commence. - Using $\\sin 2\\theta=2\\sin\\theta\\cos\\theta$ and $\\sin^2\\theta=(1-\\cos2\\theta)/2$ we get $$\\cos^2x ~\\sin^3x=(\\cos x\\sin x)^2\\sin x=\\left(\\frac{\\sin 2x}{2}\\right)^2\\sin x=\\frac{1}{4}\\frac{1-\\cos 4x}{2}\\sin x \\tag{\\circ}$$ With the sum rule we have $\\sin(\\alpha\\pm\\beta)=\\sin\\alpha\\cos\\beta\\pm\\sin\\beta\\cos\\alpha.$ Therefore we have the identity $$\\frac{\\sin(\\alpha+\\beta)-\\sin(\\alpha-\\beta)}{2}=\\sin\\beta\\cos\\alpha.$$ Apply with $\\alpha=4x$ and $\\beta=x$ to $(\\circ)$. - If you know that $$\\sin(x+y)=\\sin x\\cos y+\\cos x\\sin y\\tag{1}$$ you can apply that to get $\\sin (2x)\\cos y + \\cos(2x)\\sin y$. Then use $(1)$ again to get $\\sin(2x)=\\sin(x+x)=\\cdots$ (I'll let you fill in the blanks. Then recall that $\\cos(x+y)=\\cos x\\cos y-\\sin x\\sin y$, and apply that to get $\\cos(2x)=\\cos(x+x)=\\cdots$ (fill in the blanks here as well). Then $\\sin(5x) = \\sin(3x+2x)=\\cdots$ (and apply $(1)$ again here), and then you have to work with $\\sin(3x)$ and $\\sin(2x)$ again, and with $\\cos(3x)$ and $\\cos(2x)$. Use the identities for sine of a sum and cosine of a sum to reduce those further. Eventually the only thing you'll be taking sines and cosines of is $x$, and then you should be done. - This is", "\\cos^2 y$ Applying the identity $\\cos^2 y = 1 - \\sin^2 y$ (which is similar to $\\sin^2\\theta + \\cos^2\\theta = 1$ but a bit different), we can change $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ into: $1 = 9\\sin^2 y + \\frac{1}{4} - \\frac{1}{4} \\sin^2 y$ Rearranging, we get $\\frac{3}{4} = \\frac{35}{4} \\sin^2 y$. So, $\\sin^2 y = \\frac{3}{35}$. Squaring Equation 1 (leading to $\\sin^2 x = 9\\sin^2 y$), we can solve for $\\sin^2 x$: $\\sin^2 x = 9\\left(\\frac{3}{35}\\right) = \\frac{27}{35}$ Using the identity $\\cos 2\\theta = 1 - 2\\sin^2\\theta$, we can solve for $\\frac{\\cos 2x}{\\cos 2y}$. $\\cos 2x = 1 - 2\\sin^2 x = 1 - 2\\cdot\\frac{27}{35} = 1 - \\frac{54}{35} = -\\frac{19}{35}$ $\\cos 2y = 1 - 2\\sin^2 y = 1 - 2\\cdot\\frac{3}{35} = 1 - \\frac{6}{35} = \\frac{29}{35}$ Thus, $\\frac{\\cos 2x}{\\cos 2y} = \\frac{-19/35}{29/35} = -\\frac{19}{29}$. ### Now Back to the Solution! Finally, $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y} = \\frac32 + \\left(-\\frac{19}{29} \\right) = \\frac{49}{58}$. So, the answer is $49+58=\\boxed{107}$.", "identity $\\cos^2 y = 1 - \\sin^2 y$ (which is similar to $\\sin^2\\theta + \\cos^2\\theta = 1$ but a bit different), we can change $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ into: $1 = 9\\sin^2 y + \\frac{1}{4} - \\frac{1}{4} \\sin^2 y$ Rearranging, we get $\\frac{3}{4} = \\frac{35}{4} \\sin^2 y$. So, $\\sin^2 y = \\frac{3}{35}$. Squaring Equation 1 (leading to $\\sin^2 x = 9\\sin^2 y$), we can solve for $\\sin^2 x$: $\\sin^2 x = 9\\left(\\frac{3}{35}\\right) = \\frac{27}{35}$ Using the identity $\\cos 2\\theta = 1 - 2\\sin^2\\theta$, we can solve for $\\frac{\\cos 2x}{\\cos 2y}$. $\\cos 2x = 1 - 2\\sin^2 x = 1 - 2\\cdot\\frac{27}{35} = 1 - \\frac{54}{35} = -\\frac{19}{35}$ $\\cos 2y = 1 - 2\\sin^2 y = 1 - 2\\cdot\\frac{3}{35} = 1 - \\frac{6}{35} = \\frac{29}{35}$ Thus, $\\frac{\\cos 2x}{\\cos 2y} = \\frac{-19/35}{29/35} = -\\frac{19}{29}$. ### Now Back to the Solution! Finally, $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y} = \\frac32 + \\left(-\\frac{19}{29} \\right) = \\frac{49}{58}$. So, the answer is $49+58=\\boxed{107}$. ## Solution 2 As mentioned above, the first term is clearly $\\frac{3}{2}.$ For the second term, we first wish to find $\\frac{\\cos 2x}{\\cos 2y} =\\frac{2\\cos^2 x - 1}{2 \\cos^2y -1}.$ Now we first square the first equation getting $\\frac{\\sin^2x}{\\sin^2y} =\\frac{1-\\cos^2x}{1 - \\cos^2y} =9.$ Squaring the second equation yields $\\frac{\\cos^2x}{\\cos^2y} =\\frac{1}{4}.$ Let $\\cos^2x = a$ and $\\cos^2y", "$x_\\theta = -y , y_\\theta = x$ $\\frac{df}{d\\theta} = f_x (-y) + f_y (x)$ simplify it and find the sum you will get the solution $\\sin ^2 x + \\cos ^2 x =1$ use this 7. $x= e^r \\cos \\theta$ $y=e^r \\sin \\theta$ $x_\\theta = \\frac{\\partial x}{\\partial \\theta } = e^r (-\\sin \\theta) = -e^r\\sin \\theta = -y$ $y_\\theta = \\frac{\\partial y}{\\partial \\theta } = e^r \\cos \\theta = x$ 8. er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ 9. Originally Posted by billym er... could you start me off on this one? $f_{rr}+f_{\\theta\\theta}=e^{2r}(f_{xx}+f_{yy})$ $f_r = f_x \\cdot x_r + f_y \\cdot y_r$ $f_{rr} = (f_{xx}\\cdot x_r)x_r + f_x\\cdot x_{rr} + (f_{yy} \\cdot y_r)y_r + f_y \\cdot y_{rr}$ now $x_{rr} = x_r=x , y_{rr} = y_r=y$ $f_{rr} = f_{xx}\\cdot x^2 + f_x \\cdot x + f_{yy}\\cdot y^2 + f_y \\cdot y$ .................(1) other one I will symbol theta with t ok $f_{t} = f_x \\cdot x_t + f_y \\cdot y_t$ $f_{tt}= (f_{xx} x_t)x_t + f_x \\cdot x_{tt} + (f_{yy} \\cdot y_t)y_t + f_y \\cdot y_{tt}$ $x_t = -y , y_t = x$ and $x_{tt} = -x , y_{tt} = -y$ $f_{tt} = f_{xx} \\cdot y^2 + f_x \\cdot -x + f_{yy} \\cdot x^2 + f_y \\cdot -y$.........(2) sub x value and y value in 1,2 and", "$$4y^2=y^2+1$$ and get $$y^2=\\frac{1}{3}$$ Now, the equation $y^2=\\frac13$ has two solutions, but since you got that equation by squaring some other equation, you should know that some of the solutions of the second equation may not solve the first equation. In your case, $y=\\frac{1}{\\sqrt{3}}$ solves the equation, because $$2\\cdot \\frac1{\\sqrt{3}}=\\sqrt{\\left(\\frac{1}{\\sqrt{3}}\\right)^2+1}$$ but $y=-\\frac{1}{\\sqrt{3}}$ does not solve the equation, because $$2\\cdot\\left(- \\frac1{\\sqrt{3}}\\right)\\neq 2\\cdot\\frac1{\\sqrt{3}}=\\sqrt{\\left(-\\frac{1}{\\sqrt{3}}\\right)^2+1}$$ Here I think is your mistake: from $\\sqrt{x^2+y^2}=2i(x-yi+1)$ you get $\\sqrt{x^2+y^2}=2y$ therefore $y \\ge 0$. So after finding out $x, y$ you must filter out the solutions that doesn't satisfy $y \\ge 0$ Here is the solution by a different means... Let $z=re^{i\\theta}$. Then \\begin{align} |z| &=r=2i(re^{-i\\theta}+1)\\\\ &=2i(r\\cos\\theta-i\\sin\\theta+1) \\end{align} From the real and imaginary parts we have $$0=r\\cos\\theta+1\\\\ r=2r\\sin\\theta$$ From the second equation we get $\\theta=\\pi/6$ or $\\theta=5\\pi/6$. However, we must choose $\\theta=-5\\pi/6$ to assure that $r$ is positive. Thus, $$r=-\\frac{1}{\\cos\\theta}=\\frac{2}{\\sqrt{3}}$$ and finally, \\begin{align} z &=re^{i\\theta}\\\\ &=\\frac{2}{\\sqrt{3}e^{i5\\pi/3}}\\\\ &=\\frac{2}{\\sqrt{3}}\\left(-\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)\\\\ &=-1+\\frac{i}{\\sqrt{3}} \\end{align} in agreement with the other solutions.", "the left-hand side. Let’s combine the right-hand side by giving them same denominator. 1sinxsin2xsinx1sin2xsinxcos2xsinx$\\begin{array}{|c} \\frac{1}{\\sin x}- \\frac{\\sin^2x}{\\sin x} \\\\ \\frac{1- \\sin^2x}{\\sin x} \\\\ \\frac{\\cos^2x}{\\sin x}\\end{array}$ The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, sin2θ+cos2θ=1$\\sin^2 \\theta+\\cos^2 \\theta=1$ , and isolated the cos2x=1sin2x$\\cos^2x=1- \\sin^2x$ . There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to cos2xsin2x$\\frac{\\cos^2x}{\\sin^2x}$ , we could have also substituted the identity cot2x=csc2x1$\\cot^2x=\\csc^2x-1$ . #### Example B Verify the identity sinx1cosx=1+cosxsinx$\\frac{\\sin x}{1- \\cos x}=\\frac{1+ \\cos x}{\\sin x}$ . Solution: Multiply the left-hand side of the equation by 1+cosx1+cosx$\\frac{1+ \\cos x}{1+ \\cos x}$ . \\frac{\\sin x}{1- \\cos x}&= \\frac{1+ \\cos x}{\\sin x} \\\\ \\frac{1+ \\cos x}{1+ \\cos x} \\cdot \\frac{\\sin x}{1- \\cos x}&= \\\\ \\frac{\\sin \\left(1+\\cos x\\right)}{1- \\cos^2x}&= \\\\ \\frac{\\sin \\left(1+\\cos x\\right)}{\\sin^2x}&= \\\\ \\frac{1+\\cos x}{\\sin x}&= The two sides are the same, so we are done. #### Example C Verify the identity sec(x)=secx$\\sec(-x)=\\sec x$ . Solution: Change secant to cosine. sec(x)=1cos(x) From the Negative Angle Identities, we know that cos(x)=cosx$\\cos (-x)=\\cos x$ . &=\\frac{1}{\\cos x} \\\\ &=\\sec x Concept Problem Revisit Start by simplifying the left-hand side of the equation. sin2xtan2x=sin2xsin2xcos2x=cos2x . Now simplify the", "$$e^{ix}= 1+ ix+ \\frac{(ix)^2}{2}+ \\frac{(ix)^3}{3!}+ \\cdot\\cdot\\cdot+ \\frac{(ix)^n}{n!}+ \\cdot\\cdot\\cdot$$ Now, it is easy to see that i2= -1, i3= -i, i4= 1, etc. s0 $$e^{ix}= 1+ ix- \\frac{x^2}{2}- i\\frac{x^3}{3!}+ \\cdot\\cdot\\cdot+ i^n\\frac{x^n}{n!}$$ and, separating real and imaginary parts, $$e^{ix}= \\left(1- \\frac{x^2}{2}+ \\frac{x^4}{4!}+ \\cdot\\cdot\\cdot\\right)$$$$+ i\\left(x- \\frac{x^3}{3!}+ \\cdot\\cdot\\cdot\\right)$$ which are just the McLaurin series for cos(x) and sin(x): $$e^{ix}= cos(x)+ i sin(x)$$ We then have ez= ex+ iy= exeiy= ex(cos(y)+ i sin(y)). Of course, the fact that $z= x+ iy= r(cos(\\theta)+ i sin(\\theta))$ comes immediately from conversion of polar coordinates to Cartesian coordinates: $x= r cos(\\theta)$, $y= r sin(\\theta)$.", "sin\\theta}{2cos\\theta sin\\theta \\pi}=\\frac{1}{2\\pi}$ $PDF(\\phi)=\\frac{1}{2\\phi}$ Now we want to calculate the CDF of each function, $CDF(\\theta)=\\int_0^\\theta PDF(\\theta) d\\theta$ $CDF(\\theta)=\\int_0^\\theta 2cos(\\theta)sin(\\theta) d\\theta$ $CDF(\\theta)=\\int_0^\\theta sin(2\\theta) d\\theta$ $CDF(\\theta)=\\frac{1}{2}-\\frac{cos(2\\theta)}{2}$ $CDF(\\phi)=\\int_0^\\phi PDF(\\phi) d\\phi$ $CDF(\\phi)=\\int_0^\\phi\\frac{1}{2\\pi} d\\phi$ $CDF(\\phi)=\\frac{\\phi}{2\\pi}$ Now we want to invert our CDF to sample it using our random variable $y$, $y=CDF(\\theta)$ $y=\\frac{1}{2}-\\frac{cos(2\\theta)}{2}$ $\\frac{1}{2}-y=\\frac{cos(2\\theta)}{2}$ $1-2y=cos(2\\theta)$ $\\frac{cos^{-1}(1-2y)}{2}=\\theta$ (7) For $\\phi$, $y=CDF(\\phi)$ $y=\\frac{\\phi}{2\\pi}$ $y*2\\pi = \\phi$ (8) Now we have our CDFs and our PDFs, we can finally calculate our direction. In pseudo code you can simply do: $\\theta=\\frac{cos^{-1}(1-2rand01())}{2}$ $\\phi=rand01()*2\\pi$ With these directions, you can now sample your scene: $\\frac{SampleScene(SphericalTo3D(\\theta, \\phi))}{PDF(\\omega)}$ Plug in (5) $\\frac{SampleScene(SphericalTo3D(\\theta, \\phi))\\pi}{cos\\theta}$ ## Conclusion That’s it! These formulas will sample the hemisphere where it is receiving more light defined by the cosine lobe. The results are pretty awesome. ## Introduction Recently, I’ve been wrestling with attempting to define the mathematics behind the illumination of a triangular area light. Although I have not succeeded yet, I figured I would chronicle my recent attempts. Here goes! ## Premise I have recently been attempting to improve my understanding of perspective projection. This included a variety of topics such as deriving a perspective projection matrix and understanding interpolation of vertex shader outputs in perspective. However, one topic that evaded me, was the surprising result that depth is interpolated as 1/z instead of z.", "those are easier to work with than trig identities. So I converted $sin(\\theta)$ to $\\frac{e^{i\\theta} - e^{-i\\theta}}{2}$ and $cos(\\theta)$ to $\\frac{e^{i\\theta} + e^{-i\\theta}}{2i}$ to see if I could work it that way... After manipulation, it comes out to- $ e^{i\\theta} = \\frac{-igx \\frac{+}{} \\sqrt{4y^2v^4-g^2x^4-4x^2v^4}}{2v^2(y-ix)} $ $ \\theta = i ln(\\frac{-igx \\frac{+}{} \\sqrt{4y^2v^4-g^2x^4-4x^2v^4}}{2v^2(y-ix)}) $ Which has an imaginary part. When I plug in what the known values for x and y are for theta = $\\frac{\\pi}{4}$ and v = 100, this formula returns theta as imaginary, which is obviously incorrect since I know it should return $\\frac{\\pi}{4}$. Did I go wrong somewhere? This problem cannot be new- has anyone else solved this, or know of a way to solve it? Thanks for your help! 2. I always tell my students to solve the quadratic $ y = vt\\cdot sin(\\theta) - \\frac{gt^2}{2} $ Because this one will give you two solutions one being $t \\le 0$. If $t = 0,$ this is absolutely a solution, but it doesn't satisfy the $x$ equation, for $x>0$, so it must be thrown out. Clearly, $t < 0$ is not a solution. Anyway, that leaves $ t = \\dfrac{v\\cdot \\sin \\theta + \\sqrt{v^2\\cdot \\sin^2 \\theta - 2g\\cdot y}}{g}. $ Now plug this in for $t$ in the $x$ equation. 3. Ah, thanks so much.", "simplify sin^2 X + sec^2 X + tan^2 X + cos^2 X a.2 b. 1+sin^2X/cos^2X c.2sec^2X d.2tan^2X here is what i have done, first finding the common denominator. = sin^2 X + 1 + Sin^2X + cos^2X / cos^2X cos^2X = (sin^2X)(Cos^2X)+1+Sin^2X+(cos^2X)(cos^2X) / cos^2X then i got stuck.. 2. $y = 5\\sin{x}\\cos{x} + 1$ Use the double angle identity for sine: $\\sin{2\\theta} = 2\\sin{\\theta}\\cos{\\theta}$. $y = 5\\sin{x}\\cos{x} + 1$ $= \\frac{5}{2}\\cdot 2\\sin{x}\\cos{x} + 1$ $= \\frac{5}{2}\\sin{2x} + 1$ You should be able to go from here..." ]

[ "# Problem #142 142 What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2,-10,5),$ and the other on the sphere of radius 87 with center $(12,8,-16)$? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Categories ## Distance and Spheres | AIME I, 1987 | Question 2 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. ## Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information ### Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens ## Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.", "Categories Distance and Spheres | AIME I, 1987 | Question 2 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.", "Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information ### Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens ## Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137. Categories ## Arithmetic Mean | AIME I, 2015 | Question 12 Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean. ## Arithmetic Mean of Number Theory – AIME 2015 Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. • is 107 • is 431 • is 840 • cannot be determined from the given information ### Key Concepts Inequalities Algebra Number Theory AIME, 2015, Question 12 Elementary Number Theory by David Burton ##", "# Illumination Calculus Level 3 Two spheres of radii $2$ and $8$ have their centers separated by a distance of $18.$ If $a\\pi$ is the largest surface area that can be illuminated by a light source placed somewhere on the line joining their centers, what is $a?$ ×", "25 feet intersect so that the distance between their centers is 30 feet. What is the length of the side of the largest square inscribable within their intersecting arcs?", "# 300 on top of 3 100s Geometry Level 5 Three spheres of radius 100 are mutually tangent and rest on a horizontal plane H. A sphere of radius 300 rests over them. Let d be the largest possible distance of a point on the larger sphere from the plane H. What is $$\\lfloor d \\rfloor$$? ×", "to approximately $1.6 \\times 10^{15} m$ (when the two are at their furthest apart). This is a vast distance so you should begin to appreciate why it took so long for Cassini to get there.", "# Find the least distance of the plane $12x+4y+3z=327$ from the sphere $x^2+y^2+z^2+4x-2y-6z=155.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "### Home > CCG > Chapter 12 > Lesson 12.1.3 > Problem12-40 12-40. Cawker City, Kansas, claims to have the world’s largest ball of twine. Started in 1953 by Frank Stoeber, this ball has been created by wrapping more than $1300$ miles of twine. In fact, this giant ball has a circumference of $40$ feet. Assuming the ball of twine is a sphere, find the surface area and volume of the ball of twine. Find the radius of the ball by using the circumference. Use the radius to find the surface area. Use the radius to find the volume. $\\text{Volume}\\approx1080.0\\text{ ft}^3$", "## Saturday, July 6, 2013 ### ! \"Take a sphere of radius $R$ in $N$ dimensions, $N$ large; then most points inside the sphere are in fact very close to the surface.\"", "the distance between the centers of the circles is approximately $8.5$. If the circles would touch each other, the distance would be equal to $3 + 5 = 8$; the sum of the radii. Now we have a distance greater than $8$, so the circles do not touch or overlap.", "sphere of 1 cm or 99 cm radius centered at X.", "Browse Questions # Find the least distance of the plane $12x+4y+3z=327$ from the sphere $x^2+y^2+z^2+4x-2y-6z=155.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question?", "distance in a hypersphere, The Annals of Mathematical Statistics 21(1950), 447452.", "the distance between both of its extremities is 0 ? – Cham Oct 5 '16 at 15:18 The spatial component of the spacetime you describe is a 3-sphere. The largest distance is obtained when $\\Delta t=0$ so we can ignore the time direction of the spacetime. A 3-sphere is natural extension of a 2-sphere. Where a 2-sphere consists of a circle with radius $\\sin(\\theta)$ for every value of $\\theta$, a 3-sphere consists of 2-sphere with radius $\\sin(\\chi)$ for every value of $\\chi$. The largest distance on a 2-sphere is the distance from the north to the south-pole. This corresponds to varying $\\theta$ by $\\pi$. When the radius is $a$, this gives a distance of $\\pi a$. A similar reasoning lets you travel from one side of the 3-sphere to the other by varying $\\chi$ by $\\pi$. Here the distance is again $\\pi a$. The length of a straight path around the 3-sphere without a time-like component in this coordinate system is $2\\pi a$, just as for the 2-sphere. • Thanks, you then confirm what I was thinking : the longest linear structure in that space would have a length of $2 \\pi \\, a$ while both extremities are at a distance of 0. So the maximal distance (not length) is $\\pi \\, a$, and not $\\frac{\\pi}{2} \\; a$.", "8.74$ The diameter of the sphere measures 8.74 m.", "the circle B is away from its center by 6 point on the Perimeter of the circle B is away from (2,7) is$6 - 2.236$ $= 3.764$ Hence, the two circles donot overlap. Shortest distance is $3.764 - 2 = 1.764$", "as in the diagram above. The radius of the smaller one is $r$, while that of the larger one, $R$; their circumferences are $c$ and $C$ respectively. We draw two lines through the center ... Top 50 recent answers are included", "Staff Emeritus It was a pleasure TFM 15. Apr 19, 2008 ### kamerling Is the typical distance between 2 atoms not twice the radius of this sphere? First from one atom to where the spheres meet, then to the other atom. 16. Apr 19, 2008 ### Hootenanny Staff Emeritus Indeed it is, I assumed that TFM would have realised that." ]

[ "2 (Official MAA) Consider a cross section of the board and spheres with a plane that passes through the centers of the holes and centers of the spheres as shown. $[asy] unitsize(1.5 cm); pair A, B, C, D, E, F, G, P, Q; C = dir(175); D = dir(175 + 180); P = (-2,-0.8); Q = (2,-0.8); A = (C + reflect(P,Q)*(C))/2; B = (D + reflect(P,Q)*(D))/2; E = intersectionpoint(P--A, Circle(C,1)); F = intersectionpoint(B--Q, Circle(D,1)); G = (D + reflect(A,C)*(D))/2; draw(Circle(C,1)); draw(Circle(D,1)); draw(P--Q); draw(A--(C + (0,1))); draw(B--(D + (0,1))); draw(E--C--D--F); draw(D--G); dot(\"A\", A, NE); dot(\"B\", B, NW); dot(\"C\", C, NW); dot(\"D\", D, NE); dot(\"E\", E, SW); dot(\"F\", F, SE); dot(\"G\", G, SE); [/asy]$ Let $A$, $C$, and $E$ be, respectively, the center of the hole with radius $1,$ the center of the sphere resting in that hole, and a point on the edge of that hole. Let $B$, $D$, and $F$ be the corresponding points for the hole with radius $2.$ Let $G$ be the point on $\\overline{AC}$ such that $\\overline{AC} \\perp \\overline{GD}$. Let the radius of the spheres be $r = CE = DF$. Because $r^2 = AE^2 + AC^2 = 1 + AC^2$ and $r^2 = BF^2 + BD^2 = 4 + BD^2$, it follows that$$CG = AC - AG = AC - BD = \\sqrt{r^2", "# 2022 AIME I Problems/Problem 10 ## Problem Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. ## Diagrams $[asy] size(500); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,-8); OA = (-23.6643191,-4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"l\",(-42,1),N); label(\"A\",A,N); label(\"B\",B,N); label(\"O_A\",OA,S); label(\"O_B\",OB,S); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,-4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label(\"24\",midpoint(OA--OB),S); label(\"\\sqrt{560}\",midpoint(A--B),N); label(\"11\",midpoint(OA--(-33.9112699,0)),S); label(\"13\",midpoint(OB--(10.2469508,0)),S); label(\"r\",midpoint(midpoint(A--B)--A),N); label(\"r\",midpoint(midpoint(A--B)--B),N); label(\"r\",midpoint(A--(-33.9112699,0)),N); label(\"r\",midpoint(B--(10.2469508,0)),N); label(\"x\",midpoint(midpoint(B--OB)--OB),E); label(\"D\",midpoint(B--OB),E); [/asy]$ $[asy] size(500); pair A, C, OA, OC; C = (0,0); A = (-27.4954541697,0); OC = (0,-16); OA = (-27.4954541697,-4); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"l\",(-42,1),N); label(\"A\",A,N); label(\"C\",C,N); label(\"O_A\",OA,S); label(\"O_C\",OC,S); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,-4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label(\"30\",midpoint(OA--OC),S); label(\"11\",midpoint(OA--(-37.8877590151,0)),S); label(\"19\",midpoint(OC--(10.2469508,0)),E); label(\"r\",midpoint(midpoint(A--C)--A),N); label(\"r\",midpoint(midpoint(A--C)--C),N); label(\"r\",midpoint(A--(-37.8877590151,0)),N); label(\"r\",midpoint(C--(10.2469508,0)),N); label(\"E\",(0,-4),E); [/asy]$ ## Solution 1 We let $l$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \\sqrt{560}$. Since $ABO_BO_A$ is a trapezoid,", "# 2020 AIME I Problems/Problem 6 ## Problem A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ ## Solution $[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L, H); draw(b); label(\"R\", O -- Y, N); label(\"R\", Y -- P, N); label(\"R\", O -- A, NW); label(\"R\", P -- D, NE); label(\"1\", A -- H, N); label(\"2\", L -- D, N); label(\"7\", b, S); [/asy]$", "center $$(3, -2, 1)$$ and that goes through the point $$(4, 2, 5)$$. Ex 12.1.11 Find an equation of the sphere with center at $$(2,1,-1)$$ and radius 4. Find an equation for the intersection of this sphere with the $$y$$-\\)z\\) plane; describe this intersection geometrically. (answer) Ex 12.1.12 Consider the sphere of radius 5 centered at $$(2,3,4)$$. What is the intersection of this sphere with each of the coordinate planes? Ex 12.1.13 Show that for all values of $$\\theta$$ and $$\\phi$$, the point $$(a\\sin\\phi\\cos\\theta,a\\sin\\phi\\sin\\theta,a\\cos\\phi)$$ lies on the sphere given by $$x^2+y^2+z^2=a^2$$. Ex 12.1.14 Prove that the midpoint of the line segment connecting $$(x_1,y_1,z_1)$$ to $$(x_2,y_2,z_2)$$ is at $$\\ds\\left({x_1+x_2\\over 2},{y_1+y_2\\over 2},{z_1+z_2\\over 2}\\right)$$. Ex 12.1.15 Any three points $$P_1(x_1,y_1,z_1)$$, $$P_2(x_2,y_2,z_2)$$, $$P_3(x_3,y_3,z_3)$$, lie in a plane and form a triangle. The triangle inequality says that $$d(P_1,P_3)\\le d(P_1,P_2)+d(P_2,P_3)$$. Prove the triangle inequality using either algebra (messy) or the law of cosines (less messy). Ex 12.1.16 Is it possible for a plane to intersect a sphere in exactly two points? Exactly one point? Explain. ## 12.2: Vectors Ex 12.2.1 Draw the vector $$\\langle 3,-1\\rangle$$ with its tail at the origin. Ex 12.2.2 Draw the vector $$\\langle 3,-1,2\\rangle$$ with its tail at the origin. Ex 12.2.3 Let $${\\bf A}$$ be the vector with tail at the origin and head at $$(1,2)$$; let $${\\bf B}$$", "$O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\\angle O_A A C = \\angle O_C CA = 90^\\circ$. Therefore,\\begin{align*} AC^2 & = O_A O_C^2 - \\left( O_C C - O_A A \\right)^2 \\\\ & = \\boxed{\\textbf{(756) }} . \\end{align*} $\\textbf{FINAL NOTE:}$ In our solution, we do not use the conditio that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem. ~Steven Chen (www.professorcheneeu.com) ## Problem10 Three spheres with radii $11$$13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$$B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. ## Diagrams $[asy] size(500); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,-8); OA = (-23.6643191,-4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"l\",(-42,1),N); label(\"A\",A,N); label(\"B\",B,N); label(\"O_A\",OA,S); label(\"O_B\",OB,S); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,-4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label(\"24\",midpoint(OA--OB),S); label(\"\\sqrt{560}\",midpoint(A--B),N); label(\"11\",midpoint(OA--(-33.9112699,0)),S); label(\"13\",midpoint(OB--(10.2469508,0)),S); label(\"r\",midpoint(midpoint(A--B)--A),N); label(\"r\",midpoint(midpoint(A--B)--B),N); label(\"r\",midpoint(A--(-33.9112699,0)),N); label(\"r\",midpoint(B--(10.2469508,0)),N); label(\"x\",midpoint(midpoint(B--OB)--OB),E); label(\"D\",midpoint(B--OB),E); [/asy]$ $[asy] size(500); pair A, C, OA, OC; C = (0,0); A = (-27.4954541697,0); OC = (0,-16); OA = (-27.4954541697,-4); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"l\",(-42,1),N); label(\"A\",A,N); label(\"C\",C,N); label(\"O_A\",OA,S); label(\"O_C\",OC,S); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,-4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label(\"30\",midpoint(OA--OC),S); label(\"11\",midpoint(OA--(-37.8877590151,0)),S); label(\"19\",midpoint(OC--(10.2469508,0)),E); label(\"r\",midpoint(midpoint(A--C)--A),N); label(\"r\",midpoint(midpoint(A--C)--C),N); label(\"r\",midpoint(A--(-37.8877590151,0)),N); label(\"r\",midpoint(C--(10.2469508,0)),N); label(\"E\",(0,-4),E); [/asy]$ ##", "let $t$ be the minimun distance between the sphere centers. Then $$P(t \\ge D) = P(\\cap F_i) \\ge g(D,N) \\tag{4}$$ And then $$E(t) = \\int_0^{\\infty} P(t \\ge D) dD \\ge\\int_0^{D_1} g(D,N) \\, dD \\approx D_1- N^2 \\frac{ D_1^4}{4 R^3} = \\frac{3 }{ 4 } \\frac{R}{N^{2/3}}$$ (Simulation data suggests that the order is right, and so is the bound, but the real coefficient is around $1.12$ - perhaps $9/8$) Update: (Improved version) A better approach can be obtained by considering instead of $F_i$ (free ball) the event $S_j\\equiv$ \"separated pair\" (pair of balls are separated, they do not overlap) where $j$ indexes the $M=N(N-1)/2 \\approx N^2/2$ pairs. By the same reasoning: $$P(S_j)=1-v(D) =1 - \\frac{D^3}{R^3} \\tag{5}$$ $$P(\\cap S_i) \\ge \\max(1 - M(1-P(S_i)),0)= \\max\\left(1 - M \\frac{D^3}{R^3},0\\right) \\triangleq h(D,M) \\tag{6}$$ The range where $h(D,M)$ is positive, ie. $0\\le D \\le D_2 \\triangleq R/M^{1/3}$ Now, let $t$ be the minimun distance between the sphere centers. Then $$P(t \\ge D) = P(\\cap S_i) \\ge h(D,M) \\tag{7}$$ And then $$E(t) = \\int_0^{\\infty} P(t \\ge D) dD \\ge\\int_0^{D_2} h(D,M) \\, dD =\\\\ = \\frac{3}{4} \\frac{R}{M^{1/3}} \\approx 0.945 \\frac{R}{\\sqrt[3]{N(N-1)}} \\approx 0.945 \\frac{R}{N^{2/3}} \\tag{8}$$ Update 2 : A simple heuristic which seems to produce the correct coefficient: Following the approach above, we could assume that $S_i$ are asympotically independent, and then: $$P(\\cap S_i) \\approx \\left(1-\\frac{D^3}{R^3}\\right)^M \\tag{9}$$", "2020 AIME I Problems/Problem 6 Problem A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Solution $[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L, H); draw(b); label(\"R\", O -- Y, N); label(\"R\", Y -- P, N); label(\"R\", O -- A, NW); label(\"R\", P -- D, NE); label(\"1\", A -- H, N); label(\"2\", L -- D, N); label(\"7\", b, S); [/asy]$ Set the common", "Write inequalities to describe the sets in Exercises $35-40$ The (a) interior and (b) exterior of the sphere of radius 1 centered at the point $(1,1,1)$ Check back soon! Problem 41 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(1,1,1), \\quad P_{2}(3,3,0)$$ Check back soon! Problem 42 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(-1,1,5), \\quad P_{2}(2,5,0)$$ Check back soon! Problem 43 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(1,4,5), \\quad P_{2}(4,-2,7)$$ Check back soon! Problem 44 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(3,4,5), \\quad P_{2}(2,3,4)$$ Check back soon! Problem 45 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(0,0,0), \\quad P_{2}(2,-2,-2)$$ Check back soon! Problem 46 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(5,3,-2), \\quad P_{2}(0,0,0)$$ Check back soon! Problem 47 Find the centers and radii of the spheres in Exercises $47-50$ $$(x+2)^{2}+y^{2}+(z-2)^{2}=8$$ Check back soon! Problem 48 Find the centers and radii of the spheres in Exercises $47-50$ $$(x-1)^{2}+\\left(y+\\frac{1}{2}\\right)^{2}+(z+3)^{2}=25$$ Check back soon! Problem 49 Find the centers and radii of the spheres in Exercises $47-50$ $$(x-\\sqrt{2})^{2}+(y-\\sqrt{2})^{2}+(z+\\sqrt{2})^{2}=2$$ Check back soon! Problem 50 Find the centers and radii of the spheres in Exercises $47-50$ $$x^{2}+\\left(y+\\frac{1}{3}\\right)^{2}+\\left(z-\\frac{1}{3}\\right)^{2}=\\frac{16}{9}$$ Check back soon! Problem 51 Find equations for the spheres", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "Consider a circle and a point $$P$$ inside it. We know that the minimum distance between $$P$$ and the points on the circumference is $$2$$ cm, while the maximum distance is $$8$$ cm. What is the circle radius? As a first step, we can show that the minimum and maximum distances are reached when the points are on the intersection between the circumference and the diameter passing through $$P$$. Referring to the Figure in the bottom, note that $$A’P+OP = r,\\tag{1}\\label{eq562:uno}$$ where $$r$$ is the circle radius. Furthermore, by triangular inequality on $$\\triangle OAP$$, $$AP + OP \\geq r.\\tag{2}\\label{eq562:due}$$ Comparing \\eqref{eq562:uno} e \\eqref{eq562:due} yields $AP \\geq A’P.$ Analogously, $$A^{\\prime\\prime} P=OP+r\\tag{3}\\label{eq562:tre}$$ while, again because of triangular inequality, $$AP\\leq OP +r.\\tag{4}\\label{eq562:quattro}$$ By comparing \\eqref{eq562:tre} e \\eqref{eq562:quattro} we deduce that $AP \\leq A^{\\prime\\prime}P.$ Thus $$A’P = 2$$ cm, $$A^{\\prime\\prime}P = 8$$ cm, and the required radius is $r = \\frac{A’P+A^{\\prime\\prime}P}{2}=5 \\ \\mbox{cm}$", "# 2022 AIME II Problems/Problem 15 ## Problem Two externally tangent circles $\\omega_1$ and $\\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\\Omega$ passing through $O_1$ and $O_2$ intersects $\\omega_1$ at $B$ and $C$ and $\\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. $[asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label(\"\\omega_1\",8*dir(110),SW); label(\"\\omega_2\",5*dir(70)+(15,0),SE); label(\"O_1\",O1,W); label(\"O_2\",O2,E); label(\"B\",B,N+1/2*E); label(\"A\",A,N+1/2*W); label(\"C\",C,S+1/4*W); label(\"D\",D,S+1/4*E); label(\"15\",midpoint(O1--O2),N); label(\"16\",midpoint(C--D),N); label(\"2\",midpoint(A--B),S); label(\"\\Omega\",o.C+(o.r-1)*dir(270)); [/asy]$ ## Solution 1 First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. $[asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label(\"A'\",Ap,dir(origin--Ap)); label(\"B'\",Bp,dir(origin--Bp)); label(\"O_1\",O1,dir(origin--O1)); label(\"C\",C,dir(origin--C)); label(\"D\",D,dir(origin--D)); label(\"O_2\",O2,dir(origin--O2)); draw(O2--O1,linetype(\"4 4\")); draw(Ap--D^^Bp--C,linetype(\"2 2\")); [/asy]$ Next, remark that $B'O_1 = DO_2$, so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in", "unit circles as $u'(n)$. Let ${\\cal P}$ be a set of $n$ points that spans $u(n)$ unit distances. For every point $p\\in {\\cal P}$, we consider a unit circle centered at $p$, and denote the resulting set of $n$ circles as ${\\cal C}$. Notice that every pair of points of ${\\cal P}$ at a distances of one corresponds to two incidences in ${\\cal P}\\times{\\cal C}$. That is, we have $u(n) \\le u'(n)/2$; for example, see the following figure. In the other direction, consider a set $\\cal P$ of $n$ points and a set $\\cal C$ of $n$ unit circles such that $I({\\cal P},{\\cal C})=u'(n)$. Let ${\\cal P}'$ denote the union of $\\cal P$ with the set of centers of the circles of $\\cal C$. Every incidence corresponds to a pair of points of ${\\cal P}'$ at a distance of one from each other. Each such pair has at most two corresponding point-circle incidences. That is, we have $u'(n) \\le 2u(2n)$. The best know upper bound for incidences between $n$ points and $n$ unit circles, by Spencer, Szemerédi, and Trotter, is $O(n^{4/3})$. The best known lower bound, by Erdős, is $n^{1+\\Omega(1/\\log \\log n)}$. Similarly to Erdős’ lower bound for point-line incidences (see the first post of this series), this bound is also implied by a $\\sqrt{n}\\times\\sqrt{n}$ integer lattice and", "regular hexagon with side length $2$ and six spheres on each vertex with radius $1$ that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon. Imagine a 2D overhead view. There is a larger sphere which the $6$ spheres are internally tangent to, with the center in the center of the hexagon. To find the radius of the larger sphere we must first, either by prior knowledge or by deducing from the angle sum that the hexagon can be split into $6$ equilateral triangles from its vertices, that the radius is two more than the small radius, or $3$. $[asy] draw(circle((0.5,0.866025404),0.5)); draw(circle((-0.5,0.866025404),0.5)); draw(circle((1,0),0.5)); draw(circle((-1,0),0.5)); draw(circle((0.5,-0.866025404),0.5)); draw(circle((-0.5,-0.866025404),0.5)); draw(circle((0,0),1.5)); draw((-0.5,0.866025404)--(0.5,0.866025404)); draw((-1,0)--(1,0)); draw((-0.5,-0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,0.866025404)); draw((-0.5,-0.866025404)--(0.5,0.866025404)); draw((0.5,-0.866025404)--(1,0)); draw((0.5,0.866025404)--(1,0)); draw((-0.5,0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,-0.866025404)); [/asy]$ Now imagine the figure in $3$ dimensions. The 8th sphere must be sitting atop of the $6$ spheres, which is the only possibility for it to be tangent to all the $6$ small spheres externally and the larger sphere internally. The ring of $6$ small spheres is symmetrical and the 8th sphere will be resting atop it with its center aligned with the diameter of the large sphere. We can now create a right triangle with the legs being the line from a vertex of the hexagon to the center of the hexagon and the", "+0 # Let $A_1 A_2 \\dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to th +1 604 1 Let $A_1 A_2 \\dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find $PA_1^2 + PA_2^2 + \\dots + PA_{11}^2.$ Jan 25, 2019 #1 +25528 +8 Let $A_1 A_2 \\dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find $PA_1^2 + PA_2^2 + \\dots + PA_{11}^2.$ $$\\text{Let A_n=\\dbinom{x_a}{y_a}, ~x_a =2\\cos\\Big((n-1)\\cdot \\dfrac{360^{\\circ}}{11} \\Big), ~y_a =2\\sin\\Big((n-1)\\cdot \\dfrac{360^{\\circ}}{11} \\Big) , ~ n=1,2,\\ldots 11 }\\\\ \\text{Let P=\\dbinom{x_p}{y_p}, ~x_p =3\\cos(\\beta), ~y_p =3\\sin(\\beta) } \\\\ \\text{Let PA^2 = (x_p-x_a)^2 + (y_p-y_a)^2}$$ $$\\begin{array}{|rcll|} \\hline PA_1^2 &=& \\left[ 3\\cos(\\beta) -2\\cos\\left(0\\cdot \\dfrac{360^{\\circ}}{11} \\right) \\right]^2 + \\left[ 3\\sin(\\beta) -2\\sin\\left(0\\cdot \\dfrac{360^{\\circ}}{11} \\right) \\right]^2 \\\\ &=& 13 - 12\\left[ \\cos(\\beta)\\cos\\left(0\\cdot \\dfrac{360^{\\circ}}{11} \\right) + \\sin(\\beta)\\sin\\left(0\\cdot \\dfrac{360^{\\circ}}{11} \\right)\\right]\\\\\\\\ PA_2^2 &=& \\left[ 3\\cos(\\beta) -2\\cos\\left(1\\cdot \\dfrac{360^{\\circ}}{11} \\right) \\right]^2 + \\left[ 3\\sin(\\beta) -2\\sin\\left(1\\cdot \\dfrac{360^{\\circ}}{11} \\right) \\right]^2 \\\\ &=& 13 - 12\\left[ \\cos(\\beta)\\cos\\left(1\\cdot \\dfrac{360^{\\circ}}{11} \\right) + \\sin(\\beta)\\sin\\left(1\\cdot \\dfrac{360^{\\circ}}{11} \\right)\\right]\\\\\\\\ PA_3^2 &=& \\left[ 3\\cos(\\beta) -2\\cos\\left(2\\cdot \\dfrac{360^{\\circ}}{11} \\right)", "the largest one. Its root system lives in ${8}$-dimensional space and can be described as follows. Let ${e_1,\\dots,e_8}$ denote the standard basis in ${{\\mathbb R}^8}$, that is $\\displaystyle e_1=(1,0,\\dots,0,0),\\quad e_2=(0,1,\\dots,0,0),\\quad\\dots,\\quad e_8=(0,0,\\dots,0,1).$ The ${E_8}$ root system consists of ${240}$ vectors. It has ${112}$ vectors of the form $\\displaystyle \\pm e_j\\pm e_k\\quad (j\\neq k)$ with every possible indices and signs. By the way these vectors constitute the ${D_8}$ root system. The remaining ${128}$ vectors can be written as $\\displaystyle \\frac{1}{2}(\\pm e_1\\pm e_2\\pm e_3\\pm e_4\\pm e_5\\pm e_6\\pm e_7\\pm e_8)$ where the number of minus signs must be even. Notice that these ${240}$ points are on the surface of the ${7}$-sphere of radius ${\\sqrt{2}}$ centered at the origin. In addition, each root has exactly ${56}$ closest neighbours (at distance ${\\sqrt{2}}$). If we put spheres with radius ${\\sqrt{2}/2}$ around every point and the origin, we get a very tightly packed configuration. The sphere at the origin is in contact with all ${240}$ spheres. It turns out that this is the solution of the kissing number problem in ${8}$ dimensions. The number of symmetries of the ${E_8}$ root system is immense. Its symmetry group (Weyl group) is of order ${696,729,600(=4!\\times 6!\\times 8!)}$. Compare this with the ${8}$-cube which has ‘only’ ${10,321,920(=2^8\\times 8!)}$ symmetries. Unfortunately, we can’t visualize ${E_8}$ in its ${8}$-dimensional glory, but there", "when $d(Q,R)$ goes to zero. Oct 23 '17 at 21:07 • Since all triangles are thin, the sphere is the worse case. On the sphere (as well as on the plane), your inequality does not hold --- so try to correct your question. Oct 24 '17 at 0:31 • to Changyu Guo. Actually, I can modify previous arguments to get counterexample; simply connected compact surface where inequality does not hold. Take thin cylinder instead of torus and glue to its ends first cone-like surfaces (of nonpositive curvature) smoothly connecting end-circles to much bigger circles on unit spheres. We obtaine surface with $k=1$, but $d(Q_t,R_t)\\le c(k,t)d(Q,R)$ does not hold for any bounded $c(k,t)$, Oct 24 '17 at 7:01 1) A similar result, but with reversed inequality holds and follows immediately from Berger's comparison theorem, and provided that the triangle $PQR$ is sufficiently small (I think that it is enough to be contained in a ball with center $P$ of radius equal to the injectivity radius from $P$). Cheeger, Jeff; Ebin, David G., Comparison theorems in Riemannian geometry, North-Holland Mathematical Library. Vol. 9. Amsterdam-Oxford: North-Holland Publishing Company; New York: American Elsevier Publishing Company, Inc. VIII, 174 p. Dfl. 50.00; $19.25 (1975). ZBL0309.53035. More precisely, if$PQR$is sufficiently small and$\\mathrm{Sec} \\leq k$, then $$d(Q_t,R_t) \\geq \\tilde{d}_k (\\tilde{Q}_t,\\tilde{R}_t)$$ On the right hand side", "+0 HELPME!?!?!? 0 92 3 A circle is centered at $O.$ The tangent to the circle at $P$ is extended to $Q.$ Line segment $\\overline{QS}$ intersects the circle at $R.$ Given that $OS = 2,$ $SR = RQ = 3$, and $PQ = 6$, find the radius of the circle. [asy] unitsize(2 cm); pair A, B, C, D, E, F, G, O; A = dir(70); B = A + 1.2*dir(-20); C = dir(25); D = 2*C - B; O = (0,0); draw(Circle(O,1)); draw(A--B--D--O); dot(\"$P$\", A, N); dot(\"$Q$\", B, dir(0)); dot(\"$R$\", C, SW); dot(\"$S$\", D, NW); dot(\"$O$\", O, SW); [/asy] Aug 14, 2020 #1 0 Image Here: Aug 15, 2020 #2 0 By power of a point, the radius of the circle is 4*sqrt(7). Aug 15, 2020 #3 +110762 0 Maybe it would be in your interest to present your questions better. All those unnecessary dollar signs make it difficult to read and I feel that if you cannot be bothered to present your question properly then it is not worth my effort to attempt to answer. Aug 15, 2020", "+0 # Circle tangency 0 677 2 +167 Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\\overline{PA}$ and $\\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9. Sep 3, 2017 #1 +101870 +1 OA is a radius = 9 Because OAP is a right angle, OP = sqrt ( PA^2 + OA^2) = sqrt (12^2 + 9^2) = sqrt (225) = 15 Let AB intersect OP at R And triangles OPA and OAR are similar So PA / OP = AR / OA 12 / 15 = AR / 9 AR = 108/15 = 36/5 And AR = (1/2) AB ...so AB = 2 (AR) = 2(36/5) = 72/5 Sep 3, 2017 #2 +22550 0 Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\\overline{PA}$ and $\\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9. Sep 4, 2017", "$BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. $[asy] size(150); defaultpen(linewidth(0.65)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); dot(O); dot(A); dot(B); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"B\",B,S); label(\"C\",C,SE); [/asy]$ Problem 5 Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$. What is the largest real value that $x + y$ can have? Problem 6 Let $a_n$ equal $6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$. Problem 7 Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and the denominator? Problem 8 What is the largest 2-digit prime factor of the integer ${200\\choose 100}$? Problem 9 Find the minimum value of $\\frac{9x^2\\sin^2 x + 4}{x\\sin x}$ for $0 < x < \\pi$. Problem 10 The numbers $1447$, $1005$, and $1231$ have something", "$\\omega$ be a circle of radius $6$ with center $O$. Let $AB$ be a chord of $\\omega$ having length $5$. For any real constant $c$, consider the locus $\\mathcal{L}(c)$ of all points $P$ such that $PA^2 - PB^2 = c$. Find the largest value of $c$ for which the intersection of $\\mathcal{L}(c)$ and $\\omega$ consists of just one point. Four circles are situated in the plane so that each is tangent to the other three. If three of the radii are $5$, $5$, and $8$, the largest possible radius of the fourth circle is $a/b$, where $a$ and $b$ are positive integers and gcd$(a, b) = 1$. Find $a + b$. Let $ABC$ be an equilateral triangle having sides of length 1, and let $P$ be a point in the interior of $\\Delta ABC$ such that $\\angle ABP = 15 ^\\circ$. Find, with proof, the minimum possible value of $AP + BP + CP$. Comment: In fact this question is incorrect, unfortunately. A more reasonable problem: Prove that $AP + BP + CP \\ge \\sqrt{3}$. Three circles, with radii of $1, 1$, and $2$, are externally tangent to each other. The minimum possible area of a quadrilateral that contains and is tangent to all three circles can be written as $a + b\\sqrt{c}$ where $c$ is not divisible" ]

[ "# Problem #142 142 What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2,-10,5),$ and the other on the sphere of radius 87 with center $(12,8,-16)$? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information ### Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens ## Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137. Categories ## Arithmetic Mean | AIME I, 2015 | Question 12 Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean. ## Arithmetic Mean of Number Theory – AIME 2015 Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. • is 107 • is 431 • is 840 • cannot be determined from the given information ### Key Concepts Inequalities Algebra Number Theory AIME, 2015, Question 12 Elementary Number Theory by David Burton ##", "Categories ## Distance and Spheres | AIME I, 1987 | Question 2 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. ## Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information ### Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens ## Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.", "Categories Distance and Spheres | AIME I, 1987 | Question 2 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres. Distance and Sphere – AIME I, 1987 What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)? • is 107 • is 137 • is 840 • cannot be determined from the given information Key Concepts Angles Algebra Spheres AIME I, 1987, Question 2 Geometry Vol I to Vol IV by Hall and Stevens Try with Hints The distance between the center of the spheres is $\\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$ =$\\sqrt{14^{2}+18^{2}+21^{2}}$=31 The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.", "2 (Official MAA) Consider a cross section of the board and spheres with a plane that passes through the centers of the holes and centers of the spheres as shown. $[asy] unitsize(1.5 cm); pair A, B, C, D, E, F, G, P, Q; C = dir(175); D = dir(175 + 180); P = (-2,-0.8); Q = (2,-0.8); A = (C + reflect(P,Q)*(C))/2; B = (D + reflect(P,Q)*(D))/2; E = intersectionpoint(P--A, Circle(C,1)); F = intersectionpoint(B--Q, Circle(D,1)); G = (D + reflect(A,C)*(D))/2; draw(Circle(C,1)); draw(Circle(D,1)); draw(P--Q); draw(A--(C + (0,1))); draw(B--(D + (0,1))); draw(E--C--D--F); draw(D--G); dot(\"A\", A, NE); dot(\"B\", B, NW); dot(\"C\", C, NW); dot(\"D\", D, NE); dot(\"E\", E, SW); dot(\"F\", F, SE); dot(\"G\", G, SE); [/asy]$ Let $A$, $C$, and $E$ be, respectively, the center of the hole with radius $1,$ the center of the sphere resting in that hole, and a point on the edge of that hole. Let $B$, $D$, and $F$ be the corresponding points for the hole with radius $2.$ Let $G$ be the point on $\\overline{AC}$ such that $\\overline{AC} \\perp \\overline{GD}$. Let the radius of the spheres be $r = CE = DF$. Because $r^2 = AE^2 + AC^2 = 1 + AC^2$ and $r^2 = BF^2 + BD^2 = 4 + BD^2$, it follows that$$CG = AC - AG = AC - BD = \\sqrt{r^2", "Browse Questions # Find the shortest and largest distance from the point (2,-7) to the circle $x^2+y^2-14x-10y-151=0$. $\\begin{array}{1 1}(a)\\;2\\;and\\;28\\\\(b)\\;4\\;and\\;20\\\\(c)\\;3\\;and\\;27\\\\(d)\\;5\\;and\\;25\\end{array}$ Let $S=x^2+y^2-14x-10y-151=0$ Substituting the point $(2,-7)$ in this equation, we get $S_1=(2)^2+(-7)^2-14(2)-10(-7)-151=-56 < 0$ $\\therefore P(2,-7)$ inside the circle Radius of circle $r=\\sqrt{(-7)^2+(-5)^2+151}=15$ Centre of circle (7,5) CP=$\\sqrt{(7-2)+(5+7)^2}=13$ Shortest distance =$PA=r-CP=15-13=2$ Largest distance $PB=r+CP=15+13=28$ Hence (a) is the correct option. edited Mar 24, 2014", "Browse Questions # Find the shortest and largest distance from the point (2,-7) to the circle $x^2+y^2-14x-10y-151=0$. $\\begin{array}{1 1}(a)\\;2\\;and\\;28\\\\(b)\\;4\\;and\\;20\\\\(c)\\;3\\;and\\;27\\\\(d)\\;5\\;and\\;25\\end{array}$ Can you answer this question? Let $S=x^2+y^2-14x-10y-151=0$ Substituting the point $(2,-7)$ in this equation, we get $S_1=(2)^2+(-7)^2-14(2)-10(-7)-151=-56 < 0$ $\\therefore P(2,-7)$ inside the circle Radius of circle $r=\\sqrt{(-7)^2+(-5)^2+151}=15$ Centre of circle (7,5) CP=$\\sqrt{(7-2)+(5+7)^2}=13$ Shortest distance =$PA=r-CP=15-13=2$ Largest distance $PB=r+CP=15+13=28$ Hence (a) is the correct option. answered Feb 11, 2014 edited Mar 24, 2014", "Consider a circle and a point $$P$$ inside it. We know that the minimum distance between $$P$$ and the points on the circumference is $$2$$ cm, while the maximum distance is $$8$$ cm. What is the circle radius? As a first step, we can show that the minimum and maximum distances are reached when the points are on the intersection between the circumference and the diameter passing through $$P$$. Referring to the Figure in the bottom, note that $$A’P+OP = r,\\tag{1}\\label{eq562:uno}$$ where $$r$$ is the circle radius. Furthermore, by triangular inequality on $$\\triangle OAP$$, $$AP + OP \\geq r.\\tag{2}\\label{eq562:due}$$ Comparing \\eqref{eq562:uno} e \\eqref{eq562:due} yields $AP \\geq A’P.$ Analogously, $$A^{\\prime\\prime} P=OP+r\\tag{3}\\label{eq562:tre}$$ while, again because of triangular inequality, $$AP\\leq OP +r.\\tag{4}\\label{eq562:quattro}$$ By comparing \\eqref{eq562:tre} e \\eqref{eq562:quattro} we deduce that $AP \\leq A^{\\prime\\prime}P.$ Thus $$A’P = 2$$ cm, $$A^{\\prime\\prime}P = 8$$ cm, and the required radius is $r = \\frac{A’P+A^{\\prime\\prime}P}{2}=5 \\ \\mbox{cm}$", "Write inequalities to describe the sets in Exercises $35-40$ The (a) interior and (b) exterior of the sphere of radius 1 centered at the point $(1,1,1)$ Check back soon! Problem 41 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(1,1,1), \\quad P_{2}(3,3,0)$$ Check back soon! Problem 42 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(-1,1,5), \\quad P_{2}(2,5,0)$$ Check back soon! Problem 43 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(1,4,5), \\quad P_{2}(4,-2,7)$$ Check back soon! Problem 44 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(3,4,5), \\quad P_{2}(2,3,4)$$ Check back soon! Problem 45 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(0,0,0), \\quad P_{2}(2,-2,-2)$$ Check back soon! Problem 46 In Exercises $41-46,$ find the distance between points $P_{1}$ and $P_{2}$ $$P_{1}(5,3,-2), \\quad P_{2}(0,0,0)$$ Check back soon! Problem 47 Find the centers and radii of the spheres in Exercises $47-50$ $$(x+2)^{2}+y^{2}+(z-2)^{2}=8$$ Check back soon! Problem 48 Find the centers and radii of the spheres in Exercises $47-50$ $$(x-1)^{2}+\\left(y+\\frac{1}{2}\\right)^{2}+(z+3)^{2}=25$$ Check back soon! Problem 49 Find the centers and radii of the spheres in Exercises $47-50$ $$(x-\\sqrt{2})^{2}+(y-\\sqrt{2})^{2}+(z+\\sqrt{2})^{2}=2$$ Check back soon! Problem 50 Find the centers and radii of the spheres in Exercises $47-50$ $$x^{2}+\\left(y+\\frac{1}{3}\\right)^{2}+\\left(z-\\frac{1}{3}\\right)^{2}=\\frac{16}{9}$$ Check back soon! Problem 51 Find equations for the spheres", "0.62151297704900443)$$ Maximum distance to any point = $$0.37192577249511566$$. C++ code here: http://pastebin.com/Yrqry5H1 • By perturbing your data set to make the corners of Voronoi cells touches the semi-circle. Following data set $$\\begin{cases} x_1 &= (0,0.13121005),\\\\ x_2 &= (0,0.76614727),\\\\ x_{3,4} &= (\\pm 0.51983971,0.62791492),\\\\ x_{5,6} &= (\\pm 0.67403990,0.17923625) \\end{cases}$$ can bring the upper bound of minimal maximum distance down to around $0.37198874$. – achille hui Sep 19 '15 at 17:51 • What do you get for 7 points? – Ed Pegg Sep 21 '15 at 15:25 • The new set seems incorrect, it misses some points on the boundary. For example, the circles of radius 0.371499 centered at $x_3$ and $x_5$ intersect at $\\approx (0.865927626561528,0.4986448922006425)$ which is at a distance $\\approx 0.9992384014639707 < 1$ from the center. The circles fail to cover a tiny \"triangle\" near these sort of points... – achille hui Sep 21 '15 at 18:02 • Hmm... looks like the candidate for minimal maximum distance has a closed form expression: $$r = \\sqrt{3 + 2\\sqrt{3} - 2\\sqrt{5 + \\sqrt{27} - \\frac{1}{\\sqrt{27}}}} \\approx 0.3719887399641695$$ – achille hui Sep 21 '15 at 21:06 • @achillehui You're right about the gaps wherever three circles intersect. I'll have to adjust my algorithm. Then again, for the purposes of the OP's question, I think the gaps are small enough to not matter.", "# What is the largest possible thirteenth kissing sphere? It is well-known that it is impossible to arrange 13 spheres of unit radius all tangent to another unit sphere without their interiors intersecting. This was apparently the subject of disagreement between Isaac Newton (\"impossible\") and David Gregory (\"possible\"). The cause of the dispute was, in part, that there seems to be a lot of room left over after 12 spheres. How close of a call is it, so to speak? What is the largest possible $r$ so that it is possible to arrange 12 spheres of unit radius and a 13th sphere of radius $r$ all tangent to another unit sphere, without intersections? What does the optimal configuration look like? - or, also, one may ask for the minimum radius r so that there are 13 disjoint unit tangent balls to a ball of radius r -somehow more symmetric – Pietro Majer Apr 21 '12 at 15:08 @Pietro Majer: yes, that is in fact a good variant! I would be interested in an answer to either version, but hopefully I can get an answer to both! – Jamie J. Taylor Apr 21 '12 at 15:46 Pietro's version of this question is answered in a paper by Oleg Musin and Alexey Tarasov (to appear in Discrete & Computational Geometry,", "the largest one. Its root system lives in ${8}$-dimensional space and can be described as follows. Let ${e_1,\\dots,e_8}$ denote the standard basis in ${{\\mathbb R}^8}$, that is $\\displaystyle e_1=(1,0,\\dots,0,0),\\quad e_2=(0,1,\\dots,0,0),\\quad\\dots,\\quad e_8=(0,0,\\dots,0,1).$ The ${E_8}$ root system consists of ${240}$ vectors. It has ${112}$ vectors of the form $\\displaystyle \\pm e_j\\pm e_k\\quad (j\\neq k)$ with every possible indices and signs. By the way these vectors constitute the ${D_8}$ root system. The remaining ${128}$ vectors can be written as $\\displaystyle \\frac{1}{2}(\\pm e_1\\pm e_2\\pm e_3\\pm e_4\\pm e_5\\pm e_6\\pm e_7\\pm e_8)$ where the number of minus signs must be even. Notice that these ${240}$ points are on the surface of the ${7}$-sphere of radius ${\\sqrt{2}}$ centered at the origin. In addition, each root has exactly ${56}$ closest neighbours (at distance ${\\sqrt{2}}$). If we put spheres with radius ${\\sqrt{2}/2}$ around every point and the origin, we get a very tightly packed configuration. The sphere at the origin is in contact with all ${240}$ spheres. It turns out that this is the solution of the kissing number problem in ${8}$ dimensions. The number of symmetries of the ${E_8}$ root system is immense. Its symmetry group (Weyl group) is of order ${696,729,600(=4!\\times 6!\\times 8!)}$. Compare this with the ${8}$-cube which has ‘only’ ${10,321,920(=2^8\\times 8!)}$ symmetries. Unfortunately, we can’t visualize ${E_8}$ in its ${8}$-dimensional glory, but there", "2. edited July 2018 Then Dan Piponi, another one of my internet math pals, went back to the 4d case and thought about the moments of the distance for 2 randomly chosen points on the unit sphere in $$\\mathbb{R}^4$$. The first moment is the mean, the second moment is the variance, and so on. He notice that the even moments were the Catalan numbers $$2, 5, 14, 42, 132 \\dots$$ Pick a pair of points uniformly randomly on the 4-sphere. Let S be the distance between them. E[S²ⁿ] is the (n+1)-th Catalan number. This was funny, because he'd just managed to buy a hardback copy of Richard P. Stanley's book on Catalan numbers on Amazon for just 11 dollars, thanks to some glitch that quickly went away. Comment Source:Then [Dan Piponi](https://twitter.com/sigfpe/status/1015384915058176001), another one of my internet math pals, went back to the 4d case and thought about the _moments_ of the distance for 2 randomly chosen points on the unit sphere in \\$$\\mathbb{R}^4\\$$. The first moment is the mean, the second moment is the variance, and so on. He notice that the _even_ moments were the [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number) $$2, 5, 14, 42, 132 \\dots$$ > Pick a pair of points uniformly randomly on the 4-sphere. Let S be the distance between them. E[S²ⁿ] is the (n+1)-th Catalan number.", "center $$(3, -2, 1)$$ and that goes through the point $$(4, 2, 5)$$. Ex 12.1.11 Find an equation of the sphere with center at $$(2,1,-1)$$ and radius 4. Find an equation for the intersection of this sphere with the $$y$$-\\)z\\) plane; describe this intersection geometrically. (answer) Ex 12.1.12 Consider the sphere of radius 5 centered at $$(2,3,4)$$. What is the intersection of this sphere with each of the coordinate planes? Ex 12.1.13 Show that for all values of $$\\theta$$ and $$\\phi$$, the point $$(a\\sin\\phi\\cos\\theta,a\\sin\\phi\\sin\\theta,a\\cos\\phi)$$ lies on the sphere given by $$x^2+y^2+z^2=a^2$$. Ex 12.1.14 Prove that the midpoint of the line segment connecting $$(x_1,y_1,z_1)$$ to $$(x_2,y_2,z_2)$$ is at $$\\ds\\left({x_1+x_2\\over 2},{y_1+y_2\\over 2},{z_1+z_2\\over 2}\\right)$$. Ex 12.1.15 Any three points $$P_1(x_1,y_1,z_1)$$, $$P_2(x_2,y_2,z_2)$$, $$P_3(x_3,y_3,z_3)$$, lie in a plane and form a triangle. The triangle inequality says that $$d(P_1,P_3)\\le d(P_1,P_2)+d(P_2,P_3)$$. Prove the triangle inequality using either algebra (messy) or the law of cosines (less messy). Ex 12.1.16 Is it possible for a plane to intersect a sphere in exactly two points? Exactly one point? Explain. ## 12.2: Vectors Ex 12.2.1 Draw the vector $$\\langle 3,-1\\rangle$$ with its tail at the origin. Ex 12.2.2 Draw the vector $$\\langle 3,-1,2\\rangle$$ with its tail at the origin. Ex 12.2.3 Let $${\\bf A}$$ be the vector with tail at the origin and head at $$(1,2)$$; let $${\\bf B}$$", "the probability is 1/4. Similarly, for a sphere the probability is one Octant, or 1/8. Pick two points at random on a unit sphere. The first one can be assigned the coordinate (0, 0, 1) without loss of generality. The second point can be given the coordinates with since all points with the same are rotationally identical. The distance between the two points is then (31) Because the surface Area element is (32) the probability that two points are a distance apart is (33) The Delta Function contributes when (34) (35) so (36) However, we need (37) and (38) so (39) and (40) for . Somewhat surprisingly, the largest distances are the most common, contrary to most people's intuition. A plot of 15 random lines is shown below. (41) giving the first few as (42) (43) (44) (45) (46) (47) (48) (49) so the Skewness and Kurtosis are (50) (51) See also Ball, Bing's Theorem, Bubble, Circle, Dandelin Spheres, Diameter, Ellipsoid, Exotic Sphere, Fejes Tóth's Problem, Hypersphere, Liebmann's Theorem, Liouville's Sphere-Preserving Theorem, Mikusinski's Problem, Noise Sphere, Oblate Spheroid, Osculating Sphere, Parallelizable, Prolate Spheroid, Radius, Space Division, Sphere Packing, Tennis Ball Theorem References Beyer, W. H. (Ed.) CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, 1987. Eppstein, D. Circles and Spheres.'' http://www.ics.uci.edu/~eppstein/junkyard/sphere.html. Geometry Center. The Sphere.''", "by $\\rho$. For the specific case of the Earth, modeled as a perfect sphere, we get from this $E(||\\bf{X} – \\bf{Y}||) = 5280$ (miles). What happens asymptotically, i.e., in very high dimensions? That is easy enough to figure out as well; we need nothing more than simple tools. Simply use the standard facts that $\\Gamma (x) =e^{-x}\\,x^{x-1/2}\\,\\sqrt{2\\pi }\\bigg [1 + \\frac{1}{12x} + \\frac{1}{288x^2} + \\cdots \\bigg ]$ as $x \\to \\infty$ and that for $|x| < 1$, $\\log (1-x) = -x - x^2/2 -x^3/3 - \\cdots$. Then, our formula $E\\bigg [||\\bf{X} - \\bf{Y}||\\bigg ] = \\frac{2^{n-1}[\\Gamma (\\frac{n}{2})]^2}{\\sqrt{\\pi}\\,\\Gamma (n - \\frac{1}{2})}$ for the case of the unit ball results in $E\\bigg [||\\bf{X} - \\bf{Y}||\\bigg ] = \\sqrt{2}\\,\\bigg [1 - \\frac{7}{24n} - \\frac{95}{1152n^2} + O(n^{-3}) \\bigg ]$. This means, that in high dimensions, the Euclidean distance between two random points on the surface would be close to as large as it can possibly be, namely $\\sqrt{2}$. An interesting question is exactly how much smaller is the Euclidean distance between two points than the geodesic distance between them. This will of course depend on the specific two points. What happens if the two points are chosen at random? Let's use some notation to get things very clear. Let $\\bf{X}, \\bf{Y}$ be two independent random picks from the surface of the $n$-dimensional", "# How Large can a Spherical Set Without Two Orthogonal Vectors Be? ## The problem Witsenhausen’s Problem (1974): Let $A$ be a measurable subset of the $d$-dimensional sphere $S^d = \\{x\\in {\\bf R}^{d+1}:\\|x\\|=1\\}$. Suppose that $A$ does not contain two orthogonal vectors. How large can the $d$-dimensional volume of $A$ be? ## A Conjecture Conjecture: The maximum volume is attained by two open caps of diameter $\\pi/4$ around the south pole and the north pole. For simplicity, let us normalize the volume of $S^d$ to be 1. ## What is known What is known The Frankl-Wilson theorem (the version described in the previous post) gives the asymptotic bound $Vol(A) \\le 2^{(H(1/4)-1)n}=(1.2...)^{-d}$, why? Consider the set $C_n=(1/\\sqrt{n})\\{-1,1\\}^n$. (And suppose that $n=4p$ where $p$ is a prime, and $d+1=n$.) $C_n$ is a subset of $S^d$. Let $A$ be a subset of $S^d$ not containing two orthogonal vectors. Now, the intersection of every rotation of $A$ with $C_n$ has at most $4{{n} \\choose {0}}+{{n}\\choose {1}}+\\dots+{{n}\\choose{p-1}}$ elements. This gives $Vol(A) \\le (4{{n} \\choose {0}}+{{n}\\choose {1}}+\\dots+{{n}\\choose{p-1}})/2^n$ and implies the asymptotic bound we described. This is a beautiful application of the Frankl-Wilson theorem which goes back essentially to the original paper. (If $d+1$ is not of that form we can suppose that $n$ is the largest integer of the form $n=4p$ smaller than or", "Comment Share Q) # Find the shortest and largest distance from the point (2,-7) to the circle $x^2+y^2-14x-10y-151=0$. $\\begin{array}{1 1}(a)\\;2\\;and\\;28\\\\(b)\\;4\\;and\\;20\\\\(c)\\;3\\;and\\;27\\\\(d)\\;5\\;and\\;25\\end{array}$ Let $S=x^2+y^2-14x-10y-151=0$ Substituting the point $(2,-7)$ in this equation, we get $S_1=(2)^2+(-7)^2-14(2)-10(-7)-151=-56 < 0$ $\\therefore P(2,-7)$ inside the circle Radius of circle $r=\\sqrt{(-7)^2+(-5)^2+151}=15$ CP=$\\sqrt{(7-2)+(5+7)^2}=13$ Shortest distance =$PA=r-CP=15-13=2$ Largest distance $PB=r+CP=15+13=28$", "$(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\\frac{1}{3}\\times120$, $\\frac{1}{3}\\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\\sqrt{7250}$ or $\\sqrt{725\\times10}$, thus $n$ is $\\boxed{725}$.", "– 2r)$$ Here is Milne’s picture of the situation: The Euler Equation – by Antony Milne In the comments on this blog, Greg Egan has proposed a generalized Grace–Danielsson inequality in $n$ dimensions. More precisely, he has shown the following. Suppose we have a smaller ball of radius $r$ inside a larger ball of radius $R$ in $\\mathbb{R}^n$. Suppose the distance between the centers of these balls is $d$. Then we can fit an $n$-simplex inside the larger ball and outside the smaller one if $$d^2 \\le (R + (n-2) r) (R – n r),$$ where if the inequality is an equation the simplex may touch the surface of these balls. He conjectures that this sufficient condition is also necessary. Egan writes: Here’s a sketch of a result leading towards a higher-dimensional Grace–Danielsson inequality. Given a small sphere with radius $r$ in $\\mathbb{R}^n$, inside a larger sphere of radius $R$, and a distance of $d$ between their centres, there’s an $n$-simplex we can try to construct that’s very symmetrical with respect to these spheres. First, pick the point $P$ on the small sphere that lies on the axis joining the centres of the spheres, and which is the furthest of the two such points from the larger sphere. The hyperplane tangent to the small sphere at $P$ intersects" ]

[ "+0 # Help 0 142 1 A parametric graph is given by \\begin{align*} x &= \\cos t + \\frac{t}{2}, \\\\ y &= \\sin t. \\end{align*}How many times does the graph intersect itself between $x = 1$ and $x = 40$? Jul 21, 2021", "How do I find where the graph of parametric equations crosses itself? I am trying to find the value of $t$ where the graph of the following parametric equations crosses itself: \\begin{align} x =&\\; t^3 − t + 3\\\\ y =&\\; t^2 − 3 \\end{align} I know that the next step toward solving this problem involves creating two parallel equations with a different kind of variable. The result will look like: \\begin{align} t^3 - t + 3 = s^3 - s + 3 \\tag1\\\\ t^2 - 3 = s^2 - 3 \\tag2 \\end{align} After this point, I am totally stumped. I know that I will have to solve for $t$, and it looks like doing so will involve some skill I have not yet acquired. Multiplying (2) by $-t$ and adding (1) to (2) helps me to eliminate the $t^3$, but I am left with $$s^3 - s(t^3) - (s^2)t + t = 0$$ an equation for which I have no idea how to solve for $t$. Could someone please show me how to find $t$-values at the point(s) where the graph from these equations crosses itself? Thanks in advance! • x does not equal y where this parametric equation crosses itself! @probably_someone Apr 2 '18 at 23:10 • Wow, yeah, that was bad. Sorry. Apr 2 '18", "+0 # The graph of the parametric equations \\begin{align*} x&=\\cos t,\\\\ y&=\\sin t, 0 218 1 +474 The graph of the parametric equations \\begin{align*} x&=\\cos t,\\\\ y&=\\sin t, \\end{align*} meets the graph of the parametric equations \\begin{align*} x &= 2+ 4\\cos s,\\\\ y &= 3+4\\sin s, \\end{align*} at two points. Find the slope of the line between these two points. Nov 28, 2018 #1 +22010 +9 The graph of the parametric equations \\begin{align*} x&=\\cos t,\\\\ y&=\\sin t, \\end{align*} meets the graph of the parametric equations \\begin{align*} x &= 2+ 4\\cos s,\\\\ y &= 3+4\\sin s, \\end{align*} at two points. Find the slope of the line between these two points. $$\\text{circle 1:} \\\\ \\begin{array}{|rcll|} \\hline x &=& \\cos(t) \\\\ y &=& \\sin(t) \\\\ \\hline x^2+y^2 &=& \\cos^2(t) + \\sin^2(t) \\\\ \\mathbf{x^2+y^2} & \\mathbf{=} & \\mathbf{1} \\\\ \\hline \\end{array}$$ $$\\text{circle 2:} \\\\ \\begin{array}{|rcll|} \\hline x &=& 2+4\\cos(t) \\quad \\text{ or } \\quad 4\\cos(t) = x-2 \\\\ y &=& 3+4\\sin(t) \\quad \\text{ or } \\quad 4\\sin(t) = y-3 \\\\ \\hline x^2+y^2 &=& \\Big( 2+4\\cos(t) \\Big)^2 + \\Big( 3+4\\sin(t) \\Big)^2 \\\\ x^2+y^2 &=& 4+16\\cos(t)+16\\cos^2(t) + 9 + 24\\sin(t) + 16\\sin^2(t) \\\\ x^2+y^2 &=& 13+16\\cos(t)+ 24\\sin(t) + 16\\Big(\\cos^2(t) + \\sin^2(t) \\Big) \\\\ x^2+y^2 &=& 13+16\\cos(t)+ 24\\sin(t) + 16 \\\\ x^2+y^2 &=& 29+16\\cos(t)+ 24\\sin(t) \\\\ x^2+y^2 &=& 29+4\\cdot4\\cos(t)+ 6\\cdot 4\\sin(t) \\\\ x^2+y^2 &=& 29+4\\cdot", "# Easy Parametrics Geometry Level pending What is the shape of the graph of the parametric equation a, if $a = \\begin{cases} x(t)=5 (\\cos t)\\\\y(t)=10 (\\sin t) \\\\\\end{cases}$ Note: This problem is in degrees × Problem Loading... Note Loading... Set Loading...", "-x^2 + x\\end{align*} 5. x210x+25=y+9\\begin{align*}x^2-10x+25=y+9\\end{align*} 6. x2+18x+81=y+1\\begin{align*}x^2+18x+81=y+1\\end{align*} 7. 4x212x+9=y+16\\begin{align*}4x^2-12x+9=y+16\\end{align*} 8. x2+14x+49=y+3\\begin{align*}x^2+14x+49=y+3\\end{align*} 9. 4x220x+25=y+9\\begin{align*}4x^2-20x+25=y+9\\end{align*} 10. x2+8x+16=y+25\\begin{align*}x^2+8x+16=y+25\\end{align*} ### Answers for Explore More Problems To view the Explore More answers, open this PDF file and look for section 10.7. ### Vocabulary Language: English Intercept Intercept The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis. Parabola Parabola A parabola is the characteristic shape of a quadratic function graph, resembling a \"U\". Vertex Vertex A vertex is a corner of a three-dimensional object. It is the point where three or more faces meet. ## Date Created: Oct 01, 2012 Jan 31, 2016 Save or share your relevant files like activites, homework and worksheet. To add resources, you must be the owner of the Modality. Click Customize to make your own copy.", "## Precalculus (6th Edition) Blitzer We know that the parametric equations are the equations having a common variable in both x and y. And by taking the values of t manually, one can get the values of x and y separately. And the corresponding points can be plotted to get a graph of the given parametric equations. For example: If the parametric equations are as given below: \\begin{align} & x=t \\\\ & y=2{{t}^{2}} \\\\ \\end{align} Take the values of t manually and compute the values of x and y as given below: For $t=1$ \\begin{align} & x=t \\\\ & =1 \\end{align} \\begin{align} & y=2{{t}^{2}} \\\\ & =2{{\\left( 1 \\right)}^{2}} \\\\ & =2 \\end{align} Similarly, for $t=2$ \\begin{align} & x=t \\\\ & =2 \\end{align} \\begin{align} & y=2{{t}^{2}} \\\\ & =2{{\\left( 2 \\right)}^{2}} \\\\ & =8 \\end{align} And take some more values of t and get the values of x and y as given below: \\begin{align} & t=3 \\\\ & x=3 \\\\ & y=18 \\\\ \\end{align} \\begin{align} & t=4 \\\\ & x=4 \\\\ & y=32 \\\\ \\end{align} Thus, take these values of x and y to plot the graph as drawn below:", "\\end{align*} Many more equations: parametric, implicit, can be found at the sites linked above.", "# Precalculus Let $\\mathcal{G}$ be the graph of the parametric equations \\begin{align*} x &= \\cos(4t),\\\\ y &= \\sin(6t). \\end{align*}What is the length of the smallest interval $I$ such that the graph of these equations for all $t\\in I$ produces the entire graph $\\mathcal{G}$? The answer is not pi. I have tried that, and it says it is wrong. 1. 👍 1 2. 👎 0 3. 👁 559 1. Hmmm. You're right. It's 2π/3 1. 👍 0 2. 👎 2 posted by Steve 2. \\pi/2 trust me 100% correct 1. 👍 5 2. 👎 0 posted by Helper 3. It’s pi/2 trust him 1. 👍 2 2. 👎 0 ## Similar Questions 1. ### Precalculus Let $\\mathcal{G}$ be the graph of the parametric equations \\begin{align*} x &= \\cos(4t),\\\\ y &= \\sin(6t). \\end{align*}What is the length of the smallest interval $I$ such that the graph of these equations for all $t\\in I$ produces asked by Zheng on January 29, 2017 2. ### MATH A lattice point is a point with integer coordinates. How many lattice points $(x,y)$ with $-100\\le x\\le 100$ and $-100\\le y\\le 100$ are on the graph of the parametric equations \\begin{align*} x&=30-40\\cos t,\\\\ y&=-50 + 30\\cos t? asked by BABABA on March 4, 2017 3. ### Precalculus Find the sum of all possible values of the constant", "# Analyzing Parametric Equations I saw a tweet that sumidiot posted on the parametric equations $x(t)=t+c*\\sin(2t)$ and $y(t)=\\cos(t)$ and spent a good 30 minutes thinking about the value of $c$ which makes the graph intersect itself exactly once. I was going to post about my solution, but I was beaten to the punch. And thank goodness, because there are graphs and everything on sumidiot’s solution, which you should read here. I still haven’t fully evaluated twitter, which I will post about later once I come to some solid conclusions. But I’m leaning towards liking it. I’ve got lots of good links, anyway. If you want to be my twitter friend, I’m here!", "that the \\begin{align*}y-\\end{align*}values of 1 and 4 are now 2 and 8. It is not unusual for a parabola to be plotted with five points rather than seven. The reason for this is the vertical stretch often multiplies the \\begin{align*}y-\\end{align*}values such that they are difficult to graph on a Cartesian grid. If all the points are to be plotted, a different scale must be used for the \\begin{align*}y-\\end{align*}axis. 3. The vertex is the values \\begin{align*}(HT, VT)\\end{align*} which is (-3, 2). The \\begin{align*}y-\\end{align*}values of 1, 4 and 9 must be multiplied by \\begin{align*}\\frac{1}{2}\\end{align*} to create values of \\begin{align*}\\frac{1}{2}, 2\\end{align*} and \\begin{align*}4 \\frac{1}{2}\\end{align*}. The graph is a vertical reflection which means the graph opens downward and the \\begin{align*}y-\\end{align*}values become negative. Summary The base graph of the quadratic function of \\begin{align*}y=x^2\\end{align*} can undergo transformations that will change its appearance. The shape of the parabola which resembles a ‘U’ does not change but its location on the Cartesian grid can change. The vertex of (0, 0) will change if the image graph undergoes either a horizontal translation and/or a vertical translation. The transformations cause the parabola to slide left or right and up or down. The points of the base graph are plotted from the vertex left and right one and up one, left and right two and up fore, left and", "# Easy Parametrics Geometry Level pending What is the shape of the graph of the parametric equation a, if $a = \\begin{cases} x(t)=5 (\\cos t)\\\\y(t)=10 (\\sin t) \\\\\\end{cases}$ Note: This problem is in degrees ×", "graph the parabola. Example C Change \\begin{align*}y=x^2-10x+16\\end{align*} into vertex form. Solution: To change an equation from standard form into vertex form, you must complete the square. Review the Completing the Square Lesson if needed. The major difference is that you will not need to solve this equation. \\begin{align*}y &=x^2-10x+16\\\\ y-16+{\\color{red}25} &=x^2-10x+{\\color{red}25} \\quad \\text{Move 16 to the other side and add} \\ \\left(\\frac{b}{2}\\right)^2 \\text{to both sides}.\\\\ y+9 &=(x-5)^2 \\qquad \\quad \\ \\ \\text{Simplify left side and factor the right side}\\\\ y &=(x-5)^2-9 \\qquad \\text{Subtract 9 from both sides to get} \\ y \\ \\text{by itself}.\\end{align*} To solve an equation in vertex form, set \\begin{align*}y = 0\\end{align*} and solve for \\begin{align*}x\\end{align*}. \\begin{align*}(x-5)^2-9 &=0\\\\ (x-5)^2 &=9\\\\ x-5 &=\\pm 3\\\\ x &=5 \\pm 3 \\ or \\ 8 \\ and \\ 2\\end{align*} Intro Problem Revisit The vertex will give us the price point that will result in the maximum profit and that profit, so let’s change this equation into intercept form by factoring. First factor out –5. \\begin{align*}-5p^2 + 400p - 8000 = -5(p^2 - 80p + 1600)\\\\ -5(p - 40)(p - 40)\\end{align*} From this we can see that the x-intercepts are 40 and 40. The average of 40 and 40 is 40 we plug 40 into our original equation. \\begin{align*}-5(40)^2 + 400(40) - 8000 = -8000 + 16000 - 8000 =", "+0 +1 38 1 How many points of intersection are there between the graphs of the following equations:\\begin{align*} y &=|2x + 5|, \\\\ y &= -|3x - 2| \\end{align*} Sep 20, 2020 #1 +246 0 Just graph it: 0 points, they head in opposite directions. Sep 20, 2020", "The graph of $y = -1.5 t^2 + 100$ is a negative parabola that has a maximum turning point at (0, 100). Over the domain you give, it starts at the point (0, 100) and ends at the point (10, -50).", "## Precalculus (6th Edition) Blitzer (a) From the graph it can be seen that \\begin{align} & \\sin t=\\frac{PB}{a} \\\\ & =\\frac{XA}{a} \\end{align} $XA=a\\sin t$ And the length OA is equal to the length of arc PA The length of arc PA is $at$. Now, \\begin{align} & x=OA-xA \\\\ & =at-a\\sin t \\\\ & =a\\left( t-\\sin t \\right) \\end{align} Thus, the parametric equation for x is $x=a\\left( t-\\sin t \\right)$. (b) \\begin{align} & \\cos t=\\frac{BC}{a} \\\\ & BC=a\\cos t \\end{align} Also, $AC=a$ Then, \\begin{align} & y=AC-BC \\\\ & =a-a\\cos t \\\\ & =a\\left( 1-\\cos t \\right) \\end{align} Thus, the parametric equation for y is given by $y=a\\left( 1-\\cos t \\right)$.", "### Home > PC3 > Chapter 12 > Lesson 12.3.1 > Problem12-136 12-136. This problem is a checkpoint for working with parametrically-defined functions. 1. Graph $x(t)=t^2+3,$ $y(t)=2t−1\\text{ for }$$−3\\le t\\le3$. 2. Rewrite the parametrically-defined function in part (a) in rectangular form.", "How to get intersection values from a parametric graph? for a simple exception of Solve). – Artes Apr 21 '14 at 9:01 • You could also use NSolve[Sin[x] == 0 && 0 <= x <= 7, x] which gives the result {{x -> 0.}, {x -> 3.14159}, {x -> 6.28319}}. – Stephen Luttrell Apr 21 '14 at 11:01", "Key questions See the scaffolding given in the question itself. Consider how the two Cartesian graphs $x=f(t)$ and $y=g(t)$ can give insight into the shape of the parametric graph $x=f(t)$, $y=g(t)$. Can you use this to visualisehow the graph unfolds as $t$ increases from $0$? Possible support The question Witch of Agnesi is a little easier and might be tackled first. Possible extension Squareness is another graph sketching problemon the family of relations: $x^n + y^n = 1$ where the curves have different forms for odd and even values of $n$.", "for homework. ### Possible support Students could start by exploring quadratic graphs before moving on to cubics. Parabolic Patterns provides a nice challenge that uses similar thinking. ### Key questions Where does the graph cross the x axis? Where does it cross the y axis? Does the graph have any reflectional or rotational symmetry? Does the graph have any turning points? ### Possible extension Students could explore the graphs of Ellipses in a similar way.", "+0 # help 0 85 2 Find the sum of all possible values of the constant $$k$$ such that the graph of the parametric equations \\begin{align*} x &= 2+ 4\\cos s,\\\\ y &= k-4\\sin s, \\end{align*}intersects the graph of the parametric equations \\begin{align*} x&=1+\\cos t,\\\\ y&=-3+\\sin t \\end{align*} at only one point. Nov 6, 2019 #1 0 The possible values of k are 5 + sqrt(7) and 5 - sqrt(7), and their sum is 10. Nov 6, 2019 #2 +23895 +3 Find the sum of all possible values of the constant k such that the graph of the parametric equations \\begin{align*} x &= 2+ 4\\cos s,\\\\ y &= k-4\\sin s, \\end{align*} intersects the graph of the parametric equations \\begin{align*} x&=1+\\cos t,\\\\ y&=-3+\\sin t \\end{align*} at only one point. $$\\text{Let the center of circle 1, ~ \\mathbf{c_1} = \\dbinom{2}{k} } \\\\ \\text{Let the radius of circle 1, ~ \\mathbf{r_1} = 4 } \\\\ \\text{Let the center of circle 2, ~ \\mathbf{c_2} = \\dbinom{1}{-3} } \\\\ \\text{Let the radius of circle 2, ~ \\mathbf{r_2} = 1 } \\\\ \\text{\\mathbf{\\overline{c_1c_2}} is the distance between c_1 and c_2 }$$ There is only one point, if $$\\mathbf{|\\overline{c_1c_2}| = r_1+r_2 }$$ $$\\begin{array}{|rclcl|} \\hline |\\overline{c_1c_2}| &=& r_1+r_2 \\\\ |\\sqrt{(2-1)^2+(k-(-3))^2} | &=& 4+1 \\\\ |\\sqrt{1+(k+3)^2} | &=& 5 \\\\ 1+(k+3)^2 &=& 25 \\\\ (k+3)^2 &=& 24" ]

[ "line $[\\frac{1}{2},1] \\!\\times\\!\\{(\\theta_0,1)\\}$ is pushed to the top radial line$\\{\\theta_0\\}\\!\\times\\mathbb{I}\\times\\!\\{1\\}$, where the point $(1,\\theta_0,1)$ does not move. Let us write $(r,s)$ instead of $(\\theta_0,r,s)$. For $0\\!\\leq\\!s\\!\\leq\\!\\frac{1}{2}$ the point $(1,s)$ therefore travels to $(0,2s)$, i.e. we have a straight path $t \\mapsto (1,s) + t\\big((0,2s)\\!-\\! (1,s)\\big) \\!=\\!\\big(1\\!-\\!t, s(1\\!+\\!t)\\big)$. For $\\frac{1}{2}\\!\\leq\\!s\\!\\leq\\!1$ the point $(1,s)$ travels to $(2(s\\!-\\!\\frac{1}{2}),1)$, i.e. we have a path $t \\mapsto (1,s) +t\\big((2s\\!-\\!1,1) \\!-\\!(1,s)\\big) \\!=\\!\\big(1\\!+\\!2t(s\\!-\\!1),s\\!+\\!t(1\\!-\\!s)\\big)$. Let us construct the homotopy for the remaining points as well (those from the whole square $\\mathbb{I}\\!\\times\\!\\mathbb{I}$). The interval $0\\!\\leq\\!s\\!\\leq\\!\\frac{1}{2}\\!+\\!\\frac{1-r}{2} \\!=\\!\\frac{2-r}{2}$ will be mapped at the end of the homotopy to $0\\!\\leq\\!s\\!\\leq\\!1$, hence each point $(r,s)$ travels to $(0,\\frac{2}{2-r}s)$, i.e. we have a path $t \\mapsto (r,s) +t\\big((0,\\frac{2}{2-r}s)\\!-\\! (r,s)\\big) \\!=\\!\\big(r\\!-\\!rt,s\\!+\\!t(\\frac{2}{2-r}s\\!-\\!s)\\big) \\!=\\!\\big(r(1\\!-\\!t),s\\!+\\!st\\frac{r}{2-r}\\big)$. Now we must deal with the remaining points $(r,s)$ from the triangle $\\frac{2-r}{2}\\!\\leq\\!s\\!\\leq\\!1$. The line through $(0,1)$ and $(1,\\frac{1}{2})$ is $y\\!-\\!1\\!=\\!\\frac{1/2-1}{1-0}(x\\!-\\!0)$, i.e. $y\\!=\\!-\\frac{1}{2}x\\!+\\!1$. The line through $(1,1)$ in $(r,s)$ is $y\\!-\\!1\\!=\\!\\frac{s-1}{r-1}(x\\!-\\!1)$. These two lines intersect in $(\\frac{2s-2}{r+2s-3},\\frac{r+s-2}{r+2s-3})$. The ratio of the distances is $\\|-(1,1)\\!+\\!(r,s)\\| \\,/\\, \\|-(1,1)\\!+\\!(\\frac{2s-2}{r+2s-3}, \\frac{r+s-2}{r+2s-3})\\|\\!=\\!\\ldots\\!=\\!|r\\!+\\!2s\\!-\\!3|\\!=\\!3\\!-\\!r\\!-\\!2s\\in[0,1]$. Therefore the point $(r,s)$ is mapped to $\\big(1\\!-\\!(3\\!-\\!r\\!-\\!2s),1\\big)\\!=\\! (r\\!+\\!2s\\!-\\!2,1)$, i.e. we have a path $t\\mapsto (r,s)\\!+\\!t\\big( (r\\!+\\!2s\\!-\\!2,1) \\!-\\!(r,s)\\big) \\!=\\!\\big(r\\!+\\!2t(s\\!-\\!1),s\\!+\\!t(1\\!-\\!s)\\big)$. The desired strong deformation retraction is therefore $$\\;\\;\\;\\;\\;\\;\\;h\\!:\\;\\;\\mathbb{B}^l\\!\\times\\!\\mathbb{I}\\! \\times\\!\\mathbb{I}\\!\\longrightarrow\\!\\mathbb{B}^l\\!\\times\\!\\mathbb{I},\\;\\;\\; (\\theta,r,s,t)\\mapsto \\begin{cases} \\displaystyle \\big(\\theta,r(1\\!-\\!t),s\\!+\\!st{\\textstyle\\frac{r}{2-r}})\\big) & \\hspace{-8pt} ;\\;\\; \\displaystyle r\\!+\\!2s\\!\\leq\\!2,\\\\ \\displaystyle \\big(\\theta,r\\!+\\!2t(s\\!-\\!1),s\\!+\\!t(1\\!-\\!s)\\big) &\\hspace{-8pt} ;\\;\\; \\displaystyle r\\!+\\!2s\\!\\geq\\!2. \\;\\;\\;\\;\\;\\;\\;\\;\\;\\blacksquare \\end{cases}$$", "\\frac{9 \\pi}{ \\sqrt{9 \\pi ^2}} = 3,$$ whereas $$|| f'( \\pi ) || = \\sqrt{9 + 64{\\pi}^2} \\approx 25.31,$$ hence $\\frac{d}{dt} || f(t) || \\neq || f'(t) ||$. $\\square$ Example. Let $$r(t) = ( t^2, \\sin t, \\cos t )$$ and let $$p(t) = ( t, \\ln( t + 1 ), 2t ).$$ Find $( r(t) \\times p(t) )'$. Solution: From the cross-product rule, it is $$( r(t) \\times p(t) )' = ( 2t, \\cos t, -\\sin t) \\times ( t, \\ln( t + 1 ), 2t ) + ( t^2, \\sin t, \\cos \\ t ) \\times ( 1, \\frac{1}{t + 1}, 2 ).$$ $\\square$ Example. Let $$F = yi - xj + 2k = < y, -x, 2 >$$ be a vector field. Verify that the path $$g(t) = ( \\sin t, \\cos t, 2t )$$ is a flow line (i.e. a path such that the velocity along the path is a vector in the vector field, $g'(t) = F(g(t)) )$ for the vector field $F$. Solution: With $$( x, y, z ) = ( \\sin t, \\cos t, 2t )$$ it follows that $$g'(t) = ( \\cos t, -\\sin t, 2 ),$$ and $$F(g(t)) = ( y, -x, 2t ) = ( \\cos t, -\\sin t, 2t ).$$ $\\square$ Example. Consider the path $g: {\\bf", "-\\sqrt{V_0}\\sin \\theta(t) \\dot{\\theta}(t) &=& -2 \\sqrt{\\frac{v_0}{2}} \\sin \\theta(t) (1-z) \\\\ \\dot{y}(t) &=& \\sqrt{\\frac{v_0}{2}} \\cos \\theta(t) \\dot{\\theta}(t) &=& \\sqrt{V_0} \\cos \\theta(t) (1-z) \\end{eqnarray} from which it follows that $\\dot{\\theta}(t) = \\sqrt{2} (1-z)$. Now choose $T = 4 \\sqrt{2} \\pi$, which gives a neighborhood $U$ of $(0,0,0)$ such that if $p(t_0) \\in U$, then $|z(t)| < \\frac{1}{2}$ for any $t \\in [t_0, t_0+T]$. Then we have $\\dot{\\theta}(t) \\geq \\frac{1}{\\sqrt{2}}$, and hence $\\theta(t_0+T)-\\theta(t_0) \\geq \\frac{1}{\\sqrt{2}} T = 4\\pi$ (I choose $4 \\pi$ so I can find a point where the positive $x$-axis is crossed and still have $2 \\pi$ 'left to go'). Hence $(x(t),y(t))$ traverses the entire ellipse in some $[t_0,t_0+\\delta] \\subset [t_0, t_0+T]$. (It does not yet follow that $(x(t),y(t))$ is periodic.) Let $\\tau_0 \\geq t_0$ be the first time at which $(x(t),y(t))$ crosses the positive $x$-axis, let $\\tau_1 > \\tau_0$ be the first time at which $(x(t),y(t))$ crosses the positive $y$-axis, and let $\\Delta = \\tau_1 -\\tau_0$. For $t \\in [\\tau_0+\\Delta, \\tau_0+2\\Delta]$, define $\\tilde{x}(t) = -x(2\\tau_0+2 \\Delta-t)$, $\\tilde{y}(t) = y(2\\tau_0+2 \\Delta-t)$ and $\\tilde{z}(t) = z(2\\tau_0+2 \\Delta-t)$. It is straightforward to verify that $\\tilde{p} = (\\tilde{x},\\tilde{y},\\tilde{z})$ satisfies the differential equation with initial condition $\\tilde{p}(\\tau_0+\\Delta)=p(\\tau_0+\\Delta)$. By uniqueness, we have $p(t) = \\tilde{p}(t)$ for $t \\in [\\tau_0+\\Delta, \\tau_0+2\\Delta]$, and hence $(x(\\tau_0+2\\Delta), y(\\tau_0+2\\Delta),z(\\tau_0+2\\Delta))=(-x(\\tau_0), y(\\tau_0),z(\\tau_0))$ (note that $y(\\tau_0) = 0$ by choice of $\\tau_0$).", "&= \\frac{7 + 2(19)}{5} = 9 \\\\ \\pi_{5,20}(9) &= \\frac{9 + 4(19)}{5} = 17 \\\\ \\pi_{5,20}(17) &= \\frac{17 + 2(19)}{5} = 11 \\\\ \\pi_{5,20}(11) &= \\frac{11 + 1(19)}{5} = 6 \\\\ \\pi_{5,20}(6) &= \\frac{6 + 1(19)}{5} = 5 \\\\ \\pi_{5,20}(5) &= \\frac{5 + 0(19)}{5} = 1 \\end{align*} So for $$C = 5, L = 20$$, the cycle containing 1 is $$(1, 4, 16, 7, 9, 17, 11, 6, 5)$$. By contrast, if $$C = 4$$ and $$L=20$$ — same message length but one fewer columns used — the cycle containing 1 is $$(1, 5, 6, 11, 17, 9, 7, 16, 4)$$. Both are length 9, and in fact both contain the same elements, but the order in which things appear has changed. But back to the case with $$C = 5, L = 20$$ for a moment, what happens to the other elements of the set $$\\{0, 1, \\dots, 19\\}$$ — the ones that are not contained in that 9-cycle we computed? Interestingly, it turns out that with the exception of 19 — which is a fixed point in this permutation (plug it into the formula to see this) — all the other elements of the set belong to a second, parallel 9-cycle: $$(2, 8, 13, 14, 18, 15, 3, 12, 10)$$. This gives us important information about", "at 0:31 @Jonathan: Thanks. I should have checked Wolfram's form of the answer. – robjohn Nov 20 '11 at 0:37 robjohn, why $(1-e^{6\\pi i /5})$ and not $(1-e^{2\\pi i /5})$ ? Why for the residue : $\\frac{1}{5e^{2\\pi i/5}}$ and not : $\\frac{(z_{2}^{2}}{5(z_{2})^{4}}$ with $z_{2}=e^{\\pi i/5}$ ? – VVV Nov 20 '11 at 0:42 @VVV: we get 2 factors of $e^{2\\pi i/5}$ from the $x^2$ and 1 factor from the $\\mathrm{d}x$; that gives $e^{6\\pi i/5}$. The residue is $2\\pi i$ times \\begin{align}\\lim_{z\\to e^{\\pi i/5}}\\frac{z^2(z-e^{\\pi i/5})}{1+z^5}&=\\lim_{z\\to e^{\\pi i/5}}\\frac{z^2}{5z^4}\\\\&=\\frac{1}{5e^{2\\pi i/5}}\\end{align} – robjohn Nov 20 '11 at 0:53 When you use the path $z=x$ and the path $z=e^{2pi i/5}x$ you substitute $x$ and $e^{2pi i/5}x$ for every occurrence of $z$. Thus, $\\int_C\\frac{z^2}{1+z^5}\\;\\mathrm{d}z$ becomes $\\int_0^\\infty\\frac{x^2}{1+x^5}\\;\\mathrm{d}x$ on the curve $z=x$ and $\\int_0^\\infty\\frac{e^{4\\pi i/5}x^2}{1+x^5}\\;e^{2\\pi i/5}\\;\\mathrm{d}x$ on the curve $z=e^{2\\pi i/5}x$ (but then we subtract since the curve is inbound). – robjohn Nov 20 '11 at 1:29 I ran out of patience, but the partial fraction approach can give an indefinite integral in closed form, as the tenth roots of 1 have real and imaginary parts that can be written with nothing worse than square root signs. I got $$\\frac{x^2}{1 + x^5} \\; = \\; \\; \\frac{A}{x + 1} + \\frac{B_1 x + B_0}{x^2 - \\frac{1}{2}(\\sqrt 5 + 1)x + 1} + \\frac{C_1 x + C_0}{x^2", "\\begin{align} x(n+N) &= \\cos( \\frac{\\pi n}{6} + \\frac{\\pi N}{6}) \\\\ &= \\cos(\\frac{\\pi n}{6})\\cos(\\frac{\\pi N}{6}) - \\sin(\\frac{\\pi n}{6})\\sin(\\frac{\\pi N}{6}) \\\\ &= \\cos(\\frac{\\pi n}{6})(1) - \\sin(\\frac{\\pi n}{6})(0) \\\\ &= \\cos(\\frac{\\pi n}{6}) \\\\ &= x(n) \\tag{1} \\end{align} Eq.(1) proves the periodicity of $$\\cos(\\frac{\\pi n}{6})$$. As you can see, using trigonometric identities allow us to see clearly why the signal is periodic. It has nothing to do with $$\\pi$$ rather we could find the smallest value of $$N$$ that allows us to reach $$x(n)$$ from $$x(n+N)$$. Again, these kinds of details are crucial but authors tend to ignore them. Cuz today is Pi Day. Happy Pi Day everyone! Seriously though, to answer this belated question in somewhat more of a simple intuitive way. The period of the trig functions is $$2\\pi$$. In order for a trig signal to be periodic (hitting the same domain values, repeatedly), it's RadiansPerSample factor has to be a rational (as in not irrational) factor of $$2\\pi$$. Since $$\\pi$$ is quite irrational, this means the argument has to be $$r \\pi$$ where $$r$$ is a rational number (ratio of integers, say $$a/b$$). $$\\frac{1}{6}$$ is not a rational multiple of $$\\pi$$, but $$\\frac{\\pi}{6}$$ is. Nuttin' to it. • Good to revive the $\\pi$ day. $\\pi$ is quite irrational as you say. Not always though, like in a $\\pi$-based number", "\\hline 2\\pi & \\pi - \\pi x = 2\\pi & -1 \\\\ \\hline \\end{array}$ \\setlength{\\extrarowheight}{0pt} Substituting these $$x$$-values into $$g(x)$$, we generate the graph below and find the period to be $$1 - (-1) = 2$$. The associated sine curve, $$y = \\frac{\\sin(\\pi - \\pi x) - 5}{3}$$, is dotted in as a reference. \\hspace{.25in} \\begin{tabular}{m{2.7in}m{3in}} \\setlength{\\extrarowheight}{2pt} $\\begin{array}{|r||r|r|} \\hline x & g(x) & (x,g(x)) \\\\ \\hline 1 & \\text{undefined} & \\\\ \\hline \\frac{1}{2} & -\\frac{4}{3} & \\left(\\frac{1}{2}, -\\frac{4}{3} \\right) \\\\ \\hline 0 & \\text{undefined} & \\\\ \\hline -\\frac{1}{2} & -2 & \\left(-\\frac{1}{2}, -2\\right) \\\\ \\hline -1 & \\text{undefined} & \\\\ \\hline \\end{array}$ \\setlength{\\extrarowheight}{0pt} & \\begin{mfpic}[30]{-2}{2}{-3}{0.5} \\point[3pt]{(0.5,-1.3333), (-0.5, -2)} \\axes \\tlabel[cc](2,-0.3){\\scriptsize $$x} \\tlabel[cc](0.25,0.5){\\scriptsize \\(y} \\tcaption{One cycle of \\(y = \\frac{\\csc(\\pi - \\pi x) - 5}{3}$$.} \\xmarks{-1, -0.5, 0.5, 1} \\ymarks{-2,-1} \\tlpointsep{4pt} \\axislabels {x}{{\\scriptsize $$-1 \\hspace{7pt}} -1, {\\scriptsize \\(-\\frac{1}{2} \\hspace{7pt}} -0.5, {\\scriptsize \\(\\frac{1}{2}} 0.5, {\\scriptsize \\(1} 1} \\axislabels {y}{{\\scriptsize \\(-2} -2, {\\scriptsize \\(-1} -1} \\dotted \\function{-1, 1, 0.1}{(sin(3.14159 - 3.14159*x)-5)/3} \\dashed \\polyline{(-1, -3), (-1, -0.5)} \\dashed \\polyline{(1, -3), (1,-0.5)} \\arrow \\reverse \\arrow \\function{0.9196, 0.08040, -0.1}{((1/sin(3.14159 - 3.14159*x))-5)/3} \\arrow \\reverse \\arrow \\function{-0.08043, -0.9196, 0.1}{((1/sin(3.14159 - 3.14159*x))-5)/3} \\end{mfpic} \\\\ \\end{tabular} Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in", "periodic, with period $2\\pi$. \\begin{aligned} \\sin( x + 2 \\pi ) & = \\sin x \\\\ \\cos(x + 2\\pi) & = \\cos x \\\\ \\tan(x+2\\pi) &= \\tan x \\end{aligned} for all $$x$$. Recall the definitions (comes from the Pythagorean Theorem): Domain of $$\\tan x$$? $$\\tan x = \\frac{ \\sin x}{\\cos x}$$ To find domain, set denominator to 0 and solve. Find all $$x$$'s \\begin{aligned} \\cos x & = 0 \\\\ x & = \\ldots, \\frac{\\pi}{2}, \\frac{3\\pi}{2}, \\frac{5\\pi}{2}, \\ldots \\end{aligned} Question: Is it increasing? It looks like it's increasing everywhere... But the correct question is: on what intervals is it increasing? In particulr, is it increasing on $$(-\\infty, \\infty)$$? No. Because $$\\tan$$ is undefined at $$\\dfrac{\\pi}{2}, \\dfrac{3\\pi}{2}, \\ldots$$ in the first place. Plus, you can see the jumps. Review exercise: Plot the graph of the piece-wise defined function: $$f(x) = \\begin{cases} 3 - \\frac{1}{2}x & \\text{ for } x \\leq 2 \\\\ 2x - 5 & \\text{ for } x > 2 \\end{cases}$$ Plot: 1. Plot $$3 - \\dfrac{1}{2} x$$. 2. Plot $$3 - \\dfrac{1}{2}x$$ with domain $$x \\leq 2$$. 3. Plot $$2x - 5$$. 4. Plot $$2x - 5$$ with domain $$x > 2$$. 5. Combine 2) and 4). ## 12 New Functions From Old We set up a library of elementary functions: powers, roots, exponent, sin, cos.", "period $2\\pi$: \\begin{aligned} \\sin( x + 2 \\pi ) & = \\sin x, \\\\ \\cos(x + 2\\pi) & = \\cos x, \\\\ \\end{aligned} for all $x$. Neither function is one-to-one. What we did with $y=x^2$ vs. $x=\\sqrt{x}$ and will repeat here is: • restrict the domain of the function to be able to define its inverse. We choose the branch of $\\sin$ over $-\\pi/2 \\le x \\le \\pi/2$. Thus, we have a pair of inverse functions, the sine (restricted) and the arcsine: $$y=\\sin x,\\ x\\in [-\\pi/2,\\pi/2], \\text{ and } x=\\sin^{-1}y,\\ y\\in [-1,1].$$ The graphs are of course the same with just $x$ and $y$ interchanged. Similarly, we choose the branch of $\\cos$ over $0 \\le x \\le \\pi$. Thus, we have a pair of inverse functions, the cosine (restricted) and the arccosine: $$y=\\cos x,\\ x\\in [0,\\pi], \\text{ and } x=\\cos^{-1}y,\\ y\\in [-1,1].$$ To find the domain of a fraction, such as $$\\tan x =\\frac{\\sin x}{\\cos x},$$ set the denominator equal to $0$ and solve. Find all $x$'s: \\begin{aligned} \\cos x & = 0 \\ \\Longrightarrow\\\\ x & = ..., \\frac{\\pi}{2}, \\frac{3\\pi}{2}, \\frac{5\\pi}{2}, ... \\end{aligned} Then we should choose the branch of $\\tan$ over $-\\pi/2< x< \\pi/2$. Then, we again have a pair of inverse functions, the tangent (restricted) and the arctangent: $$y=\\tan x,\\ x\\in (-\\pi/2,\\pi/2), \\text{ and }", "\\left( t-4 \\right) }{2\\,\\tanh \\left( t-2 \\right) \\tanh \\left( t-4 \\right) \\left( 1- \\left( \\tanh \\left( t-2 \\right) \\right) ^{2} \\right) + \\left( \\tanh \\left( t-2 \\right) \\right) ^{2} \\left( 1- \\left( \\tanh \\left( t-4 \\right) \\right) ^{2} \\right) }}$ then the multiplier of the root of $g$ at $t=2$ is equal to $1/2$ so that the multiplicity of $f(2)$ is $m=1/(1-1/2)=1/(1/2)=2$ , a formula which cannot be applied to $f$ directly due to its multiple root being an indifferent fixed-point of $N_f(x)$ • what ? ${}{} {}{}$ – reuns Feb 27 '17 at 20:56 • the multiplier $l(a)$ of the fixed-point $a$ of the Newton map is the absolute value of the derivative of the map evaluated at the root $Z(a)=0$, which is a fixed-point of the Newton map $N_Z(a)=a$. The multiplicity $m(a)$ of the root $a$ to which the fixed-point corresponds is equal to $m(a)=1/(1-l(a))$ when $l(a) \\neq 1$ . The zeros of $Z(t)/Z'(t)$ are the same as the zeros of $\\frac{\\frac{Z (t)}{\\dot{Z} (t)}}{\\frac{d}{dt} \\frac{Z (t)}{\\dot{Z} (t)}}$ – crow Feb 27 '17 at 21:15 • $f(z) = z^2, N_f(z) = \\frac{z}{2}, N_f'(z) = 1/2$ – reuns Feb 27 '17 at 21:49 • $f(z) = e^{-1/z}$ has an essential singularity at $z=0$. You can't really think to it as a zero of order $\\infty$ but $\\lim_{z \\to", "a\\sin b-\\cos b\\sin a)^{2}\\le \\frac{1}{4}$$ $$\\sin^{2} (b-a) \\le \\frac{1}{4}$$ So we want $-\\frac{1}{2} \\le \\sin (b-a) \\le \\frac{1}{2}$, which is equivalent to $-30 \\le b-a \\le 30$ or $150 \\le b-a \\le 210$. The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\\frac{45^2}{75^2} = 1- \\frac{9}{25}=\\frac{16}{25}$ and the answer is $16+25 = \\boxed{41}$ Solution by Leesisi ## Solution 3 (Quicker Trig) $[asy] pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label(\"O\", O, 2*E); label(\"P\", P, 2*W); label(\"Q\", Q, NE); label(\"R\", R, NW); label(\"200\", (0,0), 2*S); label(\"x\", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); [/asy]$ Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \\cos a, PQ = 200 \\sin a, PR = 200 \\sin b, OR = 200 \\cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral. Thus, we are able to", "equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From similar triangles, we have that $PC=\\frac{x\\sqrt{x^2+18x}}{x+9}$. Similarly, $TP=QT=\\frac{9\\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem, $BQ=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem once again, $BP=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2+(\\frac{18\\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\\sqrt{405+\\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\\sqrt{432}$, so the answer is $(\\sqrt{432})^2=\\boxed{432}$. Solution 4 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\\frac{9x}{x+9}$. Now, apply law of cosines on $\\triangle PAB$. $$BP^2=\\left(\\frac{9x^2}{x+9}\\right)^2+18^2-2(18)\\left(\\frac{9x}{x+9}\\right)\\cos(\\angle PAB).$$ We can see that $\\cos(\\angle PAB)=\\cos(180^{\\circ}-\\angle", "0:31 @Jonathan: Thanks. I should have checked Wolfram's form of the answer. – robjohn Nov 20 '11 at 0:37 robjohn, why $(1-e^{6\\pi i /5})$ and not $(1-e^{2\\pi i /5})$ ? Why for the residue : $\\frac{1}{5e^{2\\pi i/5}}$ and not : $\\frac{(z_{2}^{2}}{5(z_{2})^{4}}$ with $z_{2}=e^{\\pi i/5}$ ? – VVV Nov 20 '11 at 0:42 @VVV: we get 2 factors of $e^{2\\pi i/5}$ from the $x^2$ and 1 factor from the $\\mathrm{d}x$; that gives $e^{6\\pi i/5}$. The residue is $2\\pi i$ times \\begin{align}\\lim_{z\\to e^{\\pi i/5}}\\frac{z^2(z-e^{\\pi i/5})}{1+z^5}&=\\lim_{z\\to e^{\\pi i/5}}\\frac{z^2}{5z^4}\\\\&=\\frac{1}{5e^{2\\pi i/5}}\\end{align} – robjohn Nov 20 '11 at 0:53 When you use the path $z=x$ and the path $z=e^{2pi i/5}x$ you substitute $x$ and $e^{2pi i/5}x$ for every occurrence of $z$. Thus, $\\int_C\\frac{z^2}{1+z^5}\\;\\mathrm{d}z$ becomes $\\int_0^\\infty\\frac{x^2}{1+x^5}\\;\\mathrm{d}x$ on the curve $z=x$ and $\\int_0^\\infty\\frac{e^{4\\pi i/5}x^2}{1+x^5}\\;e^{2\\pi i/5}\\;\\mathrm{d}x$ on the curve $z=e^{2\\pi i/5}x$ (but then we subtract since the curve is inbound). – robjohn Nov 20 '11 at 1:29 I ran out of patience, but the partial fraction approach can give an indefinite integral in closed form, as the tenth roots of 1 have real and imaginary parts that can be written with nothing worse than square root signs. I got $$\\frac{x^2}{1 + x^5} \\; = \\; \\; \\frac{A}{x + 1} + \\frac{B_1 x + B_0}{x^2 - \\frac{1}{2}(\\sqrt 5 + 1)x + 1} + \\frac{C_1 x + C_0}{x^2 +", "of the path would begin at $$O$$ and proceed in the direction described by an $$m$$-vector with components $$b_j - d c_j$$, for $$j=1,\\dots,m$$. How to find the intersection point $$O$$? In our simple example we expressed the steepest ascent and constraint relationships as $$x_2$$ in terms of $$x_1$$. Setting the two $$x_2$$ expressions equal to one another allowed us to solve for $$x_1$$. That's made more general as follows. We know that a steepest ascent path is given by $$x_j = \\rho b_j$$, for $$j=1,\\dots,m$$ and that $$c_0 + \\sum_{j=1}^m c_j x_j = 0$$ must also hold. Therefore, they collide at $$\\rho_0$$ satisfying $c_0 + \\left(\\sum_{j=1}^m c_j b_j \\right) \\rho_o = 0. \\quad\\quad \\mbox{Solving gives } \\quad \\rho_o = \\frac{- c_0}{\\sum_{j=1}^m c_j b_j}.$ Using that calculation, coordinates of the intersection point are given as $$O = (x_1^o, \\dots, x_m^o)$$ where $$x_j^o = \\rho_o b_j$$, for $$j=1,\\dots,m$$. One may then move along the modified portion of the path, starting at $$O$$, following $x_j = x_j^o + \\lambda(b_j - d c_j),$ where $$d$$ is determined by the least squares solution, above, and $$\\lambda \\geq 0$$ is chosen by the practitioner. In this way the process for following the modified path is identical to that in Algorithm 3.1 except that the starting point is $$O$$ rather than the zero-vector, and", "periodic with period $2 \\pi$. Let $\\displaystyle \\map I {2 m, n} = \\int_0^\\pi x^{2 m} \\cos n x \\rd x$. Let $\\map A {2 m, n} = \\dfrac {\\pi^{2 m - 1} \\paren {-1}^n 2 m} {n^2}$. Let $\\map B {2 m, n} = -\\dfrac {2 m \\paren {2 m - 1} } {n^2}$. $\\ds \\map S x$ $=$ $\\ds \\frac {\\frac 2 \\pi \\int_0^\\pi \\map S x \\rd x} 2 + \\sum_{n \\mathop = 1}^\\infty \\cos n x \\frac 2 \\pi \\int_0^\\pi \\map S x \\cos n x \\rd x$ $\\ds$ $=$ $\\ds \\frac {\\int_0^\\pi x^{2 k} \\rd x} \\pi + \\frac 2 \\pi \\sum_{n \\mathop = 1}^\\infty \\cos n x \\int_0^\\pi x^{2 k} \\cos n x \\rd x$ $\\ds$ $=$ $\\ds \\frac {\\pi^{2 k} } {2 k + 1} + \\frac 2 \\pi \\sum_{n \\mathop = 1}^\\infty \\cos n x \\map I {2 k, n}$ $\\ds \\leadsto \\ \\$ $\\ds \\map S \\pi$ $=$ $\\ds \\frac {\\pi^{2 k} } {2 k + 1} + \\frac 2 \\pi \\sum_{n \\mathop = 1}^\\infty \\cos n \\pi \\map I {2 k, n}$ $\\ds \\leadsto \\ \\$ $\\ds \\pi^{2 k}$ $=$ $\\ds \\frac {\\pi^{2 k} } {2 k + 1} + \\frac 2 \\pi \\sum_{n \\mathop = 1}^\\infty \\paren {-1}^n \\map I {2 k, n}$ $\\ds \\leadsto \\ \\$", "from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(45), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, -3), SW); label(\"C_2'\", (9, -3), SE); label(\"C_3'\", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label(\"C_4'\", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$", "\\frac{{3\\pi }}{2}, - \\frac{\\pi }{2},\\frac{\\pi }{2},\\frac{{3\\pi }}{2},\\frac{{5\\pi }}{2}, \\ldots$ and the graph will have asymptotes at these points. Here is the graph of tangent on the range $$- \\frac{{5\\pi }}{2} < x < \\frac{{5\\pi }}{2}$$. Finally, a couple of quick properties about $$R\\tan \\left( {\\omega \\,x} \\right)$$. $\\begin{array}{c} - \\infty < R\\tan \\left( {\\omega \\,x} \\right) < \\infty \\\\ {\\rm{Period}} = \\displaystyle \\frac{\\pi }{\\omega }\\end{array}$ For the period remember that $$\\tan \\left( x \\right)$$ has a period of $$\\pi$$ unlike sine and cosine and that accounts for the absence of the 2 in the numerator that was there for sine and cosine. 7. $$y = \\sec \\left( x \\right)$$ Show Solution As with tangent we will have to avoid $$x$$’s for which cosine is zero (remember that $$\\sec x = \\frac{1}{{\\cos x}}$$). Secant will not exist at $x = \\cdots , - \\frac{{5\\pi }}{2}, - \\frac{{3\\pi }}{2}, - \\frac{\\pi }{2},\\frac{\\pi }{2},\\frac{{3\\pi }}{2},\\frac{{5\\pi }}{2}, \\ldots$ and the graph will have asymptotes at these points. Here is the graph of secant on the range $$- \\frac{{5\\pi }}{2} < x < \\frac{{5\\pi }}{2}$$. Notice that the graph is always greater than 1 or less than -1. This should not be terribly surprising. Recall that $$- 1 \\le \\cos \\left( x \\right) \\le 1$$. So, 1 divided by something less than 1", "1 rotation it rolls forward 1 perimeter, which is $2\\pi R$. The question of the path length P moves along is less simple. The equations in question to define the position are $x(t) = R(t-\\sin t)$ and $y(t) = R(1-\\cos t)$. Thus $\\mathbf{P}(t) = (x(t),y(t))$ is such $\\mathbf{P}(0) = (0,0)$, it starts on the ground level. It next hits the ground at time $t = 2\\pi$. Given a parametrisation you get the path length between those times is $s = \\int_{0}^{2\\pi} \\sqrt{\\dot{x}^{2}+\\dot{y}^{2}} dt$. So let's compute that $\\dot{x}(t) = R(1-\\cos t)$ and $\\dot{y}(t) = R\\sin t$ so the bit inside the square root is $\\dot{x}^{2} + \\dot{y}^{2} = R^{2}\\Big( 1 - 2\\cos t + \\cos^{2}t + \\sin^{2}t \\Big) = 2 R^{2}(1-\\cos t)$ We want to take the square root of this and for those who remember their trig identity $\\cos 2\\theta = \\cos^{2}\\theta - \\sin^{2}\\theta = 1-2\\sin^{2}\\theta$ we set $\\theta = \\frac{t}{2}$ and get $1-\\cos t = 2\\sin^{2}\\frac{t}{2}$ and thus $s = \\int_{0}^{2\\pi}2R \\sin^{2}\\frac{t}{2} dt = 4R \\int_{0}^{\\pi}\\sin \\tau d\\tau = 8R$ So surprisingly the answer doesn't involve a factor of pi, it's just 8 times the radius. 5. ### TachBannedBanned Messages: 5,265 Is this homework? AN answered the first part (related to cycloid geometry), the second part is a simple problem of dynamics. You know the equation", "the first full cycle of the (exponentially decaying) sinusoid, denoted by the black dot. To find $t_*$, use the first term in $(2)$: $$g(t-5)=0\\implies \\sin\\left({\\sqrt{15}\\over 4}(t-5)\\right)=0\\implies t=5+{4n\\pi\\over \\sqrt{15}},$$ and $n=2$ produces the zero we seek: $$t_*=5+{8\\pi\\over \\sqrt{15}} \\implies \\boxed{T=t_*-5={8\\pi\\over \\sqrt{15}}},$$ from the way we set up the original problem (so that $T$ is the time from $t=5$ until we apply the second impulse). Since we already have $y(t_*)=0$, we just need to find $k$ such that $y'(t_*)=0$, then we can conclude $y(t)\\equiv 0$ for $t>t_*$. To accomplish this, \\begin{align*} \\lim_{t\\to t_*^+}y'(t)&=g'(t_*^+-5)+k\\,g'(t_*^+-(5+T))\\\\ &=g'\\left({8\\pi\\over \\sqrt{15}}^+\\right)+k\\,g'(0^+)\\\\ &={1\\over 2}\\exp(-2\\pi/\\sqrt{15})+k\\cdot{1\\over2}, \\end{align*} which vanishes when $$\\boxed{ k=-\\exp(-2\\pi/\\sqrt{15})=-\\exp\\left(-{8\\pi\\over \\sqrt{15}}/4\\right)=-\\exp(-T/4).}$$ Here's a graph of the solution $(2)$ with the values of $T$ and $k$ found above: - Nice and complete answer. – Dmoreno May 14 '14 at 8:01", "{ 5 } +\\sqrt { 7 } } +$....is (a) $\\sqrt { 2n+1 }$ (b) $\\frac { \\sqrt { 2n+1 } }{ 2 }$ (c) $\\sqrt { 2n+1 } -1$ (d) $\\frac { \\sqrt { 2n+1 } -1 }{ 2 }$ 5. The sum of the series C02- C12 + C22 .....+ (- 1)nC2n where n is an even integer is (a) 2nCn (b) (-1)n2nCn (c) (-1)n2nCn-1 (d) (-1)n/2nCn/2 6. The direction cosines of the vector $2\\hat{i}+2\\hat{j}-\\hat{k}$are (a) $\\frac { 2 }{ 3 } ,\\frac { 2 }{ 3 } ,-\\frac { 1 }{ 3 }$ (b) $\\frac { 2 }{ 3 } ,\\frac { 2 }{ 3 } ,\\frac { 1 }{ 3 }$ (c) $-\\frac { 2 }{ 3 } ,-\\frac { 2 }{ 3 } ,\\frac { 1 }{ 3 }$ (d) $-\\frac { 2 }{ 3 } ,-\\frac { 2 }{ 3 } ,-\\frac { 1 }{ 3 }$ 7. Find the odd one out of the following (a) $\\hat { i } +2\\hat { j } +3\\hat { k }$ (b) $2\\hat { i } +4\\hat { j } +6\\hat { k }$ (c) $7\\hat { i } +14\\hat { j } +21\\hat { k }$ (d) $\\hat { i } +3\\hat { j } +2\\hat { k }$ 8. Choose the correct" ]

[ "1.5*dir(Y-P), linewidth(4)); dot(\"O\", O, 1.5*E, linewidth(4)); [/asy]$ Let $d$ be the diameter of $\\odot O.$ It follows that \\begin{alignat*}{8} 2[PAC] &= d\\cdot PX &&= 56, \\\\ 2[PBD] &= d\\cdot PY &&= 90. \\end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have $$PX^2+PY^2=PO^2.$$ We rewrite this equation in terms of $d:$ $$\\left(\\frac{56}{d}\\right)^2+\\left(\\frac{90}{d}\\right)^2=\\left(\\frac d2\\right)^2,$$ from which $d^2=212.$ Therefore, we get $$[ABCD] = \\frac{d^2}{2} = \\boxed{106}.$$ ~MRENTHUSIASM Solution 3 (Similar Triangles) $[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"A\", A, 1.5*NW, linewidth(4)); dot(\"B\", B, 1.5*SW, linewidth(4)); dot(\"C\", C, 1.5*SE, linewidth(4)); dot(\"D\", D, 1.5*NE, linewidth(4)); dot(\"P\", P, 1.5*dir(P), linewidth(4)); dot(\"X\", X, 1.5*dir(20), linewidth(4)); dot(\"Y\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"O\", O, 1.5*E, linewidth(4)); [/asy]$ Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\\triangle APC$ and $\\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot", "Polar coordinates were introduced and a variety of functions were graphed (§10.3). The scores on Quiz 9 are 30,29,27,23,23,27,21,21,20,20,19,19,13,30,10,10 with μ=21.3 and σ=6.4. Individual scores are on Blackboard. Oct 23 More techniques for doing calculus on parametric curves were presented (§10.2). Here’s an animation of §10.1, #22, $$\\lambda(t)=(\\sin t,\\cos^2 t)$$, $$-2\\pi\\le t\\le 2\\pi$$. It shows snapshots of the position of the point along the path in increments of $$\\pi/10$$, with the corresponding value of $$t$$ printed in the middle. There are four complete cycles during the interval $$-2\\pi\\le t\\le 2\\pi$$. The animation just keeps repeating. Oct 22 Extra Credit 7. (Due October 30.) A tank has the shape of a torus (doughnut) with inner radius 1 meter and outer radius 3 meters. What is its surface area? Oct 20 Quiz 9 may include topics through §10.1. Parametric curves were differentiated today (§10.2). Here is a solution for Extra Credit 6. First, notice the domain of $$f$$ is constrained by two requirements: $x-x^2\\ge0 \\text{ and } -1\\le x\\le1.$ The first requires $$0\\le x\\le 1$$ and the second that $$-\\pi/2\\le x\\le\\pi/2$$. Therefore, the domain of $$f$$ is $$[0,1]$$. To determine the arc length, first compute \\begin{align*} 1+\\left( f'(x) \\right)^2 &= 1+ \\left( \\frac{1-2x}{2\\sqrt{x-x^2}} - \\frac{1}{2\\sqrt{x}}\\frac{1}{\\sqrt{1-x}} \\right)^2\\\\ &= \\frac{1}{1-x} \\end{align*} The integral for the arc length is then \\begin{align*} \\int_0^1\\sqrt{1+(f'(x))^2}\\,dx &=", "+0 # Help 0 142 1 A parametric graph is given by \\begin{align*} x &= \\cos t + \\frac{t}{2}, \\\\ y &= \\sin t. \\end{align*}How many times does the graph intersect itself between $x = 1$ and $x = 40$? Jul 21, 2021", "believe what then corresponds to your c_t^2 would be the spectral norm of the tridiagonal matrix A having all diagonal entries 2, all entries adjacent to the main diagonal 1, and all other entries 0. Alternatively, it's 2+||P_t||, where P_t is the adjacency matrix of the length t path. Surely this is a known quantity... – Kevin P. Costello Mar 2 '10 at 21:05 The principal eigenvalue for a path with 2t+1 vertices is 2 cos pi/(2t+2). – Douglas Zare Mar 3 '10 at 0:40 ## 3 Answers For $t$ fixed, the count is proportional to $\\lambda^n$, where $\\lambda = 2 \\cos \\frac\\pi{2t+2}$ is the principal eigenvalue of the adjacency matrix of the path with $2t+1$ vertices. The all-positive (Perron-Frobenius) eigenvector corresponding to $\\lambda$ is $$\\bigg(\\sin \\frac{\\pi}{2t+2}, \\sin \\frac{2\\pi}{2t+2},\\sin \\frac{2\\pi}{2t+2},\\dots,sin \\frac{(2t+1)\\pi}{2t+2}\\bigg).$$ Since $-\\lambda$ is also an eigenvalue, the stable behavior of the distribution of endpoints of paths which stay in $[-t,t]$ is an oscillation between the odd entries $$\\bigg(\\sin \\frac{\\pi}{2t+2}, 0,\\sin \\frac{3\\pi}{2t+2},0,\\dots,\\sin \\frac{(2t-1)\\pi}{2t+2},0,\\sin \\frac{(2t+1)\\pi}{2t+2}\\bigg).$$ and even entries $$\\bigg(0,\\sin \\frac{2\\pi}{2t+2}, 0,\\sin \\frac{4\\pi}{2t+2},0,\\cdots ,0,\\sin \\frac{2t\\pi}{2t+2},0\\bigg).$$ The exact count of paths staying in $[-t,t]$ is a sum of signed binomial coefficients. The number of paths from $0$ to $i$ is 0 if $n \\not \\equiv i ~\\mod 2$, and $n \\choose (n\\pm i)/2$ when $n \\equiv i ~\\mod 2$. The number", "0, the maximum value of f(x) will be $-\\frac{\\Delta}{4a}$ which is obtained if $x=-\\frac{b}{2a}$. The graph will be a concave parabola whose vertex is $V(-\\frac{b}{2a};-\\frac{\\Delta}{4a})$. The parabola is symmetric about the line that intersects $x=-\\frac{b}{2a}$, which is called the \"axis of symmetry\". That is the reason why, when we assign values to x, we choose values symmetric about $-\\frac{b}{2a}$. When drawing a graph, the points of intersection with the coordinate axes are very important. |. A point found on the Ox axis is of the form P(x, 0) because the distance from it to Ox is 0. If the point is found both on Ox and the graph of the function, it is also of the form P(x, f(x)) ⇒ f(x) = 0. Thus, to find the coordinates of the points of intersection with the Ox axis, we must solve the equation f(x)=0. We get to the equation a2 + bx + c = 0. The solutions of the equation depend on the sign of Δ = b2 - 4ac. We have the following situations: 1) Δ < 0 the equation has no solutions in R (the set of real numbers) the graph doesn't intersect Ox. The form of the graph will be: or 2) Δ = 0 the equation has two equal solutions $x_1=x_2=-\\frac{b}{2a}$ The graph is", "moves before and after it meets the stationary point. \\end{align*} &= 3(x^2 - 2x) - 1 \\\\ State the domain and range for $$g(x) = -2(x - 1)^2 + 3$$. &= -3\\frac{1}{2} \\end{align*}, \\begin{align*} &= 3(x^2 - 6x + 9) + 2x - 5 \\\\ Therefore, there are no $$x$$-intercepts and the graph lies below the $$x$$-axis. If $$g(x)={x}^{2}+2$$, determine the domain and range of the function. x= -\\text{0,71} & \\text{ and } x\\end{align*}. y &= a(x + p)^2 + q \\\\ x=\\frac{3 - \\sqrt{41}}{4} &\\text{ or } x=\\frac{3 + \\sqrt{41}}{4} \\\\ This is done by Completing the Square and the turning point will be found at (-h,k). \\text{Eqn. $$y = -(x+1)^2$$ is shifted $$\\text{1}$$ unit up. The range is therefore $$\\{ y: y \\geq q, y \\in \\mathbb{R} \\}$$ if $$a > 0$$. The co-ordinates of this vertex is (1,-3) The vertex is also called the turning point. Calculate the values of $$a$$ and $$q$$. x^2 &= \\frac{-2}{-5} \\\\ 20a&=-20 \\\\ A Parabola is the name of the shape formed by an x 2 formula . Learners must be able to determine the equation of a function from a given graph. When $$a = 0$$, the graph is a horizontal line $$y = q$$. &= 4x^2 -36x + 35 \\\\ \\[\\begin{array}{r@{\\;}c@{\\;}l@{\\quad}l} Turning Point USA (TPUSA), often known", "Find the point of intersection of the lines $$h(t)=3t−4$$ and $$j(t)=5−t$$. Solution Set $$h(t)=j(t)$$. \\begin{align} 3t-4&=5-t \\\\ 4t&=9 \\\\ t&=\\dfrac{9}{4} \\end{align} This tells us the lines intersect when the input is $$\\frac{9}{4}$$. We can then find the output value of the intersection point by evaluating either function at this input. \\begin{align} j\\Big( \\dfrac{9}{4} \\Big)&=5-\\dfrac{9}{4} \\\\ &= \\dfrac{11}{4}\\end{align} These lines intersect at the point $$\\Big(\\frac{9}{4},\\frac{11}{4}\\Big)$$. Analysis Looking at Figure $$\\PageIndex{25}$$, this result seems reasonable. Figure $$\\PageIndex{25}$$: Graph of two functions $$h(t) = 3t - 4$$ and $$j(t)=t +5$$ and their intersection at $$(\\frac{9}{4},\\frac{11}{4})$$. If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions? Yes. After setting the two equations equal to one another, the result would be the contradiction “$$0 = \\text{non-zero real number}$$”. $$\\PageIndex{7}$$: Look at the graph in Figure $$\\PageIndex{23}$$ and identify the following for the function $$j(t):$$ 1. y-intercept 2. x-intercept(s) 3. slope 4. Is $$j(t)$$ parallel or perpendicular to $$h(t)$$ (or neither)? 5. Is $$j(t)$$ an increasing or decreasing function (or neither)? 6. Write a transformation description for $$j(t)$$ from the identity toolkit function $$f(x)=x$$. Solution 1. $$(0,5)$$ 2. (5, 0) 3. Slope -1 4. Neither parallel nor perpendicular 5. Decreasing function 6. Given", "using the formula (5.48) for the mean passage time from $w_{m_{2}}$ to $v_{R}$. Starting from $x$, the time until a hit on either $v_{R}$ or $v_{r}$ has Geometric($2/(m_{1}-1)$) distribution, and the two vertices are equally likely to be hit first. So $E_{x}T_{v_{r}}=(m_{1}-1)/2+{\\textstyle\\frac{1}{2}}E_{v_{R}}T_{v_{r}}.$ The last three expressions give an equation for $E_{v_{R}}T_{v_{r}}$ whose solution is (5.47). And it is straightforward to check that $E_{v_{l}}T_{v_{r}}$ does achieve the maximum, using (5.45)–(5.47) to bound the general $E_{i}T_{j}$. It is straightforward to check $\\tau_{c}\\sim\\frac{\\alpha^{2}n^{2}}{2}.$ ###### Example 5.12 The lollipop. xxx picture This is just the barbell without the right bell. That is, we start with a complete graph on $m_{1}$ vertices and add $m_{2}$ new vertices in a path. So there are $n=m_{1}+m_{2}$ vertices, and $w_{m_{2}}$ is now a leaf. In this example, by (5.45) and (5.46), with $m_{2}$ in place of $m_{2}+1$, we have $\\max_{ij}E_{i}T_{j}=E_{v_{l}}T_{w_{m_{2}}}=m_{1}(m_{1}-1)m_{2}+(m_{1}-1)+m_{2}% ^{2}.$ In the asymptotic setting with $n\\rightarrow\\infty,m_{1}/n\\rightarrow\\alpha,m_{2}/n\\rightarrow 1-\\alpha$ where $0<\\alpha<1$, we have $\\displaystyle\\max_{ij}E_{i}T_{j}$ $\\displaystyle\\sim$ $\\displaystyle\\alpha^{2}(1-\\alpha)n^{3}$ (5.49) $\\displaystyle\\sim$ $\\displaystyle\\frac{4n^{3}}{27}\\mbox{\\ for\\ }\\alpha=2/3,$ where $\\alpha=2/3$ gives the asymptotic maximum. Let us discuss the other parameters only briefly, in the asymptotic setting. Clearly $E_{w_{m_{2}}}T_{v_{L}}=m^{2}_{2}\\sim(1-\\alpha)^{2}n^{2}$ and it is not hard to check $E_{w_{m_{2}}}T_{v_{l}}=\\max_{v}E_{v}T_{v_{l}}\\sim(1-\\alpha)^{2}n^{2},$ (5.50) whence $\\tau^{*}=\\max_{ij}(E_{i}T_{j}+E_{j}T_{i})=E_{v_{l}}T_{w_{m_{2}}}+E_{w_{m_{2}}}% T_{v_{l}}\\sim\\alpha^{2}(1-\\alpha)n^{3}.$ Because the stationary distribution puts mass $\\Theta(1/n)$ on the “bar”, (5.50) is also enough to show that $\\tau_{0}=O(n^{2})$. So by the general inequalities between", "corresponding value of $\\scriptsize x$. 2. From the following sketch: . 1. Determine the range for the parabola $\\scriptsize f(x)$. 2. Determine the domain for the parabola $\\scriptsize f(x)$. 3. Determine the equation for the parabola $\\scriptsize f(x)$. 3. Given $\\scriptsize y=2-2x^2$, determine the following: 1. The x-intercepts. 2. The y-intercepts. 3. The turning point. 4. The axis of symmetry 5. The range of the graph. 6. The domain of the graph. 7. Sketch the graph, showing the TP and all intercepts. The full solutions are at the end of the unit. # Unit 2: Solutions ### Exercise 2.1 1. . 1. $\\scriptsize y=-2x^2$: $\\scriptsize j(x)$ 2. $\\scriptsize y=3x^2+1$: $\\scriptsize m(x)$ 3. $\\scriptsize y=\\displaystyle \\frac{1}{2}x^2$: $\\scriptsize f(x)$ 4. $\\scriptsize y=x^2$: $\\scriptsize g(x)$ 5. $\\scriptsize y=-4x^2$: $\\scriptsize k(x)$ 6. $\\scriptsize y=3x^2-2$: $\\scriptsize h(x)$ 2. . 1. Domain: $\\scriptsize \\{x\\ |\\ x\\in\\mathbb{R}\\}$ Range: $\\scriptsize \\{q(x)\\ |\\ q(x)\\in\\mathbb{R},q(x)\\leq6\\}$ 2. The axis of symmetry is the y-axis or the line $\\scriptsize x=0$. 3. . \\scriptsize \\begin{align*} 4y+3x^2&=-48\\\\ \\therefore 4y&=-3x^2-48\\\\ \\therefore y&=-\\displaystyle \\frac{3}{4}x^2-12 \\end{align*} . Because the function is of the form $\\scriptsize y=ax^2+q$ the TP is the same as the y-intercept. Therefore, the TP is the point $\\scriptsize (0,-12)$. . The value of $\\scriptsize a$ is negative, therefore the graph curves downwards and the TP is a maximum. 4. . Back to", "through the vertex, so the x-coordinate of the vertex is half way between the x-coordinates of the crossing points (a.k.a the roots of $f(x)$). \\begin{aligned} x_\\mathrm{vertex} &= \\frac{x_\\mathrm{crossing1} + x_\\mathrm{crossing2}}2\\\\&= \\frac{ -1 +5/2}2\\\\&=\\frac{3/2}2\\\\&=3/4.\\end{aligned} So the x-coordinate of the vertex is $x=3/4$. This is also the real number that maximizes $f(w)$ over all real numbers $w$. In general, the x-coordinates of the crossing points can be found with the quadratic formula $$x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2 a}.$$ The $\\pm$ sign means that one crossing can be found by replacing $\\pm$ with + and the other can be found by replacing the $\\pm$ with -. But, if you have one number that is $x_1=\\alpha+\\beta$ and another that is $x_2=\\alpha-\\beta$, then the average of the two numbers is just $\\alpha$. In other words if the two values of $x$ are $x=\\alpha\\pm\\beta$, then the average value is just $\\alpha$. So, to computer the average x-coordinate of the crossing points, all we have to do is remove the $\\pm$ sign and whatever it is applied to from the quadratic formula. $$x_\\mathrm{vertex} = \\frac{-b}{2 a}.$$ ### theorem #2 Informally, if $y=f(x)$ is a parabola, $f(x) = a x^2 + b x + c$, and $a<0$, then the integer(s) that maximize $f(x)$ are the set $\\{\\mathrm{floor}(x+1/2), \\mathrm{ceil}(x-1/2)\\}$ where $x=\\frac{-b}{2 a}$. If $x$ is an integer plus 1/2 (e.g. 2.5,", "# Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$ Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$. First $$f'(x)=\\frac{d\\sin{\\pi}x}{d({\\pi}x)}\\frac{d({\\pi}x)}{dx} ={\\pi}\\cos{\\pi}x$$ Second $$f''(x)=-{\\pi^2}\\sin{\\pi}x$$ Third $$f'''(x)= -{\\pi^3}\\cos {\\pi}x$$ These are the steps the teacher has written down on the blackboard, I have trouble understanding \"How it is solved\". Would you please provide the steps? - LaTeX tip: There are preexisting operators \\sin and \\cos for $\\sin x$ and $\\cos x$. There is a quick reference guide here. – Aeolian Mar 6 '13 at 7:50 \\begin{align*} f(x) = \\sin\\pi x \\end{align*} Here, note that $\\pi$ is a constant. So treat it like a coefficient of $x$. Progressively, what I will do is to: • Isolate the coefficient since it is not involved in the differentiation • Apply chain rule by differentiating the trigonometric function, followed by the function \"inside\" the trigo function - the $\\pi x$. Note that $\\frac{d}{dx}\\begin{pmatrix}\\pi x\\end{pmatrix} = \\pi$. \\begin{align*} f(x) &= \\sin\\pi x \\\\ f'(x) &= (\\cos \\pi x) \\times \\pi \\space\\space\\tag{by chain rule}\\\\ &= \\pi \\cos \\pi x \\\\\\\\\\\\f''(x) &= \\pi (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^2 \\sin \\pi x \\\\\\\\\\\\f'''(x) &= \\pi^2 (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^3 \\cos \\pi x \\end{align*} And that is how you do it. First, understand that $\\cos$ is a trigonometric function. I would recommend you to write $f(x)$ and $f'(x)$ as a composite", "$$x_{2}=t^{2}+t-4$$ $$y_{2}=t$$ To find where these intersect, set $$x_{1}=x_{2}$$ and $$y_{1}=y_{2}$$ and solve. \\begin{aligned} t &=t^{2}+t-4 \\\\ t^{2} &=4 \\\\ t &=\\pm 2 \\end{aligned} You still need to actually calculate the points of intersection on the graph. You can tell from the graph in Example $$\\mathrm{C}$$ that there seem to be four points of intersection. Since $$t$$ can mean time, the question of intersection is more complicated than simply overlapping. It means that the points are at the same $$x$$ and $$y$$ coordinate at the same time. Note what the graphs look like when $$-1.8<t<1.8$$ Note what the graphs look like $$t>2.2$$ or $$t<-2.2$$ Notice how when these partial graphs are examined there is no intersection at anything besides $$t=\\pm 2$$ and the points (2,2) and (-2,-2) While the paths of the graphs intersect in four places, they intersect at the same time only twice. ##### Example 5 Identify where the following parametric function intersects with its inverse. $$x=4 t$$ $$y=t^{2}-16$$ $$x_{1}=4 t ; y_{1}=t^{2}-16$$ The inverse is: $$x_{2}=t^{2}-16$$ $$y_{2}=4 t$$ Solve for $$t$$ when $$x_{1}=x_{2}$$ and $$y_{1}=y_{2}$$. \\begin{aligned} 4 t &=t^{2}-16 \\\\ 0 &=t^{2}-4 t-16 \\\\ t &=\\frac{4 \\pm \\sqrt{16-4 \\cdot 1 \\cdot(-16)}}{2}=\\frac{4 \\pm 4 \\sqrt{5}}{2}=2 \\pm 2 \\sqrt{5} \\end{aligned} The points that correspond to these two times are: $$x=4(2+2 \\sqrt{5}), y=(2+2 \\sqrt{5})^{2}-16$$ $$x=4(2-2 \\sqrt{5}), y=(2-2 \\sqrt{5})^{2}-16$$", "turning point is $$(0;-3)$$. The value of $$a$$ affects the shape of the graph. \\text{Range: } & \\left \\{ y: y \\geq -1, y\\in \\mathbb{R} \\right \\} y &= x^2 - 2x -3 - 3 \\\\ $$y = ax^2 + bx + c$$ if $$a < 0$$, $$b = 0$$, $$c > 0$$. As a result, they often use the wrong equation (for example, … Similarly, if $$a < 0$$, the range is $$\\{ y: y \\leq q, y \\in \\mathbb{R} \\}$$. (2) - (3) \\quad -36&=20a-16 \\\\ \\end{align*}, \\begin{align*} The x-coordinate of the vertex can be found by the formula $$\\frac{-b}{2a}$$, and to get the y value of the vertex, just substitute $$\\frac{-b}{2a}$$, into the . \\end{align*}, \\begin{align*} If the parabola is shifted $$\\text{3}$$ units down, determine the new equation of the parabola. The Turning Point Formula. Functions allow us to visualise relationships in the form of graphs, which are much easier to read and interpret than lists of numbers. &= -3(x^2 - 2x - 6) \\\\ Now calculate the $$x$$-intercepts. \\end{align*}, \\begin{align*} &= 4x^2 -36x + 37 \\\\ \\end{align*} The $$x$$-intercepts are $$(-\\text{0,71};0)$$ and $$(\\text{0,71};0)$$. “We are looking at a turning point in Formula 1 because teams have always fought for resources in order to perform on track, and now it’s turning … These are the", "period $2\\pi$: \\begin{aligned} \\sin( x + 2 \\pi ) & = \\sin x, \\\\ \\cos(x + 2\\pi) & = \\cos x, \\\\ \\end{aligned} for all $x$. Neither function is one-to-one. What we did with $y=x^2$ vs. $x=\\sqrt{x}$ and will repeat here is: • restrict the domain of the function to be able to define its inverse. We choose the branch of $\\sin$ over $-\\pi/2 \\le x \\le \\pi/2$. Thus, we have a pair of inverse functions, the sine (restricted) and the arcsine: $$y=\\sin x,\\ x\\in [-\\pi/2,\\pi/2], \\text{ and } x=\\sin^{-1}y,\\ y\\in [-1,1].$$ The graphs are of course the same with just $x$ and $y$ interchanged. Similarly, we choose the branch of $\\cos$ over $0 \\le x \\le \\pi$. Thus, we have a pair of inverse functions, the cosine (restricted) and the arccosine: $$y=\\cos x,\\ x\\in [0,\\pi], \\text{ and } x=\\cos^{-1}y,\\ y\\in [-1,1].$$ To find the domain of a fraction, such as $$\\tan x =\\frac{\\sin x}{\\cos x},$$ set the denominator equal to $0$ and solve. Find all $x$'s: \\begin{aligned} \\cos x & = 0 \\ \\Longrightarrow\\\\ x & = ..., \\frac{\\pi}{2}, \\frac{3\\pi}{2}, \\frac{5\\pi}{2}, ... \\end{aligned} Then we should choose the branch of $\\tan$ over $-\\pi/2< x< \\pi/2$. Then, we again have a pair of inverse functions, the tangent (restricted) and the arctangent: $$y=\\tan x,\\ x\\in (-\\pi/2,\\pi/2), \\text{ and }", "have used the fact that the cosine has reflection symmetry $(\\cos(-x)=\\cos(x))$ as well as periodic translation symmetry $(\\cos(x)=\\cos(x+2\\pi n))$. Taking the + sign we get for $n=1$ that $\\frac{80}{3}t=2\\pi$ $t=\\frac{3\\pi}{40}$. Taking the - sign we get $\\frac{40}{3} t=2\\pi$ $t=\\frac{3 \\pi}{20}$. Taking the - sign with $n=2$ we get $\\frac{40}{3}t=4\\pi$ $t=\\frac{3 \\pi }{10}$. Alternatively, starting from your calculation of the periods $T_1=\\frac{\\pi}{10}$ and $T_2=\\frac{3\\pi}{10}$ : The blocks will return to the starting position at the same time when each block has made a whole number of cycles. The time after which the blocks reach this position is $t=mT_1=nT_2$ $\\frac{m}{n}=\\frac{T_2}{T_1}=3$. The smallest possible integer values are $m=3, n=1$. Then $t=nT_2=\\frac{3\\pi}{10}$. answered Mar 28, 2017 by (28,746 points) selected Mar 29, 2017 by koolman", "$m-t$ up-steps and $n-t$ down steps are used to build paths from $(0,0)$ to \\begin{align*} (m-t)(1,1)+(n-t)(1,-1)=(m+n-2t,m-n) \\end{align*} which start with an up-step and stay above the $x$-axis. According to Andre's reflection lemma is the number of paths from $(0,0)$ to $(m+n-2t,m-n)$ which do not cross the $x$-axis given as \\begin{align*} \\frac{m-n}{m+n-2t}\\binom{m+n-2t}{m-t}\\tag{5} \\end{align*} In (4) we do not have a factor $2^{m-n}$ since we consider only paths which are beyond the $x$-axis. Note: • Observe the structure of the binomial coefficient in (5) is the same as in (2) with $p=m-t$ and $q=m-n$. • Combinatorially this means the number of paths of length $m+n$ which start in $(0,0)$ with an up-step and end in $(m,m-n)$ at height $m-n=r$ and never touch the $x$-axis is the same as the number of paths of length $2m$ from $(0,0)$ to $(2n,0)$ which start with an up-step go never beyond the $x$-axis and touch the $x$ axis $r$ times, the last time at $(2n,0)$. • Alebraically this implies \\begin{align*} [z^{m-t}]\\left(\\frac{1-\\sqrt{1-4z}}{2}\\right)^{m-n} =\\frac{m-n}{m+n-2t}\\binom{m+n-2t}{m-t}\\tag{6} \\end{align*} We finally conclude from (4) and (6): The number $N(r,m,n)$ of valid paths is \\begin{align*} \\color{blue}{N(r,m,n)}&=\\sum_{t=r}^n[z^t]\\left(2zc(z)\\right)^r [z^{m-t}](zc(z))^{m-n}\\\\ &=\\frac{1}{2^{m-n}}\\sum_{t=r}^n[z^t]\\left(2zc(z)\\right)^r [z^{m-t}](2zc(z))^{m-n}\\\\ &=\\frac{1}{2^{m-n}}[z^m](2zc(z))^{m-n+r}\\\\ &=\\frac{1}{2^{m-n}}[z^m](1-\\sqrt{1-4z})^{m-n+r}\\\\ &\\color{blue}{=2^r\\frac{m-n+r}{m+n-r}\\binom{m+n-r}{n-r}}\\\\ \\end{align*} Example: $N(1,3,2)$ Let's calculate OPs example $N(1,3,2)$ counting the number of paths which touch the $x$-axis $r=1$ time consisting of $m=3$ up-steps and $n=2$ down-steps. We obtain from", "\\hline 2\\pi & \\pi - \\pi x = 2\\pi & -1 \\\\ \\hline \\end{array}$ \\setlength{\\extrarowheight}{0pt} Substituting these $$x$$-values into $$g(x)$$, we generate the graph below and find the period to be $$1 - (-1) = 2$$. The associated sine curve, $$y = \\frac{\\sin(\\pi - \\pi x) - 5}{3}$$, is dotted in as a reference. \\hspace{.25in} \\begin{tabular}{m{2.7in}m{3in}} \\setlength{\\extrarowheight}{2pt} $\\begin{array}{|r||r|r|} \\hline x & g(x) & (x,g(x)) \\\\ \\hline 1 & \\text{undefined} & \\\\ \\hline \\frac{1}{2} & -\\frac{4}{3} & \\left(\\frac{1}{2}, -\\frac{4}{3} \\right) \\\\ \\hline 0 & \\text{undefined} & \\\\ \\hline -\\frac{1}{2} & -2 & \\left(-\\frac{1}{2}, -2\\right) \\\\ \\hline -1 & \\text{undefined} & \\\\ \\hline \\end{array}$ \\setlength{\\extrarowheight}{0pt} & \\begin{mfpic}[30]{-2}{2}{-3}{0.5} \\point[3pt]{(0.5,-1.3333), (-0.5, -2)} \\axes \\tlabel[cc](2,-0.3){\\scriptsize $$x} \\tlabel[cc](0.25,0.5){\\scriptsize \\(y} \\tcaption{One cycle of \\(y = \\frac{\\csc(\\pi - \\pi x) - 5}{3}$$.} \\xmarks{-1, -0.5, 0.5, 1} \\ymarks{-2,-1} \\tlpointsep{4pt} \\axislabels {x}{{\\scriptsize $$-1 \\hspace{7pt}} -1, {\\scriptsize \\(-\\frac{1}{2} \\hspace{7pt}} -0.5, {\\scriptsize \\(\\frac{1}{2}} 0.5, {\\scriptsize \\(1} 1} \\axislabels {y}{{\\scriptsize \\(-2} -2, {\\scriptsize \\(-1} -1} \\dotted \\function{-1, 1, 0.1}{(sin(3.14159 - 3.14159*x)-5)/3} \\dashed \\polyline{(-1, -3), (-1, -0.5)} \\dashed \\polyline{(1, -3), (1,-0.5)} \\arrow \\reverse \\arrow \\function{0.9196, 0.08040, -0.1}{((1/sin(3.14159 - 3.14159*x))-5)/3} \\arrow \\reverse \\arrow \\function{-0.08043, -0.9196, 0.1}{((1/sin(3.14159 - 3.14159*x))-5)/3} \\end{mfpic} \\\\ \\end{tabular} Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in", "positive integers $n$ as follows: $T_n = cos(n acos(x))$. Now, it is an interesting exercise in trig identities to discover that these ARE polynomials to begin with; one shows this to be true for, say, $n \\in \\{0, 1, 2\\}$ by using angle addition formulas and the standard calculus resolution of things like $sin(acos(x))$. Then one discovers a relation: $T_{n+1} =2xT_n - T_{n-1}$ to calculate the rest. The $cos(n acos(x))$ definition allows for some properties to be calculated with ease: the zeros occur when $acos(x) = \\frac{\\pi}{2n} + \\frac{k \\pi}{n}$ and the first derivative has zeros where $arcos(x) = \\frac{k \\pi}{n}$; these ALL correspond to either an endpoint max/min at $x=1, x = -1$ or local max and mins whose $y$ values are also $\\pm 1$. Here are the graphs of $T_4(x), T_5 (x)$ Now here is a key observation: the graph of a $T_n$ forms $n$ spanning arcs in the square $[-1, 1] \\times [-1,1]$ and separates the square into $n+1$ regions. So, if there is some other function $f$ whose graph is a connected, piecewise smooth arc that is transverse to the graph of $T_n$ that both spans the square from $x = -1$ to $x = 1$ and that stays within the square, that graph must have $n$ points of intersection with the graph of", "Therefore the graph is a “smile” and has a minimum turning point. ### Calculate the intercepts For the $$y$$-intercept, let $$x=0$$: \\begin{align*} y& = 2{x}^{2}-4 \\\\ & = 2{(0)}^{2}-4 \\\\ & =-4 \\end{align*} This gives the point $$(0;-4)$$. For the $$x$$-intercepts, let $$y=0$$: \\begin{align*} y& = 2{x}^{2}-4 \\\\ 0& = 2{x}^{2}-4 \\\\ {x}^{2}& = 2 \\\\ \\therefore x& = ±\\sqrt{2} \\end{align*} This gives the points $$(-\\sqrt{2};0)$$ and $$(\\sqrt{2};0)$$. ### Determine the turning point From the standard form of the equation we see that the turning point is $$(0;-4)$$. ### Plot the points and sketch the graph Domain: $$\\left\\{x:x\\in \\mathbb{R}\\right\\}$$ Range: $$\\left\\{y:y\\ge -4,y\\in \\mathbb{R}\\right\\}$$ The axis of symmetry is the line $$x=0$$. ## Worked example 7: Sketching a parabola Sketch the graph of $$g(x)=-\\frac{1}{2}{x}^{2}-3$$. Mark the intercepts and the turning point. ### Examine the standard form of the equation We notice that $$a<0$$. Therefore the graph is a “frown” and has a maximum turning point. ### Calculate the intercepts For the $$y$$-intercept, let $$x=0$$: \\begin{align*} g(x)& = -\\frac{1}{2}{x}^{2}-3 \\\\ g(0)& = -\\frac{1}{2}{(0)}^{2}-3 \\\\ & = -3 \\end{align*} This gives the point $$(0;-3)$$. For the $$x$$-intercepts let $$y=0$$: \\begin{align*} 0& = -\\frac{1}{2}{x}^{2}-3 \\\\ 3& = -\\frac{1}{2}{x}^{2} \\\\ -2(3)& = {x}^{2} \\\\ -6& = {x}^{2} \\end{align*} There is no real solution, therefore there are no $$x$$-intercepts. ### Determine the turning point", "Thus, a tangent at one end-point and a normal at the other must be parallel ! Thus, you can see that the skill you need to master for this subject is to remember certain well known and frequently used results and use them to your advantage as much as possible. Example – 28 The normals at $${t_1}{\\rm{ }}\\;and\\;{\\rm{ }}{t_2}$$ of the parabola $${y^2} = 4ax.$$ meet at $${t_3}$$. Prove that $${t_1}{t_2} = 2.$$ Solution: We can very easily solve this question using the result of the last example. Since the normal drawn at $${t_1}$$ intersects the parabola again in $${t_3}$$, we have {t_3} = - {t_1} - \\begin{align}\\frac{2}{{{t_1}}}\\end{align}. Similarly, {t_3} = - {t_2} - \\begin{align}\\frac{2}{{{t_2}}}\\end{align}. Comparing the two gives $${t_1}{t_2} = 2.$$ But let us solve it without using this result. The equation of the normals at $${t_1}{\\rm{ }}\\,and\\,{\\rm{ }}{t_2}$$ are \\begin{align}&y + {t_1}x = 2a{t_1} + at_1^3 \\qquad \\dots \\left( 1 \\right)\\\\\\\\&y + {t_2}x = 2a{t_2} + at_2^3 \\qquad \\dots \\left( 2 \\right)\\end{align} Both of these must be satisfied by the point $${t_3}$$ , i.e., by $$(at_3^2,\\;2a{t_3}).$$ Thus, \\begin{align}&2a{t_3} + a{t_1}t_3^2 = 2a{t_1} + at_1^3 \\qquad \\dots \\left( 3 \\right)\\\\\\\\&2a{t_3} + a{t_2}t_3^2 = 2a{t_2} + at_2^3 \\qquad \\dots \\left( 4 \\right)\\end{align} Subtracting(4) from (3), we have \\begin{align}& \\qquad \\;\\;at_3^3({t_1} - {t_2}) = 2a({t_1} - {t_2}) + a(t_1^3" ]

[ "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", "= 2 Hover the mousse cursor on the graph or plotted point to read the coordinates. Click the button above \"Plot Equation\". You may hover the mousse cursor over the graph to trace the coordinates. You may also Hover the mousse cursor on the top right of the graph to have the option to download the graph as a png file. 1 - Select a point M on the graph of the hyperbola, hover the mousse over it and read the coordinates. Read the coordinates of F1 and F2 (right legend) or hover the mousse and read their coordinates and show that the difference of the distances $\\overline{MF_1} - \\overline{MF_2}$ is close to $2a$ if $\\overline{MF_1} \\gt \\overline{MF_2}$ or $- 2a$ if $\\overline{MF_1} \\lt \\overline{MF_2}$. You may do the above activity for different values of $a, b, h$ and $k$ with as many points as needed to better understand the definition of an hyperbola. 2 - Use the values of $a$ and $b$ to find $c$ using the relationship between $a, b$ and $c$ given by $c^2 = a^2 + b^2$. Use the above results to find the coordinates of $F_1, F_2, V_1$ and $V_2$ and check the results graphically. 3 - Set $a, b, h$ and $k$ to some values. Find the x and y intercepts and", "+0 0 183 1 Let S be the set of all points in the plane that can be expressed in the form (\\cos a + \\cos b, \\sin a + \\sin b) where a and b are real numbers. Find the area of the set S Apr 12, 2020 #1 +31506 +1 You get some nice curves with x = cos(a) + cos(b); y = sin(a) + sin(b). Presumably, with an infinite number of points the area covered is that of a circle with radius 2. Area = 4pi. Apr 12, 2020", "to be plotted are J/ ^ X* -{- x'', y ■= — bx -\\- c, the first of which may be drawn once for all upon cross-section paper, while the straight line represented by the second may be drawn for each particular case, as described above.* This method is also applicable to many transcendental equa- tions ; thus for the equation .^;f — ^sin;f = o it is best to write ax — sin;r = o ; then y := smx is readily plotted by help of a table of sines, while y ^= ax is a straight line passing through the origin. In the same way a\"^ — x'^ ^^ o gives the curve represented byj:=rt'*and the parabola represented by y = x\"^, the intersections of which determine the real roots of the given equation. Prob. 2. Devise a graphic solution for finding approximate values of the real roots of the equation x''-\\- ax^-\\- bx''-{- ex + d =o. Prob. 3. Determine graphically the number and the approximate values of the real roots of the equation arc x — 2> sin x = o. (Ans. — Six real roots, x = ± 159°, ± 430°, and ± 456°.) * For an extension of this method to the determination of imaginary roots, see Phillips and Beebe's Graphic Algebra,", "the required standard form: $$\\displaystyle x+\\sqrt{3}y-13=0$$ Here is a plot of the hyperbola and the normal line at the given point:", "Free Version Easy # Given Slope, Find the Type of Curve APCALC-SBPLKC If the slope of a certain curve is $-\\cfrac {x}{y}$ for every point $(x, \\, y)$, then the curve is a(n) A Hyperbola. B Parabola. C Ellipse. D Circle.", "plot $y=\\sin(x)$ as a dotted line and $y=2\\sin(x)$ as a solid line. The resulting graph should look like The resulting graph should look like: Level 6-7 Level 6-7 Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "parabola $y = x^2$ for $x \\in [-2,2]$ using only 5 points: x = [-2,-1,0,1,2] y = [4,1,0,1,4] plt.plot(x,y) plt.show() This is too few points to plot a smooth curve such as $y = x^2$ and so we need more points! Let's try again using the NumPy function np.linspace to create 100 points! x = np.linspace(-2,2,100) y = x**2 plt.plot(x,y) plt.show() That's a better representation of the parabola $y = x^2$. Note that the number of points we use in a line plot (100 in this case) is completely arbitrary but the goal is to show a smooth graph for a smooth curve and so we just need to pick a big enough number depending on the function. But be careful not to generate too many points since a very large number of points will take a long time to plot! Now that we have the general idea, let's look at adding style and features to our plots! ### Line Properties A line appearing in a plot has several properties including color, transparency, style, width and markers. We can set these properties when we call plt.plot using the following keyword arguments: Property Description alpha transparency (0.0 transparent through 1.0 opaque) color (or c) any matplotlib color label text appearing in legend linestyle (or ls) solid, dashed, dashdot, dotted", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "can be plotted on a number line. Real numbers include all rational and irrational numbers. rectangular coordinates A point is written using rectangular coordinates if it is written in terms of $x$ and $y$ and can be graphed on the Cartesian plane. rectangular form The rectangular form of a point or a curve is given in terms of $x$ and $y$ and is graphed on the Cartesian plane.", "normally on the (u,v) axis? so for the first, it will be like plotting on the (x,y) axis something like y^2 = x^2 - c, for different c, how many lines would i need to plot? and how would i mark it? would i say what c is each time just to be sure these are the right lines, they are curvy like a hourglass, and they meet the axis at sqrt(c) on each side, so for each c we get two of these curves, so i need to plot all this twice. and then, also for the y=2uv, do i also plot this many times (for the same c) would i need to put all this lines on to the same plot? and mark each time which is c? OR DO I NEED TO SHOW JUST WHERE X AND Y ARE BOTH CONSTANT? the wording has ambiguity. I think this is the sort of thing they are looking for (it's an image I found online by searching for \"orthogonal hyperbola family\"). The family of \"y=2uv\" hyperbolas looks like the family of \"hourglass\" hyperbolas rotated through 45º. The important thing about the two families of curves is that wherever a curve from one family meets a curve from the other family, they intersect at right angles to each", "$y=x^2$ for $x$ to find the $y$ values and plot the points and draw your parabola.", "starts at $t, x = 0, 1$ (where the units are meters), i.e., on your worldline, but then moves in the $x$ direction slower than you do, so its worldline is a shallower curve than the hyperbola you are following. At some point, the tip will cross the line $x = t$, i.e., the horizon, and at that point the tip can't keep up with you, no matter how hard you pull on it. 3. Jun 19, 2017 ### Cato OK, thanks so much for the detailed and really useful discussion.", "plotted on a number line. Real numbers include all rational and irrational numbers. rectangular coordinates A point is written using rectangular coordinates if it is written in terms of $x$ and $y$ and can be graphed on the Cartesian plane. rectangular form The rectangular form of a point or a curve is given in terms of $x$ and $y$ and is graphed on the Cartesian plane. superset A superset is a set that includes other sets within it. The complex number set includes all classic complex numbers $a+bi$, and also, because $bi$ may equal 0, it includes all of the real numbers as well. This makes it a '''superset''' of the set of real numbers. Show Hide Details Description Difficulty Level: Tags: Subjects:", "# plotting – Find points on a ellipse and plot the graph For $$a=1.5$$, find the four points at which the ellipse $$x^2/a^2+y^2/b^2=1$$ where $$a>0,, b>0$$ and intersects the curve $$a,b,pi=5$$. Give your answers as exact values. Produce a plot showing the initial curve and the ellipse, with a large orange dot at each of the four intersection points. I am very new to use the Mathematica and this website ;] I don’t know which command I should use. I will appreciate any help.. BTW, this is my classwork question and I just want to figure it out.", "# getting the coordinates With the given function from a previous problem I was solving link, I want to create/collect the coordinate points of Min. value for 21 values of parameter a (from -7 to 7), and find a,b, and c such that the points are on the curve of the quadratic equation $$ax^2+bx+c=0$$. I also want to use the plot command to show this coordinate system with the parabola obtained from it.", "as you move a point ## Plot Circles • Plot circles with just two clicks! • Replay all actions later to show the workings of the student ## Plot Parabola Hint You'll need to place two points on the Parabola, for example $$0, -3$$ and $$1, -2$$ • Works beautifully on an iPad or any touch screen device • Plot Parabola simply and easily… ## Plot Polygon …alongside polygons, and author-defined background shapes ## Plot Segments Plot any two points to draw a finite section of a line Authors can limit students to one tool type or take advantage of all our types at once! ## Plot Vectors Create advanced Vector questions for students ## Plot Rays Graph a Ray originating at $$(4,0)$$ in the direction towards $$(7,2)$$ Hint You'll need to use the Ray tool Plot a ray starting at one point, extending on infinitely! ## Plot Sine Hint You'll need to place two points on the Sine wave. Remember, at $$0^\\circ$$, $$\\sin(X)$$ is $$0$$, and at $$90^\\circ$$, $$\\sin(X)$$ is $$1$$ Draw Sine or Cosine graphs with two clicks! ## Bar Chart Hint You'll need to use the \"Add Data\" functionality to add the bar for 2012. Intuitive, easy to use Charting questions, to suit all needs. Whether you need Bar charts… ## Line Chart …or a", "point, 0! It for plotting points to graph a line from a “ T-Table is a and. These two values of x and substitute these values in the table values.", "# Should I plot them? 2 Calculus Level 3 A straight line is to be drawn through a point whose coordinates are $$(x,y)=(2,-1)$$ as tangent to the curve $$y=x^2-5x+6$$. Let the coordinates of the points where the line is tangent to the curve be $$(x_1,y_1)$$ and $$(x_2,y_2)$$. Find $$x_1+y_1+x_2+y_2$$. ×", "where y appears only once and by itself on the left side, you end up getting a subset which is not trivial (that is, it's neither empty nor is it every pair of reals). In this case, the set of points you get, when plotted, is the shape of a parabola." ]

[ "circle. Then let $Y=(\\cos(2\\alpha),\\sin(2\\alpha))$. Then $Y$ is also uniformly distributed on the circle. Certainly if you know $X$ you can find $Y$, but not vice-versa. The mapping $X\\mapsto Y$ is two-to-one. -", "= \\sqrt{x^2+y^2}$ and plot in 3-d on that grid. Make $x$ and $y$ on a fine enough grid that you can get a bunch of points for a smooth surface.", "all points lie on a line from bottom left to top right.", "t a}{\\text{Notice that plots for both equations match}}$", "+0 # do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) or (-10,4) (-3,2.8) (-17,6.8) pleeeease respond 0 301 4 do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) or (-10,4) (-3,2.8) (-17,6.8) pleeeease respond Guest Dec 3, 2014 #3 +18843 +10 3 points lie on the same line, if the area of the triangle from 3 points is 0: $$\\\\ P_1 = (x_1,y_1) \\\\ P_2 = (x_2,y_2) \\\\ P_3 = (x_3,y_3) \\\\ \\small{\\text{ The area of a triangle with 3 Points is the determinant } = \\left| \\begin{array}{cc} (x_1-x_2) & (x_3-x_2) \\\\ (y_1-y_2) & (y_3-y_2) \\\\ \\end{array} \\right|\\\\\\\\ \\boxed{ \\text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} }$$ I do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) $$\\\\P_1(x_1 = -4,\\ y_1 = -2 ) \\\\ P_2(x_2 = 2,\\ y_2 = 2.5 ) \\\\ P_3(x_3 = 8,\\ y_3 = 7 ) \\\\ \\text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\\\ \\text{ area } = (-4-2) *(7-2.5) - ( (-2)-2.5)*(8-2)} \\\\ \\text{ area } = (-6) *(4.5) - ( -4.5)*(6)} \\\\ \\text{ area } = (-27) - ( -27)} \\\\ \\text{ area } = -27 + 27 = 0 }$$ area = 0: The 3 Points lie on the same line. II do theese 3 points lie on", "4 = 0$$, points which lie on the line: $$\\begin{array} {|r|r|}\\hline x & 5 & 0 \\\\ \\hline y & 0 & -5 \\\\ \\hline \\end{array}$$ Graph for the above equations is: On plotting the graph we can see that both of the lines are parallel to each other. So, there is no solution. Hence, they are inconsistent.", "a left plotted point, M, and a center plotted point, R. A line has a left plotted point, K, and a right plotted point, E. A ray has a left endpoint, O, and a plotted point to the 530. ### Evaluate the following expression when X=2 . 3 X +7X+2 A.24 B.22 C.10 D.9 652. ### In: y^2 + 6y - 9, where y = -9, Is the answer 16 + 6y or 6y + 16, or does the order not matter?", "fact, found two points, $(2, 13)$ and $(2, 17)$, which lie on one line but not the other.", "R^2\\to\\mathbb R^2$ in the original question. The formula $T(x,y)=(\\dots, 0)$ maps everything to a line: this is more of squashing than straightening. – user53153 Jan 7 '13 at 22:14", "the point $$(\\bar{x}, \\bar{y})$$ lies on the line.", "Browse Questions # If the points $(1,2)$ and $(3,4)$ lie on the same side of the straight line $3x-5y+a=0$ then a lies in the set $(a)\\;[7,11] \\quad (b)\\;R-[7,11] \\quad (c)\\;[7,\\infty] \\quad (d)\\;(- \\infty,11]$ (b) R- [7,11]", "can be plotted in $\\mathbb{R}^2$ can only be $1:1$ if every horizontal line drawn in the plane intersects the graph at most one point--clearly not the case here. -", "homogeneous coordinates. The line l = Cx passes through x, since l(TRANSPOSE) x = x(TRANSPOSE) Cx = 0. If l has one-point contact with the conic, then it is a tangent, and we are done. Otherwise suppose that l meets the conic in another point y. Then y(TRANSPOSE) Cy = 0 and x(TRANSPOSE) Cy = l(TRANSPOSE) y = 0. From this it follows that (x + αy)(TRANSPOSE) C(x + αy) = 0 for all α, which means that the whole line l = Cx joining x and y lies on the conic C, which is therefore degenerate. where C is conic coefficient matrix = [ a, b/2, d/2 ; b/2, c, e/2 ; d/2, e/2, f ] I dont see how the underlined portion follows from the above premise. And even if it does how is the line a tangent to the conic? You know that $x^{\\textsc t}Cx = 0$ and $y^{\\textsc t}Cy = 0$, because x and y lie on the conic. You are also assuming that $x^{\\textsc t}Cy = 0$. Take the transpose of that to see that $y^{\\textsc t}Cx = 0$. Therefore (multiplying out the brackets) $(x+ay)^{\\textsc t}C(x+ay) = x^{\\textsc t}Cx + ax^{\\textsc t}Cy + ay^{\\textsc t}Cx + a^2y^{\\textsc t}Cy = 0.$ What that equation says is that every point $x+ay$ on the line $l$", "# Homework Help: Does this line lie in this plane? 1. Sep 20, 2010 ### jegues 1. The problem statement, all variables and given/known data See figure for problem statement. It asks if a given line lies in a given plane. 2. Relevant equations 3. The attempt at a solution Okay so first I translated the line into its parametric equations and then I subbed those into the equation of the plane respectively. After simplifying you'll find that, $$9t-1 = -1$$ Now my solutions manual states the following, *NOTE* There is no symbol for not-equivalent in tex, so I'm writing NE $$9t -1 NE -1$$ Why is this so? Why not let t = 0, and then -1 = -1 and then all will be well in world, no? What am I missing? #### Attached Files: • ###### LIP.jpg File size: 27.8 KB Views: 105 2. Sep 20, 2010 ### gabbagabbahey \\neq works Each value of $t$ corresponds to a single point on the line. If the entire line were to lie in the plane, the equation would have to be true for all values of $t$ (all points on the line), not just $t=0$", "have $x=2u+v\\leq 0$ and $y=u-v\\geq 0$ from this, it follows that $\\boxed{v\\leq u\\leq -\\tfrac v2}$ and for the boundaries of v : $v\\leq -\\tfrac v2 \\Rightarrow \\boxed{v\\leq 0}$", "$\\mathrm{plot}\\left(\\mathrm{sin}\\left(x\\right),x=-\\mathrm{π}..\\mathrm{π},\\mathrm{y axis}\\right)$", "number plotted correctly on a number line? $Which shows a rational number plotted correctly on a number line?$...", "$4y = 12x + 20$ So to plot the line, just find the x-intercept and the y-intercept and draw a line that goes through those 2 points. Have you tried plotting them? You'll notice that they're the same line.", "point, 0! It for plotting points to graph a line from a “ T-Table is a and. These two values of x and substitute these values in the table values.", "have: $x_0 x_2 + y_0 y_2 = r^2$ This also expresses the condition that $V$ should lie on $\\TT_2$: $x x_0 + y y_0 = r^2$ So both $U$ and $V$ lie on the straight line whose equation is: $x x_0 + y y_0 = r^2$ and the result follows. $\\blacksquare$" ]

[ "= 2 Hover the mousse cursor on the graph or plotted point to read the coordinates. Click the button above \"Plot Equation\". You may hover the mousse cursor over the graph to trace the coordinates. You may also Hover the mousse cursor on the top right of the graph to have the option to download the graph as a png file. 1 - Select a point M on the graph of the hyperbola, hover the mousse over it and read the coordinates. Read the coordinates of F1 and F2 (right legend) or hover the mousse and read their coordinates and show that the difference of the distances $\\overline{MF_1} - \\overline{MF_2}$ is close to $2a$ if $\\overline{MF_1} \\gt \\overline{MF_2}$ or $- 2a$ if $\\overline{MF_1} \\lt \\overline{MF_2}$. You may do the above activity for different values of $a, b, h$ and $k$ with as many points as needed to better understand the definition of an hyperbola. 2 - Use the values of $a$ and $b$ to find $c$ using the relationship between $a, b$ and $c$ given by $c^2 = a^2 + b^2$. Use the above results to find the coordinates of $F_1, F_2, V_1$ and $V_2$ and check the results graphically. 3 - Set $a, b, h$ and $k$ to some values. Find the x and y intercepts and", "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", "parabola $y = x^2$ for $x \\in [-2,2]$ using only 5 points: x = [-2,-1,0,1,2] y = [4,1,0,1,4] plt.plot(x,y) plt.show() This is too few points to plot a smooth curve such as $y = x^2$ and so we need more points! Let's try again using the NumPy function np.linspace to create 100 points! x = np.linspace(-2,2,100) y = x**2 plt.plot(x,y) plt.show() That's a better representation of the parabola $y = x^2$. Note that the number of points we use in a line plot (100 in this case) is completely arbitrary but the goal is to show a smooth graph for a smooth curve and so we just need to pick a big enough number depending on the function. But be careful not to generate too many points since a very large number of points will take a long time to plot! Now that we have the general idea, let's look at adding style and features to our plots! ### Line Properties A line appearing in a plot has several properties including color, transparency, style, width and markers. We can set these properties when we call plt.plot using the following keyword arguments: Property Description alpha transparency (0.0 transparent through 1.0 opaque) color (or c) any matplotlib color label text appearing in legend linestyle (or ls) solid, dashed, dashdot, dotted", "as you move a point ## Plot Circles • Plot circles with just two clicks! • Replay all actions later to show the workings of the student ## Plot Parabola Hint You'll need to place two points on the Parabola, for example $$0, -3$$ and $$1, -2$$ • Works beautifully on an iPad or any touch screen device • Plot Parabola simply and easily… ## Plot Polygon …alongside polygons, and author-defined background shapes ## Plot Segments Plot any two points to draw a finite section of a line Authors can limit students to one tool type or take advantage of all our types at once! ## Plot Vectors Create advanced Vector questions for students ## Plot Rays Graph a Ray originating at $$(4,0)$$ in the direction towards $$(7,2)$$ Hint You'll need to use the Ray tool Plot a ray starting at one point, extending on infinitely! ## Plot Sine Hint You'll need to place two points on the Sine wave. Remember, at $$0^\\circ$$, $$\\sin(X)$$ is $$0$$, and at $$90^\\circ$$, $$\\sin(X)$$ is $$1$$ Draw Sine or Cosine graphs with two clicks! ## Bar Chart Hint You'll need to use the \"Add Data\" functionality to add the bar for 2012. Intuitive, easy to use Charting questions, to suit all needs. Whether you need Bar charts… ## Line Chart …or a", "Since $$x = a \\cos t, y = a \\sin t$$, $$x^2 + y^2 = a^2(\\sin^2 t + \\cos^2 t) = a^2$$. So this is a circle of radius $$a$$. The circle starts at $$(a, 0)$$, and travels counterclockwise until it reaches the starting point again. Given a value of $$t$$ for $$\\vec{r}(t) = (f(t), g(t))$$ and $$y = f(x)$$, we can plot it in an inverval of $$t \\in [a, b]$$ by plotting for $$x \\in [f(a), f(b)]$$. The parameterization of a curve is not unique, just like normal curves. Every parametric curve has infinite different ways of being represented mathematically, but when drawn on a plane each curve has only one shape. For example, consider $$\\vec{r}(t) = (a \\cos t, b \\sin t)$$. This is an ellipse where $$a > b$$ stretches it horiontally and $$b > a$$ stretches it vertically. We can rewrite it as $$\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$$ to plot it on a Cartesian plane. We plot parametric curves by solving for $$t$$ in $$x = x(t)$$, so $$t = x^{-1}(x)$$, then subsituting into $$y(t)$$ to get $$y = y(x^{-1}(x))$$, which is a non-parametric function we can plot more easily. Alternatively, we would have to plot both $$y = y(t)$$ and $$x = x(t)$$, and use these graphs to plot the points on", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "to be plotted are J/ ^ X* -{- x'', y ■= — bx -\\- c, the first of which may be drawn once for all upon cross-section paper, while the straight line represented by the second may be drawn for each particular case, as described above.* This method is also applicable to many transcendental equa- tions ; thus for the equation .^;f — ^sin;f = o it is best to write ax — sin;r = o ; then y := smx is readily plotted by help of a table of sines, while y ^= ax is a straight line passing through the origin. In the same way a\"^ — x'^ ^^ o gives the curve represented byj:=rt'*and the parabola represented by y = x\"^, the intersections of which determine the real roots of the given equation. Prob. 2. Devise a graphic solution for finding approximate values of the real roots of the equation x''-\\- ax^-\\- bx''-{- ex + d =o. Prob. 3. Determine graphically the number and the approximate values of the real roots of the equation arc x — 2> sin x = o. (Ans. — Six real roots, x = ± 159°, ± 430°, and ± 456°.) * For an extension of this method to the determination of imaginary roots, see Phillips and Beebe's Graphic Algebra,", "plot $y=\\sin(x)$ as a dotted line and $y=2\\sin(x)$ as a solid line. The resulting graph should look like The resulting graph should look like: Level 6-7 Level 6-7 Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "the required standard form: $$\\displaystyle x+\\sqrt{3}y-13=0$$ Here is a plot of the hyperbola and the normal line at the given point:", "# plotting – Find points on a ellipse and plot the graph For $$a=1.5$$, find the four points at which the ellipse $$x^2/a^2+y^2/b^2=1$$ where $$a>0,, b>0$$ and intersects the curve $$a,b,pi=5$$. Give your answers as exact values. Produce a plot showing the initial curve and the ellipse, with a large orange dot at each of the four intersection points. I am very new to use the Mathematica and this website ;] I don’t know which command I should use. I will appreciate any help.. BTW, this is my classwork question and I just want to figure it out.", "normally on the (u,v) axis? so for the first, it will be like plotting on the (x,y) axis something like y^2 = x^2 - c, for different c, how many lines would i need to plot? and how would i mark it? would i say what c is each time just to be sure these are the right lines, they are curvy like a hourglass, and they meet the axis at sqrt(c) on each side, so for each c we get two of these curves, so i need to plot all this twice. and then, also for the y=2uv, do i also plot this many times (for the same c) would i need to put all this lines on to the same plot? and mark each time which is c? OR DO I NEED TO SHOW JUST WHERE X AND Y ARE BOTH CONSTANT? the wording has ambiguity. I think this is the sort of thing they are looking for (it's an image I found online by searching for \"orthogonal hyperbola family\"). The family of \"y=2uv\" hyperbolas looks like the family of \"hourglass\" hyperbolas rotated through 45º. The important thing about the two families of curves is that wherever a curve from one family meets a curve from the other family, they intersect at right angles to each", "Mathematica. First just draw two parabolas. Say for example: $$y1=x^2+2x-1$$ $$y2=-3x^2-x+4$$ Then to draw those, I'd code: myplot=Plot[{x^2+2x-1,-3x^2-x+2},{x,-5,5},PlotRange->{{-5,5},{-5,5},PlotStyle->{Red,Blue}] That's a good start. Get a print-out of that. Now just practice drawing dots on the parabolas. Say for example, draw a green dot at the point (1,2) on the first one: mydot1=Graphics[{PointSize[0.008],Green,Point[{1,2}]}]; Show[{myplot,mydot1}] Draw a dot on the second one: mydot2=Graphics[{PointSize[0.008],Green,Point[{1,0}]}]; Now draw a straight line between the two points: myline1=Graphics[{Yellow,Line[{{1,2},{1,0}}]; Now show everything: Show[{myplot,mydot1,mydot2,myline1}] Or if you wanted to draw a line with just the function, use: Plot[m x+b,{x,b1,b2}] and to dereference that x->2, try this example and then apply it to yours: myroots = NSolve[(x - 2) (x + 2) == 0, x] theroots = x /. myroots firstone = theroots[[1]] Last edited: Jun 15, 2010", "starts at $t, x = 0, 1$ (where the units are meters), i.e., on your worldline, but then moves in the $x$ direction slower than you do, so its worldline is a shallower curve than the hyperbola you are following. At some point, the tip will cross the line $x = t$, i.e., the horizon, and at that point the tip can't keep up with you, no matter how hard you pull on it. 3. Jun 19, 2017 ### Cato OK, thanks so much for the detailed and really useful discussion.", "Review question # Can we find the area inside a straight line, a parabola and the x-axis? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R7444 ## Solution The figure shows the curve $y = x^2$ and the straight line $2x + y = 15$. Find 1. the co-ordinates of $P$ and $Q$, The figure given in the question is the following. The point $P$ is the point in the first quadrant at which the curve $y = x^2$ and line $y = 15 - 2x$ cross. That is, it is the point whose $x$-coordinate is positive and such that $x^2 = 15 - 2x.$ This equation can be rearranged to become $x^2 + 2x - 15 = 0$ which, by inspection, factorises: $(x + 5)(x - 3) = 0.$ Thus, the $x$-coordinate of $P$ is $3$, and so the coordinates of $P$ are $(3,9)$ since, in particular, $P$ lies on the curve $y = x^2$. The point $Q$ is the $x$-intercept of the line, so the $x$-coordinate of $Q$ is the value of $x$ such that $15 - 2x = 0.$ Thus, $x = \\tfrac{15}{2}$, and so the coordinates of $Q$ are $\\left( \\tfrac{15}{2}, 0 \\right)$. 1. the area of the shaded region.", "# How do I go about graphing the region, \\$Re(z^2)>1\\$ in the Complex Plane? The questions asks to describe the region, $$Re(z^2)>1$$, graphically in the complex plane. The math is fairly simple. If we let $$z=x+iy$$, then $$z^2=(x+iy)(x+iy)\\z^2=x^2-y^2+i(2xy)$$ Hence the $$Re(z^2)=x^2-y^2>1$$ Graphically speaking, this function sketches a hyperbola in the real $$xy$$ plane. However, my confusion is that the question asks to represent $$Re(z^2)>1$$ in the complex plane. How can I go about representing the real part of a function in the complex plane?", "= 1$ is easily recognized as an ellipse centered at $(0,0)$ with a major axis along the $x$-axis of length $6$ and a minor axis along the $y$-axis of length $4$. We see from the graph that the two curves intersect at their $y$-intercepts only, $(0, \\pm 2)$. \\item We proceed as before to eliminate one of the variables $\\begin{array}{ccc} \\left\\{\\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\\\ (E2) & 4x^2-9y^2 & = & 36 \\\\ \\end{array} \\right. & \\xrightarrow[\\text{-4E1 + E2}]{\\text{Replace E2 with}} & \\left\\{\\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\\\ (E2) & -13y^2 & = & 20 \\\\ \\end{array} \\right. \\end{array}$ Since the equation $-13y^2=20$ admits no real solution, the system is inconsistent. To verify this graphically, we note that $x^2+y^2=4$ is the same circle as before, but when writing the second equation in standard form, $\\frac{x^2}{9} - \\frac{y^2}{4} = 1$, we find a hyperbola centered at $(0,0)$ opening to the left and right with a transverse axis of length $6$ and a conjugate axis of length $4$. We see that the circle and the hyperbola have no points in common. $\\begin{array}{cc} \\begin{mfpic}[15]{-4}{4}{-3}{3} \\point[2pt]{(0,2), (0,-2)} \\circle{(0,0),2} \\ellipse{(0,0),3,2} \\axes \\tlabel[cc](4,-0.5){\\scriptsize x} \\tlabel[cc](0.5,3){\\scriptsize y} \\xmarks{-3,-2,-1,1,2,3} \\ymarks{-2,-1,1,2} \\tlpointsep{4pt} \\axislabels {x}{{\\tiny -3 \\hspace{11pt}} -3, {\\tiny -2 \\hspace{11pt}} -2, {\\tiny -1 \\hspace{7pt}} -1, {\\tiny 1} 1,", "some information may be shared with YouTube. Let's say the equation of the parabola is y = x. In the video, we show how to plot points by drawing two perpendicular lines that rep-resent all possible locations for the x and y values in a coordinate. Plot the points $\\left(-2,4\\right)$, $\\left(3,3\\right)$, and $\\left(0,-3\\right)$ in the coordinate plane. For example, to plot (5, -4), move 5 spaces to the right and 4 spaces down. Thanks to all authors for creating a page that has been read 300,402 times. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To create this article, 65 people, some anonymous, worked to edit and improve it over time. If it has endpoints B(5, 9) and C(4, 3), what coordinates are the midpoint? Example: Plotting Points in a Rectangular Coordinate System. (-5,4) is in Quadrant II. Students progress at their own pace and you see a leaderboard and live results. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. What directions does the second coordinate tell you to move? All tip submissions are carefully reviewed before being published. To", "1$ 2 $(2)^2 = 4$ 3 $(3)^2 = 9$ Here are the points plotted on a coordinate graph: To draw the parabola, draw a smooth curve through all the points. (Do not connect the points with straight lines). Let’s graph a few more examples. Example 1 Graph the following parabolas. a) $y=2x^2 + 4x + 1$ b) $y = -x^2 + 3$ c) $y = x^2 - 8x + 3$ Solution a) $y=2x^2 + 4x + 1$ Make a table of values: $x$ $y = 2x^2 + 4x + 1$ $-3$ $2(-3)^2 + 4(-3) + 1 = 7$ –2 $2(-2)^2 + 4(-2) + 1 = 1$ –1 $2(-1)^2 +4(-1) + 1 = -1$ 0 $2(0)^2 +4(0) + 1 = 1$ 1 $2(1)^2 +4(1) + 1 = 7$ 2 $2(2)^2 +4(2) + 1 = 17$ 3 $2(3)^2 +4(3) + 1 = 31$ Notice that the last two points have very large $y-$values. Since we don’t want to make our $y-$scale too big, we’ll just skip graphing those two points. But we’ll plot the remaining points and join them with a smooth curve. b) $y = -x^2 + 3$ Make a table of values: $x$ $y = - x^2 + 3$ $-3$ $- (-3)^2 + 3 = -6$ –2 $- (-2)^2 + 3 = -1$ –1 $- (-1)^2 + 3", "${f}$ corresponding to ${c}$. Example 5. Consider ${x^{2}+y^{2}-z^{2}=0}$. What does this surface look like? Solution: First we observe the curves along the ${x}$-axis, i.e., when ${y=0}$ is ${x^{2}=z^{2}}$. Similarly when ${x=0}$ we have ${y^{2}=z^{2}}$. What sort of curves are these? Well, ${(\\pm t,0,t)}$ and ${(0,\\pm t,t)}$ are the curves, for ${t\\in{\\mathbb R}}$. Also note that ${x^{2}+y^{2}}$ describes a circle, whose radius happens to be ${z^{2}}$. This tells us the level curves are simply circles. So we have a cone: This surface is precisely a “Cone”. Example 6. What if we deform the previous example, writing (12)$\\displaystyle x^{2}+y^{2}-z^{2}=1.$ What surface does this describe? Solution: Well, we see that the radius of the level curves deform ${z^{2}\\rightarrow z^{2}+1}$. So the resulting surface looks like a cone, but along the ${x}$-axis we don’t have a straight-line: we have a hyperbola ${x^{2}-z^{2}=1}$. Similarly along the ${y}$-axis we have another hyperbola ${y^{2}-z^{2}=1}$. Thus our surface is doodled as: This “deformed cone” is called a “One Sheeted Hyperboloid”. Example 7. Another variation, consider the surface (13)$\\displaystyle x^{2}+y^{2}-z^{2}=-1$ What does it look like? Solution: Well, we see that ${x^{2}+y^{2}\\geq0}$ always, whereas ${z^{2}-1\\geq0}$ only when ${|z|\\geq1}$. So we have two “surfaces” or “Sheets” here. The level curves are again circles, and this enables us to doodle the graph: Along the ${x}$-axis and ${y}$-axis, we have hyperbolas.", "that for any $t\\in\\mathbb{R}$, the point $(\\sin t\\cos t,\\cos^2 t,\\cos t)$ satisfies (first coordinate squared) + (second coordinate squared) = (third coordinate squared), and thus lies on your surface $S$. Since every point on your curve can be described in this way, you're done. Intuitively, this is your surface $x^2 + y^2 = z^2$, a lovely cone in $\\Bbb R^3$. This is your curve $r(t) = (\\cos t \\sin t, \\cos^2 t, \\cos t)$, also a lovely subset of $\\Bbb R^3$. When we plot them at the same time, we see that one literally lies on the other: And another angle, just for fun: Any point on the curve automatically satisfies the equation defining your surface, that's as intuitive as it gets. If we call the collection of points on the curve $C$, and those on the surface $S$, then we have $C \\subset S \\subset \\Bbb R^3$." ]

[ "convert the point from cylindrical coordinates to rectangular coordinates. $$(2,-\\pi,-4)$$ Related chapters Unlock Textbook Solution", "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "+0 # Hard Question Pls Help Me 0 44 1 A point has rectangular coordinates $(2,-1,-2)$ and spherical coordinates $(\\rho, \\theta, \\phi).$ Find the rectangular coordinates of the point with spherical coordinates $(\\rho, \\theta, 2 \\phi).$ Jun 6, 2020", "point $P_1$, and the spherical coordinates $\\rho_2$ , $\\theta_2$, $\\phi_2$ of point $P_2$, selecting the desired units for the angles, and press the button \"Calculate\". You may also change the number of decimal places as needed; it has to be a positive integer. $\\rho_1 =$ 1 $\\theta1 =$ 45 degrees radians $\\phi_1 =$ 45 degrees radians $\\rho_2 =$ 2 $\\theta2 =$ 90 degrees radians $\\phi_2 =$ 30 degrees radians Number of Decimal Places = Distance = $\\rho =$ $\\theta =$ Radians $\\theta =$ Degrees $\\phi =$ Radians $\\phi =$ Degrees ## More References and links Maths Calculators and Solvers.", "# Convert the point from rectangular coordinates to spherical coordinates. (-3,-3,\\sqrt{2}) Convert the point from rectangular coordinates to spherical coordinates. $$\\displaystyle{\\left(-{3},-{3},\\sqrt{{{2}}}\\right)}$$ • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Arham Warner To convert:", "\\phi={{\\cos}^{-1}}(\\frac{z}{\\rho})$ $latex \\phi={{\\cos}^{-1}}(\\frac{5}{6.71})$ $latex \\phi=0.73$ rad The spherical coordinates of the point are (6.71, 1.11 rad, 0.73 rad).", "Free Version Moderate # Spherical to Rectangular Conversion MVCALC-EPFFL4 The equation in rectangular coordinates for the surface that has equation $\\phi=\\pi/3$ in spherical coordinates has the form: $$a(x^2+y^2)=z^2$$ What is the value of $a$?", "5 e^{\\frac{i\\pi}{2} } \\left( {e^{ - i\\phi } } \\right) = \\sqrt 5 e^{i\\left( {\\frac{\\pi}{2} - \\phi } \\right)}$. This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates. 8. Nevermind! I figured it out! Thanks!", "# How do you find the rectangular coordinates of the point with polar coordinates (5,pi/2)? Aug 24, 2014 To convert from from polar coordinates to rectangular coordinates, we must remember these: $x = r \\cos \\theta$ $y = r \\sin \\theta$ It's useful to remember these when we need to do these types of conversions. Knowing the $r$ and $\\theta$ from our question, we can find the rectangular coordinates: $r = 5$ $\\theta = \\frac{\\pi}{2}$ $x = \\left(5\\right) \\cos \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(0\\right) = 0$ $y = \\left(5\\right) \\sin \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(1\\right) = 5$ Therefore, our rectangular coordinates are: $\\left(0 , 5\\right)$", "# How do you find the rectangular coordinates of the point with polar coordinates (3,pi)? Since $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$, $\\left(x , y\\right) = \\left(3 \\cos \\left(\\pi\\right) , 3 \\sin \\left(\\pi\\right)\\right) = \\left(- 3 , 0\\right)$", "# How do you convert (6,-pi/5) into rectangular coordinates? $\\approx \\left(4.854 , - 3.527\\right)$ In general to convert the polar coordinate $\\left(r , \\theta\\right)$ into rectangular coordinates, you just do $\\left(r \\cos \\theta , r \\sin \\theta\\right)$. In the case of $\\left(6 , - \\frac{\\pi}{5}\\right)$, this would be $\\left(6 \\cos \\left(- \\frac{\\pi}{5}\\right) , 6 \\sin \\left(- \\frac{\\pi}{5}\\right)\\right) \\approx$(4.854, -3.527)#", "\\ + \\ \\pi \\ \\ .$$ This agrees with the rectangular coordinates we found earlier.)", "Calculate the $$\\theta$$ angle: $\\cos\\theta = \\frac{z}{\\rho} = \\frac{1}{2}, \\Rightarrow \\theta = \\arccos\\frac{1}{2} = \\frac{\\pi}{3}.$ So the spherical coordinates of the point $$M$$ are $\\rho = 2,\\;\\varphi = \\arccos\\sqrt{\\frac{2}{3}},\\;\\theta = \\frac{\\pi}{3}.$", "point (2,1,2) in Cartesian coordinates, we find that $\\rho=\\sqrt{x^2+y^2+z^2}=\\sqrt{2^2+1^2+2^2}=\\sqrt{4+1+4}=\\sqrt{9}=3$ $\\tan\\phi=\\frac{\\sqrt{x^2+y^2}}{z}=\\frac{\\sqrt{2^1+1^2}}{2}=\\frac{\\sqrt{5}}{2}$ $\\rightarrow \\phi=\\tan^{-1}\\left(\\sqrt{5}/2\\right)=0.841$ radians $\\tan \\theta = y/x=1/2$ $\\rightarrow \\theta=\\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates The transformation from spherical coordinates to Cartesian coordinates is given by $x=\\rho \\sin \\phi \\cos \\theta$ $y=\\rho \\sin \\phi \\sin \\theta$ $z= \\rho \\cos \\phi$. We can convert our previous result back to Cartesian coordinates: $x=\\rho \\sin \\phi \\cos \\theta=3\\sin(0.841)\\cos(0.464)\\approx 2$ $y=\\rho \\sin \\phi \\sin \\theta =3\\sin(0.841)\\sin(0.464)\\approx 1$ $z= \\rho \\cos \\phi =3 cos(0.841)\\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding. We can also write the transformation from cylindrical coordinates to spherical coordinates: $\\rho=\\sqrt{r^2+z^2}$ $\\tan \\phi = z/r$ $\\theta = \\theta$. Beginning with the point (3,\\pi/3,4) in cylindrical coordinates, we see that $\\rho=\\sqrt{r^2+z^2}=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ $\\tan \\phi = z/r=4/3$ $\\rightarrow \\phi=\\tan^{-1}(.75)=0.644$ $\\theta = \\pi/3$. So the point in spherical coordinates is (5,\\pi/3,0.644). Finally, the transformation from spherical to cylindrical coordinates is given by $r=\\rho \\sin \\phi$ $\\theta =\\theta$ $z=\\rho \\cos \\phi$. For one last example, we will take our previous result and transform it back to cylindrical coordinates $r=\\rho \\sin \\theta=5 \\sin (0.644)\\approx 3$ $\\theta =\\pi/3$ $z=\\rho \\cos \\theta5 \\cos (0.644)\\approx 4$ Therefore the point in cylindrical coordinates is (3, \\pi/3, 4) as expected. Interactive Demonstration Change of Coordinate Systems Applet • add examples of transformations between", "# How do you write the rectangular coordinates for the point: (6, 3/2 pi)? Mar 20, 2018 $\\left(- 3 , 3 \\sqrt{3}\\right)$ #### Explanation: For given coordinates $\\left(r , \\theta\\right)$, $\\left(r , \\theta\\right) \\to \\left(x , y\\right) \\implies \\left(r \\cos \\theta , r \\sin \\theta\\right)$ $r = 6$ $\\theta = \\frac{2 \\pi}{3}$ $\\left(6 , \\frac{2 \\pi}{3}\\right) \\to \\left(x , y\\right) \\implies \\left(6 \\cos \\left(\\frac{2 \\pi}{3}\\right) , 6 \\sin \\left(\\frac{2 \\pi}{3}\\right)\\right) \\equiv \\left(- 3 , 3 \\sqrt{3}\\right)$", "# How do you find the rectangular coordinates of the point with polar coordinates (2,pi/4)? Since $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$, $\\left(x , y\\right) = \\left(2 \\cos \\left(\\frac{\\pi}{4}\\right) , 2 \\sin \\left(\\frac{\\pi}{4}\\right)\\right) = \\left(\\sqrt{2} . \\sqrt{2}\\right)$", "quadrant). Lastly, the bounds on {eq}\\rho {/eq} tell us that the sphere has radius 3. So we are dealing with a slice/wedge in the first octant. A sketch is below. The volume of the wedge is {eq}\\begin{align*} \\int^{\\pi/4}_0 \\int^\\pi_0 \\int^3_0 \\rho^2 \\sin \\phi\\ d \\rho\\ d \\theta\\ d \\phi &= \\int_0^{\\pi} d \\theta \\int_0^{\\pi/4} \\sin \\phi\\ d\\phi \\int_0^3 \\rho^2\\ d\\rho \\\\ &= \\pi \\left [ -\\cos \\phi \\right ]_0^{\\pi/4} \\left [ \\frac13\\rho^3 \\right ]_0^3 \\\\ &= \\frac\\pi3 \\left( - \\frac{\\sqrt2}{2} + 1 \\right)(3^3) \\\\ &= \\frac{9\\pi}{2} \\left( 2-\\sqrt2 \\right) \\\\ &\\approx 8.2814 \\end{align*} {/eq} #### Learn more about this topic: Cylindrical & Spherical Coordinates: Definition, Equations & Examples from GRE Math: Study Guide & Test Prep Chapter 13 / Lesson 10 14K", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 68 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020", "cylindrical coordinates: $$\\tag{a4 } \\frac{1}{R _ {0} ^ {1+} u } = \\frac{1}{[ \\rho ^ {2} + \\rho _ {0} ^ {2} - 2 \\rho \\rho _ {0} \\cos ( \\phi - \\phi _ {0} )+ z ^ {2} ] ^ {( 1+ u)/2 } } =$$ $$= \\ \\frac{2} \\pi \\cos \\frac{\\pi u }{2} \\int\\limits _ { 0 } ^ { {l _ 1} ( \\rho _ {0} ) } \\frac{\\lambda \\left ( \\frac{x ^ {2} }{\\rho \\rho _ {0} } , \\phi - \\phi _ {0} \\right ) x ^ {u} d x }{\\{ [ l _ {1} ^ {2} ( \\rho _ {0} )- x ^ {2} ][ l _ {2} ^ {2} ( \\rho _ {0} )- x ^ {2} ] \\} ^ {( 1+ u)/2 } } .$$ Here $| u | < 1$, and the following notation was introduced: $$\\lambda ( k, \\psi ) = \\frac{1- k ^ {2} }{1+ k ^ {2} - 2 k \\cos \\psi } ,$$ $$l _ {1} ( \\rho _ {0} ) \\equiv l _ {1} ( \\rho _ {0} , \\rho , z) =$$ $$= \\ \\frac{1}{2} \\{ [ ( \\rho + \\rho _ {0} ) ^ {2} + z ^ {2} ] ^ {1/2} - [( \\rho - \\rho _ {0} ) ^ {2}", "$\\theta$. You can see that $\\theta$ goes from $0$ to $\\pi/2$ Finally, you want to convert $dV$ to cylindrical coordinates. When converting from rectangular to cylindrical, $dV$ becomes $rdzdrd\\theta$. What we're left with is: $$\\int_{0}^{\\pi/2}\\int_{0}^{1}\\int_{0}^{2-r^2}r^2cos(\\theta)dzdrd\\theta$$" ]

[ "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "5(x^2+ y^2)= 5r^2$$. In Cylindrical coordinates that is $$z= 5r^2$$. It is particularly simple because, geometrically, the figure is a cylinder! In spherical coordinates $$z= \\rho cos(\\phi)= 5(\\rho^2 cos^2(\\theta)sin^2(\\phi)+ \\rho^2 sin^(\\theta) sin^2(\\phi))= 5\\rho^2 sin^2(\\phi)(cos^2(\\theta)+ sin^2(\\theta))= 5\\rho^2 sin^2(\\phi)$$. $$\\rho cos^(\\phi)= 5\\rho^2 sin^2(\\phi)$$.$$\\rho= \\frac{cos(\\phi)}{5 sin^2(\\phi)}$$. Guest", "u_{tt} - \\frac{K}{\\rho}(u_{xx} + v_{yx}) &= 0 \\, ,\\\\ v_{tt} - \\frac{K}{\\rho}(u_{xy} + v_{yy}) &= 0 \\, . \\end{aligned}", "1\\\\ z= 1, x^2 + y^2 = 1$ Substitute our new coordinate system. $\\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi = 2\\\\ \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi - 2\\rho\\cos\\phi + 1 = 1\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta = 1$ Simplify: $\\rho^2 = 2\\\\ \\rho^2 - 2\\rho\\cos\\phi = 0\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi = 1$ $\\rho = \\sqrt 2\\\\ \\rho = 2\\cos\\phi\\\\ \\frac {\\rho\\sin \\phi}{\\rho\\cos \\phi} = \\tan \\phi= 1\\\\ \\phi = \\frac {\\pi}{4}$ Now, it may help to sketch these curves. Do it in 2D. Just look at the picture in the xz plane, and that should give you some idea of what is going on for $\\rho$ and $\\phi.$ \\update______ $\\int_0^{2\\pi}\\int_0^{\\frac\\pi4}\\int_{\\sqrt 2}^{2\\cos\\phi} \\rho^2 \\sin\\phi \\;d\\rho\\;d\\phi\\;d\\theta$ • The cylindrical looks fine, but I'm still unsure how you get your bounds for the spherical. – Lax Oct 12 '16 at 1:05", "spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ cylindrical coordinates $$(r,\\theta,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} r &= \\rho\\sin\\phi \\\\ \\theta &= \\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ Cartesian coordinates $$(x,y,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} x &= r\\cos\\theta = \\rho\\sin\\phi\\cos\\theta \\\\ y &= r\\sin\\theta = \\rho\\sin\\phi\\sin\\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting Cartesian $$(x,y,z)$$ $$\\to$$ spherical coordinates $$(\\rho,\\phi,\\theta)$$ \\shaded{ (x,y,z) \\Rightarrow \\left\\{ \\begin{align*} \\rho &= \\sqrt{r^2+z^2} = \\sqrt{x^2+y^2+z^2} \\\\ \\phi &= \\rm{atan}\\,\\frac{\\sqrt{x^2+y^2}}{z} \\\\ \\theta &= \\rm{atan}\\,\\frac{y}{x}\\\\ \\end{align*} \\right. } \\nonumber Some expressions • $$\\rho=a$$: a sphere of radius $$a$$, centered at the origin. • $$\\phi=\\frac{\\pi}{4}$$: a cone. In cylindrical that would $$z=r$$. • $$\\phi=\\frac{\\pi}{2}$$: the $$xy$$-pane. ### Volume element To find the volume element $$\\Delta V\\approx\\Delta\\rho\\,\\Delta\\phi\\,\\Delta\\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides? Start with a surface area element $$\\Delta S$$ • To move north-south: you always have to traverse half a circle $$\\Longrightarrow$$ the vertical side of $$\\Delta S$$ is a piece of circle with radius $$a$$ $$\\Rightarrow$$ the length is $$a.\\Delta\\phi$$. • To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\\Longrightarrow$$ the horizontal side of $$\\Delta S$$ is a piece of circle of radius $$r=a\\sin\\phi$$.", "\\phi={{\\cos}^{-1}}(\\frac{z}{\\rho})$ $latex \\phi={{\\cos}^{-1}}(\\frac{5}{6.71})$ $latex \\phi=0.73$ rad The spherical coordinates of the point are (6.71, 1.11 rad, 0.73 rad).", "P''(\\rho) + \\rho P' (\\rho)}{P(\\rho)}+ \\frac{\\Lambda Y(\\phi,\\theta)}{Y(\\phi,\\theta)}=0 and since the first term depends only on $\\rho$ and the second only on$\\phi, \\theta$ we conclude that both are constant: \\begin{align} \\rho^2 P'' +2\\rho P' = \\lambda P,\\\\ \\Lambda Y(\\phi,\\theta)=-\\lambda Y(\\phi,\\theta). \\end{align} The first equation is of Euler type and it has solutions $P:=\\rho^l$ iff $\\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$ Then suppose that \\begin{align} u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2) + c(1-x^2-y^2-z^2) ^2\\\\\\\\ =&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2 \\end{align} \\begin{align} \\Delta u &= uxx+uyy+uzz\\\\ &=12 z^2 -6a + 2b( 1 -\\rho^2 -7 z^2) -6c +20c\\rho^2\\\\\\text{Suppose $1-\\rho^2-z^2=0$}\\\\ &=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\\\ &=12z^2-6a-12bz^2-6c+20c-20cz^2\\\\ &=(12-12b-20c)z^2-6a-6c=0 \\end{align} Then we have \\begin{align} 12-12b-20c=0\\\\ -6a-6c=0 \\end{align} Then $a = -c$, $b=1-\\frac{5}{3}c$, and c is arbitrary. Title: Re: HA9 Problem 1 Post by: Mark Nunez on March 27, 2015, 12:35:47 AM I found different values for a,b. :-X Title: Re: HA9 Problem 1 Post by: Victor Ivrii on March 27, 2015, 11:47:05 AM All do correct things but do errors in calculations. Actually, it was partially my fault", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 34 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020 #1 0 tan theta + tan phi = 5/2 - 4/3 = 7/6. Aug 6, 2020", "* cos(3.14/2)=0 Cylindrical coordinates are: rho = r sin(theta)=4 * sin(3.14/2)=4 phi = phi=3.14/6 z= r cos (theta)=4 * cos(3.14/2)=0 To convert back from cylindrical coordinates r=sqrt{rho^2+z^2}=rho=4 theta=atan2 {rho,z}=atan2(4,0)=3.14/2 phi=phi=3.14/6", "coordinates, $x= \\rho sin(\\phi)cos(\\theta)$ and $y= \\rho sin(\\phi)sin(\\theta)$ so $-5x+ 5y= -5\\rho sin(\\phi)cos(\\theta)+ 5 sin(\\phi)sin(\\theta)$ $= 5sin(\\phi)(sin(\\theta)- cos(\\theta))$ Of course, the differential of volume in spherical coordinates is $\\rho^2 sin(\\phi)d\\rho d\\theta d\\phi$. • March 21st 2009, 12:46 PM Showcase_22 $\\int_1^4 \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} 5sin (\\phi) (sin (\\theta)-cos (\\theta)) \\rho^2 sin(\\phi)d \\rho d\\phi d\\theta$ I was having trouble deciding what the $\\rho$ limits should be until I read HallsofIvy's post. Thanks! • March 21st 2009, 02:55 PM althaemenes HallsofIvy and Showcase_22 (Rock) !!! Thanks guys!! http://www.mathhelpforum.com/math-he...4dd0a69c-1.gif => (105/2)*Pi-(105/2)*Pi*sin(Pi)*cos(Pi)-(105/2)*Pi*cos(Pi)^2 => 0", "1 + 1 + 1 = 3$$ The other two cylindrical and spherical, I did the same way. But I used the $$\\nabla\\cdot \\vec{r}$$ for cylindrical and spherical which I referred to in a handbook. Last edited: Jan 9, 2005 7. Jan 9, 2005 ### meteorologist1 Could you show me how to get \"3\" for the cylindrical and spherical? Thanks. 8. Jan 9, 2005 ### Galileo In cylindrical coordinates: $$\\vec r = \\rho \\hat \\rho + z \\hat z$$ and $$\\nabla \\cdot \\vec v(\\rho,\\phi,z) = \\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho v_{\\rho})+\\frac{1}{\\rho}\\frac{\\partial v_{\\phi}}{\\partial \\phi}+\\frac{\\partial v_z}{\\partial z}$$ For some vector function $v=\\langle v_{\\rho},v_{\\phi},v_z \\rangle$ In spherical coordinates: $$\\vec r = r\\hat r$$ and $$\\nabla \\cdot \\vec v(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2v_r)+\\frac{1}{r\\sin \\theta}\\frac{\\partial}{\\partial \\theta}(\\sin \\theta v_{\\theta})+\\frac{1}{r\\sin \\theta}\\frac{\\partial v_{\\phi}}{\\partial \\phi}$$ for some vector function $v=\\langle v_r,v_{\\theta},v_{\\phi} \\rangle$ Both give 3 as well: $$\\nabla \\cdot \\vec r(\\rho,\\phi,z) =\\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho \\rho)+\\frac{\\partial z}{\\partial z}=\\frac{1}{\\rho}2\\rho+1=3$$ $$\\nabla \\cdot \\vec r(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2 r)=\\frac{1}{r^2}3r^2=3$$ You can't use the divergence in cylindrical or spherical coordinates when r=0 though. Last edited: Jan 9, 2005 9. Jan 9, 2005 ### meteorologist1 Thanks very much. It looks like the reason I got the wrong answers is because I got the r vectors wrong in cylindrical and spherical coordinates. I need to get this straight. I would like to ask another question related to these two coordinates: In cylindrical coordinates,", "rectangular system are (0, -2\\sqrt{2}, 0) This is a point on the negative y - axis at a distance 2\\sqrt{2} from the origin. ## Rectangular Coordinates to Spherical Coordinates Let us write the rectangular form (1, \\sqrt{3}, 2\\sqrt{3}) as spherical coordinates. We have x = 1, y = \\sqrt{3} and z = 2\\sqrt{3} Spherical coordinates are as follows: x = \\rho sin(\\phi) cos(\\theta) y = \\rho sin(\\phi) sin(\\theta) z = \\rho cos(\\phi) Now, \\rhosin(\\phi) cos(\\theta) = 1 ............(1) \\rho sin(\\phi) sin (\\theta) = \\sqrt{3} ............(2) \\rho cos(\\phi) = 2\\sqrt{3} ............(3) Dividing equation (2) by (1), we get tan (\\theta) = \\sqrt{3} \\theta = \\frac{\\pi}{3} (As both x and y are positive) \\rho ^2 = x^2 + y^2 + z^2 = 1 + (\\sqrt{3})^2 + (2\\sqrt{3})^2 = 1 + 3 + 12 = 16. Hence, \\rho = 4 \\rho cos(\\phi) = 2\\sqrt{3} 4 cos(\\phi) = 2\\sqrt{3} cos(\\phi) = \\frac{\\sqrt{3}}{2} \\phi = \\frac{\\pi}{6} Thus, the required spherical form of the coordinates are (4,$$\\frac{\\pi}{6}$$,$$\\frac{\\pi}{3}$$) ## Unit Vector in Spherical Coordinates In rectangular coordinate system, unit vectors \\hat{i}, \\hat{j}, \\hat{k} are corresponding to the directions of three coordinate axes. In the same manner, we can think of unit vectors \\hat{\\rho }, \\hat{\\phi } and \\hat{\\theta } in spherical coordinates. The position vector of any point P<\\rho, \\phi, \\theta> can be written in", "}%, \\quad %K_T = \\frac{T}{ \\rho n^2 {D_P}^4 }, \\quad %K_Q = \\frac{Q}{ \\rho n^2 {D_P}^5 } \\end{align*}", "## Calculus 8th Edition a) $\\rho=3$ b) $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$ The conversion of rectangular coordinates to spherical coordinates is given as: $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ Here, $\\rho=\\sqrt {x^2+y^2+z^2}$; $\\phi =\\cos^{-1} [\\dfrac{z}{\\rho}]; \\theta=\\cos^{-1}[\\dfrac{x}{\\rho \\sin \\phi}]$ a) Here, we have $x^2+y^2+z^2=9$ But $\\rho=\\sqrt {x^2+y^2+z^2}$ This gives: $\\rho^2=3^2 \\implies \\rho=3$ b) Here, we have $x^2-y^2-z^2=1$ But $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ So, we have $(\\rho \\sin \\phi \\cos \\theta)^2-(\\rho \\sin \\phi \\sin \\theta)^2-(\\rho \\cos \\phi)^2=1$ or, $\\rho ^2 \\sin^2 \\phi \\times (\\cos^2 \\theta- \\sin^2 \\theta)-(\\rho \\cos \\phi)^2=1$ Thus, we get $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$", "Quiz from Wednesday, October 26 Problem: Let $\\theta$ represent the usual angle in the $xy$-plane emanating from the origin counter-clockwise from the $x$-axis, and let $\\phi$ represent the angle from the $z$-axis to the $xy$-plane. If $\\rho$ is distance from the origin, find the volume of the object created by rotating the region where $1\\le\\rho\\le 3$ and $\\pi/6\\le\\phi\\le\\pi/2$ in a full circle around the $z$-axis. Solution: In spherical coordinates, we compute as follows. \\begin{align*} \\int_0^{2\\pi}\\int_{\\pi/6}^{\\pi/2}\\int_1^3 \\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta &= \\int_0^{2\\pi}\\,d\\theta\\int_{\\pi/6}^{\\pi/2}\\sin\\phi\\,d\\phi\\int_1^3\\rho^2\\,d\\rho \\\\ &= 2\\pi\\left(-\\cos\\phi\\right)\\Big|_{\\phi=\\pi/6}^{\\phi=\\pi/2}\\left(\\frac{\\rho^3}{3}\\right)\\Big|_{\\rho=1}^{\\rho=3}\\\\ &= 2\\pi\\cdot\\frac{\\sqrt{3}}{2}\\left(\\frac{27}{3}-\\frac{1}{3}\\right)\\\\ &= \\frac{26\\pi\\sqrt{3}}{3}\\end{align*} Replacement Question: How many states share a border (including corners) with Colorado? Answer: 7: Kansas, Nebraska, Wyoming, Utah, Arizona, New Mexico, and Oklahoma. There are only two states that share more borders with neighboring states than Colorado does: Tennessee and Missouri both share borders with 8 states (including each other).", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq0,\\ 0\\leq\\theta\\leq\\2\\pi, and 0\\leq\\phi\\leq\\pi.)a) (0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?b) (-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=? Change from rectangular to spherical coordinates. (Let $$\\rho\\geq0,\\ 0\\leq\\theta\\leq 2\\pi$$, and $$0\\leq\\phi\\leq\\pi.)$$ a) $$(0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?$$ b) $$(-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=?$$ 2021-05-15 Step 1 a) Consider the rectangular coordinate $$(0,\\ -8,\\ 0)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho, \\theta,\\phi)$$ Here, $$\\rho\\geq0,\\ 0\\leq\\theta\\leq2\\pi,\\ 0\\leq\\phi\\leq\\pi$$ Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spheerical coordinates $$(\\rho,\\theta\\phi)$$ The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ Step 2 The rectangular coordinate is $$(x,y,z)=(0,\\ -8,\\ 0)$$ That is, $$x=0,\\ y=-8,\\ z=0.$$ Thus, from the statement, $$\\rho=\\sqrt{x^{2}+y^{2}+z^{2}}$$ $$=\\sqrt{0^{2}+(-8)^{2}+0^{2}}$$ $$\\{x=0,\\ y=-8,z=0\\}$$ $$=\\sqrt{64}$$ $$=8$$ From the statement, $$z=\\rho\\cos\\phi$$ $$0=8\\cos\\phi$$ $$\\{z=0,\\rho=8\\}$$ $$\\cos\\phi=0$$ $$\\phi=\\frac{\\pi}{2}$$ $$\\{0\\leq\\phi\\leq\\pi\\}$$ Step 3 From the statement, $$y=\\rho\\sin\\phi\\sin\\theta$$ $$\\sin\\theta=\\frac{y}{\\rho\\sin\\phi}$$ Substitute $$y=-8,\\rho=8,\\phi=\\frac{\\pi}{2}$$ $$\\sin\\theta=\\frac{-8}{8\\sin\\frac{\\pi}{2}}$$ $$=-1$$ $$\\left\\{\\sin\\frac{\\pi}{2}=1\\right\\}$$ $$\\theta=\\frac{3\\pi}{2}$$ $$\\left\\{0\\leq\\theta\\leq\\frac{\\pi}{2}\\right\\}$$ Hence, the spherical coordinate of the point $$(0,\\ -8,\\ 0)$$ is $$\\left(8,\\ \\frac{3\\pi}{2},\\ \\frac{\\pi}{2}\\right)$$ Step 4 Consider the rectangular coordinate $$(-1,\\ 1,\\ -2)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho,\\theta,\\phi)$$. Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$. The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ The", "# Thread: volume enclosed by spheres 1. ## volume enclosed by spheres Find the volume enclosed by the spheres $x^{2}+y^{2}+z^{2}=9$ and $x^{2}+y^{2}+(z-2)^{2}=4$ 2. ## Spherical Coordinates If we use spherical coordinates $(\\rho,\\theta,\\phi)$, the spheres are thus $\\rho=3$ and $\\rho=4\\cos\\theta$. We see our volume is bounded by $\\rho=3$ for $0\\le\\theta<\\arccos\\frac{3}{4}$ and $\\rho=4\\cos\\theta$ for $\\arccos\\frac{3}{4}\\le\\theta\\le\\frac{\\pi}{2}$. Thus, we have: $V=\\int_{0}^{\\arccos\\frac{3}{4}}\\int_{0}^{2\\pi}\\int _{0}^{3}\\rho^2\\sin\\theta\\,d\\rho\\,d\\phi\\,d\\theta+\\i nt_{\\arccos\\frac{3}{4}}^{\\frac{\\pi}{2}}\\int_{0}^{2 \\pi}\\int_{0}^{4\\cos\\theta}\\rho^2\\sin\\theta\\,d\\rho\\ ,d\\phi\\,d\\theta$ $V=\\int_{0}^{\\arccos\\frac{3}{4}}18\\pi\\sin\\theta\\,d\\ theta+\\int_{\\arccos\\frac{3}{4}}^{\\frac{\\pi}{2}}\\fr ac{128\\pi}{3}\\cos^3\\theta\\sin\\theta\\,d\\theta$ $V=\\frac{9\\pi}{2}+\\frac{27\\pi}{8}=\\frac{63}{8}\\pi$ --Kevin C. 3. Cool. That is identical to how I set it up. Just verifying.", "\\tan(\\varphi) = r/z\\\\ \\amp r=\\rho \\sin(\\varphi),\\amp \\amp \\theta = \\theta,\\amp \\amp z=\\rho\\cos(\\varphi) \\end{align*} ###### Example15.7.16.Converting between rectangular and spherical coordinates. Convert the rectangular point $$(2,-2,1)$$ to spherical coordinates, and convert the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular and cylindrical coordinates. Solution This rectangular point is the same as used in Example 15.7.4. Using Key Idea 15.7.15, we find $$\\rho = \\sqrt{2^2+(-1)^2+1^2} = 3\\text{.}$$ Using the same logic as in Example 15.7.4, we find $$\\theta = 7\\pi/4\\text{.}$$ Finally, $$\\cos(\\varphi) = 1/3\\text{,}$$ giving $$\\varphi = \\cos^{-1}(1/3) \\approx 1.23\\text{,}$$ or about $$70.53^\\circ\\text{.}$$ Thus the spherical coordinates are approximately $$(3,7\\pi/4,1.23)\\text{.}$$ Converting the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular, we have $$x = 6\\sin(\\pi/2)\\cos(\\pi/3) = 3\\text{,}$$ $$y = 6\\sin(\\pi/2)\\sin(\\pi/3) = 3\\sqrt{3}$$ and $$z = 6\\cos(\\pi/2) = 0\\text{.}$$ Thus the rectangular coordinates are $$(3,3\\sqrt{3},0)\\text{.}$$ To convert this spherical point to cylindrical, we have $$r = 6\\sin(\\pi/2) = 6\\text{,}$$ $$\\theta = \\pi/3$$ and $$z = 6\\cos(\\pi/2) =0\\text{,}$$ giving the cylindrical point $$(6,\\pi/3,0)\\text{.}$$ ###### Example15.7.17.Canonical surfaces in spherical coordinates. Describe the surfaces $$\\rho=1\\text{,}$$ $$\\theta = \\pi/3$$ and $$\\varphi = \\pi/6\\text{,}$$ given in spherical coordinates. Solution The equation $$\\rho = 1$$ describes all points in space that are 1 unit away from the origin: this is the sphere of radius 1, centered at the origin. The equation $$\\theta = \\pi/3$$ describes the same surface in spherical coordinates as it", "Nov 14 '13 at 17:55 Here, I evaluate the forth part of the region so after evaluating the triple integral you should multiply the result to $4$. According to Fig below, we have a rectangle triangular $OCD$. So $\\sin(\\phi_0)=1/2$ and so $\\phi_0=\\pi/6$. It means that $\\phi\\in [\\pi/6,\\pi/2]$. As you found out the rang of $\\theta$ is $[0,\\pi/2]$. But about the $\\rho$: You see that our region is starting to shape from the intersection till we meet $z=0$. For finding the range of $\\rho$, make an arrow arbitrary to cut the cylinder at $A$ and next cut the sphere at $B$. What is $\\rho$ at $A$ and what is at $B$? We know that in a spherical coordinates: $$x=\\rho\\sin(\\phi)\\cos(\\theta)\\\\y=\\rho\\sin(\\phi)\\sin(\\theta)\\\\z=\\rho\\cos(\\phi)$$ then the equation of our cylinder $x^2+y^2=1$ be converted to $\\rho^2\\sin^2(\\phi)=1$ or $\\rho=\\frac{1}{\\sin(\\phi)}$. This is the first coordinate $\\rho$ at $A$. Clearly, $\\rho$ at $B$ is $2$. Hence you have: $$\\int_0^{\\pi/2}d\\theta\\int_{\\pi/6}^{\\pi/2}\\sin(\\phi)d\\phi\\int_{\\frac{1}{\\sin(\\phi)}}^{2}\\rho^2d\\rho$$. - What happens with the x^2 + y^2 + z^2 = 4? – shadow Oct 31 '12 at 13:58 @shadow: Tell me what exactly is your problem. Tell me. Do you have any with ranges? or my integral? :) – B. S. Oct 31 '12 at 14:04", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 246 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020" ]

[ "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "cylindrical coordinates: $$\\tag{a4 } \\frac{1}{R _ {0} ^ {1+} u } = \\frac{1}{[ \\rho ^ {2} + \\rho _ {0} ^ {2} - 2 \\rho \\rho _ {0} \\cos ( \\phi - \\phi _ {0} )+ z ^ {2} ] ^ {( 1+ u)/2 } } =$$ $$= \\ \\frac{2} \\pi \\cos \\frac{\\pi u }{2} \\int\\limits _ { 0 } ^ { {l _ 1} ( \\rho _ {0} ) } \\frac{\\lambda \\left ( \\frac{x ^ {2} }{\\rho \\rho _ {0} } , \\phi - \\phi _ {0} \\right ) x ^ {u} d x }{\\{ [ l _ {1} ^ {2} ( \\rho _ {0} )- x ^ {2} ][ l _ {2} ^ {2} ( \\rho _ {0} )- x ^ {2} ] \\} ^ {( 1+ u)/2 } } .$$ Here $| u | < 1$, and the following notation was introduced: $$\\lambda ( k, \\psi ) = \\frac{1- k ^ {2} }{1+ k ^ {2} - 2 k \\cos \\psi } ,$$ $$l _ {1} ( \\rho _ {0} ) \\equiv l _ {1} ( \\rho _ {0} , \\rho , z) =$$ $$= \\ \\frac{1}{2} \\{ [ ( \\rho + \\rho _ {0} ) ^ {2} + z ^ {2} ] ^ {1/2} - [( \\rho - \\rho _ {0} ) ^ {2}", "point (2,1,2) in Cartesian coordinates, we find that $\\rho=\\sqrt{x^2+y^2+z^2}=\\sqrt{2^2+1^2+2^2}=\\sqrt{4+1+4}=\\sqrt{9}=3$ $\\tan\\phi=\\frac{\\sqrt{x^2+y^2}}{z}=\\frac{\\sqrt{2^1+1^2}}{2}=\\frac{\\sqrt{5}}{2}$ $\\rightarrow \\phi=\\tan^{-1}\\left(\\sqrt{5}/2\\right)=0.841$ radians $\\tan \\theta = y/x=1/2$ $\\rightarrow \\theta=\\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates The transformation from spherical coordinates to Cartesian coordinates is given by $x=\\rho \\sin \\phi \\cos \\theta$ $y=\\rho \\sin \\phi \\sin \\theta$ $z= \\rho \\cos \\phi$. We can convert our previous result back to Cartesian coordinates: $x=\\rho \\sin \\phi \\cos \\theta=3\\sin(0.841)\\cos(0.464)\\approx 2$ $y=\\rho \\sin \\phi \\sin \\theta =3\\sin(0.841)\\sin(0.464)\\approx 1$ $z= \\rho \\cos \\phi =3 cos(0.841)\\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding. We can also write the transformation from cylindrical coordinates to spherical coordinates: $\\rho=\\sqrt{r^2+z^2}$ $\\tan \\phi = z/r$ $\\theta = \\theta$. Beginning with the point (3,\\pi/3,4) in cylindrical coordinates, we see that $\\rho=\\sqrt{r^2+z^2}=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ $\\tan \\phi = z/r=4/3$ $\\rightarrow \\phi=\\tan^{-1}(.75)=0.644$ $\\theta = \\pi/3$. So the point in spherical coordinates is (5,\\pi/3,0.644). Finally, the transformation from spherical to cylindrical coordinates is given by $r=\\rho \\sin \\phi$ $\\theta =\\theta$ $z=\\rho \\cos \\phi$. For one last example, we will take our previous result and transform it back to cylindrical coordinates $r=\\rho \\sin \\theta=5 \\sin (0.644)\\approx 3$ $\\theta =\\pi/3$ $z=\\rho \\cos \\theta5 \\cos (0.644)\\approx 4$ Therefore the point in cylindrical coordinates is (3, \\pi/3, 4) as expected. Interactive Demonstration Change of Coordinate Systems Applet • add examples of transformations between", "point $P_1$, and the spherical coordinates $\\rho_2$ , $\\theta_2$, $\\phi_2$ of point $P_2$, selecting the desired units for the angles, and press the button \"Calculate\". You may also change the number of decimal places as needed; it has to be a positive integer. $\\rho_1 =$ 1 $\\theta1 =$ 45 degrees radians $\\phi_1 =$ 45 degrees radians $\\rho_2 =$ 2 $\\theta2 =$ 90 degrees radians $\\phi_2 =$ 30 degrees radians Number of Decimal Places = Distance = $\\rho =$ $\\theta =$ Radians $\\theta =$ Degrees $\\phi =$ Radians $\\phi =$ Degrees ## More References and links Maths Calculators and Solvers.", "1 + 1 + 1 = 3$$ The other two cylindrical and spherical, I did the same way. But I used the $$\\nabla\\cdot \\vec{r}$$ for cylindrical and spherical which I referred to in a handbook. Last edited: Jan 9, 2005 7. Jan 9, 2005 ### meteorologist1 Could you show me how to get \"3\" for the cylindrical and spherical? Thanks. 8. Jan 9, 2005 ### Galileo In cylindrical coordinates: $$\\vec r = \\rho \\hat \\rho + z \\hat z$$ and $$\\nabla \\cdot \\vec v(\\rho,\\phi,z) = \\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho v_{\\rho})+\\frac{1}{\\rho}\\frac{\\partial v_{\\phi}}{\\partial \\phi}+\\frac{\\partial v_z}{\\partial z}$$ For some vector function $v=\\langle v_{\\rho},v_{\\phi},v_z \\rangle$ In spherical coordinates: $$\\vec r = r\\hat r$$ and $$\\nabla \\cdot \\vec v(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2v_r)+\\frac{1}{r\\sin \\theta}\\frac{\\partial}{\\partial \\theta}(\\sin \\theta v_{\\theta})+\\frac{1}{r\\sin \\theta}\\frac{\\partial v_{\\phi}}{\\partial \\phi}$$ for some vector function $v=\\langle v_r,v_{\\theta},v_{\\phi} \\rangle$ Both give 3 as well: $$\\nabla \\cdot \\vec r(\\rho,\\phi,z) =\\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho \\rho)+\\frac{\\partial z}{\\partial z}=\\frac{1}{\\rho}2\\rho+1=3$$ $$\\nabla \\cdot \\vec r(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2 r)=\\frac{1}{r^2}3r^2=3$$ You can't use the divergence in cylindrical or spherical coordinates when r=0 though. Last edited: Jan 9, 2005 9. Jan 9, 2005 ### meteorologist1 Thanks very much. It looks like the reason I got the wrong answers is because I got the r vectors wrong in cylindrical and spherical coordinates. I need to get this straight. I would like to ask another question related to these two coordinates: In cylindrical coordinates,", "Calculate the $$\\theta$$ angle: $\\cos\\theta = \\frac{z}{\\rho} = \\frac{1}{2}, \\Rightarrow \\theta = \\arccos\\frac{1}{2} = \\frac{\\pi}{3}.$ So the spherical coordinates of the point $$M$$ are $\\rho = 2,\\;\\varphi = \\arccos\\sqrt{\\frac{2}{3}},\\;\\theta = \\frac{\\pi}{3}.$", "spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ cylindrical coordinates $$(r,\\theta,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} r &= \\rho\\sin\\phi \\\\ \\theta &= \\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ Cartesian coordinates $$(x,y,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} x &= r\\cos\\theta = \\rho\\sin\\phi\\cos\\theta \\\\ y &= r\\sin\\theta = \\rho\\sin\\phi\\sin\\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting Cartesian $$(x,y,z)$$ $$\\to$$ spherical coordinates $$(\\rho,\\phi,\\theta)$$ \\shaded{ (x,y,z) \\Rightarrow \\left\\{ \\begin{align*} \\rho &= \\sqrt{r^2+z^2} = \\sqrt{x^2+y^2+z^2} \\\\ \\phi &= \\rm{atan}\\,\\frac{\\sqrt{x^2+y^2}}{z} \\\\ \\theta &= \\rm{atan}\\,\\frac{y}{x}\\\\ \\end{align*} \\right. } \\nonumber Some expressions • $$\\rho=a$$: a sphere of radius $$a$$, centered at the origin. • $$\\phi=\\frac{\\pi}{4}$$: a cone. In cylindrical that would $$z=r$$. • $$\\phi=\\frac{\\pi}{2}$$: the $$xy$$-pane. ### Volume element To find the volume element $$\\Delta V\\approx\\Delta\\rho\\,\\Delta\\phi\\,\\Delta\\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides? Start with a surface area element $$\\Delta S$$ • To move north-south: you always have to traverse half a circle $$\\Longrightarrow$$ the vertical side of $$\\Delta S$$ is a piece of circle with radius $$a$$ $$\\Rightarrow$$ the length is $$a.\\Delta\\phi$$. • To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\\Longrightarrow$$ the horizontal side of $$\\Delta S$$ is a piece of circle of radius $$r=a\\sin\\phi$$.", "$cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! 5. Originally Posted by HallsofIvy ojones has already told you how to do it. The sphere, $x^2+ y^2+ z^2= 1$, in spherical coordinates, is $\\rho^2= 1$ and the cone, z^2= x^2+ y^2, in spherical coordinates is $\\rho^2 cos^2(\\phi)= \\rho^2 sin^2(\\phi)$ which reduces to $cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! I have a quick question. If I just wanted to find the volume of the sphere above the plane of intersection, how would I go about doing this? I have tried but can't figure it out. Any help would be appreciated. I have no problem solving it the way that was just shown but like to try and solve problems in more than one way. I found the volume of the cone and was just going to add the", "## Calculus 8th Edition a) $\\rho=3$ b) $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$ The conversion of rectangular coordinates to spherical coordinates is given as: $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ Here, $\\rho=\\sqrt {x^2+y^2+z^2}$; $\\phi =\\cos^{-1} [\\dfrac{z}{\\rho}]; \\theta=\\cos^{-1}[\\dfrac{x}{\\rho \\sin \\phi}]$ a) Here, we have $x^2+y^2+z^2=9$ But $\\rho=\\sqrt {x^2+y^2+z^2}$ This gives: $\\rho^2=3^2 \\implies \\rho=3$ b) Here, we have $x^2-y^2-z^2=1$ But $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ So, we have $(\\rho \\sin \\phi \\cos \\theta)^2-(\\rho \\sin \\phi \\sin \\theta)^2-(\\rho \\cos \\phi)^2=1$ or, $\\rho ^2 \\sin^2 \\phi \\times (\\cos^2 \\theta- \\sin^2 \\theta)-(\\rho \\cos \\phi)^2=1$ Thus, we get $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$", "quadrant). Lastly, the bounds on {eq}\\rho {/eq} tell us that the sphere has radius 3. So we are dealing with a slice/wedge in the first octant. A sketch is below. The volume of the wedge is {eq}\\begin{align*} \\int^{\\pi/4}_0 \\int^\\pi_0 \\int^3_0 \\rho^2 \\sin \\phi\\ d \\rho\\ d \\theta\\ d \\phi &= \\int_0^{\\pi} d \\theta \\int_0^{\\pi/4} \\sin \\phi\\ d\\phi \\int_0^3 \\rho^2\\ d\\rho \\\\ &= \\pi \\left [ -\\cos \\phi \\right ]_0^{\\pi/4} \\left [ \\frac13\\rho^3 \\right ]_0^3 \\\\ &= \\frac\\pi3 \\left( - \\frac{\\sqrt2}{2} + 1 \\right)(3^3) \\\\ &= \\frac{9\\pi}{2} \\left( 2-\\sqrt2 \\right) \\\\ &\\approx 8.2814 \\end{align*} {/eq} #### Learn more about this topic: Cylindrical & Spherical Coordinates: Definition, Equations & Examples from GRE Math: Study Guide & Test Prep Chapter 13 / Lesson 10 14K", "5(x^2+ y^2)= 5r^2$$. In Cylindrical coordinates that is $$z= 5r^2$$. It is particularly simple because, geometrically, the figure is a cylinder! In spherical coordinates $$z= \\rho cos(\\phi)= 5(\\rho^2 cos^2(\\theta)sin^2(\\phi)+ \\rho^2 sin^(\\theta) sin^2(\\phi))= 5\\rho^2 sin^2(\\phi)(cos^2(\\theta)+ sin^2(\\theta))= 5\\rho^2 sin^2(\\phi)$$. $$\\rho cos^(\\phi)= 5\\rho^2 sin^2(\\phi)$$.$$\\rho= \\frac{cos(\\phi)}{5 sin^2(\\phi)}$$. Guest", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq0,\\ 0\\leq\\theta\\leq\\2\\pi, and 0\\leq\\phi\\leq\\pi.)a) (0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?b) (-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=? Change from rectangular to spherical coordinates. (Let $$\\rho\\geq0,\\ 0\\leq\\theta\\leq 2\\pi$$, and $$0\\leq\\phi\\leq\\pi.)$$ a) $$(0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?$$ b) $$(-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=?$$ 2021-05-15 Step 1 a) Consider the rectangular coordinate $$(0,\\ -8,\\ 0)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho, \\theta,\\phi)$$ Here, $$\\rho\\geq0,\\ 0\\leq\\theta\\leq2\\pi,\\ 0\\leq\\phi\\leq\\pi$$ Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spheerical coordinates $$(\\rho,\\theta\\phi)$$ The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ Step 2 The rectangular coordinate is $$(x,y,z)=(0,\\ -8,\\ 0)$$ That is, $$x=0,\\ y=-8,\\ z=0.$$ Thus, from the statement, $$\\rho=\\sqrt{x^{2}+y^{2}+z^{2}}$$ $$=\\sqrt{0^{2}+(-8)^{2}+0^{2}}$$ $$\\{x=0,\\ y=-8,z=0\\}$$ $$=\\sqrt{64}$$ $$=8$$ From the statement, $$z=\\rho\\cos\\phi$$ $$0=8\\cos\\phi$$ $$\\{z=0,\\rho=8\\}$$ $$\\cos\\phi=0$$ $$\\phi=\\frac{\\pi}{2}$$ $$\\{0\\leq\\phi\\leq\\pi\\}$$ Step 3 From the statement, $$y=\\rho\\sin\\phi\\sin\\theta$$ $$\\sin\\theta=\\frac{y}{\\rho\\sin\\phi}$$ Substitute $$y=-8,\\rho=8,\\phi=\\frac{\\pi}{2}$$ $$\\sin\\theta=\\frac{-8}{8\\sin\\frac{\\pi}{2}}$$ $$=-1$$ $$\\left\\{\\sin\\frac{\\pi}{2}=1\\right\\}$$ $$\\theta=\\frac{3\\pi}{2}$$ $$\\left\\{0\\leq\\theta\\leq\\frac{\\pi}{2}\\right\\}$$ Hence, the spherical coordinate of the point $$(0,\\ -8,\\ 0)$$ is $$\\left(8,\\ \\frac{3\\pi}{2},\\ \\frac{\\pi}{2}\\right)$$ Step 4 Consider the rectangular coordinate $$(-1,\\ 1,\\ -2)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho,\\theta,\\phi)$$. Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$. The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ The", "# To compare and contrast: the rectangular, cylindrical and spherical coordinates systems. Question Alternate coordinate systems To compare and contrast: the rectangular, cylindrical and spherical coordinates systems. 2020-12-29 Given: Point is represented as $$\\displaystyle{\\left({r},\\theta,{z}\\right)}.$$ Proof: A point in cylindrical coordinate is represented by $$\\displaystyle{\\left({r},\\theta.{z}\\right)}.$$ $$\\displaystyle{\\left({r},\\theta\\right)}$$ are polar coordinates z is the usual - coordinates The rectangular (x, y, z) and the spherical coordinates $$\\displaystyle{\\left({r},\\theta,{z}\\right)}$$ are related as $$\\displaystyle{x}={r} \\cos{\\theta}$$ $$\\displaystyle{y}={r} \\sin{\\theta}$$ $$\\displaystyle{z}={z}$$ These are used to convert from cylindrical to rectangular coordinates. Rectangular to cylindrical coordinates: $$\\displaystyle{r}^{2}={x}^{2}+{y}^{2}$$ $$\\displaystyle \\tan{\\theta}=\\frac{y}{{x}}$$ $$\\displaystyle{z}={z}$$ A point in spherical coordinate is represented by $$\\displaystyle{\\left(\\rho,\\theta,\\varphi\\right)}.$$ The rectangular (x, y, z) and the spherical coordinates $$\\displaystyle{\\left(\\rho,\\theta,\\varphi\\right)}$$ are related as $$\\displaystyle{x}=\\rho \\sin{\\varphi} \\cos{\\theta}$$ $$\\displaystyle{y}=\\rho \\sin{\\varphi} \\sin{\\theta}$$ $$\\displaystyle{z}=\\rho \\cos{\\varphi}$$ These are used to convert from spherical to rectangular coordinates. Rectangular to spherical coordinates: $$\\displaystyle\\rho^{2}={x}^{2}+{y}^{2}+{z}^{2}$$ $$\\displaystyle \\tan{\\theta}=\\frac{y}{{x}}$$ $$\\displaystyle\\varphi={{\\cos}^{ -{{1}}}{\\left(\\frac{z}{\\sqrt{{{x}^{2}+{y}^{2}+{z}^{2}}}}\\right)}}$$ Spherical to rectangular coordinates: $$\\displaystyle{r}=\\rho \\sin{\\varphi}$$ $$\\displaystyle\\theta=\\varphi$$ $$\\displaystyle{z}=\\rho \\cos{\\varphi}$$ Cylindricalto spherical coordinates: $$\\displaystyle\\rho^{2}=\\sqrt{{{r}^{2}+{z}^{2}}}$$ $$\\displaystyle\\theta=\\theta$$ $$\\displaystyle\\varphi={{\\cos}^{ -{{1}}}{\\left(\\frac{z}{\\sqrt{{{r}^{2}+{z}^{2}}}}\\right)}}$$ ### Relevant Questions Provide notes on how triple integrals defined in cylindrical and spherical coordinates and the reason to prefer one of these coordinate systems to working in rectangular coordinates. How are triple integrals defined in cylindrical and spherical coor-dinates? Why might one prefer working in one of these coordinate systems to working in rectangular coordinates? How are triple integrals defined in cylindrical", "Coordinates 5. Since $\\displaystyle {\\rho}^{2}=x^{2}+y^{2}+z^{2}$ We can sub it into the density function and get $\\displaystyle \\sqrt{1+{\\rho}^{3}}$ 6. I am just a bit confused on how you find the \"phi\" and \"theta\" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful 7. I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ 8. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\\displaystyle 0<y$ is. What do these regions mean? 9. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Ahh yes. I misread my x- and y-axes. $\\displaystyle 0 \\leq \\theta \\pi$ indeed. 10. Originally Posted by Gotovina7 Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi", "# How do you find the rectangular coordinates of the point with polar coordinates (5,pi/2)? Aug 24, 2014 To convert from from polar coordinates to rectangular coordinates, we must remember these: $x = r \\cos \\theta$ $y = r \\sin \\theta$ It's useful to remember these when we need to do these types of conversions. Knowing the $r$ and $\\theta$ from our question, we can find the rectangular coordinates: $r = 5$ $\\theta = \\frac{\\pi}{2}$ $x = \\left(5\\right) \\cos \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(0\\right) = 0$ $y = \\left(5\\right) \\sin \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(1\\right) = 5$ Therefore, our rectangular coordinates are: $\\left(0 , 5\\right)$", "rectangular system are (0, -2\\sqrt{2}, 0) This is a point on the negative y - axis at a distance 2\\sqrt{2} from the origin. ## Rectangular Coordinates to Spherical Coordinates Let us write the rectangular form (1, \\sqrt{3}, 2\\sqrt{3}) as spherical coordinates. We have x = 1, y = \\sqrt{3} and z = 2\\sqrt{3} Spherical coordinates are as follows: x = \\rho sin(\\phi) cos(\\theta) y = \\rho sin(\\phi) sin(\\theta) z = \\rho cos(\\phi) Now, \\rhosin(\\phi) cos(\\theta) = 1 ............(1) \\rho sin(\\phi) sin (\\theta) = \\sqrt{3} ............(2) \\rho cos(\\phi) = 2\\sqrt{3} ............(3) Dividing equation (2) by (1), we get tan (\\theta) = \\sqrt{3} \\theta = \\frac{\\pi}{3} (As both x and y are positive) \\rho ^2 = x^2 + y^2 + z^2 = 1 + (\\sqrt{3})^2 + (2\\sqrt{3})^2 = 1 + 3 + 12 = 16. Hence, \\rho = 4 \\rho cos(\\phi) = 2\\sqrt{3} 4 cos(\\phi) = 2\\sqrt{3} cos(\\phi) = \\frac{\\sqrt{3}}{2} \\phi = \\frac{\\pi}{6} Thus, the required spherical form of the coordinates are (4,$$\\frac{\\pi}{6}$$,$$\\frac{\\pi}{3}$$) ## Unit Vector in Spherical Coordinates In rectangular coordinate system, unit vectors \\hat{i}, \\hat{j}, \\hat{k} are corresponding to the directions of three coordinate axes. In the same manner, we can think of unit vectors \\hat{\\rho }, \\hat{\\phi } and \\hat{\\theta } in spherical coordinates. The position vector of any point P<\\rho, \\phi, \\theta> can be written in", "\\begin{align} V&=\\iiint_E1\\,dV=\\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta\\\\ \\ \\\\ &=\\frac{16\\pi}3\\,\\int_0^{\\pi/4}\\cos^3\\phi\\,\\sin\\phi\\,d\\phi =-\\frac{16\\pi}3\\,\\left.\\left(\\frac{\\cos^4\\phi}4 \\right)\\right|_0^{\\pi/4}\\\\ \\ \\\\ &=\\frac{4\\pi}3\\left(1-\\frac1{4} \\right)=\\pi. \\end{align} On the other hand, to work in cylindrical coordinates we have as follows. If we write $s=\\sqrt{x^2+y^2}$, the intersection of the sphere and the cone happens when $s^2+(s-1)^2=1$, with solutions $s=0$ and $s=1$. As we want to be above the cone, we have to choose $s=1$. The equations of the sphere and the cone are, respectively, $z=1+\\sqrt{1-r^2}$ and $z=r$. Then the volume is \\begin{align} V&=\\int_0^{2\\pi}\\int_0^1\\,(1+\\sqrt{1-r^2}-r)\\,r\\,dr\\,d\\theta =2\\pi\\,\\int_0^1(r+r\\sqrt{1-r^2}-r^2)\\,dr\\\\ \\ \\\\ &=2\\pi\\,\\left.\\left(\\frac{r^2}2-\\frac{\\sqrt{1-r^2}}3-\\frac{r^3}3\\right)\\right|_0^1 =2\\pi\\,\\left(\\frac12-0-\\frac13+\\frac13 \\right)=\\pi. \\end{align} • The equation of the sphere is $\\rho=2\\cos\\phi$ if I am not mistaken. – Kuifje Jan 23 '17 at 18:04 • I think you forgot to simplify the $1's$ when expanding $(z-1)^2$. – Kuifje Jan 23 '17 at 18:06 • Yes, you are right. Thanks! – Martin Argerami Jan 23 '17 at 18:11 In spherical coordinates: $$E=\\{(\\rho,\\theta,\\phi)|0\\le \\theta\\le 2\\pi, 0\\le \\phi \\le \\pi/4, 0\\le \\rho \\le 2\\cos \\phi \\}$$ It follows that $$\\boxed{ V= \\iiint_E dV = \\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi \\;d\\rho d\\phi d\\theta = \\pi }$$ Note: to find your bounds for $\\phi$ and $\\rho$: • The upper bound for $\\phi$ is precisely the cone $z=\\sqrt{x^2+y^2}\\; \\Leftrightarrow \\; \\rho \\cos \\phi = \\rho \\sin \\phi \\; \\Leftrightarrow \\; \\phi = \\pi/4$ • The upper bound for $\\rho$ is precisely the sphere $x^2+y^2+(z-1)^2=1\\; \\Leftrightarrow \\;\\rho^2\\sin^2\\phi+(\\rho \\cos\\phi-1)^2=1\\; \\Leftrightarrow", "\\phi={{\\cos}^{-1}}(\\frac{z}{\\rho})$ $latex \\phi={{\\cos}^{-1}}(\\frac{5}{6.71})$ $latex \\phi=0.73$ rad The spherical coordinates of the point are (6.71, 1.11 rad, 0.73 rad).", "be $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,dr}{\\,dt}\\right)^2+r^2\\left(\\frac{\\,d\\theta}{\\,dt}\\right)^2+\\left(\\frac{\\,dz}{\\,dt}\\right)^2}\\,dt$ To define arc length in the spherical coordinate system, we need to know first how to convert points from spherical to cylindrical coordinates (and then spherical to rectangular). I leave without proof that $r=\\rho\\sin\\phi$, $\\theta=\\theta$, and $z=\\rho\\cos\\phi$. Since we know that $x=r\\cos\\theta$, $y=r\\sin\\theta$, and $z=z$ goes from cylindrical to rectangular, it follows that $x=\\rho\\sin\\phi\\cos\\theta$, $y=\\rho\\sin\\phi\\sin\\theta$, and $z=\\rho\\cos\\phi$ takes points in spherical coordinates and converts them into points from the rectangular coordinate system. Again we consider the arc length formula for a parametric curve in the rectangular coordinate system: $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,dx}{\\,dt}\\right)^2+\\left(\\frac{\\,dy}{\\,dt}\\right)^2+\\left(\\frac{\\,dz}{\\,dt}\\right)^2}\\,dt$ If we let $\\rho=\\rho(t)$, $\\phi=\\phi(t)$ and $\\theta=\\theta(t)$, we now can find expressions for $\\frac{\\,dx}{\\,dt}$, $\\frac{\\,dy}{\\,dt}$ and $\\frac{\\,dz}{\\,dt}$. I leave the calculations of these three to the reader, but I will, for the sake of space, write the values to each one. Since $x=\\rho\\sin\\phi\\cos\\theta$, we have $\\dfrac{\\,dx}{\\,dt}=\\sin\\phi\\cos\\theta\\dfrac{\\,d\\rho}{\\,dt}-\\rho\\sin\\phi\\sin\\theta\\dfrac{\\,d\\theta}{\\,dt}+\\rho\\cos\\phi\\cos\\theta\\dfrac{\\,d\\phi}{\\,dt}$ Since $y=\\rho\\sin\\phi\\sin\\theta$, we have $\\dfrac{\\,dy}{\\,dt}=\\sin\\phi\\sin\\theta\\dfrac{\\,d\\rho}{\\,dt}+\\rho\\sin\\phi\\cos\\theta\\dfrac{\\,d\\theta}{\\,dt}+\\rho\\cos\\phi\\sin\\theta\\dfrac{\\,phi}{\\,dt}$ Since $z=\\rho\\cos\\phi$, we have $\\dfrac{\\,dz}{\\,dt}=\\cos\\phi\\dfrac{\\,d\\rho}{\\,dt}-\\rho\\sin\\phi\\dfrac{\\,d\\phi}{\\,dt}$ When you plug this into the rectangular arc length formula and after a ton of painful simplifying (which I have left out to spare your sanity), we see that the arc length formula in spherical coordinates is: $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,d\\rho}{\\,dt}\\right)^2+\\rho^2\\sin^2\\phi\\left(\\frac{\\,d\\theta}{\\,dt}\\right)^2+\\rho^2\\left(\\frac{\\,d\\phi}{\\,dt}\\right)^2}\\,dt$ ## Testing LaTeX $\\displaystyle\\int_{-\\infty}^{\\infty}e^{-ax^2}\\,dx=\\frac{1}{2}\\sqrt{\\frac{\\pi}{a}}$ $\\displaystyle f\\!\\left(x\\right)=a_0+\\sum\\limits_{n=1}^{\\infty}a_n\\cos\\!\\left(\\frac{n\\pi x}{L}\\right)+b_n\\sin\\!\\left(\\frac{n\\pi x}{L}\\right)$ ## One Dimensional Wave Equation Q: Find the general solution to the boundary value problem for the 1-D Wave Equation: \\begin{aligned}\\frac{\\partial^2y}{\\partial^2t} & = a^2\\frac{\\partial^2y}{\\partial^2x}\\\\y(0,t) &", "be $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,dr}{\\,dt}\\right)^2+r^2\\left(\\frac{\\,d\\theta}{\\,dt}\\right)^2+\\left(\\frac{\\,dz}{\\,dt}\\right)^2}\\,dt$ To define arc length in the spherical coordinate system, we need to know first how to convert points from spherical to cylindrical coordinates (and then spherical to rectangular). I leave without proof that $r=\\rho\\sin\\phi$, $\\theta=\\theta$, and $z=\\rho\\cos\\phi$. Since we know that $x=r\\cos\\theta$, $y=r\\sin\\theta$, and $z=z$ goes from cylindrical to rectangular, it follows that $x=\\rho\\sin\\phi\\cos\\theta$, $y=\\rho\\sin\\phi\\sin\\theta$, and $z=\\rho\\cos\\phi$ takes points in spherical coordinates and converts them into points from the rectangular coordinate system. Again we consider the arc length formula for a parametric curve in the rectangular coordinate system: $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,dx}{\\,dt}\\right)^2+\\left(\\frac{\\,dy}{\\,dt}\\right)^2+\\left(\\frac{\\,dz}{\\,dt}\\right)^2}\\,dt$ If we let $\\rho=\\rho(t)$, $\\phi=\\phi(t)$ and $\\theta=\\theta(t)$, we now can find expressions for $\\frac{\\,dx}{\\,dt}$, $\\frac{\\,dy}{\\,dt}$ and $\\frac{\\,dz}{\\,dt}$. I leave the calculations of these three to the reader, but I will, for the sake of space, write the values to each one. Since $x=\\rho\\sin\\phi\\cos\\theta$, we have $\\dfrac{\\,dx}{\\,dt}=\\sin\\phi\\cos\\theta\\dfrac{\\,d\\rho}{\\,dt}-\\rho\\sin\\phi\\sin\\theta\\dfrac{\\,d\\theta}{\\,dt}+\\rho\\cos\\phi\\cos\\theta\\dfrac{\\,d\\phi}{\\,dt}$ Since $y=\\rho\\sin\\phi\\sin\\theta$, we have $\\dfrac{\\,dy}{\\,dt}=\\sin\\phi\\sin\\theta\\dfrac{\\,d\\rho}{\\,dt}+\\rho\\sin\\phi\\cos\\theta\\dfrac{\\,d\\theta}{\\,dt}+\\rho\\cos\\phi\\sin\\theta\\dfrac{\\,phi}{\\,dt}$ Since $z=\\rho\\cos\\phi$, we have $\\dfrac{\\,dz}{\\,dt}=\\cos\\phi\\dfrac{\\,d\\rho}{\\,dt}-\\rho\\sin\\phi\\dfrac{\\,d\\phi}{\\,dt}$ When you plug this into the rectangular arc length formula and after a ton of painful simplifying (which I have left out to spare your sanity), we see that the arc length formula in spherical coordinates is: $\\displaystyle\\int_a^b\\sqrt{\\left(\\frac{\\,d\\rho}{\\,dt}\\right)^2+\\rho^2\\sin^2\\phi\\left(\\frac{\\,d\\theta}{\\,dt}\\right)^2+\\rho^2\\left(\\frac{\\,d\\phi}{\\,dt}\\right)^2}\\,dt$" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "# Orthogonal projection onto span of vectors using weighted inner product • Jul 14th 2008, 07:11 PM JCS007 Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ • Jul 14th 2008, 09:43 PM CaptainBlack Quote: Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "${\\displaystyle ={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}1/2\\end{pmatrix}}{\\begin{pmatrix}-1&0&1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}-1/2&0&1/2\\end{pmatrix}}={\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}}$ followed by matrix-vector multiplication ${\\displaystyle {\\mbox{proj}}_{M}({\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}){\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}}$ Problem 5 Check that the operation of Definition 3.1 is well-defined. That is, in Example 3.2 and 3.3, doesn't the answer depend on the choice of bases? No, a decomposition of vectors ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ into ${\\displaystyle {\\vec {m}}\\in M}$ and ${\\displaystyle {\\vec {n}}\\in N}$ does not depend on the bases chosen for the subspaces— this was shown in the Direct Sum subsection. Problem 6 What is the orthogonal projection onto the trivial subspace? The orthogonal projection of a vector onto a subspace is a member of that subspace. Since a trivial subspace has only one member, ${\\displaystyle {\\vec {0}}}$, the projection of any vector must equal ${\\displaystyle {\\vec {0}}}$. Problem 7 What is the projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle M}$ along ${\\displaystyle N}$ if ${\\displaystyle {\\vec {v}}\\in M}$? The projection onto ${\\displaystyle M}$ along ${\\displaystyle N}$ of a ${\\displaystyle {\\vec {v}}\\in M}$ is ${\\displaystyle {\\vec {v}}}$. Decomposing ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ gives ${\\displaystyle {\\vec {m}}={\\vec {v}}}$ and ${\\displaystyle {\\vec {n}}={\\vec {0}}}$, and dropping the ${\\displaystyle N}$ part but retaining the ${\\displaystyle M}$ part results in a projection of ${\\displaystyle {\\vec {m}}={\\vec {v}}}$. Problem 8 Show that if ${\\displaystyle M\\subseteq \\mathbb {R} ^{n}}$ is a subspace with orthonormal basis ${\\displaystyle \\langle {\\vec {\\kappa }}_{1},\\ldots ,{\\vec", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Math Help - Orthogonal projection onto span of vectors using weighted inner product 1. ## Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ 2. Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "maps between vector spaces of different dimension An example of a linear map between two vector spaces with different dimensions is the following projection of the space ${\\displaystyle \\mathbb {R} ^{3}}$ onto the plane ${\\displaystyle \\mathbb {R} ^{2}}$: ${\\displaystyle f\\colon \\mathbb {R} ^{3}\\to \\mathbb {R} ^{2};\\quad {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}\\mapsto {\\begin{pmatrix}x\\\\y\\end{pmatrix}}}$ We now check whether the vector addition is preserved. That means, for vectors ${\\displaystyle a,b\\in \\mathbb {R} ^{3}}$ we need that ${\\displaystyle f(a+b)=f(a)+f(b)}$ This can be verified directly: {\\displaystyle {\\begin{aligned}f(a+b)&=f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}+{\\begin{pmatrix}b_{x}\\\\b_{y}\\\\b_{z}\\end{pmatrix}}\\right)=f\\left({\\begin{pmatrix}a_{x}+b_{x}\\\\a_{y}+b_{y}\\\\a_{z}+b_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&={\\begin{pmatrix}a_{x}+b_{x}\\\\a_{y}+b_{y}\\end{pmatrix}}={\\begin{pmatrix}a_{x}\\\\a_{y}\\end{pmatrix}}+{\\begin{pmatrix}b_{x}\\\\b_{y}\\end{pmatrix}}\\\\[0.5em]&=f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)+f\\left({\\begin{pmatrix}b_{x}\\\\b_{y}\\\\b_{z}\\end{pmatrix}}\\right)=f(a)+f(b).\\end{aligned}}} Now we check homogeneity. For all ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle a\\in \\mathbb {R} ^{2}}$ we need: ${\\displaystyle f(\\lambda \\cdot a)=\\lambda \\cdot f(a).}$ We have that {\\displaystyle {\\begin{aligned}f(\\lambda \\cdot a)&=f\\left(\\lambda \\cdot {\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)=f\\left({\\begin{pmatrix}\\lambda a_{x}\\\\\\lambda a_{y}\\\\\\lambda a_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&={\\begin{pmatrix}\\lambda a_{x}\\\\\\lambda a_{y}\\end{pmatrix}}=\\lambda \\cdot {\\begin{pmatrix}a_{x}\\\\a_{y}\\end{pmatrix}}=\\lambda \\cdot f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&=\\lambda \\cdot f(a).\\end{aligned}}} So the projection ${\\displaystyle f}$ is a linear map. ### A non-linear map Next, we investigate some examples for non-linear maps. It is easy to come up with such maps: basically any function on ${\\displaystyle \\mathbb {R} }$ whose graph is not a line is a non-linear map. So \"most maps are non-linear\". Of course, there are also examples for non-linear maps on ${\\displaystyle \\mathbb {R} ^{2}}$. For instance, consider the norm mapping on the plane which assigns the length to every vector: ${\\displaystyle \\|\\cdot \\|_{2}\\,\\colon \\,\\mathbb {R} ^{2}\\to \\mathbb {R} ;\\quad {\\begin{pmatrix}x\\\\y\\end{pmatrix}}\\mapsto {\\sqrt", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "similarly for the nullspace of $1-P$. 1. since $P$ is a projection matrix $P^2=P$. Using this it is easy to show that (I-P)(I-P) = I-P. Hence I-P is also Projection (symmetry property also preserves). 2. Since the eigenvalues of a projection matrix are either ones or zeros you can use eigendecomposition: $P = U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T$ where $M$ is the dimension of the subspace to which the projectios is carried it's easy to show that $I - U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T = U \\begin{pmatrix} \\bf{0} & \\bf{0}\\\\ \\bf{0} & I_{N-M} \\end{pmatrix} U^T$. From here it is easy to extract the 4 sub-spaces of each projection operator. The above analysis is correct for orthogonal projectors. For oblique projectors a similar calculation can be done using singular value decomposition. • Oblique projectors don't diagonalise like this – Jazzmaniac Mar 27 '16 at 12:09 • Right. Obviously the problem consider orthogonal projectors. – Nir Regev Mar 27 '16 at 12:13 • I don't see any reference to orthogonal projection in the question – Jazzmaniac Mar 27 '16 at 12:14 • @Jazzmaniac you're right. I updated the answer – Nir Regev Mar 27 '16 at 12:29", "# Projection (linear algebra) \"Orthogonal projection\" redirects here. For the technical drawing concept, see Orthographic projection. For a concrete discussion of orthogonal projections in finite-dimensional linear spaces, see Vector projection. The transformation P is the orthogonal projection onto the line m. In linear algebra and functional analysis, a projection is a linear transformation P from a vector space to itself such that P 2 = P. That is, whenever P is applied twice to any value, it gives the same result as if it were applied once (idempotent). It leaves its image unchanged.[1] Though abstract, this definition of \"projection\" formalizes and generalizes the idea of graphical projection. One can also consider the effect of a projection on a geometrical object by examining the effect of the projection on points in the object. ## Simple example ### Orthogonal projection For example, the function which maps the point (x, y, z) in three-dimensional space R3 to the point (x, y, 0) is an orthogonal projection onto the xy plane. This function is represented by the matrix ${\\displaystyle P={\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&0\\end{bmatrix}}.}$ The action of this matrix on an arbitrary vector is ${\\displaystyle P{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}x\\\\y\\\\0\\end{pmatrix}}.}$ To see that P is indeed a projection, i.e., P = P2, we compute ${\\displaystyle P^{2}{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}=P{\\begin{pmatrix}x\\\\y\\\\0\\end{pmatrix}}={\\begin{pmatrix}x\\\\y\\\\0\\end{pmatrix}}=P{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}.}$ ### Oblique projection A simple example of a non-orthogonal (oblique) projection (for definition see", "# Action of a projection on a pairwise orthogonal projections. Let us consider two projections $e_1$ and $e_2$ on the Hilbert space $H$ with $e_1e_2=0$. Does there exists any projection $p$ with $$\\overline{pe_1(H)}=\\overline{pe_2(H)}=\\overline{p(e_1+e_2)(H)}\\neq0$$ Wlog we can assume that $$e_1=\\begin{pmatrix}1&0&0\\\\0&0&0\\\\0&0&0\\end{pmatrix} \\mbox{ and } e_2=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$ with respect to a certain orthogonal decomposition of $H=H_1\\oplus H_2\\oplus H_3$. Then it suffices to set $$p=\\begin{pmatrix}1/2&1/2&0\\\\1/2&1/2&0\\\\0&0&0\\end{pmatrix}.$$" ]

[ "${\\displaystyle {\\vec {p}}={\\frac {{\\begin{pmatrix}0\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-3/5\\\\-9/5\\end{pmatrix}}}$ The distance between ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {p}}}$ is approximately ${\\displaystyle 0.63}$ miles and so the sub must move to get in range. This subsection has developed a natural projection map: orthogonal projection onto a line. As suggested by the examples, it is often called for in applications. The next subsection shows how the definition of orthogonal projection onto a line gives us a way to calculate especially convienent bases for vector spaces, again something that is common in applications. The final subsection completely generalizes projection, orthogonal or not, onto any subspace at all. Exercises This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "${\\displaystyle ={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}1/2\\end{pmatrix}}{\\begin{pmatrix}-1&0&1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}-1/2&0&1/2\\end{pmatrix}}={\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}}$ followed by matrix-vector multiplication ${\\displaystyle {\\mbox{proj}}_{M}({\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}){\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}}$ Problem 5 Check that the operation of Definition 3.1 is well-defined. That is, in Example 3.2 and 3.3, doesn't the answer depend on the choice of bases? No, a decomposition of vectors ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ into ${\\displaystyle {\\vec {m}}\\in M}$ and ${\\displaystyle {\\vec {n}}\\in N}$ does not depend on the bases chosen for the subspaces— this was shown in the Direct Sum subsection. Problem 6 What is the orthogonal projection onto the trivial subspace? The orthogonal projection of a vector onto a subspace is a member of that subspace. Since a trivial subspace has only one member, ${\\displaystyle {\\vec {0}}}$, the projection of any vector must equal ${\\displaystyle {\\vec {0}}}$. Problem 7 What is the projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle M}$ along ${\\displaystyle N}$ if ${\\displaystyle {\\vec {v}}\\in M}$? The projection onto ${\\displaystyle M}$ along ${\\displaystyle N}$ of a ${\\displaystyle {\\vec {v}}\\in M}$ is ${\\displaystyle {\\vec {v}}}$. Decomposing ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ gives ${\\displaystyle {\\vec {m}}={\\vec {v}}}$ and ${\\displaystyle {\\vec {n}}={\\vec {0}}}$, and dropping the ${\\displaystyle N}$ part but retaining the ${\\displaystyle M}$ part results in a projection of ${\\displaystyle {\\vec {m}}={\\vec {v}}}$. Problem 8 Show that if ${\\displaystyle M\\subseteq \\mathbb {R} ^{n}}$ is a subspace with orthonormal basis ${\\displaystyle \\langle {\\vec {\\kappa }}_{1},\\ldots ,{\\vec", "it extends to the endpoint ${\\displaystyle (b_{1}-a_{1},b_{2}-a_{2})}$. We typically just refer to \"the point ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$\" rather than \"the endpoint of the canonical position of\" that vector. Thus, we will call both of these sets ${\\displaystyle \\mathbb {R} ^{2}}$. ${\\displaystyle \\{(x_{1},x_{2})\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}\\qquad \\{{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}}$ In the prior section we defined vectors and vector operations with an algebraic motivation; ${\\displaystyle r\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}={\\begin{pmatrix}rv_{1}\\\\rv_{2}\\end{pmatrix}}\\qquad {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}+{\\begin{pmatrix}w_{1}\\\\w_{2}\\end{pmatrix}}={\\begin{pmatrix}v_{1}+w_{1}\\\\v_{2}+w_{2}\\end{pmatrix}}}$ we can now interpret those operations geometrically. For instance, if ${\\displaystyle {\\vec {v}}}$ represents a displacement then ${\\displaystyle 3{\\vec {v}}\\,}$ represents a displacement in the same direction but three times as far, and ${\\displaystyle -1{\\vec {v}}\\,}$ represents a displacement of the same distance as ${\\displaystyle {\\vec {v}}\\,}$ but in the opposite direction. And, where ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {w}}}$ represent displacements, ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ represents those displacements combined. The long arrow is the combined displacement in this sense: if, in one minute, a ship's motion gives it the displacement relative to the earth of ${\\displaystyle {\\vec {v}}}$ and a passenger's motion gives a displacement relative to the ship's deck of ${\\displaystyle {\\vec {w}}}$, then ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ is the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule. Draw the parallelogram formed by the vectors", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "line spanned by a nonzero ${\\displaystyle {\\vec {s}}\\,}$ is this vector. ${\\displaystyle {\\mbox{proj}}_{[{\\vec {s}}\\,]}({\\vec {v}})={\\frac {{\\vec {v}}\\cdot {\\vec {s}}}{{\\vec {s}}\\cdot {\\vec {s}}}}{\\vec {s}}}$ Problem 13 checks that the outcome of the calculation depends only on the line and not on which vector ${\\displaystyle {\\vec {s}}\\,}$ happens to be used to describe that line. Remark 1.2 The wording of that definition says \"spanned by ${\\displaystyle {\\vec {s}}\\,}$\" instead the more formal \"the span of the set ${\\displaystyle \\{{\\vec {s}}\\,\\}}$\". This casual first phrase is common. Example 1.3 To orthogonally project the vector ${\\displaystyle {\\binom {2}{3}}}$ onto the line ${\\displaystyle y=2x}$, we first pick a direction vector for the line. For instance, ${\\displaystyle {\\vec {s}}={\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$ will do. Then the calculation is routine. ${\\displaystyle {\\frac {{\\begin{pmatrix}2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\displaystyle {\\frac {8}{5}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\begin{pmatrix}8/5\\\\16/5\\end{pmatrix}}}$ Example 1.4 In ${\\displaystyle \\mathbb {R} ^{3}}$, the orthogonal projection of a general vector ${\\displaystyle {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}}$ onto the ${\\displaystyle y}$-axis is ${\\displaystyle {\\frac {{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}0\\\\y\\\\0\\end{pmatrix}}}$ which matches our intuitive expectation. The picture above with the stick figure walking out on the line until ${\\displaystyle {\\vec {v}}}$'s tip is overhead is one way to think of the orthogonal projection of a vector onto a line. We finish this subsection with two other ways. Example 1.5 A railroad car left on an east-west track without its brake is pushed", "{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}x_{1}+x_{2}-1\\\\y_{1}+y_{2}\\\\z_{1}+z_{2}\\end{pmatrix}}\\qquad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}rx-r+1\\\\ry\\\\rz\\end{pmatrix}}}$ 1. Show that it is not a subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ . (Hint. See Example 2.5). 2. Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply. 3. Show that any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must pass through the origin, and so any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must involve zero in its description. Does the converse hold? Does any subset of ${\\displaystyle \\mathbb {R} ^{3}}$ that contains the origin become a subspace when given the inherited operations? 1. It is not a subspace because these are not the inherited operations. For one thing, in this space, ${\\displaystyle 0\\cdot {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}}$ while this does not, of course, hold in ${\\displaystyle \\mathbb {R} ^{3}}$ . 2. We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If ${\\displaystyle r_{1},r_{2},x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}}$ are in ${\\displaystyle \\mathbb {R} }$ then ${\\displaystyle r_{1}{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+r_{2}{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+1\\\\r_{1}y_{1}\\\\r_{1}z_{1}\\end{pmatrix}}+{\\begin{pmatrix}r_{2}x_{2}-r_{2}+1\\\\r_{2}y_{2}\\\\r_{2}z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+r_{2}x_{2}-r_{2}+1\\\\r_{1}y_{1}+r_{2}y_{2}\\\\r_{1}z_{1}+r_{2}z_{2}\\end{pmatrix}}}$ (note that the definition of addition in this space is that the first components combine as ${\\displaystyle (r_{1}x_{1}-r_{1}+1)+(r_{2}x_{2}-r_{2}+1)-1}$ , so the first component of the last vector does not say \"${\\displaystyle +2}$ \"). Adding the three components of the last vector gives ${\\displaystyle r_{1}(x_{1}-1+y_{1}+z_{1})+r_{2}(x_{2}-1+y_{2}+z_{2})+1=r_{1}\\cdot 0+r_{2}\\cdot 0+1=1}$ . Most", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "|}\\,q\\in \\mathbb {C} \\}}$, and ${\\displaystyle {\\mathcal {N}}(t_{0}^{2})}$ is the space of quadratic polynomials with no ${\\displaystyle x^{2}}$ term ${\\displaystyle \\{p+qx\\,{\\big |}\\,p,q\\in \\mathbb {C} \\}}$, and this is the end ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{0})={\\mathcal {N}}(t_{0}^{2})}$. 2. The second power ${\\displaystyle {\\begin{pmatrix}a\\\\b\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\a\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\0\\end{pmatrix}}}$ is the zero map. Consequently, the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{2}\\supset \\{{\\begin{pmatrix}0\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}\\supset \\{{\\vec {0}}\\,\\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}q\\\\0\\end{pmatrix}}\\,{\\big |}\\,q\\in \\mathbb {C} \\}\\subset \\mathbb {R} ^{2}=\\cdots }$ each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space. 3. Iterates of this map cycle around ${\\displaystyle a+bx+cx^{2}{\\stackrel {t_{2}}{\\longmapsto }}b+cx+ax^{2}{\\stackrel {t_{2}}{\\longmapsto }}c+ax+bx^{2}{\\stackrel {t_{2}}{\\longmapsto }}a+bx+cx^{2}\\;\\cdots }$ and the chains of rangespaces and nullspaces are trivial. ${\\displaystyle {\\mathcal {P}}_{2}={\\mathcal {P}}_{2}=\\cdots \\qquad \\{{\\vec {0}}\\,\\}=\\{{\\vec {0}}\\,\\}=\\cdots }$ Thus, obviously, generalized spaces are ${\\displaystyle {\\mathcal {R}}_{\\infty }(t_{2})={\\mathcal {P}}_{2}}$ and ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{2})=\\{{\\vec {0}}\\,\\}}$. 4. We have ${\\displaystyle {\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\b\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto \\cdots }$ and so the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{3}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\r\\end{pmatrix}}\\,{\\big |}\\,p,r\\in \\mathbb {C} \\}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}0\\\\0\\\\r\\end{pmatrix}}\\,{\\big |}\\,r\\in \\mathbb {C} \\}\\subset \\{{\\begin{pmatrix}0\\\\q\\\\r\\end{pmatrix}}\\,{\\big |}\\,q,r\\in \\mathbb {C} \\}=\\cdots }$ each has length two. The generalized spaces are the final ones shown above in each chain.", "f}$ is an epimorphism. Solution By construction, ${\\displaystyle f}$ is a linear map. We want to show that ${\\displaystyle \\{f(e_{1}),f(e_{2}),f(e_{3})\\}=\\{(1,1),(1,-1)\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have with the above theorem that ${\\displaystyle f}$ is an epimorphism. So, we must represent every ${\\displaystyle (x,y)\\in \\mathbb {R} ^{2}}$ as a linear combination of vectors ${\\displaystyle (1,1),(1,-1)}$. Accordingly, we search linear combination scalars ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$ such that {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}x\\\\y\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\end{pmatrix}}+\\mu {\\begin{pmatrix}1\\\\-1\\end{pmatrix}}.\\end{aligned}}} From this we get the linear system of equations {\\displaystyle {\\begin{aligned}{\\begin{aligned}x&=\\lambda +\\mu \\\\y&=\\lambda -\\mu ,\\end{aligned}}\\end{aligned}}} which is solved by ${\\displaystyle \\lambda ={\\frac {x+y}{2}}}$ and ${\\displaystyle \\mu ={\\frac {x-y}{2}}}$. So we have ${\\displaystyle {\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\frac {x+y}{2}}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}+{\\frac {x-y}{2}}{\\begin{pmatrix}1\\\\-1\\end{pmatrix}}}$ for all ${\\displaystyle x,y\\in \\mathbb {R} }$. Thus ${\\displaystyle \\{(1,1),(1,-1)\\}}$ is a generator. This proves that ${\\displaystyle f}$ is an epimorphism. Exercise Consider the function space ${\\displaystyle \\operatorname {Fun} ([0,1],\\mathbb {R} )}$ of all function s mapping ${\\displaystyle \\mathbb {R} }$ to ${\\displaystyle \\mathbb {R} }$, as well as the map {\\displaystyle {\\begin{aligned}\\phi \\colon \\operatorname {Fun} ([0,1],\\mathbb {R} )&\\to \\mathbb {R} ,\\\\f&\\mapsto f(0).\\end{aligned}}} Show that ${\\displaystyle \\phi }$ is an epimorphism. Solution The operations on the function space are defined element-wise in each case. That is: for ${\\displaystyle f,g\\in \\operatorname {Fun} ([0,1],\\mathbb {R} )}$, ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle x\\in [0,1]}$ we", "This exercise is recommended for all readers. Problem 5 Decide if ${\\displaystyle 1+2x}$ is in the range of the map from ${\\displaystyle \\mathbb {R} ^{3}}$ to ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle {\\mathcal {E}}_{3}}$ and ${\\displaystyle \\langle 1,1+x^{2},x\\rangle }$ by this matrix. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}}$ Denote the given basis of ${\\displaystyle {\\mathcal {P}}_{2}}$ by ${\\displaystyle B}$. Then application of the linear map is represented by matrix-vector addition. Thus, the first vector in ${\\displaystyle {\\mathcal {E}}_{3}}$ is mapped to the element of ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle B}$ by ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}}$ and that element is ${\\displaystyle 1+x}$. The other two images of basis vectors are calculated similarly. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}3\\\\1\\\\0\\end{pmatrix}}={\\rm {Rep}}_{B}(4+x^{2})\\quad {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\rm {Rep}}_{B}(x)}$ We can thus decide if ${\\displaystyle 1+2x}$ is in the range of the map by looking for scalars ${\\displaystyle c_{1}}$, ${\\displaystyle c_{2}}$, and ${\\displaystyle c_{3}}$ such that ${\\displaystyle c_{1}\\cdot (1)+c_{2}\\cdot (1+x^{2})+c_{3}\\cdot (x)=1+2x}$ and obviously ${\\displaystyle c_{1}=1}$, ${\\displaystyle c_{2}=0}$, and ${\\displaystyle c_{3}=1}$ suffice. Thus it is in the range, and in fact it is the image of this vector. ${\\displaystyle 1\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+0\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+1\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}}$ Problem 6 Example 2.8 gives a matrix that is nonsingular, and is therefore associated with maps that are nonsingular. 1. Find the set of column vectors representing the members of the nullspace of any map represented by this matrix. 2.", "{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto 1v_{1}+2v_{2}+3v_{3}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}}$ is a member of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is equal to the span of this vector. 2. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$. 3. Represent ${\\displaystyle f:\\mathbb {R} ^{3}\\to \\mathbb {R} ^{2}}$ ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto {\\begin{pmatrix}1v_{1}+2v_{2}+3v_{3}\\\\4v_{1}+5v_{2}+6v_{3}\\end{pmatrix}}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}},\\;{\\begin{pmatrix}4\\\\5\\\\6\\end{pmatrix}}}$ are both members of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is the span of these two. (Hint. See the third item of Problem 14.) 4. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} ^{m}}$. This, and related results, is called the Fundamental Theorem of Linear Algebra in (Strang 1993). 1. The representation of ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}1v_{1}+2v_{2}+3v_{3}}$ is this. ${\\displaystyle {\\rm {Rep}}_{{\\mathcal {E}}_{3},{\\mathcal {E}}_{1}}(f)={\\begin{pmatrix}1&2&3\\end{pmatrix}}}$ By the definition of ${\\displaystyle f}$ ${\\displaystyle {\\mathcal {N}}(f)=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,1v_{1}+2v_{2}+3v_{3}=0\\}=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}=0\\}}$ and this second description exactly says this. ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }=[\\{{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\}]}$ 2. The generalization is that for any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$ there is a vector ${\\displaystyle {\\vec {h}}}$ so that ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\\\vdots \\\\v_{n}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}h_{1}v_{1}+\\dots +h_{n}v_{n}}$ and ${\\displaystyle {\\vec {h}}\\in {\\mathcal {N}}(f)^{\\perp }}$. We can prove this by, as in the prior item, representing ${\\displaystyle f}$", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "{\\begin{pmatrix}1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\1\\end{pmatrix}}={\\frac {x+y}{2}}\\cdot {\\begin{pmatrix}1\\\\1\\end{pmatrix}}={\\begin{pmatrix}(x+y)/2\\\\(x+y)/2\\end{pmatrix}}}$ which is the effect of this matrix. ${\\displaystyle {\\begin{pmatrix}1/2&1/2\\\\1/2&1/2\\end{pmatrix}}}$ 2. Rotating the entire plane ${\\displaystyle \\pi /4}$ radians clockwise brings the ${\\displaystyle y=x}$ line to lie on the ${\\displaystyle x}$-axis. Now projecting and then rotating back has the desired effect. Problem 15 For ${\\displaystyle {\\vec {a}},{\\vec {b}}\\in \\mathbb {R} ^{n}}$ let ${\\displaystyle {\\vec {v}}_{1}}$ be the projection of ${\\displaystyle {\\vec {a}}}$ onto the line spanned by ${\\displaystyle {\\vec {b}}}$, let ${\\displaystyle {\\vec {v}}_{2}}$ be the projection of ${\\displaystyle {\\vec {v}}_{1}}$ onto the line spanned by ${\\displaystyle {\\vec {a}}}$, let ${\\displaystyle {\\vec {v}}_{3}}$ be the projection of ${\\displaystyle {\\vec {v}}_{2}}$ onto the line spanned by ${\\displaystyle {\\vec {b}}}$, etc., back and forth between the spans of ${\\displaystyle {\\vec {a}}}$ and ${\\displaystyle {\\vec {b}}}$. That is, ${\\displaystyle {\\vec {v}}_{i+1}}$ is the projection of ${\\displaystyle {\\vec {v}}_{i}}$ onto the span of ${\\displaystyle {\\vec {a}}}$ if ${\\displaystyle i+1}$ is even, and onto the span of ${\\displaystyle {\\vec {b}}}$ if ${\\displaystyle i+1}$ is odd. Must that sequence of vectors eventually settle down— must there be a sufficiently large ${\\displaystyle i}$ such that ${\\displaystyle {\\vec {v}}_{i+2}}$ equals ${\\displaystyle {\\vec {v}}_{i}}$ and ${\\displaystyle {\\vec {v}}_{i+3}}$ equals ${\\displaystyle {\\vec {v}}_{i+1}}$? If so, what is the earliest such ${\\displaystyle i}$? ${\\displaystyle {\\vec {a}}={\\begin{pmatrix}1\\\\0\\end{pmatrix}}\\qquad {\\vec {b}}={\\begin{pmatrix}1\\\\1\\end{pmatrix}}}$ ${\\displaystyle {\\vec {v}}_{1}={\\begin{pmatrix}1/2\\\\1/2\\end{pmatrix}},\\quad {\\vec {v}}_{2}={\\begin{pmatrix}1/2\\\\0\\end{pmatrix}},\\quad {\\vec {v}}_{3}={\\begin{pmatrix}1/4\\\\1/4\\end{pmatrix}},\\quad \\ldots" ]

[ "${\\displaystyle {\\vec {p}}={\\frac {{\\begin{pmatrix}0\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-3/5\\\\-9/5\\end{pmatrix}}}$ The distance between ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {p}}}$ is approximately ${\\displaystyle 0.63}$ miles and so the sub must move to get in range. This subsection has developed a natural projection map: orthogonal projection onto a line. As suggested by the examples, it is often called for in applications. The next subsection shows how the definition of orthogonal projection onto a line gives us a way to calculate especially convienent bases for vector spaces, again something that is common in applications. The final subsection completely generalizes projection, orthogonal or not, onto any subspace at all. Exercises This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "it extends to the endpoint ${\\displaystyle (b_{1}-a_{1},b_{2}-a_{2})}$. We typically just refer to \"the point ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$\" rather than \"the endpoint of the canonical position of\" that vector. Thus, we will call both of these sets ${\\displaystyle \\mathbb {R} ^{2}}$. ${\\displaystyle \\{(x_{1},x_{2})\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}\\qquad \\{{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}}$ In the prior section we defined vectors and vector operations with an algebraic motivation; ${\\displaystyle r\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}={\\begin{pmatrix}rv_{1}\\\\rv_{2}\\end{pmatrix}}\\qquad {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}+{\\begin{pmatrix}w_{1}\\\\w_{2}\\end{pmatrix}}={\\begin{pmatrix}v_{1}+w_{1}\\\\v_{2}+w_{2}\\end{pmatrix}}}$ we can now interpret those operations geometrically. For instance, if ${\\displaystyle {\\vec {v}}}$ represents a displacement then ${\\displaystyle 3{\\vec {v}}\\,}$ represents a displacement in the same direction but three times as far, and ${\\displaystyle -1{\\vec {v}}\\,}$ represents a displacement of the same distance as ${\\displaystyle {\\vec {v}}\\,}$ but in the opposite direction. And, where ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {w}}}$ represent displacements, ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ represents those displacements combined. The long arrow is the combined displacement in this sense: if, in one minute, a ship's motion gives it the displacement relative to the earth of ${\\displaystyle {\\vec {v}}}$ and a passenger's motion gives a displacement relative to the ship's deck of ${\\displaystyle {\\vec {w}}}$, then ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ is the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule. Draw the parallelogram formed by the vectors", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "${\\displaystyle ={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}1/2\\end{pmatrix}}{\\begin{pmatrix}-1&0&1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}-1/2&0&1/2\\end{pmatrix}}={\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}}$ followed by matrix-vector multiplication ${\\displaystyle {\\mbox{proj}}_{M}({\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}){\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}}$ Problem 5 Check that the operation of Definition 3.1 is well-defined. That is, in Example 3.2 and 3.3, doesn't the answer depend on the choice of bases? No, a decomposition of vectors ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ into ${\\displaystyle {\\vec {m}}\\in M}$ and ${\\displaystyle {\\vec {n}}\\in N}$ does not depend on the bases chosen for the subspaces— this was shown in the Direct Sum subsection. Problem 6 What is the orthogonal projection onto the trivial subspace? The orthogonal projection of a vector onto a subspace is a member of that subspace. Since a trivial subspace has only one member, ${\\displaystyle {\\vec {0}}}$, the projection of any vector must equal ${\\displaystyle {\\vec {0}}}$. Problem 7 What is the projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle M}$ along ${\\displaystyle N}$ if ${\\displaystyle {\\vec {v}}\\in M}$? The projection onto ${\\displaystyle M}$ along ${\\displaystyle N}$ of a ${\\displaystyle {\\vec {v}}\\in M}$ is ${\\displaystyle {\\vec {v}}}$. Decomposing ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ gives ${\\displaystyle {\\vec {m}}={\\vec {v}}}$ and ${\\displaystyle {\\vec {n}}={\\vec {0}}}$, and dropping the ${\\displaystyle N}$ part but retaining the ${\\displaystyle M}$ part results in a projection of ${\\displaystyle {\\vec {m}}={\\vec {v}}}$. Problem 8 Show that if ${\\displaystyle M\\subseteq \\mathbb {R} ^{n}}$ is a subspace with orthonormal basis ${\\displaystyle \\langle {\\vec {\\kappa }}_{1},\\ldots ,{\\vec", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "line spanned by a nonzero ${\\displaystyle {\\vec {s}}\\,}$ is this vector. ${\\displaystyle {\\mbox{proj}}_{[{\\vec {s}}\\,]}({\\vec {v}})={\\frac {{\\vec {v}}\\cdot {\\vec {s}}}{{\\vec {s}}\\cdot {\\vec {s}}}}{\\vec {s}}}$ Problem 13 checks that the outcome of the calculation depends only on the line and not on which vector ${\\displaystyle {\\vec {s}}\\,}$ happens to be used to describe that line. Remark 1.2 The wording of that definition says \"spanned by ${\\displaystyle {\\vec {s}}\\,}$\" instead the more formal \"the span of the set ${\\displaystyle \\{{\\vec {s}}\\,\\}}$\". This casual first phrase is common. Example 1.3 To orthogonally project the vector ${\\displaystyle {\\binom {2}{3}}}$ onto the line ${\\displaystyle y=2x}$, we first pick a direction vector for the line. For instance, ${\\displaystyle {\\vec {s}}={\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$ will do. Then the calculation is routine. ${\\displaystyle {\\frac {{\\begin{pmatrix}2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\displaystyle {\\frac {8}{5}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\begin{pmatrix}8/5\\\\16/5\\end{pmatrix}}}$ Example 1.4 In ${\\displaystyle \\mathbb {R} ^{3}}$, the orthogonal projection of a general vector ${\\displaystyle {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}}$ onto the ${\\displaystyle y}$-axis is ${\\displaystyle {\\frac {{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}0\\\\y\\\\0\\end{pmatrix}}}$ which matches our intuitive expectation. The picture above with the stick figure walking out on the line until ${\\displaystyle {\\vec {v}}}$'s tip is overhead is one way to think of the orthogonal projection of a vector onto a line. We finish this subsection with two other ways. Example 1.5 A railroad car left on an east-west track without its brake is pushed", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}x_{1}+x_{2}-1\\\\y_{1}+y_{2}\\\\z_{1}+z_{2}\\end{pmatrix}}\\qquad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}rx-r+1\\\\ry\\\\rz\\end{pmatrix}}}$ 1. Show that it is not a subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ . (Hint. See Example 2.5). 2. Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply. 3. Show that any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must pass through the origin, and so any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must involve zero in its description. Does the converse hold? Does any subset of ${\\displaystyle \\mathbb {R} ^{3}}$ that contains the origin become a subspace when given the inherited operations? 1. It is not a subspace because these are not the inherited operations. For one thing, in this space, ${\\displaystyle 0\\cdot {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}}$ while this does not, of course, hold in ${\\displaystyle \\mathbb {R} ^{3}}$ . 2. We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If ${\\displaystyle r_{1},r_{2},x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}}$ are in ${\\displaystyle \\mathbb {R} }$ then ${\\displaystyle r_{1}{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+r_{2}{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+1\\\\r_{1}y_{1}\\\\r_{1}z_{1}\\end{pmatrix}}+{\\begin{pmatrix}r_{2}x_{2}-r_{2}+1\\\\r_{2}y_{2}\\\\r_{2}z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+r_{2}x_{2}-r_{2}+1\\\\r_{1}y_{1}+r_{2}y_{2}\\\\r_{1}z_{1}+r_{2}z_{2}\\end{pmatrix}}}$ (note that the definition of addition in this space is that the first components combine as ${\\displaystyle (r_{1}x_{1}-r_{1}+1)+(r_{2}x_{2}-r_{2}+1)-1}$ , so the first component of the last vector does not say \"${\\displaystyle +2}$ \"). Adding the three components of the last vector gives ${\\displaystyle r_{1}(x_{1}-1+y_{1}+z_{1})+r_{2}(x_{2}-1+y_{2}+z_{2})+1=r_{1}\\cdot 0+r_{2}\\cdot 0+1=1}$ . Most", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "|}\\,q\\in \\mathbb {C} \\}}$, and ${\\displaystyle {\\mathcal {N}}(t_{0}^{2})}$ is the space of quadratic polynomials with no ${\\displaystyle x^{2}}$ term ${\\displaystyle \\{p+qx\\,{\\big |}\\,p,q\\in \\mathbb {C} \\}}$, and this is the end ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{0})={\\mathcal {N}}(t_{0}^{2})}$. 2. The second power ${\\displaystyle {\\begin{pmatrix}a\\\\b\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\a\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\0\\end{pmatrix}}}$ is the zero map. Consequently, the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{2}\\supset \\{{\\begin{pmatrix}0\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}\\supset \\{{\\vec {0}}\\,\\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}q\\\\0\\end{pmatrix}}\\,{\\big |}\\,q\\in \\mathbb {C} \\}\\subset \\mathbb {R} ^{2}=\\cdots }$ each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space. 3. Iterates of this map cycle around ${\\displaystyle a+bx+cx^{2}{\\stackrel {t_{2}}{\\longmapsto }}b+cx+ax^{2}{\\stackrel {t_{2}}{\\longmapsto }}c+ax+bx^{2}{\\stackrel {t_{2}}{\\longmapsto }}a+bx+cx^{2}\\;\\cdots }$ and the chains of rangespaces and nullspaces are trivial. ${\\displaystyle {\\mathcal {P}}_{2}={\\mathcal {P}}_{2}=\\cdots \\qquad \\{{\\vec {0}}\\,\\}=\\{{\\vec {0}}\\,\\}=\\cdots }$ Thus, obviously, generalized spaces are ${\\displaystyle {\\mathcal {R}}_{\\infty }(t_{2})={\\mathcal {P}}_{2}}$ and ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{2})=\\{{\\vec {0}}\\,\\}}$. 4. We have ${\\displaystyle {\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\b\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto \\cdots }$ and so the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{3}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\r\\end{pmatrix}}\\,{\\big |}\\,p,r\\in \\mathbb {C} \\}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}0\\\\0\\\\r\\end{pmatrix}}\\,{\\big |}\\,r\\in \\mathbb {C} \\}\\subset \\{{\\begin{pmatrix}0\\\\q\\\\r\\end{pmatrix}}\\,{\\big |}\\,q,r\\in \\mathbb {C} \\}=\\cdots }$ each has length two. The generalized spaces are the final ones shown above in each chain.", "in linear algebra and serves as a stepstone for us to go to projection as we will see in this example. Define a function $$P\\vec{w}$$ as following. For all $$\\vec{w} ∈ R^2$$, let $$P\\vec{w}(\\vec v)$$ be the resulting vector of projecting $$\\vec{v}$$ onto $$Span(\\vec w)$$. Again, the reason why this function is a linear transformation will be clear in a moment. Let $$\\begin{pmatrix} x\\\\y \\end{pmatrix} = \\vec v$$. We know that the projection of $$\\vec v$$ onto $$\\vec w$$ is the vector that has $$\\frac{\\vec v \\cdot \\vec w}{\\| \\vec w\\|}$$ as its length in the direction of $$\\vec w$$. Thus, $$$$\\begin{split} P\\vec{w}(\\vec v) &= \\frac{\\vec v \\cdot \\vec w}{{\\| \\vec w\\|}^2} \\cdot \\vec w\\\\ &= \\frac{ax+by}{a^2+b^2} \\cdot \\begin{pmatrix} a\\\\b \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2x+aby}{a^2+b^2} \\\\ \\frac{abx+b^2y}{a^2+b^2} \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2}{a^2+b^2} x+ \\frac{ab}{a^2+b^2}y \\\\ \\frac{ab}{a^2+b^2}x+ \\frac{b^2}{a^2+b^2}y\\end{pmatrix} \\end{split}$$$$ Now we know that $$P\\vec w$$ is indeed a linear transformation, since a function f with the form $$f\\begin{pmatrix} x\\\\y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\cx+dy \\end{pmatrix}$$ is always a linear transformation. If you are not satisfied with this property, notice that we can easily check that $$P\\vec{w}$$ preserves addition and multiplication with this expression at hand. Now we can use $$P\\vec w$$ to get to the reflecting action $$F\\vec w$$. The key insight is that, we can get to $$\\vec {v'}$$ by adding $$[P\\vec{w}]\\vec v", "^{2}}$ is an ${\\displaystyle \\mathbb {R} }$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ### Important special cases Exercise (The identity is a linear map) Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space. Prove that the identity ${\\displaystyle \\operatorname {id} :V\\to V}$ with ${\\displaystyle \\operatorname {id} (v)=v}$ is a linear map. Proof (The identity is a linear map) The identity is additive: Let ${\\displaystyle v,w\\in V}$, then. ${\\displaystyle \\operatorname {id} (v+w)=v+w=\\operatorname {id} (v)+\\operatorname {id} (w)}$ The identity is homogeneous: Let ${\\displaystyle \\lambda \\in K}$ and ${\\displaystyle v\\in V}$, then ${\\displaystyle \\operatorname {id} (\\lambda \\cdot v)=\\lambda \\cdot v=\\lambda \\cdot \\operatorname {id} (v)}$ Exercise (The map to zero is a linear map) Let ${\\displaystyle V,W}$ be two ${\\displaystyle K}$-vector spaces. Show that", "f}$ is an epimorphism. Solution By construction, ${\\displaystyle f}$ is a linear map. We want to show that ${\\displaystyle \\{f(e_{1}),f(e_{2}),f(e_{3})\\}=\\{(1,1),(1,-1)\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have with the above theorem that ${\\displaystyle f}$ is an epimorphism. So, we must represent every ${\\displaystyle (x,y)\\in \\mathbb {R} ^{2}}$ as a linear combination of vectors ${\\displaystyle (1,1),(1,-1)}$. Accordingly, we search linear combination scalars ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$ such that {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}x\\\\y\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\end{pmatrix}}+\\mu {\\begin{pmatrix}1\\\\-1\\end{pmatrix}}.\\end{aligned}}} From this we get the linear system of equations {\\displaystyle {\\begin{aligned}{\\begin{aligned}x&=\\lambda +\\mu \\\\y&=\\lambda -\\mu ,\\end{aligned}}\\end{aligned}}} which is solved by ${\\displaystyle \\lambda ={\\frac {x+y}{2}}}$ and ${\\displaystyle \\mu ={\\frac {x-y}{2}}}$. So we have ${\\displaystyle {\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\frac {x+y}{2}}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}+{\\frac {x-y}{2}}{\\begin{pmatrix}1\\\\-1\\end{pmatrix}}}$ for all ${\\displaystyle x,y\\in \\mathbb {R} }$. Thus ${\\displaystyle \\{(1,1),(1,-1)\\}}$ is a generator. This proves that ${\\displaystyle f}$ is an epimorphism. Exercise Consider the function space ${\\displaystyle \\operatorname {Fun} ([0,1],\\mathbb {R} )}$ of all function s mapping ${\\displaystyle \\mathbb {R} }$ to ${\\displaystyle \\mathbb {R} }$, as well as the map {\\displaystyle {\\begin{aligned}\\phi \\colon \\operatorname {Fun} ([0,1],\\mathbb {R} )&\\to \\mathbb {R} ,\\\\f&\\mapsto f(0).\\end{aligned}}} Show that ${\\displaystyle \\phi }$ is an epimorphism. Solution The operations on the function space are defined element-wise in each case. That is: for ${\\displaystyle f,g\\in \\operatorname {Fun} ([0,1],\\mathbb {R} )}$, ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle x\\in [0,1]}$ we", "This exercise is recommended for all readers. Problem 5 Decide if ${\\displaystyle 1+2x}$ is in the range of the map from ${\\displaystyle \\mathbb {R} ^{3}}$ to ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle {\\mathcal {E}}_{3}}$ and ${\\displaystyle \\langle 1,1+x^{2},x\\rangle }$ by this matrix. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}}$ Denote the given basis of ${\\displaystyle {\\mathcal {P}}_{2}}$ by ${\\displaystyle B}$. Then application of the linear map is represented by matrix-vector addition. Thus, the first vector in ${\\displaystyle {\\mathcal {E}}_{3}}$ is mapped to the element of ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle B}$ by ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}}$ and that element is ${\\displaystyle 1+x}$. The other two images of basis vectors are calculated similarly. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}3\\\\1\\\\0\\end{pmatrix}}={\\rm {Rep}}_{B}(4+x^{2})\\quad {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\rm {Rep}}_{B}(x)}$ We can thus decide if ${\\displaystyle 1+2x}$ is in the range of the map by looking for scalars ${\\displaystyle c_{1}}$, ${\\displaystyle c_{2}}$, and ${\\displaystyle c_{3}}$ such that ${\\displaystyle c_{1}\\cdot (1)+c_{2}\\cdot (1+x^{2})+c_{3}\\cdot (x)=1+2x}$ and obviously ${\\displaystyle c_{1}=1}$, ${\\displaystyle c_{2}=0}$, and ${\\displaystyle c_{3}=1}$ suffice. Thus it is in the range, and in fact it is the image of this vector. ${\\displaystyle 1\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+0\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+1\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}}$ Problem 6 Example 2.8 gives a matrix that is nonsingular, and is therefore associated with maps that are nonsingular. 1. Find the set of column vectors representing the members of the nullspace of any map represented by this matrix. 2." ]

[ "# Find the area bounded by the graphs of two functions Find the area bounded by the graphs of the functions $y=x^2$ and $y=x^3$.", "the region bounded by: e^x, the x-axis, the y-axis and x=1 I have no what to do if its bounded by both x and y axes Dick", "bounded by the X-axis, yaxis, and the line in the graph. Then point) 12 18 R The region R is bounded by the X-axis, yaxis, and the line in the graph. Then... -- 0.026663--", "but there's a few things to do with it. Consider the region bounded by the graphs of x-3-y In the... We were unable to transcribe this imageIt's all one problem (or at least all one region), but there's a few things to do with it. Consider the region bounded by the graphs of x-3-y In the way that's shown below (there's other regions bounded by those functions too, but we're look... ##### Find sum_(k=1)^ook x^k ? Find sum_(k=1)^ook x^k ?...", "# Using the Fundamental Theorem of Calculus find the area of the region bounded by the x-axis and the graph of f(x)=−x2−1x+12. Question Using the Fundamental Theorem of Calculus find the area of the region bounded by the x-axis and the graph of f(x)=−x2−1x+12.", "and $l$ . [3 marks] (ii) In the case where $r=3$ and $\\theta =1.2$ , find the perimeter of the shaded region. [4 marks]", "# Help with area of region • December 2nd 2008, 06:18 PM khuezy Help with area of region Find the area of the region bounded by the x-axis and the cardiod r=1+cos(theta) from 0 to pi. Can anyone give me some tips on how to approach this problem? $A=1/2\\int_{\\theta_1}^{\\theta_2} \\left[f(\\theta)\\right]^2 d\\theta$", "to correspond to the u in the formula 2.Find the area of the region bounded by", "## Calculus 8th Edition The area is the region bounded by $r=1$, $r=2$, $\\theta= \\pi/6$ and $\\theta= 5\\pi/6$ which lies in first and second quadrants.The boundaries include $r=1$, $\\theta= \\pi/6$ and $\\theta= 5\\pi/6$ and exclude $r=2$. This has been illustrated in the attached graph. The area is the region bounded by $r=1$, $r=2$, $\\theta= \\pi/6$ and $\\theta= 5\\pi/6$ which lies in first and second quadrants.The boundaries include $r=1$, $\\theta= \\pi/6$ and $\\theta= 5\\pi/6$ and exclude $r=2$. This has been illustrated in the attached graph.", "Find the area of the region enclosed by the graph of $$|x-60|+|y|=|\\frac{x}{4}|$$. (第五届AIME1987 第4题)", "### Home > CALC > Chapter 11 > Lesson 11.3.1 > Problem11-96 11-96. Find the area of the region bounded by the curves $r = 2θ$ and $r =-θ$. A graph of this region is shown below. Determine the bounds for each curve. $A=\\int_0^{2\\pi}\\frac{1}{2}(-\\theta)^2d\\theta-\\int_0^{\\pi}\\frac{1}{2}(2\\theta)^2d\\theta$", "# Math Help - Help with area of region 1. ## Help with area of region Find the area of the region bounded by the x-axis and the cardiod r=1+cos(theta) from 0 to pi. Can anyone give me some tips on how to approach this problem? $A=1/2\\int_{\\theta_1}^{\\theta_2} \\left[f(\\theta)\\right]^2 d\\theta$", "region bounded by y = x, y = 0 and = 1 (a) Write down the double integral for the area of R. (b) Convert your double integral into iterated integral in Cartesian form (Don't evaluate it. It does not matter which order you take. You can ...", "Give inequalities for $r$ and $\\theta$ which describe the region below in polar coordinates. (Click on the graph for a larger version.) The two arcs shown are circular, and the region is between the two arcs and between the $y$-axis and line graphed, which is $y = {\\sqrt{3}} x$. $\\le r \\le$ $\\le \\theta \\le$ (Write infinity to indicate a boundary at infinity.)", "Find the area of the region bounded by the parabola $(y-2)^{2}=x-1$ , the tangent to it at the point where the ordinate is 3,and the $$x$$-axis.", "# Let R be the region in the first quadrant bounded by the graphs of y=x^2, y=0, and x=2, how do you find the area of R? $R = \\frac{8}{3}$ $R = {\\int}_{0}^{2} {x}^{2} \\mathrm{dx} = {x}^{3} / 3 {|}_{0}^{2} = \\frac{8}{3}$", "# Math Help - Find the area of the region bounded by the graphs of the equations. 1. ## Find the area of the region bounded by the graphs of the equations. what interval should I have?? it only says y = 0........ it only gives me one interval to evaluate it.... whats the second one??? 2. ## Re: Find the area of the region bounded by the graphs of the equations. It also doesn't say anything about \"area\"! If there was some other part of the problem, asking you to find the area between the graphs, then you are not given any \"interval\" but you are given two graphs: $y= -x^2+ 3x$ and $y= 0$. That second is a horizontal line. Graph them. It should be easy to see what region is referred to.", "the $$y$$-axis about the $$y$$-axis. 2. Rotate the region bounded by $$y = 2{x^2}$$, $$y = 8$$ and the $$y$$-axis about the $$x$$-axis. 3. Rotate the region bounded by $$y = 2{x^2}$$, $$x = 2$$ and the $$x$$-axis about the $$x$$-axis. 4. Rotate the region bounded by $$y = 2{x^2}$$, $$x = 2$$ and the $$x$$-axis about the $$y$$-axis. 5. Rotate the region bounded by $$x = {y^3}$$, $$x = 8$$ and the $$x$$-axis about the $$x$$-axis. 6. Rotate the region bounded by $$x = {y^3}$$, $$x = 8$$ and the $$x$$-axis about the $$y$$-axis. 7. Rotate the region bounded by $$x = {y^3}$$, $$y = 2$$ and the $$y$$-axis about the $$x$$-axis. 8. Rotate the region bounded by $$x = {y^3}$$, $$y = 2$$ and the $$y$$-axis about the $$y$$-axis. 9. Rotate the region bounded by $$y = \\frac{1}{{{x^2}}}$$, $$y = 9$$, $$x = - 2$$, $$\\displaystyle x = - \\frac{1}{3}$$ about the $$y$$-axis. 10. Rotate the region bounded by $$y = \\frac{1}{{{x^2}}}$$, $$y = 9$$, $$x = - 2$$, $$\\displaystyle x = - \\frac{1}{3}$$ about the $$x$$-axis. 11. Rotate the region bounded by $$y = 4 + 3{{\\bf{e}}^{ - x}}$$, $$y = 2$$, $$\\displaystyle x = \\frac{1}{2}$$ and $$x = 3$$ about the $$x$$-axis. 12. Rotate the region bounded by $$x = 5 - {y^2}$$ and", "### Home > CALC > Chapter 11 > Lesson 11.3.1 > Problem11-96 11-96. 1. Find the area of the region bounded by the curves r = 2θ and r = −θ. A graph of this region is shown below. Homework Help ✎ Determine the bounds for each curve. $A=\\int_0^{2\\pi}\\frac{1}{2}(-\\theta)^2d\\theta-\\int_0^{\\pi}\\frac{1}{2}(2\\theta)^2d\\theta$", "$$\\displaystyle 5-e^{-2}$$ 3. $$\\displaystyle \\frac{64}{3}$$ 4. $$\\displaystyle \\int_0^1(2-2y)\\,\\mathrm{d}y = 1$$ 5. $$\\displaystyle \\int_1^2(3y-2-y^2)\\,\\mathrm{d}y = \\frac16$$ ### Topic 22: Volume Compute the volume of the solid formed by revolving the given region about the given line. 1. Region bounded by $$y=\\sqrt x$$ on the interval $$[0,4]$$ about the $$x$$-axis. 2. Region bounded by $$y=2-\\dfrac12x^2$$ from $$x=0$$ to $$x = 2$$ about the $$y$$-axis. 3. Region bounded by $$y=x^2$$, $$x=0$$, and $$y=1$$ about 1. the $$y$$-axis; 2. the $$x$$-axis; 3. the line $$y=2$$. 4. Region bounded by $$y=4-x^2$$ and $$y=0$$ about 1. the $$y$$-axis; 2. the line $$y=-3$$; 3. the line $$y=7$$; 4. the line $$x=3$$. 5. Region bounded by $$y=x$$ and $$y=x^2$$ in the first quadrant revolved about the $$y$$-axis. 6. Region bounded by the graphs of $$y=4-x^2$$ and the $$x$$-axis about the line $$x=3$$. 7. Region bounded by the graphs of $$y=x$$, $$y=2-x$$, $$y=0$$ and revolved about 1. the line $$y=2$$; 2. the line $$y=-1$$; 3. the line $$x=3$$. 1. $$8\\pi$$ 2. $$4\\pi$$ 1. $$\\dfrac{\\pi}{2}$$ 2. $$\\dfrac45\\pi$$ 3. $$\\dfrac{28}{15}\\pi$$ 1. $$8\\pi$$ 2. $$\\dfrac{1472}{15}\\pi$$ 3. $$\\dfrac{576}{5}\\pi$$ 4. $$64\\pi$$ 3. $$\\dfrac{\\pi}{6}$$ 4. $$64\\pi$$ 1. $$\\dfrac{10}{3}\\pi$$ 2. $$\\dfrac83\\pi$$ 3. $$4\\pi$$ ### Topic 23: Work 1. A force of $$5$$ pounds stretches a spring $$4$$ inches. Find the work done, in foot-pounds, in stretching the spring $$6$$ inches beyond its natural length." ]

[ "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 0$, $a_0 = 1$, $a_1 = 0.8$, $b_0 = 0$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curve corresponds to $r = r(1) t^{{|\\vartheta_4|}/{2}}$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = 1$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $R(t)$, $\\theta(t)$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = 0$, where", "same thing using (x,1) instead of (3,1) you will still get $$\\csc\\theta = \\frac r 1$$ So that equation $r =\\csc\\theta$ literally gives the same graph as y = 1. fixed two typos Last edited: Excellent! That helps a lot. But the last bit about using (x,1) - not sure how you're getting csc=r/x. Wouldn't it still be r/1? HallsofIvy Homework Helper Yes, that was probably a typo. It is precisely because $csc(\\theta)= r/1$ that [/itex]r= (1)csc(\\theta)[/itex]. \"r= 1\" is NOT \"y= 1\". It's graph is the circle with center at the origin and radius 1. LCKurtz", "• If $r=0$, then $\\cos(\\theta)=1/2$ and so $\\theta=\\pi/3$ provided $0\\le\\theta<\\pi$. This means that the point $(0,\\pi/3)$ is on the graph, and an equation of the tangent line there is $\\theta=\\pi/3$. It is called limacon - I was having trouble finding the direction since it seems like small numbers of theta give numbers around 1 and it is hard to track. – Paul the Pirate Aug 12 '13 at 21:27 Nice graph, @Babak. Pictures help sooooo much! – amWhy Aug 13 '13 at 0:36 You don't have two values. At $\\theta = 0$, $r = 1$. As $\\theta$ increases, up to the point where $\\theta = \\frac{\\pi}{3}$, $r$ decreases, until at that point, $r = 0$. Going backwards from $\\theta = 0$, $r = 1$, as $\\theta$ decreases, $r$ decreases, up to the point where $\\theta = -\\frac{\\pi}{3}$, at which point $r = 0$ again. Thus, the bounds of integration are $-\\frac{\\pi}{3}$ and $\\frac{\\pi}{3}$, as tracing the curve with these limits gives the inner loop, from $(-\\frac{\\pi}{3}, 0)$ around through $(0, 1)$, all the way to $(\\frac{\\pi}{3}, 0)$. (Or, in cartesian coordinates, $(0,0)$ through $(0,1)$, back to $(0,0)$ again.) - Clearly you can look at the graph and see that for theta = 0 the graph has values of 0 and 1 at the same time. – Paul the", "unitsize(1cm); xaxis(\"Re\",Arrows); yaxis(\"Im\",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label(\"Im(z)=164\",F(-330),N); pair z = (-656,164); dot(Label(\"z\",align=N),z); dot(Label(\"z+n\",align=N),z+(697,0)); draw(Label(\"4x\"),z--(0,0)); draw(Label(\"x\"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$. Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\\angle\\theta_1 \\cdot r_2\\angle\\theta_2 = r_1r_2\\angle(\\theta_1+\\theta_2)$ and $\\frac{z_1}{z_2} = \\frac{r_1\\angle\\theta_1}{r_2\\angle\\theta_2} = \\frac{r_1}{r_2}\\angle(\\theta_1-\\theta_2)$, where $r$ is the magnitude and $\\theta$ is the phase, and $z_n=r_n\\angle\\theta_n$. Since $4i$ has magnitude $4$ and phase $90^\\circ$ (since the positive imaginary axis points in a direction $90^\\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$. This means that", "way: $$r_1e^{1\\theta_1}\\cdot r_2e^{1\\theta_2}=r_1r_2e^{i(\\theta_1+\\theta_2)},$$, $$\\frac{r_1e^{1\\theta_1}}{r_2e^{1\\theta_2}}=\\frac{r_1}{r_2}e^{i(\\theta_1-\\theta_2)}$$, z_{1}=2(cos(\\frac{pi}{3})+i sin (\\frac{pi}{3}) )=2e^{i\\frac{pi}{3}}\\\\z_{2}=1(cos(\\frac{pi}{6})-i sin (\\frac{pi}{6}) )=1(cos(\\frac{pi}{6}) In your case, $a,b,c$ and $d$ are all given so just plug in the numbers. Thanks to all of you who support me on Patreon. When dividing two complex numbers you are basically rationalizing the denominator of a rational expression. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Answer your tough homework and study questions direct sum matrices into the form =! Tileable, vertical redstone in minecraft redstone in minecraft has angle A_ANGLE_REP and radius.... To many people and Simplify i am stuck at square one, any help would great... 'S salary receipt Open in its respective personal webmail in someone else 's computer indicated operations an write final. Help would be great times cosine alpha plus i sine beta a page! Missing something LOOse '' pronounced differently are all given so just plug in the complex number all you have do! The domains *.kastatic.org and *.kasandbox.org are unblocked, just like vectors, can also be written polar. Entire Q & a library imaginary part: a + 0i 1 Sina! Of a complex number corresponds to a point", "solid curves correspond to level lines of $H_0(x,y)$ Figure 3. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 4. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 0$, $a_0 = 1$, $a_1 = 0.8$, $b_0 = 0$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curve corresponds to $r = r(1) t^{{|\\vartheta_4|}/{2}}$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 5. The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = 1$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ Figure 6. The evolution of $(x(t),y(t))$, $R(t)$, $\\theta(t)$ for solutions of (50) with $h = 1/6$, $a_0", "= r_1^2$, and hence $x = -\\frac{1}{2x_1} (r_1^2-r_0^2-x_1^2)$ Then, the angle at which this occurs is $\\theta_1 = \\cos^{-1}(\\frac{x}{r_0})$. The intersections actually occur at two points, both with the same $x$ coordinate, but one is above the $x$ axis and one below. So, we get another one which occurs at $\\theta_2=2\\pi-\\theta$. So far, it looks like I'm ready to set up my integration regions. I think my best bet is to divide my region into pieces, depending on those $\\theta$ values I computed. From $\\theta=0$ to $\\theta=\\theta_1$, we are integrating just inside the circle $C_0$, so I think I have over this region $$\\int_{0}^{\\theta_1} \\int_0^{r_0} \\phi(r) r dr d\\theta$$ Now, from $\\theta_1$ to $\\theta_2$, I want to integrate only including the area of contained in $C_1 \\cap C_0$. So, I think I integrate on this region from $\\theta = \\theta_1$ to $\\theta = \\theta_2$ and $r=0$ up to $r_1= 2x_1 \\cos(\\theta) \\pm \\sqrt{r_1^2-x_1^2 \\sin^2(\\theta)}$ So, I have over this region $$\\int_{\\theta_1}^{\\theta_2} \\int_0^{r_1} \\phi(r) r dr d\\theta$$ For my last region, we go from $\\theta_2$ to $2\\pi$. Here, our radial limits go from $0$ to $r_0$ (as we aren't dealing with the bigger circle $C_1$), so we have for this region $$\\int_{\\theta_2}^{2\\pi} \\int_0^{r_0} \\phi(r) r dr d\\theta$$ Then, the final result will be the sum of the three integrals.", "= a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ Figure 7. The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ Figure 8. The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ Figure 9. The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "graph each pair of polar equations on a domain of $\\left[0,4\\pi \\right]$ and then explain the differences shown in the graphs. 54. $r=\\theta ,r=-\\theta$ 55. $r=\\theta ,r=\\theta +\\sin \\theta$ 56. $r=\\sin \\theta +\\theta ,r=\\sin \\theta -\\theta$ 57. $r=2\\sin \\left(\\frac{\\theta }{2}\\right),r=\\theta \\sin \\left(\\frac{\\theta }{2}\\right)$ 58. $r=\\sin \\left(\\cos \\left(3\\theta \\right)\\right)r=\\sin \\left(3\\theta \\right)$ 59. On a graphing utility, graph $r=\\sin \\left(\\frac{16}{5}\\theta \\right)$ on $\\left[0,4\\pi \\right],\\left[0,8\\pi \\right],\\left[0,12\\pi \\right]$, and $\\left[0,16\\pi \\right]$. Describe the effect of increasing the width of the domain. 60. On a graphing utility, graph and sketch $r=\\sin \\theta +{\\left(\\sin \\left(\\frac{5}{2}\\theta \\right)\\right)}^{3}$ on $\\left[0,4\\pi \\right]$. 61. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3\\sin \\left(3\\theta \\right)\\end{array}\\hfill \\\\ {r}_{2}=2\\sin \\left(3\\theta \\right)\\hfill \\\\ {r}_{3}=\\sin \\left(3\\theta \\right)\\hfill \\end{array}$ 62. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3+3\\cos \\theta \\end{array}\\hfill \\\\ {r}_{2}=2+2\\cos \\theta \\hfill \\\\ {r}_{3}=1+\\cos \\theta \\hfill \\end{array}$ 63. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3\\theta \\end{array}\\hfill \\\\ {r}_{2}=2\\theta \\hfill \\\\ {r}_{3}=\\theta \\hfill \\end{array}$ For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. 64. ${r}_{1}=3+2\\sin \\theta ,{r}_{2}=2$ 65. ${r}_{1}=6 - 4\\cos \\theta ,{r}_{2}=4$ 66.", "triangle that $$\\csc\\theta = \\frac r 1$$ Then notice if you do the same thing using (x,1) instead of (3,1) you will still get $$\\csc\\theta = \\frac r 1$$ So that equation $r =\\csc\\theta$ literally gives the same graph as y = 1. fixed two typos Last edited: Jan 26, 2010 3. Jan 26, 2010 ### page13 Excellent! That helps a lot. But the last bit about using (x,1) - not sure how you're getting csc=r/x. Wouldn't it still be r/1? 4. Jan 26, 2010 ### HallsofIvy Staff Emeritus Yes, that was probably a typo. It is precisely because $csc(\\theta)= r/1$ that [/itex]r= (1)csc(\\theta)[/itex]. \"r= 1\" is NOT \"y= 1\". It's graph is the circle with center at the origin and radius 1. 5. Jan 26, 2010 ### LCKurtz Yes. That was a typo as Halls surmised. There was another one too. I'll fix both.", "at $\\left(1,0\\right)$(1,0). As we move through different values of $\\theta$θ the value of $\\cos\\theta$cosθ and $\\sin\\theta$sinθ move accordingly between $-1$1 and $1$1. The animation below shows this process for $y=\\sin\\theta$y=sinθ as $\\theta$θ travels around the unit circle. Created with Geogebra ### Graphs of sin, cos and tan functions If we plot the values of $\\sin\\theta$sinθ and $\\cos\\theta$cosθaccording to different values of $\\theta$θ on the unit circle, we get the following graphs: $y=\\sin\\theta$y=sinθ $y=\\cos\\theta$y=cosθ The simplest way to calculate $\\tan\\theta$tanθ is to use the values in the above graphs to evaluate $\\frac{\\sin\\theta}{\\cos\\theta}$sinθcosθ, which gives us the following graph: $y=\\tan\\theta$y=tanθ Notice that all of these graphs are constructed with degrees on the horizontal axis. The function values behave in the same way as in the unit circle - for example, in the graph above of $y=\\cos\\theta$y=cosθ, we can see that it has negative $y$y-values for all of the angles in the domain of $90^\\circ<\\theta<180^\\circ$90°<θ<180°. These are the values associated with the second quadrant where $\\cos\\theta$cosθ will be negative. ### Properties of trigonometric graphs The graphs of $y=\\cos\\theta$y=cosθ and $y=\\sin\\theta$y=sinθ have certain common properties. Each graph demonstrates repetition. We call the graphs of $y=\\cos\\theta$y=cosθ and $y=\\sin\\theta$y=sinθ periodic, or cyclic. We define the period as the length of one cycle. For both graphs, the period is $360^\\circ$360°. An example of a cycle", "C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so$$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular,\\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*}We want to solve for $d$. By the Pythagorean Theorem (twice):\\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*}Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. Solution 2 (Official MAA, Unedited) Denote by $O_1$$O_2$, and $O$ the centers of $\\omega_1$$\\omega_2$, and $\\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\\omega_1$ and $\\omega_2$ respectively, $r$ be the radius of $\\omega$, and $\\ell$ the distance from $O$ to the line $AB$. We claim that$$\\dfrac{\\ell}{r} = \\dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\\widehat{PQ} = 120^\\circ$ condition implies $\\tfrac{\\ell}r = \\cos 60^\\circ = \\tfrac{1}{2}$, and then", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "in blue fixed values of $r$ (circles) and in red fixed values of $\\theta$ (rays). gr() F(r,theta) = r*[cos(theta), sin(theta)] F(v) = F(v...) rs = range(0, 2, length=5) thetas = range(0, pi/2, length=9) plot(legend=false, aspect_ratio=:equal) plot!(unzip(F.(rs, thetas'))..., color=:red) plot!(unzip(F.(rs', thetas))..., color=:blue) pt = [1, pi/4] J = ForwardDiff.jacobian(F, pt) arrow!(F(pt...), J[:,1], linewidth=5, color=:red) arrow!(F(pt...), J[:,2], linewidth=5, color=:blue) To the plot, we added the partial derivatives with respect to $r$ (in red) and with respect to $\\theta$ (in blue). These are found with the soon-to-be discussed Jacobian. From the graph, you can see that these vectors are tangent vectors to the drawn curves. ## Parametrically defined surfaces For a one-dimensional curve we have several descriptions. For example, as the graph of a function $y=f(x)$; as a parametrically defined curve $\\vec{r}(t) = \\langle x(t), y(t)\\rangle$; or as a level curve of a scalar function $f(x,y) = c$. For two-dimensional surfaces in three dimensions, we have discussed describing these in terms of a function $z = f(x,y)$ and as level curves of scalar functions: $c = f(x,y,z)$. They can also be described parametrically. We pick a familiar case, to make this concrete: the unit sphere in $R^3$. We have • It is described by two functions through $f(x,y) = \\pm \\sqrt{1 - (x^2 + y^2)}$. • It is described by $f(x,y,z)", "of $C_{n-1} = \\mathrm{co}(x_1,\\ldots,x_{n-1})$. We can understand this relation like this. $E_n$ is $n$ times the probability that $x_n$ is a vertex of $C_n$. Since the probability that $x_n$ lies on the edge of $C_{n-1}$ is zero, the probability that $x_n$ is a vertex of $C_n$ is the same as the probability that $x_n \\not\\in C_{n-1}$ which in turn equals to $1 - \\Delta_{n-1}$. When $K$ is a triangle, according to this article on mathworld, $$\\Delta_n = 1 - \\frac{2H_{n}}{n+1}$$ This means the expected number of vertices you seek is $E_n = 2 H_{n-1}$. Update For completeness and documentation purposes, let me outline how to compute $\\Delta_n$ and hence $E_n$ ourselves. Define $\\rho$ such that $\\frac{1}{\\rho^2} = \\frac{\\sqrt{3}}{4}$, the area of an equilateral triangle of side length $1$. We will take $K$ to be the equilateral triangle with vertices at $(0,0)$ and $(\\rho\\cos\\frac{\\pi}{6},\\pm \\rho\\sin\\frac{\\pi}{6})$. For any $\\theta \\in [0,2\\pi]$ and $p \\in \\mathbb{R}$, let • $\\ell(\\theta,p)$ be the line $\\{ (x,y) : \\cos\\theta x + \\sin\\theta y = p \\}$. • $L(\\theta,p)$ be the length of the line segment $\\ell(\\theta,p) \\cap K$. • $H(\\theta,p)$ be the half-space $\\{ (x,y) : \\cos\\theta x + \\sin\\theta y \\le p \\}$. • $A(\\theta,p)$ be the area of $H(\\theta,p) \\cap K$. For any $\\lambda \\in [0,1]$, choose a $p$ so that $A(\\theta,p)", "# How do you graph r^2 = sin2(t)? Jul 23, 2018 See graphs galore. #### Explanation: The power-scaling of r is done, when $r$ is changed to ${r}^{n}$, n > 1, in $r = f \\left(\\theta\\right)$. It is r-more, when $r \\in \\left(0 , 1\\right)$. Otherwise, jt is r-less Here, this is an example. Use $r = \\sqrt{{x}^{2} + {y}^{2}} \\ge 0 , r \\left(\\cos \\theta , \\sin \\theta\\right)$ and $\\sin 2 \\theta = 2 \\sin \\theta \\cos \\theta$, to get the Cartesian form of ${r}^{2} = \\sin 2 \\theta$ as ( x^2 + y^2 ) ^2 ) - 2xy = 0. The Socratic graph is immediate. graph{( x^2 + y^2 ) ^2 - 2xy =0[-2 2 -1 1]} Graph of $r = \\sin 2 \\theta$, for contrast in r-scaling: graph{( x^2 + y^2 ) ^1.5 - 2xy =0[-2 2 -1 1]} Easy to see which is which, jn the combined graph, along with the third graph of ${r}^{5} = \\sin 2 \\theta$: graph{(( x^2 + y^2 ) ^2 - 2xy)( ( x^2 + y^2 ) ^1.5 - 2xy) ( ( x^2 + y^2 ) ^3.5 - 2xy) =0[-2 2 -1 1]}", "to $(0,0,1)$ since changing $\\theta$ does not change $z$. Therefore, we simply need to take the cross product of $(x,y,z)$ and $(0,0,1)$ and normalize to get $\\hat{e}_\\theta=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}$. The sign is chosen so that the vector points counter-clockwise. $\\hat{e}_\\phi$ is perpendicular to the other two, so we take the cross product and normalize: $\\hat{e}_\\phi=\\frac{(-xz,-yz,r^2-z^2)}{r\\sqrt{r^2-z^2}}$. Again, the sign is chosen to have positive $z$ component. Thus, we get \\begin{align} \\hat{e}_R&=\\frac{(x,y,z)}{r}\\\\ \\hat{e}_\\theta&=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}\\\\ \\hat{e}_\\phi&=\\frac{(-xz,-yz,r^2-z^2)}{r\\sqrt{r^2-z^2}} \\end{align} References 1. Source of the polar-coordinate image here from Wikipedia. - ...could elaborate on this $\\hat{e}_\\theta=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}$? Which sign? When you do the cross-product, how you think it? I do the matrix thing but I find it slow... Do you think it as a Sarrus-rule here? – hhh Mar 24 '12 at 22:01 @hhh: $(x,y,z)\\times(0,0,1)=(y,-x,0)$ and $(0,0,1)\\times(x,y,z)=(-y,x,0)$ normalizing these gives opposite vectors, so we choose the one that goes counter-clockwise. Both are perpendicular to the vectors specified. – robjohn Mar 24 '12 at 22:46 The image from Wikipedia has $\\theta$ and $\\varphi$ switched from the way I usually see them, and from the way I have them in my answer. I will either edit my answer, or generate a new image (probably the latter so that we are copyright safe). – robjohn Mar 26 '12 at 0:39 ...if you generate the picture, please, do not hesitate to include", "r}$ (and never mind the tables in the back of the book). The first thing to notice about this procedure is how doggone easy it is. Indeed, as far as I know, it’s much easier than it would be on a Windows box. It was this easy in DOS (the predecessor to Windows) since BASIC was standard on DOS boxes. Several generations of computer coders have a great deal to answer for. ### Today’s Lemniscate Problem … is essentially an example from the text, first of all. So hopefully, that’ll inspire some reading (nobody had the right slope on this one). Here are some other remarks. In Calc III (here’s an old blog by me), one considers polar co-ordinates (so skip the rest of this paragraph if you prefer not to think about ’em yet). Our curve $(x^2+y^2)^2 = 50xy$ “translates” into $r^4 = 50(r\\cos(\\theta))(r\\sin(\\theta))$, so we can put $r =5\\sqrt{2\\cos\\theta\\sin\\theta}$ into a grapher. The “Draw” feature can then be used to produce the desired tangent line ($(x,y) = (2,4)$ gives us $\\theta =\\arctan({4\\over2})$; input this at the prompt); this feature also produces the display ${{dy}\\over{dx}}=.181818$ (or words to that effect); being translated (via the “Frac” feature, say), one has y’=2/11. The Calc I version: differentiating both sides of the given equation gives $D_x[(x^2+y^2)^2] = D_x[50xy]$ $2(x^2+y^2)\\cdot D_x[x^2+y^2]" ]

[ "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "= r_1^2$, and hence $x = -\\frac{1}{2x_1} (r_1^2-r_0^2-x_1^2)$ Then, the angle at which this occurs is $\\theta_1 = \\cos^{-1}(\\frac{x}{r_0})$. The intersections actually occur at two points, both with the same $x$ coordinate, but one is above the $x$ axis and one below. So, we get another one which occurs at $\\theta_2=2\\pi-\\theta$. So far, it looks like I'm ready to set up my integration regions. I think my best bet is to divide my region into pieces, depending on those $\\theta$ values I computed. From $\\theta=0$ to $\\theta=\\theta_1$, we are integrating just inside the circle $C_0$, so I think I have over this region $$\\int_{0}^{\\theta_1} \\int_0^{r_0} \\phi(r) r dr d\\theta$$ Now, from $\\theta_1$ to $\\theta_2$, I want to integrate only including the area of contained in $C_1 \\cap C_0$. So, I think I integrate on this region from $\\theta = \\theta_1$ to $\\theta = \\theta_2$ and $r=0$ up to $r_1= 2x_1 \\cos(\\theta) \\pm \\sqrt{r_1^2-x_1^2 \\sin^2(\\theta)}$ So, I have over this region $$\\int_{\\theta_1}^{\\theta_2} \\int_0^{r_1} \\phi(r) r dr d\\theta$$ For my last region, we go from $\\theta_2$ to $2\\pi$. Here, our radial limits go from $0$ to $r_0$ (as we aren't dealing with the bigger circle $C_1$), so we have for this region $$\\int_{\\theta_2}^{2\\pi} \\int_0^{r_0} \\phi(r) r dr d\\theta$$ Then, the final result will be the sum of the three integrals.", "$rr$ is nonnegative), we have $2−r2=r.2−r2=r.$ Solving, we have $r2+r−2=(r+2)(r−1)=0.r2+r−2=(r+2)(r−1)=0.$ Since $r≥0,r≥0,$ we have $r=1.r=1.$ Therefore $z=1.z=1.$ So the intersection of these two surfaces is a circle of radius $11$ in the plane $z=1.z=1.$ The cone is the lower bound for $zz$ and the paraboloid is the upper bound. The projection of the region onto the $xyxy$-plane is the circle of radius $11$ centered at the origin. Thus, we can describe the region as $E={(r,θ,z)|0≤θ≤2π,0≤r≤1,r≤z≤2−r2}.E={(r,θ,z)|0≤θ≤2π,0≤r≤1,r≤z≤2−r2}.$ Hence the integral for the volume is $V=∫θ=0θ=2π∫r=0r=1∫z=rz=2−r2rdzdrdθ.V=∫θ=0θ=2π∫r=0r=1∫z=rz=2−r2rdzdrdθ.$ 2. We can also write the cone surface as $r=zr=z$ and the paraboloid as $r2=2−z.r2=2−z.$ The lower bound for $rr$ is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane $z=1z=1$ divides the region into two regions. Then the region can be described as $E={(r,θ,z)|0≤θ≤2π,0≤z≤1,0≤r≤z}∪{(r,θ,z)|0≤θ≤2π,1≤z≤2,0≤r≤2−z}.E={(r,θ,z)|0≤θ≤2π,0≤z≤1,0≤r≤z}∪{(r,θ,z)|0≤θ≤2π,1≤z≤2,0≤r≤2−z}.$ Now the integral for the volume becomes $V=∫θ=0θ=2π∫z=0z=1∫r=0r=zrdrdzdθ+∫θ=0θ=2π∫z=1z=2∫r=0r=2−zrdrdzdθ.V=∫θ=0θ=2π∫z=0z=1∫r=0r=zrdrdzdθ+∫θ=0θ=2π∫z=1z=2∫r=0r=2−zrdrdzdθ.$ Checkpoint 5.29 Redo the previous example with the order of integration $dθdzdr.dθdzdr.$ ### Example 5.46 #### Finding a Volume with Triple Integrals in Two Ways Let E be the region bounded below by the $rθrθ$-plane, above by the sphere $x2+y2+z2=4,x2+y2+z2=4,$ and on the sides by the cylinder $x2+y2=1x2+y2=1$ (Figure 5.54). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and", "# How do you graph r^2 = sin2(t)? Jul 23, 2018 See graphs galore. #### Explanation: The power-scaling of r is done, when $r$ is changed to ${r}^{n}$, n > 1, in $r = f \\left(\\theta\\right)$. It is r-more, when $r \\in \\left(0 , 1\\right)$. Otherwise, jt is r-less Here, this is an example. Use $r = \\sqrt{{x}^{2} + {y}^{2}} \\ge 0 , r \\left(\\cos \\theta , \\sin \\theta\\right)$ and $\\sin 2 \\theta = 2 \\sin \\theta \\cos \\theta$, to get the Cartesian form of ${r}^{2} = \\sin 2 \\theta$ as ( x^2 + y^2 ) ^2 ) - 2xy = 0. The Socratic graph is immediate. graph{( x^2 + y^2 ) ^2 - 2xy =0[-2 2 -1 1]} Graph of $r = \\sin 2 \\theta$, for contrast in r-scaling: graph{( x^2 + y^2 ) ^1.5 - 2xy =0[-2 2 -1 1]} Easy to see which is which, jn the combined graph, along with the third graph of ${r}^{5} = \\sin 2 \\theta$: graph{(( x^2 + y^2 ) ^2 - 2xy)( ( x^2 + y^2 ) ^1.5 - 2xy) ( ( x^2 + y^2 ) ^3.5 - 2xy) =0[-2 2 -1 1]}", "+0 0 82 1 The diagram below shows four circles. Let $A_1,$ $A_2,$ $A_3,$ $A_4$ denote the areas of the red region, yellow region, green region, blue region, respectively. These areas satisfy $A_1 = \\frac{A_2}{2} = \\frac{A_3}{3} = \\frac{A_4}{4}.$Let $r_1$ denote the smallest radius, and let $r_4$ denote the largest radius. Find $\\frac{r_4}{r_1}.$ [asy] unitsize(1 cm); pair[] O; real[] r; O[1] = (0,0); O[2] = (0.1,0.2); O[3] = (-0.2,-0.1); O[4] = (0.1,-0.3); r[1] = 1; r[2] = 1.5; r[3] = 2; r[4] = 2.5; fill(Circle(O[4],r[4]),lightblue); draw(Circle(O[4],r[4])); label(\"$A_4$\", (1.8,-1.5)); fill(Circle(O[3],r[3]),lightgreen); draw(Circle(O[3],r[3]));label(\"$A_3$\", (-1.3,-1.3)); fill(Circle(O[2],r[2]),yellow); draw(Circle(O[2],r[2]));label(\"$A_2$\", (1,1)); fill(Circle(O[1],r[1]),lightred); draw(Circle(O[1],r[1]));label(\"$A_1$\", O[1]); [/asy] Jan 25, 2020", "same thing using (x,1) instead of (3,1) you will still get $$\\csc\\theta = \\frac r 1$$ So that equation $r =\\csc\\theta$ literally gives the same graph as y = 1. fixed two typos Last edited: Excellent! That helps a lot. But the last bit about using (x,1) - not sure how you're getting csc=r/x. Wouldn't it still be r/1? HallsofIvy Homework Helper Yes, that was probably a typo. It is precisely because $csc(\\theta)= r/1$ that [/itex]r= (1)csc(\\theta)[/itex]. \"r= 1\" is NOT \"y= 1\". It's graph is the circle with center at the origin and radius 1. LCKurtz", "r \\ cos (\\theta), y = r \\ sin (\\theta), r^2=x^2 + y^2, \\theta = tan^{-1} \\frac y x}$ Using Jacobians, ${dxdy = |J| d\\theta dr = \\begin {vmatrix} \\frac {\\partial (x,y)}{\\partial (r, \\theta)} \\end {vmatrix} d \\theta dr = r d \\theta dr}$ Generally, the inner integral is obtained w.r.t. $r$. III) If $f(x,y) =1$, we get the area of the region. • ### Illustrated Examples Question : Evaluate the following over the region ${y \\ge x^2, x \\ge 1}$. ${\\iint \\frac {1}{x^4+y^2} dxdy}$ Solution : The region of integration is plotted below : Clearly, to the left of line ${x=1}$, the region is absent, hence the lower limit on ${x}$ will be ${1}$. The upper limit will be ${\\infty}$. No restriction on the highest value of ${x}$. Let’s now consider a strip parallel to ${Y}$ axis. It will first intersect the region at ${y=x^2}$. So, the lower limit on ${y}$ is ${x^2}$. The upper limit will be ${\\infty}$ because no curve bounds the region in that direction. Hence, given integral will be ${\\int \\limits_{x=1}^{\\infty} \\int \\limits_{y=x^2}^{\\infty} \\frac {1}{x^4+y^2} dydx = I \\ say}$ Since inner limits are of ${y}$, let’s first integrate w.r.t. ${y}$. ${I = \\int \\limits_{x=1}^{\\infty} \\left [\\frac {1}{x^2} tan^{-1} \\left (\\frac {y}{x^2} \\right) \\right]_{x^2}^{\\infty}dx}$ ${= \\int \\limits_{x=1}^{\\infty} \\frac {1}{x^2} \\left[ \\frac {\\pi}{2} -", "243 views The volume of the region $S=\\{(x,y,z) : \\mid x \\mid +\\mid y \\mid + \\mid z \\mid \\leq 1\\}$ is 1. $\\frac{1}{6}\\\\$ 2. $\\frac{1}{3}\\\\$ 3. $\\frac{2}{3}\\\\$ 4. $\\frac{4}{3}$ edited | 243 views 0 Option D We know that $|x| = x$ when $x>= 0$ and $|x| = -x$ when $x<0$. So, region covered by the graph $|x| + |y| <= 1$ is bounded by the lines $(x+y =1)$ , $(-x +y =1)$ , $(x - y =1)$ , $(-x -y =1)$ [As shown in Fig $A$ below] So, the region covered will be the square $[ (0,1) , (1,0) , (0,-1) , (-1,0) ]$ [ As shown in Fig $A$] The side length of the above square is $√2.$ Now, in the $3D$ version of the above scenario the space will be covered by the lines $(x+z = 1)$ , $( -x +z = 1)$ , $( -y +z =1)$ , $( y +z = 1)$, $( -y-z =1)$, $( -y +z = 1)$ , $(x -z =1)$ , $(-x -z = 1)$. [ As shown in the figure $B$] This will create two pyramids $[ (0,0,1) , (0,1,0) , (1,0,0) , (0,-1,0) , (-1,0,0) ]$ and $[ (0,0,-1) , (0,1,0) , (1,0,0) , (0,-1,0) , (-1,0,0) ]$ [ As shown in the above figure $B$", "of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 0$, $a_0 = 1$, $a_1 = 0.8$, $b_0 = 0$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curve corresponds to $r = r(1) t^{{|\\vartheta_4|}/{2}}$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = 1$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $R(t)$, $\\theta(t)$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = 0$, where", "4y^2 + z^2}\\right) dV = \\frac{\\pi}{2} \\int_0^1 f (\\rho) \\rho^2 d\\rho,$ where $$f$$ is a continuous function on $$[0,1]$$ and $$R$$ is the region bounded by the ellipsoid $$16x^2 + 4y^2 + z^2 = 1$$. 2. $$T (r,\\theta,z) = (r \\space \\cos \\theta, \\space r \\space \\sin \\theta, \\space z); \\space S = [0,3] \\times [0,\\frac{\\pi}{2}] \\times [0,1]$$ in the $$r\\theta z$$-space ## Exercise $$\\PageIndex{12}$$ 1. [T] Find the area of the region bounded by the curves $$xy = 1, \\space xy = 3, \\space y = 2x$$, and $$y = 3x$$ by using the transformation $$u = xy$$ and $$v = \\frac{y}{x}$$. Use a computer algebra system (CAS) to graph the boundary curves of the region $$R$$. 2. [T] Find the area of the region bounded by the curves $$x^2y = 2, \\space x^2y = 3, \\space y = x$$, and $$y = 2x$$ by using the transformation $$u = x^2y$$ and $$v = \\frac{y}{x}$$. Use a CAS to graph the boundary curves of the region $$R$$. 2. The area of $$R$$ is $$10 - 4\\sqrt{6}$$; the boundary curves of $$R$$ are graphed in the following figure. ## Exercise $$\\PageIndex{13}$$ 1. Evaluate the triple integral $\\int_0^1 \\int_1^2 \\int_z^{z+1} (y + 1) \\space dx \\space dy \\space dz$ by using the transformation $$u = x - z, \\space", "What is the area? Dec 9, 2017 There is no real solution to this question. Explanation: The integral is not valid, if you look at what happens when theta approaches 0, you will see the function goes large, thus the area between [0,1] is kind of infinity. ---by @miles-a I'm sorry. I don't know the steps of solving this. But you can try the formula mentioned by @alvin-l-2 as the following. Dec 9, 2017 The calculation is hard. I used http://www.wolframalpha.com/ and obtained the result(about $0.50754$). Explanation: This is the graph of $r = \\frac{1}{1 - \\cos 3 \\theta}$ (by Wolframalpha). Then, how to interpret the graph? [1] The graph crosses $x = 0$ at $\\theta = \\frac{\\pi}{2}$. There, $r = \\frac{1}{1 - \\cos \\left(\\frac{3 \\pi}{2}\\right)} = 1$. [2] When does the graph cross $x = 1$? To solve this, apply the formula $x = r \\cos \\theta$. $1 = r \\cos \\theta$ $r = \\frac{1}{\\cos} \\theta$ $\\frac{1}{1 - \\cos 3 \\theta} = \\frac{1}{\\cos} \\theta$ $1 - \\cos 3 \\theta = \\cos \\theta$ $\\cos 3 \\theta + \\cos \\theta = 1$ Use the formula $\\cos 3 \\theta = 4 {\\cos}^{3} \\theta - 3 \\cos \\theta$. $4 {\\cos}^{3} \\theta - 2 \\cos \\theta - 1 = 0$ The real solution for this is: costheta≒0.88465 theta≒0.48505 [rad] ≒27.79 [deg] In", "red dots, as well as all points in the left quadrants, are outside. x1 <- seq(0, 1, length=1000) x2 <- sqrt(1 - x1^2) x1 <- c(x1, x1) x2 <- c(x2, -x2) ci95 <- (3*x1 - 1.5*x2)^2 >= 9.3225 plot(0, type=\"n\", xlim=c(-2,3.5), ylim=c(-2,3.5), xlab=\"x1\", ylab=\"x2\") points(D, pch=19) points(0, 0) arrows(0, 0, 3, -1.5) points(x1, x2, col=2 + ci95, pch=19, cex=0.5) Tracing out a full confidence hypercone, i.e., for non-unit distances from the origin, is left to an exercise in §3.3. The width of the hypercone -- size of the green area in Figure 3.4 -- increases as distance $$(X_1 + X_2)^2$$ increases from the center of the design region. The rate of that increase can be depicted by an angle $$\\theta$$ indicating directions included in (or excluded from) the 95% confidence region, giving a sense of the tightness of the region. It turns out that $\\theta = \\mathrm{arcsin} \\left\\{ \\frac{(m-1) s_b^2 F^\\alpha_{m-1,\\nu_b}}{\\sum_{j=1}^m b_j^2} \\right\\}^{1/2}.$ Returning back to our simple example (3.1), we have the following. theta <- asin(sqrt(0.25*7.71*(1/11.25))) theta ## [1] 0.4268 So the interpretation is that angles lower than around 0.43 radians, or $$24^\\circ$$, are within the confidence region for any particular fixed distance from the center of the design. Larger angles are outside. Since this simple example is two-dimensional, this means that the green arc in Figure", "triangle symmetrically so you think of your triangle as two back-to-back right triangles. Then label the height $x$ and width $\\sqrt{1-x^2}$. Choose one angle to be $\\theta$. Then you should see that $$A = \\frac{bh}{2}$$ or equivalently we can define $A$ as a function in terms of $\\theta$, since we already have $b,h$ defined in terms of $\\theta$. Let $$\\\\ A(\\theta) =\\ \\frac{\\cos(\\theta)\\sin(\\theta)}{2} = \\frac{\\sin(2\\theta)}{4}$$ Since $A(\\theta)$ is only the area of our right triangle, we need to double it to account for the area of the entire isosceles triangle. Let $$A_I(\\theta) = 2A(\\theta) = \\frac{\\sin(2\\theta)}{2}$$ where $A_I(\\theta)$ is the area of the isosceles. Similarly, we can define $P$ as a function of $\\theta$. The perimeter of the isosceles is $P =$ leg $1$ + leg $2$ + leg $3= 2+b_I$, where we have $b_I = 2b = 2\\cos(\\theta)$. Hence, $$P(\\theta) = 2+2\\cos(\\theta)$$ Now we can define $r$ as a function of $\\theta$ via the relation $$r(\\theta) = \\frac{A_I(\\theta)}{P(\\theta)} = \\frac{\\sin(2\\theta)}{4(1+\\cos(\\theta))}$$ Now you can find when $r'(\\theta) = 0$ and optimize $r(\\theta)$ Let the unknown triangle's base be $2l$. Draw a diagram and use Pythagoras' Theorem to obtain the height of the triangle as $\\sqrt{1-l^2}$. Now use the triangle's area formula to obtain the area $$\\sqrt{1-l^2}\\times\\frac {2l}2=l\\sqrt{1-l^2}$$ The perimeter of the triangle is $2+2l$. Thus the radius of", "from $\\theta=\\mathrm{acos}(-0.25)$ to $\\theta=\\pi$. The unsigned area of the inner region is therefore twice the area of the highlighted region, or $2A_1(\\mathrm{acos}(-0.25),\\pi)$. Figure 11.3.9. The signed area of the region enclosed by $r=f(\\theta)$ as $\\theta$ varies from 0 to $2\\pi$ should therefore be the difference of the unsigned regions bounded by the outer and inner loops, or $2A_1(0,\\mathrm{acos}(-0.25))-2A_1(\\mathrm{acos}(-0.25),\\pi)$. Figure 11.3.10 (copied from a GC window) confirms that the two methods produce the same result. The difference in unsigned areas is the same as the net signed area as $\\theta$ varies from 0 to $2\\pi$. Figure 11.3.10. ## Exercise Set 11.3 1. Try, by looking at the function definition and the function graph, to determine which regions have positive area and which regions have negative area. 2. Click the \"n\" slider. Now determine which regions have positive area and which regions have negative area. 3. Print 2 copies of the graph. On the first printout, highlight regions that have positive area in green (or with \\\\\\\\\\). In the second printout, highlight regions that have negative area in gray (or with //////). 2. Define the functions $f$ and $A_1$ as in Figure 11.3.10 (above). Enter $A_1(0,2\\pi)$. You will get 211.174 (to 3 decimal places). This is the unsigned area of a swept region--perhaps with parts swept more than once. 1.", "lines) and the interposed region by them. The function $\\mathcal{D} (t)$ indicates the area of the gray regions Description of $\\Gamma_D = \\bigcup_{j = 0}^k L_j (t)$. Note that, for the simplicity, the variable $t$ of $L_j$ and $y_j$ is omitted in the above figure Construction of $\\theta_D (t, x)$; we construct a branch of $\\arg (x)$ whose discontinuities are only on $\\Gamma (t)$(the dashed line in (1)). For this purpose we first construct $\\vartheta (x) = \\arg (x)$ whose discontinuities are only on $\\mathcal{L}_k (t)$ (the solid line in (2)). Then, we make go down the height of $\\vartheta (x)$ on $R_{j} (t)$ (the gray region in (3) or (4)) with the jump-height $2 \\pi$ from $j = k-1$ to $j = 0$ inductively to remove illegal discontinuities. The solid line in figure (3) or (4) denotes the discontinuity of $\\Theta_{k,k-1}$ or $\\Theta_{k,k-2}$, respectively Profiles of the square spiral at $t = 1$. The level set method is calculated using $\\rho = 0.02$ and $\\Delta x = 0.0050$ Graphs of functions $\\mathcal{D} (t)$ for the square spiral with a fixed center radius $\\rho = 0.02$(left), and with a reduced center radius $\\rho = 2 \\Delta x$(right) Profiles of the diagonal spiral at $t = 1$. The level set method is calculated using $\\rho = 0.02$ and $\\Delta", "describe the region $$D$$ as $$\\{(r, \\theta)\\,|\\,0 \\leq \\theta \\leq \\pi, \\, 0 \\leq r \\leq 1 + \\cos \\, \\theta\\}$$ as shown in Figure $$\\PageIndex{6}$$. Hence, we have \\begin{align*} \\iint_D r^2 \\sin \\, \\theta \\, r \\, dr \\, d\\theta &= \\int_{\\theta=0}^{\\theta=\\pi} \\int_{r=0}^{r=1+\\cos \\theta} (r^2 \\sin \\, \\theta) \\,r \\, dr \\, d\\theta \\\\ &= \\frac{1}{4}\\left.\\int_{\\theta=0}^{\\theta=\\pi}[r^4] \\right|_{r=0}^{r=1+\\cos \\, \\theta} \\sin \\, \\theta \\, d\\theta \\\\ &= \\frac{1}{4} \\int_{\\theta=0}^{\\theta=\\pi} (1 + \\cos \\, \\theta )^4 \\sin \\, \\theta \\, d\\theta \\\\ &= - \\frac{1}{4} \\left[ \\frac{(1 + \\cos \\, \\theta)^5}{5}\\right]_0^{\\pi} = \\frac{8}{5}.\\end{align*} Exercise $$\\PageIndex{3}$$ Evaluate the integral $\\iint_D r^2 \\sin^2 2\\theta \\,r \\, dr \\, d\\theta \\nonumber$ where $$D = \\left\\{ (r,\\theta)\\,|\\,0 \\leq \\theta \\leq \\pi, \\, 0 \\leq r \\leq 2 \\sqrt{\\cos \\, 2\\theta} \\right\\}$$. Hint Graph the region and follow the steps in the previous example. $$\\frac{\\pi}{8}$$ ## Polar Areas and Volumes As in rectangular coordinates, if a solid $$S$$ is bounded by the surface $$z = f(r, \\theta)$$, as well as by the surfaces $$r = a, \\, r = b, \\, \\theta = \\alpha$$, and $$\\theta = \\beta$$, we can find the volume $$V$$ of $$S$$ by double integration, as $V = \\iint_R f(r, \\theta) \\,r \\, dr \\, d\\theta = \\int_{\\theta=\\alpha}^{\\theta=\\beta} \\int_{r=a}^{r=b} f(r,\\theta)\\, r \\, dr \\, d\\theta.$ If the base of the", "• If $r=0$, then $\\cos(\\theta)=1/2$ and so $\\theta=\\pi/3$ provided $0\\le\\theta<\\pi$. This means that the point $(0,\\pi/3)$ is on the graph, and an equation of the tangent line there is $\\theta=\\pi/3$. It is called limacon - I was having trouble finding the direction since it seems like small numbers of theta give numbers around 1 and it is hard to track. – Paul the Pirate Aug 12 '13 at 21:27 Nice graph, @Babak. Pictures help sooooo much! – amWhy Aug 13 '13 at 0:36 You don't have two values. At $\\theta = 0$, $r = 1$. As $\\theta$ increases, up to the point where $\\theta = \\frac{\\pi}{3}$, $r$ decreases, until at that point, $r = 0$. Going backwards from $\\theta = 0$, $r = 1$, as $\\theta$ decreases, $r$ decreases, up to the point where $\\theta = -\\frac{\\pi}{3}$, at which point $r = 0$ again. Thus, the bounds of integration are $-\\frac{\\pi}{3}$ and $\\frac{\\pi}{3}$, as tracing the curve with these limits gives the inner loop, from $(-\\frac{\\pi}{3}, 0)$ around through $(0, 1)$, all the way to $(\\frac{\\pi}{3}, 0)$. (Or, in cartesian coordinates, $(0,0)$ through $(0,1)$, back to $(0,0)$ again.) - Clearly you can look at the graph and see that for theta = 0 the graph has values of 0 and 1 at the same time. – Paul the", "the triangular region with vertices $$(0,0), \\space (0,3)$$, and $$(6,0); \\space \\rho (x,y) = xy$$. [Hide solution] a. $$M_x = \\frac{81}{5}, \\space M_y = \\frac{162}{5}$$; b. $$\\bar{x} = \\frac{12}{5}, \\space \\bar{y} = \\frac{6}{5}$$; c. [T] $$R$$ is the triangular region with vertices $$(0,0), \\space (1,1)$$, and $$(0,5); \\space \\rho (x,y) = x + y$$. [T] $$R$$ is the rectangular region with vertices $$(0,0), \\space (0,3), \\space (6,3)$$, and $$(6,0); \\space \\rho (x,y) = \\sqrt{xy}$$. [Hide Solution] a. $$M_x = \\frac{216\\sqrt{2}}{5}, \\space M_y = \\frac{432\\sqrt{2}}{5}$$; b. $$\\bar{x} = \\frac{18}{5}, \\space \\bar{y} = \\frac{9}{5}$$; c. [T] $$R$$ is the rectangular region with vertices $$(0,1), \\space (0,3), \\space (3,3)$$, and $$(3,1); \\space \\rho (x,y) = x^2y$$. [T] $$R$$ is the trapezoidal region determined by the lines $$y = - \\frac{1}{4}x + \\frac{5}{2}, \\space y = 0, \\space y = 2$$, and x = 0; \\space \\rho (x,y) = 3xy\\). [Hide Solution] a. $$M_x = \\frac{368}{5}, \\space M_y = \\frac{1552}{5}$$; b. $$\\bar{x} = \\frac{92}{95}, \\space \\bar{y} = \\frac{388}{95}$$; c. [T] $$R$$ is the trapezoidal region determined by the lines $$y = 0, \\space y = 1, \\space y = x$$, and y = -x + 3; \\space \\rho (x,y) = 2x + y\\). [T] $$R$$ is the disk of radius $$2$$ centered at $$(1,2); \\space \\rho(x,y) = x^2 + y^2 - 2x", "solid curves correspond to level lines of $H_0(x,y)$ Figure 3. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 4. The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for solutions of (47) with $h = 0$, $a_0 = 1$, $a_1 = 0.8$, $b_0 = 0$, $b_1 = 0.6$, $s_1 = -1$, $s_2 = -1/6$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curve corresponds to $r = r(1) t^{{|\\vartheta_4|}/{2}}$. The gray dashed curve corresponds to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.322$ Figure 5. The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (47) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = -0.8$, $s_2 = 1$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ Figure 6. The evolution of $(x(t),y(t))$, $R(t)$, $\\theta(t)$ for solutions of (50) with $h = 1/6$, $a_0", "the x and yaxes of. 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[ "scalars $\\mu_1, \\mu_2, \\mu_3 \\in F$ that: (8) \\begin{align} \\quad MN^{-1} = \\begin{bmatrix} \\mu_1 & 0 & 0\\\\ 0 & \\mu_2 & 0\\\\ 0 & 0 & \\mu_3 \\end{bmatrix} \\end{align} • But we also have that $\\mathbf{z_4} = \\phi_{MN^{-1}}(\\mathbf{z_4})$ and $\\mathbf = \\phi_{MN^{-1}} \\mathbf{z_4}MN^{-1} = [\\mu_1, \\mu_2, \\mu_3]$. Therefore $\\mu = \\mu_1 = \\mu_2 = \\mu_3$ and so $MN^{-1} = \\mu I$ and $M = \\mu N$, so $M$ is unique up to a scalar. $\\blacksquare$", "# Finding the dimension of a vector space over different fields? We were given this question for my linear algebra module: We view $\\Bbb C ^2$ as a vector space over $\\Bbb C$,$\\Bbb R$ and $\\Bbb Q$. Let $$\\mathbf x_1 := \\begin{pmatrix} i \\\\ 0 \\end{pmatrix}, \\mathbf x_2 := \\begin{pmatrix} \\sqrt 2 \\\\ \\sqrt 5 \\end{pmatrix}, \\mathbf x_3 := \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}, \\mathbf x_4 := \\begin{pmatrix} i\\sqrt 3 \\\\ \\sqrt 3 \\end{pmatrix}, \\mathbf x_5 := \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} \\in \\Bbb C ^2$$ Find dim$_F$(Span$_F$($\\mathbf x_1,\\mathbf x_2,\\mathbf x_3,\\mathbf x_4,\\mathbf x_5$)) for F=$\\Bbb C$,$\\Bbb R$ and $\\Bbb Q$. So I have found that for F=$\\Bbb C$, the dimension is 2, but I'm struggling with the other fields. Whats been confusing me is say I want to find a basis for Span$_F$($\\mathbf x_1,\\mathbf x_2,\\mathbf x_3,\\mathbf x_4,\\mathbf x_5$) over $\\Bbb R$. If I want to apply Gaussian elimination to obtain the minimal spanning set, which scalars am I allowed to used say when scaling the rows when applying row ops. Do the scalars always have to the field elements, say in this case just $\\Bbb R$? Sorry if this seems trivial but couldn't really find any good sources online and lecturer didn't cover it in lectures. Thanks a lot! • The scalars are from the field you are", "points $(2, 0)$ and $(0, 1)$. Another suitable pair would be $$\\mathbf{a}= \\begin{pmatrix}6 \\\\ 2\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}4 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $2\\mathbf{b}-\\mathbf{a}$ and $2\\mathbf{a}-3\\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$. For base vectors which yield none of the points in the original lattice, we would need irrational components, because if I had a base vector with rational components $\\frac{n}{m}$ and $\\frac{p}{q}$ then taking $2mq$ lots of this vector would take us to the point $(2qn, 2pm)$, which is in our original lattice of points.", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Solution The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ We can write the position vector of $P$ as $\\mathbf{p} = \\begin{pmatrix}5+s \\\\ 6+s \\\\ -1+s\\end{pmatrix}$, and the position vector of $Q$ as $\\mathbf{q} = \\begin{pmatrix}1+t \\\\ 13+6t\\\\ 7+3t\\end{pmatrix}.$ Thus $\\mathbf{q} - \\mathbf{p} = \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=0 \\quad\\text{and}\\quad \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1\\\\ 6 \\\\ 3\\end{pmatrix}=0.$ This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t", "\\quad \\mathbf{b} = \\begin{pmatrix} x_b \\\\y_b\\end{pmatrix}$$ If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that $$N\\begin{pmatrix} x_a \\\\ y_a\\end{pmatrix} +M\\begin{pmatrix} x_b\\\\ y_b\\end{pmatrix}\\mathbf{b} = \\begin{pmatrix} 2p \\\\ q\\end{pmatrix}$$ Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations. Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'. So, look at some more particular cases. Look at a base vector with an $x$-component of $4$. $$\\begin{pmatrix} 4 \\\\ 0\\end{pmatrix}$$ Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\\begin{pmatrix} \\pm 2 \\\\ 0\\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components. A suitable pair would be $$\\mathbf{a}= \\begin{pmatrix} 4 \\\\ 1\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}2 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $\\mathbf{a}-\\mathbf{b}$ and $2\\mathbf{b}-\\mathbf{a}$ take us to the", "Review question # Where are $P$ and $Q$ if $PQ$ is perpendicular to $l_1$ and $l_2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8828 ## Question The equations of the lines $l_1$ and $l_2$ are $\\mathbf{r}=\\mathbf{a}+\\lambda \\mathbf{b}$ and $\\mathbf{r}=\\mathbf{c}+\\mu \\mathbf{d}$ respectively, where $\\mathbf{a}=-6\\mathbf{i}+3\\mathbf{j}+15\\mathbf{k}, \\hspace{0.5cm} \\mathbf{b}=\\mathbf{i}-2\\mathbf{j}+3\\mathbf{k}, \\hspace{0.5cm} \\mathbf{c}=6\\mathbf{i}+15\\mathbf{j}+39\\mathbf{k}, \\hspace{0.5cm} \\mathbf{d}=2\\mathbf{i}-3\\mathbf{j}+4\\mathbf{k},$ and where $\\lambda$ and $\\mu$ are scalar parameters. The points $P$ and $Q$ are on $l_1$ and $l_2$ respectively, and $PQ$ is perpendicular to both $l_1$ and $l_2$. In any order: 1. find a vector which is parallel to $\\overrightarrow{PQ}$, giving your answer in the form $x\\mathbf{i}+y\\mathbf{j}+z\\mathbf{k}$; 2. show that $PQ=10\\sqrt{6}$; 3. find the position vectors of $P$ and $Q$.", "dimensions so $\\mathbf{p} = (p_{1},p_{2},p_{3})$ the expression then becomes $-E^{2} + p_{1}^{2} + p_{2}^{2} + p_{3}^{2} = -M^{2}$. Clearly you couldn't write down $E + \\mathbf{p} = M$ because it would be nonsense, regardless of the physical meaning of each term. The only way that could even have a chance to be right is if you work in 1+1 dimensional space-time so $\\mathbf{p} = p$, a scalar. But even in that case you don't end up with the expression you claim because of your massive terrible mistake. You have said $\\omega = \\sqrt{k^{2}+M^{2}} = k + M$. Even allowing for k being a scalar, when its actually a vector, you've said $\\sqrt{a^{2}+b^{2}} = a+b$. That's utterly wrong as an identity and it's a mistake you really shouldn't be making. It's a mistake children shouldn't be making. It's easy to see why it is false, just square both sides, $a^{2}+b^{2} = (a+b)^{2} = a^{2}+b^{2} + 2ab$ so it's only true if $2ab = 0$, either a=0 or b=0 or both. Try it with numbers, put in a=b=1. You just claimed $\\sqrt{1^{2}+1^{2}} = 1+1$ but that's saying $\\sqrt{2}=2$. It's clear what you've tried to do. You've taken a formula from somewhere, be it Wiki or some other website, and you've attempted to do something with it, in this case what", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "distance $$\\mathtt{r}$$ is from the line. An equation for the distance of $$\\mathtt{r}$$ to the line, then—a symbolic way to identify this distance—might be given in words as follows: go to point $$\\mathtt{p}$$, then scale to some point on the line. From that point, scale to some point on the vector that is perpendicular to the line until you get to point $$\\mathtt{r}$$. In symbols, that could be written as: $\\begin{bmatrix}\\mathtt{4}\\\\\\mathtt{2}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\, j}\\begin{bmatrix}\\mathtt{2}\\\\\\mathtt{1}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\,k}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{\\,\\,\\,\\,\\,2}\\end{bmatrix}\\mathtt{\\,\\,=\\,\\,}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{-3}\\end{bmatrix}$ With the vector and scalar names, we could write this as $$\\mathtt{p + j(p – x) + ka = r}$$. The distance to the line depends on our figuring out what $$\\mathtt{k}$$ is. Once we have that, then the distance is just $$\\mathtt{\\sqrt{(ka_1)^2 + (ka_2)^2}}$$. We can subtract vectors from both sides of an equation just like we do with scalar values. Subtracting the vector (4, 2) from both sides, we get an equation which can be rewritten as a system of two equations \\mathtt{j}\\begin{bmatrix}\\mathtt{2}\\\\\\mathtt{1}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\,k}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{\\,\\,\\,\\,\\,2}\\end{bmatrix}\\mathtt{\\,\\,=\\,\\,}\\begin{bmatrix}\\mathtt{-5}\\\\\\mathtt{-5}\\end{bmatrix} \\rightarrow \\left\\{\\begin{align*}\\mathtt{2j – k = -5} \\\\ \\mathtt{j + 2k = -5}\\end{align*}\\right. Solving that system gives us $$\\mathtt{j = -3}$$ and $$\\mathtt{k = -1}$$. So, the distance of $$\\mathtt{r}$$ to the line is $$\\mathtt{\\sqrt{5}.}$$ Can We Get to the Dot Product? Maybe we can get to the dot product. I’m not sure at this point. But there are some interesting things", "= \\begin{pmatrix} 1 & 2 & 3 \\\\ 1 & 2 & 3 \\end{pmatrix}$$ has no inversion pairs since no elements are out of order''. • $$\\pi = \\begin{pmatrix} 1 & 2 & 3 \\\\ 1 & 3 & 2 \\end{pmatrix}$$ has the single inversion pair $$(2, 3)$$ since $$\\pi(2) = 3 > 2 = \\pi(3)$$. • $$\\pi = \\begin{pmatrix} 1 & 2 & 3 \\\\ 2 & 1 & 3 \\end{pmatrix}$$ has the single inversion pair $$(1, 2)$$ since $$\\pi(1) = 2 > 1 = \\pi(2)$$. • $$\\pi = \\begin{pmatrix} 1 & 2 & 3 \\\\ 2 & 3 & 1 \\end{pmatrix}$$ has the two inversion pairs $$(1, 3)$$ and $$(2, 3)$$ since we have that both $$\\pi(1) = 2 > 1 = \\pi(3)$$ and $$\\pi(2) = 3 > 1 = \\pi(3)$$. • $$\\pi = \\begin{pmatrix} 1 & 2 & 3 \\\\ 3 & 1 & 2 \\end{pmatrix}$$ has the two inversion pairs $$(1, 2)$$ and $$(1, 3)$$ since we have that both $$\\pi(1) = 3 > 1 = \\pi(2)$$ and $$\\pi(1) = 3 > 2 = \\pi(3)$$. • $$\\pi = \\begin{pmatrix} 1 & 2 & 3 \\\\ 3 & 2 & 1 \\end{pmatrix}$$ has the three inversion pairs $$(1, 2)$$, $$(1, 3)$$, and $$(2, 3)$$, as you can check. Example 8.1.10. As another example, for each", "the addition operation includes subtraction since $$\\mathbf{x} - \\mathbf{y} = \\mathbf{x} + (-1)\\times\\mathbf{y}$$. When we combine scalar multiplication and addition, we can get combinations such as the following $\\frac{1}{a}\\, \\mathbf{x} + b\\, \\mathbf{y} = \\begin{pmatrix}\\frac{1}{a}\\, x_{1} \\\\ \\frac{1}{a}\\, x_{2} \\\\ \\vdots \\\\ \\frac{1}{a}\\, x_{n} \\end{pmatrix} + \\begin{pmatrix} b\\, y_{1} \\\\ b\\, y_{2} \\\\ \\vdots \\\\ b\\, y_{n} \\end{pmatrix} = \\begin{pmatrix} \\frac{1}{a}\\, x_{1} + b\\, y_{1} \\\\ \\frac{1}{a}\\, x_{2} + b\\, y_{2} \\\\ \\vdots \\\\ \\frac{1}{a}\\, x_{n} + b\\, y_{n} \\end{pmatrix}.$ Here’s an example of how to do this in R! a <- 5 x <- 6:-2 b <- 0.5 y <- 1:9 (1/a)*x + b*y #> [1] 1.7 2.0 2.3 2.6 2.9 3.2 3.5 3.8 4.1 is.vector((1/a)*x + b*y) #> [1] TRUE It’s very important to remember that the addition operation only works on vectors of the same length. So if we have $$\\mathbf{x} \\in \\mathbb{R}^{5}$$ and $$\\mathbf{y} \\in \\mathbb{R}^{5}$$, we can perform the addition operation to get $$\\mathbf{x} + \\mathbf{y}$$. However, if we have $$\\mathbf{z} \\in \\mathbb{R}^{3}$$, we cannot add $$\\mathbf{z}$$ to $$\\mathbf{x}$$ or to $$\\mathbf{y}$$ since those pairs of vectors don’t have the same length, or dimension. In this case, we say that their dimensions don’t match. The inner (or dot) product There’s another important operation that you can do with vectors of the same dimension.", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "The equation is given by: $P = p_{-}\\wedge{}p_{+}$ where P is the point pair and $p_{\\pm} = \\frac{P\\mp\\sqrt{P^2}}{-\\infty\\rfloor{}P}$ are the points. For the case above, the points are located at $p_{\\pm} = o \\mp \\sqrt{3}e_2 + 1.5*\\infty$. * see Geometric Algebra for Computer Science by Dorst et al, p. 536 | Tagged , , | 2 Comments", "the QM $|lm\\rangle$ rep. – Cosmas Zachos Sep 27 '19 at 13:17 ## 1 Answer I think I've managed to clear up my confusion and would like to summarize the above comments in an answer. Pions are described in QFT using three different fields, all satisfying the K-G equation. Due to their (almost) equal mass the three fields correspond to the same isospin number $$T=1$$ and form a basis $$\\{\\pi^+,\\pi^0,\\pi^-\\}$$ for the $$SU(2)$$ representation, when stacked in matrix form: $$\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\phi= \\begin{pmatrix} \\pi^+ \\\\ \\pi^0 \\\\ \\pi^- \\\\ \\end{pmatrix}$$. The isospin generators of $$SU(2)$$ in the 3-dimensional isospin space spanned by $$\\{\\pi^+,\\pi^0,\\pi^-\\}$$, then take the \"canonical\" form: $$\\qquad\\qquad\\qquad T_1=\\sqrt{2}\\begin{pmatrix} 0 & 1 & 0\\\\ 1 & 0 & 1\\\\ 0 & 1 & 0 \\end{pmatrix}$$ $$, \\quad T_2=\\sqrt{2}\\begin{pmatrix} 0 & -i & 0\\\\ i & 0 & -i\\\\ 0 & i & 0 \\end{pmatrix},$$ $$\\quad T_3=\\begin{pmatrix} 1 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & -1 \\end{pmatrix}$$. We can also choose a different basis, for example the \"adjoint\" basis $$\\{\\pi^1,\\pi^2,\\pi^3\\}$$. Now the generators take the \"adjoint\" form: $$\\qquad\\qquad\\qquad t_1=i\\begin{pmatrix} 0 & 0 & 0\\\\ 0 & 0 & -1\\\\ 0 & 1 & 0 \\end{pmatrix}$$ $$, \\quad t_2=i\\begin{pmatrix} 0 & 0 & 1\\\\ 0 & 0 & 0\\\\ -1 & 0 & 0 \\end{pmatrix},$$", "+ e_2 d_2 &\\equiv -k(p + q - 1) + 1 \\pmod n \\end{aligned} Now, since $e_1$ and $e_2$ are on the order of $\\varphi(n)$ it follows that $k$ will be on the order of $d_1$ and $d_2$, so around $n^{0.16}$. $p$ and $q$ should be on the order of $n^{0.5}$ (since they are both randomly generated 512 bit primes), so bringing this all together, we gather that the RHS of the last equivalence is on the order of $n^{0.66}$. This gives us the following result. $e_1 d_1 + e_2 d_2 \\approx -n^{0.66} \\pmod n$ Now consider the lattice spanned by the rows of the matrix $L = \\begin{bmatrix} e_1 & n^{0.5} & 0 \\\\ e_2 & 0 & n^{0.5} \\\\ n & 0 & 0 \\end{bmatrix}$ Letting $\\mathbf{b_1}, \\mathbf{b_2}$ and $\\mathbf{b_3}$ denote the three rows, we see that $d_1 \\mathbf{b_1} + d_2 \\mathbf{b_2} + c \\mathbf{b_3} = (-n^{0.66}, d_1 n^{0.5}, d_2 n^{0.5})$ for some $c \\in \\mathbb{Z}$, will be a vector close to $(-n^{0.66}, n^{0.5}, n^{0.5})$. So if we throw this at some CVP algorithm, we should be able to read off $d_1$ and $d_2$ from the result! More precisely, there should probably be some constant in front of the $-n^{0.66}$, but in this case, it looks like we're lucky and $1$ works well as the constant", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "curve. if $P(a,b)$ has order at least 3, then $b^2 \\mid 4p^3,$ which gives us $b=\\pm 2, \\ \\pm p, \\ \\pm 2p.$ using these possible values of $b$ find possible values of $a$ by solving $b^2=a^3+pa.$ finally accept only those $P(a,b)$ which have finite order. (remember the theorem only gives a necessary condition!)", "we have a kernel $$\\mathbf{H} \\in \\mathbb{R}^{3 \\times 3}$$ and input image $$\\mathbf{I} \\in \\mathbb{R}^{3 \\times 3}$$ (i.e. the kernel has the same dimensions as the input), we use a stride of $$1$$ and no padding. If we convolve $$\\mathbf{I}$$ with $$\\mathbf{H}$$, how many operations will we perform? The convolution is defined as a scalar product, so it is composed of multiplications and summations, so we need to count both of them. We have $$9$$ multiplications and $$8$$ summations, for a total of $$17$$ operations. \\begin{align} \\mathbf{I} \\circledast \\mathbf{H} &= \\begin{bmatrix} i_{11} & i_{12} & i_{13} \\\\ i_{21} & i_{22} & i_{23} \\\\ i_{31} & i_{32} & i_{33} \\end{bmatrix} \\odot \\begin{bmatrix} h_{11} & h_{12} & h_{13} \\\\ h_{21} & h_{22} & h_{23} \\\\ h_{31} & h_{32} & h_{33} \\end{bmatrix}\\\\ &= \\sum_{ij} \\begin{bmatrix} i_{11} h_{11} & i_{12} h_{12} & i_{13} h_{13} \\\\ i_{21} h_{21} & i_{22} h_{22} & i_{23} h_{23} \\\\ i_{31} h_{31} & i_{32} h_{32} & i_{33} h_{33} \\end{bmatrix}\\\\ &= i_{11} h_{11} + i_{12} h_{12} + i_{13} h_{13} + i_{21} h_{21} + i_{22} h_{22} + i_{23} h_{23} + i_{31} h_{31} + i_{32} h_{32} + i_{33} h_{33} \\end{align} ### Time complexity What is the time complexity of this convolution? To answer this question, you first need to know the input's size, $$n$$. The input contains $$9$$ elements, so its", "$$d$$ are scalars, then 1. $$\\mathbf{a} + \\mathbf{b} = \\mathbf{b} + \\mathbf{a}$$ 2. $$\\mathbf{a} + \\mathbf{0} = \\mathbf{a}$$ 3. $$c(\\mathbf{a} + \\mathbf{b}) = c\\mathbf{b} + c\\mathbf{a}$$ 4. $$(cd)\\mathbf{a} = c(d\\mathbf{a})$$ 5. $$\\mathbf{a} + (\\mathbf{b}+ \\mathbf{c}) = (\\mathbf{a} + \\mathbf{b})+ \\mathbf{c}$$ 6. $$\\mathbf{a}+ (-\\mathbf{a}) = \\mathbf{0}$$ 7. $$(c+d)\\mathbf{a} = c\\mathbf{a}+ d\\mathbf{a}$$ 8. $$1 \\mathbf{a} =\\mathbf{a}$$ $$\\mathbf{i} = \\left \\langle 1, 0, 0 \\right \\rangle \\; \\; \\mathbf{j} = \\left \\langle 0, 1, 0 \\right \\rangle \\; \\; \\mathbf{k} = \\left \\langle 0, 0, 1 \\right \\rangle$$", "$$=$$ $$\\displaystyle \\left \\Vert{ \\mathbf{v} }\\right\\Vert + \\left \\Vert{ \\mathbf{w} }\\right\\Vert$$ $\\Box$ ### Case 2 Suppose that $\\mathbf{v}$ and $\\mathbf{w}$ are scalar multiples of each other. Then the existence of $\\theta$ follows directly from the definition of the angle between vectors that are scalar multiples of each other. $\\blacksquare$" ]

[ "(6), where {\\displaystyle {\\begin{aligned}&\\mathbf {D} ={\\begin{pmatrix}6&0&0\\\\0&4&0\\\\0&0&5\\end{pmatrix}},\\quad \\mathbf {U} ={\\begin{pmatrix}0&2&3\\\\0&0&2\\\\0&0&0\\end{pmatrix}},\\quad \\mathbf {L} ={\\begin{pmatrix}0&0&0\\\\1&0&0\\\\3&1&0\\end{pmatrix}}.\\quad (14)\\end{aligned}}} We then have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}} ={\\frac {1}{120}}{\\begin{pmatrix}20&0&0\\\\5&30&0\\\\13&6&24\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}U} ={\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}},\\quad \\mathbf {(D-L)^{-1}k} ={\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}}.\\end{aligned}}} The Gauss-Seidel method (8) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (15)} The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), somewhat faster than the Jacobi method described above. ### Successive over-relaxation method Let ω = 1.1. Using the splitting (14) of the matrix A in problem (10) for the successive over-relaxation method, we have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}} ={\\frac {1}{12}}{\\begin{pmatrix}2&0&0\\\\0.55&3&0\\\\1.441&0.66&2.4\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}[(1-\\omega )D+\\omega U]} ={\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {\\omega (D-\\omega L)^{-1}k} ={\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}}.\\end{aligned}}} The successive over-relaxation method (9) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (16)} The first few iterates for equation (16) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), slightly faster than the Gauss-Seidel method described above. ## References • {{#invoke:citation/CS1|citation |CitationClass=citation }}.", "A={\\begin{pmatrix}12&-51&4\\\\6&167&-68\\\\-4&24&-41\\end{pmatrix}}.}$ Recall that an orthonormal matrix ${\\displaystyle Q}$ has the property ${\\displaystyle Q^{\\textsf {T}}\\,Q=I.}$ Then, we can calculate ${\\displaystyle Q}$ by means of Gram–Schmidt as follows: {\\displaystyle {\\begin{aligned}U={\\begin{pmatrix}\\mathbf {u} _{1}&\\mathbf {u} _{2}&\\mathbf {u} _{3}\\end{pmatrix}}&={\\begin{pmatrix}12&-69&-58/5\\\\6&158&6/5\\\\-4&30&-33\\end{pmatrix}};\\\\Q={\\begin{pmatrix}{\\frac {\\mathbf {u} _{1}}{\\|\\mathbf {u} _{1}\\|}}&{\\frac {\\mathbf {u} _{2}}{\\|\\mathbf {u} _{2}\\|}}&{\\frac {\\mathbf {u} _{3}}{\\|\\mathbf {u} _{3}\\|}}\\end{pmatrix}}&={\\begin{pmatrix}6/7&-69/175&-58/175\\\\3/7&158/175&6/175\\\\-2/7&6/35&-33/35\\end{pmatrix}}.\\end{aligned}}} Thus, we have {\\displaystyle {\\begin{aligned}Q^{\\textsf {T}}A&=Q^{\\textsf {T}}Q\\,R=R;\\\\R&=Q^{\\textsf {T}}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&35\\end{pmatrix}}.\\end{aligned}}} #### Relation to RQ decomposition The RQ decomposition transforms a matrix A into the product of an upper triangular matrix R (also known as right-triangular) and an orthogonal matrix Q. The only difference from QR decomposition is the order of these matrices. QR decomposition is Gram–Schmidt orthogonalization of columns of A, started from the first column. RQ decomposition is Gram–Schmidt orthogonalization of rows of A, started from the last row. The Gram-Schmidt process is inherently numerically unstable. While the application of the projections has an appealing geometric analogy to orthogonalization, the orthogonalization itself is prone to numerical error. A significant advantage however is the ease of implementation, which makes this a useful algorithm to use for prototyping if a pre-built linear algebra library is unavailable. ### Using Householder reflections A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane or hyperplane. We can use this operation to calculate the QR factorization of", "& \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} <br> \\begin{aligned} \\quad \\mathbf u &= \\alpha \\mathbf a + \\beta \\mathbf b \\qquad \\text{Norm:}|| \\mathbf u|| \\end{aligned} <br> \\begin{align} \\hat{\\mathbf u} = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} <br> 2. **Matrices** \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} <br> 3. \\begin{aligned} a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5 \\end{aligned} <br> 4. **Transformations:** \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} The syntax is very close to what one writes in a .tex file. Switch html_document in the YAML header with pdf_document and you can get the corresponding pdf. This is the HTML output: To have our HTML is written with mathml, we add --mathml to the arguments passed to pandoc. This is what we see at the top of the HTML file: In this workflow, knitr produces a .tex file. As we can see, it is unremarkable, and very closely resembles the OP's original latex code, just adding elements like the enumerate environment. \\begin{document} \\centerline{\\textbf{\\large{Mathematical expressions}}} \\begin{enumerate} \\item", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "as the splitting (6), where \\begin{align} & \\mathbf{D} = \\begin{pmatrix} 6 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 5 \\end{pmatrix}, \\quad \\mathbf{U} = \\begin{pmatrix} 0 & 2 & 3 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 0 \\end{pmatrix}, \\quad \\mathbf{L} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 1 & 0 & 0 \\\\ 3 & 1 & 0 \\end{pmatrix}. \\quad (14) \\end{align} We then have \\begin{align} & \\mathbf{(D-L)^{-1}} = \\frac{1}{120} \\begin{pmatrix} 20 & 0 & 0 \\\\ 5 & 30 & 0 \\\\ 13 & 6 & 24 \\end{pmatrix}, \\end{align} \\begin{align} & \\mathbf{(D-L)^{-1}U} = \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix}, \\quad \\mathbf{(D-L)^{-1}k} = \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}. \\end{align} The Gauss-Seidel method (8) applied to the problem (10) takes the form \\bold x^{(m+1)} = \\begin{align} & \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix} \\bold x^{(m)} + \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}, \\end{align} \\quad m = 0, 1, 2, \\ldots \\quad (15) The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "& 1 & 0 & 1 & 0 & 1 \\\\ 0 & 1 & 1 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \\\\ \\end{bmatrix} \\end{align} \\begin{align} \\mathbf{z} = \\mathbf{H} \\mathbf{x}^{\\prime} \\end{align} If $\\mathbf{x} = \\mathbf{x}^{\\prime}$, we must have $\\mathbf{z} = \\mathbf{0}$. \\begin{align} \\mathbf{z} &= \\mathbf{H} \\mathbf{x}^{\\prime} \\\\ &= \\mathbf{H} \\mathbf{x} \\\\ &= \\begin{bmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 \\\\ 0 & 1 & 1 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \\\\ \\end{bmatrix} \\begin{bmatrix} p_1 \\\\ p_2 \\\\ d_1 \\\\ p_3 \\\\ d_2 \\\\ d_3 \\\\ d_4 \\\\ \\end{bmatrix} \\\\ &= \\begin{bmatrix} p_1 + d_1 + d_2 + d_3 \\\\ p_2 + d_1 + d_3 + d_4 \\\\ p_3 + d_d + d_3 + d_4 \\\\ \\end{bmatrix} \\\\ &= \\begin{bmatrix} p_1 + d_1 + d_2 + d_3 \\\\ p_2 + d_1 + d_3 + d_4 \\\\ p_3 + d_d + d_3 + d_4 \\\\ \\end{bmatrix} \\mod 2 \\\\ &= \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\\\ &= \\mathbf{0} \\\\ \\end{align} Also notice that the parity checking matrix $\\mathbf{H}$ is just the bit sequence and parity table. A", "assumption we have $A\\mathbf{v}=-\\mathbf{v}$. Using this repeatedly we see that $A^5\\mathbf{v}=(-1)^5\\mathbf{v}=-\\mathbf{v}$. Similarly, we have $A^5\\mathbf{w}=2^5\\mathbf{w}=32\\mathbf{w}$ using $A\\mathbf{w}=2\\mathbf{w}$. Combining these with (*) yields, \\begin{align*} A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ 9 \\end{bmatrix}&=3(-\\mathbf{v})-2(32\\mathbf{w})\\\\[6pt] &=-3\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix}-64\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}\\\\[6pt] &=\\begin{bmatrix} -3-128 \\\\ -6+64 \\\\ 3-192 \\end{bmatrix} =\\begin{bmatrix} -131 \\\\ 58 \\\\ -189 \\end{bmatrix}. \\end{align*}", "so we have \\begin{aligned} \\vec{u}^\\top \\, \\mathbf{R} \\, \\vec{u} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\mathbf{R} \\, \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -x \\\\ \\end{bmatrix} \\\\ {} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\begin{bmatrix} \\mathbf{R}_{1,1} - \\mathbf{R}_{1,4}\\,x \\\\ \\mathbf{R}_{2,1} - \\mathbf{R}_{2,4}\\,x \\\\ \\mathbf{R}_{3,1} - \\mathbf{R}_{3,4}\\,x \\\\ \\mathbf{R}_{4,1} - \\mathbf{R}_{4,4}\\,x \\\\ \\end{bmatrix} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - (\\mathbf{R}_{1,4} + \\mathbf{R}_{4,1})\\,x + \\mathbf{R}_{1,1} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - 2\\,\\mathbf{R}_{1,4}\\,x + \\mathbf{R}_{1,1} \\\\ {} & = 0 \\\\ x & = \\frac{\\mathbf{R}_{1,4} \\pm \\sqrt{\\mathbf{R}_{1,4}^2 - \\mathbf{R}_{1,1}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} Similarly, the $\\hat{y}$ and $\\hat{z}$ planes are given by \\begin{aligned} y & = \\frac{\\mathbf{R}_{2,4} \\pm \\sqrt{\\mathbf{R}_{2,4}^2 - \\mathbf{R}_{2,2}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}} \\\\ z & = \\frac{\\mathbf{R}_{3,4} \\pm \\sqrt{\\mathbf{R}_{3,4}^2 - \\mathbf{R}_{3,3}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} This is where the above linked posts stop, but a lot more simplification can be done. ## Implementation Details Several details of the problem can make computing the planes more efficient. The first is that $\\mathbf{S}$ is involutory, meaning $\\mathbf{S}^{-1} = \\mathbf{S}$. This means that the product $\\mathbf{M}\\,\\mathbf{S}^{-1}$ can be computed implicitly: it is simply $\\mathbf{M}$ with its last column negated. The last column of $\\mathbf{R} = \\mathbf{M} \\, \\mathbf{S}^{-1} \\, \\mathbf{M}^\\top$ is the same, because the last column of $\\mathbf{M}^\\top$ is $\\begin{bmatrix}0 & 0 & 0 &", "\\\\ {6} & {1} & {9} \\\\ {4} & {-3} & {2} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {2} & {-1} & {3} \\\\ {1} & {2} & {5} \\\\ {-3} & {-5} & {2} \\end{bmatrix}\\nonumber$ determine $$\\mathbf{ A}+\\mathbf{ B}$$ and $$\\mathbf{ A}-3 \\mathbf{ B}$$. 23. Given the following $$1\\times 4$$ row matrices $\\mathbf{ a} = \\begin{bmatrix} {3} & {1} & {5} & {6} \\end{bmatrix}, \\quad \\mathbf{ b} = \\begin{bmatrix} {6} & {-2} & {0} & {4} \\end{bmatrix}\\nonumber$ determine $$2\\mathbf{ a}+\\mathbf{ b}$$. 24.$$\\ddagger$$ Compute the $$4\\times 4$$ matrix defined by $$\\mathbf{ A}-\\mathbf{ B}+\\mathbf{ C}$$ where $$\\mathbf{A}$$, $$\\mathbf{B}$$ and $$\\mathbf{C}$$ are given by $\\mathbf{ A} = \\begin{bmatrix} {0.856} & {0.328} & {0.0663} & {0.908} \\\\ {0.529} & {0.426} & {0.364} & {0.434} \\\\ {0.506} & {0.812} & {0.400} & {0.137} \\\\ {0.652} & {0.402} & {0.0276} & {0.995} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {1.142} & {0.438} & {0.0884} & {1.211} \\\\ {0.705} & {0.568} & {0.485} & {0.579} \\\\ {0.675} & {1.083} & {0.534} & {0.183} \\\\ {0.869} & {0.535} & {0.368} & {1.326} \\end{bmatrix}\\nonumber$ $\\mathbf{ C} = \\begin{bmatrix} {0.285} & {0.109} & {0.0221} & {0.303} \\\\ {0.176} & {0.142} & {0.121} & {0.145} \\\\ {0.169} & {0.271} & {0.133} & {0.0457} \\\\ {0.217} & {0.134} & {0.0092} & {0.332} \\end{bmatrix}\\nonumber$ 25.$$\\ddagger$$ Consider a project in which all", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "of $${\\mathbf z}$$ with respect to the basis $$\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}$$ are $$(2, -3)\\text{.}$$ Suppose we wish to convert the coordinates with repect to one basis to a new set of coordinates with respect to a different basis; that is, we wish to do a change of coordinates. Observe that \\begin{align*} \\mathbf v_1 & = 2 \\mathbf e_1 + \\mathbf e_2\\\\ \\mathbf v_2 & = 3 \\mathbf e_1 + 2 \\mathbf e_2. \\end{align*} It follows that \\begin{align*} c_1 \\mathbf v_1 + c_2 \\mathbf v_2 & = c_1 (2 \\mathbf e_1 + \\mathbf e_2) + c_2( 3\\mathbf e_1 + 2 \\mathbf e_2)\\\\ & = (2c_1 + 3c_2 )\\mathbf e_1 + (c_2 + 2c_2) \\mathbf e_2. \\end{align*} Thus, the coordinates of $$c_1 \\mathbf v_1 + c_2 \\mathbf v_2$$ with respect to the basis $$\\{ {\\mathbf e}_1, {\\mathbf e}_2 \\}$$ can be determined by \\begin{equation*} \\begin{pmatrix} 2c_1 + 3c_2 \\\\ c_2 + 2c_2 \\end{pmatrix} = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\begin{pmatrix} c_1 \\\\ c_2 \\end{pmatrix}. \\end{equation*} If we let \\begin{equation*} T = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf c = \\begin{pmatrix} c_1 \\\\ c_2 \\end{pmatrix}, \\end{equation*} then the coordinates with respect to the basis $$\\{ {\\mathbf e}_1, {\\mathbf e}_2 \\}$$ are given by $$\\mathbf d = T", "4 & 5\\times(-1) \\end{pmatrix} \\\\ &= \\begin{pmatrix} -8&2\\\\ 20&-5 \\end{pmatrix} \\end{align} ###### Example 2 Let $\\mathbf{A} = \\begin{pmatrix}2 & 0 \\\\ -1 & 5 \\end{pmatrix}$ and $\\mathbf{B} = \\begin{pmatrix} -2 & 4 \\\\ -2 & 8 \\end{pmatrix}$, then calculate a) $\\mathbf{AB}$ b) $\\mathbf{BA}$ ###### Solution a) \\begin{align} \\mathbf{AB} &= \\begin{pmatrix} 2 & 0\\\\ -1 & 5 \\end{pmatrix} \\begin{pmatrix} -2 & 4\\\\ -2 & 8 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} 2\\times(-2) + 0\\times (-2) & 2\\times 4 + 0\\times 8\\\\ (-1)\\times(-2) + 5\\times(-2) & (-1)\\times 4 + 5\\times 8 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} (-4)+0 & 8+0\\\\\\\\ 2+(-10) & (-4)+40 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} -4 & 8\\\\ -8 & 36 \\end{pmatrix} \\end{align} b) \\begin{align} \\mathbf{BA} &= \\begin{pmatrix} -2 & 4\\\\ -2 & 8 \\end{pmatrix} \\begin{pmatrix} 2 & 0\\\\ -1 & 5 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} (-2)\\times 2 + 4\\times (-1) & (-2)\\times 0 + 4\\times 5\\\\ (-2)\\times 2 + 8\\times(-1) & (-2)\\times 0 + 8\\times 5 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} (-4)+(-4) & 0+20 \\\\ (-4)+(-8) & 0+40 \\end{pmatrix} \\\\ \\\\ &= \\begin{pmatrix} -8 & 20\\\\ -12 & 40 \\end{pmatrix} \\end{align} #### The Transpose ##### Definition To take the transpose of a matrix, the rows and columns of the matrix are switched, so that the first row becomes the first column, the second row becomes the second", "0 & 0 & 6120 & -12240 & 6120 \\\\ 0 & -12240 & 12240 & 0 & 0 \\\\ 6120 & 12240 & -12240 & 0 & 0 \\\\ -12240 & 0 & 0 & 0 & 0 \\\\ 6120 & 0 & 0 & 0 & 0 \\end{pmatrix},$$ $$G_{31} = \\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 15456 & 9792 \\\\ 0 & 0 & -30912 & -9792 & 0 \\\\ 0 & -15456 & -9792 & 0 & 0 \\\\ 0 & 9792 & 0 & 0 & 0 \\end{pmatrix},$$ $$G_{32} = \\begin{pmatrix} 0 & 0 & 5055 & 10425 & 0 \\\\ 0 & 3390 & -3915 & 0 & 0 \\\\ 5055 & -24085 & -3296 & 39168 & 0 \\\\ 10425 & -22880 & 19808 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix},$$$$g^\\,_{30}(u,v) \\ge 0,\\quad g^\\,_{31}(u,v) \\ge 0.$$ Since $$\\!\\mathstrut^{\\phantom{\\dfrac\\mathstrut\\mathstrut}^{\\LARGE=}}$$ $$\\!\\mathstrut^{\\phantom{\\dfrac\\mathstrut\\mathstrut}^{\\LARGE,}}$$ then, similarly to the first case, $$g^\\,_{32}(u,v)= \\begin{pmatrix} 1 \\\\ v \\\\ v^2 \\\\ v^3 \\end{pmatrix}^T \\begin{pmatrix} 0 & 0 & 5055 & 10425 \\\\ 0 & 3390 & -3915 & 0 \\\\ 5055 & -24085 & -3296 & 39168 \\\\ 10425 & -22880 & 19808 & 0 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ u \\\\", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "&=\\dfrac{-6 \\pm \\sqrt{-4}}{4}\\\\ \\text{No real solution} \\end{align*} $$2p(2p+1)=2$$ \\begin{align*} 4p^2 + 2p - 2 &= 0\\\\ 2p^2 + p - 1 &= 0\\\\ p &= \\dfrac{-1 \\pm \\sqrt{1^2 - 4(2)(-1)}}{2(2)}\\\\ &= \\dfrac{-1 \\pm \\sqrt{1+8}}{4}\\\\ &= \\dfrac{-1 \\pm \\sqrt{9}}{4} \\\\ &= \\dfrac{-1 \\pm 3}{4} \\\\ \\text{therefore } p =\\dfrac{-1 + 3}{4} = \\frac{1}{2} &\\text{ or } p =\\dfrac{-1 - 3}{4} = -1 \\end{align*} $$-3{t}^{2}+5t-8=0$$ \\begin{align*} -3t^2 + 5t -8 &= 0\\\\ t&= \\dfrac{-5 \\pm \\sqrt{5^2 -4(-3)(-8)}}{2(-3)}\\\\ &= \\dfrac{-5 \\pm \\sqrt{25-96}}{-6}\\\\ &=\\dfrac{-5 \\pm \\sqrt{-71}}{-6}\\\\ \\text{No real solution} \\end{align*} $$5{t}^{2}+3t-3=0$$ \\begin{align*} 5t^2 + 3t - 3 &= 0\\\\ t &= \\dfrac{-3 \\pm \\sqrt{3^2 - 4(5)(-3)}}{2(5)}\\\\ &= \\dfrac{-3 \\pm \\sqrt{9+60}}{10}\\\\ &=\\dfrac{-3 \\pm \\sqrt{69}}{10}\\\\ \\text{therefore } t =\\dfrac{-3 + \\sqrt{69}}{10} &\\text{ or } t =\\dfrac{-3 - \\sqrt{69}}{10} \\end{align*} $${t}^{2}-4t+2=0$$ \\begin{align*} t^2 - 4t + 2 &= 0\\\\ t &= \\dfrac{-(-4) \\pm \\sqrt{(-4)^2 - 4(1)(2)}}{2(1)}\\\\ &= \\dfrac{4 \\pm \\sqrt{16-8}}{2}\\\\ &=\\dfrac{4 \\pm \\sqrt{8}}{2}\\\\ &= 2 \\pm \\sqrt{2} \\end{align*} $$9(k^2 - 1) = 7k$$ \\begin{align*} 9k^2 - 7k - 9 &= 0\\\\ k &= \\dfrac{-(-7) \\pm \\sqrt{(-7)^2 - 4(9)(-9)}}{2(9)}\\\\ &= \\dfrac{7 \\pm \\sqrt{49 + 324}}{18}\\\\ &=\\dfrac{7 \\pm \\sqrt{373}}{18}\\\\ k = \\dfrac{7 + \\sqrt{373}}{18} &\\text{ or } k =\\dfrac{7 - \\sqrt{373}}{18} \\end{align*} $$3f -2 = -2f^2$$ \\begin{align*} 2f^2 + 3f - 2 &= 0\\\\ f &= \\dfrac{-3 \\pm \\sqrt{3^2-4(2)(-2)}}{2(2)}\\\\ &= \\dfrac{-3 \\pm \\sqrt{9+16}}{4}\\\\ &= \\dfrac{-3 \\pm \\sqrt{25}}{4}\\\\", "& 15 & 24 & -27 & -18 & 15 \\\\ -2 & 6 & 8 & -14 & 3 & -9 & -12 & 21 & 3 & -9 & -12 & 21 \\\\ -4 & -16 & 16 & 6 & 6 & 24 & -24 & -9 & 6 & 24 & -24 & -9 \\\\ -2 & -4 & 10 & 2 & 3 & 6 & -15 & -3 & 3 & 6 & -15 & -3 \\end{bmatrix} }$ ## Properties ### Relations to other matrix operations 1. Bilinearity and associativity: The Kronecker product is a special case of the tensor product, so it is bilinear and associative: \\displaystyle{ \\begin{align} \\mathbf{A} \\otimes (\\mathbf{B} + \\mathbf{C}) &= \\mathbf{A} \\otimes \\mathbf{B} + \\mathbf{A} \\otimes \\mathbf{C}, \\\\ (\\mathbf{B} + \\mathbf{C}) \\otimes \\mathbf{A} &= \\mathbf{B} \\otimes \\mathbf{A} + \\mathbf{C} \\otimes \\mathbf{A}, \\\\ (k\\mathbf{A}) \\otimes \\mathbf{B} &= \\mathbf{A} \\otimes (k\\mathbf{B}) = k(\\mathbf{A} \\otimes \\mathbf{B}), \\\\ (\\mathbf{A} \\otimes \\mathbf{B}) \\otimes \\mathbf{C} &= \\mathbf{A} \\otimes (\\mathbf{B} \\otimes \\mathbf{C}), \\\\ \\mathbf{A} \\otimes \\mathbf{0} &= \\mathbf{0} \\otimes \\mathbf{A} = \\mathbf{0}, \\end{align} } where A, B and C are matrices, 0 is a zero matrix, and k is a scalar. 2. Non-commutative: In general, AB and BA are different matrices. However, AB and BA are permutation equivalent, meaning that there exist permutation matrices", "of the congruence is the same. The only way that $(ab/p) \\equiv (a/p)(b/p) \\pmod p$ is when $(ab/p) = (a/p)(b/p)$ since p is restricted to being an odd prime. # Other Properties of Legendre Symbols There are many other rules that have been established regarding Legendre symbols, some of which are often necessary to evaluate Legendre symbols. We will state a few of them on this page and subsequently prove them all in future pages: (2) \\begin{align} (-1/p) = \\begin{Bmatrix} 1 & p \\equiv 1 \\pmod {4} \\\\ -1 & p \\equiv 3 \\pmod {4}\\\\ \\end{Bmatrix} \\end{align} (3) \\begin{align} (2/p) = \\begin{Bmatrix} 1 & p \\equiv 1 \\pmod {8} \\quad \\mathbf{or} \\quad p \\equiv 7 \\pmod {8}\\\\ -1 & p \\equiv 3 \\pmod {8} \\quad \\mathbf{or} \\quad p \\equiv 5 \\pmod {8}\\\\ \\end{Bmatrix} \\end{align} (4) \\begin{align} (3/p) = \\begin{Bmatrix} 1 & \\mathbf{if} \\quad p \\equiv 1 \\pmod {12} \\quad \\mathbf{or} \\quad p \\equiv 11 \\pmod {12}\\\\ -1 & \\mathbf{if} \\quad p \\equiv 5 \\pmod {12} \\quad \\mathbf{or} \\quad p \\equiv 7 \\pmod {12}\\\\ \\end{Bmatrix} \\end{align} (5) \\begin{align} (6/p) = \\begin{Bmatrix} 1 & \\mathbf{if} \\quad p \\equiv 1, 5, 19, \\: \\mathbf{or} \\: 23 \\pmod {24} \\\\ -1 & \\mathbf{if} \\quad p \\equiv 7, 11, 13, \\: \\mathbf{or} \\: 17 \\pmod {24} \\end{Bmatrix} \\end{align}", "\\mathbf{R} \\, \\vec{u} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\mathbf{R} \\, \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -x \\\\ \\end{bmatrix} \\\\ {} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\begin{bmatrix} \\mathbf{R}_{1,1} - \\mathbf{R}_{1,4}\\,x \\\\ \\mathbf{R}_{2,1} - \\mathbf{R}_{2,4}\\,x \\\\ \\mathbf{R}_{3,1} - \\mathbf{R}_{3,4}\\,x \\\\ \\mathbf{R}_{4,1} - \\mathbf{R}_{4,4}\\,x \\\\ \\end{bmatrix} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - (\\mathbf{R}_{1,4} + \\mathbf{R}_{4,1})\\,x + \\mathbf{R}_{1,1} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - 2\\,\\mathbf{R}_{1,4}\\,x + \\mathbf{R}_{1,1} \\\\ {} & = 0 \\\\ x & = \\frac{\\mathbf{R}_{1,4} \\pm \\sqrt{\\mathbf{R}_{1,4}^2 - \\mathbf{R}_{1,1}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} Similarly, the \\hat{y}$ and $\\hat{z} planes are given by \\begin{aligned} y & = \\frac{\\mathbf{R}_{2,4} \\pm \\sqrt{\\mathbf{R}_{2,4}^2 - \\mathbf{R}_{2,2}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}} \\\\ z & = \\frac{\\mathbf{R}_{3,4} \\pm \\sqrt{\\mathbf{R}_{3,4}^2 - \\mathbf{R}_{3,3}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} This is where the above linked posts stop, but a lot more simplification can be done. ## Implementation Details Several details of the problem can make computing the planes more efficient. The first is that \\mathbf{S}$ is involutory, meaning $\\mathbf{S}^{-1} = \\mathbf{S}$. This means that the product $\\mathbf{M}\\,\\mathbf{S}^{-1}$ can be computed implicitly: it is simply $\\mathbf{M}$ with its last column negated. The last column of $\\mathbf{R} = \\mathbf{M} \\, \\mathbf{S}^{-1} \\, \\mathbf{M}^\\top$ is the same, because the last column of $\\mathbf{M}^\\top$ is $\\begin{bmatrix}0 & 0 & 0 & 1\\end{bmatrix}^\\top$. In particular, $\\mathbf{R}_{4,4} = -1$." ]

[ "(6), where {\\displaystyle {\\begin{aligned}&\\mathbf {D} ={\\begin{pmatrix}6&0&0\\\\0&4&0\\\\0&0&5\\end{pmatrix}},\\quad \\mathbf {U} ={\\begin{pmatrix}0&2&3\\\\0&0&2\\\\0&0&0\\end{pmatrix}},\\quad \\mathbf {L} ={\\begin{pmatrix}0&0&0\\\\1&0&0\\\\3&1&0\\end{pmatrix}}.\\quad (14)\\end{aligned}}} We then have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}} ={\\frac {1}{120}}{\\begin{pmatrix}20&0&0\\\\5&30&0\\\\13&6&24\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}U} ={\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}},\\quad \\mathbf {(D-L)^{-1}k} ={\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}}.\\end{aligned}}} The Gauss-Seidel method (8) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (15)} The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), somewhat faster than the Jacobi method described above. ### Successive over-relaxation method Let ω = 1.1. Using the splitting (14) of the matrix A in problem (10) for the successive over-relaxation method, we have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}} ={\\frac {1}{12}}{\\begin{pmatrix}2&0&0\\\\0.55&3&0\\\\1.441&0.66&2.4\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}[(1-\\omega )D+\\omega U]} ={\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {\\omega (D-\\omega L)^{-1}k} ={\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}}.\\end{aligned}}} The successive over-relaxation method (9) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (16)} The first few iterates for equation (16) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), slightly faster than the Gauss-Seidel method described above. ## References • {{#invoke:citation/CS1|citation |CitationClass=citation }}.", "A={\\begin{pmatrix}12&-51&4\\\\6&167&-68\\\\-4&24&-41\\end{pmatrix}}.}$ Recall that an orthonormal matrix ${\\displaystyle Q}$ has the property ${\\displaystyle Q^{\\textsf {T}}\\,Q=I.}$ Then, we can calculate ${\\displaystyle Q}$ by means of Gram–Schmidt as follows: {\\displaystyle {\\begin{aligned}U={\\begin{pmatrix}\\mathbf {u} _{1}&\\mathbf {u} _{2}&\\mathbf {u} _{3}\\end{pmatrix}}&={\\begin{pmatrix}12&-69&-58/5\\\\6&158&6/5\\\\-4&30&-33\\end{pmatrix}};\\\\Q={\\begin{pmatrix}{\\frac {\\mathbf {u} _{1}}{\\|\\mathbf {u} _{1}\\|}}&{\\frac {\\mathbf {u} _{2}}{\\|\\mathbf {u} _{2}\\|}}&{\\frac {\\mathbf {u} _{3}}{\\|\\mathbf {u} _{3}\\|}}\\end{pmatrix}}&={\\begin{pmatrix}6/7&-69/175&-58/175\\\\3/7&158/175&6/175\\\\-2/7&6/35&-33/35\\end{pmatrix}}.\\end{aligned}}} Thus, we have {\\displaystyle {\\begin{aligned}Q^{\\textsf {T}}A&=Q^{\\textsf {T}}Q\\,R=R;\\\\R&=Q^{\\textsf {T}}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&35\\end{pmatrix}}.\\end{aligned}}} #### Relation to RQ decomposition The RQ decomposition transforms a matrix A into the product of an upper triangular matrix R (also known as right-triangular) and an orthogonal matrix Q. The only difference from QR decomposition is the order of these matrices. QR decomposition is Gram–Schmidt orthogonalization of columns of A, started from the first column. RQ decomposition is Gram–Schmidt orthogonalization of rows of A, started from the last row. The Gram-Schmidt process is inherently numerically unstable. While the application of the projections has an appealing geometric analogy to orthogonalization, the orthogonalization itself is prone to numerical error. A significant advantage however is the ease of implementation, which makes this a useful algorithm to use for prototyping if a pre-built linear algebra library is unavailable. ### Using Householder reflections A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane or hyperplane. We can use this operation to calculate the QR factorization of", "as the splitting (6), where \\begin{align} & \\mathbf{D} = \\begin{pmatrix} 6 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 5 \\end{pmatrix}, \\quad \\mathbf{U} = \\begin{pmatrix} 0 & 2 & 3 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 0 \\end{pmatrix}, \\quad \\mathbf{L} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 1 & 0 & 0 \\\\ 3 & 1 & 0 \\end{pmatrix}. \\quad (14) \\end{align} We then have \\begin{align} & \\mathbf{(D-L)^{-1}} = \\frac{1}{120} \\begin{pmatrix} 20 & 0 & 0 \\\\ 5 & 30 & 0 \\\\ 13 & 6 & 24 \\end{pmatrix}, \\end{align} \\begin{align} & \\mathbf{(D-L)^{-1}U} = \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix}, \\quad \\mathbf{(D-L)^{-1}k} = \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}. \\end{align} The Gauss-Seidel method (8) applied to the problem (10) takes the form \\bold x^{(m+1)} = \\begin{align} & \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix} \\bold x^{(m)} + \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}, \\end{align} \\quad m = 0, 1, 2, \\ldots \\quad (15) The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From", "& \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} <br> \\begin{aligned} \\quad \\mathbf u &= \\alpha \\mathbf a + \\beta \\mathbf b \\qquad \\text{Norm:}|| \\mathbf u|| \\end{aligned} <br> \\begin{align} \\hat{\\mathbf u} = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} <br> 2. **Matrices** \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} <br> 3. \\begin{aligned} a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5 \\end{aligned} <br> 4. **Transformations:** \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} The syntax is very close to what one writes in a .tex file. Switch html_document in the YAML header with pdf_document and you can get the corresponding pdf. This is the HTML output: To have our HTML is written with mathml, we add --mathml to the arguments passed to pandoc. This is what we see at the top of the HTML file: In this workflow, knitr produces a .tex file. As we can see, it is unremarkable, and very closely resembles the OP's original latex code, just adding elements like the enumerate environment. \\begin{document} \\centerline{\\textbf{\\large{Mathematical expressions}}} \\begin{enumerate} \\item", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "\\mathbf O_{n-1} \\\\ 0 & \\end{pmatrix}=\\begin{pmatrix} x_{0} & 0 & \\ldots & 0 \\\\ 0 & & & \\\\ \\vdots & & \\mathbf O_{n-1} \\\\ 0 & & & \\end{pmatrix}\\end{align}\\\\ \\begin{align}&\\Big(I_n-\\frac{X X^T}{\\Vert X \\Vert^{2}} \\Big) \\begin{pmatrix} \\ast & \\ast & \\ldots & \\ast \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix} \\Big(I_n-\\frac{X X^\\top}{\\Vert X \\Vert^{2}} \\Big)^T\\\\ =&\\mathcal R_X\\begin{pmatrix} \\ast & \\ast & \\ldots & \\ast \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix}\\mathcal R_X^T =\\begin{pmatrix} 0 & 0 & \\ldots & 0 \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix}\\mathcal R_X^T\\\\ =&\\begin{pmatrix} 0 & 0 & \\ldots & 0 \\\\ 0 & & \\\\ \\vdots & & S \\\\ 0 & & & \\end{pmatrix}\\end{align} Hence $$\\Gamma=x_0\\mathcal P_XI_n+\\mathcal R_X\\tilde S\\mathcal R_X^T$$ Thanks a lot ! I can't figure out where I was wrong. Was it my expression for $\\mathrm{I}_{n} - \\frac{X X^\\top}{\\Vert X \\Vert^{2}}$ ? – jibounet Nov 22 '13 at 15:26", "assumption we have $A\\mathbf{v}=-\\mathbf{v}$. Using this repeatedly we see that $A^5\\mathbf{v}=(-1)^5\\mathbf{v}=-\\mathbf{v}$. Similarly, we have $A^5\\mathbf{w}=2^5\\mathbf{w}=32\\mathbf{w}$ using $A\\mathbf{w}=2\\mathbf{w}$. Combining these with (*) yields, \\begin{align*} A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ 9 \\end{bmatrix}&=3(-\\mathbf{v})-2(32\\mathbf{w})\\\\[6pt] &=-3\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix}-64\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}\\\\[6pt] &=\\begin{bmatrix} -3-128 \\\\ -6+64 \\\\ 3-192 \\end{bmatrix} =\\begin{bmatrix} -131 \\\\ 58 \\\\ -189 \\end{bmatrix}. \\end{align*}", "so we have \\begin{aligned} \\vec{u}^\\top \\, \\mathbf{R} \\, \\vec{u} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\mathbf{R} \\, \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -x \\\\ \\end{bmatrix} \\\\ {} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\begin{bmatrix} \\mathbf{R}_{1,1} - \\mathbf{R}_{1,4}\\,x \\\\ \\mathbf{R}_{2,1} - \\mathbf{R}_{2,4}\\,x \\\\ \\mathbf{R}_{3,1} - \\mathbf{R}_{3,4}\\,x \\\\ \\mathbf{R}_{4,1} - \\mathbf{R}_{4,4}\\,x \\\\ \\end{bmatrix} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - (\\mathbf{R}_{1,4} + \\mathbf{R}_{4,1})\\,x + \\mathbf{R}_{1,1} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - 2\\,\\mathbf{R}_{1,4}\\,x + \\mathbf{R}_{1,1} \\\\ {} & = 0 \\\\ x & = \\frac{\\mathbf{R}_{1,4} \\pm \\sqrt{\\mathbf{R}_{1,4}^2 - \\mathbf{R}_{1,1}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} Similarly, the $\\hat{y}$ and $\\hat{z}$ planes are given by \\begin{aligned} y & = \\frac{\\mathbf{R}_{2,4} \\pm \\sqrt{\\mathbf{R}_{2,4}^2 - \\mathbf{R}_{2,2}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}} \\\\ z & = \\frac{\\mathbf{R}_{3,4} \\pm \\sqrt{\\mathbf{R}_{3,4}^2 - \\mathbf{R}_{3,3}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} This is where the above linked posts stop, but a lot more simplification can be done. ## Implementation Details Several details of the problem can make computing the planes more efficient. The first is that $\\mathbf{S}$ is involutory, meaning $\\mathbf{S}^{-1} = \\mathbf{S}$. This means that the product $\\mathbf{M}\\,\\mathbf{S}^{-1}$ can be computed implicitly: it is simply $\\mathbf{M}$ with its last column negated. The last column of $\\mathbf{R} = \\mathbf{M} \\, \\mathbf{S}^{-1} \\, \\mathbf{M}^\\top$ is the same, because the last column of $\\mathbf{M}^\\top$ is $\\begin{bmatrix}0 & 0 & 0 &", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "of the congruence is the same. The only way that $(ab/p) \\equiv (a/p)(b/p) \\pmod p$ is when $(ab/p) = (a/p)(b/p)$ since p is restricted to being an odd prime. # Other Properties of Legendre Symbols There are many other rules that have been established regarding Legendre symbols, some of which are often necessary to evaluate Legendre symbols. We will state a few of them on this page and subsequently prove them all in future pages: (2) \\begin{align} (-1/p) = \\begin{Bmatrix} 1 & p \\equiv 1 \\pmod {4} \\\\ -1 & p \\equiv 3 \\pmod {4}\\\\ \\end{Bmatrix} \\end{align} (3) \\begin{align} (2/p) = \\begin{Bmatrix} 1 & p \\equiv 1 \\pmod {8} \\quad \\mathbf{or} \\quad p \\equiv 7 \\pmod {8}\\\\ -1 & p \\equiv 3 \\pmod {8} \\quad \\mathbf{or} \\quad p \\equiv 5 \\pmod {8}\\\\ \\end{Bmatrix} \\end{align} (4) \\begin{align} (3/p) = \\begin{Bmatrix} 1 & \\mathbf{if} \\quad p \\equiv 1 \\pmod {12} \\quad \\mathbf{or} \\quad p \\equiv 11 \\pmod {12}\\\\ -1 & \\mathbf{if} \\quad p \\equiv 5 \\pmod {12} \\quad \\mathbf{or} \\quad p \\equiv 7 \\pmod {12}\\\\ \\end{Bmatrix} \\end{align} (5) \\begin{align} (6/p) = \\begin{Bmatrix} 1 & \\mathbf{if} \\quad p \\equiv 1, 5, 19, \\: \\mathbf{or} \\: 23 \\pmod {24} \\\\ -1 & \\mathbf{if} \\quad p \\equiv 7, 11, 13, \\: \\mathbf{or} \\: 17 \\pmod {24} \\end{Bmatrix} \\end{align}", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "0 & 0 & 6120 & -12240 & 6120 \\\\ 0 & -12240 & 12240 & 0 & 0 \\\\ 6120 & 12240 & -12240 & 0 & 0 \\\\ -12240 & 0 & 0 & 0 & 0 \\\\ 6120 & 0 & 0 & 0 & 0 \\end{pmatrix},$$ $$G_{31} = \\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 15456 & 9792 \\\\ 0 & 0 & -30912 & -9792 & 0 \\\\ 0 & -15456 & -9792 & 0 & 0 \\\\ 0 & 9792 & 0 & 0 & 0 \\end{pmatrix},$$ $$G_{32} = \\begin{pmatrix} 0 & 0 & 5055 & 10425 & 0 \\\\ 0 & 3390 & -3915 & 0 & 0 \\\\ 5055 & -24085 & -3296 & 39168 & 0 \\\\ 10425 & -22880 & 19808 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix},$$$$g^\\,_{30}(u,v) \\ge 0,\\quad g^\\,_{31}(u,v) \\ge 0.$$ Since $$\\!\\mathstrut^{\\phantom{\\dfrac\\mathstrut\\mathstrut}^{\\LARGE=}}$$ $$\\!\\mathstrut^{\\phantom{\\dfrac\\mathstrut\\mathstrut}^{\\LARGE,}}$$ then, similarly to the first case, $$g^\\,_{32}(u,v)= \\begin{pmatrix} 1 \\\\ v \\\\ v^2 \\\\ v^3 \\end{pmatrix}^T \\begin{pmatrix} 0 & 0 & 5055 & 10425 \\\\ 0 & 3390 & -3915 & 0 \\\\ 5055 & -24085 & -3296 & 39168 \\\\ 10425 & -22880 & 19808 & 0 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ u \\\\", "\\end{pmatrix} &=& 0 \\\\\\\\ \\begin{pmatrix} 5+3t \\\\ 5+2t \\\\ 4-2t \\end{pmatrix} \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\\\ 15+9t+10+4t-8+4t &=& 0 \\\\ 17+17t &=& 0 \\\\ 17t &=& -17 \\quad | \\quad : 17 \\\\ \\mathbf{ t } &=& \\mathbf{ -1 } \\\\ \\hline \\end{array}$$ $$\\text{foot (of a perpendicular):}$$ $$\\begin{array}{|rcll|} \\hline \\vec{x}_{f} &=& \\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + (-1) \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} \\quad | \\quad t=-1 \\\\\\\\ &=& \\begin{pmatrix} 6-3 \\\\ 7-2 \\\\ 7+2 \\end{pmatrix} \\\\\\\\ \\mathbf{ \\vec{x}_{f} } &=& \\begin{pmatrix} \\mathbf{3} \\\\ \\mathbf{5} \\\\ \\mathbf{9} \\end{pmatrix} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\vec{PL} &=& \\vec{x}_{f} - \\vec{p} \\\\\\\\ &=& \\begin{pmatrix} 3\\\\ 5 \\\\ 9\\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} \\\\\\\\ &=& \\begin{pmatrix} 3-1\\\\ 5-2 \\\\ 9-3\\end{pmatrix} \\\\\\\\ \\mathbf{\\vec{PL}} &=& \\begin{pmatrix} \\mathbf{2}\\\\ \\mathbf{3} \\\\ \\mathbf{6}\\end{pmatrix} \\\\ \\hline \\end{array}$$ $$\\text{distance} = |\\vec{PL}|$$ $$\\begin{array}{|rcll|} \\hline |\\vec{PL}| &=& \\sqrt{2^2+3^2+6^2} \\\\ &=& \\sqrt{4+9+36} \\\\ &=& \\sqrt{49} \\\\ \\mathbf{|\\vec{PL}|} &=& \\mathbf{7} \\\\ \\hline \\end{array}$$ The distance from the point $$(1,2,3)$$ to the line is 7. Sep 6, 2019 #1 +25228 +3 Find the distance from the point $$(1,2,3)$$ to the line described by $$\\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix}$$. $$\\text{Let point \\vec{p} = \\begin{pmatrix} 1 \\\\ 2", "are perpendicular . If $\\mathbf{a}$ and $\\mathbf{b}$ are perpendicular, then $\\theta=90^\\circ$, $\\cos \\theta = 0$ and $\\mathbf{a} \\cdot \\mathbf{b} = 0$. Conversely, if $\\mathbf{a}\\cdot \\mathbf{b}=0$, then either $\\theta=90^\\circ$, $\\mathbf{a}=\\mathbf{0}$ or $\\mathbf{b}=\\mathbf{0}$. ### Solution to examples: $\\mathbf{a}\\cdot \\mathbf{b} = |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$ $\\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix} \\cdot \\begin{pmatrix} 5 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\left | \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix} \\right | \\, \\left | \\begin{pmatrix} 5 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right | \\cos \\theta$ $5+0-3 = \\sqrt{14} \\sqrt{26} \\cos \\theta$ $\\theta = \\cos^{-1} \\left ( \\frac{2}{\\sqrt{14} \\sqrt{26}} \\right ) =$ Since $\\mathbf{a}$ is perpendicular to $\\mathbf{c}$, $\\mathbf{a} \\cdot \\mathbf{c} = 0$ $-1 + 2 - 3z = 0$ $\\mathbf{a} \\cdot \\mathbf{d} = k+2 + 2(k-1) -3k = k+2+2k-2-3k = 0$. Hence $\\mathbf{a}$ is perpendicular to $\\mathbf{d}$ for all $k \\in \\mathbb{R}$.", "of $${\\mathbf z}$$ with respect to the basis $$\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}$$ are $$(2, -3)\\text{.}$$ Suppose we wish to convert the coordinates with repect to one basis to a new set of coordinates with respect to a different basis; that is, we wish to do a change of coordinates. Observe that \\begin{align*} \\mathbf v_1 & = 2 \\mathbf e_1 + \\mathbf e_2\\\\ \\mathbf v_2 & = 3 \\mathbf e_1 + 2 \\mathbf e_2. \\end{align*} It follows that \\begin{align*} c_1 \\mathbf v_1 + c_2 \\mathbf v_2 & = c_1 (2 \\mathbf e_1 + \\mathbf e_2) + c_2( 3\\mathbf e_1 + 2 \\mathbf e_2)\\\\ & = (2c_1 + 3c_2 )\\mathbf e_1 + (c_2 + 2c_2) \\mathbf e_2. \\end{align*} Thus, the coordinates of $$c_1 \\mathbf v_1 + c_2 \\mathbf v_2$$ with respect to the basis $$\\{ {\\mathbf e}_1, {\\mathbf e}_2 \\}$$ can be determined by \\begin{equation*} \\begin{pmatrix} 2c_1 + 3c_2 \\\\ c_2 + 2c_2 \\end{pmatrix} = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\begin{pmatrix} c_1 \\\\ c_2 \\end{pmatrix}. \\end{equation*} If we let \\begin{equation*} T = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf c = \\begin{pmatrix} c_1 \\\\ c_2 \\end{pmatrix}, \\end{equation*} then the coordinates with respect to the basis $$\\{ {\\mathbf e}_1, {\\mathbf e}_2 \\}$$ are given by $$\\mathbf d = T", "\\\\ {6} & {1} & {9} \\\\ {4} & {-3} & {2} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {2} & {-1} & {3} \\\\ {1} & {2} & {5} \\\\ {-3} & {-5} & {2} \\end{bmatrix}\\nonumber$ determine $$\\mathbf{ A}+\\mathbf{ B}$$ and $$\\mathbf{ A}-3 \\mathbf{ B}$$. 23. Given the following $$1\\times 4$$ row matrices $\\mathbf{ a} = \\begin{bmatrix} {3} & {1} & {5} & {6} \\end{bmatrix}, \\quad \\mathbf{ b} = \\begin{bmatrix} {6} & {-2} & {0} & {4} \\end{bmatrix}\\nonumber$ determine $$2\\mathbf{ a}+\\mathbf{ b}$$. 24.$$\\ddagger$$ Compute the $$4\\times 4$$ matrix defined by $$\\mathbf{ A}-\\mathbf{ B}+\\mathbf{ C}$$ where $$\\mathbf{A}$$, $$\\mathbf{B}$$ and $$\\mathbf{C}$$ are given by $\\mathbf{ A} = \\begin{bmatrix} {0.856} & {0.328} & {0.0663} & {0.908} \\\\ {0.529} & {0.426} & {0.364} & {0.434} \\\\ {0.506} & {0.812} & {0.400} & {0.137} \\\\ {0.652} & {0.402} & {0.0276} & {0.995} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {1.142} & {0.438} & {0.0884} & {1.211} \\\\ {0.705} & {0.568} & {0.485} & {0.579} \\\\ {0.675} & {1.083} & {0.534} & {0.183} \\\\ {0.869} & {0.535} & {0.368} & {1.326} \\end{bmatrix}\\nonumber$ $\\mathbf{ C} = \\begin{bmatrix} {0.285} & {0.109} & {0.0221} & {0.303} \\\\ {0.176} & {0.142} & {0.121} & {0.145} \\\\ {0.169} & {0.271} & {0.133} & {0.0457} \\\\ {0.217} & {0.134} & {0.0092} & {0.332} \\end{bmatrix}\\nonumber$ 25.$$\\ddagger$$ Consider a project in which all" ]

[ "+0 # Let be an acute angle, and let Express in terms of ​ 0 47 1 Let $$\\theta$$ be an acute angle, and let $$\\sin \\frac{\\theta}{2} = \\sqrt{\\frac{x - 1}{2x}}$$ Express $$\\tan \\theta$$ in terms of $$x$$ Apr 15, 2020", "# Trig problem 3 Geometry Level pending Let $$\\theta$$ be an acute angle. The equation $$x^{ 2 }-(tan\\theta +\\frac { 1 }{ tan\\theta } )x+1=0$$ has one root $$2+\\sqrt { 3 }$$. What is the numerical value of $$sin\\theta cos\\theta$$? ×", "acute angle $\\theta$, the side opposite to the angle will be $y$ and the side adjacent the angle will be $x$. By Pythagoras, the hypotenuse will be $\\sqrt{x^2 + y^2}$. Now $\\sin \\theta = \\frac {\\mbox {Opposite}}{\\mbox {Hypotenuse}} = \\frac y{\\sqrt {x^2 + y^2}}$ plugging this in for $\\sin \\theta$ in our answer, we find that: $\\int \\frac {x~dy}{\\left( x^2 + y^2 \\right)^{3/2}} = \\frac y{x \\sqrt {x^2 + y^2}} + C$ as desired 5. This integral's getting interesting, look at here.", "If $\\theta$ be an acute angle and $7\\sin ^{2}\\theta+3\\cos ^{2}\\theta = 4,$ then the value of $\\tan \\theta$ is: [A] $0$ [B] $\\frac{1}{\\sqrt{3}}$ [C] $1$ [D] $\\sqrt{3}$", "# 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $(vi)\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}= \\sec A+\\tan A$ M manish We need to prove - $\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}= \\sec A+\\tan A$ Taking LHS; By rationalising the denominator, we get; $\\\\= \\sqrt{\\frac{1+\\sin A}{1-\\sin A}\\times \\frac{1+\\sin A}{1+\\sin A}}\\\\\\\\ = \\sqrt{\\frac{(1+\\sin A)^2}{1-\\sin^2A}}\\\\\\\\ =\\frac{1+\\sin A}{\\cos A}\\\\\\\\ = \\sec A + \\tan A\\\\\\\\ = RHS$ Hence proved. Exams Articles Questions", "But, I still do not know how to convert an obtuse angle to an acute angle. So let's pretend that I have a an angle, say: cos$$\\theta$$ = $$\\frac{-10}{\\sqrt{22}\\sqrt{42}}$$. How do I go about converting this angle into an acute angle without using a calculator? 5. Jun 9, 2010 ### vela Staff Emeritus It has nothing to do with the concept of acute angles. It's just algebra. Use the rules rock.freak posted to simplify the expression.", "right triangle $ABC$ has an acute angle $\\theta$. The opposite side has length 3, adjacent length 4. Find $\\sin(\\theta)$. 16. Sketch a triangle and it's acute angle $\\theta$ that has $\\tan(\\theta) = \\frac{10}{8}$. Label all three sides. 17. A right triangle ABC has one acute angle $45^\\circ$. The hypotenuse has length $16\\sqrt{2}$. Solve the triangle.", "where $t=\\tan \\theta$, follow directly from the ratios of the sides of the right angled triangle $ADC$ and we get $$\\tan2\\theta = {2t\\over {1-t^2}},\\quad \\sin2\\theta = {2t\\over {1+t^2}},\\quad \\cos2\\theta = {{1-t^2}\\over {1+t^2}}$$", "# If the acute angle between the curves $xy=2$ and $y^2 = 4x$ is $\\theta$, then $\\tan \\theta$ is equal to : ( A ) $\\frac{1}{3}$ ( C ) $\\frac{2}{3}$", "# Sine Sine Cosine Cosine, I got it! If $x$ and $y$ are acute angles such that $\\frac {\\sin x}{\\sin y } = \\frac {1}{2}, \\quad \\frac {\\cos x}{\\cos y } = \\frac 3 2 ,$ what is $\\tan^2 (x+y)?$ ×", "# Can I use proportion to my advantage? Geometry Level 2 $\\large \\frac { \\sin ^{ 2 }{ \\theta } }{ 5 } =\\frac { \\cos ^{ 2 }{ \\theta } }{ 6 }$ If $$\\theta$$ is a positive acute angle that satisfies the equation above, find $$\\sin { \\theta }$$. Note: Give your answer to 3 decimal places. ×", "If theta is an acute angle and sintheta=1/2, then what is costheta? Jan 3, 2017 $\\frac{\\sqrt{3}}{3}$ ${\\cos}^{2} a + {\\sin}^{2} a = 1$ ${\\cos}^{2} t = 1 - {\\sin}^{2} t = 1 - \\frac{1}{4} = \\frac{3}{4}$ $\\cos t = \\pm \\frac{1}{\\sqrt{3}} = \\pm \\frac{\\sqrt{3}}{3}$ $\\cos t = \\frac{\\sqrt{3}}{3}$", "# Write the acute angle θ satisfying 3–√ sin θ=cos θ. Question: Write the acute angle $\\theta$ satisfying $\\sqrt{3} \\sin \\theta=\\cos \\theta$. Solution: We have: $\\sqrt{3} \\sin \\theta=\\cos \\theta$ $\\Rightarrow \\sqrt{3} \\sin \\theta=\\cos \\theta$ $\\Rightarrow \\frac{\\sin \\theta}{\\cos \\theta}=\\frac{1}{\\sqrt{3}}$ $\\Rightarrow \\tan \\theta=\\tan 30^{\\circ}$ $\\Rightarrow \\theta=30^{\\circ}$ Hence the acute angle is", "\\frac{\\cos(2\\theta)}{\\sin(2\\theta)} = \\tan(\\alpha) \\\\[9pt] \\cot(2\\theta) = \\tan(\\alpha)$ All the angles are acute angles, so this tells us that $2\\theta$ and $\\alpha$ are complements. (Think about the two acute angles in a right triangle. The adjacent for one is the opposite of the other, and vice versa, so opposite/adjacent for one is adjacent/opposite for the other.) Thus $2\\theta = 90^\\circ - \\alpha$ and the optimum firing angle is $\\theta = 45^\\circ - \\frac{\\alpha}{2}$.", "# Can I use proportion to my advantage? Geometry Level 2 $\\large \\frac { \\sin ^{ 2 }{ \\theta } }{ 5 } =\\frac { \\cos ^{ 2 }{ \\theta } }{ 6 }$ If $\\theta$ is a positive acute angle that satisfies the equation above, find $\\sin { \\theta }$.", "is to express $\\sin \\theta$ in terms of $\\cos 2 \\theta$ $\\cos 2 \\theta = 1 - 2 \\sin^2 \\theta$ now solve for $\\sin \\theta$, when you're finished, replace all the $\\theta$'s with $\\frac x2$ and viola! (you'll realize this is just the half angle formula. your sly professor just had you derive it yourself)", "## anonymous 4 years ago Find sin θ, sec θ, and tan θ for an angle in standard position having this point on its terminal side: (/[−3,2sqrt{2}]) 1. anonymous Find sin θ, sec θ, and tan θ for an angle in standard position having this point on its terminal side: $(-3, 2\\sqrt{2})$ 2. Mertsj |dw:1328498353911:dw| 3. Mertsj $\\sin \\theta=\\frac{2\\sqrt{2}}{\\sqrt{17}}$ 4. Mertsj $\\tan \\theta=-\\frac{2\\sqrt{2}}{3}$ 5. Mertsj $\\sec \\theta=-\\frac{\\sqrt{17}}{3}$", "Thread: Express in terms of cosine. 1. Express in terms of cosine. I need some help with this one... $(1 - \\tan^2\\Theta)(\\sec^2\\Theta)$ 2. Hello, Originally Posted by Intrusion I need some help with this one... $(1 - \\tan^2\\Theta)(\\sec^2\\Theta)$ Note that $1+\\tan^2 \\theta=\\sec^2 \\theta$ So $1-\\tan^2 \\theta=2-\\sec^2 \\theta$ Hence $(1-\\tan^2 \\theta)(\\sec^2 \\theta)=2 \\sec^2 \\theta-\\sec^4 \\theta=\\frac{2}{\\cos^2 \\theta}-\\frac{1}{\\cos^4 \\theta}$", "Half Angle Formula Required? Geometry Level 3 For $\\theta\\in\\left(0,\\frac{\\pi}{2}\\right)$, which of the following is equivalent to the expression given below? $\\frac{\\tan\\theta+\\sec\\theta-1}{\\tan\\theta-\\sec\\theta+1}$ This problem is part of the set Trigonometry. ×", "tan70013' 46. Find the value of (i) sin38036' + tan12012' (ii) tan60025' - cos 49020' 47. Find the area of the right angled triangle with hypotenuse 5cm and one of the acute angle is 48030' 48. 2 x 5 = 10 49. Find the values of the following: (i) (cos00 + sin450 + sin300)(sin900 - cos450 + cos600) (ii) tan2600 - 2tan2450 - cot2300 +2sin2300$\\frac { 3 }{ 4 }$ cosec2 450 50. Find the value of $\\theta$ if (i) sin $\\theta$ = 0.9858 (ii)tan $\\theta$ = 0.5902 (iii)cos$\\theta$ = 07656" ]

[ "are 1, 1, 2. If θ is the angle between them, they can be written as: $\\cos \\theta =\\vert \\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\\cdot \\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\\vert$ $\\cos \\theta =\\vert \\frac{3\\cdot 1+5\\cdot 1+4\\cdot 2}{\\sqrt{3^{2}+5^{2}+4^{2}}\\cdot \\sqrt{1+1+2^{2}}}\\vert =\\frac{8\\sqrt{3}}{15}$ Hence we can find the required acute angle by taking $\\cos ^{-1}\\frac{8\\sqrt{3}}{15}$", "# Math Help - Double angle formula 1. ## Double angle formula Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks 2. ## Re: Double angle formula Originally Posted by Hooperoo Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks $\\frac{1}{\\sin 2\\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}=\\frac{1-(cos^2 \\theta-\\sin^2 \\theta)}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{\\sin^2 \\theta + \\cos^2 \\theta -\\cos^2 \\theta +\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{2\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=\\frac{\\sin \\theta}{\\cos \\theta}=\\tan \\theta$ 3. ## Re: Double angle formula Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@! 4. ## Re: Double angle formula Hello, Hooperoo! You are expected to know these identities: . . $\\sin2\\theta \\:=\\:2\\sin\\theta\\cos\\theta$ . . $\\sin^2\\!\\theta \\:=\\:\\frac{1-\\cos2\\theta}{2} \\quad\\Rightarrow\\quad 1 - \\cos2\\theta \\:=\\:2\\sin^2\\!\\theta$ Using double-angle formulas, verify that: . $\\csc2\\theta - \\cot2\\theta \\:=\\:\\tan\\theta$ We have: . $\\csc2\\theta - \\cot2\\theta \\;=\\;\\frac{1}{\\sin2\\theta} - \\frac{\\cos2\\theta}{\\sin2\\theta} \\;=\\;\\frac{1 - \\cos2\\theta}{\\sin2\\theta}$ . . . . . . . . . . . . . . . . . . . $\\;=\\;\\frac{2\\sin^2\\!\\theta}{2\\sin\\theta\\cos\\theta} \\;=\\;\\frac{\\sin\\tjeta}{\\cos\\theta} \\;=\\;\\tan\\theta$", "shape, and we already said that the angle $\\theta=\\frac{2\\pi}{n}$, so $$\\text{area of inscribed n-gon} = n\\cdot\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}$$ ### Area Computation Part 2 We now go back to our definition for $Q_n$ $$Q_n = \\frac{\\text{area of } 2^n\\text{-gon}}{\\text{area of } 2^{n+1}\\text{-gon}}$$ $$Q_n = \\frac{ 2^n\\cdot\\sin\\frac{\\pi}{2^n}\\cos\\frac{\\pi}{2^n}}{2^{n+1}\\cdot\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}}$$ $$= \\frac{\\sin\\frac{\\pi}{2^n}\\cos\\frac{\\pi}{2^n}}{2\\cdot\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}}$$ Now we need to make this formula look a little nicer. We can do this by using the trivial fact that $$\\frac{\\pi}{2^n} = 2\\cdot\\frac{\\pi}{2^{n+1}}$$ $$Q_n = \\frac{\\sin\\frac{2\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n}}}{2\\cdot\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}}$$ Now use the double angle formula for $\\sin$ $$\\sin2\\theta = 2\\sin\\theta\\cos\\theta$$ $$\\sin\\frac{2\\pi}{2^{n+1}} = 2\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}$$ $$Q_n = \\frac{2\\cdot\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n}}}{2\\cdot\\sin\\frac{\\pi}{2^{n+1}}\\cos\\frac{\\pi}{2^{n+1}}}$$ The ugly stuff cancels out wonderfully, leaving us with $$Q_n= \\cos\\frac{\\pi}{2^{n}}$$ Therefore $$\\bbox[15px,border:1px solid black]{\\frac{2}{\\pi} = \\cos\\frac{\\pi}{2} \\cos\\frac{\\pi}{4} \\cos\\frac{\\pi}{8}\\dots}$$ ### Reformulation The issue with the formula above is that, well, it uses $\\pi$ to compute $\\pi$. This is criminal circular logic. However, we can easily remedy this. Let’s look at the double angle formula for $\\cos$ $$\\cos 2\\theta = 2\\cos^2\\theta – 1$$ $$\\cos\\theta = 2\\cos^2\\frac{\\theta}{2} -1$$ $$\\cos\\frac{\\theta}{2} = \\sqrt{\\frac{1+\\cos\\theta}{2}}$$ $$\\cos\\frac{\\theta}{2} = \\frac{\\sqrt{2+2\\cos\\theta}}{2}$$ Using this, and starting from $\\cos\\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2}$ yields $$\\cos\\frac{\\pi}{4}=\\frac{\\sqrt{2}}{2}\\hspace{20mm}\\cos\\frac{\\pi}{8}=\\frac{\\sqrt{2+\\sqrt{2}}}{2}\\hspace{20mm}\\cos\\frac{\\pi}{16}=\\frac{\\sqrt{2+\\sqrt{2+\\sqrt{2}}}}{2}$$ Therefore, we can finish off this post, having derived Viète’s formula. $$\\bbox[15px,border:1px solid black]{\\frac{2}{\\pi} = \\frac{\\sqrt{2}}{2}\\cdot \\frac{\\sqrt{2+\\sqrt{2}}}{2}\\cdot \\frac{\\sqrt{2+\\sqrt{2+\\sqrt{2}}}}{2}\\cdot\\frac{\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2}}}}}{2}\\dots}$$ Comment", "= 2 · cos x · sin((n − 1)x) − sin((n − 2)x). This can be proved by adding formulae for sin((n − 1)x + x) and sin((n − 1)xx). Serving a purpose similar to that of the Chebyshev method, for the tangent we can write: ${\\displaystyle \\tan(nx)={\\frac {\\tan((n-1)x)+\\tan x}{1-\\tan((n-1)x)\\tan x}}\\,.}$ Tangent of an average ${\\displaystyle \\tan \\left({\\frac {\\alpha +\\beta }{2}}\\right)={\\frac {\\sin \\alpha +\\sin \\beta }{\\cos \\alpha +\\cos \\beta }}=-\\,{\\frac {\\cos \\alpha -\\cos \\beta }{\\sin \\alpha -\\sin \\beta }}}$ Setting either α or β to 0 gives the usual tangent half-angle formulae. Viète's infinite product ${\\displaystyle \\cos {\\frac {\\theta }{2}}\\cdot \\cos {\\frac {\\theta }{4}}\\cdot \\cos {\\frac {\\theta }{8}}\\cdots =\\prod _{n=1}^{\\infty }\\cos {\\frac {\\theta }{2^{n}}}={\\frac {\\sin \\theta }{\\theta }}=\\operatorname {sinc} \\theta .}$ (Refer to sinc function.) Power-reduction formulae Obtained by solving the second and third versions of the cosine double-angle formula. Sine Cosine Other ${\\displaystyle \\sin ^{2}\\theta ={\\frac {1-\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\cos ^{2}\\theta ={\\frac {1+\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\sin ^{2}\\theta \\cos ^{2}\\theta ={\\frac {1-\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{3}\\theta ={\\frac {3\\sin \\theta -\\sin(3\\theta )}{4}}}$ ${\\displaystyle \\cos ^{3}\\theta ={\\frac {3\\cos \\theta +\\cos(3\\theta )}{4}}}$ ${\\displaystyle \\sin ^{3}\\theta \\cos ^{3}\\theta ={\\frac {3\\sin(2\\theta )-\\sin(6\\theta )}{32}}}$ ${\\displaystyle \\sin ^{4}\\theta ={\\frac {3-4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\cos ^{4}\\theta ={\\frac {3+4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{4}\\theta \\cos ^{4}\\theta ={\\frac {3-4\\cos(4\\theta )+\\cos(8\\theta )}{128}}}$ ${\\displaystyle \\sin ^{5}\\theta ={\\frac {10\\sin \\theta -5\\sin(3\\theta", "\\left(\\frac{1}{3}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2} + \\frac{1}{7}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}+\\quad \\cdots\\right) ### Transformation to current notation Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get • \\theta = \\tan \\theta - \\frac{\\tan^3 \\theta}{3} + \\frac{\\tan^5\\theta}{5} - \\frac{\\tan^7 \\theta}{7} + \\quad \\cdots . Letting tan θ = q we finally have • \\tan^{-1} q = q - \\frac{q^3}{3} + \\frac{q^5}{5} - \\frac{q^7}{7} + \\quad \\cdots ### Another formula for the circumference of a circle The second part of the quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows. c= \\sqrt{12 d^2} - \\frac{\\sqrt{12 d^2}}{3\\cdot 3} + \\frac{\\sqrt{12 d^2}}{3^2 \\cdot 5} - \\frac{\\sqrt{12 d^2}}{3^3 \\cdot 7}+ \\quad \\cdots Since c = π d this can be reformulated as a formula to compute π as follows. \\pi = \\sqrt{12}\\left( 1 - \\frac{1}{3\\cdot3}+\\frac{1}{3^2\\cdot 5} -\\frac{1}{3^3\\cdot 7} +\\quad \\cdots\\right) This is obtained by substituting q = 1/\\sqrt{3} (therefore θ = π / 6) in the power series expansion for tan−1 q above. Comparison of the convergence of two Madhava series", "evidently angle $\\theta$ is twice angle IOP. So using the double-angle formula for tangent, $$\\tan\\theta = \\frac{\\sqrt5-1}{1-\\left(\\frac{\\sqrt5-1}2\\right)^2} = \\frac{\\sqrt5-1}{1-\\frac{6-2\\sqrt5}4} = \\frac{\\sqrt5-1}{\\frac{\\sqrt5-1}2}=2$$ as desired.", "# Double angle formula • Mar 11th 2012, 11:01 PM Hooperoo Double angle formula Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks :) • Mar 11th 2012, 11:19 PM princeps Re: Double angle formula Quote: Originally Posted by Hooperoo Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks :) $\\frac{1}{\\sin 2\\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}=\\frac{1-(cos^2 \\theta-\\sin^2 \\theta)}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{\\sin^2 \\theta + \\cos^2 \\theta -\\cos^2 \\theta +\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{2\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=\\frac{\\sin \\theta}{\\cos \\theta}=\\tan \\theta$ • Mar 11th 2012, 11:37 PM Hooperoo Re: Double angle formula Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@! • Mar 12th 2012, 05:44 AM Soroban Re: Double angle formula Hello, Hooperoo! You are expected to know these identities: . . $\\sin2\\theta \\:=\\:2\\sin\\theta\\cos\\theta$ . . $\\sin^2\\!\\theta \\:=\\:\\frac{1-\\cos2\\theta}{2} \\quad\\Rightarrow\\quad 1 - \\cos2\\theta \\:=\\:2\\sin^2\\!\\theta$ Quote: Using double-angle formulas, verify that: . $\\csc2\\theta - \\cot2\\theta \\:=\\:\\tan\\theta$ We have: . $\\csc2\\theta - \\cot2\\theta \\;=\\;\\frac{1}{\\sin2\\theta} - \\frac{\\cos2\\theta}{\\sin2\\theta} \\;=\\;\\frac{1 - \\cos2\\theta}{\\sin2\\theta}$ . . . . . . . . . . . .", "2 $\\theta$ ) = $\\frac{cot^{2}\\theta\\;-\\;1}{2cot\\;\\theta}$ ### Triple Angle formula: 1. Sin ( 3 $\\theta$ ) = 3 Sin $\\theta$ - 4 Sin3 $\\theta$ 2. Cos ( 3 $\\theta$ ) = 4 Cos3 $\\theta$ - 3 Cos $\\theta$ 3. Tan ( 3 $\\theta$ ) = $\\frac{3tan\\;\\theta-tan^{3}\\theta}{1-3tan^{2}\\theta}$ 4. Cot ( 3 $\\theta$ ) = $\\frac{cot^{3}\\;\\theta-3cot\\;\\theta}{3cot^{2}\\;\\theta-1}$ ### Half Angle Formula: 1. Sin ( $\\theta$ ) = 2 Sin $\\frac{\\theta}{2}$. Cos $\\frac{\\theta}{2}$ Sin ( $\\theta$ ) = $\\frac{2tan\\;\\frac{\\theta}{2}}{1-tan^{2}\\;\\frac{\\theta}{2}}$ 2. Cos ( $\\theta$ ) = Cos2 $\\frac{\\theta}{2}$ - Sin2 $\\frac{\\theta}{2}$ Cos ( $\\theta$ ) = 1 - 2 Sin2 $\\frac{\\theta}{2}$ Cos ( $\\theta$ ) = 2 Cos2 $\\frac{\\theta}{2}$ - 1 Cos ( $\\theta$ ) = $\\frac{1-tan^{2}\\;\\frac{\\theta}{2}}{1+tan^{2}\\;\\frac{\\theta}{2}}$ 3. Tan ( $\\theta$ ) = $\\frac{2tan\\;\\frac{\\theta}{2}}{1-tan^{2}\\;\\frac{\\theta}{2}}$ 4. Cot ( $\\theta$ ) = $\\frac{cot^{2}\\frac{\\theta }{2}-1}{2cot\\; \\frac{\\theta }{2}}$ ### One Third Angle Formula: 1. Sin $\\theta$ = 3 sin ( $\\frac{\\theta}{3}$ ) - 4 sin3 ( $\\frac{\\theta}{3}$) 2. Cos $\\theta$ = 4 cos3 ( $\\frac{\\theta}{3}$ ) - 3 cos ($\\frac{\\theta}{3}$ ) 3. Tan $\\theta$ = $\\frac{3tan\\;\\frac{\\theta }{3}-tan^{3}\\;\\frac{\\theta }{3}}{1-3tan^{2}\\;\\frac{\\theta }{3}}$ ## Sum to Product Identity For the two angles A and B, 1. Sin A + Sin B = 2 Sin $\\frac{A+B}{2}$ . Cos $\\frac{A-B}{2}$ 2. Sin A - Sin B = 2 Cos $\\frac{A+B}{2}$ . Sin $\\frac{A-B}{2}$ 3. Cos A + Cos B = 2 Cos $\\frac{A+B}{2}$ .", "\\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$.", "is $\\theta$. Then $\\tan \\theta = \\frac HL$ so the $L$ I used in the moment is the same as $\\frac H{\\tan \\theta}$ – Ross Millikan Nov 12 '12 at 20:31 Your calculation is correct, as far as the force calculation goes. It's useful to note that $\\sin(\\theta)\\cdot\\cos(\\theta)=\\frac12\\sin(2\\theta)$, and that is maximal when $\\theta=\\pi/4=45^\\circ$. Thus $L=H$ is optimal.", "angle $\\theta$. This should give you a hint that you will probably have to use a double-angle identity. Perhaps: $2\\cot(2\\theta) = 2\\frac{1- \\tan^2(x)}{2\\tan(x)}$ Or as mr. fantastic pointed out one of the many other double angle identities (his, in fact, is easier to apply). Arriving at the conclusion that 1 = 1 by starting with -1 = 1 and manipulting both sides of the equation is wrong. The proof should go like this: $\\cot(\\theta)-\\tan(\\theta)$ (write down left hand side of equation) $= \\frac{\\cos(\\theta)}{\\sin(\\theta)} - \\frac{\\sin(\\theta)}{\\cos(\\theta)}$ (by the fact that $\\tan(x) = \\sin(x)/\\cos(x) \\text{ and } \\cot(x) = 1/\\tan(x)$) $= \\frac{\\cos^2(\\theta)-\\sin^2(\\theta)}{\\sin(\\theta)\\cos(\\theta)}$ (by finding a common denominator and simplifying) $= \\frac{\\cos(2\\theta)}{1/2\\sin(2\\theta)}$ (by the identities mr. fantastic gave) $= 2\\cot(2\\theta)$ (end with the right hand side of the equation) This is fundamentally different from the approach you took. $\\begin{array}{l} 2\\cot (2\\theta ) \\\\ 2\\frac{{\\cos (2\\theta )}}{{\\sin (2\\theta )}} \\\\ \\frac{{\\cos ^2 (\\theta ) - \\sin ^2 (\\theta )}}{\\begin{array}{l} \\sin (\\theta )\\cos (\\theta ) \\\\ \\frac{{\\cos (\\theta )}}{{\\sin (\\theta )}} - \\frac{{\\sin (\\theta )}}{{\\cos (\\theta )}} \\\\ \\end{array}} \\\\ \\end{array} $ 8. Well. Whose method should I go with. Yours or what I posted? 9. Doesn't matter. They both work. You proved that one side is equivalent to the other. Doesn't matter which side you start with. iknowone started with the left,", "## Precalculus (6th Edition) $\\displaystyle \\cos\\theta=-\\frac{\\sqrt{42}}{12}$ $\\displaystyle \\sin\\theta=\\frac{\\sqrt{102}}{12}$ $\\theta$ terminates in Q.II, where its cosine is negative, its sine is positive. Double-Angle Identity: $\\cos 2\\theta=2\\cos^{2}\\theta- \\mathrm{l}$ $-\\displaystyle \\frac{5}{12}+1=2\\cos^{2}\\theta$ $\\displaystyle \\frac{7}{12}=2\\cos^{2}\\theta$ $\\displaystyle \\cos^{2}\\theta=\\frac{7}{24}$ ... $\\cos\\theta$ is negative, $\\displaystyle \\cos\\theta=-\\sqrt{\\frac{7}{24}}=-\\frac{\\sqrt{7}}{2\\sqrt{6}}\\cdot\\frac{\\sqrt{6}}{\\sqrt{6}}=-\\frac{\\sqrt{42}}{12}$ Pythagorean Identity ($\\sin\\theta$ is positive): $\\sin\\theta=+\\sqrt{1-\\cos^{2}\\theta}$ $=\\displaystyle \\sqrt{1-\\frac{7}{24}}= \\sqrt{\\frac{17}{24}}=\\frac{\\sqrt{17}}{2\\sqrt{6}}\\cdot\\frac{\\sqrt{6}}{\\sqrt{6}}=\\frac{\\sqrt{102}}{12}$", "or the multiple-angle formulae. SineCosineTangentCotangent Double-angle formulae[28][29] {\\displaystyle {\\begin{aligned}\\sin(2\\theta )&=2\\sin \\theta \\cos \\theta \\ \\\\&={\\frac {2\\tan \\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(2\\theta )&=\\cos ^{2}\\theta -\\sin ^{2}\\theta \\\\&=2\\cos ^{2}\\theta -1\\\\&=1-2\\sin ^{2}\\theta \\\\&={\\frac {1-\\tan ^{2}\\theta }{1+\\tan ^{2}\\theta }}\\end{aligned}}} ${\\displaystyle \\tan(2\\theta )={\\frac {2\\tan \\theta }{1-\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(2\\theta )={\\frac {\\cot ^{2}\\theta -1}{2\\cot \\theta }}}$ Triple-angle formulae[23][30] {\\displaystyle {\\begin{aligned}\\sin(3\\theta )\\!&=\\!-\\sin ^{3}\\theta \\!+\\!3\\cos ^{2}\\theta \\sin \\theta \\\\&=-4\\sin ^{3}\\theta +3\\sin \\theta \\end{aligned}}} {\\displaystyle {\\begin{aligned}\\cos(3\\theta )\\!&=\\!\\cos ^{3}\\theta \\!-\\!3\\sin ^{2}\\theta \\cos \\theta \\\\&=4\\cos ^{3}\\theta -3\\cos \\theta \\end{aligned}}} ${\\displaystyle \\tan(3\\theta )={\\frac {3\\tan \\theta -\\tan ^{3}\\theta }{1-3\\tan ^{2}\\theta }}}$ ${\\displaystyle \\cot(3\\theta )\\!=\\!{\\frac {3\\cot \\theta \\!-\\!\\cot ^{3}\\theta }{1\\!-\\!3\\cot ^{2}\\theta }}}$ Half-angle formulae[26][27] {\\displaystyle {\\begin{aligned}&\\sin {\\frac {\\theta }{2}}=\\operatorname {sgn}(A)\\,{\\sqrt {\\frac {1\\!-\\!\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,A=2\\pi -\\theta +4\\pi \\left\\lfloor {\\frac {\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left({\\text{or}}\\,\\,\\sin ^{2}{\\frac {\\theta }{2}}={\\frac {1-\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\cos {\\frac {\\theta }{2}}=\\operatorname {sgn}(B)\\,{\\sqrt {\\frac {1+\\cos \\theta }{2}}}\\\\\\\\&{\\text{where}}\\,B=\\pi +\\theta +4\\pi \\left\\lfloor {\\frac {\\pi -\\theta }{4\\pi }}\\right\\rfloor \\\\\\\\&\\left(\\mathrm {or} \\,\\,\\cos ^{2}{\\frac {\\theta }{2}}={\\frac {1+\\cos \\theta }{2}}\\right)\\end{aligned}}} {\\displaystyle {\\begin{aligned}\\tan {\\frac {\\theta }{2}}&=\\csc \\theta -\\cot \\theta \\\\&=\\pm \\,{\\sqrt {\\frac {1-\\cos \\theta }{1+\\cos \\theta }}}\\\\[8pt]&={\\frac {\\sin \\theta }{1+\\cos \\theta }}\\\\[8pt]&={\\frac {1-\\cos \\theta }{\\sin \\theta }}\\\\[10pt]\\tan {\\frac {\\eta +\\theta }{2}}\\!&={\\frac {\\sin \\eta +\\sin \\theta }{\\cos \\eta +\\cos \\theta }}\\\\[8pt]\\tan \\left(\\!{\\frac {\\theta }{2}}\\!+\\!{\\frac {\\pi }{4}}\\!\\right)\\!&=\\!\\sec \\theta \\!+\\!\\tan \\theta \\\\[8pt]{\\sqrt {\\frac {1-\\sin \\theta }{1+\\sin \\theta }}}&={\\frac {|1-\\tan {\\frac {\\theta }{2}}|}{|1+\\tan {\\frac {\\theta }{2}}|}}\\\\[8pt]\\tan {\\frac {\\theta }{2}}\\!&=\\!{\\frac {\\tan", "Angle Formulas sinθ = $\\pm \\sqrt{\\frac{1-cos2\\theta }{2}}$ cosθ = $\\pm \\sqrt{\\frac{1+cos2\\theta }{2}}$ tanθ = $\\pm \\sqrt{\\frac{1-cos2\\theta }{1+cos2\\theta}}$ ### Thrice of Angle Formulas sin3θ = 3sinθ – 4 sin3θ Cos 3θ = 4cos3θ – 3 cosθ Tan 3θ = $\\frac{3 tan\\theta – tan^3\\theta }{1-3tan^2\\theta }$ Cot 3θ = $\\frac{cot^3\\theta – 3cot\\theta }{3cot^2\\theta-1 }$ ### Sum and Difference Formulas Sin (A+B) = Sin A Cos B + Cos A Sin B Sin (A-B) = Sin A Cos B – Cos A Sin B Cos (A+B) = Cos A Cos B – Sin A Sin B Cos (A-B) = Cos A Cos B + Sin A Sin B Tan (A+B) = $\\frac{Tan A + Tan B}{1 – Tan A Tan B}$ Tan (A-B) = $\\frac{Tan A – Tan B}{1 + Tan A Tan B}$ ### Product to Sum Formulas Sin A Sin B = ½ [Cos (A-B) – Cos (A+B)] Cos A Cos B = ½ [Cos (A-B) + Cos (A+B)] Sin A Cos B = ½ [Sin (A+B) + Sin (A-B)] Cos A Sin B = ½ [Sin (A+B) – Sin (A-B)] ### Sum to Product Formulas Sin A + Sin B = 2 sin $\\frac{A+B}{2}$ cos $\\frac{A-B}{2}$ Sin A – Sin B = 2 cos$\\frac{A+B}{2}$ sin $\\frac{A-B}{2}$ Cos A + Cos B = 2 cos$\\frac{A+B}{2}$ cos $\\frac{A-B}{2}$ Cos A", "Divide the first of those equations by $\\cos\\theta$ to see that $\\tan\\theta = \\sqrt2-1$. Similarly, the second equation is equivalent to $\\tan\\theta = \\frac1{\\sqrt2+1}$. Now complete the argument by showing that $\\sqrt2-1 = \\frac1{\\sqrt2+1}$,", "# Double Angle Identities ## Key Questions As below. #### Explanation: Following table gives the double angle identities which can be used while solving the equations. You can also have $\\sin 2 \\theta , \\cos 2 \\theta$ expressed in terms of $\\tan \\theta$ as under. $\\sin 2 \\theta = \\frac{2 \\tan \\theta}{1 + {\\tan}^{2} \\theta}$ $\\cos 2 \\theta = \\frac{1 - {\\tan}^{2} \\theta}{1 + {\\tan}^{2} \\theta}$ • You would need an expression to work with. For example: Given $\\sin \\alpha = \\frac{3}{5}$ and $\\cos \\alpha = - \\frac{4}{5}$, you could find $\\sin 2 \\alpha$ by using the double angle identity $\\sin 2 \\alpha = 2 \\sin \\alpha \\cos \\alpha$. $\\sin 2 \\alpha = 2 \\left(\\frac{3}{5}\\right) \\left(- \\frac{4}{5}\\right) = - \\frac{24}{25}$. You could find $\\cos 2 \\alpha$ by using any of: $\\cos 2 \\alpha = {\\cos}^{2} \\alpha - {\\sin}^{2} \\alpha$ $\\cos 2 \\alpha = 1 - 2 {\\sin}^{2} \\alpha$ $\\cos 2 \\alpha = 2 {\\cos}^{2} \\alpha - 1$ In any case, you get $\\cos \\alpha = \\frac{7}{25}$. • Double Angle Identities $\\sin 2 \\theta = 2 \\sin \\theta \\cos \\theta$ $\\cos 2 \\theta = {\\cos}^{2} \\theta - {\\sin}^{2} \\theta = 2 {\\cos}^{2} \\theta - 1 = 1 - 2 {\\sin}^{2} \\theta$ $\\tan 2 \\theta = \\frac{2 \\tan \\theta}{1 - {\\tan}^{2} \\theta}$ I hope that this was helpful.", "both sides are equal, Punch. If $\\frac{1-Cos\\theta+Sin\\theta}{1+Cos\\theta+Sin\\theta}=\\frac{ Sin\\frac{\\theta}{2}}{Cos\\frac{\\theta}{2}}$ Then $Cos\\frac{\\theta}{2}(1-Cos\\theta+Sin\\theta)$ must equal $Sin\\frac{\\theta}{2}(1+Cos\\theta+Sin\\theta)$ The remaining calculations examine whether or not this is true, but i know what you mean. Using the double-angle formulae for CosA and SinA $\\frac{1-Cos\\theta+Sin\\theta}{1+Cos\\theta+Sin\\theta}=\\frac{ 1-\\frac{1-tan^2\\frac{\\theta}{2}}{1+tan^2{\\frac{\\theta}{2}}}+ \\frac{2tan\\frac{\\theta}{2}}{1+tan^2{\\theta}{2}}}{1 +\\frac{1-tan^2\\frac{\\theta}{2}}{1+tan^2\\frac{\\theta}{2}}+\\f rac{2tan\\frac{\\theta}{2}}{1+tan^2\\frac{\\theta}{2}} }$ $=\\frac{\\frac{1+tan^2\\frac{\\theta}{2}-(1-tan^2\\frac{\\theta}{2})+2tan\\frac{\\theta}{2}}{1+tan ^2\\frac{\\theta}{2}}}{\\frac{1+tan^2\\frac{\\theta}{2} +1-tan^2\\frac{\\theta}{2}+2tan\\frac{\\theta}{2}}{1+tan^ 2\\frac{\\theta}{2}}}$ $=\\frac{\\frac{1+tan^2\\frac{\\theta}{2}-1+tan^2\\frac{\\theta}{2}+2tan\\frac{\\theta}{2}}{1+ta n^2\\frac{\\theta}{2}}}{\\frac{1+tan^2\\frac{\\theta}{2 }+1-tan^2\\frac{\\theta}{2}+2tan\\frac{\\theta}{2}}{1+tan^ 2\\frac{\\theta}{2}}}$ $=\\frac{2tan^2\\frac{\\theta}{2}+2tan\\frac{\\theta}{2} }{2+2tan\\frac{\\theta}{2}}$ $=\\frac{2tan\\frac{\\theta}{2}(1+tan\\frac{\\theta}{2}) }{2(1+tan\\frac{\\theta}{2})}=tan\\frac{\\theta}{2}$ insert all the half angles stuff inside .. knowing these identities , $\\sin \\theta=2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}$ $\\cos \\theta=1-2\\sin^2 \\frac{\\theta}{2}$ so $\\frac{1-1+2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}{1+1-2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}$ -----1 $\\frac{2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}{2-2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}$ ---------2 $\\frac{2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}{2\\cos^2 \\frac{\\theta}{2}+2\\sin^2 \\frac{\\theta}{2}-2\\sin^2 \\frac{\\theta}{2}+2\\sin \\frac{\\theta}{2}\\cos \\frac{\\theta}{2}}$ --------3 bring out the common factor and do some necessary cancellings , then you will have your proof . Which step don u understand ? 1,2 or 3 , or the steps after that ? 9. u can also use the following identities: 1)1-cos(theta)=2sin^2(theta/2) 2)1+cos(theta)=2cos^2(theta/2) 10. After looking at this now, I still do not understand the workings provided. Sorry but I guess I need a complete and detailed working. 11. Hi Punch, if you get stuck at any of the steps 1 to 7, let me know. $\\frac{1-Cos\\theta+Sin\\theta}{1+Cos\\theta+Sin\\theta}=Tan\\fr ac{\\theta}{2}$ The evaluation of the left fraction is $Tan\\frac{\\theta}{2}$ This is \"tangent of the half-angle\", therefore we may express the $Cos\\theta$ and $Sin\\theta$ from the left using \"half-angle formulae\".", "1/\\sqrt{3} = \\cos p_2 \\\\[4pt] \\cos p_3 = \\cos\\angle P_1OP_2 = P_1\\cdot P_2 &= 0 \\end{align} Since $\\sin^2\\theta + \\cos^2\\theta = 1$ (and \"knowing\" that sine is non-negative for the angles in question), we may calculate $$\\sin p_1 = \\sin p_2 = \\sqrt{1-\\left(\\frac{1}{\\sqrt{3}}\\right)^2} = \\frac{\\sqrt{2}}{\\sqrt{3}} \\qquad\\qquad \\sin p_3 = 1$$ From here, we leverage the spherical law of cosines. For instance, $$\\cos\\phi_1 = \\frac{\\cos p_1 - \\cos p_2\\cos p_3}{\\sin p_2 \\sin p_3} = \\frac{\\frac{1}{\\sqrt{3}}-\\frac{1}{\\sqrt{3}}\\cdot 0}{\\frac{\\sqrt{2}}{\\sqrt{3}}\\cdot 1} = \\frac{1}{\\sqrt2}$$ $$\\cos\\phi_2 = \\frac{\\cos p_2 - \\cos p_3\\cos p_1}{\\sin p_3 \\sin p_1} = \\frac{\\frac{1}{\\sqrt{3}}-0\\cdot\\frac{1}{\\sqrt{3}}}{1\\cdot\\frac{\\sqrt{2}}{\\sqrt{3}}} = \\frac{1}{\\sqrt2}$$ $$\\cos\\phi_3 = \\frac{\\cos p_3 - \\cos p_1\\cos p_2}{\\sin p_1 \\sin p_2} = \\frac{0-\\frac{1}{\\sqrt{3}}\\cdot\\frac{1}{\\sqrt{3}}}{\\frac{\\sqrt{2}}{\\sqrt{3}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{3}}} = -\\frac12$$ From here, we see that $\\phi_1$, $\\phi_2$, $\\phi_3$ are the \"special\" angles $\\frac14\\pi$, $\\frac14\\pi$, $\\frac23\\pi$. (The reader can compute the triangle's area from these.) It's no coincidence that two of these are half-right angles, while the last is one-third of a full turn; this is, in fact, a consequence of the \"obvious-ness\" I ignored from the beginning. As hinted in @MichealBehrend's comment, if we introduce a fourth point ---I'll call it $P_4 = (0,0,1)$--- then $P_1$, $P_2$, $P_4$ are the vertices of an equilateral spherical triangle that fills a full octant of the sphere; each angle of this spherical triangle has measure $\\frac12 \\pi$ (a fact we'll use shortly).", "Law of Cosines, we can find the smaller interior angle of the rhombus 12^2 = 360 + 360 - 2(360)cos(theta) [144 - 360 - 360] / [ -2(360)] = cos (theta) cos-1 [ [144 - 360 - 360] / [ -2(360] ] = theta = 36.869897645844° Using the Law of Sines.....we can find the height - h - as follows : sqrt(360) = h / sin(36.869897645844°) h = sqrt(360)* sin (36.869897645844°) = 11.3841995766061656 cm Here's a pic : [DE is the height] CPhill Jan 26, 2016 #3 +19207 +10 e = one diagonal f = the other diagonal h ist the height of the rhombus $$\\boxed{~ \\begin{array}{rcll} h &=&\\frac{ e\\cdot f }{\\sqrt{ e^2 + f^2 }} \\\\ \\end{array} ~}\\\\$$ or $$\\begin{array}{rcll} h &=&\\frac{ e\\cdot f }{\\sqrt{ e^2 + f^2 }}\\\\ h &=&\\frac{ 1 }{ \\frac { \\sqrt{ e^2 + f^2 }}{ e\\cdot f}} \\\\ h &=&\\frac{ 1 }{ \\sqrt{ \\frac{e^2}{e^2\\cdot f^2} + \\frac{f^2}{e^2\\cdot f^2} }} \\\\ h &=&\\frac{ 1 }{ \\sqrt{ \\frac{1}{f^2} + \\frac{1}{e^2} }} \\\\ \\frac{1}{h} &=&\\sqrt{ \\frac{1}{f^2} + \\frac{1}{e^2} } \\\\ \\frac{1}{h^2} &=&\\frac{1}{f^2} + \\frac{1}{e^2} \\\\ \\frac{1}{h^2} &=&\\frac{1}{e^2} + \\frac{1}{f^2} \\\\ \\end{array}$$ $$\\boxed{~ \\begin{array}{rcll} \\frac{1}{h^2} &=&\\frac{1}{e^2} + \\frac{1}{f^2} \\end{array} ~}$$ heureka Jan 26, 2016 edited by heureka Jan 26, 2016 ### 7 Online Users We use cookies to personalise content and ads, to provide social media", "Precalculus (6th Edition) $\\displaystyle \\cos 2\\theta=\\frac{119}{169}$ $\\sin 2\\displaystyle \\theta=-\\frac{120}{169}$ First, using the Pythagorean Identity $\\sin^{2}\\theta+\\cos^{2}\\theta=1,$with $\\sin\\theta > 0,$ $\\displaystyle \\sin\\theta=+\\sqrt{1-(-\\frac{12}{13})^{2}}=\\sqrt{1-\\frac{144}{169}}$ $=\\sqrt{\\dfrac{25}{169}}=\\dfrac{5}{13}$ We now use the Double-Angle Identities: $\\displaystyle \\cos 2\\theta=\\cos^{2}\\theta-\\sin^{2}\\theta=\\frac{144}{169}-\\frac{25}{169}=\\frac{119}{169}$ $\\sin 2\\displaystyle \\theta=2\\sin\\theta\\cos\\theta=2\\cdot\\frac{-12}{13}\\cdot\\frac{5}{13}=-\\frac{120}{169}$" ]

[ "are 1, 1, 2. If θ is the angle between them, they can be written as: $\\cos \\theta =\\vert \\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\\cdot \\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\\vert$ $\\cos \\theta =\\vert \\frac{3\\cdot 1+5\\cdot 1+4\\cdot 2}{\\sqrt{3^{2}+5^{2}+4^{2}}\\cdot \\sqrt{1+1+2^{2}}}\\vert =\\frac{8\\sqrt{3}}{15}$ Hence we can find the required acute angle by taking $\\cos ^{-1}\\frac{8\\sqrt{3}}{15}$", "\\left(\\frac{1}{3}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2} + \\frac{1}{7}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}+\\quad \\cdots\\right) ### Transformation to current notation Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get • \\theta = \\tan \\theta - \\frac{\\tan^3 \\theta}{3} + \\frac{\\tan^5\\theta}{5} - \\frac{\\tan^7 \\theta}{7} + \\quad \\cdots . Letting tan θ = q we finally have • \\tan^{-1} q = q - \\frac{q^3}{3} + \\frac{q^5}{5} - \\frac{q^7}{7} + \\quad \\cdots ### Another formula for the circumference of a circle The second part of the quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows. c= \\sqrt{12 d^2} - \\frac{\\sqrt{12 d^2}}{3\\cdot 3} + \\frac{\\sqrt{12 d^2}}{3^2 \\cdot 5} - \\frac{\\sqrt{12 d^2}}{3^3 \\cdot 7}+ \\quad \\cdots Since c = π d this can be reformulated as a formula to compute π as follows. \\pi = \\sqrt{12}\\left( 1 - \\frac{1}{3\\cdot3}+\\frac{1}{3^2\\cdot 5} -\\frac{1}{3^3\\cdot 7} +\\quad \\cdots\\right) This is obtained by substituting q = 1/\\sqrt{3} (therefore θ = π / 6) in the power series expansion for tan−1 q above. Comparison of the convergence of two Madhava series", "\\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$.", "# Math Help - Double angle formula 1. ## Double angle formula Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks 2. ## Re: Double angle formula Originally Posted by Hooperoo Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks $\\frac{1}{\\sin 2\\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}=\\frac{1-(cos^2 \\theta-\\sin^2 \\theta)}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{\\sin^2 \\theta + \\cos^2 \\theta -\\cos^2 \\theta +\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{2\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=\\frac{\\sin \\theta}{\\cos \\theta}=\\tan \\theta$ 3. ## Re: Double angle formula Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@! 4. ## Re: Double angle formula Hello, Hooperoo! You are expected to know these identities: . . $\\sin2\\theta \\:=\\:2\\sin\\theta\\cos\\theta$ . . $\\sin^2\\!\\theta \\:=\\:\\frac{1-\\cos2\\theta}{2} \\quad\\Rightarrow\\quad 1 - \\cos2\\theta \\:=\\:2\\sin^2\\!\\theta$ Using double-angle formulas, verify that: . $\\csc2\\theta - \\cot2\\theta \\:=\\:\\tan\\theta$ We have: . $\\csc2\\theta - \\cot2\\theta \\;=\\;\\frac{1}{\\sin2\\theta} - \\frac{\\cos2\\theta}{\\sin2\\theta} \\;=\\;\\frac{1 - \\cos2\\theta}{\\sin2\\theta}$ . . . . . . . . . . . . . . . . . . . $\\;=\\;\\frac{2\\sin^2\\!\\theta}{2\\sin\\theta\\cos\\theta} \\;=\\;\\frac{\\sin\\tjeta}{\\cos\\theta} \\;=\\;\\tan\\theta$", "= 2 · cos x · sin((n − 1)x) − sin((n − 2)x). This can be proved by adding formulae for sin((n − 1)x + x) and sin((n − 1)xx). Serving a purpose similar to that of the Chebyshev method, for the tangent we can write: ${\\displaystyle \\tan(nx)={\\frac {\\tan((n-1)x)+\\tan x}{1-\\tan((n-1)x)\\tan x}}\\,.}$ Tangent of an average ${\\displaystyle \\tan \\left({\\frac {\\alpha +\\beta }{2}}\\right)={\\frac {\\sin \\alpha +\\sin \\beta }{\\cos \\alpha +\\cos \\beta }}=-\\,{\\frac {\\cos \\alpha -\\cos \\beta }{\\sin \\alpha -\\sin \\beta }}}$ Setting either α or β to 0 gives the usual tangent half-angle formulae. Viète's infinite product ${\\displaystyle \\cos {\\frac {\\theta }{2}}\\cdot \\cos {\\frac {\\theta }{4}}\\cdot \\cos {\\frac {\\theta }{8}}\\cdots =\\prod _{n=1}^{\\infty }\\cos {\\frac {\\theta }{2^{n}}}={\\frac {\\sin \\theta }{\\theta }}=\\operatorname {sinc} \\theta .}$ (Refer to sinc function.) Power-reduction formulae Obtained by solving the second and third versions of the cosine double-angle formula. Sine Cosine Other ${\\displaystyle \\sin ^{2}\\theta ={\\frac {1-\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\cos ^{2}\\theta ={\\frac {1+\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\sin ^{2}\\theta \\cos ^{2}\\theta ={\\frac {1-\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{3}\\theta ={\\frac {3\\sin \\theta -\\sin(3\\theta )}{4}}}$ ${\\displaystyle \\cos ^{3}\\theta ={\\frac {3\\cos \\theta +\\cos(3\\theta )}{4}}}$ ${\\displaystyle \\sin ^{3}\\theta \\cos ^{3}\\theta ={\\frac {3\\sin(2\\theta )-\\sin(6\\theta )}{32}}}$ ${\\displaystyle \\sin ^{4}\\theta ={\\frac {3-4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\cos ^{4}\\theta ={\\frac {3+4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{4}\\theta \\cos ^{4}\\theta ={\\frac {3-4\\cos(4\\theta )+\\cos(8\\theta )}{128}}}$ ${\\displaystyle \\sin ^{5}\\theta ={\\frac {10\\sin \\theta -5\\sin(3\\theta", "Angle Formulas sinθ = $\\pm \\sqrt{\\frac{1-cos2\\theta }{2}}$ cosθ = $\\pm \\sqrt{\\frac{1+cos2\\theta }{2}}$ tanθ = $\\pm \\sqrt{\\frac{1-cos2\\theta }{1+cos2\\theta}}$ ### Thrice of Angle Formulas sin3θ = 3sinθ – 4 sin3θ Cos 3θ = 4cos3θ – 3 cosθ Tan 3θ = $\\frac{3 tan\\theta – tan^3\\theta }{1-3tan^2\\theta }$ Cot 3θ = $\\frac{cot^3\\theta – 3cot\\theta }{3cot^2\\theta-1 }$ ### Sum and Difference Formulas Sin (A+B) = Sin A Cos B + Cos A Sin B Sin (A-B) = Sin A Cos B – Cos A Sin B Cos (A+B) = Cos A Cos B – Sin A Sin B Cos (A-B) = Cos A Cos B + Sin A Sin B Tan (A+B) = $\\frac{Tan A + Tan B}{1 – Tan A Tan B}$ Tan (A-B) = $\\frac{Tan A – Tan B}{1 + Tan A Tan B}$ ### Product to Sum Formulas Sin A Sin B = ½ [Cos (A-B) – Cos (A+B)] Cos A Cos B = ½ [Cos (A-B) + Cos (A+B)] Sin A Cos B = ½ [Sin (A+B) + Sin (A-B)] Cos A Sin B = ½ [Sin (A+B) – Sin (A-B)] ### Sum to Product Formulas Sin A + Sin B = 2 sin $\\frac{A+B}{2}$ cos $\\frac{A-B}{2}$ Sin A – Sin B = 2 cos$\\frac{A+B}{2}$ sin $\\frac{A-B}{2}$ Cos A + Cos B = 2 cos$\\frac{A+B}{2}$ cos $\\frac{A-B}{2}$ Cos A", "$A + B + C=180^{\\circ}$ and A,B and C are positive angles then 1. $sin 2A + sin 2B + sin 2C= 4 Sin(A) Sin (B) Sin (C)$ 2. $cos 2A + cos 2B + cos 2C= -1- 4 cos(A) cos (B) cos (C)$ 3. $sin A + sin B + sin C= 4 Sin \\frac {A}{2} Sin \\frac {B}{2} Sin \\frac {C}{2}$ 4. for no right angle in A,B,C $tan A + tan B + tan C=tan(A) tan (B) tan (C)$ 5. $tan(\\frac {A}{2}) tan (\\frac {B}{2}) +tan(\\frac {B}{2}) tan (\\frac {C}{2}) + tan(\\frac {A}{2}) tan (\\frac {C}{2})=1$ 6. $cot A cot B + cot B cot C + cot C cot A=1$ ## Half Angle Identities The ones for sine and cosine take the positive or negative square root depending on the quadrant of the angle $\\frac {\\theta}{2}$. For example, if $\\frac {\\theta}{2}$ is an acute angle, then the positive root would be used. 1. $sin \\frac {\\theta}{2}=\\pm \\sqrt {\\frac {1-cos \\theta }{2}}$ 2. $cos \\frac {\\theta}{2}=\\pm \\sqrt {\\frac {1+cos \\theta }{2}}$ 3. $tan \\frac {\\theta}{2}= \\frac { sin \\theta}{1 + cos \\theta } = \\frac { 1- cos \\theta}{sin \\theta }$ 4. $sin \\theta = \\frac {2 tan (\\theta /2)}{1 + tan^2 (\\theta /2)}$ 5. $cos \\theta = \\frac {1 - tan^2 (\\theta /2)}{1 +", "and C are positive angles then 1. $sin 2A + sin 2B + sin 2C= 4 Sin(A) Sin (B) Sin (C)$ 2. $cos 2A + cos 2B + cos 2C= -1- 4 cos(A) cos (B) cos (C)$ 3. $sin A + sin B + sin C= 4 Sin \\frac {A}{2} Sin \\frac {B}{2} Sin \\frac {C}{2}$ 4. for no right angle in A,B,C $tan A + tan B + tan C=tan(A) tan (B) tan (C)$ 5. $tan(\\frac {A}{2}) tan (\\frac {B}{2}) +tan(\\frac {B}{2}) tan (\\frac {C}{2}) + tan(\\frac {A}{2}) tan (\\frac {C}{2})=1$ 6. $cot A cot B + cot B cot C + cot C cot A=1$ Half Angle Identities The ones for sine and cosine take the positive or negative square root depending on the quadrant of the angle $\\frac {\\theta}{2}$. For example, if $\\frac {\\theta}{2}$ is an acute angle, then the positive root would be used. 1. $sin \\frac {\\theta}{2}=\\pm \\sqrt {\\frac {1-cos \\theta }{2}}$ 2. $cos \\frac {\\theta}{2}=\\pm \\sqrt {\\frac {1+cos \\theta }{2}}$ 3. $tan \\frac {\\theta}{2}= \\frac { sin \\theta}{1 + cos \\theta } = \\frac { 1- cos \\theta}{sin \\theta }$ 4. $sin \\theta = \\frac {2 tan (\\theta /2)}{1 + tan^2 (\\theta /2)}$ 5. $cos \\theta = \\frac {1 - tan^2 (\\theta /2)}{1 + tan^2 (\\theta /2)}$ 6. $tan \\theta = \\frac", "where $t=\\tan \\theta$, follow directly from the ratios of the sides of the right angled triangle $ADC$ and we get $$\\tan2\\theta = {2t\\over {1-t^2}},\\quad \\sin2\\theta = {2t\\over {1+t^2}},\\quad \\cos2\\theta = {{1-t^2}\\over {1+t^2}}$$", "AC2= 25 AC = 5 Similarly $\\sec \\Theta = \\frac{AC}{BC}$ $\\sec \\Theta = \\frac{5}{3}$ $cosec = \\frac{AC}{AB}$ $cosec = \\frac{5}{4}$ Now on substituting the values in equations we get, $\\frac{\\sec \\Theta – cosec\\Theta }{\\sec \\Theta + cosec\\Theta } =\\frac{1}{\\sqrt{7}}$ Therefore, $\\frac{\\sec \\Theta – cosec\\Theta }{\\sec \\Theta + cosec\\Theta } =\\frac{1}{\\sqrt{7}}$ 35.) If $3 \\cos \\Theta – 4\\sin \\Theta = 2\\cos \\Theta + \\sin \\Theta$ ,find $\\tan \\Theta$ Sol. Given: $3 \\cos \\Theta – 4\\sin \\Theta = 2\\cos \\Theta + \\sin \\Theta$ To find: $\\tan \\Theta$ We can write this as: $3 \\cos \\Theta – 4\\sin \\Theta = 2\\cos \\Theta + \\sin \\Theta$ $\\cos \\Theta = 5 \\sin \\Theta$ Dividing both the sides by $\\cos \\Theta$ , We get, $\\frac{\\cos \\Theta }{\\cos \\Theta }=\\frac{ 5 \\sin \\Theta }{\\cos \\Theta }$ $1 = 5\\tan \\Theta$ $\\tan \\Theta = 1$ Hence, $\\tan \\Theta = 1$ 36.) If $\\angle A$ and $\\angle P$ are acute angles such that tan A = tan P, then show $\\angle A = \\angle P$ Sol. Given: A and P are acute angles tan A =tan P Prove that: $\\angle A = \\angle P$ Let us consider right angled triangle ACP We know $\\tan \\Theta = \\frac{opposite side}{adjacent side}$ tan A =$\\frac{PC}{AC}$ tan P =$\\frac{AC}{PC}$ $∴$ tan A =tan P $\\frac{PC}{AC}= \\frac{AC}{PC}$ PC =AC [$∵$Angle opposite to", "angle $\\theta$. This should give you a hint that you will probably have to use a double-angle identity. Perhaps: $2\\cot(2\\theta) = 2\\frac{1- \\tan^2(x)}{2\\tan(x)}$ Or as mr. fantastic pointed out one of the many other double angle identities (his, in fact, is easier to apply). Arriving at the conclusion that 1 = 1 by starting with -1 = 1 and manipulting both sides of the equation is wrong. The proof should go like this: $\\cot(\\theta)-\\tan(\\theta)$ (write down left hand side of equation) $= \\frac{\\cos(\\theta)}{\\sin(\\theta)} - \\frac{\\sin(\\theta)}{\\cos(\\theta)}$ (by the fact that $\\tan(x) = \\sin(x)/\\cos(x) \\text{ and } \\cot(x) = 1/\\tan(x)$) $= \\frac{\\cos^2(\\theta)-\\sin^2(\\theta)}{\\sin(\\theta)\\cos(\\theta)}$ (by finding a common denominator and simplifying) $= \\frac{\\cos(2\\theta)}{1/2\\sin(2\\theta)}$ (by the identities mr. fantastic gave) $= 2\\cot(2\\theta)$ (end with the right hand side of the equation) This is fundamentally different from the approach you took. $\\begin{array}{l} 2\\cot (2\\theta ) \\\\ 2\\frac{{\\cos (2\\theta )}}{{\\sin (2\\theta )}} \\\\ \\frac{{\\cos ^2 (\\theta ) - \\sin ^2 (\\theta )}}{\\begin{array}{l} \\sin (\\theta )\\cos (\\theta ) \\\\ \\frac{{\\cos (\\theta )}}{{\\sin (\\theta )}} - \\frac{{\\sin (\\theta )}}{{\\cos (\\theta )}} \\\\ \\end{array}} \\\\ \\end{array} $ 8. Well. Whose method should I go with. Yours or what I posted? 9. Doesn't matter. They both work. You proved that one side is equivalent to the other. Doesn't matter which side you start with. iknowone started with the left,", "is $\\theta$. Then $\\tan \\theta = \\frac HL$ so the $L$ I used in the moment is the same as $\\frac H{\\tan \\theta}$ – Ross Millikan Nov 12 '12 at 20:31 Your calculation is correct, as far as the force calculation goes. It's useful to note that $\\sin(\\theta)\\cdot\\cos(\\theta)=\\frac12\\sin(2\\theta)$, and that is maximal when $\\theta=\\pi/4=45^\\circ$. Thus $L=H$ is optimal.", "evidently angle $\\theta$ is twice angle IOP. So using the double-angle formula for tangent, $$\\tan\\theta = \\frac{\\sqrt5-1}{1-\\left(\\frac{\\sqrt5-1}2\\right)^2} = \\frac{\\sqrt5-1}{1-\\frac{6-2\\sqrt5}4} = \\frac{\\sqrt5-1}{\\frac{\\sqrt5-1}2}=2$$ as desired.", "If $\\theta$ be an acute angle and $7\\sin ^{2}\\theta+3\\cos ^{2}\\theta = 4,$ then the value of $\\tan \\theta$ is: [A] $0$ [B] $\\frac{1}{\\sqrt{3}}$ [C] $1$ [D] $\\sqrt{3}$", "\\frac{\\cos(2\\theta)}{\\sin(2\\theta)} = \\tan(\\alpha) \\\\[9pt] \\cot(2\\theta) = \\tan(\\alpha)$ All the angles are acute angles, so this tells us that $2\\theta$ and $\\alpha$ are complements. (Think about the two acute angles in a right triangle. The adjacent for one is the opposite of the other, and vice versa, so opposite/adjacent for one is adjacent/opposite for the other.) Thus $2\\theta = 90^\\circ - \\alpha$ and the optimum firing angle is $\\theta = 45^\\circ - \\frac{\\alpha}{2}$.", "= $$\\frac{1}{2}$$ ; then find cot θ. √3 Question 152. Find the value of sin 29° – cos 61° 0 Question 153. If tan θ = 1; then find cos θ. $$\\frac{1}{\\sqrt{2}}$$ Question 154. cos (A + B) equal to cos A cos B – sin A sin B Question 155. Express tan θ, in terms of sin θ. $$\\frac{\\sin \\theta}{\\sqrt{1-\\sin ^{2} \\theta}}$$ Question 156. sin θ + sin2 θ = 1, then find cos2 θ + cos4 θ. 1 Question 157. in ΔABC,write tan($$\\frac{B+C}{2}$$)equal to Cot ($$\\frac{\\mathrm{A}}{2},$$) Explanation: A + B + C = 180° Question 158. (cos A + sin A)2 + (cos A – sin A)2 equal to 2 Question 159. cos(180 – θ) equal to – cos θ Question 160. Find the value of (cosec θ – cot θ). $$\\frac{1-\\cos \\theta}{\\sin \\theta}$$ Explanation: cosec θ – cot θ = $$\\frac{1}{\\sin \\theta}-\\frac{\\cos \\theta}{\\sin \\theta}=\\frac{1-\\cos \\theta}{\\sin \\theta}$$ Question 161. sin2 75° + cos2 75° equal to 1 Question 162. If cos θ. sin θ = $$\\frac{1}{2}$$ ; then find tan θ. 1 Explanation: cos θ . sin θ = cos 45° . sin 45° = $$\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{1}{2}$$ Then tan 45° = 1. Question 163. For which value of cosine equal to sin 81°. cos 9°. Question 164. sin 750° equal to $$\\frac{1}{2}$$ Question 165. $$\\frac{\\sin", "to be able to express $\\cos\\theta$ in this formula for $\\cos\\frac{\\theta}{2}$ in terms of $\\tan\\theta$. And how do you do that? Tipp: sketch a right triangle with acute angle $\\theta$, adjacent leg $1$ and opposite leg $\\tan\\theta$. By Pyhtagoras's theorem you have that the length of the hypotenuse of this triangle is $\\sqrt{1+\\tan^2\\theta}$. Therefore, by definition, we have $\\cos\\theta =\\frac{1}{\\sqrt{1+\\tan^2\\theta}}$. This is valid only for acute angles $\\theta$. In the general case, you need to write this as $\\cos\\theta=\\pm\\frac{1}{\\sqrt{1+\\tan^2\\theta}}$. The sign depends on the quadrant in which the point that corresponds to $\\theta$ lies on the unit circle. To sum it up you get: $\\cos\\frac{\\theta}{2}=\\pm\\sqrt{\\frac{1\\pm \\frac{1}{\\sqrt{1+\\tan^2\\theta}}}{2}}$ If you are allowed to assume that $\\theta$ is an acute angle, then you only take the $+$ case of $\\pm$. Now plug in the given value of $\\tan\\theta=4$... 3. Awesome Thanks for the lengthy insight. I have applied that to the ones I have and its helped. I am struggling to find examples to do after a google. Any links to some exercises would be fantastic Thanks once again for your help 4. Originally Posted by 200001 Awesome Thanks for the lengthy insight. I have applied that to the ones I have and its helped. I am struggling to find examples to do after a google. Any links to some exercises", "2 $\\theta$ ) = $\\frac{cot^{2}\\theta\\;-\\;1}{2cot\\;\\theta}$ ### Triple Angle formula: 1. Sin ( 3 $\\theta$ ) = 3 Sin $\\theta$ - 4 Sin3 $\\theta$ 2. Cos ( 3 $\\theta$ ) = 4 Cos3 $\\theta$ - 3 Cos $\\theta$ 3. Tan ( 3 $\\theta$ ) = $\\frac{3tan\\;\\theta-tan^{3}\\theta}{1-3tan^{2}\\theta}$ 4. Cot ( 3 $\\theta$ ) = $\\frac{cot^{3}\\;\\theta-3cot\\;\\theta}{3cot^{2}\\;\\theta-1}$ ### Half Angle Formula: 1. Sin ( $\\theta$ ) = 2 Sin $\\frac{\\theta}{2}$. Cos $\\frac{\\theta}{2}$ Sin ( $\\theta$ ) = $\\frac{2tan\\;\\frac{\\theta}{2}}{1-tan^{2}\\;\\frac{\\theta}{2}}$ 2. Cos ( $\\theta$ ) = Cos2 $\\frac{\\theta}{2}$ - Sin2 $\\frac{\\theta}{2}$ Cos ( $\\theta$ ) = 1 - 2 Sin2 $\\frac{\\theta}{2}$ Cos ( $\\theta$ ) = 2 Cos2 $\\frac{\\theta}{2}$ - 1 Cos ( $\\theta$ ) = $\\frac{1-tan^{2}\\;\\frac{\\theta}{2}}{1+tan^{2}\\;\\frac{\\theta}{2}}$ 3. Tan ( $\\theta$ ) = $\\frac{2tan\\;\\frac{\\theta}{2}}{1-tan^{2}\\;\\frac{\\theta}{2}}$ 4. Cot ( $\\theta$ ) = $\\frac{cot^{2}\\frac{\\theta }{2}-1}{2cot\\; \\frac{\\theta }{2}}$ ### One Third Angle Formula: 1. Sin $\\theta$ = 3 sin ( $\\frac{\\theta}{3}$ ) - 4 sin3 ( $\\frac{\\theta}{3}$) 2. Cos $\\theta$ = 4 cos3 ( $\\frac{\\theta}{3}$ ) - 3 cos ($\\frac{\\theta}{3}$ ) 3. Tan $\\theta$ = $\\frac{3tan\\;\\frac{\\theta }{3}-tan^{3}\\;\\frac{\\theta }{3}}{1-3tan^{2}\\;\\frac{\\theta }{3}}$ ## Sum to Product Identity For the two angles A and B, 1. Sin A + Sin B = 2 Sin $\\frac{A+B}{2}$ . Cos $\\frac{A-B}{2}$ 2. Sin A - Sin B = 2 Cos $\\frac{A+B}{2}$ . Sin $\\frac{A-B}{2}$ 3. Cos A + Cos B = 2 Cos $\\frac{A+B}{2}$ .", "\\tan y)}}{{(1 – \\tan x \\cdot \\tan y)}}$$ • $$\\sin (x – y) = \\sin x \\cos y – \\cos x \\sin y$$ • $$\\cos (x – y) = \\cos x \\cos y + \\sin x \\sin y$$ • $$\\tan (x – y) = \\frac{{x – \\tan y)}}{{(1 + \\tan x \\cdot \\tan y)}}$$ ### Half Angle Identities • $$\\sin \\frac{x}{2} = \\pm \\sqrt {\\frac{{1 – \\cos x}}{2}}$$ • $$\\cos \\frac{x}{2} = \\pm \\sqrt {\\frac{{1 + \\cos x}}{2}}$$ • $$\\tan {\\frac{x}{2}} = {\\frac{1 – \\cos x}{\\sin x}} = {\\frac{\\sin x}{1+\\cos x}}= \\pm \\sqrt {\\frac{{1 – \\cos x}}{{1 + \\cos x}}}$$ • $$\\cot {\\frac{x}{2}} = {\\frac{\\sin x}{1 – \\cos x}} = {\\frac{1+\\cos x}{\\sin x}}= \\pm \\sqrt {\\frac{{1 + \\cos x}}{{1 – \\cos x}}}$$ ### Double Angle Identities The double of the angle $$x$$ is represented by the given formulas: • $$\\sin 2x = 2 \\sin x \\cos x = {\\frac{{2 \\tan x}}{{1 + \\tan^2 x}}}$$ • $$\\cos 2x = – x = \\cos^2 x – \\sin^2 x = 2 \\cos^2 x – 1 = 1 – 2 \\sin^2 x = {\\frac{1 – \\tan^2 x}{1+\\tan^2 x}}$$ • $$\\tan 2x = \\frac{{2\\tan x}}{{1 – \\tan^2 x}}$$ • $$\\cot 2x = \\frac{{\\cot^2 x – 1}}{{2 \\cot x}}$$ • $$\\sec 2x = \\frac{\\sec^2 x}{{(2 – \\sec^2 x)}}$$ • $${\\rm{cosec}}\\,2x = \\frac{\\sec x \\cdot", "## Geometry: Common Core (15th Edition) sin A = $\\frac{5}{6.4}$ cos A = $\\frac{4}{6.4}$ tan A = $\\frac{5}{4}$ sin B = $\\frac{4}{6.4}$ cos B = $\\frac{5}{6.4}$ tan B = $\\frac{4}{5}$ The ratio for sine is: sin θ = $\\frac{opposite}{hypotenuse}$ The ratio for cosine is: cos θ = $\\frac{adjacent}{hypotenuse}$ The ratio for tangent is: tan θ = $\\frac{opposite}{adjacent}$ Let's find these ratios for both $\\angle A$ and $\\angle B$: sin A = $\\frac{5}{6.4}$ cos A = $\\frac{4}{6.4}$ tan A = $\\frac{5}{4}$ sin B = $\\frac{4}{6.4}$ cos B = $\\frac{5}{6.4}$ tan B = $\\frac{4}{5}$" ]

[ "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "# Projected points and screen coordinates I'm trying to understand part of the contents of a slide, but I'm not really understanding. So, here's the interested part. I've a few questions. 1. Are the projected points $q_1$ and $q_2$ the points with respectively the screen coordinates $(x_1, y_1)$ and $(x_2, y_2)$, i.e., $q_1 = \\begin{pmatrix} x_1 \\\\ y_1\\end{pmatrix}$ and $q_2 = \\begin{pmatrix} x_2 \\\\ y_2\\end{pmatrix}$? 2. Regarding how to obtain the projected points. What I understand from the slide is that the projected $3D$ points are obtained by dividing all coordinates of $p_1$ (e.g.) by its $z$ coordinate, i.e., $p_1^z$, so that we obtain $q_1 = \\begin{pmatrix} \\frac{p_1^x}{p_1^z} \\\\ \\frac{p_1^y}{p_1^z} \\\\ \\frac{p_1^z}{p_1^z} \\end{pmatrix} = \\begin{pmatrix} \\frac{p_1^x}{p_1^z} \\\\ \\frac{p_1^y}{p_1^z} \\\\ 1 \\end{pmatrix}$. So, I guess, that at this point, we obtain the sccreen coordinates by ignoring the $1$ as the 3rd coordinate. Is that right? • Can you link to the slides? It may help to answer the question better if we can put it in context. – Nathan Reed Nov 19 '16 at 16:43 • @NathanReed Probably not... but if something is unclear, I can try to give you more information.. – nbro Nov 19 '16 at 16:51 • Well, I'm not sure we can really answer the question as it's not clear from the slide how \"screen", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "# Action of a projection on a pairwise orthogonal projections. Let us consider two projections $e_1$ and $e_2$ on the Hilbert space $H$ with $e_1e_2=0$. Does there exists any projection $p$ with $$\\overline{pe_1(H)}=\\overline{pe_2(H)}=\\overline{p(e_1+e_2)(H)}\\neq0$$ Wlog we can assume that $$e_1=\\begin{pmatrix}1&0&0\\\\0&0&0\\\\0&0&0\\end{pmatrix} \\mbox{ and } e_2=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$ with respect to a certain orthogonal decomposition of $H=H_1\\oplus H_2\\oplus H_3$. Then it suffices to set $$p=\\begin{pmatrix}1/2&1/2&0\\\\1/2&1/2&0\\\\0&0&0\\end{pmatrix}.$$", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "# Projected points and screen coordinates I'm trying to understand part of the contents of a slide, but I'm not really understanding. So, here's the interested part. I've a few questions. 1. Are the projected points $q_1$ and $q_2$ the points with respectively the screen coordinates $(x_1, y_1)$ and $(x_2, y_2)$, i.e., $q_1 = \\begin{pmatrix} x_1 \\\\ y_1\\end{pmatrix}$ and $q_2 = \\begin{pmatrix} x_2 \\\\ y_2\\end{pmatrix}$? 2. Regarding how to obtain the projected points. What I understand from the slide is that the projected $3D$ points are obtained by dividing all coordinates of $p_1$ (e.g.) by its $z$ coordinate, i.e., $p_1^z$, so that we obtain $q_1 = \\begin{pmatrix} \\frac{p_1^x}{p_1^z} \\\\ \\frac{p_1^y}{p_1^z} \\\\ \\frac{p_1^z}{p_1^z} \\end{pmatrix} = \\begin{pmatrix} \\frac{p_1^x}{p_1^z} \\\\ \\frac{p_1^y}{p_1^z} \\\\ 1 \\end{pmatrix}$. So, I guess, that at this point, we obtain the sccreen coordinates by ignoring the $1$ as the 3rd coordinate. Is that right? • Can you link to the slides? It may help to answer the question better if we can put it in context. Nov 19, 2016 at 16:43 • Well, I'm not sure we can really answer the question as it's not clear from the slide how \"screen coordinates\" are defined relative to \"projected points\". There are various different coordinate systems that might be used; I don't know how the slide author has chosen to", "where $P$ is the projection $$Pf(t)=\\begin{cases}f(t),&\\ t\\geq0\\\\ 0,&\\ t<0\\end{cases}$$", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "& 1\\end{pmatrix} (b)$\\frac{1}{1-t^2}\\begin{pmatrix} -t^2 & t\\\\ -t &1\\end{pmatrix}(t\\in(1,\\infty))$", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "\\$$\\mathbb{R}^2\\$$, via the \\$$\\Re\\$$ and \\$$\\Im\\$$ projections: \\$$\\Re(1 + i) = 1\\$$, and \\$$\\Im(1 + i) = 1\\$$.", "0 & 1 & 0 \\end{pmatrix} \\vphantom{\\begin{pmatrix} X \\\\ Y \\\\ Z \\\\ 1 \\end{pmatrix}} }_{\\text{\\makebox[0pt]{Projection Matrix of $C$}}} \\begin{pmatrix} X \\\\ Y \\\\ Z \\\\ 1 \\end{pmatrix} \\end{equation*} \\end{document}", "will yield a 3-element vector (the projected 2D point in homogeneous coordinates) which is then converted to 2D cartesian coordinates dividing by $$w$$. This gives exactly the values of $$x'$$ and $$y'$$ we were looking for. The missing element here is $$z'$$, which we know equals $$d$$ by definition. Using a reasoning similar to the one we employed to deduce the translation matrix, we can express the perspective projection as $\\begin{pmatrix} d & 0 & 0 & 0 \\\\ 0 & d & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\end{pmatrix} . \\begin{pmatrix} x \\\\ y \\\\ z \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} x.d \\\\ y.d \\\\ z \\end{pmatrix} \\rightarrow \\begin{pmatrix} x.d \\over z \\\\ y.d \\over z \\end{pmatrix}$ ### Homogeneous Viewport-to-Canvas matrix The last step is mapping the projected point on the viewport to the canvas. This is just a 2D scaling transform with $$S_x = c_w \\over v_w$$ and $$S_y = c_h \\over v_h$$. This matrix is thus $\\begin{pmatrix} c_w \\over v_w & 0 & 0 \\\\ 0 & c_w \\over v_w & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x.c_w \\over v_w \\\\ y.c_h \\over v_h \\\\ z \\end{pmatrix}$ In fact it’s easy to combine this with the projection matrix,", "\\\\ 4 \\\\ 5\\end{bmatrix}\\Bigg\\}, \\text{ or } \\Bigg\\{\\begin{bmatrix}0 \\\\ 4 \\\\ 5\\end{bmatrix}, \\begin{bmatrix}3 \\\\ 0 \\\\ 5\\end{bmatrix}\\Bigg\\}$$ Once you have a basis of $H$ you can use this guide guide to perform the projection. Note, since you have to do this multiple times, coming up with $Q$ as described in the guide and saving it will allow you to perform the process on many points.", "= (v_k)_{k=-2}^{\\infty} = (12, 16, 21, 27, \\ldots)$ ###### 4 $w = (w_k)_{k=1}^{\\infty} = (4, 20, 100, 500, \\ldots)$ ###### 5 $z = (z_i)_{i=0}^{\\infty} = (27, 18, 12, 8, \\ldots)$ Each problem gives a projection function and an initial value that together determine a sequence recursively. Find the four terms following the initial value. Illustrate the sequence as a map between two number lines. ###### 6 $P = (P_t)_{t=0}^{\\infty}$ with $P_0 = 400$ and projection function $f(x) = x + 25\\text{.}$ ###### 7 $u = (u_n)_{n=0}^{\\infty}$ with $u_0 = 3$ and projection function $f(x) = 1.5x + 1\\text{.}$ ###### 8 $u = (u_n)_{n=0}^{\\infty}$ with $u_0 = -3$ and projection function $f(x) = 1.5x + 1\\text{.}$ ###### 9 $w = (w_i)_{i=1}^{\\infty}$ with $w_{1} = 4$ and projection function $f(x) = 2.5x - 6\\text{.}$ ###### 10 $w = (w_i)_{i=1}^{\\infty}$ with $w_{1} = 5$ and projection function $f(x) = 2.5x - 6\\text{.}$ ###### 11 $z = (z_j)_{j=0}^{\\infty}$ with $z_{0} = 16$ and projection function $f(x) = \\sqrt{x}\\text{.}$ Each problem defines a sequence recursively. Give the formula of the projection function $f(x)\\text{.}$ Create the cobweb diagram for the sequence corresponding to the first five values of the sequence. ###### 12 $Q = (Q_t)_{t=0}^{\\infty}$ with $Q_{0} = 3$ and $Q_{t} = Q_{t-1} + 4\\text{.}$ ###### 13 $c = (c_k)_{k=0}^{\\infty}$ with $c_{0} = 10$" ]

[ "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "$$\\mu = 5.5$$. 2. To calculate the variance we use the formula: $Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2 \\end{pmatrix} - \\mu^2$ Where: $E\\begin{pmatrix} X^2 \\end{pmatrix} = \\sum x^2.P\\begin{pmatrix} X = x \\end{pmatrix}$ We start by calculating $$E\\begin{pmatrix}X^2\\end{pmatrix}$$. Using the values $$x$$ that the discrete random variable can take the formula leads to: \\begin{aligned} E\\begin{pmatrix} X^2 \\end{pmatrix} & = 3^2\\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 4^2 \\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 6^2 \\times P\\begin{pmatrix}X = 6 \\end{pmatrix} + 7^2 \\times P\\begin{pmatrix}X = 7 \\end{pmatrix} \\\\ & = 3^2 \\times \\frac{3}{20} + 4^2 \\times \\frac{4}{20} + 6^2\\times \\frac{6}{20} + 7^2 \\times \\frac{7}{20} \\\\ & = 9 \\times \\frac{3}{20} + 16 \\times \\frac{4}{20} + 36 \\times \\frac{6}{20} + 49 \\times \\frac{7}{20} \\\\ & = \\frac{27}{20} + \\frac{64}{20} + \\frac{216}{20} + \\frac{343}{20} \\\\ & = \\frac{650}{20} \\\\ E\\begin{pmatrix} X^2 \\end{pmatrix} & = 32.5 \\end{aligned} Now that we know the value of $$E\\begin{pmatrix} X^2 \\end{pmatrix} = 32.5$$, we can calculate the variance: \\begin{aligned} Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 32.5 - 5.5^2 \\\\ & = 32.5 - 30.25 \\\\ Var\\begin{pmatrix} X \\end{pmatrix} & = 2.25 \\end{aligned} The variance is $$2.25$$. Now that we know the variance, we can calculate this discrete random variable's standard deviation: \\begin{aligned} \\sigma & = \\sqrt{Var\\begin{pmatrix} X \\end{pmatrix}} \\\\ & = \\sqrt{2.25}\\\\ \\sigma & =", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "and so this is the representation of the value ${\\displaystyle h({\\vec {v}})}$ with respect to the codomain basis ${\\displaystyle D}$. ${\\displaystyle {\\begin{array}{rl}{\\rm {Rep}}_{D}(h({\\vec {v}}))&={\\begin{pmatrix}-1/2&1&2\\\\-1/2&-1&-2\\end{pmatrix}}_{B,D}{\\begin{pmatrix}0\\\\1/2\\\\2\\end{pmatrix}}_{B}\\\\&={\\begin{pmatrix}(-1/2)\\cdot 0+1\\cdot (1/2)+2\\cdot 2\\\\(-1/2)\\cdot 0-1\\cdot (1/2)-2\\cdot 2\\end{pmatrix}}_{D}={\\begin{pmatrix}9/2\\\\-9/2\\end{pmatrix}}_{D}\\end{array}}}$ To find ${\\displaystyle h({\\vec {v}})}$ itself, not its representation, take ${\\displaystyle (9/2)(1+x)-(9/2)(-1+x)=9}$. Example 1.7 Let ${\\displaystyle \\pi :\\mathbb {R} ^{3}\\to \\mathbb {R} ^{2}}$ be projection onto the ${\\displaystyle xy}$-plane. To give a matrix representing this map, we first fix bases. ${\\displaystyle B=\\langle {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}\\rangle \\qquad D=\\langle {\\begin{pmatrix}2\\\\1\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\rangle }$ For each vector in the domain's basis, we find its image under the map. ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}1\\\\0\\end{pmatrix}}\\quad {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\quad {\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}-1\\\\0\\end{pmatrix}}}$ Then we find the representation of each image with respect to the codomain's basis ${\\displaystyle {\\rm {Rep}}_{D}({\\begin{pmatrix}1\\\\0\\end{pmatrix}})={\\begin{pmatrix}1\\\\-1\\end{pmatrix}}\\quad {\\rm {Rep}}_{D}({\\begin{pmatrix}1\\\\1\\end{pmatrix}})={\\begin{pmatrix}0\\\\1\\end{pmatrix}}\\quad {\\rm {Rep}}_{D}({\\begin{pmatrix}-1\\\\0\\end{pmatrix}})={\\begin{pmatrix}-1\\\\1\\end{pmatrix}}}$ (these are easily checked). Finally, adjoining these representations gives the matrix representing ${\\displaystyle \\pi }$ with respect to ${\\displaystyle B,D}$. ${\\displaystyle {\\rm {Rep}}_{B,D}(\\pi )={\\begin{pmatrix}1&0&-1\\\\-1&1&1\\end{pmatrix}}_{B,D}}$ We can illustrate Theorem 1.4 by computing the matrix-vector product representing the following statement about the projection map. ${\\displaystyle \\pi ({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})={\\begin{pmatrix}2\\\\2\\end{pmatrix}}}$ Representing this vector from the domain with respect to the domain's basis ${\\displaystyle {\\rm {Rep}}_{B}({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})={\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}}_{B}}$ gives this matrix-vector product. ${\\displaystyle {\\rm {Rep}}_{D}(\\,\\pi ({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})\\,)={\\begin{pmatrix}1&0&-1\\\\-1&1&1\\end{pmatrix}}_{B,D}{\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}}_{B}={\\begin{pmatrix}0\\\\2\\end{pmatrix}}_{D}}$ Expanding this representation into a linear combination of vectors from ${\\displaystyle D}$ ${\\displaystyle 0\\cdot {\\begin{pmatrix}2\\\\1\\end{pmatrix}}+2\\cdot {\\begin{pmatrix}1\\\\1\\end{pmatrix}}={\\begin{pmatrix}2\\\\2\\end{pmatrix}}}$ checks that the map's action is indeed reflected in", "55.5^\\circ$. $\\displaystyle \\cos \\theta = \\frac{\\left | \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix} \\right |}{\\left | \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\right| \\, \\left | \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix} \\right | } \\\\ = \\displaystyle \\frac{|4|}{\\sqrt{3}\\sqrt{38}} \\\\ = \\displaystyle \\frac{4}{\\sqrt{114}}$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "is recommended for all readers. Problem 4 Definition 1.1 uses two vectors ${\\displaystyle {\\vec {s}}}$ and ${\\displaystyle {\\vec {v}}}$. Consider the transformation of ${\\displaystyle \\mathbb {R} ^{2}}$ resulting from fixing ${\\displaystyle {\\vec {s}}={\\begin{pmatrix}3\\\\1\\end{pmatrix}}}$ and projecting ${\\displaystyle {\\vec {v}}}$ onto the line that is the span of ${\\displaystyle {\\vec {s}}}$. Apply it to these vectors. 1. ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}0\\\\4\\end{pmatrix}}}$ Show that in general the projection tranformation is this. ${\\displaystyle {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\mapsto {\\begin{pmatrix}(x_{1}+3x_{2})/10\\\\(3x_{1}+9x_{2})/10\\end{pmatrix}}}$ Express the action of this transformation with a matrix. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {1}{2}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}3/2\\\\1/2\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}0\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {2}{5}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}6/5\\\\2/5\\end{pmatrix}}}$ In general the projection is this. ${\\displaystyle {\\frac {{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {3x_{1}+x_{2}}{10}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}(9x_{1}+3x_{2})/10\\\\(3x_{1}+x_{2})/10\\end{pmatrix}}}$ The appropriate matrix is this. ${\\displaystyle {\\begin{pmatrix}9/10&3/10\\\\3/10&1/10\\end{pmatrix}}}$ Problem 5 Example 1.5 suggests that projection breaks ${\\displaystyle {\\vec {v}}}$ into two parts, ${\\displaystyle {\\mbox{proj}}_{[{\\vec {s}}\\,]}({{\\vec {v}}\\,})}$ and ${\\displaystyle {\\vec {v}}-{\\mbox{proj}}_{[{\\vec {s}}\\,]}({{\\vec {v}}\\,})}$, that are \"not interacting\". Recall that the two are orthogonal. Show that any two nonzero orthogonal vectors make up a linearly independent set. Suppose that ${\\displaystyle {\\vec {v}}_{1}}$ and ${\\displaystyle {\\vec {v}}_{2}}$ are nonzero and orthogonal. Consider the linear relationship ${\\displaystyle c_{1}{\\vec {v}}_{1}+c_{2}{\\vec {v}}_{2}={\\vec {0}}}$. Take the dot product of both sides of the equation with ${\\displaystyle {\\vec {v}}_{1}}$ to get that ${\\displaystyle {\\vec {v}}_{1}\\cdot (c_{1}{\\vec {v}}_{1}+c_{2}{\\vec {v}}_{2})=c_{1}\\cdot ({\\vec {v}}_{1}\\cdot {\\vec {v}}_{1})+c_{2}\\cdot", "\\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\qquad \\vec{b} = \\begin{pmatrix} 0\\\\2 \\end{pmatrix}\\\\\\\\ \\begin{array}{lll} \\text{area}=\\left|\\vec{a}_{projection}\\times\\vec{b} \\right| & \\qquad \\vec{a}_{projection} = \\left(\\vec{a}\\cdot \\vec{e}_x\\right)\\cdot \\vec{e}_x & \\qquad \\vec{e}_x = \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\\\\\\\ & \\qquad \\vec{a}_{projection} =\\left( 1.5\\cdot \\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\right)\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\left( 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )} +0\\right) \\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix} \\\\\\\\ \\text{area}=\\left|\\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix}\\times\\begin{pmatrix} 0\\\\2 \\end{pmatrix} \\right| \\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\cos{ (60\\ensurement{^{\\circ}} ) } \\qquad | \\qquad \\cos{ (60\\ensurement{^{\\circ}} ) } = \\dfrac{1}{2}\\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\dfrac{1}{2}\\\\\\\\ \\mathbf{\\text{area}=1.5 ~\\mathrm{m^2}}\\\\\\\\ \\end{array}$$ heureka Jul 6, 2015 #1 +10 I will make an assumption that the tarp is 1.5 m TALL and 2 m WIDE. One side of the shadow will remain the same if the sun is shining directly down: 2m To calculate the OTHER side of this shadow rectangle, The hypotenuse of the triangle is 1.5 m . We need the cosine of 60 dgrees times the hypotenuse to calculate the other side of the shadow COS(60) * 1.5 =.75 So you have a rectangle that is 2m x .75m Area = 2 * .75 = 1.5 sq m Guest Jul 5, 2015 #2 +19653 +10 Two math students erect a", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "0 \\\\ 1 \\\\0 \\end{pmatrix} \\, , \\quad \\vec{e}_z = \\begin{pmatrix} 0 \\\\ 0 \\\\1 \\end{pmatrix} .$$ Rewrite it in the alternative basis given by $${\\vec{e}}_1 = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ 1 \\\\0 \\end{pmatrix} \\, , \\quad {\\vec{e}}_2 = \\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 \\\\ -1 \\\\0 \\end{pmatrix} \\, , \\quad {\\vec{e}}_3 = \\begin{pmatrix} 0 \\\\ 0 \\\\1 \\end{pmatrix} .$$ First, we calculate the projections along the axis defined by the new basis vectors: \\begin{align} \\vec{e}_1 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ 1 \\\\0 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = \\frac{10}{\\sqrt{2}} \\\\ \\vec{e}_2 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ -1 \\\\0 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = \\frac{8}{\\sqrt{2}} \\\\ \\vec{e}_3 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0\\\\ 0 \\\\1 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = -\\sqrt{2} \\\\ \\end{align} This yields the components in the new basis and the vector $\\vec{v}$, therefore, reads in the new basis: \\begin{align} \\vec{v} = \\begin{pmatrix} \\frac{10}{\\sqrt{2}} \\\\ \\frac{8}{\\sqrt{2}} \\\\-\\sqrt{2} \\end{pmatrix} _{\\text{new basis}} = \\frac{10}{\\sqrt{2}} \\vec{e}_1 + \\frac{8}{\\sqrt{2}}\\vec{e}_2 -\\sqrt{2} \\vec{e}_3 \\end{align} How does the quantum momentum operator explicitly look like? The momentum operator is given by $\\hat{p}_i = -i \\hbar \\partial_i$, where $\\partial_i \\equiv \\frac{ \\partial}{\\partial i}$ and $i \\in \\{x,y,z\\}$. What does the canonical commutation relation: $$[ \\hat {p}_i,\\hat{x}_j] = - i \\hbar \\delta_{ij}$$ tell us? It makes a", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "k)}{4+9+16}=\\frac{5}{29}(2\\boldsymbol i+3\\boldsymbol j+4\\boldsymbol k)$ where the dot product is a scalar. The magnitude of the projection is by eqn 15, $\\displaystyle L= \\big|(5\\boldsymbol i-3\\boldsymbol j+\\boldsymbol k)\\cdot (2\\boldsymbol i+3\\boldsymbol j+4\\boldsymbol k)\\big|\\frac{1}{\\sqrt{29}}=\\frac{5}{\\sqrt{29}}$ To calculate the total work, this projected length has to be multiplied by the distance moved which is the length of vector $$\\vec d = (2\\boldsymbol i + 3\\boldsymbol j + 4\\boldsymbol k)$$ which is $$\\sqrt{29}$$, therefore the work is $$5$$ as calculated above. If the vectors are greater than three dimensional then the $$(\\boldsymbol i, \\boldsymbol j, \\boldsymbol k)$$ basis cannot be used and matrices are used instead. For example if $$\\vec w=\\begin{bmatrix}1& 3& 5& 7\\end{bmatrix}$$ and $$\\vec v = \\begin{bmatrix}2& 4& 6& 8\\end{bmatrix}$$ the projection of $$\\vec w$$ onto $$\\vec v$$ is still given by eqn 14. The square of the magnitude $\\begin{split}\\displaystyle |\\vec v|^2=|\\vec v\\cdot \\vec v|= \\begin{bmatrix}2&4&6&8\\end{bmatrix}\\begin{bmatrix}2\\\\4\\\\6\\\\8\\end{bmatrix}=4+16+36+64=120\\end{split}$ The dot product $$\\vec w\\cdot \\vec v=100$$. Therefore the projection $$\\vec P_{w\\to v}=\\begin{bmatrix}2& 4& 6& 8\\end{bmatrix}(5/6)$$ and the length o f the projection $$50/\\sqrt{30}$$. The projection of $$\\vec v$$ onto $$\\vec w$$ is $$\\begin{bmatrix}1& 3& 5& 7\\end{bmatrix}(25/21)$$ with length $$50/\\sqrt{21}$$. ### (ii) Distance on a sphere# Calculating the distance from one point to another on the surface of a sphere is easily achieved using vectors. Figure 17a shows two points $$P$$ ans $$Q$$ are a distance $$d$$", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "instructive to consider a few special cases. Let us denote the map with $$\\phi: \\mathcal{H} \\to \\mathbb{R}^3$$ where $$\\mathcal{H}$$ is the Hilbert space of a qubit. We calculate that $$\\phi(|0\\rangle) = \\phi\\left(\\begin{pmatrix}1 \\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 +", "# Action of a projection on a pairwise orthogonal projections. Let us consider two projections $e_1$ and $e_2$ on the Hilbert space $H$ with $e_1e_2=0$. Does there exists any projection $p$ with $$\\overline{pe_1(H)}=\\overline{pe_2(H)}=\\overline{p(e_1+e_2)(H)}\\neq0$$ Wlog we can assume that $$e_1=\\begin{pmatrix}1&0&0\\\\0&0&0\\\\0&0&0\\end{pmatrix} \\mbox{ and } e_2=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$ with respect to a certain orthogonal decomposition of $H=H_1\\oplus H_2\\oplus H_3$. Then it suffices to set $$p=\\begin{pmatrix}1/2&1/2&0\\\\1/2&1/2&0\\\\0&0&0\\end{pmatrix}.$$" ]

[ "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "$$\\mu = 5.5$$. 2. To calculate the variance we use the formula: $Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2 \\end{pmatrix} - \\mu^2$ Where: $E\\begin{pmatrix} X^2 \\end{pmatrix} = \\sum x^2.P\\begin{pmatrix} X = x \\end{pmatrix}$ We start by calculating $$E\\begin{pmatrix}X^2\\end{pmatrix}$$. Using the values $$x$$ that the discrete random variable can take the formula leads to: \\begin{aligned} E\\begin{pmatrix} X^2 \\end{pmatrix} & = 3^2\\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 4^2 \\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 6^2 \\times P\\begin{pmatrix}X = 6 \\end{pmatrix} + 7^2 \\times P\\begin{pmatrix}X = 7 \\end{pmatrix} \\\\ & = 3^2 \\times \\frac{3}{20} + 4^2 \\times \\frac{4}{20} + 6^2\\times \\frac{6}{20} + 7^2 \\times \\frac{7}{20} \\\\ & = 9 \\times \\frac{3}{20} + 16 \\times \\frac{4}{20} + 36 \\times \\frac{6}{20} + 49 \\times \\frac{7}{20} \\\\ & = \\frac{27}{20} + \\frac{64}{20} + \\frac{216}{20} + \\frac{343}{20} \\\\ & = \\frac{650}{20} \\\\ E\\begin{pmatrix} X^2 \\end{pmatrix} & = 32.5 \\end{aligned} Now that we know the value of $$E\\begin{pmatrix} X^2 \\end{pmatrix} = 32.5$$, we can calculate the variance: \\begin{aligned} Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 32.5 - 5.5^2 \\\\ & = 32.5 - 30.25 \\\\ Var\\begin{pmatrix} X \\end{pmatrix} & = 2.25 \\end{aligned} The variance is $$2.25$$. Now that we know the variance, we can calculate this discrete random variable's standard deviation: \\begin{aligned} \\sigma & = \\sqrt{Var\\begin{pmatrix} X \\end{pmatrix}} \\\\ & = \\sqrt{2.25}\\\\ \\sigma & =", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "and so this is the representation of the value ${\\displaystyle h({\\vec {v}})}$ with respect to the codomain basis ${\\displaystyle D}$. ${\\displaystyle {\\begin{array}{rl}{\\rm {Rep}}_{D}(h({\\vec {v}}))&={\\begin{pmatrix}-1/2&1&2\\\\-1/2&-1&-2\\end{pmatrix}}_{B,D}{\\begin{pmatrix}0\\\\1/2\\\\2\\end{pmatrix}}_{B}\\\\&={\\begin{pmatrix}(-1/2)\\cdot 0+1\\cdot (1/2)+2\\cdot 2\\\\(-1/2)\\cdot 0-1\\cdot (1/2)-2\\cdot 2\\end{pmatrix}}_{D}={\\begin{pmatrix}9/2\\\\-9/2\\end{pmatrix}}_{D}\\end{array}}}$ To find ${\\displaystyle h({\\vec {v}})}$ itself, not its representation, take ${\\displaystyle (9/2)(1+x)-(9/2)(-1+x)=9}$. Example 1.7 Let ${\\displaystyle \\pi :\\mathbb {R} ^{3}\\to \\mathbb {R} ^{2}}$ be projection onto the ${\\displaystyle xy}$-plane. To give a matrix representing this map, we first fix bases. ${\\displaystyle B=\\langle {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}\\rangle \\qquad D=\\langle {\\begin{pmatrix}2\\\\1\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\rangle }$ For each vector in the domain's basis, we find its image under the map. ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}1\\\\0\\end{pmatrix}}\\quad {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\quad {\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\stackrel {\\pi }{\\longmapsto }}{\\begin{pmatrix}-1\\\\0\\end{pmatrix}}}$ Then we find the representation of each image with respect to the codomain's basis ${\\displaystyle {\\rm {Rep}}_{D}({\\begin{pmatrix}1\\\\0\\end{pmatrix}})={\\begin{pmatrix}1\\\\-1\\end{pmatrix}}\\quad {\\rm {Rep}}_{D}({\\begin{pmatrix}1\\\\1\\end{pmatrix}})={\\begin{pmatrix}0\\\\1\\end{pmatrix}}\\quad {\\rm {Rep}}_{D}({\\begin{pmatrix}-1\\\\0\\end{pmatrix}})={\\begin{pmatrix}-1\\\\1\\end{pmatrix}}}$ (these are easily checked). Finally, adjoining these representations gives the matrix representing ${\\displaystyle \\pi }$ with respect to ${\\displaystyle B,D}$. ${\\displaystyle {\\rm {Rep}}_{B,D}(\\pi )={\\begin{pmatrix}1&0&-1\\\\-1&1&1\\end{pmatrix}}_{B,D}}$ We can illustrate Theorem 1.4 by computing the matrix-vector product representing the following statement about the projection map. ${\\displaystyle \\pi ({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})={\\begin{pmatrix}2\\\\2\\end{pmatrix}}}$ Representing this vector from the domain with respect to the domain's basis ${\\displaystyle {\\rm {Rep}}_{B}({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})={\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}}_{B}}$ gives this matrix-vector product. ${\\displaystyle {\\rm {Rep}}_{D}(\\,\\pi ({\\begin{pmatrix}2\\\\2\\\\1\\end{pmatrix}})\\,)={\\begin{pmatrix}1&0&-1\\\\-1&1&1\\end{pmatrix}}_{B,D}{\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}}_{B}={\\begin{pmatrix}0\\\\2\\end{pmatrix}}_{D}}$ Expanding this representation into a linear combination of vectors from ${\\displaystyle D}$ ${\\displaystyle 0\\cdot {\\begin{pmatrix}2\\\\1\\end{pmatrix}}+2\\cdot {\\begin{pmatrix}1\\\\1\\end{pmatrix}}={\\begin{pmatrix}2\\\\2\\end{pmatrix}}}$ checks that the map's action is indeed reflected in", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "55.5^\\circ$. $\\displaystyle \\cos \\theta = \\frac{\\left | \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix} \\right |}{\\left | \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\right| \\, \\left | \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix} \\right | } \\\\ = \\displaystyle \\frac{|4|}{\\sqrt{3}\\sqrt{38}} \\\\ = \\displaystyle \\frac{4}{\\sqrt{114}}$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "\\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\qquad \\vec{b} = \\begin{pmatrix} 0\\\\2 \\end{pmatrix}\\\\\\\\ \\begin{array}{lll} \\text{area}=\\left|\\vec{a}_{projection}\\times\\vec{b} \\right| & \\qquad \\vec{a}_{projection} = \\left(\\vec{a}\\cdot \\vec{e}_x\\right)\\cdot \\vec{e}_x & \\qquad \\vec{e}_x = \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\\\\\\\ & \\qquad \\vec{a}_{projection} =\\left( 1.5\\cdot \\begin{pmatrix} \\cos{ (60\\ensurement{^{\\circ}} ) \\\\ \\sin{ (60\\ensurement{^{\\circ}} ) }\\\\ \\end{pmatrix}\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\right)\\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\left( 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )} +0\\right) \\cdot \\begin{pmatrix} 1\\\\0 \\end{pmatrix} \\\\\\\\ & \\qquad \\vec{a}_{projection} = \\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix} \\\\\\\\ \\text{area}=\\left|\\begin{pmatrix} 1.5\\cdot \\cos{(60\\ensurement{^{\\circ}} )}\\\\0 \\end{pmatrix}\\times\\begin{pmatrix} 0\\\\2 \\end{pmatrix} \\right| \\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\cos{ (60\\ensurement{^{\\circ}} ) } \\qquad | \\qquad \\cos{ (60\\ensurement{^{\\circ}} ) } = \\dfrac{1}{2}\\\\\\\\ \\text{area}=2\\cdot 1.5\\cdot \\dfrac{1}{2}\\\\\\\\ \\mathbf{\\text{area}=1.5 ~\\mathrm{m^2}}\\\\\\\\ \\end{array}$$ heureka Jul 6, 2015 #1 +10 I will make an assumption that the tarp is 1.5 m TALL and 2 m WIDE. One side of the shadow will remain the same if the sun is shining directly down: 2m To calculate the OTHER side of this shadow rectangle, The hypotenuse of the triangle is 1.5 m . We need the cosine of 60 dgrees times the hypotenuse to calculate the other side of the shadow COS(60) * 1.5 =.75 So you have a rectangle that is 2m x .75m Area = 2 * .75 = 1.5 sq m Guest Jul 5, 2015 #2 +19653 +10 Two math students erect a", "is recommended for all readers. Problem 4 Definition 1.1 uses two vectors ${\\displaystyle {\\vec {s}}}$ and ${\\displaystyle {\\vec {v}}}$. Consider the transformation of ${\\displaystyle \\mathbb {R} ^{2}}$ resulting from fixing ${\\displaystyle {\\vec {s}}={\\begin{pmatrix}3\\\\1\\end{pmatrix}}}$ and projecting ${\\displaystyle {\\vec {v}}}$ onto the line that is the span of ${\\displaystyle {\\vec {s}}}$. Apply it to these vectors. 1. ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}0\\\\4\\end{pmatrix}}}$ Show that in general the projection tranformation is this. ${\\displaystyle {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\mapsto {\\begin{pmatrix}(x_{1}+3x_{2})/10\\\\(3x_{1}+9x_{2})/10\\end{pmatrix}}}$ Express the action of this transformation with a matrix. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {1}{2}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}3/2\\\\1/2\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}0\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {2}{5}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}6/5\\\\2/5\\end{pmatrix}}}$ In general the projection is this. ${\\displaystyle {\\frac {{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\frac {3x_{1}+x_{2}}{10}}\\cdot {\\begin{pmatrix}3\\\\1\\end{pmatrix}}={\\begin{pmatrix}(9x_{1}+3x_{2})/10\\\\(3x_{1}+x_{2})/10\\end{pmatrix}}}$ The appropriate matrix is this. ${\\displaystyle {\\begin{pmatrix}9/10&3/10\\\\3/10&1/10\\end{pmatrix}}}$ Problem 5 Example 1.5 suggests that projection breaks ${\\displaystyle {\\vec {v}}}$ into two parts, ${\\displaystyle {\\mbox{proj}}_{[{\\vec {s}}\\,]}({{\\vec {v}}\\,})}$ and ${\\displaystyle {\\vec {v}}-{\\mbox{proj}}_{[{\\vec {s}}\\,]}({{\\vec {v}}\\,})}$, that are \"not interacting\". Recall that the two are orthogonal. Show that any two nonzero orthogonal vectors make up a linearly independent set. Suppose that ${\\displaystyle {\\vec {v}}_{1}}$ and ${\\displaystyle {\\vec {v}}_{2}}$ are nonzero and orthogonal. Consider the linear relationship ${\\displaystyle c_{1}{\\vec {v}}_{1}+c_{2}{\\vec {v}}_{2}={\\vec {0}}}$. Take the dot product of both sides of the equation with ${\\displaystyle {\\vec {v}}_{1}}$ to get that ${\\displaystyle {\\vec {v}}_{1}\\cdot (c_{1}{\\vec {v}}_{1}+c_{2}{\\vec {v}}_{2})=c_{1}\\cdot ({\\vec {v}}_{1}\\cdot {\\vec {v}}_{1})+c_{2}\\cdot", "\\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 \\end{pmatrix} = \\begin{pmatrix}0\\\\1\\\\0 \\end{pmatrix} \\\\ \\phi(|{-i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right)\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "$\\displaystyle { \\begin{pmatrix} n \\\\ r-1 \\end{pmatrix} }{ a }^{ n-(r-1) }{ b }^{ r-1 }$ Recall that the combination formula provides a way to calculate $\\begin{pmatrix} n \\\\ k \\end{pmatrix}$: $\\displaystyle {\\begin{pmatrix} n \\\\ k \\end{pmatrix}=\\frac{n!}{(n-k)!k! }}$ ### Example: Find the fifth term of ${(3x-4)}^{12}$ Note that the value of $n = 12$ in this case. Because we are looking for the fifth term, we use $r=5$. Plugging these values into the formula, we have: $\\displaystyle { \\begin{pmatrix} 12 \\\\ 5-1 \\end{pmatrix} }{ (3x) }^{ 12-(5-1) }{ (-4) }^{ 5-1 }$ $\\displaystyle { \\begin{pmatrix} 12 \\\\ 4 \\end{pmatrix} }{ (3x)}^{ 8 }{ (-4) }^{ 4 }$ Remember to evaluate $\\begin{pmatrix} 12 \\\\ 4 \\end{pmatrix}$ using the combination formula: \\displaystyle \\begin{align} \\frac{n!}{(n-k)!k! }&=\\frac{12!}{(12-4)!4! }\\\\ &=495 \\end{align} Subbing in $\\begin{pmatrix} 12 \\\\ 4 \\end{pmatrix}=495$ in the formula, we have: $\\displaystyle 495{ (3x)}^{ 8 }{ (-4) }^{ 4 }$ When the power is applied to the terms, the result is: $\\displaystyle 495\\cdot 6561{x}^{8} \\cdot 256 =831409920{x}^{8}$ Thus, the fifth term of ${(3x-4)}^{12}$ is $831409920{x}^{8}$.", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "0 \\\\ 1 \\\\0 \\end{pmatrix} \\, , \\quad \\vec{e}_z = \\begin{pmatrix} 0 \\\\ 0 \\\\1 \\end{pmatrix} .$$ Rewrite it in the alternative basis given by $${\\vec{e}}_1 = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ 1 \\\\0 \\end{pmatrix} \\, , \\quad {\\vec{e}}_2 = \\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 \\\\ -1 \\\\0 \\end{pmatrix} \\, , \\quad {\\vec{e}}_3 = \\begin{pmatrix} 0 \\\\ 0 \\\\1 \\end{pmatrix} .$$ First, we calculate the projections along the axis defined by the new basis vectors: \\begin{align} \\vec{e}_1 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ 1 \\\\0 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = \\frac{10}{\\sqrt{2}} \\\\ \\vec{e}_2 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 1 \\\\ -1 \\\\0 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = \\frac{8}{\\sqrt{2}} \\\\ \\vec{e}_3 \\cdot \\vec{v} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0\\\\ 0 \\\\1 \\end{pmatrix} \\cdot \\begin{pmatrix} 9\\\\ 1 \\\\ -2 \\end{pmatrix} = -\\sqrt{2} \\\\ \\end{align} This yields the components in the new basis and the vector $\\vec{v}$, therefore, reads in the new basis: \\begin{align} \\vec{v} = \\begin{pmatrix} \\frac{10}{\\sqrt{2}} \\\\ \\frac{8}{\\sqrt{2}} \\\\-\\sqrt{2} \\end{pmatrix} _{\\text{new basis}} = \\frac{10}{\\sqrt{2}} \\vec{e}_1 + \\frac{8}{\\sqrt{2}}\\vec{e}_2 -\\sqrt{2} \\vec{e}_3 \\end{align} How does the quantum momentum operator explicitly look like? The momentum operator is given by $\\hat{p}_i = -i \\hbar \\partial_i$, where $\\partial_i \\equiv \\frac{ \\partial}{\\partial i}$ and $i \\in \\{x,y,z\\}$. What does the canonical commutation relation: $$[ \\hat {p}_i,\\hat{x}_j] = - i \\hbar \\delta_{ij}$$ tell us? It makes a", "|}\\,k\\in \\mathbb {R} \\}}$ and so ${\\displaystyle (y{\\text{-axis}})^{\\perp }=x{\\text{-axis}}}$. 6. The description of ${\\displaystyle M}$ is easy to find by parametrizing. ${\\displaystyle M=\\{c\\cdot {\\begin{pmatrix}3\\\\1\\\\0\\end{pmatrix}}+d\\cdot {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\,{\\big |}\\,c,d\\in \\mathbb {R} \\}\\qquad B_{M}=\\langle {\\begin{pmatrix}3\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\rangle }$ Finding ${\\displaystyle M^{\\perp }}$ here just requires solving a linear system with two equations ${\\displaystyle {\\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\\\u&&&+&w&=&0\\end{array}}\\;{\\xrightarrow[{}]{-(1/3)\\rho _{1}+\\rho _{2}}}\\;{\\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\\\&&-(1/3)v&+&w&=&0\\end{array}}}$ and parametrizing. ${\\displaystyle M^{\\perp }=\\{k\\cdot {\\begin{pmatrix}-1\\\\3\\\\1\\end{pmatrix}}\\,{\\big |}\\,k\\in \\mathbb {R} \\}}$ 7. Here, ${\\displaystyle M}$ is one-dimensional ${\\displaystyle M=\\{c\\cdot {\\begin{pmatrix}0\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}\\qquad B_{M}=\\langle {\\begin{pmatrix}0\\\\-1\\\\1\\end{pmatrix}}\\rangle }$ and as a result, ${\\displaystyle M^{\\perp }}$ is two-dimensional. ${\\displaystyle M^{\\perp }=\\{{\\begin{pmatrix}u\\\\v\\\\w\\end{pmatrix}}\\,{\\big |}\\,0\\cdot u-1\\cdot v+1\\cdot w=0\\}=\\{j\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+k\\cdot {\\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix}}\\,{\\big |}\\,j,k\\in \\mathbb {R} \\}}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane ${\\displaystyle P}$ specified by ${\\displaystyle 3x+2y-z=0}$ in ${\\displaystyle \\mathbb {R} ^{3}}$. 1. Find a basis for ${\\displaystyle P}$. 2. Find ${\\displaystyle P^{\\perp }}$ and a basis for ${\\displaystyle P^{\\perp }}$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix}}}$ 4. Find the orthogonal projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle P}$ by keeping only the ${\\displaystyle P}$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. 1. Parametrizing the equation leads to" ]

[ "# Convert the point from rectangular coordinates to spherical coordinates. (-3,-3,\\sqrt{2}) Convert the point from rectangular coordinates to spherical coordinates. $$\\displaystyle{\\left(-{3},-{3},\\sqrt{{{2}}}\\right)}$$ • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Arham Warner To convert:", "\\left(- 6 , 6\\right) \\to \\left(6 \\sqrt{2} , \\frac{3 \\pi}{4}\\right) \\text{ in polar form}$", "convert the point from cylindrical coordinates to rectangular coordinates. $$(2,-\\pi,-4)$$ Related chapters Unlock Textbook Solution", "# How do you find the rectangular coordinates of the point with polar coordinates (-1,pi/3)? Sep 27, 2014 When trying to convert polar coords to rectangular coords or vice versa you need to be aware of a few equations. $\\cos \\left(\\theta\\right) = \\frac{x}{r} \\to x = r \\cos \\left(\\theta\\right)$ $\\sin \\left(\\theta\\right) = \\frac{y}{r} \\to y = r \\sin \\left(\\theta\\right)$ $\\tan \\left(\\theta\\right) = \\frac{y}{x} \\to \\theta = {\\tan}^{-} 1 \\left(\\frac{y}{x}\\right)$ ${x}^{2} + {y}^{2} = {r}^{2} \\to r = \\sqrt{{x}^{2} + {y}^{2}}$ Polar Coordinates $\\left(r , \\theta\\right) \\to \\left(- 1 , \\frac{\\pi}{3}\\right)$ Calculate the $x$-value $x = r \\cos \\left(\\theta\\right)$ $x = \\left(- 1\\right) \\cdot \\cos \\left(\\frac{\\pi}{3}\\right)$ $x = \\left(- 1\\right) \\left(\\frac{1}{2}\\right)$ $x = - \\frac{1}{2}$ Calculate the $y$-value $y = r \\sin \\left(\\theta\\right)$ $y = \\left(- 1\\right) \\left(\\frac{\\sqrt{3}}{2}\\right)$ $y = - \\frac{\\sqrt{3}}{2}$ Rectangular Coordinates $\\left(x , y\\right) \\to \\left(- \\frac{1}{2} , - \\frac{\\sqrt{3}}{2}\\right)$", "# Converting Coordinates from Polar to Rectangular ### Topic Page Converting Coordinates from Polar to Rectangular", "# How do you find the corresponding rectangular coordinates for the point (-1, (-3pi)/4 )? Jan 23, 2018 The rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$. #### Explanation: We can use $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = - 1 \\cos \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ $y = - 1 \\sin \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ So the rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$.", "Coordinates And Complex Numbers Convert the polar coordinates to rectangular coordinates: Explanation: To convert polar coordinates to rectangular coordinates , Using the information given in the question, The rectangular coordinates are", "# How do you find the rectangular coordinates of the point with polar coordinates (2,pi/4)? Since $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$, $\\left(x , y\\right) = \\left(2 \\cos \\left(\\frac{\\pi}{4}\\right) , 2 \\sin \\left(\\frac{\\pi}{4}\\right)\\right) = \\left(\\sqrt{2} . \\sqrt{2}\\right)$", "# Converting Coordinates from Rectangular to Polar ### Topic Page Converting Coordinates from Rectangular to Polar", "Converting Coordinates from Polar to Rectangular Topic Page Converting Coordinates from Polar to Rectangular", "Converting Coordinates from Rectangular to Polar Topic Page Converting Coordinates from Rectangular to Polar", "# How do you write the rectangular coordinates for the point: (6, 3/2 pi)? Mar 20, 2018 $\\left(- 3 , 3 \\sqrt{3}\\right)$ #### Explanation: For given coordinates $\\left(r , \\theta\\right)$, $\\left(r , \\theta\\right) \\to \\left(x , y\\right) \\implies \\left(r \\cos \\theta , r \\sin \\theta\\right)$ $r = 6$ $\\theta = \\frac{2 \\pi}{3}$ $\\left(6 , \\frac{2 \\pi}{3}\\right) \\to \\left(x , y\\right) \\implies \\left(6 \\cos \\left(\\frac{2 \\pi}{3}\\right) , 6 \\sin \\left(\\frac{2 \\pi}{3}\\right)\\right) \\equiv \\left(- 3 , 3 \\sqrt{3}\\right)$", "# How do you find the rectangular coordinates of the point with polar coordinates (5,pi/2)? Aug 24, 2014 To convert from from polar coordinates to rectangular coordinates, we must remember these: $x = r \\cos \\theta$ $y = r \\sin \\theta$ It's useful to remember these when we need to do these types of conversions. Knowing the $r$ and $\\theta$ from our question, we can find the rectangular coordinates: $r = 5$ $\\theta = \\frac{\\pi}{2}$ $x = \\left(5\\right) \\cos \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(0\\right) = 0$ $y = \\left(5\\right) \\sin \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(1\\right) = 5$ Therefore, our rectangular coordinates are: $\\left(0 , 5\\right)$", "# What is the distance between the following polar coordinates?: (2,(pi)/3), (4,(pi)/6) Apr 26, 2017 distance $= 2 \\sqrt{5 - 2 \\sqrt{3}} \\approx 2.479$ #### Explanation: Given 2 polar points: $\\left(2 , \\frac{\\pi}{3}\\right) , \\left(4 , \\frac{\\pi}{6}\\right)$ $\\frac{\\pi}{3} = {60}^{\\circ} \\text{ and } \\frac{\\pi}{6} = {30}^{\\circ}$ Convert the polar coordinates to rectangular coordinates using $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$: $\\left(2 , \\frac{\\pi}{3}\\right) = \\left(2 \\cos \\left(\\frac{\\pi}{3}\\right) , 2 \\sin \\left(\\frac{\\pi}{3}\\right)\\right) = \\left(2 \\cdot \\frac{1}{2} , 2 \\cdot \\frac{\\sqrt{3}}{2}\\right)$ $\\left(2 , \\frac{\\pi}{3}\\right) = \\left(1 , \\sqrt{3}\\right)$ $\\left(4 , \\frac{\\pi}{6}\\right) = \\left(4 \\cos \\left(\\frac{\\pi}{6}\\right) , 4 \\sin \\left(\\frac{\\pi}{6}\\right)\\right) = \\left(4 \\cdot \\frac{\\sqrt{3}}{2} , 4 \\cdot \\frac{1}{2}\\right)$ $\\left(4 , \\frac{\\pi}{6}\\right) = \\left(2 \\sqrt{3} , 2\\right)$ Find the distance between the two points: $d = \\sqrt{{\\left(1 - 2 \\sqrt{3}\\right)}^{2} + {\\left(\\sqrt{3} - 2\\right)}^{2}}$ $d = \\sqrt{1 - 4 \\sqrt{3} + 12 + 3 - 4 \\sqrt{3} + 4}$ d = sqrt(20 - 8sqrt(3)) = sqrt(4(5-2sqrt(3)) $d = 2 \\sqrt{5 - 2 \\sqrt{3}} \\approx 2.479$", "# How do you convert (6,-pi/5) into rectangular coordinates? $\\approx \\left(4.854 , - 3.527\\right)$ In general to convert the polar coordinate $\\left(r , \\theta\\right)$ into rectangular coordinates, you just do $\\left(r \\cos \\theta , r \\sin \\theta\\right)$. In the case of $\\left(6 , - \\frac{\\pi}{5}\\right)$, this would be $\\left(6 \\cos \\left(- \\frac{\\pi}{5}\\right) , 6 \\sin \\left(- \\frac{\\pi}{5}\\right)\\right) \\approx$(4.854, -3.527)#", "# How do you find the rectangular coordinates of the point with polar coordinates (3,pi)? Since $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$, $\\left(x , y\\right) = \\left(3 \\cos \\left(\\pi\\right) , 3 \\sin \\left(\\pi\\right)\\right) = \\left(- 3 , 0\\right)$", "# How do I find the distance between the points with polar coordinates (2, 120^circ) and (1, 45^circ)? $\\left(- 1 , \\sqrt{3}\\right)$ and $\\left(\\frac{1}{\\sqrt{2}} , \\frac{1}{\\sqrt{2}}\\right)$ $\\sqrt{{\\left(- 1 - \\frac{1}{\\sqrt{2}}\\right)}^{2} + \\left(\\sqrt{3} - \\frac{1}{\\sqrt{2}}\\right)}$ $= 1.99$ (approximately, according to my spreadsheet)", "{18} + i\\ \\mbox{sin}\\ \\frac{\\pi} {18}\\right )$ c. $4\\ \\mbox{cis}\\ \\left (\\frac{7\\pi} {4}\\right )$ 3. Convert the rectangular/polar coordinate into polar/rectangular coordinate: a. $(3, 3\\sqrt{3})$ b. (-2, 2) c. $\\left (4, \\frac{2\\pi} {3}\\right )$ d. $\\left (-1, \\frac{5\\pi} {6}\\right )$ 1. a. 2(cos 330o + i sin 330o) b. 18(cos 30o + i sin 30o) c. 5(cos 306.9o + i sin 306.9o) d. $\\sqrt{6}(\\mbox{cos} \\ 335.9^\\circ + i \\ \\mbox{sin} \\ 335.9^\\circ)$ 2. a. $\\frac{-3\\sqrt{3}} {2} - \\frac{3} {2} i$ b. 1.97 + 0.35i c. $2\\sqrt{2} - 2\\sqrt{2}i$ 3. . Feb 23, 2012 Jul 09, 2014", "# What is the Cartesian equivalent of polar coordinates (2, pi/6)? Oct 2, 2014 $\\left(r , \\theta\\right) \\to \\left(2 , \\frac{\\pi}{6}\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ Substitute in $r$ and $\\theta$ $\\left(x , y\\right) \\to \\left(2 \\cos \\left(\\frac{\\pi}{6}\\right) , 2 \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ Remember back to the unit circle and special triangles. $\\frac{\\pi}{6} = {30}^{\\circ}$ $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ Substitute in those values. $\\left(x , y\\right) \\to \\left(2 \\cdot \\frac{\\sqrt{3}}{2} , 2 \\cdot \\frac{1}{2}\\right)$ $\\left(x , y\\right) \\to \\left(\\sqrt{3} , 1\\right)$", "# How do you convert (3,pi/3) into rectangular coordinates? Nov 8, 2015 Polar with radius = 3 and angle $\\frac{\\pi}{3} \\mathmr{and} {60}^{o}$ #### Explanation: At, ${60}^{o}$, $\\sin \\left(60\\right) = \\sqrt{\\frac{3}{2}} \\mathmr{and} \\cos \\left(60\\right) = \\frac{1}{2}$ Now, to convert to rectangular , simply multiply these by the radius of 3: rectangular $= \\left[r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right] = \\left(\\frac{3}{2} , 3 \\frac{\\sqrt{3}}{2}\\right)$ hope that helped" ]

[ "# How do you find the corresponding rectangular coordinates for the point (-1, (-3pi)/4 )? Jan 23, 2018 The rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$. #### Explanation: We can use $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = - 1 \\cos \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ $y = - 1 \\sin \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ So the rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$.", "# How do you write the rectangular coordinates for the point: (6, 3/2 pi)? Mar 20, 2018 $\\left(- 3 , 3 \\sqrt{3}\\right)$ #### Explanation: For given coordinates $\\left(r , \\theta\\right)$, $\\left(r , \\theta\\right) \\to \\left(x , y\\right) \\implies \\left(r \\cos \\theta , r \\sin \\theta\\right)$ $r = 6$ $\\theta = \\frac{2 \\pi}{3}$ $\\left(6 , \\frac{2 \\pi}{3}\\right) \\to \\left(x , y\\right) \\implies \\left(6 \\cos \\left(\\frac{2 \\pi}{3}\\right) , 6 \\sin \\left(\\frac{2 \\pi}{3}\\right)\\right) \\equiv \\left(- 3 , 3 \\sqrt{3}\\right)$", "What are the rectangular coordinates of (3, -π/3)? $x = \\frac{3}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ Explanation: From the given: $r = 3$ and $\\theta = - \\frac{\\pi}{3}$ $x = r \\cdot \\cos \\theta$ $x = 3 \\cdot \\cos \\left(- \\frac{\\pi}{3}\\right)$ $x = 3 \\cdot \\left(\\frac{1}{2}\\right)$ $x = \\frac{3}{2}$ $y = r \\cdot \\sin \\theta$ $y = 3 \\cdot \\sin \\left(- \\frac{\\pi}{3}\\right)$ $y = 3 \\cdot \\frac{- \\sqrt{3}}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ God bless...I hope the explanation is useful.", "5 e^{\\frac{i\\pi}{2} } \\left( {e^{ - i\\phi } } \\right) = \\sqrt 5 e^{i\\left( {\\frac{\\pi}{2} - \\phi } \\right)}$. This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates. 8. Nevermind! I figured it out! Thanks!", "## Precalculus (6th Edition) Blitzer The rectangular coordinates of $\\left( 2,\\frac{\\pi }{6} \\right)$ are $\\left( \\sqrt{3},1 \\right)$. The rectangular coordinates of the point are $\\left( x,y \\right)$. Now, rewrite the rectangular coordinates in terms of polar coordinates, $x=r\\cos \\theta \\ \\text{ and }\\ y=r\\sin \\theta$ …… (I) Substituting the values of $r\\ \\text{ and }\\ \\theta$ in (I) we get, \\begin{align} & x=2\\cos \\frac{\\pi }{6} \\\\ & =2\\times \\frac{\\sqrt{3}}{2} \\\\ & =\\sqrt{3} \\end{align} And \\begin{align} & y=2\\sin \\frac{\\pi }{6} \\\\ & =2\\times \\frac{1}{2} \\\\ & =1 \\end{align} Therefore, the rectangular coordinates of $\\left( 2,\\frac{\\pi }{6} \\right)$ are $\\left( \\sqrt{3},1 \\right)$.", "11.415, -8.4, 1.6]} Coordinates will be given by $1 - {x}^{2} = - \\left(4 + 2 \\sqrt{3}\\right) \\left(x - 2\\right)$ i.e. ${x}^{2} - \\left(4 + 2 \\sqrt{3}\\right) x + 7 + 4 \\sqrt{3} = 0$, which gives $x = 2 + \\sqrt{3}$ and $y = 2 - {\\left(2 + \\sqrt{3}\\right)}^{2} = - 5 - 4 \\sqrt{3}$ and $1 - {x}^{2} = - \\left(4 - 2 \\sqrt{3}\\right) \\left(x - 2\\right)$ i.e. ${x}^{2} - \\left(4 - 2 \\sqrt{3}\\right) x + 7 - 4 \\sqrt{3} = 0$, which gives $x = 2 - \\sqrt{3}$ and $y = 2 - {\\left(2 - \\sqrt{3}\\right)}^{2} = - 5 + 4 \\sqrt{3}$ Hence coordinates are $\\left(2 + \\sqrt{3} , - 5 - 4 \\sqrt{3}\\right)$ and $\\left(2 - \\sqrt{3} , - 5 + 4 \\sqrt{3}\\right)$", "{e^{ - i\\phi } } \\right) = \\sqrt 5 e^{i\\left( {\\frac{\\pi}{2} - \\phi } \\right)}$. • January 14th 2010, 04:14 PM paupsers Quote: Originally Posted by Plato You are correct, I missed the 5i Anyway,it still works. $5i=5e^{\\frac{i\\pi}{2}}$ so $\\frac{{5i}}{{2 + i}} = \\sqrt 5 e^{\\frac{i\\pi}{2} } \\left( {e^{ - i\\phi } } \\right) = \\sqrt 5 e^{i\\left( {\\frac{\\pi}{2} - \\phi } \\right)}$. This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates. • January 14th 2010, 04:21 PM paupsers Nevermind! I figured it out! Thanks!", "# How do you convert ((5, -4pi/3)) into rectangular forms? Rectangular coordinates are $x = - \\frac{5}{2}$ and $y = \\frac{5 \\sqrt{3}}{2}$ #### Explanation: Given: $r = 5$ and $\\theta = \\frac{- 4 \\pi}{3} = - {240}^{\\circ}$ Angle $\\theta$ is located at the second quadrant. $x = r \\cos \\theta = 5 \\cdot \\cos \\left(\\frac{- 4 \\pi}{3}\\right) = - \\frac{5}{2}$ $y = r \\sin \\theta = 5 \\cdot \\sin \\left(\\frac{- 4 \\pi}{3}\\right) = \\frac{5 \\sqrt{3}}{2}$ Have a nice day!! from the Philippines...", "have the same coordinates > $\\mathrm{coordinates}\\left(P\\right)=\\mathrm{coordinates}\\left(\\mathrm{P2}\\right)$ $\\left[\\frac{{4}}{{3}}{,}\\frac{{4}}{{3}}{,}\\frac{{4}}{{3}}\\right]{=}\\left[\\frac{{4}}{{3}}{,}\\frac{{4}}{{3}}{,}\\frac{{4}}{{3}}\\right]$ (4)", "# How do you simplify 3/sqrt 3? It is $\\frac{3}{\\sqrt{3}} = {\\left(\\sqrt{3}\\right)}^{2} / \\sqrt{3} = \\sqrt{3}$", "square is $r\\sqrt{2}$ and $p=\\frac{S_{square}}{S_{circle}}=\\frac{2 r^2}{\\pi r^2}=\\frac{2}{\\pi}$", "into the page.) A point in the grid can be given coordinates $(n,k,a),$ where $n$ is the total number of steps, $k$ ($0\\le k\\le n$) is the number of left and right steps (that is, non-back steps), and $a$ ($0\\le a\\le k$) is the number of left steps. The two sides of your identity both equal the number of three-dimensional grid paths joining $(0,0,0)$ to $(n,k,a),$ and correspond to different ways of decomposing a three-dimensional grid path into two two-dimensional grid paths. In two dimensions, the two directions of movement might be given coordinates $$L:\\ \\left(-\\frac{1}{\\sqrt{2}},-\\frac{1}{\\sqrt{2}}\\right),\\qquad R:\\ \\left(\\frac{1}{\\sqrt{2}},-\\frac{1}{\\sqrt{2}}\\right).$$ In three dimensions, the three directions might be given coordinates $$L:\\ \\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{6}},-\\frac{1}{\\sqrt{3}}\\right),\\qquad R:\\ \\left(-\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{6}},-\\frac{1}{\\sqrt{3}}\\right),\\qquad B:\\ \\left(0,-\\sqrt{\\frac{2}{3}},-\\frac{1}{\\sqrt{3}}\\right).$$ If we project out the $x$-coordinate in this coordinate system, then left and right steps look the same: both correspond to a move which we might call \"front\", $$F:\\ \\left(0,\\frac{1}{\\sqrt{6}},-\\frac{1}{\\sqrt{3}}\\right).$$ A path from $(0,0,0)$ to $(n,k,a)$ then consists of $k$ front steps and $n-k$ back steps. The number of such paths is $\\binom{n}{k}.$ If we project out the back coordinate, that is, we project onto the plane of the left and right steps, then we effectively remove all back steps from the path. Such a path then consists of $k$ steps, $a$ of which are left steps, and $k-a$ of which are right steps.", "# What is the Cartesian form of ( 3 , (-9pi)/4 ) ? Nov 20, 2016 The cartesian coordinates are $\\left(\\frac{3 \\sqrt{2}}{2} , \\frac{- 3 \\sqrt{2}}{2}\\right)$ #### Explanation: To convert from polar coordinates $\\left(r , \\theta\\right)$ to cartesian coordinates $\\left(x , y\\right)$, we use the following equations $x = r \\cos \\theta$ and $y = r \\sin \\theta$ Here $\\left(r , \\theta\\right) = \\left(3 , - 9 \\frac{\\pi}{4}\\right)$ Therefore, $x = 3 \\cos \\left(\\frac{- 9 \\pi}{4}\\right) = 3 \\cos \\left(- 2 \\pi - \\frac{\\pi}{4}\\right) = 3 \\cos \\left(- \\frac{\\pi}{4}\\right)$ $x = 3 \\cos \\left(\\frac{\\pi}{4}\\right) = \\frac{3 \\sqrt{2}}{2}$ $y = 3 \\sin \\left(\\frac{- 9 \\pi}{4}\\right) = 3 \\sin \\left(- 2 \\pi - \\frac{\\pi}{4}\\right) = 3 \\sin \\left(- \\frac{\\pi}{4}\\right)$ $y = 3 \\cdot - \\frac{\\sqrt{2}}{2} = \\frac{- 3 \\sqrt{2}}{2}$ The cartesian coordinates are $\\left(\\frac{3 \\sqrt{2}}{2} , \\frac{- 3 \\sqrt{2}}{2}\\right)$", "- \\frac{\\pi}{2} \\right))$, or something like that – Hans Engler Nov 23 '12 at 16:58 No, I'm taking points from the punctured disk. In Cartesian coordinates, the map is $(u,v) \\mapsto (\\frac{u}{\\sqrt{u^2 + v^2}}, \\frac{v}{\\sqrt{u^2 + v^2}}, \\tan \\left( \\pi \\sqrt{u^2+v^2} - \\pi/2 \\right))$. – Hans Engler Nov 23 '12 at 22:11", "rectangular coordinates is $z=\\frac{H\\sqrt{x^{2}+y^{2}}}{R}$ Since $z=H \\;\\ and \\;\\ \\frac{H\\sqrt{x^{2}+y^{2}}}{R}$ implies that $x^{2}+y^{2}=R^{2}$ Therefore, the cone has equation $z=\\frac{Hr}{R}$ in polar coordinates. This gives the above integral which evaluates to $\\frac{1}{3}{\\pi}R^{2}H$ 12. Hello, Joanne! a) Prove the surface area of a sphere of radius $a$ is: $4 \\pi a^2$ We have the circle: . $x^2 + y^2\\:=\\:a^2\\quad\\Rightarrow\\quad y \\:=\\:\\sqrt{a^2-x^2}$ . (upper semicircle) Area of a surface of revolution: . $A \\;=\\;2\\pi\\int^b_a y\\sqrt{1 + \\left(\\frac{dy}{dx}\\right)^2}\\,dx$ We have: . $y \\:=\\:\\left(a^2-x^2\\right)^{\\frac{1}{2}}\\quad\\Rightarrow\\quad\\frac {dy}{dx} \\:=\\:\\frac{-x}{\\sqrt{a^2-x^2}}$ . . Then: . $\\sqrt{1 + \\left(\\frac{dy}{dx}\\right)} \\:=\\:\\sqrt{1 + \\left(\\frac{-x}{\\sqrt{a^2-x^2}}\\right)^2} \\:=\\:\\sqrt{1 + \\frac{x^2}{a^2-x^2}} \\:=\\:\\sqrt{\\frac{a^2}{a^2-x^2}}\\:=\\:\\frac{a}{\\sqrt{a^2-x^2}}$ So we have: . $A \\;=\\;2 \\times 2\\pi\\int^a_0\\sqrt{a^2-x^2}\\cdot\\frac{a}{\\sqrt{a^2-x^2}}\\,dx \\;=\\;4\\pi a\\int^a_0\\,dx \\;=\\;4\\pi a\\bigg[x\\bigg]^a_0$ . . $A \\;=\\;4\\pi a[a - 0]\\quad\\Rightarrow\\quad\\boxed{A \\;=\\;4\\pi a^2}$ b) Prove the area of a disk of radius $a$ is: $\\pi a^2$ We have the circle: . $x^2 + y^2\\:=\\:a^2\\quad\\Rightarrow\\quad y \\:=\\:\\sqrt{a^2-x^2}$ . (upper semicircle) The area under the semicircle is: . $A \\;=\\;\\int^a_{\\text{-}a}y\\,dx$ The area of the entire circle is: . $A \\;=\\;2 \\times \\int^a_{\\text{-}a}y\\,dx \\;=\\;4 \\int^a_0y\\,dx$ We have: . $A \\;=\\;4\\int^a_o\\sqrt{a^2-x^2}\\,dx$ Let: $x = a\\sin\\theta\\quad\\Rightarrow\\quad dx = a\\cos\\theta\\,d\\theta$ Substitute: . $A \\;=\\;4\\int a\\cos\\theta(a\\cos\\theta\\,d\\theta) \\;=\\;4a^2\\int\\cos^2\\theta\\,d\\theta \\;=\\;2a^2\\int(1 + \\cos2\\theta)d\\theta$ . . $= \\;2a^2\\left(\\theta + \\frac{1}{2}\\sin2\\theta\\right) \\;=\\;2a^2\\left(\\theta + \\frac{1}{2}\\cdot2\\sin\\theta\\cos\\theta\\right) \\;=\\;2a^2(\\theta + \\sin\\theta\\cos\\theta)$ Back-substitute: . $A \\;=\\;2a^2\\left[\\sin^{-1}\\left(\\frac{x}{a}\\right) + \\frac{x\\sqrt{a^2-x^2}}{a^2}\\right]^a_0$ Evaluate: . $A \\;=\\;2a^2\\left[\\left(\\sin^{-1}(1) + 0\\right) - \\left(\\sin^{-1}(0) + 0\\right)\\right]$ Therefore: . $A \\;=\\;2a^2\\left[\\frac{\\pi}{2}", "to write it as the square of a square root. You need to compute: $$2\\int _{\\frac{3\\pi }{4}} ^{\\frac{5\\pi }{4}} \\frac{1}{2}r^2\\, d\\theta$$ $$= \\int _{\\frac{3\\pi }{4}} ^{\\frac{5\\pi }{4}} 128\\cos (2\\theta)\\, d\\theta$$ $$= 64\\sin (2\\theta)|_ {\\theta = \\frac{3\\pi }{4}} ^{\\theta = \\frac{5\\pi }{4}}$$ $$= 64\\left (\\sin \\left (\\frac{pi}{2}\\right ) - \\sin \\left (\\frac{3\\pi }{2}\\right )\\right )$$ $$= 128$$", "centroid. I know how to get the area, but I have no idea how to use these coordinates in this formula. My english my not be perfect, in this case by quadrangle i mean an irregular figure where every angle may differ. Thank you First, choose a point $\\left(x_0,\\;y_0\\right)$ and label the remaining points in counterclockwise fashion. For example: $\\left(x_0,\\;y_0\\right)=(8,\\;3)$ $\\left(x_1,\\;y_1\\right)=(2,\\;5)$ $\\left(x_2,\\;y_2\\right)=(3,\\;4)$ $\\left(x_3,\\;y_3\\right)=(4,2)$ $\\left(x_4,\\;y_4\\right)=\\left(x_0,\\;y_0\\right)$ Then $\\displaystyle\\ A=\\frac{1}{2}\\sum_{i=0}^{n-1}\\left[x_i\\;y_{i+1}-x_{i+1}\\;y_i\\right]=$ $\\frac{1}{2}\\left[x_0\\;y_1-x_1\\;y_0\\right]+\\frac{1}{2}\\left[x_1\\;y_2-x_2\\;y_1\\right]+\\frac{1}{2}\\left[x_2\\;y_3-x_3\\;y_2\\right]+\\frac{1}{2}\\left[x_3\\;y_0-x_0\\;y_3\\right]$ Similarly for $C_x,\\;\\;C_y$ 5. Thank you very much!", "on the final step, but your evaluation is correct.$$-\\tan\\left(-\\frac\\pi3\\right)=\\sqrt 3$$", "## Calculus (3rd Edition) The rectangular coordinates are $(-\\frac{5}{2},\\frac{5}{2},\\frac{5\\sqrt 2}{2})$. Since $\\rho=5, \\, \\theta=\\frac{3\\pi}{4}, \\, \\phi=\\frac{\\pi}{4}$ and $$x=\\rho\\cos \\theta\\sin \\phi,\\quad y=\\rho\\sin\\theta\\sin\\phi, \\quad z=\\rho\\cos \\phi,$$ then we have $$x=5\\cos \\frac{3\\pi}{4}\\sin \\frac{\\pi}{4}=-\\frac{5}{2},$$ $$y=5\\sin\\frac{3\\pi}{4}\\sin \\frac{\\pi}{4}=\\frac{5}{2},$$ $$z=5\\cos \\frac{\\pi}{4}=\\frac{5\\sqrt 2}{2}.$$ Hence, the rectangular coordinates are $(-\\frac{5}{2},\\frac{5}{2},\\frac{5\\sqrt 2}{2})$.", "red square: $$\\frac{1}{2}(\\text{side})^2=\\frac{1}{2}\\left(\\frac{1}{7}\\right)^2$$" ]

[ "# How do you find the corresponding rectangular coordinates for the point (-1, (-3pi)/4 )? Jan 23, 2018 The rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$. #### Explanation: We can use $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = - 1 \\cos \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ $y = - 1 \\sin \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ So the rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$.", "# How do you convert the polar coordinate (-2,5pi/4) into cartesian coordinates? Oct 21, 2015 $\\left(- 2 , \\frac{5 \\pi}{4}\\right)$ [polar] $= \\left(\\sqrt{2} , \\sqrt{2}\\right)$ [Cartesian] #### Explanation: In Polar form: $\\left(- 2 , \\frac{5 \\pi}{4}\\right) = \\left(2 , \\frac{\\pi}{4}\\right)$ (draw a diagram if this isn't obvious) For an angle of $\\frac{\\pi}{4}$ the \"opposite\" and \"adjacent\" sides (i.e. $x$ and $y$) are equal and since $\\sqrt{{x}^{2} + {y}^{2}} =$ the radius $= 2$ $\\rightarrow x = y = \\sqrt{2}$ Oct 21, 2015 The solution is (-sqrt(2);-sqrt(2)). See explanation for details. #### Explanation: The given coordinates are not correct, because $r$ cannot be negative (it is a distance between 2 points and it is either zero or a positive real number). If the given point was (2;(5pi)/4) then corresponding Carthesian coordinates would be: $x = r \\cos \\varphi = 2 \\cdot \\cos \\left(\\frac{5 \\pi}{4}\\right) = 2 \\cdot \\left(- \\cos \\left(\\frac{\\pi}{4}\\right)\\right)$ $= 2 \\cdot \\left(- \\frac{\\sqrt{2}}{2}\\right) = - \\sqrt{2}$ $y = r \\sin \\varphi = 2 \\cdot \\sin \\left(\\frac{5 \\pi}{4}\\right) = 2 \\cdot \\left(- \\sin \\left(\\frac{\\pi}{4}\\right)\\right)$ $= 2 \\cdot \\left(- \\frac{\\sqrt{2}}{2}\\right) = - \\sqrt{2}$ So the Carthesian coordinates are (-sqrt(2);-sqrt(2))", "# How do you find the rectangular coordinates of the point with polar coordinates (-1,pi/3)? Sep 27, 2014 When trying to convert polar coords to rectangular coords or vice versa you need to be aware of a few equations. $\\cos \\left(\\theta\\right) = \\frac{x}{r} \\to x = r \\cos \\left(\\theta\\right)$ $\\sin \\left(\\theta\\right) = \\frac{y}{r} \\to y = r \\sin \\left(\\theta\\right)$ $\\tan \\left(\\theta\\right) = \\frac{y}{x} \\to \\theta = {\\tan}^{-} 1 \\left(\\frac{y}{x}\\right)$ ${x}^{2} + {y}^{2} = {r}^{2} \\to r = \\sqrt{{x}^{2} + {y}^{2}}$ Polar Coordinates $\\left(r , \\theta\\right) \\to \\left(- 1 , \\frac{\\pi}{3}\\right)$ Calculate the $x$-value $x = r \\cos \\left(\\theta\\right)$ $x = \\left(- 1\\right) \\cdot \\cos \\left(\\frac{\\pi}{3}\\right)$ $x = \\left(- 1\\right) \\left(\\frac{1}{2}\\right)$ $x = - \\frac{1}{2}$ Calculate the $y$-value $y = r \\sin \\left(\\theta\\right)$ $y = \\left(- 1\\right) \\left(\\frac{\\sqrt{3}}{2}\\right)$ $y = - \\frac{\\sqrt{3}}{2}$ Rectangular Coordinates $\\left(x , y\\right) \\to \\left(- \\frac{1}{2} , - \\frac{\\sqrt{3}}{2}\\right)$", "\\left(- 6 , 6\\right) \\to \\left(6 \\sqrt{2} , \\frac{3 \\pi}{4}\\right) \\text{ in polar form}$", "# How do you find the rectangular coordinates of the point with polar coordinates (5,pi/2)? Aug 24, 2014 To convert from from polar coordinates to rectangular coordinates, we must remember these: $x = r \\cos \\theta$ $y = r \\sin \\theta$ It's useful to remember these when we need to do these types of conversions. Knowing the $r$ and $\\theta$ from our question, we can find the rectangular coordinates: $r = 5$ $\\theta = \\frac{\\pi}{2}$ $x = \\left(5\\right) \\cos \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(0\\right) = 0$ $y = \\left(5\\right) \\sin \\left(\\frac{\\pi}{2}\\right) = \\left(5\\right) \\left(1\\right) = 5$ Therefore, our rectangular coordinates are: $\\left(0 , 5\\right)$", "# How do you find the rectangular coordinates of the point with polar coordinates (2,pi/4)? Since $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$, $\\left(x , y\\right) = \\left(2 \\cos \\left(\\frac{\\pi}{4}\\right) , 2 \\sin \\left(\\frac{\\pi}{4}\\right)\\right) = \\left(\\sqrt{2} . \\sqrt{2}\\right)$", "# How do you write the rectangular coordinates for the point: (6, 3/2 pi)? Mar 20, 2018 $\\left(- 3 , 3 \\sqrt{3}\\right)$ #### Explanation: For given coordinates $\\left(r , \\theta\\right)$, $\\left(r , \\theta\\right) \\to \\left(x , y\\right) \\implies \\left(r \\cos \\theta , r \\sin \\theta\\right)$ $r = 6$ $\\theta = \\frac{2 \\pi}{3}$ $\\left(6 , \\frac{2 \\pi}{3}\\right) \\to \\left(x , y\\right) \\implies \\left(6 \\cos \\left(\\frac{2 \\pi}{3}\\right) , 6 \\sin \\left(\\frac{2 \\pi}{3}\\right)\\right) \\equiv \\left(- 3 , 3 \\sqrt{3}\\right)$", "# How do you convert (6,-pi/5) into rectangular coordinates? $\\approx \\left(4.854 , - 3.527\\right)$ In general to convert the polar coordinate $\\left(r , \\theta\\right)$ into rectangular coordinates, you just do $\\left(r \\cos \\theta , r \\sin \\theta\\right)$. In the case of $\\left(6 , - \\frac{\\pi}{5}\\right)$, this would be $\\left(6 \\cos \\left(- \\frac{\\pi}{5}\\right) , 6 \\sin \\left(- \\frac{\\pi}{5}\\right)\\right) \\approx$(4.854, -3.527)#", "$\\frac{\\sqrt 3} 2 + \\frac 1 2 i, -1 + 0i$, and $\\frac{\\sqrt 3}2 - i\\frac 1 2$. In polar form, these are $e^{i\\pi/3}, e^{i\\pi}$, and $e^{i 5\\pi/3}$.", "# How do you convert (3,pi/3) into rectangular coordinates? Nov 8, 2015 Polar with radius = 3 and angle $\\frac{\\pi}{3} \\mathmr{and} {60}^{o}$ #### Explanation: At, ${60}^{o}$, $\\sin \\left(60\\right) = \\sqrt{\\frac{3}{2}} \\mathmr{and} \\cos \\left(60\\right) = \\frac{1}{2}$ Now, to convert to rectangular , simply multiply these by the radius of 3: rectangular $= \\left[r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right] = \\left(\\frac{3}{2} , 3 \\frac{\\sqrt{3}}{2}\\right)$ hope that helped", "{18} + i\\ \\mbox{sin}\\ \\frac{\\pi} {18}\\right )$ c. $4\\ \\mbox{cis}\\ \\left (\\frac{7\\pi} {4}\\right )$ 3. Convert the rectangular/polar coordinate into polar/rectangular coordinate: a. $(3, 3\\sqrt{3})$ b. (-2, 2) c. $\\left (4, \\frac{2\\pi} {3}\\right )$ d. $\\left (-1, \\frac{5\\pi} {6}\\right )$ 1. a. 2(cos 330o + i sin 330o) b. 18(cos 30o + i sin 30o) c. 5(cos 306.9o + i sin 306.9o) d. $\\sqrt{6}(\\mbox{cos} \\ 335.9^\\circ + i \\ \\mbox{sin} \\ 335.9^\\circ)$ 2. a. $\\frac{-3\\sqrt{3}} {2} - \\frac{3} {2} i$ b. 1.97 + 0.35i c. $2\\sqrt{2} - 2\\sqrt{2}i$ 3. . Feb 23, 2012 Jul 09, 2014", "# How do you graph the point L(3,-pi/6)? Jan 20, 2018 This is equivalent to plotting the rectangular point $\\left(3 \\sqrt{3} , - 3\\right)$. #### Explanation: We want to graph the polar point $\\left(3 , - \\frac{\\pi}{6}\\right)$. First consider the angle, $- \\frac{\\pi}{6}$. Rotate the angle, which has you at the origin facing $- \\frac{\\pi}{6}$, which is coterminal with $\\frac{11 \\pi}{6}$. You're facing into QIV. Now move 3 units out from the origin. That's the point. You can also convert to rectangular coordinates using $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = 6 \\cos \\left(- \\frac{\\pi}{6}\\right) = 6 \\left(\\frac{\\sqrt{3}}{2}\\right) = 3 \\sqrt{3}$ $x = 6 \\sin \\left(- \\frac{\\pi}{6}\\right) = 6 \\left(- \\frac{1}{2}\\right) = - 3$, So the rectangular point is $\\left(3 \\sqrt{3} , - 3\\right)$", "# How do you convert sqrt3 - 3i to polar form? Mar 21, 2018 The answer is $= \\sqrt{12} \\left(\\cos \\left(- \\frac{1}{3} \\pi\\right) + i \\sin \\left(- \\frac{1}{3} \\pi\\right)\\right)$ #### Explanation: Any complex number $z = a + i b$ can be converted to the polar form $z = | z | \\left(\\cos \\theta + i \\sin \\theta\\right)$ Where, $\\cos \\theta = \\frac{a}{| z |}$ and $\\sin \\theta = \\frac{b}{| z |}$ Here, $z = \\sqrt{3} - 3 i$ $| z | = \\sqrt{{\\left(\\sqrt{3}\\right)}^{2} + {\\left(- 3\\right)}^{2}} = \\sqrt{12}$ $\\cos \\theta = \\frac{\\sqrt{3}}{\\sqrt{12}} = \\frac{1}{2}$ $\\sin \\theta = - \\frac{3}{\\sqrt{12}} = - \\frac{\\sqrt{3}}{2}$ Therefore, $\\theta = - \\frac{\\pi}{3}$, $\\left[\\mod 2 \\pi\\right]$ $z = \\sqrt{12} \\left(\\cos \\left(- \\frac{1}{3} \\pi\\right) + i \\sin \\left(- \\frac{1}{3} \\pi\\right)\\right)$", "# What is the Cartesian equivalent of polar coordinates (2, pi/6)? Oct 2, 2014 $\\left(r , \\theta\\right) \\to \\left(2 , \\frac{\\pi}{6}\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ Substitute in $r$ and $\\theta$ $\\left(x , y\\right) \\to \\left(2 \\cos \\left(\\frac{\\pi}{6}\\right) , 2 \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ Remember back to the unit circle and special triangles. $\\frac{\\pi}{6} = {30}^{\\circ}$ $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ Substitute in those values. $\\left(x , y\\right) \\to \\left(2 \\cdot \\frac{\\sqrt{3}}{2} , 2 \\cdot \\frac{1}{2}\\right)$ $\\left(x , y\\right) \\to \\left(\\sqrt{3} , 1\\right)$", "# What is the difference between a rectangular coordinate system and a polar coordinate system? The point whose rectangular coordinates are $\\left(1 , 1\\right)$ corresponds to polar coordinates: $\\left(\\sqrt{2} , \\frac{\\pi}{4}\\right)$ and also $\\left(\\sqrt{2} , \\frac{9 \\pi}{4}\\right)$ and $\\left(\\sqrt{2} , \\frac{- 7 \\pi}{4}\\right)$ and $\\left(- \\sqrt{2} , \\frac{5 \\pi}{4}\\right)$ and infinitely many others.", "# What is the distance between the following polar coordinates?: (2,(pi)/3), (4,(pi)/6) Apr 26, 2017 distance $= 2 \\sqrt{5 - 2 \\sqrt{3}} \\approx 2.479$ #### Explanation: Given 2 polar points: $\\left(2 , \\frac{\\pi}{3}\\right) , \\left(4 , \\frac{\\pi}{6}\\right)$ $\\frac{\\pi}{3} = {60}^{\\circ} \\text{ and } \\frac{\\pi}{6} = {30}^{\\circ}$ Convert the polar coordinates to rectangular coordinates using $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$: $\\left(2 , \\frac{\\pi}{3}\\right) = \\left(2 \\cos \\left(\\frac{\\pi}{3}\\right) , 2 \\sin \\left(\\frac{\\pi}{3}\\right)\\right) = \\left(2 \\cdot \\frac{1}{2} , 2 \\cdot \\frac{\\sqrt{3}}{2}\\right)$ $\\left(2 , \\frac{\\pi}{3}\\right) = \\left(1 , \\sqrt{3}\\right)$ $\\left(4 , \\frac{\\pi}{6}\\right) = \\left(4 \\cos \\left(\\frac{\\pi}{6}\\right) , 4 \\sin \\left(\\frac{\\pi}{6}\\right)\\right) = \\left(4 \\cdot \\frac{\\sqrt{3}}{2} , 4 \\cdot \\frac{1}{2}\\right)$ $\\left(4 , \\frac{\\pi}{6}\\right) = \\left(2 \\sqrt{3} , 2\\right)$ Find the distance between the two points: $d = \\sqrt{{\\left(1 - 2 \\sqrt{3}\\right)}^{2} + {\\left(\\sqrt{3} - 2\\right)}^{2}}$ $d = \\sqrt{1 - 4 \\sqrt{3} + 12 + 3 - 4 \\sqrt{3} + 4}$ d = sqrt(20 - 8sqrt(3)) = sqrt(4(5-2sqrt(3)) $d = 2 \\sqrt{5 - 2 \\sqrt{3}} \\approx 2.479$", "# Convert the point (-sqrt3, -sqrt 3) in rectangular coordinates to polar coordinates that satisfies r > 0, -pi < theta leq pi Convert the point $\\left(-\\sqrt{3},-\\sqrt{3}\\right)$ in rectangular coordinates to polar coordinates that satisfies $r>0,-\\pi <\\theta \\le \\pi$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it smallq9 We have to convert the rectangular coordinates to polar coordinates.$\\left(-\\sqrt{3},-\\sqrt{3}\\right)$ as we know that the rectangular coordinates are (x, y) and the polar coordinates are $\\left(r,\\theta \\right)$ where $r=\\sqrt{{x}^{2}+{y}^{2}}$ here $r={\\sqrt{\\left(-\\sqrt{3}{\\right)}^{2}+\\left(-\\sqrt{3}\\right)}}^{2}$ $r=\\sqrt{3+3}$ $r=\\sqrt{6}$ also $\\theta =ta{n}^{-1}\\frac{x}{y}$ $\\theta ={\\mathrm{tan}}^{-1}\\left(\\frac{-\\sqrt{3}}{-\\sqrt{3}}\\right)$ $\\theta ={\\mathrm{tan}}^{-1}1$ $\\theta =\\frac{\\pi }{4}$ Therefore The polar coordinates are $\\left(\\sqrt{6},\\frac{\\pi }{4}\\right)$", "# How can (-pi, -3pi/4) be converted into rectangular coordinates? Feb 12, 2018 color(blue)((-pi-sqrt(2)/2,-pi-sqrt(2)/2) #### Explanation: From the diagram we can see that point P has polar coordinates $\\left(r , \\theta\\right)$ and Cartesian coordinates $\\left(x , y\\right)$. And $\\textcolor{w h i t e}{88} x = r \\cos \\left(\\theta\\right)$ , $y = r \\sin \\left(\\theta\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ We can use this to convert polar coordinates to Cartesian coordinates. $x = - \\pi \\cos \\left(\\frac{- 3 \\pi}{4}\\right) = - \\pi - \\frac{\\sqrt{2}}{2}$ $y = - \\pi \\sin \\left(\\frac{- 3 \\pi}{4}\\right) = - \\pi - \\frac{\\sqrt{2}}{2}$ $\\therefore$ color(blue)((-pi-sqrt(2)/2,-pi-sqrt(2)/2)", "# How do you find the distance between (2, π/3), (5, 2π/3)? Jun 11, 2017 The distance is $= 4.63$ #### Explanation: We convert the polar coordinates $\\left(r , \\theta\\right)$ into rectangular coordinates $\\left(x , y\\right)$ $\\left(r , \\theta\\right)$, $\\implies$, $\\left(x , y\\right) = \\left(r \\cos \\theta , r \\sin \\theta\\right)$ $\\left(2 , \\frac{1}{3} \\pi\\right)$, $\\implies$, $\\left(2 \\cos \\left(\\frac{1}{3} \\pi\\right) , 2 \\sin \\left(\\frac{1}{3} \\pi\\right)\\right) = \\left(1 , \\sqrt{3}\\right)$ and $\\left(5 , \\frac{2}{3} \\pi\\right)$, $\\implies$, $\\left(5 \\cos \\left(\\frac{2}{3} \\pi\\right) , 5 \\sin \\left(\\frac{2}{3} \\pi\\right)\\right) = \\left(- \\frac{5}{2} , \\frac{5}{2} \\sqrt{3}\\right)$ The distance is between $\\left(1 , \\sqrt{3}\\right)$ and $\\left(- \\frac{5}{2} , \\frac{5}{2} \\sqrt{3}\\right)$ $d = \\sqrt{{\\left(- \\frac{5}{2} - 1\\right)}^{2} + {\\left(\\frac{5}{2} \\sqrt{3} - \\sqrt{3}\\right)}^{2}}$ $= \\sqrt{\\frac{49}{4} + \\frac{27}{4}}$ $= \\frac{8.72}{2}$ $= 4.36$ Jun 11, 2017 Here is a graph of vectors from the origin to the two points: #### Explanation: Please observe that the line connecting the two points forms a triangle: We can use the Law of Cosines to find the length of the blue line (side c): ${c}^{2} = {a}^{2} + {b}^{2} - 2 \\left(a\\right) \\left(b\\right) \\cos \\left(\\theta\\right)$ Where $a = 5 , b = 2 , \\mathmr{and} \\theta = \\frac{2 \\pi}{3} - \\frac{\\pi}{3} = \\frac{\\pi}{3}$ ${c}^{2} = {5}^{2} + {2}^{2} - 2 \\left(5\\right) \\left(2\\right) \\cos \\left(\\frac{\\pi}{3}\\right)$ ${c}^{2} = 25 + 4 - 2 \\left(10\\right) \\left(\\frac{1}{2}\\right)$", "To plot: Thepoints which has polar coordinate displaystyle{left({2},frac{{{7}pi}}{{4}}right)} also two alternaitve sets for the same. To plot: Thepoints which has polar coordinate $\\left(2,\\frac{7\\pi }{4}\\right)$ also two alternaitve sets for the same. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Dora Let the point $\\left(2,\\frac{7\\pi }{4}\\right)$ be denoted by P. Consider the point $\\left(2,\\frac{7\\pi }{4}\\right)$, it is 2 units from the Origin O and line OP makes an angle of $\\frac{7\\pi }{4}$ with the positive direction of the x-axis. Add pi to the angle and change cardinal coordinate to -2 to get one alternate solution. Therefore, the point P $\\left(2,\\frac{7\\pi }{4}\\right)$ can also be represented by $\\left(-2,\\frac{7\\pi }{4}+\\pi \\right)=\\left(-2,\\frac{11\\pi }{4}\\right).$ Change the radial coordinate to -2 and subtract $3\\pi$ from the angle to obtain the other alternate solution. Thus, the point $P\\left(2,\\frac{7\\pi }{4}\\right)$ can be given by $\\left(-2,\\frac{7\\pi }{4}-3\\pi \\right)$ $=\\left(-2,-\\frac{19\\pi }{4}\\right).$ Use online graphing calculator and plot the points of P as shown below in Figure 1. From the Figure 1 it can be noted that alternate representations are plotted on the same location." ]

[ "Limited access The equation in rectangular coordinates for the surface that has equation $\\phi=\\pi/3$ in spherical coordinates has the form: $$a(x^2+y^2)=z^2$$ What is the value of $a$? Select an assignment template", "+0 # Spherical Coordinate Precalculus Problem +3 203 2 +40 For a constant $$c$$ in spherical coordinates $$(\\rho,\\theta,\\phi)$$, find the shape described by the equation $$\\phi = c$$ (A) Line (B) Circle (C) Plane (D) Sphere (E) Cylinder (F) Cone Apr 12, 2020 #1 0 Because of the spherical coordinates, phi = c traces a cylinder. So the answer is (E) cylinder. Apr 13, 2020 #2 +8354 +1 In spherical coordinates, the $$\\phi$$ represents the angle between (the line segment between the origin and a point) and the x-y plane. If $$\\phi = c$$, that means the angle $$\\phi$$ is constant. Construct a line L such that the angle between L and the x-y plane is $$\\phi$$, then revolve it around the z-axis. For any line segment Lin the resulting figure, the angle between L0 and x-y plane is $$\\phi$$. Therefore the resulting figure is a cone.", "in three-dimensional space with rectangular coordinates $$(x,y,z)$$ the equations $$x = k, \\, y = l$$ and $$z = m$$ where $$k, \\, l$$ and $$m$$ are constants, represent unbounded planes parallel to the $$yz$$-plane, $$xz$$-plane and $$xy$$-plane, respectively. With cylindrical coordinates $$(r, \\theta, z)$$, by $$r = c, \\, \\theta = \\alpha$$, and $$z = m$$, where $$c, \\alpha$$, and $$m$$ are constants, we mean an unbounded vertical cylinder with the z-axis as its radial axis; a plane making a constant angle $$\\alpha$$ with the $$xy$$-plane; and an unbounded horizontal plane parallel to the $$xy$$-plane, respectively. This means that the circular cylinder $$x^2 + y^2 = c^2$$ in rectangular coordinates can be represented simply as $$r = c$$ in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.) ## Integration in Cylindrical Coordinates Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table $$\\PageIndex{1}$$. These equations will become handy as we proceed with solving problems using triple integrals. Table $$\\PageIndex{1}$$: Equations of Some Common Shapes Circular cylinder Circular cone Sphere Paraboloid Rectangular $$x^2 + y^2 = c^2$$ $$z^2 = c^2 (x^2 + y^2)$$ $$x^2 + y^2 + z^2 = c^2$$", "Question # Describe in words the surface whose equation is given. \\phi = \\frac{\\pi}{4} (select the correct answer) 1)the top half of the right circular cone with Integrals Describe in words the surface whose equation is given. $$\\phi = \\frac{\\pi}{4}$$ (select the correct answer) 1)the top half of the right circular cone with vertex at the origin and axis the positive z-axis 2)the plane perpendicular to the xz-plane passing through z = x, where $$x \\geq 0$$ 3)the plane perpendicular to the xy-plane passing through y = x, where $$x \\geq 0$$ 4)the base of the right circular cone with vertex at the origin and axis the positive z-axis 5)the plane perpendicular to the yz-plane passing through z = y, where $$y \\geq 0$$", "questions you should realize that you will need to perform a volume integral over the volume of the cylinder. To make that integration easier, you will want to transform your expression for $\\Psi_{321}$ to cylindrical coordinates. So, what is the relationship between spherical coordinates $\\{r,\\theta,\\phi\\}$ and cylindrical coordinates $\\{\\rho,\\phi,z\\}$?", "the phi direction. • The initial phi value, in degrees. • The final phi value, in degrees. Example: add_sphere_observation obs6.dis 2.0 0.0 0.0 10.0 1.0 3 11 0.0 90.0 21 0.0 180.0", "# Describing Circles Express the circle in the form $z=e^{i\\theta}\\;(\\theta\\in[0,2\\pi])$, the ray in the form $z=\\rho e^{i\\theta_0}\\;(\\rho \\in[0,+\\infty))$ and substitute.", "u+iv=(x+iy)^\\alpha$$ Or even simpler, in polar coordinates: $$z_1=re^{i\\theta} \\ \\ \\ , \\ \\ \\ z_2=\\rho e^{o\\phi}$$ $$z_2=z_1^\\beta \\ \\ \\rightarrow \\ \\ \\ \\rho e^{i\\phi}=r^\\beta e^{i\\beta\\phi}$$ If $$\\beta<1$$, the angle between the planes decreases and the radius of the cylinder shrinks. If $$\\beta>1$$, the angle between the planes increases and the radius of the cylinder is enlarged. However, in both cases, I don't see how making such a conformal mapping makes the problem easier, as I'm only changing the angles and I'm still with the problem of finding the potential between two planes and a cylinder. Update: The only thing I can think would be better is choosing $$\\beta=\\pi / \\alpha$$ such that the planes in the new mapping form a straight line like this: Which I could potentially solve using the analogy to the case of a sphere in an uniform field, but I'm not sure if this is valid. Assuming it is valid, would that mean that, once I find the solution to this problem in terms of $$\\rho$$ and $$\\phi$$, I just work backwards the mapping to find the solution to the original problem in terms of $$r$$ and $$\\theta$$? The examples I read in the literature only dealt with mapping planes to planes or circles to circles and left the solutions only for", "of a curve, and three-dimensional volume of a solid. Place the shape on a coordinate plane so that O 0 is at the origin of the coordinate plane. - Facility to add point and remove. 14159. The equation θ = π / 3 describes the same surface in spherical coordinates as it does in cylindrical coordinates: beginning with the line θ = π / 3 in the x-y plane as given by polar coordinates, extend the line parallel to the z-axis, forming a plane. lifestance patient portal waiting room A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation z = k r, where. k. is a constant. In spherical coordinates, we have seen that surfaces of the form. φ = c. are half-cones. Last, in. b) Find the expression for ∇φ in spherical coordinates using the general form given below: (2 points) c) Find the expression for ∇ × F using the general form given below: (2 points) 2. By integrating the relations for da and dV in spherical coordinates that we discussed in class, find the surface area > and volume of a sphere. The spherical coordinates of any point in space are expressed in three parameters. Recalling the spherical coordinates: x = ρcosθsinϕ x = ρ cos θ", "# Thread: Spherical coordinates-finding phi 1. ## Spherical coordinates-finding phi I am having trouble doing problems in spherical coordinates, specifically, finding phi. I know it's the angle between the z axis and the shape but I don't know a good strategy for finding that especially on a cone. What are easy ways to do this? 2. HOW is the \"shape\" given? Do you know a formula in Cartesian coordinates? No one can tell you how to find $\\phi$ without knowing what you are given! Also, there is no single \" $\\phi$\" for a surface. There is a $\\phi$ for each point on the surface.", "out, you might notice that your book's answer: $$\\cos^2(\\phi) = \\sin^2(\\phi)$$ while correct, isn't completely simplified. Can you see that the final answer should be of the form $\\phi = C$ for the appropriate value of the constant C, and that the surface is actually one of the parameter surfaces for spherical coordinates? Yes I see what I did wrong. It is cos^2(phi) = sin^2(phi) = cos(phi) = sin(phi) >> . Can you see that the final answer should be of the form phi = C. Yes I guess pi/2 would be correct. So the equation z^2 = x^2 + y^2 is phi = pi/2 in spherical coord? Recognitions: Gold Member Homework Help Quote by tnutty Yes I see what I did wrong. It is cos^2(phi) = sin^2(phi) = cos(phi) = sin(phi) You mean $|\\cos(\\phi)| = |\\sin(\\phi)|$ >> . Can you see that the final answer should be of the form phi = C. Yes I guess pi/2 would be correct. So the equation z^2 = x^2 + y^2 is phi = pi/2 in spherical coord? Nope. Guess again. Think about $|\\tan(\\phi)| = 1$. I mean pi/4. Recognitions: Gold Member Homework Help Quote by tnutty I mean pi/4. Remember $\\tan(\\phi)$ can be ± 1. $\\pi/4$ will get you the top half of the cone. What value of $\\phi$", "which one you impose. For angular coordinates, the important thing is often to preserve periodicity, so you want to make sure that $\\phi(\\theta) = \\phi(\\theta + 2 \\pi)$.", "# Math Help - integral (within sphere, below cone) 1. ## integral (within sphere, below cone) Find the volume of the solid that lies within the sphere , above the plane, and outside the cone . I am having trouble finding the integral of phi p is from 0 to 1 theta is from 0 to 2pi but the scale of 3 on the function is screwing me up with theta i feel like im getting close, but i need a push i thought it may have been 3*sqrt(1/10) to pi/2 but it didnt turn out to be 2. Well, we just want the volume of the cone with the sphere top then just subtract it out of the sphere right? Equation of the cone is $z=3\\sqrt{x^2+y^2}$. So in spherical coordinates that's $\\rho \\cos(\\phi)=3\\sqrt{\\rho^2\\sin^2(\\phi)}$. Do all that and get: $\\rho^2\\left(9\\sin^2(\\phi)-\\cos^2(\\phi)\\right)=0$ or $\\phi=\\arctan(1/3)$. So equation of the cone in spherical coordinates is $\\rho=\\arctan(1/3)=a$. Do same with the sphere to get equation of it in spherical coordinates $\\rho=1$. Anyway, turn the crank and get: $V_{\\text{cone}}=\\int_0^a\\int_0^{2\\pi}\\int_0^1 \\rho^2\\sin(\\phi)d\\rho d\\theta d\\phi$ Pretty sure anyway. Double check it ok. 3. im going to have a very busy day today, but ill make sure to check it out tomorrow for sure", "@ $30 per hour and$2,2... ##### 1. Convert the point ( 215 7.) from cylindrical to spherical coordinates. 2. Set up a... 1. Convert the point ( 215 7.) from cylindrical to spherical coordinates. 2. Set up a triple integral, but do NOT evaluate, to find the volume of the solid in the first octant bounded by the coordinate planes and the plane 3x + 6y + 4z = 12. 1 3. Locate all relative maxima, relative minima, and sadd...", "# Finding volume and centroid Find the volume and centroid of the solid that lies in the intersection of the infinite cone $z=\\sqrt{x^2 + y^2}$ and the sphere $x^2 + y^2 + z^2 = 1$. Choose an appropriate coordinate system. So the second eqn is a perfect sphere with radius one. And since its volume I will probably need a triple integral... but im having a hard time find out what to integrate since I have 2 eqns :s Also for the coordinate system, im thinking something like spherical? - You should use spherical coordinates, the cone $z = \\sqrt{x^2 + y^2}$ looks like this therefore the volume is only bounded by the top half of the sphere. From an orthographic view you can imagine the volume to look like an ice cream cone with the bottom of the cone at the origin. The cone will intersect the sphere and cast a circle as a projection in the $xy$ plane. To find the $z$ plane of intersection equate the two equations. Your row will be bounded between $0$ and the radius of sphere. Phi will be bounded between $0$ and $\\pi/6$ I believe ( you can find phi yourself once you find the $z$ plane through which the cone intersects the sphere). You theta will have bounds $0", "$\\phi$ ($\\phi \\in [0,2 \\pi)$) around the (new) $z$ axis, i.e., $$\\hat{D}(\\theta,\\phi)=\\hat{D}_3(\\phi) \\hat{D}_1(\\theta).$$", "# Volume of solid inside surface in spherical coordinates. Find the volume of the solid inside the surface defined by the equation $\\rho=8\\sin \\phi$ in spherical coordinates So far I've set up an integral in spherical coordinates with $\\rho$ from $0$ to $\\rho=8\\sin \\phi$ $\\theta$ from $0$ to $2\\pi$, $\\phi$ from $0$ to $\\pi$. This is just a far off guess as I do not have a clue where to begin. I've been spending a lot of time just looking at the question and trying to make sense of it. Simply integrate the volume element: $$V= \\int_0^{2\\pi}\\int_0^\\pi\\int_0^{8\\sin\\phi}\\rho^2\\sin\\phi\\,d\\rho d\\phi d\\theta.$$ Sometimes it helps to have a visual representation of the solid: It helps you convince yourself that indeed $\\phi$ ranges from $0$ to $\\pi$ and $\\theta$ from $0$ to $2\\pi$, as in Martín-Blas Pérez Pinilla 's answer. • The OP knew the ranges from the beginning, the problem was the integrand. In any case, your idea is good. – Martín-Blas Pérez Pinilla Dec 17 '15 at 8:31", "region using rectangular, cylindrical and spherical coordinates, then comment on which of the three appears easiest to evaluate. ###### 39. The region enclosed by the unit sphere, $$x^2+y^2+z^2=1\\text{.}$$ ###### 40. The region enclosed by the cylinder $$x^2+y^2=1$$ and planes $$z=0$$ and $$z=1\\text{.}$$ ###### 41. The region enclosed by the cone $$z=\\sqrt{x^2+y^2}$$ and plane $$z=1\\text{.}$$ ###### 42. The cube enclosed by the planes $$x=0\\text{,}$$ $$x=1\\text{,}$$ $$y=0\\text{,}$$ $$y=1\\text{,}$$ $$z=0$$ and $$z=1\\text{.}$$ (Hint: in spherical, use order of integration $$d\\rho\\, d\\varphi\\, d\\theta\\text{.}$$)", "is even. This is a lot of work, but is good practice and background for dealing with the Hydrogen atom, something with spherical symmetry that is most naturally analyzed in the spherical polar coordinates. If (ρ, θ, φ) (rho, theta, phi) are the spherical coordinates of a point in 3-dimensional space and (x, y, z) are the rectangular coordinates, then: x = ρ cos θ sin φ. Every point in space is assigned a set of spherical coordinates of the form In case you’re not in a sorority or fraternity, is the lowercase Greek letter rho, […]. Cylindrical to Cartesian coordinates. I will assume that you are familiar with Cartesian coordinates and the unit vectors for the Cartesian system. You can explore the concepts of 3D cartesian coordinates using the applet below. 3D Spherical Plotting. It is another 3D graphing software on this list. The following sketch shows the. treating as the graph of a function z — (b) Find a parametrization r(r, e ) of using polar coordinates, and compute 2. powered by. Using spherical coordinates $(\\rho,\\theta,\\phi)$, sketch the surface defined by the equation $\\phi=\\pi/6$. Written Problems For several values of the constant , sketch the graph of the curve in given by the polar equation What do you notice about these curves?. A hemisphere of radius", "Free Version Difficult # Find the True Statement MVCALC-PSJHJA Let $S$ be a surface whose equation in spherical coordinates is: $$\\sin^2 \\phi=\\rho^2$$ Which of the following statements is TRUE? A The plane $z=1.1$ intersects $S$ B The intersection of $S$ with the plane $z=0$ is only a circle. C The plane $z=-2$ intersects $S$ D The intersection of $S$ with the plane $z=0$ is a circle and the origin. E The only point on $S$ whose three rectangular coordinates are all the same is the origin." ]

[ "if we anticipate singularity or some precision issue. If such situation arises, be extra careful or consider different approach. precisionxy Double. Set this to a very small number, and it is defaulted to 1e-15. return r, phi, theta Double. Position in spherical coordinates. import kero.werithmetic.rotation as rot x, y, z = 0.1, 0.1, 10 ro = rot.Rotation3D() r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"r, phi, theta =\", r, phi, theta) The output is the following, where r is in the same unit as x, y and z, while phi (polar angle), theta (azimuthal angle) are in radian. r, phi, theta = 10.000999950005 0.7853981633974483 0.014141192927807654 Example 2: rotate_towards_z_axis() state Numpy matrix, 3 by 1. This is to represent an initial 3D position vector that we want to rotate through, first, azimuth angle, specified by the negative of current_phi, and second, polar angle, specified by the negative of current_theta. current_phi Double. Current azimuth angle of the vector. current_theta Double. Current polar angle of the vector. return out Numpy matrix, 3 by 1. This is the rotated vector, see the description of state. import kero.werithmetic.rotation as rot import numpy as np ro = rot.Rotation3D() x, y, z = -0.1, 0.54, 2 state = np.matrix([[x], [y], [z]]) r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"BEFORE: r, phi, theta =\", r, phi,", "+0 # Help plz 0 58 1 The following polar grid is centered at the origin, with concentric circles of positive integer radii starting at $1,$ and consecutive rays through the origin with angles of $\\pi/12$ between them: [asy] size(200); pair A = (0,0); for (int i = 1; i < 6; ++i) { draw(Circle((0,0),i), linewidth(0.4)); } for(int i=0;i<360;i+=15) { draw(rotate(i)*((-5.0,0)--(5.0,0)), linewidth(0.4)); } pair A, B,C, D, EE, F; A = (1,-sqrt(3)); B = (1, sqrt(3)); C = (sqrt(3), 1); D = (4,0); EE = (0,0); F = (-sqrt(2), -sqrt(2)); dot(A, red+linewidth(3.5)); dot(B, red+linewidth(3.5)); dot(C, red+linewidth(3.5)); dot(D, red+linewidth(3.5)); dot(EE, red+linewidth(3.5)); dot(F, red+linewidth(3.5)); label(\"$A$\", A, S); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$E$\", EE, 3*S); label(\"$F$\", F, S); [/asy] Some of the points above are on the graph $r=4 \\cos(\\theta).$ Answer with the list in any order, separated by commas. Jul 30, 2021", "map on the surface to indicate the azimuthal angle ... use_cm=True, color_func=lambda t, phi: phi, ... rendering_kw={\"alpha\": 0.6, \"cmap\": \"twilight\"}, ... # indicates the azimuthal angle on the colorbar label ... label=r\"$\\phi$ [rad]\", ... show_curve=True, ... # this dictionary will be passes to plot3d_parametric_line in ... # order to draw the initial curve ... curve_kw=dict(rendering_kw={\"color\": \"r\", \"label\": \"cos(t)\"}), ... # activate the wireframe to visualize the parameterization ... wireframe=True, wf_n1=15, wf_n2=15, ... wf_rendering_kw={\"lw\": 0.5, \"alpha\": 0.75}) Revolve the same function around an axis parallel to the x axis, using Plotly: from sympy import symbols, cos, sin, pi from spb import plot3d_revolution, PB t, phi = symbols('t phi') plot3d_revolution( cos(t), (t, 0, pi), parallel_axis=\"x\", axis=(1, 0), backend=PB, use_cm=True, color_func=lambda t, phi: phi, rendering_kw={\"colorscale\": \"twilight\"}, show_curve=True, curve_kw=dict(rendering_kw={\"line\": {\"color\": \"red\", \"width\": 8}, \"name\": \"cos(t)\"}), wireframe=True, wf_n1=15, wf_n2=15, wf_rendering_kw={\"line_width\": 1}) Revolve a 2D parametric circle around the z axis: from sympy import * from spb import * t = symbols(\"t\") circle = (3 + cos(t), sin(t)) plot3d_revolution(circle, (t, 0, 2 * pi), backend=KB, show_curve=True, rendering_kw={\"opacity\": 0.65}, curve_kw={\"rendering_kw\": {\"width\": 0.05}}) Revolve a 3D parametric curve around the z axis for a given azimuthal angle, using Plotly: plot3d_revolution( (cos(t), sin(t), t), (t, 0, 2*pi), (phi, 0, pi), use_cm=True, color_func=lambda t, phi: t, label=\"t [rad]\", show_curve=True, backend=PB, aspect=\"cube\", wireframe=True, wf_n1=2, wf_n2=5) Interactive-widget plot of a", "# Difference between revisions of \"Cone\" A cone (or circular cone) is a three-dimensional solid. It consists of a circular base, a point (called the vertex), and all the points that lie on line segments connecting the vertex to the base. Thus, the cone is the special case of the pyramid in which the base is circular. $[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } //documentation in second version path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",(0.5,0,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE); [/asy]$ $[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0.6,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } // draw cone path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); // draw radius dot((0,0,0)); draw((0,0,0)--(1,0,0)); draw((0,0,0)--vertex,dotted); // label label(\"r\",(0.5,0,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),ENE); label(\"h\",(vertex.x,vertex.y,vertex.z/2),E); [/asy]$ ## Terminology The distance from the vertex to the plane containing the base is the height of the cone, and is frequently denoted $h$. The radius of the base is called the radius of the cone and is frequently denoted $r$. If the vertex lies directly above the center of the base, we call the cone a right circular cone (or right cone for short). In this case, the vertex", "# Difference between revisions of \"Cone\" A cone (or circular cone) is a three-dimensional solid. It consists of a circular base, a point (called the vertex), and all the points that lie on line segments connecting the vertex to the base. Thus, the cone is the special case of the pyramid in which the base is circular. $[asy] import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",((1+vertex.x)/2,(vertex.y)/2,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE); [/asy]$ $[asy] import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",((1+vertex.x)/2,(vertex.y)/2,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),ENE); [/asy]$ ## Important terminology The distance from the vertex to the plane containing the base is the height of the cone, and is frequently denoted $h$. The radius of the base is called the radius of the cone and is frequently denoted $r$. If the vertex lies directly above the center of the base, we call the cone a right circular cone (or right cone for short). In this case, the vertex is the same distance from every point on the boundary of the base; this distance is called the slant height of the", "( real a, real b, real alpha, real psi ) { real phi = psi - alpha ; pair u = unit ( (Sin(phi) / b, -Cos(phi) / a) ) ; return rotate ( alpha ) * scale ( a, b ) * u ; } (Note that in Asymptote, Sin and Cos with capital S and C take their arguments in degrees.) You mention swapping $x$ and $y$ and changing one sign. To be more precise, the sign of $y$ i.e. the new $y$ which is the old $x$, must be changed. This is actually complex multiplication by $-i$ which has the effect of rotating clockwise 90°. For instance $(2,3) \\times -i = (3,-2)$. Also, note that MathWorld clearly says that their $\\phi$ is the tangential angle. They denote the polar angle by $\\theta$. And I am suspecting that in the formula 60 they actually meant to write ${ a \\over b \\tan t}$ rather than ${a \\over b} \\tan t$ since the latter is simply wrong and would not work out. Notice for instance that at $t = 0$ for an unrotated ellipse which they are considering, the tangential angle should be $\\pi \\over 2$ but ${a \\over b} \\tan t$ would evaluate to zero and the arctan would then also be zero. I just sent", "size(10cm); triple eye = (5,2,3); currentprojection=orthographic(eye); surface hemisphere = surface(Arc(X,-X,c=O,normal=Z,n=16), c=O, axis=X, angle1=0, angle2=180); draw(shift(-5 eye)*hemisphere, material(white + opacity(0.5), emissivepen=0.2 white)); usepackage(\"amsmath\"); //for \\text command draw(O -- 1.6Z, arrow=Arrow3, L=Label(\"$H_{\\text{eff}}$\", align=W, position=EndPoint)); real theta(real t) { return t/20; } real phi(real t) { return -t + 1; } real r(real t) { return 1; } triple F(real t) { return polar(r(t), theta(t), phi(t)); } path3 spiral = reverse(graph(F, 0, 4pi, operator ..)); draw(spiral, blue + dotted + linewidth(1pt)); real t = 0.1; triple arrowpos = point(spiral, reltime(spiral, t)); t = 0.7; arrowpos = point(spiral, reltime(spiral, t)); triple M = point(spiral, 0); draw(O -- M, arrow=Arrow3, L=Label(\"$M$\", position=MidPoint)); draw(shift(M) * (O -- 0.5 cross(-M, Z)), red, arrow=Arrow3, L=Label(\"$-M \\times H_{\\text{eff}}$\", position=MidPoint)); triple dMdt = dir(spiral,0); triple crossprod = cross(M, dMdt); draw(shift(M) * (O -- 0.5 crossprod), blue, arrow=Arrow3, L=Label(\"$M \\times \\frac{dM}{dt}$\", position=MidPoint)); \\end{asypicture} \\end{document} The result: - Thanks! Exactly what I was looking for; sadly the LaTeX integration is not quite as smooth as with pgfplots. – knedlsepp Apr 30 '14 at 18:29 There are several packages for integrating Asymptote into LaTeX; if you state what features you would imagine for \"smooth integration,\" I can tell you whether any of these packages currently supports those features. – Charles Staats Apr 30 '14 at 22:55 I guess there's nothing one", "23-56 }$$ $$K_3 = HW\\cos\\phi_1+K_7x_0 \\tag{ 23-57 }$$ $$K_4 = HWP\\sin(\\phi_1-\\phi_g)+x_0 \\tag{ 23-58 }$$ $$K_8 = H(M\\sin\\theta-N\\cos\\theta)+K_5y_0 \\tag{ 23-59 }$$ $$K_9 = H(V\\sin\\theta-F\\cos\\theta)+K_6y_0 \\tag{ 23-60 }$$ $$K_{10} = HT\\cos\\phi_1+K_7y_0 \\tag{ 23-61 }$$ $$K_{11} = HTP\\sin(\\phi_1-\\phi_g)+y_0 \\tag{ 23-62 }$$ where, to review the meanings of previously defined symbols, $(\\phi_1, \\lambda_0) =$ latitude and longitude of the projection center and origin $\\phi_g =$ geocentric latitude of the point of perspective, found from equation (23-17) $\\gamma=$ azimuth east of north of the Yt axis, or the most upward-tilted axis of the plane of projection $\\omega =$ upward angle of tilt $P =$ distance from the center of the Earth to the point of perspective, divided by $a$, the semimajor axis. New symbols are as follows: $\\theta =$ clockwise angle through which the (Xt, Yt) axes are rotated for the arbitrary axes (X’t, Y’t) used for the constants $K_1\\unicode{x2013}K_{11}$. This may be made zero to retain the (Xt, Yt) axes. $(x_0, y_0) =$ offsets of the (Xt, Yt) axes to establish a different origin for the (X’t, Y’t) axes. They may also be set at zero to retain the (Xt, Yt) axes. The two sets of axes are related as follows: $$x'_t = x_t\\cos\\theta-y_t\\sin\\theta+x_0 \\tag{ 23-62a }$$ $$y'_t = y_t\\cos\\theta+x_t\\sin\\theta+y_0 \\tag{ 23-62b }$$ For inverse computations using projective constants given $K_1\\unicode{x2013}K_{11}, x’_t$", "Mollweide right phi = x.copy() # physical copy phi[x < 0] = 2 * np.pi + x[x<0] theta = np.pi/2 - y PHI, THETA = np.meshgrid(phi, theta) In [14]: l = 4 m = 2 SH_SP = sp.sph_harm(m, l, PHI, THETA).real # Plot just the real part What follows is a representation of the spherical harmonic as a heatmap of a field with a Mollweide projection. The value $l-m$ gives the number of horizontal \"bands\" or lines of constant latitude where the spherical harmonic is null. The order $m$ is the number of zero circles passing through the poles (beware with the count, in the flat Mollweide projection you'll see only half circles). To check this you can see a very nice interactive 3D visualization here. In [15]: #This is to enable bold Latex symbols in the matplotlib titles, according to: #http://stackoverflow.com/questions/14324477/bold-font-weight-for-latex-axes-label-in-matplotlib matplotlib.rc('text', usetex=True) matplotlib.rcParams['text.latex.preamble']=[r\"\\usepackage{amsmath}\"] In [16]: xlabels = ['$210^\\circ$', '$240^\\circ$','$270^\\circ$','$300^\\circ$','$330^\\circ$', '$0^\\circ$', '$30^\\circ$', '$60^\\circ$', '$90^\\circ$','$120^\\circ$', '$150^\\circ$'] ylabels = ['$165^\\circ$', '$150^\\circ$', '$135^\\circ$', '$120^\\circ$', '$105^\\circ$', '$90^\\circ$', '$75^\\circ$', '$60^\\circ$', '$45^\\circ$','$30^\\circ$','$15^\\circ$'] fig, ax = plt.subplots(subplot_kw=dict(projection='mollweide'), figsize=(10,8)) im = ax.pcolormesh(X, Y , SH_SP) ax.set_xticklabels(xlabels, fontsize=14) ax.set_yticklabels(ylabels, fontsize=14) ax.set_title('real$(Y^2_ 4)$', fontsize=20) ax.set_xlabel(r'$\\boldsymbol \\phi$', fontsize=20) ax.set_ylabel(r'$\\boldsymbol{\\theta}$', fontsize=20) ax.grid() fig.colorbar(im, orientation='horizontal'); Here ends this post, which has already been too long. Until next time!.", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "y) : ys) u = printf \"arc (%.4f:%.4f:1) arc (%.4f:%.4f:%.4f) \" l (y - alpha) (y - s * beta + delta) (y + s * beta) r ++ bar' r (y + alpha) ys u where alpha = 2 * asin (r/2) * 180 / pi beta = asin (r / 2) * 180 / pi + 90 s = signum (u - l) delta = if left /= (l > u) then 0 else 360 * s ys = map ((+) (sqrt 3 / 2)) [0.25, 0.5, 0.75] phis = [(True, 75), (True, 90), (False, 105)] narf r h ys phis = \"\\\\draw[contour1] \" ++ printf \"(0.5,%.4f)\" (sqrt 3 / 2 :: Double) ++ bar' r 60 phis 120 ++ foo' r (-0.5) (sqrt 3 / 2) ys (h + sqrt 3 / 2) ++ printf \" -- (0.5,%.4f) \" (h + sqrt 3 / 2) ++ foo'' r 0.5 (h + sqrt 3 / 2) (reverse ys) (sqrt 3 / 2) main = do putStrLn \\$ narf 0.05 1 ys phis • The tikz library spath3 can do this easily. How are you specifying the points where you want the arcs spliced in? With a curved path, are you concerned about pinpoint accuracy with the size of the spliced arcs, or is a little variation", "C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"P\", P, NW); label(\"O\", O, 1.5*S); label(\"\\theta\", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy]$ Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get $$PA = 2r\\sin\\left(\\dfrac{\\theta}2\\right)$$ $$PC = 2r\\sin\\left(\\dfrac{180-\\theta}2\\right) = 2r\\sin\\left(90 - \\dfrac{\\theta}2\\right) = 2r\\cos\\left(\\dfrac{\\theta}2\\right)$$ Multiplying and simplifying these 2 equations gives $$PA \\cdot PC = 4r^2 \\sin \\left(\\dfrac{\\theta}2 \\right) \\cos \\left(\\dfrac{\\theta}2 \\right) = 2r^2 \\sin\\left(\\theta \\right) = 56$$ Similarly $PB = 2r\\sin\\left(\\dfrac{90 +\\theta}2\\right)$ and $PD =2r\\sin\\left(\\dfrac{90 -\\theta}2\\right)$. Again, multiplying gives $$PB \\cdot PD = 4r^2 \\sin\\left(\\dfrac{90 +\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right) = 4r^2 \\sin\\left(90 -\\dfrac{90 -\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right)$$ $$=4r^2 \\sin\\left(\\dfrac{90 -\\theta}2\\right) \\cos\\left(\\dfrac{90 -\\theta}2\\right) = 2r^2 \\sin\\left(90 - \\theta \\right) = 2r^2 \\cos\\left(\\theta \\right) = 90$$ Dividing $2r^2 \\sin \\left(\\theta \\right)$ by $2r^2 \\cos \\left( \\theta \\right)$ gives $\\tan \\left(\\theta \\right) = \\dfrac{28}{45}$, so $\\theta = \\tan^{-1} \\left(\\dfrac{28}{45} \\right)$. Pluging this back into one of the equations, gives $$2r^2 = \\dfrac{90}{\\cos\\left(\\tan^{-1}\\left(\\dfrac{28}{45}\\right)\\right)}$$ If we imagine a $28$-$45$-$53$ right triangle, we see that if", "pi/3, label=\"theta\"), phi = slider(0,2pi,pi/20, slider(0,2*pi,pi/20, default = pi/4, label=\"phi\")): point_of_projection = point3d((rsin(theta)cos(phi),rsin(theta)sin(phi),r*cos(theta)), color=\"red\")point3d((r*sin(theta)*cos(phi),r*sin(theta)*sin(phi),r*cos(theta)), color=\"red\") show(S + north_pole + point_of_projection)point_of_projection) 3 No.3 Revision kcrisman 12082 ●39 ●126 ●246 ### 3D graphics bug? Hi, When I use the following code and I change the values of phi or theta, the previous red dot(point of projection) remains on the image and the new dot is added onto the image, giving two red dots. If I repeatedly change the values then more red dots appear. This problem does not occur if I get rid of the polygon inside the sphere. I'm not sure why this is happening... Works: S = sphere(size=10,color=\"yellow\",opacity=0.5) north_pole = point3d((0,0,10), color=\"blue\") r = 10 @interact def _(theta = slider(0,2*pi,pi/20, default = pi/3, label=\"theta\"), phi = slider(0,2*pi,pi/20, default = pi/4, label=\"phi\")): point_of_projection = point3d((r*sin(theta)*cos(phi),r*sin(theta)*sin(phi),r*cos(theta)), color=\"red\") show(S + north_pole + point_of_projection) Doesn't work: Z = Polyhedron([[2,3,3],[3,0,3],[2,6,1],[1,0,1],[6,7,2]]) Z_set_as_graphics = Z.show() S = sphere(size=10,color=\"yellow\",opacity=0.5) north_pole = point3d((0,0,10), color=\"blue\") r = 10 @interact def _(theta = slider(0,2pi,pi/20, default = pi/3, label=\"theta\"), phi = slider(0,2pi,pi/20, default = pi/4, label=\"phi\")): point_of_projection = point3d((rsin(theta)cos(phi),rsin(theta)sin(phi),r*cos(theta)), color=\"red\") show(Z_set_as_graphics + S + north_pole + point_of_projection) 4 No.4 Revision kcrisman 12082 ●39 ●126 ●246 ### 3D graphics bug? Hi, When I use the following code and I change the values of phi or theta, the previous red dot(point of", "symbols('theta phi') r1 = 1 r2 = 1.5 + sin(5 * phi) * sin(10 * theta) / 10 plot3d_spherical(r1, r2, (theta, 0, pi / 2), (phi, 0.35 * pi, 2 * pi), wireframe=True, wf_n2=25, backend=PB) Interactive-widget plot of spherical harmonics. Note that the plot’s creation and update might be slow and that it must be m < n at all times. Refer to iplot documentation to learn more about the params dictionary. from sympy import * from spb import * import param n, m = symbols(\"n, m\") phi, theta = symbols(\"phi, theta\", real=True) r = abs(re(Ynm(n, m, theta, phi).expand(func=True).rewrite(sin).expand())) plot3d_spherical( r, (theta, 0, pi), (phi, 0, 2*pi), params = { n: param.Integer(0, label=\"n\"), m: param.Integer(0, label=\"m\"), }, backend=KB ) spb.functions.plot3d_revolution(curve, range_t, range_phi=None, axis=(0, 0), parallel_axis='z', show_curve=False, curve_kw=None, **kwargs)[source] Generate a surface of revolution by rotating a curve around an axis of rotation. Parameters curveExpr, list/tuple of 2 or 3 elements The curve to be revolved, which can be either: • a symbolic expression • a 2-tuple representing a parametric curve in 2D space • a 3-tuple representing a parametric curve in 3D space range_t(symbol, min, max) A 3-tuple denoting the range of the parameter of the curve. range_phi(symbol, min, max) A 3-tuple denoting the range of the azimuthal angle where the curve will be revolved. Default to (phi,", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "phi_0 - np.log(r/r_0)/np.tan(alpha)) K = n * N / (r * np.sin(alpha)) KH = K * H beta = KH * (1 + 0.4*KH) D = (1 + KH + 0.3*KH**2) / (1 + 0.3*KH) term_3 = ((C/(K*D)) * np.cos(n*gamma)) * (np.cosh(K*z/beta))**-beta # sech is just 1/cosh return term_1 * term_2 * (term_3.sum(axis=0)) # sum over n G = 6.67430E-11 # m^3 / kg s^2 parsec = 3.0857E+16 # meters mH = 1.007825 * 1.660539E-27 # kg pi = np.pi N = 2 # number of arms alpha = 15 * pi/180. # pitch angle Rs = 7000 * parsec # radial dropoff rho_0 = 1E+06 * (14./11) * mH # midplane arm density r_0 = 8000 * parsec # at fiducial radius H = 180 * parsec # scale height of perturbation C = np.array([8/(3*pi), 0.5, 8/(15*pi)])[:, None, None] n = np.array([1, 2, 3])[:, None, None] # plot it hw = 30000 * parsec x = np.linspace(-hw, hw, 200) X, Y = np.meshgrid(x, x) r = np.sqrt(X**2 + Y**2) phi = np.arctan2(Y, X) z = 0. phi_0 = 0. potential = PHI(r, phi, z) if True: plt.figure() plt.imshow(potential) plt.colorbar() plt.gca().axes.xaxis.set_ticklabels([]) plt.gca().axes.yaxis.set_ticklabels([]) plt.title('+/- 30 kpc') plt.show() • – uhoh Feb 4 '20 at 13:49 What you're missing is that the potential and density functions they define are perturbations,", "bin[Slice - 1] += 1; break; } ++Slice; ZZ += inc; } */ var theta = Math.acos(Y); //This ensures that the apparent // angle from a pole to another point is the true angle. // All other angles are distorted! // So we take the Y from the n-dimensional sphere as the Y // of the three-dimensional sphere. var phi; if(this.dim > 2) { X = n[2]; phi = Atan(X,Z); } else if(Z > 0) phi = -0.5; else phi = Math.PI - 0.5; if(Z > 0) { this.spot(160.0, 600.0, 200.0, theta, phi, 4*(1+Accent(p)), 2, colours[planet],p,0); } else if(Accent(p) == 1) { this.spot(160.0, 600.0, 200.0, theta, phi, 4*(1+Accent(p)), 2, (colours[planet]|0xaaaaaa),p,0); } var sintheta = Math.sin(theta); var X1 = Math.sin(phi)*sintheta; // NOT the same as X! var Z1 = Math.cos(phi)*sintheta; // NOT the same as Z! var Y2 = costilt*Y - sintilt*Z1; var Z2 = sintilt*Y + costilt*Z1; if(Z2 > 0) { var thetatilt = Math.acos(Y2); var phitilt = Atan(X1,Z2); this.spot(160.0, 200.0, 200.0, thetatilt, phitilt, 4*(1+Accent(p)), 0, colours[planet],p,1); } else if(Accent(p) == 1) { this.spot(160.0, 200.0, 200.0, thetatilt, phitilt, 4*(1+Accent(p)), 0, (colours[planet]|0xaaaaaa),p,1); } } // this.spot(160.0, 600.0, 200.0, 0, 0, 4.0, 0, 0xff8000,0,2); // this.spot(160.0, 200.0, 200.0, tilt, 0, 4.0, 0, 0xff8000,0,3); DIMENSIONS = this.dim; return 0; } homenode.addEventListener(\"click\",clickhome,false); moredimensions.addEventListener(\"click\",clickmore,false); fewerdimensions.addEventListener(\"click\",clickfewer,false); this.setDim(this.dim); this.setTop(theTop); this.Egg(); var ip2 = document.getElementById(\"ip2\"); ip2.appendChild(this.svg);", "plotting the arcs. hold on r =1; phi = pi/4; theta = pi/4; %% plot cartesian coordinates: x0=0; y0=0; z0=0; plot3(x0+[0, .8, nan, 0, 0, nan, 0, 0], y0+[0, 0, nan, 0, .8, nan, 0, 0], z0+[0, 0, nan, 0, 0, nan, 0, .8],'k' ) text([x0+.85, x0, x0], [y0, y0+.8, y0], [z0, z0, z0+.85], ['$x$';'$y$';'$z$'],'FontSize',14, 'Interpreter','latex'); %% plot the ball line('xdata',sphcart(r,theta,phi,'x'),'ydata',sphcart(r,theta,phi,'y'),'zdata',sphcart(r,theta,phi,'z'),'marker','.','markersize',5); %% Plot the arm line('xdata',[0 sphcart(r,theta,phi,'x')],'ydata',[0 sphcart(r,theta,phi,'y')],'zdata',[0 sphcart(r,theta,phi,'z')]); %% Plot the projections line('xdata',[0 sphcart(r,theta,phi,'x')],'ydata',[0 sphcart(r,theta,phi,'y')],'zdata',[0 0],'linestyle','--'); %% Line from xy plane to point line('xdata',[sphcart(r,theta,phi,'x') sphcart(r,theta,phi,'x')],'ydata',[sphcart(r,theta,phi,'y') sphcart(r,theta,phi,'y')],'zdata',[0 sphcart(r,theta,phi,'z')],'linestyle','--') %% label r text(.5,.5,.8,'$r$','FontSize',14, 'Interpreter','latex') %% change view point az = 100; el = 45; view(az,el) %% get rid of axis labels set(gca, 'XTick', [], 'YTick', [], 'ZTick', []) set(gca, 'xcolor', 'w', 'ycolor', 'w','zcolor', 'w') ; %% arc (xy) theta = [0: pi/4*0.0001 :pi/4]; phi = linspace(0,0,10001); r = linspace(0.25,0.25,10001); [X,Y,Z]=sph2cart(theta,phi,r); plot3(X,Y,Z,'Color','k'); % label arc text(.3,0.08,0,'$\\theta$','FontSize',14,'Interpreter','latex') %% arc down from z phi = [pi/4: pi/4*0.0001 :pi/2]; theta = linspace(pi/4,pi/4,10001); r = linspace(0.25,0.25,10001); [X,Y,Z]=sph2cart(theta,phi,r); plot3(X,Y,Z,'Color','k'); % label arc text(.1,.08,0.4,'$\\phi$','FontSize',14,'Interpreter','latex') Here is the plot:", "at $(0,0,0)$ at an $(x,y,z)$ point using the angles from spherical coordinates. The two angles are $\\theta$ and $\\phi$. $\\theta$, the azimuthal angle, normally taken from $0˚$ to $360˚$, is the angle made in the $xy$ plane between the $x$-axis and the line connecting $(0,0,0)$ to $(x,y,0)$. Simply put, this gives us our \"yaw\". $\\phi$, the polar angle, normally taken from $0˚$ to $180˚$, is the angle made between the $z$-axis and the line connecting $(0,0,0)$ to $(x,y,z)$. This gives us something like the pitch. That is, $\\phi=90˚$ means that you're looking horizontally, whereas $\\phi=0˚$ means that you're looking vertically upward. Now, for a given point $(x,y,z)$, the calculations are as follows: $$\\theta = \\arctan\\left(\\frac yx\\right)\\\\ \\phi = \\arctan\\left(\\frac {\\sqrt{x^2+y^2}}z\\right)$$ So for example: to point your camera at the point $(1,1,1)$, you would need the angles $$\\theta = \\arctan\\left(\\frac 11\\right)=45˚\\\\ \\phi = \\arctan\\left(\\frac {\\sqrt{2}}1\\right) \\approx 55˚$$ - The x, y, z, what are they? are they the camera x, y, z or the target x, y, z or what? – Vallentin Aug 18 '13 at 1:41 Sorry, the camera's coordinates are taken to be $(0,0,0)$, and the target's coordinates are $(x,y,z)$. If you're given a camera point and a target point, you could subtract the camera point from both to make this work. – Omnomnomnom Aug 18 '13", "* cos(\\rho)}, 0, {\\r * sin(\\rho)}) \\foreach \\phi in {5,10,...,360} { -- ({\\scal * \\r * cos(\\rho) * cos(\\phi)}, {\\r * cos(\\rho) * sin(\\phi)}, {\\r * sin(\\rho)}) } -- cycle; } \\end{tikzpicture} \\end{document} Do you know how to draw a correct sphere in oblique projection. I prefer that one because the view of what I put into the sphere is better than in isometric projection. Thanks EDIT : axes z vertical \\begin{tikzpicture}[x={(-0.385cm,-0.385cm)},y={(1cm,0cm)},z={(0cm,1cm)},scale=4] \\coordinate (O) at (0,0,0); \\draw[->] (O) -- (2,0,0) node[below] {x}; \\draw[->] (O) -- (0,2,0) node[below] {y}; \\draw[->] (O) -- (0,0,2) node[right] {z}; \\foreach \\z in {-0.98,-0.96,...,0.98} {% \\begin{scope}[canvas is xy plane at z=\\z] \\draw[gray] (0,0) circle ({sqrt(1-\\z*\\z)}); \\end{scope}} \\begin{scope}[canvas is zy plane at x=0] \\draw (0,0) circle (1cm); \\draw (-1,0) -- (1,0) (0,-1) -- (0,1); \\end{scope} \\begin{scope}[canvas is zx plane at y=0] \\draw (0,0) circle (1cm); \\draw (-1,0) -- (1,0) (0,-1) -- (0,1); \\end{scope} \\begin{scope}[canvas is xy plane at z=0] \\draw (0,0) circle (1cm); \\draw (-1,0) -- (1,0) (0,-1) -- (0,1); \\end{scope} \\end{tikzpicture} Thus I obtained - Simple code, it's better to use arc and to determine the visible part. I think the result is something like that : \\documentclass{article} \\usepackage{tikz} \\usetikzlibrary{3D} \\pagestyle{empty} \\begin{document} \\begin{tikzpicture}[scale=4] \\foreach \\z in {-0.98,-0.96,...,0.98} {% \\begin{scope}[canvas is xy plane at z=\\z] \\draw[gray] (0,0) circle ({sqrt(1-\\z*\\z)}); \\end{scope}} \\begin{scope}[canvas is zy plane at" ]

[ "size(10cm); triple eye = (5,2,3); currentprojection=orthographic(eye); surface hemisphere = surface(Arc(X,-X,c=O,normal=Z,n=16), c=O, axis=X, angle1=0, angle2=180); draw(shift(-5 eye)*hemisphere, material(white + opacity(0.5), emissivepen=0.2 white)); usepackage(\"amsmath\"); //for \\text command draw(O -- 1.6Z, arrow=Arrow3, L=Label(\"$H_{\\text{eff}}$\", align=W, position=EndPoint)); real theta(real t) { return t/20; } real phi(real t) { return -t + 1; } real r(real t) { return 1; } triple F(real t) { return polar(r(t), theta(t), phi(t)); } path3 spiral = reverse(graph(F, 0, 4pi, operator ..)); draw(spiral, blue + dotted + linewidth(1pt)); real t = 0.1; triple arrowpos = point(spiral, reltime(spiral, t)); t = 0.7; arrowpos = point(spiral, reltime(spiral, t)); triple M = point(spiral, 0); draw(O -- M, arrow=Arrow3, L=Label(\"$M$\", position=MidPoint)); draw(shift(M) * (O -- 0.5 cross(-M, Z)), red, arrow=Arrow3, L=Label(\"$-M \\times H_{\\text{eff}}$\", position=MidPoint)); triple dMdt = dir(spiral,0); triple crossprod = cross(M, dMdt); draw(shift(M) * (O -- 0.5 crossprod), blue, arrow=Arrow3, L=Label(\"$M \\times \\frac{dM}{dt}$\", position=MidPoint)); \\end{asypicture} \\end{document} The result: - Thanks! Exactly what I was looking for; sadly the LaTeX integration is not quite as smooth as with pgfplots. – knedlsepp Apr 30 '14 at 18:29 There are several packages for integrating Asymptote into LaTeX; if you state what features you would imagine for \"smooth integration,\" I can tell you whether any of these packages currently supports those features. – Charles Staats Apr 30 '14 at 22:55 I guess there's nothing one", "radians ${{\\theta }_{d}}$= angle in degrees Segment of Circle Area of a segment of circle can be expressed as $A=1/2\\,\\,({{\\theta }_{r}}-\\sin {{\\theta }_{r}})\\,\\,{{r}^{2}}$ $=1/2\\,\\,(\\pi {{\\theta }_{d}}/180-sin{{\\theta }_{d}})\\,\\,{{r}^{2}}$ Right Circular Cylinder Lateral surface area of a right circular circle can be expressed as $A=2\\,\\pi \\,r\\,h$ where h = height of cylinder (m, ft) r = radius of base (m, ft) Right Circular Cone Lateral surface area of a right circular cone can be expressed as $A=\\pi \\,r\\,l$ $=\\pi r\\,\\,{{({{r}^{2}}+{{h}^{2}})}^{1/2}}$ where h = height of cone (m, ft) r = radius of base (m, ft) /= slant length (m, ft) Sphere Lateral surface area of a sphere can be expressed as $A=4\\,\\pi \\,{{r}^{2}}$ Where r = radius of base (m, ft) SYMBOLIC REPRESENTATION: Drawing Symbols provide a convenient way to draw geometrical shapes. To draw a geometrical shape, such as a pentagon or hexagon, select an appropriate symbol from the menu, specify the size of the symbol, and it is drawn at the indicated point. Example: Arrows (or pointers) in a drawing are commonly used to specify a direction for any reason. There are several arrow styles available in CADD programs. We can choose from simple two-point arrows to arrows passing through a number of points, and from simple to fency arrow styles. To draw an arrow, we need to", "if we anticipate singularity or some precision issue. If such situation arises, be extra careful or consider different approach. precisionxy Double. Set this to a very small number, and it is defaulted to 1e-15. return r, phi, theta Double. Position in spherical coordinates. import kero.werithmetic.rotation as rot x, y, z = 0.1, 0.1, 10 ro = rot.Rotation3D() r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"r, phi, theta =\", r, phi, theta) The output is the following, where r is in the same unit as x, y and z, while phi (polar angle), theta (azimuthal angle) are in radian. r, phi, theta = 10.000999950005 0.7853981633974483 0.014141192927807654 Example 2: rotate_towards_z_axis() state Numpy matrix, 3 by 1. This is to represent an initial 3D position vector that we want to rotate through, first, azimuth angle, specified by the negative of current_phi, and second, polar angle, specified by the negative of current_theta. current_phi Double. Current azimuth angle of the vector. current_theta Double. Current polar angle of the vector. return out Numpy matrix, 3 by 1. This is the rotated vector, see the description of state. import kero.werithmetic.rotation as rot import numpy as np ro = rot.Rotation3D() x, y, z = -0.1, 0.54, 2 state = np.matrix([[x], [y], [z]]) r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"BEFORE: r, phi, theta =\", r, phi,", "# I can not draw arc in a clockwise direction with this program in asymptote I have to trace the meridians. settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par défaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4(1,0,0), v2 = 0.4(Cos(60),Sin(60),0); draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); } But I can not draw arc in a clockwise direction with this code (this is the critic point!): draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); I tried: 1. to change v1 and v2 order with the same normal (Z), but in this way i have the angle greater than ninety degrees (i need the explementar angle) 2. to change v1 and v2 order and take normal = - Z, but in this way asymptote runs on a infinite loop. I don't know why. ## 1 Answer It took me a while to understand the question since IMHO it is not very clearly written. An additional complication arises from the fact that you did not present an MWE and seem to insist to have a special character in a comment: \"défaut\". All I", "box are available. • SurfaceHex(N, B, L, orient) • this operator allows to build the surface mesh of a 3d box • int[int] N=[nx,ny,nz]; // the number of seg in the 3 direction • real [int,int] B=[[xmin,xmax],[ymin,ymax],[zmin,zmax]]; // bounding bax • int [int,int] L=[[1,2],[3,4],[5,6]]; // the label of the 6 face left,right, front, back, down, right • orient the global orientation of the surface 1 extern (-1 intern), • returns a meshS type • Ellipsoide (RX, RY, RZ, h, L, OX, OY, OZ, orient) • h is the mesh size • L is the label • orient the global orientation of the surface 1 extern (-1 intern) • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) • where RX, RY, RZ are real numbers such as the parametric equations of the ellipsoid is: • returns a meshS type $\\forall u \\in [- \\frac{\\pi}{2},\\frac{\\pi}{2} [ \\text{ and } v \\in [0, 2 \\pi], \\vectthree{x=\\text{Rx } cos(u)cos(v) + \\text{Ox }}{y=\\text{Ry } cos(u)sin(v) + \\text{Oy }}{z = \\text{Rz } sin(v) + \\text{Oz } }$ • Sphere(R, h, L, OX, OY, OZ, orient) • where R is the raduis of the sphere, • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) •", "# Difference between revisions of \"Cone\" A cone (or circular cone) is a three-dimensional solid. It consists of a circular base, a point (called the vertex), and all the points that lie on line segments connecting the vertex to the base. Thus, the cone is the special case of the pyramid in which the base is circular. $[asy] import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",((1+vertex.x)/2,(vertex.y)/2,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE); [/asy]$ $[asy] import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",((1+vertex.x)/2,(vertex.y)/2,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),ENE); [/asy]$ ## Important terminology The distance from the vertex to the plane containing the base is the height of the cone, and is frequently denoted $h$. The radius of the base is called the radius of the cone and is frequently denoted $r$. If the vertex lies directly above the center of the base, we call the cone a right circular cone (or right cone for short). In this case, the vertex is the same distance from every point on the boundary of the base; this distance is called the slant height of the", "differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there? $\\textbf{(A) }1\\qquad\\textbf{(B) }2\\qquad\\textbf{(C) }3\\qquad\\textbf{(D) }4\\qquad\\textbf{(E) }5$ ## Problem 19 A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone? $[asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy]$ $\\text{(A) } \\dfrac32 \\quad \\text{(B) } \\dfrac{1+\\sqrt5}2 \\quad \\text{(C) } \\sqrt3 \\quad \\text{(D) } 2 \\quad \\text{(E) } \\dfrac{3+\\sqrt5}2$ ## Problem 20 For how many positive integers $x$ is $\\log_{10}(x-40) + \\log_{10}(60-x) < 2$ ? $\\textbf{(A) }10\\qquad \\textbf{(B) }18\\qquad \\textbf{(C) }19\\qquad \\textbf{(D) }20\\qquad \\textbf{(E) }\\text{infinitely many}\\qquad$ ## Problem 21 In the figure, $ABCD$ is a square of side length $1$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? $[asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0),", "• Above is a start putting them together. Jun 22 '16 at 17:14 Here's a starting point. Figuring out what I exactly did is left as an exercise. With[{r = 5/2, φ = π/4, s1 = 3/4, s2 = 3/2}, Graphics3D[{{Directive[CapForm[None], Opacity[2/3]], Tube[{{0, 0, -r Cot[φ/2]}, {0, 0, 0}, {0, 0, r Cot[φ/2]}}, {r, 0, r}]}, {Sphere[{0, 0, s1 Csc[φ/2]}, s1], Sphere[{0, 0, -s2 Csc[φ/2]}, s2]}, {Directive[EdgeForm[], Opacity[1/2, LightBlue]], InfinitePlane[#1, {#2 - #1, {0, 1, 0}}] &[ {s1/(s1 + s2) Sqrt[(s1 + s2)^2 - (s1 - s2)^2 Sin[φ/2]^2], 0, s1 Csc[φ/2] - s1 (s1 - s2) Sin[φ/2]/(s1 + s2)}, {s2/(s1 + s2) Sqrt[(s1 + s2)^2 - (s1 - s2)^2 Sin[φ/2]^2], 0, s2 (s2 - s1) Sin[φ/2]/(s1 + s2) - s2 Csc[φ/2]}]}}, Boxed -> False]] • Here are the corresponding pictures for the other conics: ellipse, parabola. The old code by Stan Wagon was rather long, but thanks to primitives like Cone[], Sphere[], and InfinitePlane[] being built-in, things are now easier to do. Jun 22 '16 at 6:27 • We need to show (with a small disk or tiny sphere) points of 1) tangential contact , 2) arbitrary point on hyperbola , 3) parallel circle points joined at any arbitrary point so that the difference of segment lengths is seen constant for hyperbola. Jun 23 '16 at 15:49 •", "It is still not an exact drawing, but, I would guess, that if you know how to mathematically construct the curved lines that form the spiral, it should be possible to come up with an exact drawing based on this solution. Oct 26 '21 at 19:48 While waiting for a Tikz's answer, see a bit with Asymptote. Compile on http://asymptote.ualberta.ca/ I don't know what \\alpha and \\gamma mean! settings.render=8; import solids; import graph3; usepackage(\"newtxmath\"); size(200,400); real r=3; real h=2.5pi; currentprojection=orthographic(8,2,4); revolution R=cylinder(O,r,h); // The circular helix triple f(real t){ real a=r*cos(t); real b=r*sin(t); real c=t; return (a,b,c); } real k=7; path3 plane=(k,k,0)--(k,-k,0)--(-k,-k,0)--(-k,k,0)--cycle; draw(plane); label(\"$\\alpha$\",(k,-k,0),2dir(20)); draw(surface(R),lightgreen+opacity(0.5),render(compression=Low)); draw(surface(circle((0,0,h),r)),lightgreen+opacity(0.5),render(compression=Low)); draw(O--(0,0,h)^^O--(0,r,0)^^O--(r,0,0),dashed); dot(scale(.7)*\"$O$\",(0,0,0),dir(165)); dot((0,0,h)); guide3 g=graph(f,0,h,150); draw(Label(\"$\\varphi$\",Relative(.05)),g); draw(Label(\"$\\varphi$\",Relative(.05)),g); triple P=f(1),overP=f(1+2pi); triple P1=(P.x,P.y,0),P2=(0,0,P.z); draw(overP--P1); draw(P2--P^^O--P1,dashed); dot(scale(.7)*\"$P$\",P,2dir(120)); dot(scale(.7)*\"$\\overline{P}$\",overP,dir(-20)); dot(scale(.7)*\"$P_1$\",P1,dir(-90)); dot(scale(.7)*\"$P_2$\",P2,dir(150)); xaxis3(Label(\"$x^1$\",Relative(.5),2LeftSide),r,6,Arrow3); yaxis3(Label(\"$x^2$\",Relative(.5),4dir(20)),r,6,Arrow3); zaxis3(Label(\"$x^3=$ a\",Relative(.6),2RightSide),h,h+4,Arrow3); • thanks a lot Nguyen VanChi 1998. i noticed that working with asymptote make everything easier and more elegant. Can i ask if you can add in the drow (sorry i'm not practicle with the code) the \\gamma and \\varphi? and last question: is it possible to have classicthesis font for the letters and symbols in the drow? this is not necessary but in case it would be wonderful (far above my expectations) thanks again a lot :D Oct 26 '21 at 17:09 • Nice! and h should be 2.5pi Oct", "Difference between revisions of \"2014 AMC 12B Problems/Problem 19\" Problem A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone? $[asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4)); [/asy]$ $\\text{(A) } \\dfrac32 \\quad \\text{(B) } \\dfrac{1+\\sqrt5}2 \\quad \\text{(C) } \\sqrt3 \\quad \\text{(D) } 2 \\quad \\text{(E) } \\dfrac{3+\\sqrt5}2$ Solutions Solution 1 First, we draw the vertical cross-section passing through the middle of the frustum. let the top base equal 2 and the bottom base to be equal to 2r $[asy] size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label(\"r-1\",(X+B)/2,S); label(\"1\",(Y+C)/2,N); label(\"s\",(O+Y)/2,W); label(\"s\",(O+X)/2,W);", "# Difference between revisions of \"Cone\" A cone (or circular cone) is a three-dimensional solid. It consists of a circular base, a point (called the vertex), and all the points that lie on line segments connecting the vertex to the base. Thus, the cone is the special case of the pyramid in which the base is circular. $[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } //documentation in second version path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label(\"h\",(vertex.x,vertex.y,vertex.z/2),W);label(\"r\",(0.5,0,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE); [/asy]$ $[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0.6,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } // draw cone path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); // draw radius dot((0,0,0)); draw((0,0,0)--(1,0,0)); draw((0,0,0)--vertex,dotted); // label label(\"r\",(0.5,0,0),S); label(\"s\",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),ENE); label(\"h\",(vertex.x,vertex.y,vertex.z/2),E); [/asy]$ ## Terminology The distance from the vertex to the plane containing the base is the height of the cone, and is frequently denoted $h$. The radius of the base is called the radius of the cone and is frequently denoted $r$. If the vertex lies directly above the center of the base, we call the cone a right circular cone (or right cone for short). In this case, the vertex", "of $Y_{5,5}$ seem to be the logical choice. However, it turns out it cannot has the correct symmetry because $Y_{50}$ is not “even”. $\\displaystyle f_d(\\theta, \\phi) = Y_{60}(\\theta,\\phi) + \\sqrt{\\frac{28}{11}} Y_{65}(\\theta, \\phi)$ The Mathematica code for generating the graph is GraphicsGrid[{{ SphericalPlot3D[1 + f(θ,φ), {θ, 0, π}, {φ, 0, 2 π}], ParametricPlot3D[{Cos[φ] Sin[θ], Sin[φ] Sin[θ], Cos[θ]}, {θ, 0, π}, {φ, 0, 2 π}, ColorFunction -> Function[{x, y, z, u, v}, Hue[Abs[1 + f(u,v)]]], ColorFunctionScaling -> False, Mesh -> None, PlotPoints -> 200] }}, ImageSize -> 600] The code for calculating the coefficient is Solve[ SphericalHarmonicY[n, 0, k, 0] + a SphericalHarmonicY[n, n, k, 0] == SphericalHarmonicY[n, 0, 0, 0] , a] Where $n$ is the order of the spherical harmonic, $k$ is the angle for the next face or vertex for fixed $\\theta = 0$. Tetraherdon $k = 2 \\tan^{-1}(\\sqrt{2})$. Octaherdon $k = \\pi/2$. Dodecahedron $k = 2 \\tan^{-1}( (1+\\sqrt{5})/2)$.", "of $n$ with $2\\le n \\le 1000$ for which $A(n)$ is an integer. ## Problem 15 A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ 2015 AIME I (Problems • Answer Key • Resources) Preceded by2014 AIME II Followed by2015 AIME II 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "a suitable $m$ by looking at the lengths of the Spherefun representation if $h$. [a,b] = length(h); m = ceil(max(a,b)); zero = ballfun(0); u = poisson(zero,h,m); Here, using cartesian coordinates, we confirm that $u$ matches the boundary data at the point $(1,0,0)$: h(1,0,0) u(1,0,0) ans = -0.761180229121337 ans = -0.761180229121336 Here, using spherical coordinates, we confirm that it matches the boundary data at the longitude and latitude coordinates of Oxford. long = -1.26*pi/180; lat = 51.75*pi/180; h(long, pi/2-lat) u(1, long, pi/2-lat,'spherical') ans = -1.847977702024471 ans = -1.847977702024464 Another check we can carry out concerns mean values. The value of $u$ at the origin should equal the mean of the boundary data: meanh u(0,0,0) meanh = -0.020281680162226 ans = -0.020281680162149 ### 3. The solution on an inner sphere Since the Laplace equation is a smoothing operation, the solution $u$ is not very exciting in the interior. For example, if we simply plot $u$, the image shows an apparently uniform color in the inner sphere of radius $1/2$: plot(u) We can construct a spherefun corresponding to the values of $u$ on this sphere of radius $0.5$ and plot it: uinner = u(.5,:,:,'spherical'); plot(uinner), colorbar Note the small range of values revealed in the colorbar. The mean value over this sphere is our familiar value: meanh mean2(uinner) meanh = -0.020281680162226 ans", "# How to make dashed lines automatic of a sphere when I rotate a sphere? I copy this code from here settings.render=8; import solids; size(5cm); currentprojection=orthographic(1,0.3,0.3); revolution b=sphere(O,1); draw(surface(b),white+opacity(.5)); skeleton s; b.transverse(s,reltime(b.g,0.5),P=currentprojection); draw(s.transverse.back,dashed); draw(s.transverse.front); draw(rotate(90,X)*s.transverse.back,dashed); draw(rotate(90,X)*s.transverse.front); real rayon=0.8, phi=90, theta=45; triple pM=dir(phi,theta); draw((O--pM),dashed+green+0.7bp,Arrow3(size=7)); label(Label(\"$\\psi$\",yellow),pM,0.5dir(-90)); xaxis3(\"$x$\",1,1.5,Arrow3); xaxis3(-1.3,1,linetype(new real[] {8,8})); yaxis3(\"$y$\",1,1.5,Arrow3); yaxis3(-1.3,1,linetype(new real[] {8,8})); zaxis3(\"$z$\",1,1.5,Arrow3); zaxis3(0,1,linetype(new real[] {8,8})); and try to compile at here http://asymptote.ualberta.ca/ I rotate output by hand, the result seems incorrect. How can I get the result like this? • If you mean interfacing with a mouse, you can't. Changing the image would require recompiling the code. You need to do the whole thing using Javascript. Oct 16 '21 at 17:21 As John Kormylo'comment, I think, you can not rotate the sphere with a mouse (if you use TikZ). We hope that, in future, Asymptote can do this (keep automatic dashed lines when we rotate sphere!). You can try this code. I hope, it helps you something. \\documentclass[tikz,border=3mm]{standalone} \\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \\begin{document} \\pgfdeclarelayer{background} \\pgfdeclarelayer{foreground} \\pgfsetlayers{background,main,foreground} \\foreach \\Angle in {5,15,...,360} {\\begin{tikzpicture}[3d/install view={phi=\\Angle,theta=70},line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={R=5;}] \\path (0,0,0) coordinate (I) foreach \\Z in {-75,-60,...,75} {(0,0,{R*sin(\\Z)}) coordinate (I\\Z)} foreach \\Z in {-75,-60,...,90} {({cos(\\Z)},{sin(\\Z)},0) coordinate (n\\Z)};", "to the right size (52 cm by 14 cm), it is glued around a cardboard cylinder (radius 8 cm). A home made conical mirror (base radius 7 cm, height 12 cm) is put inside the cylinder at the center. The anamorphic image on the cylinder wall is reflected as the undeformed original by the conical mirror. Here is the result: (the center is hidden by a coin resting at the top of the cone since anamorphic maps of points close to the cone center are off at infinite height on the cylinder wall)Another application of the function is to use one of the many popular curves (\".....-like curve\" ) that can be extracted using Interpreter Interpreter[\"PopularCurve\"][\"bunny-like curve\"]; bugsbunnyPrimitives = First@Cases[ First[ParametricPlot[ Entity[\"PopularCurve\", \"BunnyCurve\"][\"ParametricEquations\"][ t], {t, 0, 30 \\[Pi]}]] /. {x_?NumericQ, y_?NumericQ} :> {x - 85, y - 50}/800, _Line, \\[Infinity]]; The anamorphic map is created by applying anaCone2Cylinder to the point coordinates: anaBunny = Map[anaCone2Cylinder[#, 1.755, 1.25] &, bugsbunnyPrimitives, {2}]; Animate[Graphics3D[ Rotate[{{Opacity[.2], White, Cylinder[{{0, 0, .25}, {0, 0, 1}}, 1.25]}, AbsoluteThickness[3], Red, anaBunny}, \\[Phi], {0, 0, 1}], Boxed -> False], {\\[Phi], 0, 2 \\[Pi]}] This is the developed cylinder: developRules = {x_?NumericQ, y_?NumericQ, z_?NumericQ} :> {ArcTan[x, y], z}; developed = anaBunny /. developRules; DeleteCases[ Partition[developed[[1]], 2, 1], _?(EuclideanDistance @@ # > 1 &)]; Graphics[{Red, AbsoluteThickness[3], Line /@ %},", "map on the surface to indicate the azimuthal angle ... use_cm=True, color_func=lambda t, phi: phi, ... rendering_kw={\"alpha\": 0.6, \"cmap\": \"twilight\"}, ... # indicates the azimuthal angle on the colorbar label ... label=r\"$\\phi$ [rad]\", ... show_curve=True, ... # this dictionary will be passes to plot3d_parametric_line in ... # order to draw the initial curve ... curve_kw=dict(rendering_kw={\"color\": \"r\", \"label\": \"cos(t)\"}), ... # activate the wireframe to visualize the parameterization ... wireframe=True, wf_n1=15, wf_n2=15, ... wf_rendering_kw={\"lw\": 0.5, \"alpha\": 0.75}) Revolve the same function around an axis parallel to the x axis, using Plotly: from sympy import symbols, cos, sin, pi from spb import plot3d_revolution, PB t, phi = symbols('t phi') plot3d_revolution( cos(t), (t, 0, pi), parallel_axis=\"x\", axis=(1, 0), backend=PB, use_cm=True, color_func=lambda t, phi: phi, rendering_kw={\"colorscale\": \"twilight\"}, show_curve=True, curve_kw=dict(rendering_kw={\"line\": {\"color\": \"red\", \"width\": 8}, \"name\": \"cos(t)\"}), wireframe=True, wf_n1=15, wf_n2=15, wf_rendering_kw={\"line_width\": 1}) Revolve a 2D parametric circle around the z axis: from sympy import * from spb import * t = symbols(\"t\") circle = (3 + cos(t), sin(t)) plot3d_revolution(circle, (t, 0, 2 * pi), backend=KB, show_curve=True, rendering_kw={\"opacity\": 0.65}, curve_kw={\"rendering_kw\": {\"width\": 0.05}}) Revolve a 3D parametric curve around the z axis for a given azimuthal angle, using Plotly: plot3d_revolution( (cos(t), sin(t), t), (t, 0, 2*pi), (phi, 0, pi), use_cm=True, color_func=lambda t, phi: t, label=\"t [rad]\", show_curve=True, backend=PB, aspect=\"cube\", wireframe=True, wf_n1=2, wf_n2=5) Interactive-widget plot of a", "draw( arc(O,Aq,expi(pi/2,f)),thinblack ); draw( arc(O,Aqf,expi(pi/2,f+df) ),thinblack); //theta arcs draw(O--A,thinblack); draw(O--Aq,thinblack); draw(\"$\\theta$\", arc(O,0.25*length(A)*Z,0.25*A),thinblack,Arrow3); draw(L=Label(\"$d\\theta$\",position=Relative(0.5),align=NE) ,arc(O,0.66*A,0.66*Aq),thinblack ); // inner surface triple rin(pair t) { return r*expi(t.x,t.y);} surface inner=surface(rin,(q,f),(q+dq,f+df),16,16); draw(inner,emissive(gray+opacity(0.33))); //part of a nearly transparent sphere to help see perspective surface sphere=surface(rin,(0,0),(pi/2,pi/2),16,16); draw(sphere,emissive(gray+opacity(0.125))); // dr and rdtheta labels triple V= Af + 0.5*(Arf-Af); draw(L=Label(\"$dr$\",position=Relative(1.1)), V--(1.5*V.x,1.5*V.y,V.z),dotted); triple U=expi(q+0.5*dq,f); draw(L=Label(\"$rd\\theta$\",position=Relative(1.1)), r*U ---r*(1.66*U.x,1.66*U.y,U.z),dotted ); \\end{asy} \\end{document} • This is beautiful! Nevertheless, I can't resist a couple of points. The dashed line below dtheta should be solid, and there should be another line (solid, from this point of view) to the fourth corner of the surface. (The original diagram gets away with omitting this because that line would be blocked from view; but from your point of view, it's not.) – Charles Staats Feb 15 '14 at 0:24 • One of the great things about this answer is that it's easy to experiment with changing the parameters and the point of view. The one exception is the inner surface, which will work better from many points of view if you replace gray+opacity(0.25) by emissive(gray+opacity(0.25)). – Charles Staats Feb 15 '14 at 0:28 • One final point, for those interesting in playing around with this excellent diagram: the projection used in the original diagram is called obliqueX in Asymptote. – Charles Staats Feb 15 '14 at", "+0 0 58 1 Consider point $A$ in the Cartesian plane below (the grid lines occur every unit): [asy] size(200); import TrigMacros; rr_cartesian_axes(-7,7,-7,7,complexplane=false,usegrid=true); pair A = (2,3); dot(A, red+ linewidth(3.5)); label(\"$A$\", A, S); [/asy] Say that $A$ is at the polar coordinates of $(r, \\theta)$. If the Cartesian coordinates of the point with polar coordinates $\\left(2r, \\theta + \\pi/2 \\right)$ are $(x_1, y_1),$ and the Cartesian coordinates of the point with polar coordinates $\\left(-r, -\\theta \\right)$ are $(x_2, y_2)$, enter $x_1, y_1, x_2, y_2$ in that order. Jul 30, 2021 $(x_1, y_1, x_2, y_2) = (-4, 6, -3, -2)$", "// spherexplane.asy // // run // asy -f pdf -render=4 -noprc spherexplane.asy // to get a standalone raster spherexplane.pdf // import solids; size(8cm); size3(100,100); currentprojection=orthographic(camera=(66,40,-9),zoom=0.9); pen linePen=darkblue+1.3bp; pen dotPen= darkblue+3bp; pen dashPen=1bp+linetype(new real[]{4,3})+linecap(0); // Eqn of the sphere (x - 1)^2 + (y + 1)^ 2 + (z - 2)^ 2 - 25 = 0 triple O=(1,-1,2); real R=5; // Eqn of the plane 2 x - 2 y + z - 15 = 0 triple fp(real x, real y){return (x,y,- 2 x + 2 y + 15);} triple Np=unit((2,-2,1)); // plane normal triple A=fp(0,0); // any point on the plane triple C=O+dot(A-O,Np)*Np; // center of the circle cross section real d=abs(C-O); real r=sqrt(R^2-d^2); guide3 baseArc=Arc(O,O+Np*R,O-Np*R,normal=cross(Z,Np)); revolution b=revolution(O,baseArc,axis=Np); // spherical surface skeleton s; real t=acos(d/R)/pi; // fraction of the arc length at the cutting point b.transverse(s,reltime(b.g,t),P=currentprojection); guide3 circCut=s.transverse.back[0] & s.transverse.front[0] & cycle; triple D=relpoint(circCut,0.6); draw(surface(b),paleblue+opacity(0.3)); draw(surface(circCut),orange+opacity(0.3)); draw(s.transverse.front,linePen); draw(s.transverse.back, dashPen); draw(O--C--D, dashPen); dot(\"$O$\",O,dotPen); dot(Label(\"$C$\",unit(C-O)),C,dotPen); dot(\"$D$\",D,dotPen); xaxis3(xmin=0,xmax=1,red,above=true); yaxis3(ymin=0,ymax=1,deepgreen,above=true); zaxis3(zmin=0,zmax=1,blue,above=true); • +1, I am still learning your asymptote answers. – user213378 Oct 31 '20 at 11:33 • Thank you very much. – minhthien_2016 Oct 31 '20 at 12:30 • @g.kov Does asymptote always draw the hidden parts dashed automatically? – minhthien_2016 Dec 19 '20 at 14:59 • @minhthien_2016: I don't think it does. – g.kov Dec 19 '20 at 16:44 •" ]

[ "Browse Questions # If the distance between planes , $\\;4x - 2y -4z+1=0\\;$ and $\\;4x - 2y -4z+d=0\\;$ is 7 , then d is : $(a)\\;41\\;or\\;-42 \\qquad(b)\\;42\\;or\\;-43\\qquad(c)\\;-41\\;or\\;43\\qquad(d)\\;-42\\;or\\;44$", "# Distance between the two planes: 2x+3y+4z=4 and 4x+6y+8z = 12 is: Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, Watch 1000+ concepts & tricky questions explained! 2 units 4 units 8 units (2)/(sqrt(29)) units", "N \\right\\|}}$ . That is the distance between the planes.", "Math Help - The distance between two planes 1. The distance between two planes The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes? Attempted Solution: First, I drew a triangle. Then I used the law of cosines. $c^2 = a^2 + b^2 -2ac(cos \\theta)$ Then put in the $a$, $c$, and $cos \\theta)$ values. $c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$ $c^2 = 289 + 400 - 680(cos 110^0)$ $c^2 = 679.33$ ${\\sqrt{c^2}} = {\\sqrt{679.33}}$ $c = 21.06$ So the answer should be 21.06 km. But, according to the answer key, it's 30.36 km. I cannot figure out where I erred. Could someone please help? 2. The part where you went wrong is that the Law of Cosine you used isn't correct. $c^2 = 17^2 + 20^2 - 2{\\color{red}ac}(cos 110^o)$ It should be ab. Let A, B, and C be the vertices of a triangle and a, b, and c be the sides of the same triangle. Side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C. The Law of Cosines state that: $a^2 = b^2 + c^2 - 2bc\\cos{A}$ $b^2 = a^2", "Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is ### Question Asked by a Student from EXXAMM.com Team Q 1805391268. Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is A 2 units B 4 units C 8 units D 2/sqrt(29) units #### HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) #### Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis", "## anonymous 5 years ago What is the distance between the planes x-y+2z=1 and -3x+3y-6z=5 ? $|ax +by +cz +d|/\\sqrt{a ^{2}+b ^{2}+c ^{2}}$", "# Find the least distance of the plane $12x+4y+3z=327$ from the sphere $x^2+y^2+z^2+4x-2y-6z=155.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "Free Version Moderate # Distance between Planes MVCALC-GVJOEO Which number is the closest to the distance between the parallel planes given by the equations: $$4x + 6y - 8z - 10 = 0$$ $$6x + 9y -12z - 20 = 0$$ A $5$ B $0.31$ C $5/29$ D $\\sqrt{261}$ E $\\sqrt{261}/5$", "= (10, 4, -6)$ Determine the dihedral angle between the two planes 1. $9x + 17y - 4z - 7 = 0$ and $-17x + 24y + 14z + 2 = 0$ 2. $2x - 4y + 10z - 11 = 0$ and $2x - 9y + 4z + 12 = 0$ 3. $-7x + 20y + 6z + 4 = 0$ and $-19x - 3y + z + 5 = 0$ 4. $5x - 8y + 20z - 5 = 0$ and $6x + y + 19z - 7 = 0$ 5. $14x + 11y - 5z - 16 = 0$ and $11x - 13y + 8z + 4 = 0$ 6. $-10x + 9y + 2z + 8 = 0$ and $21x + 7y - 4z + 15 = 0$", "# Write the distance between the parallel planes 2x - y + 3z = 4 and 2x - y + 3z = 18. 55.6k views Write the distance between the parallel planes 2x - y + 3z = 4 and 2x - y + 3z = 18. +1 vote by (60.9k points) selected Required distance", "The Distance Between Parallel Planes The Distance Between Parallel Planes We will first define what it means for two lines to be parallel, and then learn how to compute the distance between such planes. Definition: Two planes $\\Pi_1$ and $\\Pi_2$ are said to be Parallel if the normal of $\\Pi_1$ ($n_1 = (a_1, b_1, c_1)$ is a scalar multiple $k$ of the normal of $\\Pi_2$ ($n_2 = (ka_2, kb_2, kc_2)$). For example, consider the planes $\\Pi_1: 2x + 4y + 6z + 1 = 0$ and $\\Pi_2: 4x + 8y + 12z + 6 = 0$. We can easily pull off the norms of these two planes to get that $n_1 = (2, 4, 6)$ and $n_2 = (4, 8, 12)$. Clearly $2n_1 = n_2$, so $\\Pi_1 \\parallel \\Pi_2$. Finding the distance between two parallel planes is relatively easily. First, suppose we have two planes $\\Pi_1$ and $\\Pi_2$. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes. For example, consider the planes $\\Pi_1: 2x + 3y + 4z -3 = 0$ and $\\Pi_2: -4x -6y -8z + 8 = 0$. These planes are parallel. First let's select an arbitrary", "Browse Questions # Find the least distance of the plane $12x+4y+3z=327$ from the sphere $x^2+y^2+z^2+4x-2y-6z=155.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question?", "= 5 and π 2: 6x -3y +2z = 8 required! Because the length of another side is required or iGoogle it faces the corner projection this... The distance between two vectors from expert Advanced Math tutors Please find the angle ( in radians and find the angle between two planes calculator! In radians and degrees ) between find the angle between two planes calculator two curves y=2x 2 and y=x 2-4x+4 arccos √2. For your website, blog, Wordpress, Blogger, or neither Blogger, find the angle between two planes calculator. From the cif file multiplication sign, so 5x is find the angle between two planes calculator to 5 x. This find the angle between two planes calculator site and wrote all the mathematical theory, online exercises, formulas and calculators 2 6x! Between VC and the line is called a dihedral angle is the angle between find the angle between two planes calculator given. 5 and π 2: 6x -3y +2z = 8 show the work expert Advanced tutors! ( y\\ ) Math find the angle between two planes calculator Please find the acute angle determined by the equation 2x plus 2y minus z 10. ) '' in your calculator to get the free primat.org angle between the planes is to! Numbers or fractions in this straight - this", "Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes? Attempted Solution: First, I drew a triangle. Then I used the law of cosines. $c^2 = a^2 + b^2 -2ac(cos \\theta)$ Then put in the $a$, $c$, and $cos \\theta)$ values. $c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$ $c^2 = 289 + 400 - 680(cos 110^0)$ $c^2 = 679.33$ ${\\sqrt{c^2}} = {\\sqrt{679.33}}$ $c = 21.06$", "## Elementary Geometry for College Students (5th Edition) The distance from the plane to touch down is $\\approx 1376.8$ ft. 1. Use the sine law to solve for $x$ $\\frac{x}{sin90} = \\frac{120}{sin5}$ $x = \\frac{120sin90}{sin5}$ $x = \\frac{120}{0.087155...}$ $x = 1376.845...$ ft $x \\approx 1376.8$ ft", "Want to ask us a question? Click here Browse Questions Ad 0 votes # Find the least distance of the plane $12x+4y+3z=327$ from the sphere $x^2+y^2+z^2+4x-2y-6z=155.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question? 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 1 answer 0 votes 0 answers 0 votes 0 answers", "we have $l = \\dfrac{{\\left| {{d_2} - {d_1}} \\right|}}{{\\sqrt {{a^2} + {b^2} + {c^2}} }}$ This is the required distance between the two planes", "Distance between two planes in space To calculate the distance between any two planes, it is necessary to bear in mind its relative position: • If the planes coincide or are secant, the distance between them is zero, $$\\text{d}(\\pi, \\pi') = 0$$. • If the planes are parallel, the distance between them is calculated taking any point of one plane and calculating the distance between this choosen point and the other plane. $$\\text{d}(\\pi,\\pi') = \\text{d}(P,\\pi') = \\text{d}(\\pi,P')$$$where $$P\\in\\pi$$ and $$P'\\in\\pi'$$. Find the distance between the following planes: $$\\pi: 2x - 4y + 4z +3 = 0 \\qquad \\pi': x - 2y + 2z -1 = 0$$$ We verify that the planes are parallel: $$\\dfrac{2}{1}=\\dfrac{-4}{-2}=\\dfrac{4}{2}$$$Yes, they are. Therefore, we can take the point $$P'= (1, 0, 0)$$ belonging to $$\\pi'$$ and do: $$\\text{d}(\\pi,\\pi')=\\text{d}(P',\\pi) = \\dfrac{|2\\cdot1-4\\cdot0+4\\cdot0+3|}{\\sqrt{2^2+(-4)^2+4^2}}= \\dfrac{5}{6}$$$ Another good way of calculating the distance between parallel planes. If we have them expressed as follows: $$\\pi: Ax + By + Cz + D = 0 \\qquad \\pi': Ax + By + Cz + D' = 0$$$It consists in using its distance to the origin of coordinates, which allows us to obtain the following expression: $$\\text{d}(\\pi,\\pi') = \\dfrac{|D-D|}{\\sqrt{A^2+B^2+C^2}}$$$", "# Calculate distance from plane to parallel plane in O using vector and normal I'm trying to figure out what's the best method to get the distance between two planes where i have the normalized vector of the plane and a point in the plane. What I want to do is to create a parallel plane where 0,0,0 is in and then calculate the distance between the two planes. My thought was that it's actually the same like the distance between the plane and 0,0,0 which I tried just to calculate as the dot product between the normalized vector and the given point. e.g.: (The values are that strange because they are from a 3D camera system) { 0.00102464424908972; -1.5806471007608384E-4; -0.0014898099362794807} // normalized vector (plane normal) { 1.2168973210867913; -0.2862541104934054; -0.39146720489920095} //given point 0.0018753452463538794 //calculated distance ? I'm not sure but this looks not like the result I expected. Is it the right formula? I tried to use the Hessesche Normalform Update: somehow I droped my normalize function and the normal was not normalized. now my results look much better. Thankf for all answers. - Computing the dot product between the plane normal and the position vector of the given point strikes me as the right solution. If you imagine the plane and the position vector $x$ from the", "the distance of $4 m$ up the plane, so $W = 41.578 N \\cdot 4 m = 166.31 N m$" ]

[ "The Distance Between Parallel Planes The Distance Between Parallel Planes We will first define what it means for two lines to be parallel, and then learn how to compute the distance between such planes. Definition: Two planes $\\Pi_1$ and $\\Pi_2$ are said to be Parallel if the normal of $\\Pi_1$ ($n_1 = (a_1, b_1, c_1)$ is a scalar multiple $k$ of the normal of $\\Pi_2$ ($n_2 = (ka_2, kb_2, kc_2)$). For example, consider the planes $\\Pi_1: 2x + 4y + 6z + 1 = 0$ and $\\Pi_2: 4x + 8y + 12z + 6 = 0$. We can easily pull off the norms of these two planes to get that $n_1 = (2, 4, 6)$ and $n_2 = (4, 8, 12)$. Clearly $2n_1 = n_2$, so $\\Pi_1 \\parallel \\Pi_2$. Finding the distance between two parallel planes is relatively easily. First, suppose we have two planes $\\Pi_1$ and $\\Pi_2$. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes. For example, consider the planes $\\Pi_1: 2x + 3y + 4z -3 = 0$ and $\\Pi_2: -4x -6y -8z + 8 = 0$. These planes are parallel. First let's select an arbitrary", "+ by_0 + cz_0) = \\frac{ax_0 + by_0 + cz_0}{\\sqrt{a^2 + b^2 + c^2}}$ (taking the positive since it is required that $d' >= o$) Since $ai + bj + ck$ is a normal vector to the plane and, $x_0i + y_0j + z_0k$ is the position vector of point $Q$ $d' = \\frac{ax_0 + by_0 + cz_0}{\\sqrt{a^2 + b^2 + c^2}} = \\frac{|n \\cdot \\vec{\\mathbf{OQ}|} }{|n|}$ This is the formula from a point to a plane (in this case the origin). b) Find the position vector of the point $S$ that is closest to the origin on the plane, with equation in the primed form of part (a). My solution to part (b): From part (a) it was determined that $d' = \\frac{ax_0 + by_0 + cz_0}{\\sqrt{a^2 + b^2 + c^2}} = \\frac{|n \\cdot \\vec{\\mathbf{OQ}|} }{|n|}$ , is the distance from the plane to the origin. And, in the primed form $\\vec{\\mathbf{n}} = a'i + b'j + c'k$ is a normal vector to the plane Since $\\vec{\\mathbf{n}} \\cdot \\vec{\\mathbf{n}} = |\\vec{\\mathbf{n}}|^2 = a'^2 + b'^2 + c'^2 = 1$ so $|\\vec{\\mathbf{n}}| = 1$ $\\vec{\\mathbf{n}}$ is the unit normal vector to $a'x + b'y + c'z = d'$ The position vector $\\vec{\\mathbf{S}}$ is then: $d'\\vec{\\mathbf{n}} = d' (a'i + b'j + c'k)$ Is this a sufficient answer? Or", "can further simplify the formula into ${\\displaystyle D={\\frac {|ax_{1}+by_{1}+cz_{1}+d|}{\\sqrt {a^{2}+b^{2}+c^{2}}}}}$, thus finishing our deduction. There are other ways to write down the formula. For people who prefer simpler notations, the following formulae are also ways to express the distance: ${\\displaystyle D={\\frac {|\\mathbf {n} \\ \\cdot \\ {\\overrightarrow {P_{0}P}}|}{\\|\\mathbf {n} \\|}}}$ or ${\\displaystyle D=\\mathrm {comp} _{\\mathbf {n} }{\\overrightarrow {P_{0}P}}}$ ${\\displaystyle \\blacksquare }$ After we do some further \"investigation\" on the distance formula, we can find that only the equation of the plane and one point are needed to calculate the distance between them, which is very convenient because the fewer things we need to solve a problem, the more convenient the problem is. When we try to find the distance between two parallel planes, we just need one point on one plane and the equation for the other plane to solve. Distance between two skew lines (continue) Assume that lines ${\\displaystyle L_{1},L_{2}}$ are skew lines, we can calculate the distance by assuming that those lines belong in two parallel planes ${\\displaystyle P_{1},P_{2}}$ respectively. Then, the problem changes from solving the distance between two skew lines to solving the distance between two parallel planes. We still need to know the normal vector of both planes. We can simply apply the cross product on the two direction vectors of both lines:", "The Distance Between a Plane and a Point # The Distance Between a Plane and a Point Sometimes it may be important to find the distance $D$, between some point $P_0$ and a plane in the form $ax + by + cz + d = 0$. The formula for the distance between a plane $\\Pi: ax + by + cz + d = 0$ and a point is $P(x_0, y_0, z_0)$ is as follows: (1) \\begin{align} D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}} \\end{align} We will now go forth to prove this formula. Theorem 1: Let $\\Pi$ be a plane in $\\mathbb{R}^3$ and $P_0(x_0, y_0, z_0)$ be a point in $\\mathbb{R}^3$. Then the distance $D$ between $\\Pi$ and $P_0$ is given by $D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}$. • Proof: Let $Q(x_1, y_1, z_1)$ be any point on the plane $\\Pi$ and let $P_0 = (x_0, y_0, z_0)$ be our point of interest. Let $\\vec{QP_0} = (x_0 - x_1, y_0 - y_1, z_0 - z_1)$. Recall that this plane can be defined with a vector that is perpendicular to the plane known as a normal. Let $n = (a, b, c)$ be a normal to our plane $\\Pi$. • The orthogonal", "of 2 points in is! Between parallel planes: the trick here is to reduce it to the first.! To find the distance formula related rate at one particular point in the 2D 3D! Normal N = I + 2j − k so they are parallel point! Can be found using the distance formula is derived from the other I 'm going to fill in..., 3 ), two-dimensional surface that extends infinitely far in mathematics, a plane gives a distance... Just evaluate at -- so this will just be 1 times 2 minus 2 --... 2J − k so they are parallel as an exercise the angle between the planes x 2y! Planes within the same − k so they are parallel = I + 2j − k they. It to the plane obviously the distance between two planes is calculated in vector form and in two-dimensional! Given two points and is given by the formula user57927 Jul 21 '16 at$. Such as the distance formula to calculate the shortest distance between two lines post, we extend to. Pick any point on the first plane, say, P = ( 4, 0 0. Lesson is to reduce it to the plane derived in coordinate Geometry Theorem! In space is a formula that is used to find their distance between two", "# Angle Between Two Planes - Planes & Angles In 3d geometry, position vectors are used to denote the position or location of a point with respect to the origin. The plane, as we know, is a 3d object formed by stacks of lines kept side by side. A 3d space can have an infinite number of planes aligned to one another at an infinite number of angles. In this article, we will derive a general formula for the calculation of angle between two planes in 3d space. ## How to Calculate the Angle Between Two Planes? The angle between two planes is generally calculated with the knowledge of angle between their normal. In other words, the angle between normal to two planes is the angle between the two planes. This can be understood quite clearly from the below figure: Let $\\vec{n_{1}}$ and $\\vec{n_{2}}$ be the two normal to the planes aligned to each other at an angle θ. The equation of two planes can be given by: $\\vec{r}$.$\\vec{n_{1}}$ = d1 $\\vec{r}$.$\\vec{n_{2}}$ = d2 From the above figure, we learnt that the angle between the two planes is equal to the angle between their normal, thus, ## Calculation of Angle Between Two plane in the Cartesian Plane Let A1 x + B1 y + C1z + D1 =", "and $\\mathbf{C}$ respectively so that the deviation angle of this normals from the plane is minimal. In the figure the green vector is the normalized vector sum of $\\hat {\\mathbf{n}}_\\mathrm{A}$ and $\\hat {\\mathbf{n}}_\\mathrm{C}$. The plane normal $\\hat {\\mathbf{z}}$ (blue vector) is perpendicular to the green vector and the vector from $\\mathbf{A}$ to $\\mathbf{C}$. Note: The center of the earth lies normally not on this plane but is nearby. ## Problem with 3D Points To find the center of a circle defined by 3 points, all points must lay on the same plane. The plane is calculated so that the points $\\mathbf{A}$ and $\\mathbf{C}$ lie on the plane. The middle point $\\mathbf{B}$ does not have to be exactly on the plane. It is projected perpendicular to the plane and becomes the 2D point $\\mathbf{B}^\\prime$. Most of the calculations are the same for both plane variants. The only difference is how the local coordinate system of the plane is calculated. For the origin of the plane in both variants the point $\\mathbf{A}$ is chosen. ### a) Plane through the Center of the Earth First we define the plane by a local coordinate system with the origin at $\\mathbf{A}$, X-axes from the center of the earth in the direction to point $\\mathbf{A}$, Z-axes perpendicular to the vectors $\\mathbf{A}$ and $\\mathbf{C}$ and", "have ax 1 + by 1 + cz 1 + d 2 = 0 ⇒ ax 1 + by 1 + cz 1 = −d 2. If ax + by + cz + d 1 = 0 and ax + by + cz + d 2 = 0 be equation of two parallel planes. Let be a vector between points on the two lines. So, the line and the plane are neither orthogonal nor parallel. two parallel planes, say. => z = -d1 / c1 Click here to toggle editing of individual sections of the page (if possible). The two planes need to be parallel to each other to calculate their distance. Notify administrators if there is objectionable content in this page. Then, the distance between them is. If two planes are parallel, their normal vectors are also parallel. The condition for two planes to be parallel is:-. Please write to us at [email protected] to report any issue with the above content. Check out how this page has evolved in the past. We will first define what it means for two lines to be parallel, and then learn how to compute the distance between such planes. Find a point in any one plane such that the distance from that point to the other plane that will be the", "to the plane is: D=$$\\mid \\frac{ax_{0}+by_{0}+cz_{0}+d}{\\sqrt{a^{2}+b^{2}+c^{2}}$$ 8. Sep 29, 2011 ### cyt91 Got it! Yes. I did not pick a point on the first plane through which the normal vector passes. Thanks.", "the distance between the planes $x + 2y + 3z + 4 = 0$ and $2x + 4y + 6z + 5 = 0$. We note that these two planes are parallel since their normals, $(1, 2, 3)$ and $(2, 4, 6)$ are scalar multiples of each other. Consider the point $(-1, 0, -1)$ on the first plane. Then from theorem 2, we have that the distance between these two planes are: (3) \\begin{align} \\quad \\mathrm{distance} = \\frac{\\mid 2(-1) + 4(0) + 6(-1) + 5 \\mid}{\\sqrt{2^2 + 4^2 + 6^2}} = \\frac{ \\mid -2 - 6 + 5 \\mid}{\\sqrt{56}} = \\frac{3}{\\sqrt{56}} \\end{align}", "SEE HERE The vector $\\vec{N}=\\vec{D}\\times\\vec{E}$ the normal for the plane you want. Since the OP has solved the problem, I will point out that alternatively, if $\\vec R_1 = \\vec P +t\\vec D$ and $\\vec R_2 = \\vec Q + s\\vec E$ then the midpoint between them is $$\\vec M = \\frac 1 2(\\vec R_1 + \\vec R_2) = \\frac 1 2(\\vec P + \\vec Q) + \\frac 1 2 t\\vec D + \\frac 1 2 s\\vec E$$which is a parametric representation of the \"mid\" plane if $\\vec D$ and $\\vec E$ are not parallel. Page 2 of 2 First 12", "file attachments for this page. Experience. Proof. The formula for the distance between two points in space is a natural extension of this formula. The direction vector of the plane orthogonal to the given lines is collinear or coincides with their direction vectors that is. Let’s check this. So, the line and the plane are neither orthogonal nor parallel. Doing a plane to plane distance is not good. See pages that link to and include this page. The bisector planes of the angles between the planes. How to check if a given point lies inside or outside a polygon? The distance between two parallel planes ax + by + cz + d 1 = 0 and ax + by + cz + d 2 = 0 is given by . Click here to edit contents of this page. We use cookies to ensure you have the best browsing experience on our website. contributed. Now figure out the distance between the two planes using this formula. We will now use the formula $D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}$ in order to calculate the distance between both planes: \\begin{align} D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}} \\\\ D = \\frac{\\mid -4(0) + -6(0)", "Distance between parallel planes The distance between parallel planes is a simple concept of which we are going to write its formula and we will see an example. Formula for distance between parallel planes Vamos a redactar las ecuaciones de los siguientes planos paralelos para poder redactar la fórmula:The peculiarity that parallel planes have is that in the coefficients of the unknowns x, y and z they will have the same value and the only thing that is different is the value of the independent term, let’s see at the equation of the plane 1 and plane 2 respectively: ax + by + cz = d_{1} ax + by + cz = d_{2} Once we have understood that the independent term is the only thing that changes, we are going to show the formula to calculate the distance of the parallel planes: D = \\cfrac{\\left| d_{2} - d_{1}\\right|}{\\sqrt{a^{2} + b^{2} + c^{2}}} Example of distance between parallel planes Calculate the distance between the planes: (1) \\ \\ x + y + z = 4 (2) \\ \\ 2x + 2y + 2z = 6 As you can see, the coefficients of the unknowns do not have the same values, so to solve this we can multiply equation 1 by 2 or we can divide equation 2 by 2.", "* y + c1 * z + d1 = 0 and P2: a2 * x + b2 * y + c2 * z + d2 = 0. close, link Click here to edit contents of this page. code. The distance from this point to the other plane is the distance between the planes. Thus, the distance between two parallel lines is given by – d = | \\vec{PT} |. I then measure a plane on the non-datum feature dimension the distance and parallelism per the print. Distance between two parallel planes. $\\vec n\\centerdot \\vec v = 0 + 0 + 8 = 8 \\ne 0$ The two vectors aren’t orthogonal and so the line and plane aren’t parallel. View wiki source for this page without editing. DISTANCE PLANE-PLANE (3D). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}, Creative Commons Attribution-ShareAlike 3.0 License. We will now use the formula D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}} in order to calculate the distance between both planes: \\begin{align} D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}", "# Calculate distance from plane to parallel plane in O using vector and normal I'm trying to figure out what's the best method to get the distance between two planes where i have the normalized vector of the plane and a point in the plane. What I want to do is to create a parallel plane where 0,0,0 is in and then calculate the distance between the two planes. My thought was that it's actually the same like the distance between the plane and 0,0,0 which I tried just to calculate as the dot product between the normalized vector and the given point. e.g.: (The values are that strange because they are from a 3D camera system) { 0.00102464424908972; -1.5806471007608384E-4; -0.0014898099362794807} // normalized vector (plane normal) { 1.2168973210867913; -0.2862541104934054; -0.39146720489920095} //given point 0.0018753452463538794 //calculated distance ? I'm not sure but this looks not like the result I expected. Is it the right formula? I tried to use the Hessesche Normalform Update: somehow I droped my normalize function and the normal was not normalized. now my results look much better. Thankf for all answers. - Computing the dot product between the plane normal and the position vector of the given point strikes me as the right solution. If you imagine the plane and the position vector $x$ from the", "(-m)(\\frac{-c_{1}}{m}) – c_{2} \\right |}{\\sqrt{1 + m^{2}}}$$, $$\\Rightarrow d$$ $$= \\frac{\\left | c_{1} – c_{2} \\right |}{\\sqrt{1 + m^{2}}}$$. Distance Between Two Parallel Planes. It is equivalent to the length of the vertical distance from any point on one of the lines to another line. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes. Formula for distance between parallel lines is If we consider the general form of the equation of straight line, and the lines are given by: Then, the distance between them is given by: $$d$$ = $$\\frac{|C_1 ~- ~C_2|}{√A^2~ +~ B^2}$$. Alternatively we can find the distance between two parallel lines as follows: Considers two parallel lines. T. tigerleo. Your email address will not be published. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. We know that slopes of two parallel lines are equal. Distance between two parallel lines. Consider line L and point P in a coordinate plane. The required distance d will be PA – PB. Your email address will", "< 6$, this number should be just a little over $2$, and the correct answer is $\\boxed{C}$. Note that the equation above for the distance from a point to a plane is a 3D analogue of the 2D case of the distance formula, where you take the distance from a point to a plane. In the 2D case, both $c$ and $C$ are set equal to $0$.", "another point. Then the distance of $$Q$$ from $$\\Pi$$ is $d(Q,\\Pi)=\\frac{|\\vec{PQ}\\cdot\\mathbf n|}{|\\mathbf n|}.$ Warning. Don't forget to take the absolute value in the nominator. Distances can't be negative. Example. Let $$\\Pi$$ be the plane $$-2x+3y+z=4$$, and $$Q(-1,0,-1)$$ some other point. To calculate the distance of $$Q$$ from $$\\Pi$$, we need to find a point $$P$$ on $$\\Pi$$, and a normal vector. Setting $$y=z=0$$, we see that we need $$x=-2$$ to get a solution to the standard equation. That is, $$P(-2,0,0)$$ is a point on $$\\Pi$$. A normal vector $$\\mathbf n=(-2,3,4)$$ can be read off the standard equation. We have $$\\vec{PQ}=(1,0,-1)$$, therefore the distance is $d(Q,\\Pi)=\\frac{|\\mathbf n\\cdot\\vec{PQ}|}{|\\mathbf n|}=\\frac{|-6|}{\\sqrt{29}}=\\frac{6}{29}.$ ## Matrix multiplication Example. Let $$A=\\left(\\begin{smallmatrix}2 & -1 \\\\ 3 & -2 \\end{smallmatrix}\\right)$$. Let's find $$A^11$$. Well, first let's calculate $A^2=\\begin{pmatrix}1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ Since $$A^2=I$$, we can simplify the calculation: $A^11=A^{10+1}=(A^2)^5A=IA=A.$", "Angle sum property of triangles) Thus, △CQD ~ △AQP by AAA similarity condition. $$\\frac{CQ}{AQ}$$ = $$\\frac{QD}{QP}$$ = $$\\frac{CD}{AP}$$ [∵ Ratio of corresponding sides of similar triangles are equal] $$\\frac{CQ}{AQ}$$ = $$\\frac{QD}{QP}$$ CQ . PQ = QA . QD [Q.E.D] Question 4. △ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) $$\\frac{CA}{PA}$$ = $$\\frac{BC}{MP}$$ Given: △ABC; ∠B = 90° AAMP; ∠M = 90° R.T.P : i) △ABC ~ △AMP Proof: In △ABC and △AMP ∠B = ∠M [each 90° given] ∠A = ∠A [common] Hence, ∠C = ∠P [∵ Angle sum property of triangles] ∴ △ABC ~ △AMP (by A.A.A. similarity) ii) △ABC ~ △AMP (already proved) $$\\frac{AB}{AM}$$ = $$\\frac{BC}{MP}$$ = $$\\frac{CA}{PA}$$ [∵ Ratio of corresponding sides of similar triangles are equal] $$\\frac{CA}{PA}$$ = $$\\frac{BC}{MP}$$ Question 5. An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1$$\\frac{1}{2}$$ hour? Given: Speed of the first plane due north = 1000 kmph. Speed of the second plane due west = 1200 kmph. Distance = Speed × Time Distance travelled by the first plane in 1$$\\frac{1}{2}$$", "angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. ???\\frac{a_1}{b_1}=\\frac{a_2}{b_2}=\\frac{a_3}{b_3}??? An easier way to find the angle between two vectors is the dot product formula(A.B=|A|x|B|xcos(X)) let vector A be 2i and vector be 3i+4j. ?a\\langle a_1,a_2,a_3\\rangle??? (its length). The angle between two planes is equal to the angle determined by the normal vectors of the planes. The line FC and the plane ABCD form a right angle. Step-by-step math courses covering Pre-Algebra through Calculus 3. mixing problems, math, learn online, online course, online math, math online, fundamental theorem of calculus, Parallel perpendicular and angle between planes. In other words the angle between normal to two planes is the angle between the two planes. First we’ll find the normal vectors of the given planes. The angle between AF and the plane is $$x$$. ?a\\langle 3,-1,2\\rangle??? Refers to movement where the angle between two bones decreases and on the horizontal plane. Ex 11.3, 12 Find the angle between the planes whose vector equations are 𝑟 ⃗ . 2.852x 22 − 4x 2 − 1.296 = 0. So the plane equation are: 1.674x + y + z + D = 0 And 0.271x − y − z + D = 0. Finding the angle between two" ]

[ "# Help understand related rates problem: calculating the derivative of the distance function I am having a particular problem understanding related rates problems. I think this is a good example to show my two issues: Two planes are flying towards the same point. The first is going to the east, and the other to the north. The first goes at a speed of $450 km/h$ and the other at $600km/h$. When the two planes are at $150km$ and $200km$ respectively from the collision point, at what rate is the distance between them changing? Well. Some data: The distance between the first plane and the collision point is $x$, so: $$\\frac{dx}{dy} = 450km/h$$ Whereas the distance between the second plane and the collision point is $y$: $$\\frac{dy}{dt}=600km/h$$ The distance between the two planes is $z$, and it can be calculated as $$z^2=x^2+y^2$$ We want to find the rate of change of $z$, that is $$\\frac{dz}{dt} = \\ ?$$ The previous formula is the distance between planes, but since we want the rate of change, we need its derivative... $$2z\\cdot\\frac{dz}{dt} = 2x\\cdot\\frac{dx}{dt} + 2y\\cdot\\frac{dy}{dt}$$ Now, this particular step is what confuses me. I understand that we need the derivative of the distance function, but what I don't grasp is: the derivative with respect to what? To $x$? To $y$? To", "The Distance Between Parallel Planes The Distance Between Parallel Planes We will first define what it means for two lines to be parallel, and then learn how to compute the distance between such planes. Definition: Two planes $\\Pi_1$ and $\\Pi_2$ are said to be Parallel if the normal of $\\Pi_1$ ($n_1 = (a_1, b_1, c_1)$ is a scalar multiple $k$ of the normal of $\\Pi_2$ ($n_2 = (ka_2, kb_2, kc_2)$). For example, consider the planes $\\Pi_1: 2x + 4y + 6z + 1 = 0$ and $\\Pi_2: 4x + 8y + 12z + 6 = 0$. We can easily pull off the norms of these two planes to get that $n_1 = (2, 4, 6)$ and $n_2 = (4, 8, 12)$. Clearly $2n_1 = n_2$, so $\\Pi_1 \\parallel \\Pi_2$. Finding the distance between two parallel planes is relatively easily. First, suppose we have two planes $\\Pi_1$ and $\\Pi_2$. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes. For example, consider the planes $\\Pi_1: 2x + 3y + 4z -3 = 0$ and $\\Pi_2: -4x -6y -8z + 8 = 0$. These planes are parallel. First let's select an arbitrary", "## Elementary Geometry for College Students (5th Edition) The distance from the plane to touch down is $\\approx 1376.8$ ft. 1. Use the sine law to solve for $x$ $\\frac{x}{sin90} = \\frac{120}{sin5}$ $x = \\frac{120sin90}{sin5}$ $x = \\frac{120}{0.087155...}$ $x = 1376.845...$ ft $x \\approx 1376.8$ ft", "the distance between the planes $x + 2y + 3z + 4 = 0$ and $2x + 4y + 6z + 5 = 0$. We note that these two planes are parallel since their normals, $(1, 2, 3)$ and $(2, 4, 6)$ are scalar multiples of each other. Consider the point $(-1, 0, -1)$ on the first plane. Then from theorem 2, we have that the distance between these two planes are: (3) \\begin{align} \\quad \\mathrm{distance} = \\frac{\\mid 2(-1) + 4(0) + 6(-1) + 5 \\mid}{\\sqrt{2^2 + 4^2 + 6^2}} = \\frac{ \\mid -2 - 6 + 5 \\mid}{\\sqrt{56}} = \\frac{3}{\\sqrt{56}} \\end{align}", "* y + c1 * z + d1 = 0 and P2: a2 * x + b2 * y + c2 * z + d2 = 0. close, link Click here to edit contents of this page. code. The distance from this point to the other plane is the distance between the planes. Thus, the distance between two parallel lines is given by – d = | \\vec{PT} |. I then measure a plane on the non-datum feature dimension the distance and parallelism per the print. Distance between two parallel planes. $\\vec n\\centerdot \\vec v = 0 + 0 + 8 = 8 \\ne 0$ The two vectors aren’t orthogonal and so the line and plane aren’t parallel. View wiki source for this page without editing. DISTANCE PLANE-PLANE (3D). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}, Creative Commons Attribution-ShareAlike 3.0 License. We will now use the formula D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}} in order to calculate the distance between both planes: \\begin{align} D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}", "Distance between two planes in space To calculate the distance between any two planes, it is necessary to bear in mind its relative position: • If the planes coincide or are secant, the distance between them is zero, $$\\text{d}(\\pi, \\pi') = 0$$. • If the planes are parallel, the distance between them is calculated taking any point of one plane and calculating the distance between this choosen point and the other plane. $$\\text{d}(\\pi,\\pi') = \\text{d}(P,\\pi') = \\text{d}(\\pi,P')$$$where $$P\\in\\pi$$ and $$P'\\in\\pi'$$. Find the distance between the following planes: $$\\pi: 2x - 4y + 4z +3 = 0 \\qquad \\pi': x - 2y + 2z -1 = 0$$$ We verify that the planes are parallel: $$\\dfrac{2}{1}=\\dfrac{-4}{-2}=\\dfrac{4}{2}$$$Yes, they are. Therefore, we can take the point $$P'= (1, 0, 0)$$ belonging to $$\\pi'$$ and do: $$\\text{d}(\\pi,\\pi')=\\text{d}(P',\\pi) = \\dfrac{|2\\cdot1-4\\cdot0+4\\cdot0+3|}{\\sqrt{2^2+(-4)^2+4^2}}= \\dfrac{5}{6}$$$ Another good way of calculating the distance between parallel planes. If we have them expressed as follows: $$\\pi: Ax + By + Cz + D = 0 \\qquad \\pi': Ax + By + Cz + D' = 0$$$It consists in using its distance to the origin of coordinates, which allows us to obtain the following expression: $$\\text{d}(\\pi,\\pi') = \\dfrac{|D-D|}{\\sqrt{A^2+B^2+C^2}}$$$", "Distance between parallel planes The distance between parallel planes is a simple concept of which we are going to write its formula and we will see an example. Formula for distance between parallel planes Vamos a redactar las ecuaciones de los siguientes planos paralelos para poder redactar la fórmula:The peculiarity that parallel planes have is that in the coefficients of the unknowns x, y and z they will have the same value and the only thing that is different is the value of the independent term, let’s see at the equation of the plane 1 and plane 2 respectively: ax + by + cz = d_{1} ax + by + cz = d_{2} Once we have understood that the independent term is the only thing that changes, we are going to show the formula to calculate the distance of the parallel planes: D = \\cfrac{\\left| d_{2} - d_{1}\\right|}{\\sqrt{a^{2} + b^{2} + c^{2}}} Example of distance between parallel planes Calculate the distance between the planes: (1) \\ \\ x + y + z = 4 (2) \\ \\ 2x + 2y + 2z = 6 As you can see, the coefficients of the unknowns do not have the same values, so to solve this we can multiply equation 1 by 2 or we can divide equation 2 by 2.", "file attachments for this page. Experience. Proof. The formula for the distance between two points in space is a natural extension of this formula. The direction vector of the plane orthogonal to the given lines is collinear or coincides with their direction vectors that is. Let’s check this. So, the line and the plane are neither orthogonal nor parallel. Doing a plane to plane distance is not good. See pages that link to and include this page. The bisector planes of the angles between the planes. How to check if a given point lies inside or outside a polygon? The distance between two parallel planes ax + by + cz + d 1 = 0 and ax + by + cz + d 2 = 0 is given by . Click here to edit contents of this page. We use cookies to ensure you have the best browsing experience on our website. contributed. Now figure out the distance between the two planes using this formula. We will now use the formula $D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}}$ in order to calculate the distance between both planes: \\begin{align} D = \\frac{\\mid ax_0 + by_0 + cz_0 + d \\mid}{\\sqrt{a^2 + b^2 + c^2}} \\\\ D = \\frac{\\mid -4(0) + -6(0)", "# Math Help - Distance between two parallel planes 1. ## Distance between two parallel planes Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . 2. Originally Posted by knuckles1234 Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . Given a point $Pp,q)\" alt=\"Pp,q)\" /> and a line $\\ell:Ax+By+C=0$ then the distance between the point and the line is $\\frac{|pA+qB+C|}{\\sqrt{A^2+B^2}}$. Take any point on one of the lines and find the distance to the other line. 3. To find the distance between two planes, use the formula: $D=\\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\\sqrt{a^{2}+b^{2 }+c^{2}}}$ This is the formula for the distance between a point and a plane. To find the distance between the planes, choose an arbitrary point on one of the planes. Set y=z=0 in x-2y-2z-12=0 and solve for x. Then, use that as your point.", "# Distance between two parallel planes • Nov 13th 2009, 09:08 AM knuckles1234 Distance between two parallel planes Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . • Nov 13th 2009, 09:16 AM Plato Quote: Originally Posted by knuckles1234 Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . Given a point $P:(p,q)$ and a line $\\ell:Ax+By+C=0$ then the distance between the point and the line is $\\frac{|pA+qB+C|}{\\sqrt{A^2+B^2}}$. Take any point on one of the lines and find the distance to the other line. • Nov 13th 2009, 09:18 AM galactus To find the distance between two planes, use the formula: $D=\\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\\sqrt{a^{2}+b^{2 }+c^{2}}}$ This is the formula for the distance between a point and a plane. To find the distance between the planes, choose an arbitrary point on one of the planes. Set y=z=0 in x-2y-2z-12=0 and solve for x. Then, use that as your point.", "Math Help - The distance between two planes 1. The distance between two planes The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes? Attempted Solution: First, I drew a triangle. Then I used the law of cosines. $c^2 = a^2 + b^2 -2ac(cos \\theta)$ Then put in the $a$, $c$, and $cos \\theta)$ values. $c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$ $c^2 = 289 + 400 - 680(cos 110^0)$ $c^2 = 679.33$ ${\\sqrt{c^2}} = {\\sqrt{679.33}}$ $c = 21.06$ So the answer should be 21.06 km. But, according to the answer key, it's 30.36 km. I cannot figure out where I erred. Could someone please help? 2. The part where you went wrong is that the Law of Cosine you used isn't correct. $c^2 = 17^2 + 20^2 - 2{\\color{red}ac}(cos 110^o)$ It should be ab. Let A, B, and C be the vertices of a triangle and a, b, and c be the sides of the same triangle. Side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C. The Law of Cosines state that: $a^2 = b^2 + c^2 - 2bc\\cos{A}$ $b^2 = a^2", "# Calculate distance from plane to parallel plane in O using vector and normal I'm trying to figure out what's the best method to get the distance between two planes where i have the normalized vector of the plane and a point in the plane. What I want to do is to create a parallel plane where 0,0,0 is in and then calculate the distance between the two planes. My thought was that it's actually the same like the distance between the plane and 0,0,0 which I tried just to calculate as the dot product between the normalized vector and the given point. e.g.: (The values are that strange because they are from a 3D camera system) { 0.00102464424908972; -1.5806471007608384E-4; -0.0014898099362794807} // normalized vector (plane normal) { 1.2168973210867913; -0.2862541104934054; -0.39146720489920095} //given point 0.0018753452463538794 //calculated distance ? I'm not sure but this looks not like the result I expected. Is it the right formula? I tried to use the Hessesche Normalform Update: somehow I droped my normalize function and the normal was not normalized. now my results look much better. Thankf for all answers. - Computing the dot product between the plane normal and the position vector of the given point strikes me as the right solution. If you imagine the plane and the position vector $x$ from the", "Browse Questions # If the distance between planes , $\\;4x - 2y -4z+1=0\\;$ and $\\;4x - 2y -4z+d=0\\;$ is 7 , then d is : $(a)\\;41\\;or\\;-42 \\qquad(b)\\;42\\;or\\;-43\\qquad(c)\\;-41\\;or\\;43\\qquad(d)\\;-42\\;or\\;44$", "# Angle Between Two Planes - Planes & Angles In 3d geometry, position vectors are used to denote the position or location of a point with respect to the origin. The plane, as we know, is a 3d object formed by stacks of lines kept side by side. A 3d space can have an infinite number of planes aligned to one another at an infinite number of angles. In this article, we will derive a general formula for the calculation of angle between two planes in 3d space. ## How to Calculate the Angle Between Two Planes? The angle between two planes is generally calculated with the knowledge of angle between their normal. In other words, the angle between normal to two planes is the angle between the two planes. This can be understood quite clearly from the below figure: Let $\\vec{n_{1}}$ and $\\vec{n_{2}}$ be the two normal to the planes aligned to each other at an angle θ. The equation of two planes can be given by: $\\vec{r}$.$\\vec{n_{1}}$ = d1 $\\vec{r}$.$\\vec{n_{2}}$ = d2 From the above figure, we learnt that the angle between the two planes is equal to the angle between their normal, thus, ## Calculation of Angle Between Two plane in the Cartesian Plane Let A1 x + B1 y + C1z + D1 =", "# Thread: Distance between two parallel planes 1. ## Distance between two parallel planes Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . 2. Originally Posted by knuckles1234 Find the shortest distance between the following two parallel planes: x -2y -2z -12 = 0 and x -2y -2z -6 = 0 . Given a point $\\displaystyle Pp,q)$ and a line $\\displaystyle \\ell:Ax+By+C=0$ then the distance between the point and the line is $\\displaystyle \\frac{|pA+qB+C|}{\\sqrt{A^2+B^2}}$. Take any point on one of the lines and find the distance to the other line. 3. To find the distance between two planes, use the formula: $\\displaystyle D=\\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\\sqrt{a^{2}+b^{2 }+c^{2}}}$ This is the formula for the distance between a point and a plane. To find the distance between the planes, choose an arbitrary point on one of the planes. Set y=z=0 in x-2y-2z-12=0 and solve for x. Then, use that as your point.", "can further simplify the formula into ${\\displaystyle D={\\frac {|ax_{1}+by_{1}+cz_{1}+d|}{\\sqrt {a^{2}+b^{2}+c^{2}}}}}$, thus finishing our deduction. There are other ways to write down the formula. For people who prefer simpler notations, the following formulae are also ways to express the distance: ${\\displaystyle D={\\frac {|\\mathbf {n} \\ \\cdot \\ {\\overrightarrow {P_{0}P}}|}{\\|\\mathbf {n} \\|}}}$ or ${\\displaystyle D=\\mathrm {comp} _{\\mathbf {n} }{\\overrightarrow {P_{0}P}}}$ ${\\displaystyle \\blacksquare }$ After we do some further \"investigation\" on the distance formula, we can find that only the equation of the plane and one point are needed to calculate the distance between them, which is very convenient because the fewer things we need to solve a problem, the more convenient the problem is. When we try to find the distance between two parallel planes, we just need one point on one plane and the equation for the other plane to solve. Distance between two skew lines (continue) Assume that lines ${\\displaystyle L_{1},L_{2}}$ are skew lines, we can calculate the distance by assuming that those lines belong in two parallel planes ${\\displaystyle P_{1},P_{2}}$ respectively. Then, the problem changes from solving the distance between two skew lines to solving the distance between two parallel planes. We still need to know the normal vector of both planes. We can simply apply the cross product on the two direction vectors of both lines:", "find that distance first find the normal vector of those planes - it is the cross product of directional vectors of the given lines. For the normal vector of the form (A, B, C) equations representing the planes are: $Ax + By + Cz + D_1 = 0$ $Ax + By + Cz + D_2 = 0$ Take coordinates of a point lying on the first line and solve for D1. Similarly for the second line and D2. The distance we're looking for is: $$d = \\frac{|D_1 - D_2|}{\\sqrt{A^2 + B^2 + C^2}}$$ -", "calculate the distance between points and in the second plane is a flat, two-dimensional surface that infinitely! S an online Geometry tool requires coordinates of 2 points in the complex,. Vector: distance between two points in the complex plane, say, P = ( 4, 0.. Correctly, the distance formula is positive on one side of the plane obviously the distance as the between! The length of that over, and I have n't put these guys.! An absolute distance have to determine this related rate at one particular point in the two-dimensional coordinate! Plane obviously the distance between two points on Rectangular coordinate plane is actually the length of two! Have to determine this related rate at one particular point in the complex plane planes have N. Requires coordinates of 2 points in the xy-plane can be found using the distance them. Fill it in -- plus 3 times something, minus 5 this rate! Side of the two complex numbers have distance between two planes formula put these guys in points in two-dimensional... Any two points on Rectangular coordinate plane = I + 2j − k so are... Between parallel planes: the trick here is to calculate the distance between this point to line... 10 Maths Chapter 7 find their distance the understanding of the important covered! Within the", "(-m)(\\frac{-c_{1}}{m}) – c_{2} \\right |}{\\sqrt{1 + m^{2}}}$$, $$\\Rightarrow d$$ $$= \\frac{\\left | c_{1} – c_{2} \\right |}{\\sqrt{1 + m^{2}}}$$. Distance Between Two Parallel Planes. It is equivalent to the length of the vertical distance from any point on one of the lines to another line. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes. Formula for distance between parallel lines is If we consider the general form of the equation of straight line, and the lines are given by: Then, the distance between them is given by: $$d$$ = $$\\frac{|C_1 ~- ~C_2|}{√A^2~ +~ B^2}$$. Alternatively we can find the distance between two parallel lines as follows: Considers two parallel lines. T. tigerleo. Your email address will not be published. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. We know that slopes of two parallel lines are equal. Distance between two parallel lines. Consider line L and point P in a coordinate plane. The required distance d will be PA – PB. Your email address will", "we apply the second line$g(2) = 2 + g(1)$. Now 1 is odd so we use the third line$g(1) = 1 + g(0)$. and Finally 0 is the first line,$g(0) = 1$. Putting it all together,$g(0) = 1$,$g(1) = 2$,$g(2) = 4$and$g(4) = 6$. Think you'd be able to try out for$g(7)$and$g(12)$? I think I'd discuss (ii) with you since it's quite weird. ## Question 11 Do you recall an idea about$\\mathbf{r}\\cdot \\hat{\\mathbf{n}}$for equation of planes? Let's take a look at the plane (a separate example) $$\\mathbf{r} \\cdot \\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}= 5$$ The number 5 on the RHS has not much meaning except that points on the plane have to satisfy the equation to get 5. But let's transform this equation to the$\\mathbf{r}\\cdot \\hat{\\mathbf{n}}$form. Since$| \\mathbf{n} | = 3$, we divide our equation of the plane by 3 on both sides to get $$\\mathbf{r} \\cdot \\frac{\\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}}{3}= \\frac{5}{3}$$ Now$\\frac{5}{3}$has a meaning: it is the distance between the plane and the origin is$\\frac{5}{3}$. If a plane has equation$\\mathbf{r}\\cdot \\hat{\\mathbf{n}} = k$, then$k$is the perpendicular distance between the plane and the origin. In your earlier working you should have found an equation of$p$similar to$\\mathbf{r} \\cdot \\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}= -1$. Converting to the form we discussed we get$\\mathbf{r}" ]

[ "4 & 2 \\\\\\\\ \\end{vmatrix}$$", "\\begin{vmatrix} -1 & -2 & 1\\\\ -3 & -2 & 1\\\\ 2& 2 & 1\\\\ \\end{vmatrix}$$ Δ = $$\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit", "\\end{vmatrix}\\:\\:\\:$ is $2$ $2e$ $2 \\pi$ $2i$", "and 9x - 5y + 8 =0 We have $\\begin{vmatrix} 2 & -3 & 5\\\\ 3 & 4 & -7\\\\ 9 & -5 & 8\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line", "Comment Share Q) # Solve for $x$ if $\\begin{vmatrix} 1 & 4 & 4 \\\\ 1 & -2 & 1 \\\\ 1 & 2x & x^2 \\end{vmatrix} =0$ $\\begin{array}{1 1} (A) x=2,-1 \\\\(B) x=-2,1 \\\\ (C) x=-2,-1 \\\\ none\\;of \\;the \\;above \\end{array}$", "$|A| =3\\begin{vmatrix} 1 & -2 \\\\[0.3em] 3 & 1\\end{vmatrix}$ $+4\\begin{vmatrix} 1 & -2 \\\\[0.3em] 2 & 1 \\end{vmatrix}$ $+5\\begin{vmatrix} 1 & 1 \\\\[0.3em] 2 & 3 \\end{vmatrix}$$\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\ =3(7)+4(5)+5(1)\\\\ 21+20+5=46$", "& -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 2 \\cdot \\begin{vmatrix} 0 & 3 & -4 \\\\ -5 & -8 & 3 \\\\ 0 & 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot \\begin{vmatrix} 3 & -4 \\\\ 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot (-18 -(-20)) = 20.$$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book. If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$\\det A = \\begin{vmatrix} 5 & -7 & 2 & 2 \\\\ 0 & 3 & 0 & -4 \\\\ -5 & -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 5 \\begin{vmatrix} 3 & 0 & -4 \\\\ -8 & 0 & 3 \\\\ 5 & 0 & -6 \\\\ \\end{vmatrix} -(-7) \\begin{vmatrix} 0 & 0 & -4 \\\\ -5 & 0 & 3", "+ \\begin{vmatrix} 1&1\\\\ -4 & 2\\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k*(5-6)+2+4 &=& 0\\\\ -k &=& -6 \\\\ \\mathbf{k} &=& \\mathbf{6} \\\\ \\hline \\end{array}$$ Aug 3, 2019", "& 4 & 7\\end{vmatrix} = 2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 & 0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix}$$ Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get $$2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 &0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix} = 2(-1)^{2+1}\\begin{vmatrix} 2 & 3 \\\\ 4 & 7\\end{vmatrix} = -2\\begin{vmatrix} 2 & 3 \\\\ 0 & 1\\end{vmatrix}$$ where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.", "2 &1 &5 \\end{vmatrix} &=\\begin{vmatrix} 2 &0 &0 \\\\ 0 &3 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 2 &0 &0 \\\\ 0 &0 &\\underline{0} \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &1 &0 \\\\ 4 &0 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &\\underline{0} \\\\ 2 &0 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &0 &-1 \\\\ 4 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} + \\begin{vmatrix} 0 &0 &-1 \\\\ 0 &3 &0 \\\\ 2 &0 &0 \\end{vmatrix} \\end{array}$ We can bring out the scalars. $\\begin{array}{rl} &=(2)(3)(5)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(2)(\\underline{0})(1)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &0 &1 \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+(1)(4)(5)\\begin{vmatrix} 0 &1 &0 \\\\ 1 &0 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(1)(\\underline{0})(2)\\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &1 \\\\ 1 &0 &0 \\end{vmatrix} \\\\ &\\quad+(-1)(4)(1)\\begin{vmatrix} 0 &0 &1 \\\\ 1 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} +(-1)(3)(2)\\begin{vmatrix} 0 &0 &1 \\\\ 0 &1 &0 \\\\ 1 &0 &0 \\end{vmatrix} \\end{array}$ To finish, we evaluate those six determinants by row-swapping them to the identity matrix, keeping track of the resulting sign changes. $\\begin{array}{rl} &=30\\cdot (+1)+0\\cdot (-1) \\\\ &\\quad+20\\cdot (-1)+0\\cdot (+1) \\\\ &\\quad -4\\cdot (+1)-6\\cdot (-1)=12 \\end{array}$ That example illustrates the key idea. We've", "2 & 5 \\end{vmatrix} \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} = \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ and continue encryption as follows: $\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix}$ The matrix K is invertible, hence K − 1 exists such that KK − 1 = K − 1K = I2. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = 3 \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = \\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}$ $HIAT \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix} \\to \\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ Then we compute $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix} = \\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix},$ $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} = \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix}$ Therefore $\\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix}, \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ E \\end{vmatrix}, \\begin{vmatrix} L \\\\ P \\end{vmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts n2 plaintext/ciphertext character pairs can set up a", "Note that $$\\det \\begin{vmatrix} 1 & 1 & 1 \\\\ 1 & -1 & 1 \\\\ -1 & 3 & -1 \\\\ \\end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace", "2& 2& ...& 3 \\end{vmatrix}=\\begin{vmatrix} 1 & -1& 0&...& 0\\\\ 0& 1& -1& ...& 0\\\\ 0& 0& 1& ... &0\\\\ 2& 2& 2&...& 3 \\end{vmatrix}$$", "C_z \\end{vmatrix}$ 6. Aug 25, 2016 ### RJLiberator Ah. Okay guys, thank you for your help. The information I was unaware of was that it was a 3x3 determinant. I know calculated brute force both sides of 1.16 and arrived at the result.", "\\\\ x_2 & y_2 & z_2 & 1 \\\\ x_3 & y_3 & z_3 & 1 \\\\ \\end{vmatrix}$$", "(n+3)^2 & (n+4)^2 \\\\ \\end{vmatrix} =-4(-1)(-1)(2) = -8$", "n & n+1 \\ 2 & 3 & … & n & n & n+1 \\ … & … & … & … & … & … \\ 2 & n & … & n-1 & n & n+1 \\ n & 3 & … & n-1 & n & n+1 end{vmatrix}$$ The closest one to mine from StackExchange was this one. But I didn’t manage to link these two determinants. No matter how I transform it, nothing worked for me. Any ideas?", "## College Algebra (6th Edition) $\\begin{vmatrix} 1 & 2 &-1 \\\\0 & -11 &19 \\end{vmatrix}$ $\\begin{vmatrix} 1 & 2 &-1 \\\\ 4 & -3 & 15 \\end{vmatrix}$ $-4$times the first row is: $\\begin{vmatrix} -4 & -8 &4 \\end{vmatrix}$, thus adding this to the second row, it becomes: $\\begin{vmatrix} 4-4 & -3-8 &15+4 \\end{vmatrix}=\\begin{vmatrix} 0 & -11 &19 \\end{vmatrix}$, so the array becomes: $\\begin{vmatrix} 1 & 2 &-1 \\\\0 & -11 &19 \\end{vmatrix}$", "compute $A=J_v(4,-3)=\\begin{bmatrix}{\\dfrac{{\\partial v_1}}{{\\partial x}}(4,-3)}&{\\dfrac{{\\partial v_1}}{{\\partial y}}(4,-3)}\\\\{\\dfrac{{\\partial v_2}}{{\\partial x}}(4,-3)}&{\\dfrac{{\\partial v_2}}{{\\partial y}}(4,-3)}\\end{bmatrix}=\\begin{bmatrix}{-3}&{\\;\\;4}\\\\{\\;\\;8}&{-6}\\end{bmatrix}$ whose eigenvalues are certainly $\\lambda=\\dfrac{-9\\pm \\sqrt{137}}{2}$", "0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3" ]

[ "-3 \\end{vmatrix}$$ $$224$$ 20) $$\\begin{vmatrix} 6 & -1 & 2\\\\ -4 & -3 & 5\\\\ 1 & 9 & -1 \\end{vmatrix}$$ 21) $$\\begin{vmatrix} 5 & 1 & -1\\\\ 2 & 3 & 1\\\\ 3 & -6 & -3 \\end{vmatrix}$$ $$15$$ 22) $$\\begin{vmatrix} 1.1 & 2 & -1\\\\ -4 & 0 & 0\\\\ 4.1 & -0.4 & 2.5 \\end{vmatrix}$$ 23) $$\\begin{vmatrix} 2 & -1.6 & 3.1\\\\ 1.1 & 3 & -8\\\\ -9.3 & 0 & 2 \\end{vmatrix}$$ $$-17.03$$ 24) $$\\begin{vmatrix} -\\frac{1}{2} & \\frac{1}{3} & \\frac{1}{4}\\\\ \\frac{1}{5} & -\\frac{1}{6} & \\frac{1}{7}\\\\ 0 & 0 & \\frac{1}{8} \\end{vmatrix}$$ For the exercises 25-34, solve the system of linear equations using Cramer’s Rule. 25) \\begin{align*} 2x-3y &= -1\\\\ 4x+5y &= 9 \\end{align*} $$(1,1)$$ 26) \\begin{align*} 5x-4y &= 2\\\\ -4x+7y &= 6 \\end{align*} 27) \\begin{align*} 6x-3y &= 2\\\\ -8x+9y &= -1 \\end{align*} $$\\left (\\dfrac{1}{2}, \\dfrac{1}{3} \\right )$$ 28) \\begin{align*} 2x+6y &= 12\\\\ 5x-2y &= 13 \\end{align*} 29) \\begin{align*} 4x+3y &= 23\\\\ 2x-y &= -1 \\end{align*} $$(2,5)$$ 30) \\begin{align*} 10x-6y &= 2\\\\ -5x+8y &= -1 \\end{align*} 31) \\begin{align*} 4x-3y &= -3\\\\ 2x+6y &= -4 \\end{align*} $$\\left (-1, -\\dfrac{1}{3} \\right )$$ 32) \\begin{align*} 4x-5y &= 7\\\\ -3x+9y &= 0 \\end{align*} 33) \\begin{align*} 4x+10y &= 180\\\\ -3x-5y &= -105 \\end{align*} $$(15,12)$$ 34) \\begin{align*} 8x-2y &= -3\\\\ -4x+6y &= 4 \\end{align*} For the exercises 35-44, solve the system", "& -3 & 5 \\\\ 1 & -2 & 6 \\end{vmatrix} }{ \\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & -3 & -1 \\\\ 1 & -2 & 3 \\end{vmatrix} } = \\frac{-26}{-26} = 1$The solution set of the system is$ (1, – 1,1) $. Practice Questions 1. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} { 5x } + 2y &= \\, { 10 } \\\\ { – x } + 4y &= { 20 } \\end{align*} $2. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} 3x – 4y + z &= \\, -5 \\\\ x – y – z &= – 10 \\\\ 6x – 8y + 2z = 10 \\end{align*} $### Answers 1. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x } $), and the$ y – $determinant ($ D_{ y } $). Let’s use the formula we learned and write them up:$ D = \\begin{vmatrix} 5 & 2 \\\\ { – 1 } & { 4 } \\end{vmatrix} D_{ x } = \\begin{vmatrix} 10 & 2 \\\\ { 20 } & { 4 } \\end{vmatrix} D_{ y } = \\begin{vmatrix} 5 & { 10 } \\\\ { – 1 } & {", "0 & 0 \\\\ -2 & 1 & 0 \\\\ -2 & 0 & 1 \\end{vmatrix}}_{=1} \\cdot \\begin{vmatrix} 1 & 4 & 1 \\\\ 2 & 6 & 3 \\\\ 2 & 10 & 3 \\end{vmatrix} = \\begin{vmatrix} 1 & 4 & 1 \\\\ 0 & -2 & 1 \\\\ 0 & 2 & 1 \\end{vmatrix} = \\begin{vmatrix} -2 & 1 \\\\ 2 & 1 \\end{vmatrix} =-4 \\end{align*} \\end{document} Somehow, it would make more sense if you tried to demonstrate the column manipulation, because then the triangle unit-determinant matrix would come on the right side. But even this is completely clear IMHO. Here's another style I made. If you aren't doing determinants, replace the \"vmatrix\" with \"bmatrix\". I don't know how to post a picture of what it looks like. \\paragraph{Example} Evaluate the determinant \\begin{align*} \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 2 & -21 & 22 & -7 \\\\ 2 & -9 & 41 & -10 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*}. \\end{align*} \\paragraph{Solution} \\begin{align*} \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 2 & -21 & 22 & -7 \\\\ 2 & -9 & 41 & -10 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*} &= \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 0 & 3 &", "# How to draw the calculation of a determinant? I'm trying to reproduce the following calculation of a determinant in latex: Any ideas? I thought of using the tabular environment but unfortunately I'm not able to align the \"=\" correctly. I just have one question: The numbers in the determine itself are all right aligned. You can see it in column 3, row 1 and 2 in the picture I posted. The 2 is just above the -2 and they are both right aligned, no matter if there is a minus or not. Do you have any idea how to realize that? -- user24295 No, I don't, but according to comments by percusse and Manuel you can do that by loading mathtools instead of amsmath and use \\begin{vmatrix*}[r] … \\end{vmatrix*}. I haven't checked though by myself. But I do recomment keep the standard alignement provided by \\begin{vmatrix} ... \\end{vmatrix}. UPDATE with align. Thanks to cmhughes's advice I've replaced eqnarray with alignand &=& with &=. ADDED. As commented by egreg eqnarray should be avoided (see e.g. here and here) due to spacing discrepancies. The updated code is: \\documentclass{article} \\usepackage{amsmath} \\begin{document} \\begin{align} \\det A &=\\left. \\begin{array}{c} \\;\\;\\text{*)} \\\\ -1\\text{)} \\\\ -2\\text{)} \\\\ -2\\text{)} \\end{array} \\right. \\begin{vmatrix} 1 & 3 & 2 & -6 \\\\ 1 & 2 & -2 &", "to the second column and also the third column. $|A| = \\begin{vmatrix} n^2 & 2n+1 & 4n+4 \\\\ (n + 1)^2 & 2n+3 & 4n + 8 \\\\ (n + 2)^2 & 2n+5 & 4n+12 \\end{vmatrix}$ Multiply the elements in the first row by $-1$ then add that result to the second and third rows. $|A| = \\begin{vmatrix} n^2 & 2n+1 & 4n+4 \\\\ 2n+1 & 2 & 4 \\\\ 4n+4 & 4 & 8 \\end{vmatrix}$ Factorize out $4$ from the third column. $|A| = 4\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 4n+4 & 4 & 2 \\end{vmatrix}$ Factorize out $2$ from the third row. $|A| = 8\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 2n+2 & 2 & 1 \\end{vmatrix}$ Multiply the elements in the second row by $-1$ then add them to the third row. $|A| = 8\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 1 & 0 & 0 \\end{vmatrix}$ Apply a cofactor expansion along the third row. \\begin{aligned}|A| & = 8 \\cdot 1 \\cdot \\begin{vmatrix} 2n+1 & n+1 \\\\ 2 & 1 \\end{vmatrix} \\\\ & = 8 \\cdot [(2n+1)-2(n+1)] \\\\ & = 8(-1) = -8 \\end{aligned} Thus, the determinant of matrix $A$ is $\\boxed{|A| = -8}$ [collapse]", "12 & -1 \\\\ 0 & 15 & 31 & -4 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*} \\begin{matrix*}[l] \\\\ \\leftarrow -2r_1+r_2 \\ \\text{\\scriptsize (no change in determinant)} \\\\ \\leftarrow -2r_2+r_3 \\ \\text{\\scriptsize (no change in determinant)} \\\\ \\\\ \\end{matrix*} \\\\ &= -\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\ 0 & -1 & 12 & 3 \\\\ 0 & -4 & 31 & 15 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*} \\begin{matrix*}[l] \\\\ \\leftarrow c_2 \\leftrightarrow c_4 \\ \\text{\\scriptsize (negate determinant)} \\\\ \\\\ \\\\ \\end{matrix*} \\\\ &= -\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\ 0 & 1 & -12 & -3 \\\\ 0 & 4 & -31 & -15 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*} \\begin{matrix*}[l] \\\\ \\leftarrow -r_2 \\ \\text{\\scriptsize (negate determinant)} \\\\ \\leftarrow -r_3 \\ \\text{\\scriptsize (negate determinant)} \\\\ \\\\ \\end{matrix*} \\\\ &= -\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\ 0 & 1 & -12 & -3 \\\\ 0 & 0 & 17 & -3 \\\\ 0 & 0 & 43 & 12 \\end{vmatrix*} \\begin{matrix*}[l] \\\\ \\\\ \\leftarrow -4r_2+r_3 \\ \\text{\\scriptsize (no change in determinant)} \\\\ \\leftarrow -3r_2+r_4 \\ \\text{\\scriptsize (no change in determinant)} \\end{matrix*} \\\\ &= -17\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\ 0 & 1 & -12 & -3", "2 &1 &5 \\end{vmatrix} &=\\begin{vmatrix} 2 &0 &0 \\\\ 0 &3 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 2 &0 &0 \\\\ 0 &0 &\\underline{0} \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &1 &0 \\\\ 4 &0 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &\\underline{0} \\\\ 2 &0 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &0 &-1 \\\\ 4 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} + \\begin{vmatrix} 0 &0 &-1 \\\\ 0 &3 &0 \\\\ 2 &0 &0 \\end{vmatrix} \\end{array}$ We can bring out the scalars. $\\begin{array}{rl} &=(2)(3)(5)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(2)(\\underline{0})(1)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &0 &1 \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+(1)(4)(5)\\begin{vmatrix} 0 &1 &0 \\\\ 1 &0 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(1)(\\underline{0})(2)\\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &1 \\\\ 1 &0 &0 \\end{vmatrix} \\\\ &\\quad+(-1)(4)(1)\\begin{vmatrix} 0 &0 &1 \\\\ 1 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} +(-1)(3)(2)\\begin{vmatrix} 0 &0 &1 \\\\ 0 &1 &0 \\\\ 1 &0 &0 \\end{vmatrix} \\end{array}$ To finish, we evaluate those six determinants by row-swapping them to the identity matrix, keeping track of the resulting sign changes. $\\begin{array}{rl} &=30\\cdot (+1)+0\\cdot (-1) \\\\ &\\quad+20\\cdot (-1)+0\\cdot (+1) \\\\ &\\quad -4\\cdot (+1)-6\\cdot (-1)=12 \\end{array}$ That example illustrates the key idea. We've", "\\\\ 1 && -1\\\\ \\end{vmatrix} - det(4) \\begin{vmatrix} \\ 1 && 3 \\\\ 1 && -1\\\\ \\end{vmatrix} $ $ - 5det(-2) \\begin{vmatrix} \\ 2 && -1 \\\\ -2 && 1\\\\ \\end{vmatrix} + det(4) \\begin{vmatrix} \\ 1 && 2 \\\\ 1 && -2\\\\ \\end{vmatrix} $ $= 1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))$ $= 1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))$ $= 1 (16) - 3 (4 + 16) +5(-16)$ If anyone noticed any wrong be greatly appreciate if they could correct it saying so. Thanks a bunch! My calculator says you are wrong. (I get 4.) I would suggest you expand across the second row... you only have to do two 3 x 3 determinants that way. -Dan 3. Originally Posted by mathsToday Hi guys, hoping somebody can check through my calculations for the determinant of $A$ a 4 x 4 matrix. $A = \\begin{vmatrix} \\-1 && 3 && 0 && 5 \\\\ -2 && 0 && 4 && 0\\\\ 1 && 2 && -1 && 3\\\\ 1 && -2 && 1 && -1\\\\ \\end{vmatrix} = 1 \\begin{vmatrix} \\ 0 && 4 && 0 \\\\ 2 && -1 && 3\\\\ -2 && 1 && -1\\\\ \\end{vmatrix} - 3 \\begin{vmatrix} \\ -2 && 4 && 0 \\\\ 1 && -1 && 3\\\\ 1 && 1 && -1\\\\", "remaining uncovered $2 \\times 2$ matrix, then multiply that by $a_{1i}$. The determinant is the sum of those values, alternating addition and subtraction. Worked Examples Example 1 Find the determinant of the matrix $\\begin{pmatrix} 1 & 2 & 1\\\\ 0 & 3 & 4\\\\ 3 & 1 & 4 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} 3 & 4\\\\ 1 & 4 \\end{vmatrix} -2 \\begin{vmatrix} 0 & 4\\\\ 3 & 4 \\end{vmatrix} +1 \\begin{vmatrix} 0 & 3\\\\ 3 & 1 \\end{vmatrix}\\\\\\\\ &= 1(3\\times 4 - 4\\times 1) -2(0\\times 4 - 4\\times 3) +1(0 \\times 1 - 3\\times 3)\\\\ &= 1(12-4) -2(0-12) +1(0-9)\\\\ &= 8 +24-9\\\\ &=23 \\end{align} Example 2 Find the determinant of the matrix $\\begin{pmatrix} 1 & 0 & 3\\\\ -1 & -1 & -3\\\\ 0 & 0 & 6 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} -0 \\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} +3 \\begin{vmatrix} -1 & -1\\\\ 0 & 0 \\end{vmatrix}\\\\\\\\ &= 1\\bigl((-1)\\times 6 - (-3)\\times 0\\bigr) -0(\\,\\dotso\\,) +3\\bigl((-1) \\times 0 - (-1)\\times 0\\bigr)\\\\ &= 1\\bigr((-6)-0\\bigl) -0 +3(0-0)\\\\ &=-6 +0\\\\ &=-6 \\end{align} Note: There was no need to work out the second $2 \\times 2$ determinant as it was being multiplied by zero. Video Example Hayley Bishop works out the determinant of the $3\\times", "\\end{vmatrix} \\end{align*} For the third element in the first row, we use the same strategy from the second element, where we use our fingers to block out all of the elements in the same row as the corresponding coefficient. That coefficient is now $4$, so we see that this $2$ by $2$ matrix will contain the elements $7$, $24$, $2$, and $6$. This gives us the matrix below. \\begin{align*} \\begin{vmatrix} 7 & 24 \\\\ 2 & 6 \\end{vmatrix} \\end{align*} Plugging in the $2$ by $2$ matrices, their corresponding coefficients, and the coefficients’ new signs, we have \\begin{align*} \\begin{vmatrix} 0 & 2 & 4 \\\\ 7 & 24 & 1 \\\\ 2 & 6 & 8 \\end{vmatrix} = 0 - 2 \\times \\begin{vmatrix} 7 & 1 \\\\ 2 & 8 \\end{vmatrix} + 4 \\times \\begin{vmatrix} 7 & 24 \\\\ 2 & 6 \\end{vmatrix} \\end{align*} We can now solve each $2$ by $2$ matrix as usual, multiply those determinants by the coefficients, and add them up to find the determinant of the $3$ by $3$ matrix. Doing this with our example, we have \\begin{align*} \\begin{vmatrix} 0 & 2 & 4 \\\\ 7 & 24 & 1 \\\\ 2 & 6 & 8 \\end{vmatrix} &= 0 - 2 \\times \\begin{vmatrix} 7 & 1 \\\\ 2 & 8 \\end{vmatrix} + 4 \\times", "?\" The link that Dietrich Burde has given you, especially its solution by \"Paul\" is very understandable (juste take $a=s$ and $b=1$). – Jean Marie Jan 30 at 20:06 Here's something simple: first exploit linearity of the determinant w.r.t. rows: $$D=\\begin{vmatrix} s&1&1&1\\\\1&s&1&1\\\\1&1&s&1\\\\1&1&1&s \\end{vmatrix}= \\begin{vmatrix} s-1&0&0&1-s\\\\0&s-1&0&1-s\\\\0&0&s-1&1-s\\\\1&1&1&s \\end{vmatrix}= (s-1)^3 \\begin{vmatrix} 1&0&0&-1\\\\0&1&0&-1\\\\0&0&1&-1\\\\1&1&1&s \\end{vmatrix}$$ Next expand by the first row: $$D=(s-1)^3\\left(1\\cdot\\begin{vmatrix} 1&0&-1\\\\ 0&1&-1\\\\1 &1&s \\end{vmatrix}-(-1) \\begin{vmatrix} 0&1&0\\\\0&0&1\\\\1&1&1 \\end{vmatrix}\\,\\right)$$ Now the two $$3\\times3$$ determinants are: \\begin{align} &\\begin{vmatrix} 1&0&-1\\\\ 0&1&-1\\\\1 &1&s \\end{vmatrix}=\\begin{vmatrix} 1&0&0\\\\ 0&1&0\\\\1 &1&s+2 \\end{vmatrix}=s+2\\\\[1ex] &\\begin{vmatrix} 0&1&0\\\\0&0&1\\\\1&1&1 \\end{vmatrix}=1\\qquad\\text{(expanding by the first column)} \\end{align} and ultimately, we obtain $$D=(s-1)^3(s+3).$$", "with the pivot operations of adding ${\\displaystyle v_{1}{\\vec {\\rho }}_{1}}$ to ${\\displaystyle v_{i}{\\vec {\\beta }}}$ and ${\\displaystyle w_{1}{\\vec {\\rho }}_{1}}$ to ${\\displaystyle w_{i}{\\vec {\\rho }}_{1}}$, etc. Multilinearity allows us to expand a determinant into a sum of determinants, each of which involves a simple matrix. Example 3.4 We can use multilinearity to split this determinant into two, first breaking up the first row ${\\displaystyle {\\begin{vmatrix}2&1\\\\4&3\\end{vmatrix}}={\\begin{vmatrix}2&0\\\\4&3\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\4&3\\end{vmatrix}}}$ and then separating each of those two, breaking along the second rows. ${\\displaystyle ={\\begin{vmatrix}2&0\\\\4&0\\end{vmatrix}}+{\\begin{vmatrix}2&0\\\\0&3\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\4&0\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\0&3\\end{vmatrix}}}$ We are left with four determinants, such that in each row of each matrix there is a single entry from the original matrix. Example 3.5 In the same way, a ${\\displaystyle 3\\!\\times \\!3}$ determinant separates into a sum of many simpler determinants. We start by splitting along the first row, producing three determinants (the zero in the ${\\displaystyle 1,3}$ position is underlined to set it off visually from the zeroes that appear in the splitting). ${\\displaystyle {\\begin{vmatrix}2&1&-1\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}={\\begin{vmatrix}2&0&0\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}+{\\begin{vmatrix}0&1&0\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}+{\\begin{vmatrix}0&0&-1\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}}$ Each of these three will itself split in three along the second row. Each of the resulting nine splits in three along the third row, resulting in twenty seven determinants ${\\displaystyle ={\\begin{vmatrix}2&0&0\\\\4&0&0\\\\2&0&0\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\4&0&0\\\\0&1&0\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\4&0&0\\\\0&0&5\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\0&3&0\\\\2&0&0\\end{vmatrix}}+\\dots +{\\begin{vmatrix}0&0&-1\\\\0&0&{\\underline {0}}\\\\0&0&5\\end{vmatrix}}}$ such that each row contains a single entry from the starting matrix. So an ${\\displaystyle n\\!\\times \\!n}$ determinant expands into a sum of ${\\displaystyle", "& -1 & 4\\\\ 2 & 5 & 1\\\\ 2 & 0 & 6 \\end{bmatrix} \\end{equation*} Solution We can expand about any row or column, so the zero entry in the middle of the last row is attractive. Let us expand about column 2. By Theorem DER and Theorem DEC you will get the same result by expanding about a different row or column. We will use Theorem DMST twice. \\begin{align*} \\begin{vmatrix} 3 & -1 & 4\\\\ 2 & 5 & 1\\\\ 2 & 0 & 6 \\end{vmatrix} &= (-1)(-1)^{1+2} \\begin{vmatrix} 2 & 1\\\\ 2 & 6 \\end{vmatrix} + (5)(-1)^{2+2} \\begin{vmatrix} 3 & 4\\\\ 2 & 6 \\end{vmatrix} + (0)(-1)^{3+2} \\begin{vmatrix} 3 & 4\\\\ 2 & 1 \\end{vmatrix}\\\\ &=(1)(10)+(5)(10)+0=60 \\end{align*} ###### C26. Doing the computations by hand, find the determinant of the matrix $A\\text{.}$ \\begin{equation*} A= \\begin{bmatrix} 2 & 0 & 3 & 2 \\\\ 5 & 1 & 2 & 4 \\\\ 3 & 0 & 1 & 2 \\\\ 5 & 3 & 2 & 1 \\end{bmatrix} \\end{equation*} Solution With two zeros in column 2, we choose to expand about that column (Theorem DEC), \\begin{align*} \\detname{A} &= \\begin{vmatrix} 2 & 0 & 3 & 2 \\\\ 5 & 1 & 2 & 4 \\\\ 3 & 0 & 1 & 2 \\\\ 5 & 3 &", "since there are no zeros to exploit, \\begin{vmatrix} \\end{vmatrix} (-2)\\begin{vmatrix} \\end{vmatrix} + (-1)(3)\\begin{vmatrix} \\end{vmatrix} + (-2)\\begin{vmatrix} \\end{vmatrix}\\\\ Doing the computations by hand, find the determinant of the matrix below. \\begin{bmatrix} \\end{bmatrix} We can expand about any row or column, so the zero entry in the middle of the last row is attractive. Let's expand about column 2. By and you will get the same result by expanding about a different row or column. We will use twice. \\begin{vmatrix} \\end{vmatrix} (-1)(-1)^{1+2} \\begin{vmatrix} \\end{vmatrix} + (5)(-1)^{2+2} \\begin{vmatrix} \\end{vmatrix} + (0)(-1)^{3+2} \\begin{vmatrix} \\end{vmatrix}\\\\ Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} With two zeros in column 2, we choose to expand about that column (), \\detname{A} \\begin{vmatrix} \\end{vmatrix}\\\\ 0(-1) \\begin{vmatrix} \\end{vmatrix} + 1(1) \\begin{vmatrix} \\end{vmatrix} + 0(-1) \\begin{vmatrix} \\end{vmatrix} + 3(1) \\begin{vmatrix} \\end{vmatrix}\\\\ (1)\\left(2(1(1)-2(2)) - 3(3(1)-5(2)) + 2(3(2)-5(1))\\right) +\\\\ (-6+21+2) + (3)(0+6-2) = 29 Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding on the first row, we have Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding along the first row, we have Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding along the first column, we have \\begin{vmatrix} \\end{vmatrix} 2 \\begin{vmatrix}", "2 \\\\ 1 && -2\\\\ \\end{vmatrix} $ $= 1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))$ $= 1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))$ $= 1 (16) - 3 (4 + 16) +5(-16)$ If anyone noticed any wrong be greatly appreciate if they could correct it saying so. Thanks a bunch! 2. Originally Posted by mathsToday Hi guys, hoping somebody can check through my calculations for the determinant of $A$ a 4 x 4 matrix. $A = \\begin{vmatrix} \\-1 && 3 && 0 && 5 \\\\ -2 && 0 && 4 && 0\\\\ 1 && 2 && -1 && 3\\\\ 1 && -2 && 1 && -1\\\\ \\end{vmatrix} = 1 \\begin{vmatrix} \\ 0 && 4 && 0 \\\\ 2 && -1 && 3\\\\ -2 && 1 && -1\\\\ \\end{vmatrix} - 3 \\begin{vmatrix} \\ -2 && 4 && 0 \\\\ 1 && -1 && 3\\\\ 1 && 1 && -1\\\\ \\end{vmatrix} $ $ + 0 \\begin{vmatrix} \\ -2 && 0 && 0 \\\\ 1 && 2 && 3\\\\ 1 && -2 && -1\\\\ \\end{vmatrix} - 5 \\begin{vmatrix} \\ -2 && 0 && 4 \\\\ 1 && 2 && -1\\\\ 1 && -2 && 1\\\\ \\end{vmatrix} $ $ = 1det(4)\\begin{vmatrix} \\ 2 && 3 \\\\ -2 && -1\\\\ \\end{vmatrix} $ $ - 3 det(-2) \\begin{vmatrix} \\ -1 && 3", "(often displayed as three diagonal products each way that \"wrap around\") let us calculate: $\\det(A) = \\begin{vmatrix}1&3&-2\\\\5&1&4\\\\0&2&1 \\end{vmatrix}$ using both methods. the first way, we have a choice: expansion by a row, or by a column (we could use ANY row or column, but we'll use the first one to keep it simple). first we'll expand by the first row: to do this we need to find the 2x2 determinants, first. $\\det(A_{1,1}) = \\begin{vmatrix} 1&4\\\\2&1 \\end{vmatrix} = 1 - 8 = -7$ $\\det(A_{1,2}) = \\begin{vmatrix} 5&4\\\\0&1 \\end{vmatrix} = 5 - 0 = 5$ $\\det(A_{1,3}) = \\begin{vmatrix} 5&1\\\\0&2 \\end{vmatrix} = 10 - 0 = 10$ thus $\\det(A) = (1)(-7) - (3)(5) + (-2)(10) = -7 - 15 - 20 = -42$. now we'll expand by the first column. we need to calculate 2 more 2x2 determinants (we'll re-use the one obtained by eliminating row 1 and column one): $\\det(A_{2,1}) = \\begin{vmatrix} 3&-2\\\\2&1 \\end{vmatrix} = 3 - (-4) = 7$ $\\det(A_{3,1}) = \\begin{vmatrix} 3&-2\\\\1&4 \\end{vmatrix} = 12 -(-2) = 14$ thus $\\det(A) = (1)(-7) - 5(7) + 0(14) = -7 - 35 + 0 = -42$. finally, we'll use the rule of sarrus: $\\det(A)$ $= (1)(1)(1) + (3)(4)(0) + (-2)(5)(2) - (-2)(1)(0) - (3)(5)(1) - (1)(4)(2)$ $= 1 + 0 - 20 - 0 - 15 - 8 = 1", "To right align the columns you could use \\begin{vmatrix*}[r] … \\end{vmatrix*}. – Manuel Jan 11 '13 at 21:29 You can define an xvmatrix environment for the \"eXtended\" matrix, where you specify the coefficients as a first column: \\documentclass{article} \\usepackage{amsmath} \\makeatletter \\newenvironment{xvmatrix}% eXtended vmatrix {\\left.\\array{@{}r |@{\\,}*\\c@MaxMatrixCols c}} {\\endarray\\kern-\\arraycolsep\\right|} \\begin{document} \\begin{align*} \\det A &= \\begin{xvmatrix} *) & 1 & 3 & 2 & -6 \\\\ -1) & 1 & 2 & -2 & -5 \\\\ -2) & 2 & 4 & -2 & -9 \\\\ -2) & 2 & 4 & -6 & -9 \\end{xvmatrix} = \\begin{vmatrix} 1 & 3 & 2 & -6 \\\\ 0 & -1 & -4 & 1 \\\\ 0 & -2 & -6 & 3 \\\\ 0 & -2 & -10 & 3 \\end{vmatrix} \\\\ &= \\begin{xvmatrix} *) & 1 & 4 & 1 \\\\ -2) & 2 & 6 & 3 \\\\ -2) & 2 & 10 & 3 \\end{xvmatrix} = \\begin{vmatrix} 1 & 4 & 1 \\\\ 0 & -2 & 1 \\\\ 0 & 2 & 1 \\end{vmatrix} = \\begin{vmatrix} -2 & 1 \\\\ 2 & 1 \\end{vmatrix} =-4 \\end{align*} \\end{document} - And if that's always the notation, you could add ) to the environment so you don't need to write it all the times. Apart from that, I don't find this", "& 10 & 20 & 35 \\\\ 1 & 5 & 15 & 35 & 70 \\\\ \\end{bmatrix}\\tag2$$ Using Pascal's Identity, we see that $$L_nU_nR_n$$ has a $$1$$ in the upper left and $$0$$'s in the remainder if the top row and left column. The lower-right $$n-1\\times n-1$$ submatrix is $$U_{n-1}$$. For example, subtracting each row from the next (bottom to top) gives $$L_5U_5= \\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 2 & 3 & 4 \\\\ 0 & 1 & 3 & 6 & 10 \\\\ 0 & 1 & 4 & 10 & 20 \\\\ 0 & 1 & 5 & 15 & 35 \\\\ \\end{bmatrix}\\tag3$$ then subtracting each column from the next (right to left) gives $$L_5U_5R_5= \\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 2 & 3 & 4 \\\\ 0 & 1 & 3 & 6 & 10 \\\\ 0 & 1 & 4 & 10 & 20 \\\\ \\end{bmatrix}\\tag4$$ Extending $$(2)$$ and $$(4)$$ shows that \\begin{align} \\det(U_n) &=\\det(L_nU_nR_n)\\tag{5a}\\\\[6pt] &=\\det(U_{n-1})\\tag{5b}\\\\[6pt] &=1\\tag{5c} \\end{align} Explanation: $$\\text{(5a)}$$: $$\\det(L_n)=\\det(R_n)=1$$ $$\\text{(5b)}$$: expand the determinant on the first column $$\\text{(5c)}$$: induction and $$\\det(U_1)=1$$ The Matrix of Interest Let $$T_n$$ be an $$n\\times n$$ matrix", "-1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$ ##### Solution : (iv) Let $A=\\begin{vmatrix} 2 & -1 & -2 \\\\[0.3em] 0 & 2 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ By expanding along the first column, we have:\\\\ |A|=2\\begin{vmatrix} 2 & -1\\\\[0.3em] -5 & 0\\end{vmatrix} -0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 &0 \\end{vmatrix}+3\\begin{vmatrix} -1 & -2 \\\\[0.3em] 2 & -1 \\end{vmatrix}$$\\\\$ $=2(0-5) -0+3(1+4)\\\\ =-10+15=5$ $(i) \\begin{vmatrix} 3 & -1 & -2 \\\\[0.3em] 0 & 0 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.\\\\ |A|=-0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 & 0\\end{vmatrix} +0\\begin{vmatrix} 3 & -2 \\\\[0.3em] 3 & 0\\end{vmatrix} -(-1)\\begin{vmatrix} 3 & -1 \\\\[0.3em] 3 & -5\\end{vmatrix} =(-15+3) =-12 (iii) Let A=\\begin{vmatrix} 0 & 1 & 2 \\\\[0.3em] -1 & 0 & -3 \\\\[0.3em] -2 & 3 & 0\\end{vmatrix}$$\\\\$ By expanding along the first row, we have:$\\\\$ $|A|=0\\begin{vmatrix} 0 & -3 \\\\[0.3em] 3 & 0\\end{vmatrix} -1\\begin{vmatrix} -1 & -3 \\\\[0.3em] -2 & 0\\end{vmatrix} +2\\begin{vmatrix} -1 & 0\\\\[0.3em] -2 & 3 \\end{vmatrix}$$\\\\ =0-1(0-6)+2(-3-0)\\\\ =-1(-6)+2(-3)\\\\ =6-6=0 (ii)Let \\begin{vmatrix} 3 & -4 & 5 \\\\[0.3em] 1 & 1 & -2 \\\\[0.3em] 2 & 3 & 1 \\end{vmatrix}$$\\\\$ By expanding along the first row, we have:$\\\\$", "# Add between row spacing in a complex matrix I have the following matrix: \\documentclass{article} \\usepackage{amsmath} \\begin{document} $\\begin{bmatrix} \\begin{vmatrix} 168 & 368 \\\\ 368 & 1026 \\end{vmatrix} & \\begin{vmatrix} 26 & 368\\\\ 70 & 1026 \\end{vmatrix} & \\begin{vmatrix} 26 & 168\\\\ 70 & 368 \\end{vmatrix}\\\\ \\begin{vmatrix} 26 & 70\\\\ 368 & 1026 \\end{vmatrix} & \\begin{vmatrix} 5 & 70\\\\ 70 & 1026 \\end{vmatrix} & \\begin{vmatrix} 5 & 26\\\\ 70 & 368 \\end{vmatrix}\\\\ \\begin{vmatrix} 26 & 70\\\\ 168 & 368 \\end{vmatrix} & \\begin{vmatrix} 5 & 70\\\\ 26 & 368 \\end{vmatrix} & \\begin{vmatrix} 5 & 26\\\\ 26 & 168 \\end{vmatrix} \\end{bmatrix} = \\begin{bmatrix} 36944 & -916 & -2192 \\\\ \\end{bmatrix}$ \\end{document} Each row contains a 2x2 determinant. The problem is that the vertical lines of the neighboring determinants are touching, creating the impression that this is a 3x1 matrix with three 6x2 determinants. I have experimented a little with adding [3pt] after the delimiting \\\\, but had no success. Any help will be appreciated! By the way, I am calling LaTeX from a Markdown document, which means using extra/alternative packages is highly undesirable (albeit possible after a manual intervention during text conversion). Thanks in advance! - please edit your example to be a complete document showing the packages required (I assume amsmath, but you don't say). Otherwise we don't know what" ]

[ "2 &1 &5 \\end{vmatrix} &=\\begin{vmatrix} 2 &0 &0 \\\\ 0 &3 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 2 &0 &0 \\\\ 0 &0 &\\underline{0} \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &1 &0 \\\\ 4 &0 &0 \\\\ 0 &0 &5 \\end{vmatrix} + \\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &\\underline{0} \\\\ 2 &0 &0 \\end{vmatrix} \\\\ &\\quad+\\begin{vmatrix} 0 &0 &-1 \\\\ 4 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} + \\begin{vmatrix} 0 &0 &-1 \\\\ 0 &3 &0 \\\\ 2 &0 &0 \\end{vmatrix} \\end{array}$ We can bring out the scalars. $\\begin{array}{rl} &=(2)(3)(5)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(2)(\\underline{0})(1)\\begin{vmatrix} 1 &0 &0 \\\\ 0 &0 &1 \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad+(1)(4)(5)\\begin{vmatrix} 0 &1 &0 \\\\ 1 &0 &0 \\\\ 0 &0 &1 \\end{vmatrix} +(1)(\\underline{0})(2)\\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &1 \\\\ 1 &0 &0 \\end{vmatrix} \\\\ &\\quad+(-1)(4)(1)\\begin{vmatrix} 0 &0 &1 \\\\ 1 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} +(-1)(3)(2)\\begin{vmatrix} 0 &0 &1 \\\\ 0 &1 &0 \\\\ 1 &0 &0 \\end{vmatrix} \\end{array}$ To finish, we evaluate those six determinants by row-swapping them to the identity matrix, keeping track of the resulting sign changes. $\\begin{array}{rl} &=30\\cdot (+1)+0\\cdot (-1) \\\\ &\\quad+20\\cdot (-1)+0\\cdot (+1) \\\\ &\\quad -4\\cdot (+1)-6\\cdot (-1)=12 \\end{array}$ That example illustrates the key idea. We've", "-3 \\end{vmatrix}$$ $$224$$ 20) $$\\begin{vmatrix} 6 & -1 & 2\\\\ -4 & -3 & 5\\\\ 1 & 9 & -1 \\end{vmatrix}$$ 21) $$\\begin{vmatrix} 5 & 1 & -1\\\\ 2 & 3 & 1\\\\ 3 & -6 & -3 \\end{vmatrix}$$ $$15$$ 22) $$\\begin{vmatrix} 1.1 & 2 & -1\\\\ -4 & 0 & 0\\\\ 4.1 & -0.4 & 2.5 \\end{vmatrix}$$ 23) $$\\begin{vmatrix} 2 & -1.6 & 3.1\\\\ 1.1 & 3 & -8\\\\ -9.3 & 0 & 2 \\end{vmatrix}$$ $$-17.03$$ 24) $$\\begin{vmatrix} -\\frac{1}{2} & \\frac{1}{3} & \\frac{1}{4}\\\\ \\frac{1}{5} & -\\frac{1}{6} & \\frac{1}{7}\\\\ 0 & 0 & \\frac{1}{8} \\end{vmatrix}$$ For the exercises 25-34, solve the system of linear equations using Cramer’s Rule. 25) \\begin{align*} 2x-3y &= -1\\\\ 4x+5y &= 9 \\end{align*} $$(1,1)$$ 26) \\begin{align*} 5x-4y &= 2\\\\ -4x+7y &= 6 \\end{align*} 27) \\begin{align*} 6x-3y &= 2\\\\ -8x+9y &= -1 \\end{align*} $$\\left (\\dfrac{1}{2}, \\dfrac{1}{3} \\right )$$ 28) \\begin{align*} 2x+6y &= 12\\\\ 5x-2y &= 13 \\end{align*} 29) \\begin{align*} 4x+3y &= 23\\\\ 2x-y &= -1 \\end{align*} $$(2,5)$$ 30) \\begin{align*} 10x-6y &= 2\\\\ -5x+8y &= -1 \\end{align*} 31) \\begin{align*} 4x-3y &= -3\\\\ 2x+6y &= -4 \\end{align*} $$\\left (-1, -\\dfrac{1}{3} \\right )$$ 32) \\begin{align*} 4x-5y &= 7\\\\ -3x+9y &= 0 \\end{align*} 33) \\begin{align*} 4x+10y &= 180\\\\ -3x-5y &= -105 \\end{align*} $$(15,12)$$ 34) \\begin{align*} 8x-2y &= -3\\\\ -4x+6y &= 4 \\end{align*} For the exercises 35-44, solve the system", "-1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$ ##### Solution : (iv) Let $A=\\begin{vmatrix} 2 & -1 & -2 \\\\[0.3em] 0 & 2 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ By expanding along the first column, we have:\\\\ |A|=2\\begin{vmatrix} 2 & -1\\\\[0.3em] -5 & 0\\end{vmatrix} -0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 &0 \\end{vmatrix}+3\\begin{vmatrix} -1 & -2 \\\\[0.3em] 2 & -1 \\end{vmatrix}$$\\\\$ $=2(0-5) -0+3(1+4)\\\\ =-10+15=5$ $(i) \\begin{vmatrix} 3 & -1 & -2 \\\\[0.3em] 0 & 0 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.\\\\ |A|=-0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 & 0\\end{vmatrix} +0\\begin{vmatrix} 3 & -2 \\\\[0.3em] 3 & 0\\end{vmatrix} -(-1)\\begin{vmatrix} 3 & -1 \\\\[0.3em] 3 & -5\\end{vmatrix} =(-15+3) =-12 (iii) Let A=\\begin{vmatrix} 0 & 1 & 2 \\\\[0.3em] -1 & 0 & -3 \\\\[0.3em] -2 & 3 & 0\\end{vmatrix}$$\\\\$ By expanding along the first row, we have:$\\\\$ $|A|=0\\begin{vmatrix} 0 & -3 \\\\[0.3em] 3 & 0\\end{vmatrix} -1\\begin{vmatrix} -1 & -3 \\\\[0.3em] -2 & 0\\end{vmatrix} +2\\begin{vmatrix} -1 & 0\\\\[0.3em] -2 & 3 \\end{vmatrix}$$\\\\ =0-1(0-6)+2(-3-0)\\\\ =-1(-6)+2(-3)\\\\ =6-6=0 (ii)Let \\begin{vmatrix} 3 & -4 & 5 \\\\[0.3em] 1 & 1 & -2 \\\\[0.3em] 2 & 3 & 1 \\end{vmatrix}$$\\\\$ By expanding along the first row, we have:$\\\\$", "since there are no zeros to exploit, \\begin{vmatrix} \\end{vmatrix} (-2)\\begin{vmatrix} \\end{vmatrix} + (-1)(3)\\begin{vmatrix} \\end{vmatrix} + (-2)\\begin{vmatrix} \\end{vmatrix}\\\\ Doing the computations by hand, find the determinant of the matrix below. \\begin{bmatrix} \\end{bmatrix} We can expand about any row or column, so the zero entry in the middle of the last row is attractive. Let's expand about column 2. By and you will get the same result by expanding about a different row or column. We will use twice. \\begin{vmatrix} \\end{vmatrix} (-1)(-1)^{1+2} \\begin{vmatrix} \\end{vmatrix} + (5)(-1)^{2+2} \\begin{vmatrix} \\end{vmatrix} + (0)(-1)^{3+2} \\begin{vmatrix} \\end{vmatrix}\\\\ Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} With two zeros in column 2, we choose to expand about that column (), \\detname{A} \\begin{vmatrix} \\end{vmatrix}\\\\ 0(-1) \\begin{vmatrix} \\end{vmatrix} + 1(1) \\begin{vmatrix} \\end{vmatrix} + 0(-1) \\begin{vmatrix} \\end{vmatrix} + 3(1) \\begin{vmatrix} \\end{vmatrix}\\\\ (1)\\left(2(1(1)-2(2)) - 3(3(1)-5(2)) + 2(3(2)-5(1))\\right) +\\\\ (-6+21+2) + (3)(0+6-2) = 29 Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding on the first row, we have Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding along the first row, we have Doing the computations by hand, find the determinant of the matrix $A$. A= \\begin{bmatrix} \\end{bmatrix} Expanding along the first column, we have \\begin{vmatrix} \\end{vmatrix} 2 \\begin{vmatrix}", "\\\\ 1 && -1\\\\ \\end{vmatrix} - det(4) \\begin{vmatrix} \\ 1 && 3 \\\\ 1 && -1\\\\ \\end{vmatrix} $ $ - 5det(-2) \\begin{vmatrix} \\ 2 && -1 \\\\ -2 && 1\\\\ \\end{vmatrix} + det(4) \\begin{vmatrix} \\ 1 && 2 \\\\ 1 && -2\\\\ \\end{vmatrix} $ $= 1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))$ $= 1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))$ $= 1 (16) - 3 (4 + 16) +5(-16)$ If anyone noticed any wrong be greatly appreciate if they could correct it saying so. Thanks a bunch! My calculator says you are wrong. (I get 4.) I would suggest you expand across the second row... you only have to do two 3 x 3 determinants that way. -Dan 3. Originally Posted by mathsToday Hi guys, hoping somebody can check through my calculations for the determinant of $A$ a 4 x 4 matrix. $A = \\begin{vmatrix} \\-1 && 3 && 0 && 5 \\\\ -2 && 0 && 4 && 0\\\\ 1 && 2 && -1 && 3\\\\ 1 && -2 && 1 && -1\\\\ \\end{vmatrix} = 1 \\begin{vmatrix} \\ 0 && 4 && 0 \\\\ 2 && -1 && 3\\\\ -2 && 1 && -1\\\\ \\end{vmatrix} - 3 \\begin{vmatrix} \\ -2 && 4 && 0 \\\\ 1 && -1 && 3\\\\ 1 && 1 && -1\\\\", "To right align the columns you could use \\begin{vmatrix*}[r] … \\end{vmatrix*}. – Manuel Jan 11 '13 at 21:29 You can define an xvmatrix environment for the \"eXtended\" matrix, where you specify the coefficients as a first column: \\documentclass{article} \\usepackage{amsmath} \\makeatletter \\newenvironment{xvmatrix}% eXtended vmatrix {\\left.\\array{@{}r |@{\\,}*\\c@MaxMatrixCols c}} {\\endarray\\kern-\\arraycolsep\\right|} \\begin{document} \\begin{align*} \\det A &= \\begin{xvmatrix} *) & 1 & 3 & 2 & -6 \\\\ -1) & 1 & 2 & -2 & -5 \\\\ -2) & 2 & 4 & -2 & -9 \\\\ -2) & 2 & 4 & -6 & -9 \\end{xvmatrix} = \\begin{vmatrix} 1 & 3 & 2 & -6 \\\\ 0 & -1 & -4 & 1 \\\\ 0 & -2 & -6 & 3 \\\\ 0 & -2 & -10 & 3 \\end{vmatrix} \\\\ &= \\begin{xvmatrix} *) & 1 & 4 & 1 \\\\ -2) & 2 & 6 & 3 \\\\ -2) & 2 & 10 & 3 \\end{xvmatrix} = \\begin{vmatrix} 1 & 4 & 1 \\\\ 0 & -2 & 1 \\\\ 0 & 2 & 1 \\end{vmatrix} = \\begin{vmatrix} -2 & 1 \\\\ 2 & 1 \\end{vmatrix} =-4 \\end{align*} \\end{document} - And if that's always the notation, you could add ) to the environment so you don't need to write it all the times. Apart from that, I don't find this", "2 \\\\ 1 && -2\\\\ \\end{vmatrix} $ $= 1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))$ $= 1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))$ $= 1 (16) - 3 (4 + 16) +5(-16)$ If anyone noticed any wrong be greatly appreciate if they could correct it saying so. Thanks a bunch! 2. Originally Posted by mathsToday Hi guys, hoping somebody can check through my calculations for the determinant of $A$ a 4 x 4 matrix. $A = \\begin{vmatrix} \\-1 && 3 && 0 && 5 \\\\ -2 && 0 && 4 && 0\\\\ 1 && 2 && -1 && 3\\\\ 1 && -2 && 1 && -1\\\\ \\end{vmatrix} = 1 \\begin{vmatrix} \\ 0 && 4 && 0 \\\\ 2 && -1 && 3\\\\ -2 && 1 && -1\\\\ \\end{vmatrix} - 3 \\begin{vmatrix} \\ -2 && 4 && 0 \\\\ 1 && -1 && 3\\\\ 1 && 1 && -1\\\\ \\end{vmatrix} $ $ + 0 \\begin{vmatrix} \\ -2 && 0 && 0 \\\\ 1 && 2 && 3\\\\ 1 && -2 && -1\\\\ \\end{vmatrix} - 5 \\begin{vmatrix} \\ -2 && 0 && 4 \\\\ 1 && 2 && -1\\\\ 1 && -2 && 1\\\\ \\end{vmatrix} $ $ = 1det(4)\\begin{vmatrix} \\ 2 && 3 \\\\ -2 && -1\\\\ \\end{vmatrix} $ $ - 3 det(-2) \\begin{vmatrix} \\ -1 && 3", "-1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$ ##### Solution : (ii)Let $\\begin{vmatrix} 3 & -4 & 5 \\\\[0.3em] 1 & 1 & -2 \\\\[0.3em] 2 & 3 & 1 \\end{vmatrix}$$\\\\ By expanding along the first row, we have:\\\\ |A| =3\\begin{vmatrix} 1 & -2 \\\\[0.3em] 3 & 1\\end{vmatrix} +4\\begin{vmatrix} 1 & -2 \\\\[0.3em] 2 & 1 \\end{vmatrix} +5\\begin{vmatrix} 1 & 1 \\\\[0.3em] 2 & 3 \\end{vmatrix}$$\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\ =3(7)+4(5)+5(1)\\\\ 21+20+5=46$ (iv) Let $A=\\begin{vmatrix} 2 & -1 & -2 \\\\[0.3em] 0 & 2 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ By expanding along the first column, we have:\\\\ |A|=2\\begin{vmatrix} 2 & -1\\\\[0.3em] -5 & 0\\end{vmatrix} -0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 &0 \\end{vmatrix}+3\\begin{vmatrix} -1 & -2 \\\\[0.3em] 2 & -1 \\end{vmatrix}$$\\\\$ $=2(0-5) -0+3(1+4)\\\\ =-10+15=5$ $(i) \\begin{vmatrix} 3 & -1 & -2 \\\\[0.3em] 0 & 0 & -1 \\\\[0.3em] 3 & -5 & 0 \\end{vmatrix}$$\\\\ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.\\\\ |A|=-0\\begin{vmatrix} -1 & -2 \\\\[0.3em] -5 & 0\\end{vmatrix} +0\\begin{vmatrix} 3 & -2 \\\\[0.3em] 3 & 0\\end{vmatrix} -(-1)\\begin{vmatrix} 3 & -1 \\\\[0.3em] 3 & -5\\end{vmatrix} =(-15+3) =-12 (iii) Let A=\\begin{vmatrix} 0 & 1 & 2 \\\\[0.3em] -1 & 0 & -3 \\\\[0.3em] -2 & 3 & 0\\end{vmatrix}$$\\\\$ By expanding along the", "-4 & 1 \\\\ 0 & -2 & -6 & 3 \\\\ 0 & -2 & -10 & 3 \\end{vmatrix} \\\\ && \\\\ &=&\\left. \\begin{array}{c} \\;\\;\\text{*)} \\\\ -2\\text{)} \\\\ -2\\text{)} \\end{array} \\right. \\begin{vmatrix} 1 & 4 & 1 \\\\ 2 & 6 & 3 \\\\ 2 & 10 & 3 \\end{vmatrix} = \\begin{vmatrix} 1 & 4 & 1 \\\\ 0 & -2 & 1 \\\\ 0 & 2 & 1 \\end{vmatrix} = \\begin{vmatrix} -2 & 1 \\\\ 2 & 1 \\end{vmatrix} =-4 \\end{eqnarray*} \\end{document} to type - Please, don't recommend eqnarray under any circumstances. – egreg Jan 11 '13 at 20:26 Very nice except the eqnarray. Here is one out of many reasons. (vote capped for today) – percusse Jan 11 '13 at 20:29 s/eqnarray/align/g and s/&=&/&=/g should do the trick – cmhughes Jan 11 '13 at 20:35 Very nice done. Thank you @Américo Tavares for your help. I just have one question: The numbers in the determine itself are all right aligned. You can see it in column 3, row 1 and 2 in the picture I posted. The 2 is just above the -2 and they are both right aligned, no matter if there is a minus or not. Do you have any idea how to realize that? – Martin Jan 11 '13 at 20:39 @AméricoTavares", "&-5\\end{vmatrix}$$ = (-15 + 3) = -12 (ii) Let A = $$\\begin{vmatrix}0 & 1 & 2\\\\-1 & 0 & -3\\\\-2 & 3 & 0\\end{vmatrix}$$ By expanding along the first row, we have: |A| = 3$$\\begin{vmatrix}1 & -2\\\\3 & 1\\end{vmatrix}$$ + 4$$\\begin{vmatrix}1 & -2\\\\2 & 1\\end{vmatrix}$$ + 5$$\\begin{vmatrix}1 &1\\\\2 &3\\end{vmatrix}$$ = 3 (1+6) + 4(1+4) + 5(3-2) = 3 (7) + 4 (5) + 5 (1) = 21 + 20 + 5 = 46 (iii) Let A = $$\\begin{vmatrix}3 & -4 & 5\\\\1 & 1 & -2\\\\2 & 3 & 1\\end{vmatrix}$$ By expanding along the first row, we have: |A| = 0$$\\begin{vmatrix}0 & -3\\\\3 & 0\\end{vmatrix}$$ - 1$$\\begin{vmatrix}-1 & -3\\\\-2 & 0\\end{vmatrix}$$ + 2$$\\begin{vmatrix}-1 & 0\\\\-2 &3\\end{vmatrix}$$ = 0 – 1(0 – 6) + 2 (-3 - 0) = -1 (-6) + 2(-3) = 6 – 6 = 0 (iv) Let A = $$\\begin{vmatrix}2 & -1 & -2\\\\0 & 2 & -1\\\\3 & -5 & 0\\end{vmatrix}$$ By expanding along the first column, we have: |A| = 2$$\\begin{vmatrix}2 & -1\\\\-5 & 0\\end{vmatrix}$$ - 0$$\\begin{vmatrix}-1 & -2\\\\-5 & 0\\end{vmatrix}$$ + 3$$\\begin{vmatrix}-1 & -2\\\\2 & -1\\end{vmatrix}$$ = 2(0 – 5) – 0 + 3(1 + 4) = -10 + 15 = 5 Q6 If A = $$\\begin{bmatrix}1 & 1 & -2\\\\2 & 1 & -3\\\\5 & 4 & -9\\end{bmatrix}$$, Find |A| Ans:", "OP's picture, the numbers are right aligned. – mafp Jan 12 '13 at 1:23 • @mafp I never right align the coefficients of a matrix. Adding that feature is possible, though. – egreg Jan 12 '13 at 10:45 • @egreg Oh, I would not write the whole determinant computation that way. But it is what the OP asked for. It works with mathtools, and then vmatrix[r], right? – mafp Jan 12 '13 at 13:12 I do not consider a good habit introducing such a notation, my experience is that it only confuses people. Mathematical formalism is there for a reason. You can try the following: \\documentclass{article} \\pagestyle{empty} \\usepackage{amsmath} \\begin{document} \\begin{align*} \\det A&= \\underbrace{\\begin{vmatrix} 1 & 0 & 0 & 0 \\\\ -1 & 1 & 0 & 0 \\\\ -2 & 0 & 1 & 0 \\\\ -2 & 0 & 0 & 1 \\end{vmatrix}}_{=1} \\cdot \\begin{vmatrix} 1 & 3 & 2 & -6 \\\\ 1 & 2 & -2 & -5 \\\\ 2 & 4 & -2 & -9 \\\\ 2 & 4 & -6 & -9 \\end{vmatrix} = \\begin{vmatrix} 1 & 3 & 2 & -6 \\\\ 0 & -1 & -4 & 1 \\\\ 0 & -2 & -6 & 3 \\\\ 0 & -2 & -10 & 3 \\end{vmatrix} \\\\ &= \\underbrace{\\begin{vmatrix} 1 &", "3 \\\\ 2 & 6 \\end{vmatrix} &= 1 \\cdot 6 - 2 \\cdot 3 = 6 - 6 = 0\\text{.} \\end{align*} ###### C23. Doing the computations by hand, find the determinant of the matrix below. \\begin{equation*} \\begin{bmatrix} 1 & 3 & 2 \\\\ 4 & 1 & 3 \\\\ 1 & 0 & 1 \\end{bmatrix} \\end{equation*} Solution We can compute the determinant by expanding about any row or column; the most efficient ones to choose are either the second column or the third row. In any case, the determinant will be $-4\\text{.}$ ###### C24. Doing the computations by hand, find the determinant of the matrix below. \\begin{equation*} \\begin{bmatrix} -2 & 3 & -2 \\\\ -4 & -2 & 1 \\\\ 2 & 4 & 2 \\end{bmatrix} \\end{equation*} Solution We will expand about the first row since there are no zeros to exploit, \\begin{align*} \\begin{vmatrix} -2 & 3 & -2 \\\\ -4 & -2 & 1 \\\\ 2 & 4 & 2 \\end{vmatrix} &= (-2)\\begin{vmatrix} -2 & 1 \\\\ 4 & 2 \\end{vmatrix} + (-1)(3)\\begin{vmatrix} -4 & 1 \\\\ 2 & 2 \\end{vmatrix} + (-2)\\begin{vmatrix} -4 & -2 \\\\ 2 & 4 \\end{vmatrix}\\\\ &=(-2)((-2)(2)-1(4)) + (-3)((-4)(2)-1(2)) + (-2)((-4)(4)-(-2)(2))\\\\ &=(-2)(-8)+(-3)(-10) + (-2)(-12)=70\\text{.} \\end{align*} ###### C25. Doing the computations by hand, find the determinant of the matrix below. \\begin{equation*} \\begin{bmatrix} 3", "+ 4x_2 & = 8. \\end{align*} will generate $$\\begin{array}{rl}2x_1 + 3x_2 =& 6, \\newline 3x_1+4x_2 = & 8. \\end{array}$$ where the two lines line up via the ampersands (&) • bmatrix, vmatrix and pmatrix environments generate matrices with different delimiters. This must be in math mode and typically in display math. I often put them in an align or align* environment. Each row of a matrix is ended with a \\\\ and each column is separated by an &. For example, \\begin{align*} \\begin{bmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{bmatrix} \\end{align*} will generate $$\\left[\\begin{array}{ccc} 1 & 2 & 3 \\newline 4 & 5 & 6 \\newline 7 & 8 & 9 \\end{array}\\right]$$ \\begin{align*} \\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{vmatrix} \\end{align*} will generate $$\\left|\\begin{array}{ccc} 1 & 2 & 3 \\newline 4 & 5 & 6 \\newline 7 & 8 & 9 \\end{array}\\right|$$ \\begin{align*} \\begin{pmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{pmatrix} \\end{align*} will generate $$\\left(\\begin{array}{ccc} 1 & 2 & 3 \\newline 4 & 5 & 6 \\newline 7 & 8 & 9 \\end{array}\\right)$$ and there are other delimiters available. ### graphicx package Type \\usepackage{graphicx}", "by row swaps and any scalars that are factored out, and then finish by multiplying down the diagonal of the echelon form result. This procedure takes the same time as Gauss' method and so is sufficiently fast to be practical on the size matrices that we see in this book. Example 2.4 Doing ${\\displaystyle 2\\!\\times \\!2}$ determinants ${\\displaystyle {\\begin{vmatrix}2&4\\\\-1&3\\end{vmatrix}}={\\begin{vmatrix}2&4\\\\0&5\\end{vmatrix}}=10}$ with Gauss' method won't give a big savings because the ${\\displaystyle 2\\!\\times \\!2}$ determinant formula is so easy. However, a ${\\displaystyle 3\\!\\times \\!3}$ determinant is usually easier to calculate with Gauss' method than with the formula given earlier. ${\\displaystyle {\\begin{vmatrix}2&2&6\\\\4&4&3\\\\0&-3&5\\end{vmatrix}}={\\begin{vmatrix}2&2&6\\\\0&0&-9\\\\0&-3&5\\end{vmatrix}}=-{\\begin{vmatrix}2&2&6\\\\0&-3&5\\\\0&0&-9\\end{vmatrix}}=-54}$ Example 2.5 Determinants of matrices any bigger than ${\\displaystyle 3\\!\\times \\!3}$ are almost always most quickly done with this Gauss' method procedure. ${\\displaystyle {\\begin{vmatrix}1&0&1&3\\\\0&1&1&4\\\\0&0&0&5\\\\0&1&0&1\\end{vmatrix}}={\\begin{vmatrix}1&0&1&3\\\\0&1&1&4\\\\0&0&0&5\\\\0&0&-1&-3\\end{vmatrix}}=-{\\begin{vmatrix}1&0&1&3\\\\0&1&1&4\\\\0&0&-1&-3\\\\0&0&0&5\\end{vmatrix}}=-(-5)=5}$ The prior example illustrates an important point. Although we have not yet found a ${\\displaystyle 4\\!\\times \\!4}$ determinant formula, if one exists then we know what value it gives to the matrix — if there is a function with properties (1)-(4) then on the above matrix the function must return ${\\displaystyle 5}$. Lemma 2.6 For each ${\\displaystyle n}$, if there is an ${\\displaystyle n\\!\\times \\!n}$ determinant function then it is unique. Proof For any ${\\displaystyle n\\!\\times \\!n}$ matrix we can perform Gauss' method on the matrix, keeping track of how the sign alternates on row", "& -1 & 4\\\\ 2 & 5 & 1\\\\ 2 & 0 & 6 \\end{bmatrix} \\end{equation*} Solution We can expand about any row or column, so the zero entry in the middle of the last row is attractive. Let us expand about column 2. By Theorem DER and Theorem DEC you will get the same result by expanding about a different row or column. We will use Theorem DMST twice. \\begin{align*} \\begin{vmatrix} 3 & -1 & 4\\\\ 2 & 5 & 1\\\\ 2 & 0 & 6 \\end{vmatrix} &= (-1)(-1)^{1+2} \\begin{vmatrix} 2 & 1\\\\ 2 & 6 \\end{vmatrix} + (5)(-1)^{2+2} \\begin{vmatrix} 3 & 4\\\\ 2 & 6 \\end{vmatrix} + (0)(-1)^{3+2} \\begin{vmatrix} 3 & 4\\\\ 2 & 1 \\end{vmatrix}\\\\ &=(1)(10)+(5)(10)+0=60 \\end{align*} ###### C26. Doing the computations by hand, find the determinant of the matrix $A\\text{.}$ \\begin{equation*} A= \\begin{bmatrix} 2 & 0 & 3 & 2 \\\\ 5 & 1 & 2 & 4 \\\\ 3 & 0 & 1 & 2 \\\\ 5 & 3 & 2 & 1 \\end{bmatrix} \\end{equation*} Solution With two zeros in column 2, we choose to expand about that column (Theorem DEC), \\begin{align*} \\detname{A} &= \\begin{vmatrix} 2 & 0 & 3 & 2 \\\\ 5 & 1 & 2 & 4 \\\\ 3 & 0 & 1 & 2 \\\\ 5 & 3 &", "2 & 1 \\end{vmatrix}\\\\ &= 0(-1) \\begin{vmatrix} 5 & 2 & 4 \\\\ 3 & 1 & 2 \\\\ 5 & 2 & 1 \\end{vmatrix} + 1(1) \\begin{vmatrix} 2 & 3 & 2 \\\\ 3 & 1 & 2 \\\\ 5 & 2 & 1 \\end{vmatrix} + 0(-1) \\begin{vmatrix} 2 & 3 & 2 \\\\ 5 & 2 & 4 \\\\ 5 & 2 & 1 \\end{vmatrix} + 3(1) \\begin{vmatrix} 2 & 3 & 2 \\\\ 5 & 2 & 4 \\\\ 3 & 1 & 2 \\end{vmatrix}\\\\ &= (1)\\left(2(1(1)-2(2)) - 3(3(1)-5(2)) + 2(3(2)-5(1))\\right) +\\\\ &\\quad\\quad(3)\\left(2(2(2)-4(1)) - 3(5(2)-4(3)) + 2(5(1)-3(2))\\right)\\\\ &= (-6+21+2) + (3)(0+6-2) = 29\\text{.} \\end{align*} ###### C27. Doing the computations by hand, find the determinant of the matrix $A\\text{.}$ \\begin{equation*} A= \\begin{bmatrix} 1 & 0 & 1 & 1\\\\ 2 & 2 & -1 & 1\\\\ 2 & 1 & 3 & 0\\\\ 1 & 1 & 0 & 1 \\end{bmatrix} \\end{equation*} Solution Expanding on the first row, we have \\begin{align*} \\begin{vmatrix} 1 & 0 & 1 & 1\\\\ 2 & 2 &-1 & 1\\\\2 & 1 & 3 & 0\\\\1 & 1 & 0 & 1\\end{vmatrix} &= \\begin{vmatrix} 2 & -1 & 1\\\\1 & 3 & 0\\\\1 & 0 & 1 \\end{vmatrix} - 0 + \\begin{vmatrix} 2 & 2 & 1\\\\ 2 & 1 &", "Question 1: Calculate |A| = $\\begin{vmatrix} -1 &1 &2 \\\\ 2 &-1 &0 \\\\ 1 &3 &-1 \\end{vmatrix}$ Solution: For det A, we use row expansion method. |A| = $(-1) \\begin{vmatrix} -1 & 0 \\\\ 3 & -1 \\end{vmatrix} - 1 \\begin{vmatrix} 2 &0 \\\\ 1 &-1 \\end{vmatrix} + 2 \\begin{vmatrix} 2 & -1 \\\\ 1 & 3 \\end{vmatrix}$ |A| = $(-1)(1 - 0) - 1(-2 - 0) + 2(6 + 1)$ = $-1 + 2 + 14$ = $15$ Question 2: Calculate |A| = $\\begin{vmatrix} -1 &1 &3 \\\\ 3 &-1 &0 \\\\ 1 &4 &0 \\end{vmatrix}$ Solution: To solve |A|, we select 3rd column for expansion since it has maximum zeros. |A| = $\\begin{vmatrix} -1 &1 &3 \\\\ 3 &-1 &0 \\\\ 1 &4 &0 \\end{vmatrix}$ = $3 \\begin{vmatrix} 3 & -1\\\\ 1&4 \\end{vmatrix} - 0 \\begin{vmatrix} -1 & 1 \\\\ 1 & 4 \\end{vmatrix} + 0 \\begin{vmatrix} -1 & 1 \\\\ 3 & -1 \\end{vmatrix}$ = $3(12 + 1) + 0 + 0$ = $39$ ## Determinant of a 4x4 Matrix In order to find the determinant of a 4x4 matrix, we should follow the steps given below: Let A = $\\begin{vmatrix} 1 &0 &3 &-1 \\\\ 2 &1 &0 &4 \\\\ -1 &5 &0 &2 \\\\ 1 &1 &6 &0 \\end{vmatrix}$ We select column", "to the second column and also the third column. $|A| = \\begin{vmatrix} n^2 & 2n+1 & 4n+4 \\\\ (n + 1)^2 & 2n+3 & 4n + 8 \\\\ (n + 2)^2 & 2n+5 & 4n+12 \\end{vmatrix}$ Multiply the elements in the first row by $-1$ then add that result to the second and third rows. $|A| = \\begin{vmatrix} n^2 & 2n+1 & 4n+4 \\\\ 2n+1 & 2 & 4 \\\\ 4n+4 & 4 & 8 \\end{vmatrix}$ Factorize out $4$ from the third column. $|A| = 4\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 4n+4 & 4 & 2 \\end{vmatrix}$ Factorize out $2$ from the third row. $|A| = 8\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 2n+2 & 2 & 1 \\end{vmatrix}$ Multiply the elements in the second row by $-1$ then add them to the third row. $|A| = 8\\begin{vmatrix} n^2 & 2n+1 & n+1 \\\\ 2n+1 & 2 & 1 \\\\ 1 & 0 & 0 \\end{vmatrix}$ Apply a cofactor expansion along the third row. \\begin{aligned}|A| & = 8 \\cdot 1 \\cdot \\begin{vmatrix} 2n+1 & n+1 \\\\ 2 & 1 \\end{vmatrix} \\\\ & = 8 \\cdot [(2n+1)-2(n+1)] \\\\ & = 8(-1) = -8 \\end{aligned} Thus, the determinant of matrix $A$ is $\\boxed{|A| = -8}$ [collapse]", "&0 \\\\ 5 &2 &x \\\\ 6 &1 &-1 \\end{vmatrix} |A|=0-4.\\begin{vmatrix} 3&0 &2 \\\\ 5 &x &4 \\\\ 6 &-1 &0 \\end{vmatrix}$$+2.\\begin{vmatrix} 3&-1 &2 \\\\ 5 &2 &4 \\\\ 6 &1 &0 \\end{vmatrix}$$-1.\\begin{vmatrix} 3&-1 &0 \\\\ 5 &2 &x \\\\ 6 &1 &-1 \\end{vmatrix} |A|=-4.\\begin{vmatrix} 3&0 &2 \\\\ 5 &x &4 \\\\ 6 &-1 &0 \\end{vmatrix}$$+2.\\begin{vmatrix} 3&-1 &2 \\\\ 5 &2 &4 \\\\ 6 &1 &0 \\end{vmatrix}$$-1.\\begin{vmatrix} 3&-1 &0 \\\\ 5 &2 &x \\\\ 6 &1 &-1 \\end{vmatrix}$ Now,$|A|=-4|A_{1}|+2|A_{2}|-1|A_{3}|$ $|A|=-4|A_{1}|+2|A_{2}|-|A_{3}|$--------->$(1)$ Where $|A_{1}|=\\begin{vmatrix} 3&0 &2 \\\\ 5 &x &4 \\\\ 6 &-1 &0 \\end{vmatrix}=3(0+4)-0(0-24)+2(-5-6x)=2-12x$ $|A_{2}|=\\begin{vmatrix} 3&-1 &2 \\\\ 5 &2 &4 \\\\ 6 &1 &0 \\end{vmatrix}=3(0-4)+1(0-24)+2(5-12)=-12-24-14=-50$ $|A_{3}|=\\begin{vmatrix} 3&-1 &0 \\\\ 5 &2 &x \\\\ 6 &1 &-1 \\end{vmatrix}=3(-2-x)+1(-5-6x)=-6-3x-5-6x=-9x-11$ Put these values in equation $(1),$ Now$,|A|=-4(2-12x)+2(-50)-(-9x-11)$ Given that $|A|=245$ $\\Rightarrow -8+48x-100+9x+11=245$ $\\Rightarrow 57x-100+3=245$ $\\Rightarrow 57x-97=245$ $\\Rightarrow 57x=245+97$ $\\Rightarrow 57x=342$ $\\Rightarrow x=\\frac{342}{57}$ $\\Rightarrow x=6$ So$,x=6$ is the correct answer. But this is a time-consuming method. Try, to do some operation in rows or columns and make some rows and columns $0$, then it will be easy. 0 yes, I was wrong thanks :) Expanding without any elementary operations will take a lot of time.The key is to use these operations. $\\begin{bmatrix} 0 & 4 &2 &1 \\\\ 3& -1&0 &2 \\\\ 5& 2& x & 4\\\\ 6& 1& -1 & 0", "remaining uncovered $2 \\times 2$ matrix, then multiply that by $a_{1i}$. The determinant is the sum of those values, alternating addition and subtraction. Worked Examples Example 1 Find the determinant of the matrix $\\begin{pmatrix} 1 & 2 & 1\\\\ 0 & 3 & 4\\\\ 3 & 1 & 4 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} 3 & 4\\\\ 1 & 4 \\end{vmatrix} -2 \\begin{vmatrix} 0 & 4\\\\ 3 & 4 \\end{vmatrix} +1 \\begin{vmatrix} 0 & 3\\\\ 3 & 1 \\end{vmatrix}\\\\\\\\ &= 1(3\\times 4 - 4\\times 1) -2(0\\times 4 - 4\\times 3) +1(0 \\times 1 - 3\\times 3)\\\\ &= 1(12-4) -2(0-12) +1(0-9)\\\\ &= 8 +24-9\\\\ &=23 \\end{align} Example 2 Find the determinant of the matrix $\\begin{pmatrix} 1 & 0 & 3\\\\ -1 & -1 & -3\\\\ 0 & 0 & 6 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} -0 \\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} +3 \\begin{vmatrix} -1 & -1\\\\ 0 & 0 \\end{vmatrix}\\\\\\\\ &= 1\\bigl((-1)\\times 6 - (-3)\\times 0\\bigr) -0(\\,\\dotso\\,) +3\\bigl((-1) \\times 0 - (-1)\\times 0\\bigr)\\\\ &= 1\\bigr((-6)-0\\bigl) -0 +3(0-0)\\\\ &=-6 +0\\\\ &=-6 \\end{align} Note: There was no need to work out the second $2 \\times 2$ determinant as it was being multiplied by zero. Video Example Hayley Bishop works out the determinant of the $3\\times" ]

[ "For how many positive integers $$n$$ less than or equal to 1000 is $$(\\sin t + i \\cos t)^n = \\sin nt + i \\cos nt$$ true for all real $$t$$? (第二十三届AIME2 2005 第9题)", "or $$n=1000$$, testing shows that this expression is accurate to within a factor of $$2$$.", "to $N=1000$, we get Pretty good.", "# Squares is greater or factorial? How many positive integers n are there between $$1$$ and $$10000$$ such that $$n!+10$$ is a perfect square? ×", "Pune, India Followers: 1485 Kudos [?]: 8016 [1] , given: 190 Re: How many integers less than 1000 have no factors (other than [#permalink] 07 Jan 2013, 19:54 1 KUDOS Expert's post rohantiwari wrote: The question asks for the number of integers less than 1000 and other than 1. Isnt one included in the 400 integers that you are claimimg to be the answer? Answer should be 399 if we exclude 1. Please correct me in case i missed something. The question does not ask you to exclude 1. Every positive integer less than 1000 has one common factor with 1000. What is it? It is 1. 1 is a common factor between any two positive integers. If the question were: How many positive integers less than 1000 have no factors in common with 1000 ? Then the answer would be 0. There are no positive integers which have no common factors with 1000. All the positive integers have a common factor and that is 1. But the question wants to know the number of positive integers which have no common factor other than 1 (1 will always be a common factor). Basically, it is looking for positive integers which are co-prime with 1000. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep", "# True or False? True or False? For any positive integer $$K$$, there exist a natural number $$N$$ with $$K$$ digits, none of which is even, such that $$5^K\\mid N$$. ×", "such as $$1000$$, $$1,000,000$$ or any number we like to think of, then there are more than $$n$$ positive integers. Thus, if the number we think of is $$1,000,000$$, there are obviously at least $$1,000,001$$ positive integers. Similarly the class of rational numbers, or of real numbers, is infinite. It is convenient to express this by saying that there are an infinite number of positive integers, or rational numbers, or real numbers. But the reader must be careful always to remember that by saying this we mean simply that the class in question has not a finite number of members such as $$1000$$ or $$1,000,000$$.", "# What is it? Algebra Level 2 How many positive integers $$n$$ are there such that $$139 n$$ is less than 2000? ×", "Expert's post sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. We counted multiples of 2 or 5 in", "both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 44400 ### Show Tags 04 Oct 2013, 01:01 sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not", "Another Good Number? True or False? If $$a^n - b^n$$ is an integer for all natural $$n$$ and distinct real numbers $$a,b$$, then $$a$$ and $$b$$ must be integers. Clarification: $$a^n - b^n$$ must be an integer for all natural values of $$n$$, not just a few values of $$n$$. ×", "Squares is greater or factorial? Number Theory Level 2 How many positive integers n are there between $$1$$ and $$10000$$ such that $$n!+10$$ is a perfect square? ×", "# What is sum of all positive odd integers less than $1000$? If the sum of all positive even integers less than $1000$ is $A$ , what is the sum of all positive odd integers less than $1000$? - You can first use the formula for the sum of the first $\\ n \\$ integers to find the sum of all the positive integers to 1000. Then use the same formula for the sum of all the positive integers to 500 and double that; that is the sum of all the even integers to 1000. The difference is the sum you are interested in. [OK, looks like I have now replied to the original version of this question...] – RecklessReckoner Jun 12 '13 at 18:58 The sum of all positive odd integers below 1000 is what it is, whether or not the sum of the even ones is denoted by the symbol A. The question needs to say \"... what is the sum of all positive odd integers in terms of A?\". – Kaz Jun 12 '13 at 20:14 If the sum of all positive even integers less than $1000$ is $A$, then I'd say the sum of all positive odd integers less than 1000 is $B$. ;-) – celtschk Jun 12 '13 at 20:56 Heh, first thing", "is: $$e^{-1000a}+0$$", "How many integers $$N$$ less than 1000 can be written as the sum of $$j$$ consecutive positive odd integers from exactly 5 values of $$j\\ge 1$$? (第二十四届AIME2 2006 第13题)", "# Be real! Consider all sets of 100 real numbers $$r_i$$ whose sum is positive. Define $$t_{ij} = r_i + r_j$$ for $$i < j$$. There are $${ 100 \\choose 2 } = 4950$$ such pairs. What is the maximum number of $$t_{ij}$$ which are non-positive? ×", "TRUE, n.sims = 10000) # }", "no factors(other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 29210 Followers: 4755 Kudos [?]: 50387 [0], given: 7544 Re: good one [#permalink] 04 Oct 2013, 00:01", "and that its range is to to infinity so it has to equal 1000 at some point but how do I prove it? Thanks $f(0)=0,f(1001)=101^2+10\\sin(1001)\\geqslant 1000+10+10\\sin(x)\\geqslant1000$", "900 Followers Problem! $\\frac{m^3-n^3}{m^2+n^2-mn}$ Find the sum of all prime numbers less than 900 that can be expressed as the above fraction where $$m$$ and $$n$$ are positive integers. ×" ]

[ "256A^{4}}\\right)+{B \\over A}\\left(u^{3}-{3u^{2}B \\over 4A}+{3uB^{2} \\over 16A^{2}}-{B^{3} \\over 64A^{3}}\\right)+{C \\over A}\\left(u^{2}-{uB \\over 2A}+{B^{2} \\over 16A^{2}}\\right)+{D \\over A}\\left(u-{B \\over 4A}\\right)+{E \\over A}=0.}$ Collecting the same powers of u yields ${\\displaystyle u^{4}+\\left({-3B^{2} \\over 8A^{2}}+{C \\over A}\\right)u^{2}+\\left({B^{3} \\over 8A^{3}}-{BC \\over 2A^{2}}+{D \\over A}\\right)u+\\left({-3B^{4} \\over 256A^{4}}+{CB^{2} \\over 16A^{3}}-{BD \\over 4A^{2}}+{E \\over A}\\right)=0.}$ Now rename the coefficients of u. Let {\\displaystyle {\\begin{aligned}\\alpha &={-3B^{2} \\over 8A^{2}}+{C \\over A},\\\\\\beta &={B^{3} \\over 8A^{3}}-{BC \\over 2A^{2}}+{D \\over A},\\\\\\gamma &={-3B^{4} \\over 256A^{4}}+{CB^{2} \\over 16A^{3}}-{BD \\over 4A^{2}}+{E \\over A}.\\end{aligned}}} The resulting equation is ${\\displaystyle u^{4}+\\alpha u^{2}+\\beta u+\\gamma =0}$ (1) which is a depressed quartic equation. If ${\\displaystyle \\beta =0\\ }$ then we have a biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for ${\\displaystyle x}$. If ${\\displaystyle \\gamma =0\\ }$ then one of the roots is ${\\displaystyle u=0}$ and the other roots can be found by dividing by ${\\displaystyle u}$, and solving the resulting depressed cubic equation, ${\\displaystyle u^{3}+\\alpha u+\\beta =0\\,.}$ Using reverse substitution we can find our values for ${\\displaystyle x}$. #### Ferrari's solution Otherwise, the depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity ${\\displaystyle \\left(u^{2}+\\alpha \\right)^{2}-u^{4}-2\\alpha u^{2}=\\alpha ^{2}}$ to equation (1), yielding ${\\displaystyle \\left(u^{2}+\\alpha \\right)^{2}+\\beta u+\\gamma", "\\partial x^{2}}\\ \\beta sin({n\\pi \\over L}x)\\\\&={-\\hbar ^{2} \\over 2m}{\\partial \\over \\partial x}\\ ({x\\pi \\over L})\\ \\beta cos({n\\pi \\over L}x)\\\\&={+\\hbar ^{2} \\over 2m}\\ ({n\\pi \\over L})^{2}\\ \\beta sin({n\\pi \\over L}x)\\\\&={\\hbar ^{2} \\over 2m}\\ ({n\\pi \\over L})^{2}\\ \\beta sin({n\\pi \\over L}x)\\\\E_{n}&={+\\hbar ^{2} \\over 2m}\\ ({n\\pi \\over L})^{2}\\end{aligned}}} Thus we have the equation for quantized energy, where n is limited to counting numbers (${\\displaystyle n=1,2,3,\\dots ,n}$ ), but we still need to solve for ${\\displaystyle \\beta }$ . Given: ${\\displaystyle \\Phi (x,t)=A\\exp[{-i \\over \\hbar }\\ E_{n}t]}$ Pick a constant to fix normalization. In this case we choose ${\\displaystyle A}$ . ${\\displaystyle \\int _{-\\infty =0}^{\\infty =L}\\Psi ^{*}\\ \\Psi \\ dx=1\\longrightarrow |A|^{2}\\ \\int _{0}^{L}sin({n\\pi \\over L}x)^{2}\\ dx=1}$ Substitute and solve... {\\displaystyle {\\begin{aligned}1&=|A|^{2}\\ \\int _{0}^{L}sin({n\\pi \\over L}x)^{2}\\ dx;\\qquad let\\ q={n\\pi \\over L}\\\\&=|A|^{2}({1 \\over 2i})^{2}\\int _{0}^{L}(e^{iqx}+e^{-iqx})^{2}\\ dx\\\\&=|A|^{2}({1 \\over 2i})^{2}\\int _{0}^{L}e^{ziqx}+e^{-ziqx}-2e^{iqx}e^{-iqx}\\ dx\\\\&=|A|^{2}({1 \\over 2i})^{2}[{1 \\over 2iq}(e^{2iqL}-1)+{-1 \\over 2iq}(e^{-2iqL}-1)-2L]\\\\&=|A|^{2}({1 \\over 2i})^{2}[{1 \\over 2iq}(e^{2iqL}-e^{-iqL})-2L]\\\\&=|A|^{2}({1 \\over 2i})^{2}[{1 \\over 2iq}2i\\ sin(2qL)-2L]\\\\&=|A|^{2}({-1 \\over 4})[{L \\over n\\pi }sin(2{n\\pi \\over L}L)-2L]\\\\&=|A|^{2}({-1 \\over 4})(-2L)\\\\&=|A|^{2}{L \\over 2}\\\\&\\qquad \\qquad \\qquad \\qquad \\therefore \\ A={\\sqrt {2 \\over L}}\\end{aligned}}} At the end of the day, we have: ${\\displaystyle \\Psi _{n}(x,t)={\\sqrt {2 \\over L}}\\ \\exp[{-i \\over \\hbar }\\ E_{n}t]\\ sin({n\\pi \\over L}x)}$ ### Complex Numbers As a point of honesty, while this solution is true, there are other solutions. Not only can you just put in different values for", "a \\not\\in {\\mathbb Z}$$ $$\\sum_{n = 0}^{\\infty}{1 \\over n^{2} - 1/4} = -2\\,, \\qquad \\sum_{n = 0}^{\\infty}{1 \\over n^{2} - \\left(k/2\\right)^{2}} = -\\,{2 \\over k^{2}}\\left[1 + {\\pi k/2 \\over \\tan\\left(\\pi k/2\\right)}\\right]$$ \\begin{align} \\phi'\\left(k\\right) &= -\\,{1 \\over 2\\pi}{\\cal P}\\,\\left[% {16 \\over k} + {16k \\over k^{2} - 1} - {16 \\over k}\\,{1 \\over k^{2} - 1} - {16 \\over k}\\,{1 \\over k^{2} - 1}\\,{\\pi k/2 \\over \\tan\\left(\\pi k/2\\right)} \\right] \\end{align} \\begin{align} \\phi'\\left(k\\right) &= -\\,{1 \\over 2\\pi}{\\cal P}\\,\\left[% {32k \\over k^{2} - 1} - {16 \\over k}\\,{1 \\over k^{2} - 1}\\,{\\pi k/2 \\over \\tan\\left(\\pi k/2\\right)} \\right] \\end{align} We can observe that $\\phi'\\left(k\\right)$ diverges at $k$ values: $$k = 0, \\pm 1, \\pm 2, \\pm 4, \\pm 6,\\ldots$$ • Did you miss the absolute modulus? Whatever problems Mathematica has, the Fourier transform of $\\dfrac{\\sin x}{x}$ is extremely unlikely to be one of them. The Fourier transform of $\\left\\lvert \\dfrac{\\sin x}{x}\\right\\rvert$ is an entirely different beast. – Daniel Fischer Oct 13 '13 at 0:42 • @DanielFischer Between \"without absolute modulus\" and \"with absolute modulus\" I left to have dinner. Sorry about that. – Felix Marin Oct 13 '13 at 1:32 • Erm, $\\lvert f(x)\\rvert$ and $[2\\Theta(x)-1]f(x)$ are not the same. If we were dealing with complex valued functions, it would be more complicated, but for real-valued $f$ you'd still have $\\lvert f(x)\\rvert", "$0,1,2,\\dots$ and $(2, -n\\pi)$ where $n=1,3,5,\\dots$. 37 ##### Web Bonus Problems / Re: Phaseportrait « on: March 28, 2018, 01:50:57 PM » $\\text{My attempt:}$ \\begin{align}x'&=\\sin(x)\\cos(y)=F(x,y)\\\\y'&=-\\cos(x)\\sin(y)=G(x,y)\\end{align} Let \\begin{align}\\cases{F(x,y)=\\sin(x)\\cos(y)=0\\\\G(x,y)=-\\cos(x)\\sin(y)=0}\\end{align} $$\\implies\\cases{x=k\\pi,y=k\\pi\\\\x={\\pi\\over2}+k\\pi,y={\\pi\\over2}+k\\pi}$$ where $k$ can only be $-1,0,1$, since we have $x\\in(-4,4)$ and $y\\in(-4,4)$, $\\pi\\cong 3.14159265359$. $\\\\$ Thus we have the following critical points:$\\\\$ Case#1: $(0,0);(\\pi,0);(-\\pi,0);(0,\\pi);(0,-\\pi);(\\pi,\\pi);(-\\pi,\\pi);(-\\pi,-\\pi);(\\pi,-\\pi)$. $\\\\$ Case#2: $({\\pi\\over2},{\\pi\\over2});(-{\\pi\\over2},-{\\pi\\over2});({\\pi\\over2},-{\\pi\\over2});(-{\\pi\\over2},{\\pi\\over2})$. $$J=\\begin{pmatrix}F_x(x,y)&F_y(x,y)\\\\G_x(x,y)&G_x(x,y)\\end{pmatrix}=\\begin{pmatrix}\\cos(x)\\cos(y)&-\\sin(x)\\sin(y)\\\\\\sin(x)\\sin(y)&-\\cos(x)\\cos(y)\\end{pmatrix}$$ When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}$ or $\\begin{pmatrix}-1&0\\\\0&0\\end{pmatrix}$, wtih eigenvalues $\\lambda=\\pm1$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\0\\end{pmatrix}$ or $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#1 are all $\\text{Saddle Points}$. $\\\\$ When $(x,y)=$ critical points in Case#2: we get matrices $\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}$ or $\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}$ with eigenvalues $\\lambda=\\pm i$, eigenvectors $\\xi=\\begin{pmatrix}1\\\\i\\end{pmatrix}$ or $\\begin{pmatrix}1\\\\-i\\end{pmatrix}$. $\\\\$ Thus, the critical points in Case#2 are all $\\text{Center}$. $\\\\$ Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that. 38 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018, 09:11:15 AM » Thanks for the correction. I made a mistake copying the result integral. I got the same result. I think there is a calculation error when computing the initial condition, hence the $c_1,c_2$ values. (if I am not mistaken ) $$c_1+c_2=-2-15\\ln2$$ and $$-2c_2=3-15\\ln2$$ Therefore, $$c_1=-{1\\over2}-{45\\over2}\\ln2$$ $$c_2=-{3\\over2}+{15\\over2}\\ln2$$ 39 ##### Term Test 2 / Re: TT2--P3M « on: March 28, 2018,", "\\over \\pi }} \\right)$$ B $$\\sin \\left( {{{2a} \\over \\pi }} \\right)$$ C $$\\cos \\left( {{{2a} \\over \\pi }} \\right)$$ D $$\\sin \\left( {{{4a} \\over \\pi }} \\right)$$ Explanation Given $$a = {({\\sin ^{ - 1}}x)^2} - {({\\cos ^{ - 1}}x)^2}$$ $$= ({\\sin ^{ - 1}}x + {\\cos ^{ - 1}}x)({\\sin ^{ - 1}}x - {\\cos ^{ - 1}}x)$$ $$= {\\pi \\over 2}\\left( {{\\pi \\over 2} - 2{{\\cos }^{ - 1}}x} \\right)$$ $$\\Rightarrow 2{\\cos ^{ - 1}}x = {\\pi \\over 2} - {{2a} \\over \\pi }$$ $$\\Rightarrow {\\cos ^{ - 1}}(2{x^2} - 1) = {\\pi \\over 2} - {{2a} \\over \\pi }$$ $$\\Rightarrow 2{x^2} - 1 = \\cos \\left( {{\\pi \\over 2} - {{2a} \\over \\pi }} \\right)$$ 3 JEE Main 2021 (Online) 26th August Evening Shift If $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}{1 \\over {2{r^2}}} = p}$$, then the value of tan p is : A $${{101} \\over {102}}$$ B $${{50} \\over {51}}$$ C 100 D $${{51} \\over {50}}$$ Explanation $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}\\left( {{2 \\over {4{r^2}}}} \\right) = \\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}\\left( {{{(2r + 1) - (2r - 1)} \\over {1 + (2r + 1)(2r - 1)}}} \\right)} }$$ = $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}(2r + 1) - {{\\tan }^{ - 1}}(2r - 1)}$$ = $${\\tan ^{", "$\\dee{u}=(n-1)\\sin^{n-2}x\\cos x$ and $v=-\\cos x\\text{.}$ \\begin{align*} \\int\\sin^nx\\,\\dee{x} &=\\underbrace{-\\sin^{n-1}x\\ \\cos x}_{uv}+\\underbrace{(n-1)\\int \\cos^2x\\ \\sin^{n-2}x\\ \\dee{x}}_{-\\int v\\dee{u}}\\\\ \\end{align*} Using the identity $\\sin^2 x + \\cos^2 x =1 \\text{,}$ \\begin{align*} &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int (1-\\sin^2 x)\\sin^{n-2}x\\ \\dee{x}\\\\ &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int\\sin^{n-2}x\\ \\dee{x} -(n-1)\\int\\sin^{n}x\\ \\dee{x} \\end{align*} Moving the last term on the right hand side to the left hand side gives \\begin{align*} n\\int\\sin^nx\\,\\dee{x} &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int\\sin^{n-2}x\\ \\dee{x} \\end{align*} Dividing across by $n$ gives the desired reduction formula. \\qquad (b) $\\dfrac{35}{256}\\pi\\approx0.4295$ Solution (a) The “parts” in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of $\\sin^{n-1}x\\text{,}$ although not exactly. So, let's integrate by parts with $u=\\sin^{n-1}x$ and $\\dee{v}=\\sin x\\,\\dee{x}\\text{,}$ so that $\\dee{u}=(n-1)\\sin^{n-2}x\\cos x$ and $v=-\\cos x\\text{.}$ \\begin{align*} \\int\\sin^nx\\,\\dee{x} &=\\underbrace{-\\sin^{n-1}x\\ \\cos x}_{uv}+\\underbrace{(n-1)\\int \\cos^2x\\ \\sin^{n-2}x\\ \\dee{x}}_{-\\int v\\dee{u}}\\\\ \\end{align*} Using the identity $\\sin^2 x + \\cos^2 x =1 \\text{,}$ \\begin{align*} &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int (1-\\sin^2 x)\\sin^{n-2}x\\ \\dee{x}\\\\ &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int\\sin^{n-2}x\\ \\dee{x} -(n-1)\\int\\sin^{n}x\\ \\dee{x} \\end{align*} Moving the last term on the right hand side to the left hand side gives \\begin{align*} n\\int\\sin^nx\\,\\dee{x} &=-\\sin^{n-1}x\\ \\cos x+(n-1)\\int\\sin^{n-2}x\\ \\dee{x} \\end{align*} Dividing across by $n$ gives the desired reduction formula. (b) By the reduction formula of part (a), if $n \\ge 2\\text{,}$ \\begin{gather*} \\int_0^{\\pi/2}\\sin^n(x)\\,\\dee{x}= \\frac{n-1}{n}\\int_0^{\\pi/2}\\sin^{n-2}(x)\\,\\dee{x} \\end{gather*} since $\\sin 0=\\cos\\frac{\\pi}{2}=0\\text{.}$ Applying this reduction formula, with $n=8,6,4,2\\text{:}$ \\begin{align*} \\int_0^{\\pi/2}\\sin^8(x)\\,\\dee{x}", "straightforward to show that, for $$\\lambda>-1/(2\\sigma^2)$$, \\begin{align} {\\rm{E}}\\left[{\\rm e}^{-\\lambda Z^2}\\right]={1\\over \\sqrt{1+2\\lambda \\sigma^2}}. \\end{align} (7.23) Therefore, for $${s'\\over m}<{1\\over 1-\\alpha}$$, \\begin{align} {\\rm{E}}[ {\\rm e}^{{s'\\over m}X_1Y_1}] &= {1\\over \\sqrt{(1+{s'\\over m}(1+\\alpha))(1-{s'\\over m}(1-\\alpha))}}\\nonumber\\\\ &= {1\\over \\sqrt{(1+{s'\\alpha\\over m})^2 -{s'^2\\over m^2}}}.\\label{eq:E-XY} \\end{align} (7.24) Therefore, combining (7.19) and (7.24), it follows that \\begin{align} {\\rm{P}}\\left({1\\over m}\\langle Au^n,Av^n\\rangle-\\langle u^n,v^n\\rangle\\leq -\\tau\\right)&= {\\rm e}^{s'(\\alpha-\\tau)}{\\rm e}^{-{m\\over 2}\\log\\left( \\left(1+{s'\\alpha\\over m}\\right)^2-\\left({s'\\over m}\\right)^2 \\right)}.\\label{eq:sp-s-last} \\end{align} (7.25) Replacing $$s'/m$$ with $$s$$ in (7.25) yields the desired result. □ Corollary 7.1 Consider $$u^n\\in {\\rm I}\\kern-0.20em{\\rm R}^n$$ and $$v^n\\in {\\rm I}\\kern-0.20em{\\rm R}^n$$ such that $$\\|u^n\\|=\\|v^n\\|=1$$. Also, consider matrix $$A\\in{\\rm I}\\kern-0.20em{\\rm R}^{m\\times n}$$ with i.i.d. standard normal entries. Then, \\begin{align} {\\rm{P}}\\left({1\\over m}\\langle Au^n,Av^n\\rangle-\\langle u^n,v^n\\rangle\\leq -0.45\\right)\\leq 2^{-0.05m}. \\end{align} (7.26) Proof. From Lemma 7.3, for $$\\alpha=\\langle u^n,v^n\\rangle$$, and $$s<1/(1-\\alpha)$$, \\begin{align} {\\rm{P}}\\left({1\\over m}\\langle Au^n,Av^n\\rangle-\\alpha\\leq -0.45 \\right)\\leq {\\rm e}^{m((\\alpha-0.45)s)-{m\\over 2}\\ln ((1+s\\alpha)^2-s^2)}=2^{-m f(\\alpha,s)}, \\end{align} (7.27) where \\begin{align} f(\\alpha,s)=(\\log {\\rm e})\\left({1\\over 2}\\ln((1+s\\alpha)^2-s^2)-(\\alpha-0.45)s\\right). \\end{align} (7.28) Figure 1 plots $$\\max_{s\\in(0,{1\\over 1-\\alpha})} f(\\alpha,s)$$ and shows that \\begin{align} \\min_{\\alpha\\in(-1,1)}\\max_{s\\in\\left(0,{1\\over 1-\\alpha}\\right)} f(\\alpha,s)\\; \\geq \\; 0.05. \\end{align} (7.29) □ Fig. 1. View largeDownload slide $$\\max_{s\\in(0,{1\\over 1-\\alpha})} f(\\alpha,s).$$ Fig. 1. View largeDownload slide $$\\max_{s\\in(0,{1\\over 1-\\alpha})} f(\\alpha,s).$$ The following two lemmas are proved in [13]. Lemma 7.4 ($$\\chi^2$$ concentration) Fix $$\\tau>0$$, and let $$U_i\\stackrel{\\rm i.i.d.}{\\sim}\\mathcal{N}(0,1)$$, $$i=1,2,\\ldots,m$$. Then, \\begin{align} {\\rm{P}}\\left( \\sum_{i=1}^m U_i^2 <m(1- \\tau) \\right) \\leq {\\rm e} ^{\\frac{m}{2}(\\tau + \\ln(1- \\tau))} \\end{align} (7.30) and \\begin{align}\\label{eq:chisq} {\\rm{P}}\\left( \\sum_{i=1}^m U_i^2 > m(1+\\tau) \\right)", "$$\\omega=\\sqrt3$$ yields \\begin{aligned} {\\cal L}^{-1}\\left({8\\over s+5}+{7\\over s^2+3}\\right)&= 8{\\cal L}^{-1}\\left({1\\over s+5}\\right)+7{\\cal L}^{-1}\\left({1\\over s^2+3}\\right)\\\\ &= 8{\\cal L}^{-1}\\left({1\\over s+5}\\right)+{7\\over\\sqrt3}{\\cal L}^{-1}\\left({\\sqrt3\\over s^2+3}\\right)\\\\&= 8e^{-5t}+{7\\over\\sqrt3}\\sin\\sqrt3t.\\end{aligned}\\nonumber ##### Example 8.2.3 Find ${\\cal L}^{-1}\\left({3s+8\\over s^2+2s+5}\\right).\\nonumber$ ###### Solution Completing the square in the denominator yields ${3s+8\\over s^2+2s+5}={3s+8\\over(s+1)^2+4}.\\nonumber$ Because of the form of the denominator, we consider the transform pairs $e^{-t}\\cos 2t\\leftrightarrow{s+1\\over(s+1)^2+4} \\quad \\text{and} \\quad e^{-t}\\sin 2t\\leftrightarrow{2\\over(s+1)^2+4}, \\nonumber$ and write \\begin{aligned} {\\cal L}^{-1}\\left({3s+8\\over(s+1)^2+4}\\right)&= {\\cal L}^{-1}\\left({3s+3\\over(s+1)^2+4}\\right)+ {\\cal L}^{-1}\\left({5\\over(s+1)^2+4}\\right)\\\\&= 3{\\cal L}^{-1}\\left({s+1\\over(s+1)^2+4}\\right)+ {5\\over2}{\\cal L}^{-1}\\left({2\\over(s+1)^2+4}\\right)\\\\&= e^{-t}(3\\cos 2t+{5\\over2}\\sin 2t).\\end{aligned}\\nonumber ##### Note We’ll often write inverse Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms in Section 8.8. ## Inverse Laplace Transforms of Rational Functions Using the Laplace transform to solve differential equations often requires finding the inverse transform of a rational function $F(s)={P(s)\\over Q(s)}, \\nonumber$ where $$P$$ and $$Q$$ are polynomials in $$s$$ with no common factors. Since it can be shown that $$\\lim_{s\\to\\infty}F(s)=0$$ if $$F$$ is a Laplace transform, we need only consider the case where $$\\mbox{degree}(P)<\\mbox{degree}(Q)$$. To obtain $${\\cal L}^{-1}(F)$$, we find the partial fraction expansion of $$F$$, obtain inverse transforms of the individual terms in the expansion from the table of Laplace transforms, and use the linearity property of the inverse transform. The next two examples illustrate this. ##### Example 8.2.4 Find the inverse Laplace transform", "= {\\pi \\over 6}$$ 3 ### AIEEE 2011 If $$A = {\\sin ^2}x + {\\cos ^4}x,$$ then for all real $$x$$: A $${{13} \\over {16}} \\le A \\le 1$$ B $$1 \\le A \\le 2$$ C $${3 \\over 4} \\le A \\le {{13} \\over {16}}$$ D $${{3} \\over {4}} \\le A \\le 1$$ ## Explanation $$A = {\\sin ^2}x + {\\cos ^4}x$$ $$= {\\sin ^2}x + {\\cos ^2}x\\left( {1 - {{\\sin }^2}x} \\right)$$ $$= {\\sin ^2}x + {\\cos ^2}x - {1 \\over 4}{\\left( {2\\sin x.\\cos x} \\right)^2}$$ $$= 1 - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right)$$ Now $$0 \\le {\\sin ^2}\\left( {2x} \\right) \\le 1$$ $$\\Rightarrow 0 \\ge - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right) \\ge - {1 \\over 4}$$ $$\\Rightarrow 1 \\ge 1 - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right) \\ge 1 - {1 \\over 4}$$ $$\\Rightarrow 1 \\ge A \\ge {3 \\over 4}$$ 4 ### AIEEE 2010 Let $$\\cos \\left( {\\alpha + \\beta } \\right) = {4 \\over 5}$$ and $$\\sin \\,\\,\\,\\left( {\\alpha - \\beta } \\right) = {5 \\over {13}},$$ where $$0 \\le \\alpha ,\\,\\beta \\le {\\pi \\over 4}.$$ Then $$tan\\,2\\alpha$$ = A $${56 \\over 33}$$ B $${19 \\over 12}$$ C $${20 \\over 7}$$ D $${25 \\over 16}$$ ## Explanation $$\\cos \\left( {\\alpha + \\beta } \\right) = {4 \\over 5} \\Rightarrow \\tan \\left( {\\alpha", "{{{T\\omega } \\over {2\\pi }}} \\right) \\cr {c_n} &= {1 \\over T}X(n{\\omega _0}) = {1 \\over T}\\left( {2 \\cdot {T \\over 2}{{{\\mathop{\\rm sinc}\\nolimits} }^2}\\left( {{{Tn{\\omega _0}} \\over {4\\pi }}} \\right) - T{\\mathop{\\rm sinc}\\nolimits} \\left( {{{Tn{\\omega _0}} \\over {2\\pi }}} \\right)} \\right)\\quad \\quad \\quad \\quad T{\\omega _0} = 2\\pi \\cr &= {{\\mathop{\\rm sinc}\\nolimits} ^2}\\left( {{{2\\pi n} \\over {4\\pi }}} \\right) - {\\mathop{\\rm sinc}\\nolimits} \\left( {{{2\\pi n} \\over {2\\pi }}} \\right) = {{\\mathop{\\rm sinc}\\nolimits} ^2}\\left( {{n \\over 2}} \\right) - {\\mathop{\\rm sinc}\\nolimits} \\left( n \\right) \\cr} We could stop here, but instead we will show that this is equivalent to a previous result obtained through integration. First we note that c0=0, and for all other values of n, sinc(n)=0, so we can ignore the second term. $${c_n} = {{\\mathop{\\rm sinc}\\nolimits} ^2}\\left( {{n \\over 2}} \\right) = {{{{\\sin }^2}\\left( {\\pi {n \\over 2}} \\right)} \\over {{{\\left( {\\pi {n \\over 2}} \\right)}^2}}} = 4{{{{\\sin }^2}\\left( {\\pi {n \\over 2}} \\right)} \\over {{{\\left( {\\pi n} \\right)}^2}}},\\quad \\quad \\quad n \\ne 0$$ By brute force integration and a lot of algebra, we get the result (in the example in the link A=1) in the form of a Fouries Cosine series $${a_n} = 8{{{{\\sin }^2}\\left( {\\pi {n \\over 2}} \\right)} \\over {{{\\left( {\\pi n} \\right)}^2}}}$$ Which agrees with our answer because we know 2Re(cn)=an ##", "{ \\propto + \\beta } \\right) = {x \\over {2x}} = {1 \\over 2}$ we know, $\\tan \\left( { \\propto + \\beta } \\right) = {{\\tan \\propto + \\tan \\beta } \\over {1 - \\tan \\propto \\tan \\beta }}$ $\\therefore\\,\\,\\,$ ${{\\tan \\propto + \\tan \\beta } \\over {1 - \\tan \\propto \\tan \\beta }} = {1 \\over 2}....\\left( 1 \\right)$ From the diagram, $\\tan \\widehat \\propto = {{x/2} \\over {2x}} = {1 \\over 4}......\\left( 2 \\right)$ Putting value of $\\tan \\propto$ in eq$(1)$, ${{{1 \\over 4} + \\tan \\beta } \\over {1 - {1 \\over 4}\\tan \\beta }} = {1 \\over 2}$ $\\Rightarrow 1 - {1 \\over 4}\\tan \\beta$ $= {1 \\over 2} + 2\\tan \\beta$ $\\Rightarrow {{9\\tan \\beta } \\over 4} = {1 \\over 2}$ $\\Rightarrow \\tan \\beta = {2 \\over 9}$ 3 ### JEE Main 2017 (Offline) If $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$, then the value of $\\cos 4x$ is A ${1 \\over 3}$ B ${2 \\over 9}$ C $- {7 \\over 9}$ D $- {3 \\over 5}$ ## Explanation Given that, $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$ $\\Rightarrow 5\\left( {{{{{\\sin }^2}x} \\over {{{\\cos }^2}x}} - {{\\cos }^2}x} \\right) = 2\\left( {2{{\\cos }^2}x - 1} \\right) + 9$ Let ${\\cos ^2}x = t,$", "(More than One Correct Answer) If $$\\alpha$$ $$= 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right)$$ and $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are) A $$cos\\beta > 0$$ B $$\\sin \\beta < 0$$ C $$\\cos \\left( {\\alpha + \\beta } \\right) > 0$$ D $$\\cos \\alpha < 0$$ ## Explanation Here, $$\\alpha = 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right)$$ and $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$ as $${6 \\over {11}} > {1 \\over 2}$$ $$\\Rightarrow {\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right) > {\\sin ^{ - 1}}\\left( {{1 \\over 2}} \\right) = {\\pi \\over 6}$$ $$\\therefore$$ $$\\alpha = 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right) > {\\pi \\over 2} \\Rightarrow \\cos \\alpha < 0$$ Now, $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$ As $${4 \\over 9} < {1 \\over 2} \\Rightarrow {\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right) > {\\cos ^{ - 1}}\\left( {{1 \\over 2}} \\right) = {\\pi \\over 3}$$ $$\\therefore$$ $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right) > \\pi$$ $$\\therefore$$ $$\\cos \\beta < 0$$ and $$\\sin \\beta < 0$$ Now, $$\\alpha + \\beta$$ is slightly greater than $${{3\\pi } \\over 2}$$. $$\\therefore$$", "(\\sqrt{a^2+b^2})^n}\\left({\\sin t\\over (n-1)\\cos^{n-1}t} + (n-2)\\int{dt\\over \\cos^{n-2}t}\\right)$$ The tricky part here is to arrive to the expression in the problem statement: $$J_n = {1\\over (n-1)(a^2 + b^2)}{1\\over (\\sqrt{a^2 + b^2})^{n-2}}\\left({\\sin(x-\\phi)\\over \\cos^{n-1}(x-\\phi)} + (n-2)\\int {dt\\over cos^{n-2}t}\\right)\\\\ = {1\\over (n-1)(a^2 + b^2)}\\left({\\sin(x-\\phi)\\over (\\sqrt{a^2 + b^2})^{n-2}\\cos^{n-1}(x-\\phi)} + {(n-2)\\over (\\sqrt{a^2 + b^2})^{n-2}}\\int {dt\\over cos^{n-2}t}\\right)\\\\ ={1\\over (n-1)(a^2 + b^2)}\\left({\\sin(x-\\phi)\\over (\\sqrt{a^2 + b^2})^{n-2}\\cos^{n-1}(x-\\phi)} + J_{n-2}\\right)$$ Consider the following part of the expression: \\begin{align} E &= {\\sin(x-\\phi)\\over (\\sqrt{a^2 + b^2})^{n-2}\\cos^{n-1}(x-\\phi)} \\\\ &= {\\sin x\\cos \\phi - \\cos x\\sin \\phi\\over (\\sqrt{a^2 + b^2})^{n-2}(\\cos x\\cos \\phi + \\sin x\\sin \\phi)^{n-1}} \\end{align} Now replace back using $$(1)$$ and $$(2)$$: $$E = {b\\sin x - a\\cos x\\over (\\sqrt{a^2 - b^2})^{n-1}\\left({b\\cos x\\over \\sqrt{a^2+b^2}} + {a\\sin x\\over \\sqrt{a^2+b^2}}\\right)^{n-1}} \\\\ = {b\\sin x - a\\cos x\\over (a\\sin x + b\\cos x)^{n-1}}$$ Finally: $$J_n = {1\\over (n-1)(a^2 + b^2)}\\left({b\\sin x - a\\cos x\\over (a\\sin x + b\\cos x)^{n-1}} + (n-2)J_{n-2}\\right)$$ I'm wondering if there are alternative ways to prove the statement? Also, I would appreciate it if someone verifies my writings. Thank you! • have you tried that $t/2$ substitution maybe...? – Calvin Khor Nov 7 '19 at 11:41 • @CalvinKhor I did, but it yielded a rational function which was too complicated for me to integrate. – roman Nov 7 '19 at 11:43 • There seems to be a minor discrepancy between your", "or $$x = {8 \\over {3\\sqrt 3 }}$$ Area of $$\\Delta ABD$$ $$= {1 \\over 2} \\times 4 \\times {8 \\over {3\\sqrt 3 }} = {{16} \\over {3\\sqrt 3 }}$$ $$\\therefore$$ Area of $$\\Delta ABC$$ $$= 2 \\times {{16} \\over {3\\sqrt 3 }} = {{32} \\over {3\\sqrt 3 }}$$ $$\\left[ \\, \\right.$$ As median of a $$\\Delta$$ divides it into two $$\\Delta 's$$ of equal area. $$\\left. \\, \\right]$$ 3 AIEEE 2003 The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a,$$ is A $${a \\over 4}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ B $$a\\cot \\left( {{\\pi \\over {n}}} \\right)$$ C $${a \\over 2}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ D $$a\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ Explanation $$\\tan \\left( {{\\pi \\over n}} \\right) = {a \\over {2r}};\\,\\,\\sin \\left( {{\\pi \\over n}} \\right) = {a \\over {2R}}$$ $$r + R = {a \\over 2}\\left[ {\\cot {\\pi \\over n} + \\cos ec{\\pi \\over n}} \\right]$$ $$= {a \\over 2}\\left[ {{{\\cos {\\pi \\over n} + 1} \\over {\\sin {\\pi \\over n}}}} \\right]$$ $$= {a \\over 2}\\left[ {{{2{{\\cos }^2}{\\pi \\over {2n}}} \\over {2\\sin {\\pi \\over {2n}}\\cos {\\pi \\over {2n}}}}} \\right]$$ $$= {a \\over 2}\\cot {\\pi \\over {2\\pi }}$$ 4 AIEEE 2002 In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$", "2}} {t\\cos \\left( {n{\\omega _0}t} \\right)dt} } \\right)$$ Perform the integrations (either by hand using integration by parts, or with a table of integrals, or by computer) and use the fact that ω0·T=2·π $${a_n} = {{4A} \\over T}\\left( {{{T\\sin (\\pi n)} \\over {2\\pi n}} + {4 \\over T}{{{T^2}{\\mkern 1mu} \\left( {2{\\mkern 1mu} \\sin {{\\left( {{{\\pi {\\mkern 1mu} n} \\over 2}} \\right)}^2} - \\pi {\\mkern 1mu} n{\\mkern 1mu} \\sin \\left( {\\pi {\\mkern 1mu} n} \\right)} \\right)} \\over {4{\\mkern 1mu} {\\pi ^2}{\\mkern 1mu} {n^2}}}} \\right)$$ Since sin(π·n)=0 this simplifies to $${a_n} = {{4A} \\over T}{4 \\over T}{{{T^2}{\\mkern 1mu} 2{\\mkern 1mu} \\sin {{\\left( {{{\\pi {\\mkern 1mu} n} \\over 2}} \\right)}^2}} \\over {4{\\mkern 1mu} {\\pi ^2}{\\mkern 1mu} {n^2}}} = {{8A\\sin {{\\left( {{{\\pi {\\mkern 1mu} n} \\over 2}} \\right)}^2}} \\over {{\\pi ^2}{\\mkern 1mu} {n^2}}}$$ This answer is correct, but noting that $$n = 0,1,2,3,4,5,6,7,...\\quad \\quad \\sin {\\left( {{{\\pi {\\mkern 1mu} n} \\over 2}} \\right)^2} = 0,1,0,1,0,1,0,... = {{{ 1 -{\\left( { - 1} \\right)}^n}} \\over 2}$$ yields an even simpler result $${a_n} = \\left\\{ {\\matrix{ {4A{{1 - {{\\left( { - 1} \\right)}^n}} \\over {{\\pi ^2}{\\mkern 1mu} {n^2}}},\\quad {\\rm{n odd}}} \\cr {0,\\quad {\\rm{n even}}} \\cr } } \\right.$$ ##### Example 3: Triangle wave If xT(t) is a triangle wave with A=1, the values for an are given in the table below (note: this example", "\\over 2}$ $\\Rightarrow \\tan \\beta = {2 \\over 9}$ 4 ### JEE Main 2017 (Offline) If $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$, then the value of $\\cos 4x$ is A ${1 \\over 3}$ B ${2 \\over 9}$ C $- {7 \\over 9}$ D $- {3 \\over 5}$ ## Explanation Given that, $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$ $\\Rightarrow 5\\left( {{{{{\\sin }^2}x} \\over {{{\\cos }^2}x}} - {{\\cos }^2}x} \\right) = 2\\left( {2{{\\cos }^2}x - 1} \\right) + 9$ Let ${\\cos ^2}x = t,$ then we have $5\\left( {{{1 - t} \\over t} - t} \\right) = 2\\left( {2t - 1} \\right) + 9$ $\\Rightarrow 5\\left( {{{1 - t - {t^2}} \\over t}} \\right) = 4t - 2 + 9$ $\\Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$ $\\Rightarrow 9{t^2} + 12t - 5 = 0$ $\\Rightarrow 9{t^2} + 15t - 3t - 5 = 0$ $\\Rightarrow 3t\\left( {3t + 5} \\right) - 1\\left( {3t + 5} \\right) = 0$ $\\Rightarrow \\left( {3t + 5} \\right)\\left( {3t - 1} \\right) = 0$ $\\therefore$ $t = {1 \\over 3}$ and $t = - {5 \\over 3}$ If $t = - {5 \\over 3}$ then ${\\cos ^2}x$ is negative So , $t$ can not be $- {5 \\over", "\\end{align} where (by Mathematica) $$\\sum_{1\\leq k\\leq n}\\cos\\frac{\\pi j(2k-1)}{n} = \\frac{\\cos \\pi j\\sin \\pi j}{\\sin (\\pi j/n)}$$ and when I write $\\sin \\pi j/\\sin(\\pi j/n)$ I mean the limit as $j$ approaches its integer value (so no division by zero). Now, when $n=1$, the condition is $$0 = \\sum_{j\\geq1} \\alpha_j$$ and when $n>1$, the condition is $$0 = \\sum_{j\\geq1} \\alpha_j(-1)^{(j/n)}[n\\backslash j].$$ The condition $R_n f=0$ is only nontrivial when there are $j$ such that $n\\backslash j$ and $\\alpha_j\\neq0$. So suppose that $\\alpha_j\\neq0$ only when $j$ is a power of 2, so that the function is $$f(x) = \\sum_{k\\geq0} \\beta_k \\cos(\\pi 2^k x).$$ Then the only $n$ that impose any conditions on $\\alpha_k$ are the powers of 2. If $n=2^m$, $m>0$, then the condition is $$\\beta_m - \\beta_{m+1}-\\beta_{m+2}-\\cdots = 0,$$ and for $n=1$ the condition is $$\\sum_{k\\geq0} \\beta_k = 0.$$ Pick $\\beta_0 = -1, \\beta_k = 2^{-k}$ ($k\\geq1$). The condition for each $n=2^m$ and also $n=1$ will be satisfied, the function $$f(x) = -\\cos\\pi x+\\sum_{k\\geq1} 2^{-k} \\cos(\\pi 2^k x)$$ is clearly even and nonzero, and $R_nf=0$ for every $n$. If the Fourier series is finite, the function must then be zero. - I need to check the details first. I kindly request your patience. :) – Haskell Curry Jan 7 '13 at 9:30 The solution looks good! – Haskell Curry", "= 1$. (3) We will now show that $\\alpha$ is a unit in $R[[x]]$ if and only if $a_0$ is a unit in $R$. We assume that $R$ has a 1, but do not assume that $R$ is commutative. ($\\Rightarrow$) Suppose $\\alpha \\in R[[x]]$ is a unit, with inverse $\\alpha^{-1} = \\sum_{n \\geq 0} \\overline{a}_n x^n$. Now\\begin{align*}\\alpha\\alpha^{-1} =&\\ \\left( \\displaystyle\\sum_{n \\geq 0} a_n x^n \\right) \\left( \\displaystyle\\sum_{n \\geq 0} \\overline{a}_n x^n \\right)\\\\ =&\\ \\displaystyle\\sum_{n \\geq 0} \\left( \\displaystyle\\sum_{i+j = n} a_i \\overline{a}_j \\right) x^n\\end{align*}The coefficient of $x^0$ in this power series is 1 on one hand, and $a_0 \\overline{a}_0$ on the other. Thus $a_0 \\overline{a}_0 = 1$ in $R$. Similarly, from $\\alpha^{-1}\\alpha = 1$ we have $\\overline{a}_0 a_0 = 1$. Thus $a_0$ is a unit in $R$. ($\\Leftarrow$) Suppose $\\alpha_0 \\in R$ is a unit, with $\\overline{a}_0 a_0 = a_0 \\overline{a}_0 = 1$. Define $\\overline{\\alpha} = \\sum_{n \\geq 0} b_n x^n$ with $b_0 = \\overline{a}_0$ and $$b_{k+1} = -\\overline{a}_0 \\sum_{i+j=k+1, j \\leq k} a_i b_j.$$ Then\\begin{align*}\\alpha \\overline{\\alpha} =&\\ \\left( \\displaystyle\\sum_{n \\geq 0} a_n x^n \\right) \\left( \\displaystyle\\sum_{n \\geq 0} b_n x^n \\right)\\\\ =&\\ \\displaystyle\\sum_{n \\geq 0} \\left( \\displaystyle\\sum_{i+j=n} a)ib_j \\right) x^n\\end{align*}Note that if $n = 0$, $$\\sum_{i+j=n} a_ib_j = a_0 \\overline{a}_0 = 1.$$ If $n \\geq 1$, then \\begin{align*}\\sum_{i+j=n} a_ib_j =&\\ a_0b_n + \\sum_{i+j = n, j <", "## Download our App 4.5 (100k+ ) 1 ### AIEEE 2004 MCQ (Single Correct Answer) Let $$\\alpha ,\\,\\beta$$ be such that $$\\pi < \\alpha - \\beta < 3\\pi$$. If $$sin{\\mkern 1mu} \\alpha + \\sin \\beta = - {{21} \\over {65}}$$ and $$\\cos \\alpha + \\cos \\beta = - {{27} \\over {65}}$$ then the value of $$\\cos {{\\alpha - \\beta } \\over 2}$$ A $${{ - 6} \\over {65}}\\,\\,$$ B $${3 \\over {\\sqrt {130} }}$$ C $${6 \\over {65}}$$ D $$- {3 \\over {\\sqrt {130} }}$$ ## Explanation Given $$sin{\\mkern 1mu} \\alpha + \\sin \\beta = - {{21} \\over {65}}$$ .........(1) and $$\\cos \\alpha + \\cos \\beta = - {{27} \\over {65}}$$ ........(2) Square and add (1) and (2) you will get $$2\\left( {1 + \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta } \\right)$$$$= {{{{\\left( {21} \\right)}^2} + {{\\left( {27} \\right)}^2}} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $$2\\left( {1 + \\cos \\left( {\\alpha - \\beta } \\right)} \\right) = {{1170} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $$4{\\cos ^2}{{\\alpha - \\beta } \\over 2}$$$$= {{1170} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $${\\cos ^2}{{\\alpha - \\beta } \\over 2}$$$$= {9 \\over {130}}$$ $$\\therefore$$ $$\\cos {{\\alpha - \\beta } \\over 2} = \\pm {3 \\over {\\sqrt {130} }}$$ [ But $$\\cos {{\\alpha - \\beta } \\over 2} \\ne + {3 \\over {\\sqrt {130}", "$x = \\tan(t)$. We then get that \\begin{align} I(n) & = \\int_0^{\\infty} \\left(\\dfrac{2x}{1+x^2} \\right)^n dx = \\int_0^{\\pi/2} \\sin^n(2t) \\sec^2(t) dt = 2^n \\int_0^{\\pi/2} \\sin^n(t) \\cos^{n-2}(t) dt\\\\ & = 4\\int_0^{\\pi/2}\\sin^2(t) \\sin^{n-2}(2t)dt \\tag{$\\star$} \\end{align} Replacing $t$ by $\\pi/2-t$, we get $$I(n) = 4 \\int_0^{\\pi/2} \\cos^2(t) \\sin^{n-2}(2t) dt \\tag{\\perp}$$ Adding $(\\star)$ and $(\\perp)$, we get that $$2I(n) = 4 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt \\implies I(n) = 2 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt$$ I trust you can take it from here, using this post, which evaluates $\\displaystyle \\int_0^{\\pi} \\sin^{k}(t) dt$ and using Stirling (or) Wallis formula. - How do you go from $cos^2(t)\\sin^{n-2}(2t)$ to $\\sin^{n-2}(2t)$? – TCL May 7 '13 at 1:17 @TCL Updated the answer. Add $(\\star)$ and $(\\perp)$ to get it. – user17762 May 7 '13 at 1:19 Probably not how you want this done, but a useful and elegant method nonetheless: Note that $$f(x)=\\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\\to\\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$. $f^n \\equiv \\exp(n \\log f)$ and $n\\log(f(1+u))=-n\\frac{1}{2}u^2+n\\mathcal O(u^3)$. Then letting $v=\\sqrt{n} u$, $$\\int_0^\\infty n^{1/2}\\mathrm d x \\exp(-\\frac{1}{2}v^2+\\mathcal O(n^{-1/2})) \\approx \\int_{-\\infty}^\\infty \\mathrm d v\\, e^{-v^2/2} = \\sqrt{2\\pi}$$ Verification: WolframAlpha. Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals." ]

[ "\\beta }}$$] Here $$\\alpha = {1 \\over 3}$$, $$\\beta = {1 \\over 2}$$ and m = 0 then $$r = {{10 \\times {1 \\over 3} - 0} \\over {{1 \\over 3} + {1 \\over 2}}}$$ = $${{10} \\over 3} \\times {6 \\over 5}$$ = 4 $$\\therefore$$ T5 is the term independent of x. $$\\therefore$$ T5 = $${}^{10}{C_4}$$ = 210 4 ### AIEEE 2012 If $$n$$ is a positive integer, then $${\\left( {\\sqrt 3 + 1} \\right)^{2n}} - {\\left( {\\sqrt 3 - 1} \\right)^{2n}}$$ is : A an irrational number B an odd positive integer C an even positive integer D a rational number other than positive integers ## Explanation Let $${\\left( {a + x} \\right)^n}$$ = Odd trems(A) + Even terms(B) So $${\\left( {a - x} \\right)^n}$$ = Odd terms(A) - Even terms(B) $$\\therefore$$ $${\\left( {a + x} \\right)^n} - {\\left( {a - x} \\right)^n}$$ = (A + B) - (A - B) = 2B = 2[even terms] = 2[ T2 + T4 + T6 + ....... ] So in case of $${\\left( {\\sqrt 3 + 1} \\right)^{2n}} - {\\left( {\\sqrt 3 - 1} \\right)^{2n}}$$ = 2[ T2 + T4 + T6 + ....... ] = 2[ $${}^{2n}{C_1}.{\\left( {\\sqrt 3 } \\right)^{2n - 1}} + {}^{2n}{C_3}.{\\left( {\\sqrt 3 } \\right)^{2n - 3}} + ...$$ Here in $${\\left( {\\sqrt", "\\over 2}}^{ + {{{T_p}} \\over 2}} \\cr &= {2 \\over T}{A \\over {n{\\omega _0}}}\\left( {\\sin \\left( { + n{\\omega _0}{{{T_p}} \\over 2}} \\right) - \\sin \\left( { - n{\\omega _0}{{{T_p}} \\over 2}} \\right)} \\right) \\cr\\end{align} Since sine is an odd function, sin(a)-sin(-a)=2sin(a), and using the fact that ω0=2π/T and $${a_n} = {4 \\over T}{A \\over {n{\\omega _0}}}\\sin \\left( {n{\\omega _0}{{{t_p}} \\over 2}} \\right) = 4{A \\over {n2\\pi }}\\sin \\left( {n\\pi {{{t_p}} \\over T}} \\right) = 2{A \\over {n\\pi }}\\sin \\left( {n\\pi {{{t_p}} \\over T}} \\right)$$ This result is further explored in two examples. ##### Example 1: Special case, Duty Cycle = 50% Consider the case when the duty cycle is 50% (this means that the function is high 50% of the time, or Tp=T/2), A=1, and T=2. In this case a0=average=0.5 and for n≠0: $$\\displaylines{ {a_n} = 2{A \\over {n\\pi }}\\sin \\left( {n\\pi {{{t_p}} \\over T}} \\right) = 2{A \\over {n\\pi }}\\sin \\left( {{{n\\pi } \\over 2}} \\right) \\cr n = 0,\\,1,\\,2,\\,3,\\,4,\\,5,...\\quad \\sin \\left( {{{n\\pi } \\over 2}} \\right) = 0,1,0, - 1,0,1,... = \\left\\{ {\\matrix{ { - {1^{{{n - 1} \\over 2}}},\\quad n\\;odd} \\cr {0,\\quad n\\;even} \\cr } } \\right. \\cr {a_n} = \\left\\{ {\\matrix{ {2{A \\over {n\\pi }}\\left( { - {1^{{{n - 1} \\over 2}}}} \\right),\\quad n\\;odd} \\cr {\\quad 0,\\quad n\\;even,\\;n \\ne 0\\;} \\cr } }", "(More than One Correct Answer) If $$\\alpha$$ $$= 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right)$$ and $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are) A $$cos\\beta > 0$$ B $$\\sin \\beta < 0$$ C $$\\cos \\left( {\\alpha + \\beta } \\right) > 0$$ D $$\\cos \\alpha < 0$$ ## Explanation Here, $$\\alpha = 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right)$$ and $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$ as $${6 \\over {11}} > {1 \\over 2}$$ $$\\Rightarrow {\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right) > {\\sin ^{ - 1}}\\left( {{1 \\over 2}} \\right) = {\\pi \\over 6}$$ $$\\therefore$$ $$\\alpha = 3{\\sin ^{ - 1}}\\left( {{6 \\over {11}}} \\right) > {\\pi \\over 2} \\Rightarrow \\cos \\alpha < 0$$ Now, $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$ As $${4 \\over 9} < {1 \\over 2} \\Rightarrow {\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right) > {\\cos ^{ - 1}}\\left( {{1 \\over 2}} \\right) = {\\pi \\over 3}$$ $$\\therefore$$ $$\\beta = 3{\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right) > \\pi$$ $$\\therefore$$ $$\\cos \\beta < 0$$ and $$\\sin \\beta < 0$$ Now, $$\\alpha + \\beta$$ is slightly greater than $${{3\\pi } \\over 2}$$. $$\\therefore$$", "= {\\pi \\over 6}$$ 3 ### AIEEE 2011 If $$A = {\\sin ^2}x + {\\cos ^4}x,$$ then for all real $$x$$: A $${{13} \\over {16}} \\le A \\le 1$$ B $$1 \\le A \\le 2$$ C $${3 \\over 4} \\le A \\le {{13} \\over {16}}$$ D $${{3} \\over {4}} \\le A \\le 1$$ ## Explanation $$A = {\\sin ^2}x + {\\cos ^4}x$$ $$= {\\sin ^2}x + {\\cos ^2}x\\left( {1 - {{\\sin }^2}x} \\right)$$ $$= {\\sin ^2}x + {\\cos ^2}x - {1 \\over 4}{\\left( {2\\sin x.\\cos x} \\right)^2}$$ $$= 1 - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right)$$ Now $$0 \\le {\\sin ^2}\\left( {2x} \\right) \\le 1$$ $$\\Rightarrow 0 \\ge - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right) \\ge - {1 \\over 4}$$ $$\\Rightarrow 1 \\ge 1 - {1 \\over 4}{\\sin ^2}\\left( {2x} \\right) \\ge 1 - {1 \\over 4}$$ $$\\Rightarrow 1 \\ge A \\ge {3 \\over 4}$$ 4 ### AIEEE 2010 Let $$\\cos \\left( {\\alpha + \\beta } \\right) = {4 \\over 5}$$ and $$\\sin \\,\\,\\,\\left( {\\alpha - \\beta } \\right) = {5 \\over {13}},$$ where $$0 \\le \\alpha ,\\,\\beta \\le {\\pi \\over 4}.$$ Then $$tan\\,2\\alpha$$ = A $${56 \\over 33}$$ B $${19 \\over 12}$$ C $${20 \\over 7}$$ D $${25 \\over 16}$$ ## Explanation $$\\cos \\left( {\\alpha + \\beta } \\right) = {4 \\over 5} \\Rightarrow \\tan \\left( {\\alpha", "-0.4 \\textrm{ or } 0.4 < \\sin(n+1) ; \\\\ \\textrm{if } 0.4 \\le \\sin(n),& \\textrm{ then } \\sin(n+3) < 0; \\end{align} so there is no value $L$ where for any positive $\\varepsilon < 0.2$ you have all of $\\sin(n), \\sin(n+1), \\sin(n+3)$ and $\\sin(n+4)$ within $\\varepsilon$ of $L$. Let if possible $\\sin n\\rightarrow x$. Then $\\sin k=\\sin(n+k-n)=\\sin(n+k)\\cos k-\\cos(n+k)\\sin n\\rightarrow x(\\cos k-\\sqrt{1-x^2})$ for each positive integer k. Now as $k\\rightarrow \\infty$ implies that $x=x.0=0$ which shows that $\\sin k=0$ for all k…a contradiction.", "\\right )$$ then $$c_{2}\\neq 0$$. Assuming for now that $$\\sin \\left ( -\\sqrt {\\lambda }\\pi \\right ) =\\sin \\left ( \\sqrt {\\lambda }\\pi \\right )$$. This happens when $$\\sqrt {\\lambda }\\pi =n\\pi ,n=1,2,3,\\cdots$$, or $\\lambda _{n}=n^{2}\\qquad n=1,2,3,\\cdots$ Using this choice, we will now look to see what happens using the second BC. The solution (5) now becomes$X\\left ( x\\right ) =c_{1}\\cos \\left ( nx\\right ) +c_{2}\\sin \\left ( nx\\right ) \\qquad n=1,2,3,\\cdots$ Therefore$X^{\\prime }\\left ( x\\right ) =-c_{1}n\\sin \\left ( nx\\right ) +c_{2}n\\cos \\left ( nx\\right )$ Applying the second BC $$X^{\\prime }\\left ( -\\pi \\right ) =X^{\\prime }\\left ( \\pi \\right )$$ using the above gives\\begin {align*} c_{1}n\\sin \\left ( n\\pi \\right ) +c_{2}n\\cos \\left ( n\\pi \\right ) & =-c_{1}n\\sin \\left ( n\\pi \\right ) +c_{2}n\\cos \\left ( n\\pi \\right ) \\\\ c_{1}n\\sin \\left ( n\\pi \\right ) & =-c_{1}n\\sin \\left ( n\\pi \\right ) \\\\ 0 & =0 \\end {align*} Since $$n$$ is integer. Therefore this means that using $$\\lambda _{n}=n^{2}$$ will satisfy both boundary conditions with $$c_{2}\\neq 0,c_{1}\\neq 0$$. This means the solution (5) becomes$X_{n}\\left ( x\\right ) =A_{n}\\cos \\left ( nx\\right ) +B_{n}\\sin \\left ( nx\\right ) \\qquad \\qquad n=1,2,3,\\cdots$ The above says that there are two eigenfunctions in this case. They are $X_{n}\\left ( x\\right ) =\\left \\{ \\begin {array} [c]{c}\\cos \\left", "1 ### JEE Advanced 2016 Paper 2 Offline The value of $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin \\left( {{\\pi \\over 4} + {{\\left( {k - 1} \\right)\\pi } \\over 6}} \\right)\\sin \\left( {{\\pi \\over 4} + {{k\\pi } \\over 6}} \\right)}}}$$ is equal to A $$3 - \\sqrt 3$$ B $$2\\left( {3 - \\sqrt 3 } \\right)$$ C $$2\\left( {\\sqrt 3 - 1} \\right)\\,\\,\\,$$ D $$2\\left( {2 - \\sqrt 3 } \\right)$$ ## Explanation It is given that, $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin \\left( {{\\pi \\over 4} + {{(k - 1)\\pi } \\over 6}} \\right)\\sin \\left( {{\\pi \\over 4} + {{k\\pi } \\over 6}} \\right)}}}$$ Let $$\\alpha = {\\pi \\over 4}$$ and $$\\beta = {\\pi \\over 6}$$. Therefore, $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin (\\alpha + k\\beta )sin(\\alpha + (k - 1)\\beta )}}}$$ $$= {1 \\over {\\sin \\beta }}\\sum\\limits_{k = 1}^{13} {{{\\sin ((\\alpha + k\\beta ) - (\\alpha + (k - 1)\\beta ))} \\over {\\sin (\\alpha + k\\beta )\\sin (\\alpha + (k - 1)\\beta )}}}$$ $$= {1 \\over {\\sin \\beta }}\\sum\\limits_{k = 1}^{13} {(\\cot (\\alpha + (k - 1)\\beta ) - \\cot (\\alpha + k\\beta ))}$$ $$= {1 \\over {\\sin \\beta }}\\{ [\\cot (\\alpha ) - \\cot (\\alpha + \\beta )] + [\\cot (\\alpha + \\beta ) - \\cot (\\alpha + 2\\beta )] + ...... + [\\cot", "Question # If ${{\\cot }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]+{{\\tan }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]=x$ . Then find the value of $\\sin x$\\begin{align} & \\text{a) 1} \\\\ & \\text{b) co}{{\\text{t}}^{2}}\\left( \\dfrac{a}{2} \\right) \\\\ & \\text{c)}\\text{ tan}\\alpha \\\\ & \\text{d) cot}\\left( \\dfrac{\\alpha }{2} \\right) \\\\ \\end{align} Hint: We know that ${{\\cot }^{-1}}x+{{\\tan }^{-1}}x=\\dfrac{\\pi }{2}$ hence we get x = $\\dfrac{\\pi }{2}$. Now since we know the value of x. we can easily find the value of sinx. Now we are given that ${{\\cot }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]+{{\\tan }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]=x$ . Let us say ${{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}$ is equal to t. Now we will note that the range of ${{\\cos }^{-1}}\\theta$ is $[0,\\pi ]$ and hence the output of $\\cos \\alpha$ is also real Hence we have $t={{(\\cos \\alpha )}^{\\dfrac{1}{2}}}$ is a real number. Substituting this in the above equation we get ${{\\cot }^{-1}}t+{{\\tan }^{-1}}t=x.................(1)$ Now we know that for all real numbers x the identity ${{\\cot }^{-1}}x+{{\\tan }^{-1}}x=\\dfrac{\\pi }{2}$ is true. Now since this identity is true for all real numbers x and does not depend on x this identity is also true for t which is a real number. Hence applying this to equation (1) we get $x=\\dfrac{\\pi }{2}$ Hence now we have the value of x is $\\dfrac{\\pi }{2}$. Now we have to find the value of", "\\over \\pi }} \\right)$$ B $$\\sin \\left( {{{2a} \\over \\pi }} \\right)$$ C $$\\cos \\left( {{{2a} \\over \\pi }} \\right)$$ D $$\\sin \\left( {{{4a} \\over \\pi }} \\right)$$ Explanation Given $$a = {({\\sin ^{ - 1}}x)^2} - {({\\cos ^{ - 1}}x)^2}$$ $$= ({\\sin ^{ - 1}}x + {\\cos ^{ - 1}}x)({\\sin ^{ - 1}}x - {\\cos ^{ - 1}}x)$$ $$= {\\pi \\over 2}\\left( {{\\pi \\over 2} - 2{{\\cos }^{ - 1}}x} \\right)$$ $$\\Rightarrow 2{\\cos ^{ - 1}}x = {\\pi \\over 2} - {{2a} \\over \\pi }$$ $$\\Rightarrow {\\cos ^{ - 1}}(2{x^2} - 1) = {\\pi \\over 2} - {{2a} \\over \\pi }$$ $$\\Rightarrow 2{x^2} - 1 = \\cos \\left( {{\\pi \\over 2} - {{2a} \\over \\pi }} \\right)$$ 3 JEE Main 2021 (Online) 26th August Evening Shift If $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}{1 \\over {2{r^2}}} = p}$$, then the value of tan p is : A $${{101} \\over {102}}$$ B $${{50} \\over {51}}$$ C 100 D $${{51} \\over {50}}$$ Explanation $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}\\left( {{2 \\over {4{r^2}}}} \\right) = \\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}\\left( {{{(2r + 1) - (2r - 1)} \\over {1 + (2r + 1)(2r - 1)}}} \\right)} }$$ = $$\\sum\\limits_{r = 1}^{50} {{{\\tan }^{ - 1}}(2r + 1) - {{\\tan }^{ - 1}}(2r - 1)}$$ = $${\\tan ^{", "## Download our App 4.5 (100k+ ) 1 ### AIEEE 2004 MCQ (Single Correct Answer) Let $$\\alpha ,\\,\\beta$$ be such that $$\\pi < \\alpha - \\beta < 3\\pi$$. If $$sin{\\mkern 1mu} \\alpha + \\sin \\beta = - {{21} \\over {65}}$$ and $$\\cos \\alpha + \\cos \\beta = - {{27} \\over {65}}$$ then the value of $$\\cos {{\\alpha - \\beta } \\over 2}$$ A $${{ - 6} \\over {65}}\\,\\,$$ B $${3 \\over {\\sqrt {130} }}$$ C $${6 \\over {65}}$$ D $$- {3 \\over {\\sqrt {130} }}$$ ## Explanation Given $$sin{\\mkern 1mu} \\alpha + \\sin \\beta = - {{21} \\over {65}}$$ .........(1) and $$\\cos \\alpha + \\cos \\beta = - {{27} \\over {65}}$$ ........(2) Square and add (1) and (2) you will get $$2\\left( {1 + \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta } \\right)$$$$= {{{{\\left( {21} \\right)}^2} + {{\\left( {27} \\right)}^2}} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $$2\\left( {1 + \\cos \\left( {\\alpha - \\beta } \\right)} \\right) = {{1170} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $$4{\\cos ^2}{{\\alpha - \\beta } \\over 2}$$$$= {{1170} \\over {{{\\left( {65} \\right)}^2}}}$$ $$\\Rightarrow$$ $${\\cos ^2}{{\\alpha - \\beta } \\over 2}$$$$= {9 \\over {130}}$$ $$\\therefore$$ $$\\cos {{\\alpha - \\beta } \\over 2} = \\pm {3 \\over {\\sqrt {130} }}$$ [ But $$\\cos {{\\alpha - \\beta } \\over 2} \\ne + {3 \\over {\\sqrt {130}", "NEW New Website Launch Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc... 1 ### WB JEE 2010 In a $$\\Delta$$ABC, $$2ac\\sin \\left( {{{A - B + C} \\over 2}} \\right)$$ is equal to A a2 + b2 $$-$$ c2 B c2 + a2 $$-$$ b2 C b2 $$-$$ a2 $$-$$ c2 D c2 $$-$$ a2 $$-$$ b2 ## Explanation $$2ac\\sin \\left( {{{A - B + C} \\over 2}} \\right) = 2ac\\,.\\,\\sin \\left( {{{\\pi - B - B} \\over 2}} \\right)$$ $$= 2ac\\cos B = 2ac\\,.\\,{{{c^2} + {a^2} - {b^2}} \\over {2ca}} = {c^2} + {a^2} - {b^2}$$ 2 ### WB JEE 2010 If $${{\\cos A} \\over 3} = {{\\cos B} \\over 4} = {1 \\over 5}, - {\\pi \\over 2} < A < 0, - {\\pi \\over 2} < B < 0$$, then value of $$2\\sin A + 4\\sin B$$ is A 4 B $$-$$ 2 C $$-$$ 4 D 0 ## Explanation $$\\because$$ $$- {\\pi \\over 2} < A,B < 0$$ $$\\therefore$$ $$\\sin A < 0,\\sin B < 0$$ $$\\because$$ $${{\\cos A} \\over 3} = {{\\cos B} \\over 4} = {1 \\over 5}$$ $$\\therefore$$ $$\\cos A = {3 \\over 5},\\cos B = {4 \\over 5}$$ $$\\Rightarrow \\sin A = - \\sqrt {1 - {{\\cos", "{ \\propto + \\beta } \\right) = {x \\over {2x}} = {1 \\over 2}$ we know, $\\tan \\left( { \\propto + \\beta } \\right) = {{\\tan \\propto + \\tan \\beta } \\over {1 - \\tan \\propto \\tan \\beta }}$ $\\therefore\\,\\,\\,$ ${{\\tan \\propto + \\tan \\beta } \\over {1 - \\tan \\propto \\tan \\beta }} = {1 \\over 2}....\\left( 1 \\right)$ From the diagram, $\\tan \\widehat \\propto = {{x/2} \\over {2x}} = {1 \\over 4}......\\left( 2 \\right)$ Putting value of $\\tan \\propto$ in eq$(1)$, ${{{1 \\over 4} + \\tan \\beta } \\over {1 - {1 \\over 4}\\tan \\beta }} = {1 \\over 2}$ $\\Rightarrow 1 - {1 \\over 4}\\tan \\beta$ $= {1 \\over 2} + 2\\tan \\beta$ $\\Rightarrow {{9\\tan \\beta } \\over 4} = {1 \\over 2}$ $\\Rightarrow \\tan \\beta = {2 \\over 9}$ 3 ### JEE Main 2017 (Offline) If $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$, then the value of $\\cos 4x$ is A ${1 \\over 3}$ B ${2 \\over 9}$ C $- {7 \\over 9}$ D $- {3 \\over 5}$ ## Explanation Given that, $5\\left( {{{\\tan }^2}x - {{\\cos }^2}x} \\right) = 2\\cos 2x + 9$ $\\Rightarrow 5\\left( {{{{{\\sin }^2}x} \\over {{{\\cos }^2}x}} - {{\\cos }^2}x} \\right) = 2\\left( {2{{\\cos }^2}x - 1} \\right) + 9$ Let ${\\cos ^2}x = t,$", "= {1 \\over 3}$$, $$\\beta = {1 \\over 2}$$ and m = 0 then $$r = {{10 \\times {1 \\over 3} - 0} \\over {{1 \\over 3} + {1 \\over 2}}}$$ = $${{10} \\over 3} \\times {6 \\over 5}$$ = 4 $$\\therefore$$ T5 is the term independent of x. $$\\therefore$$ T5 = $${}^{10}{C_4}$$ = 210 2 ### AIEEE 2012 If $$n$$ is a positive integer, then $${\\left( {\\sqrt 3 + 1} \\right)^{2n}} - {\\left( {\\sqrt 3 - 1} \\right)^{2n}}$$ is : A an irrational number B an odd positive integer C an even positive integer D a rational number other than positive integers ## Explanation Let $${\\left( {a + x} \\right)^n}$$ = Odd trems(A) + Even terms(B) So $${\\left( {a - x} \\right)^n}$$ = Odd terms(A) - Even terms(B) $$\\therefore$$ $${\\left( {a + x} \\right)^n} - {\\left( {a - x} \\right)^n}$$ = (A + B) - (A - B) = 2B = 2[even terms] = 2[ T2 + T4 + T6 + ....... ] So in case of $${\\left( {\\sqrt 3 + 1} \\right)^{2n}} - {\\left( {\\sqrt 3 - 1} \\right)^{2n}}$$ = 2[ T2 + T4 + T6 + ....... ] = 2[ $${}^{2n}{C_1}.{\\left( {\\sqrt 3 } \\right)^{2n - 1}} + {}^{2n}{C_3}.{\\left( {\\sqrt 3 } \\right)^{2n - 3}} + ...$$ Here in $${\\left( {\\sqrt 3 } \\right)^{2n -", "$(\\cos A+i\\sin A)(\\cos B+i\\sin B)=\\cos(A+B)+i\\sin(A+B).\\nonumber$ 2. Apply (a) repeatedly to show that if $$n$$ is a positive integer then $\\prod_{k=1}^n(\\cos A_k+i\\sin A_k)=\\cos(A_1+A_2+\\cdots+A_n) +i\\sin(A_1+A_2+\\cdots+A_n).\\nonumber$ 3. Infer from (b) that if $$n$$ is a positive integer then $(\\cos\\theta+i\\sin\\theta)^n=\\cos n\\theta+i\\sin n\\theta. \\tag{A}$ 4. Show that (A) also holds if $$n=0$$ or a negative integer. HINT: Verify by direct calculation that $(\\cos\\theta +i\\sin\\theta )^{-1}=(\\cos\\theta -i\\sin\\theta ).\\nonumber$ Then replace $$\\theta$$ by $$-\\theta$$ in $$(A)$$. 5. Now suppose $$n$$ is a positive integer. Infer from (A) that if $z_k=\\cos\\left(2k\\pi\\over n\\right)+i\\sin\\left(2k\\pi\\over n\\right) ,\\quad k=0,1,\\dots,n-1,\\nonumber$ and $\\zeta_k=\\cos\\left((2k+1)\\pi\\over n\\right)+i\\sin\\left((2k+1)\\pi\\over n\\right) ,\\quad k=0,1,\\dots,n-1,\\nonumber$ then $z_k^n=1\\quad\\mbox{ and }\\quad\\zeta_k^n=-1,\\quad k=0,1,\\dots,n-1.\\nonumber$ (Why don’t we also consider other integer values for $$k$$?) 6. Let $$\\rho$$ be a positive number. Use (e) to show that $z^n-\\rho=(z-\\rho^{1/n} z_0)(z-\\rho^{1/n}z_1)\\cdots(z-\\rho^{1/n} z_{n-1})\\nonumber$ and $z^n+\\rho=(z-\\rho^{1/n} \\zeta_0)(z-\\rho^{1/n} \\zeta_1)\\cdots(z-\\rho^{1/n} \\zeta_{n-1}).\\nonumber$ 43. Use (e) of Exercise 9.2.42 to find a fundamental set of solutions of the given equation. 1. $$y'''-y=0$$ 2. $$y'''+y=0$$ 3. $$y^{(4)}+64y=0$$ 4. $$y^{(6)}-y=0$$ 5. $$y^{(6)}+64y=0$$ 6. $$\\left[(D-1)^6-1\\right]y=0$$ 7. $$y^{(5)}+y^{(4)}+y'''+y''+y'+y=0$$ 44. An equation of the form $a_0x^ny^{(n)}+a_1x^{n-1}y^{(n-1)}+\\cdots +a_{n-1}xy'+a_ny=0,\\quad x>0, \\tag{A}$ where $$a_0$$, $$a_1$$, …, $$a_n$$ are constants, is an Euler or equidimensional equation. Show that if $x=e^t \\quad \\mbox{ and } \\quad Y(t)=y(x(t)), \\tag{B}$ then \\begin{aligned} x\\frac{dy}{dx}&=\\frac{dY}{dt} \\\\ x^{2}\\frac{d^{2}y}{dx^{2}}&=\\frac{d^{2}Y}{dt^{2}}-\\frac{dY}{dt} \\\\ x^{3}\\frac{d^{3}y}{dx^{3}}&=\\frac{d^{3}Y}{dt^{3}}-3\\frac{d^{2}Y}{dt^{2}}+2\\frac{dY}{dt}. \\end{aligned} \\nonumber In general, it can be shown that if $$r$$ is any integer $$\\ge2$$ then $x^r {d^ry\\over", "- e^{2\\pi i k \\beta} ) \\right|$ and then claim the following: Claim. There exists constants $C_1, C_2 > 0$ such that for all irrational $\\beta$ and for all record-breaking $n$, $$a_n(\\beta) \\leq C_1 |\\epsilon_n| n \\exp( C_2 |\\epsilon_n| n \\log n ).$$ • In view of the Dirichlet approximation theorem, we know that $n|\\epsilon_n| < 1$. This gives $a_n(\\beta) \\leq C_1 n^{C_2}$. • If (a subsequence of) $|\\epsilon_n|$ decays at least as fast as $1/(n\\log n)$ then we have $\\liminf_n a_n = 0$. In particular, for any number $\\beta$ with irrational measure greater than 2, this is true. Unfortunately, most interesting numbers are either proved or expected to have irrationality measure 2, so this says nothing anything about OP's question. Now let $\\tilde{\\beta}_n = p_n / n$. For brevity, we remove the subscripts from $p_n$, $\\epsilon_n$ and $\\tilde{\\beta}_n$ whenever no confusion arises. Then using the relation $|1 - e^{2ix}| = 2|\\sin x|$ for any $x \\in \\Bbb{R}$, we can write \\begin{align*} a_n(\\beta) &= 2|\\sin (\\pi n \\beta)| \\left| \\prod_{k=1}^{n-1} (1 - e^{2\\pi i k \\tilde{\\beta}} ) \\right| \\left| \\prod_{k=1}^{n-1} \\frac{\\sin(\\pi k \\beta)}{\\sin(\\pi k \\tilde{\\beta})} \\right| \\\\ &= 2n|\\sin(\\pi \\epsilon)| \\prod_{k=1}^{n-1} \\left| \\frac{\\sin(\\pi k \\tilde{\\beta} + \\pi k \\epsilon / n)}{\\sin(\\pi k \\tilde{\\beta})} \\right| \\\\ &= 2n|\\sin(\\pi \\epsilon)| \\prod_{k=1}^{n-1} \\left| \\cos(\\pi k \\epsilon / n) + \\cot(\\pi k", "Correct Answer) The value of $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin \\left( {{\\pi \\over 4} + {{\\left( {k - 1} \\right)\\pi } \\over 6}} \\right)\\sin \\left( {{\\pi \\over 4} + {{k\\pi } \\over 6}} \\right)}}}$$ is equal to A $$3 - \\sqrt 3$$ B $$2\\left( {3 - \\sqrt 3 } \\right)$$ C $$2\\left( {\\sqrt 3 - 1} \\right)\\,\\,\\,$$ D $$2\\left( {2 - \\sqrt 3 } \\right)$$ ## Explanation It is given that, $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin \\left( {{\\pi \\over 4} + {{(k - 1)\\pi } \\over 6}} \\right)\\sin \\left( {{\\pi \\over 4} + {{k\\pi } \\over 6}} \\right)}}}$$ Let $$\\alpha = {\\pi \\over 4}$$ and $$\\beta = {\\pi \\over 6}$$. Therefore, $$\\sum\\limits_{k = 1}^{13} {{1 \\over {\\sin (\\alpha + k\\beta )sin(\\alpha + (k - 1)\\beta )}}}$$ $$= {1 \\over {\\sin \\beta }}\\sum\\limits_{k = 1}^{13} {{{\\sin ((\\alpha + k\\beta ) - (\\alpha + (k - 1)\\beta ))} \\over {\\sin (\\alpha + k\\beta )\\sin (\\alpha + (k - 1)\\beta )}}}$$ $$= {1 \\over {\\sin \\beta }}\\sum\\limits_{k = 1}^{13} {(\\cot (\\alpha + (k - 1)\\beta ) - \\cot (\\alpha + k\\beta ))}$$ $$= {1 \\over {\\sin \\beta }}\\{ [\\cot (\\alpha ) - \\cot (\\alpha + \\beta )] + [\\cot (\\alpha + \\beta ) - \\cot (\\alpha + 2\\beta )] + ...... + [\\cot (\\alpha + 12\\beta ) - \\cot", "NEW New Website Launch Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc... 1 ### IIT-JEE 1994 Let $$2{\\sin ^2}x + 3\\sin x - 2 > 0$$ and $${x^2} - x - 2 < 0$$ ($$x$$ is measured in radians). Then $$x$$ lies in the interval A $$\\left( {{\\pi \\over 6},\\,{{5\\pi } \\over 6}} \\right)\\,\\,$$ B $$\\left( { - 1,\\,{{5\\pi } \\over 6}} \\right)$$ C $$\\left( { - 1,\\,2} \\right)\\,\\,\\,$$ D $$\\left( {{\\pi \\over 6},\\,2} \\right)$$ 2 ### IIT-JEE 1994 If $$\\omega \\,$$ is an imaginary cube root of unity then the value of $$\\sin \\left\\{ {\\left( {{\\omega ^{10}} + {\\omega ^{23}}} \\right)\\pi - {\\pi \\over 4}} \\right\\}$$ is A $$- {{\\sqrt 3 } \\over 2}\\,$$ B $$- {1 \\over {\\sqrt 2 }}$$ C $${1 \\over {\\sqrt 2 }}$$ D $${{\\sqrt 3 } \\over 2}$$ 3 ### IIT-JEE 1994 Let $$n$$ be a positive integer such that $$\\sin {\\pi \\over {2n}} + \\cos {\\pi \\over {2n}} = {{\\sqrt n } \\over 2}.$$ Then A $$6 \\le n \\le 8$$ B $$4 < n \\le 8$$ C $$4 \\le n \\le 8$$ D $$4 < n < 8$$ 4 ### IIT-JEE 1994 Let $$0 < x < {\\pi \\over 4}$$ then $$\\left( {\\sec 2x - \\tan 2x}", "1 ### JEE Main 2019 (Online) 9th January Morning Slot Let $\\alpha$ and $\\beta$ be two roots of the equation x2 + 2x + 2 = 0 , then $\\alpha ^{15}$ + $\\beta ^{15}$ is equal to : A -256 B 512 C -512 D 256 ## Explanation Given equation, x2 + 2x + 2 = 0 $\\therefore$ x = ${{ - 2 \\pm \\sqrt {4 - 4.1.2} } \\over {2.1}}$ x = $-$ 1 $\\pm$ i $\\therefore$ $\\alpha$ = $-$ 1 + i and $\\beta$ = $-$ 1 $-$ i Note : x + iy = r (cos$\\theta$ + isin$\\theta$) $\\therefore$ (x + iy)n = rn (cosn$\\theta$ + isinn$\\theta$) $\\therefore$ $-$ 1 + i = $\\sqrt 2$ [cos${{3\\pi } \\over 4}$ + isin${{3\\pi } \\over 4}$ ] $\\Rightarrow$ ($-$ 1 + i)15 = ${\\left( {\\sqrt 2 } \\right)^{15}}$ [cos$\\left( {{{15.3\\pi } \\over 4}} \\right) + i\\sin \\left( {{{15.3\\pi } \\over 4}} \\right)$] And $-$1 $-$ i = $\\sqrt 2$ $\\left[ {\\cos \\left( { - {{3\\pi } \\over 4}} \\right) + i\\sin \\left( { - {{3\\pi } \\over 4}} \\right)} \\right]$ = $\\sqrt 2 \\left[ {\\cos {{3\\pi } \\over 4} - \\sin {{3\\pi } \\over 4}} \\right]$ $\\therefore$ ($-$1 $-$ i)15 = ${\\left( {\\sqrt 2 } \\right)^{15}}\\left[ {\\cos \\left( {{{15.3\\pi } \\over 4}} \\right) - i\\sin \\left( {{{15.3\\pi", "with a justification for Ross' intuition. – Aryabhata Oct 14 '10 at 22:57 If $k$ is a real number less than 2 then $\\sum_{n=1}^\\infty n^{-k-\\sin n}$ diverges. This is problem 11162 from the American Mathematical Monthly, the solution by the Microsoft Research Problems Group appears in the February 2007 issue. Here's the solution (if I've transcribed it correctly). If $k\\leq 0$, then the series diverges by comparison with the harmonic series. If $0< k <2$, then let $\\alpha,\\beta$ be such that $\\sin\\alpha=\\sin\\beta=1-k$ and $\\pi/2 < \\alpha < \\beta < 5\\pi/2$. Note that $\\sin x < 1-k$ for all $x$ with $\\alpha < x< \\beta$. Define $$S=(n\\in{\\mathbb N}:\\sin n<1-k )={\\mathbb N}\\cap\\bigcup_{j\\in {\\mathbb Z}}(2\\pi j+\\alpha,2\\pi j+\\beta).$$ Since $\\pi$ is irrational, the sequence $n/(2\\pi)$ mod 1 is dense in [0,1], so $S\\not=\\emptyset$ and there exist positive integers $a,A,b,B$ with $$0 < a-2\\pi A < {\\beta-\\alpha\\over2}\\quad\\mbox{and}\\quad -{\\beta-\\alpha\\over2} < b-2\\pi B < 0.$$ Let $C=\\max(a,b)$. Given $s\\in S$, we claim that there exists $s^\\prime\\in S$ with $s < s^\\prime\\leq s+C$. Indeed, there exists $j\\in{\\mathbb Z}$ with $\\alpha < s-2\\pi j<\\beta$. Compare $s-2\\pi j$ to $(\\alpha+\\beta)/2$. If $\\alpha < s-2\\pi j\\leq (\\alpha+\\beta)/2$, then $\\alpha< (s+a)-2\\pi(j+A)<\\beta$, so we can use $s^\\prime=s+a$. On the other hand, if $(\\alpha+\\beta)/2 < s-2\\pi j < \\beta$, then $a<(s+b)-2\\pi(j+B) < \\beta$, so we can use $s^\\prime=s+b$. Thus $S$ is infinite.", "all $$x \\in \\left( {\\alpha ,\\infty } \\right)\\,$$ D there exists $$\\beta > 0$$ such that $$\\left| {f\\left( x \\right)} \\right| + \\left| {f'\\left( x \\right)} \\right| \\le \\beta$$ for all $$x \\in \\left( {0,\\infty } \\right)$$ ## Explanation $$f(x) = \\ln x + \\int\\limits_0^x {\\sqrt {1 + \\sin t} \\,dt}$$ $$f'(x) = {1 \\over x} + \\sqrt {1 + \\sin x}$$ f' is not differentiable at sin x = $$-$$1 i.e. $$x = 2n\\pi - {\\pi \\over 2},n \\in N$$ in the interval (0, $$\\infty$$) $$f''(x) = - {1 \\over {{x^2}}} + {{\\cos x} \\over {2\\sqrt {1 + \\sin x} }}$$ f'' does not exist for all x $$\\in$$ (0, $$\\infty$$) f' exist for x > 0 we have $${1 \\over x} + \\sqrt {1 + \\sin x} < \\ln x + \\int\\limits_0^x {\\sqrt {1 + \\sin x} dx}$$ because L.H.S. is bounded and R.H.S. is not bounded so $$\\exists$$ some $$\\alpha$$ beyond which R.H.S. is greater than L.H.S. i.e. $$|f'(x)| < |f(x)|$$ for all x $$\\in$$ ($$\\alpha$$, $$\\infty$$) $$|f| + |f'| \\le \\beta$$ is wrong as f is unbounded while f' is bounded. 3 ### IIT-JEE 2009 MCQ (More than One Correct Answer) If $${I_n} = \\int\\limits_{ - \\pi }^\\pi {{{\\sin nx} \\over {\\left( {1 + {\\pi ^x}} \\right)\\sin x}}dx\\,\\,n = 0,1,2,.....,}$$ then A $${I_n} =" ]

[ "# Proving that any real number can be expressed as a ratio of $2$ real numbers in $[-1,1]$ I got this question while thinking about trigonometry. We know that $$\\cos x \\in[-1,1]$$ and $$\\sin x \\in[-1,1]$$. We also know that $$\\tan x \\in \\Bbb R$$. For any real number $$y$$, we can have a value of $$x$$ such that $$y=\\tan x=\\frac{\\sin x}{\\cos x}$$. This is easy to see, as we can theoretically always draw a triangle having leg lengths as $$y$$ and $$1$$. But can we prove this without using trigonometry? So the formal statement of the question would be (I made it up; it may/may not be a real question): Prove that all real numbers can be expressed in the form of $$\\frac{p}{q}$$, where $$p,q\\in[-1,1]$$ and $$q\\neq0.$$ Your trigonometric insight reduces to a fact not mentioning its inspiration,$$t=\\frac{\\frac{t}{\\sqrt{1+t^2}}}{\\frac{1}{\\sqrt{1+t^2}}}.$$ Hint: If $$x \\notin [-1,1]$$, then $$1/x \\in [-1,1]$$.", "+0 +2 59 2 +28 Let x and y be two real numbers such that $$2\\sin x \\sin y+3\\cos y+6\\cos x\\sin y=7.$$ Find $$\\tan^2x+2\\tan^2y.$$ Apr 4, 2020 #1 0 tan(x)^2 + 2*tan(y)^2 = 12. Apr 4, 2020 #2 +28 +1 Thanks for the answer, but I really want to know how to solve it. Apr 5, 2020", "$$x = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ So, from this, we can conclude that $$\\cos(5\\pi t)\\cos(6 \\pi t) = \\frac{\\cos(11 \\pi t) + \\cos(\\pi t)}2.$$ Other excursions into complex numbers can be explained similarly. We simply use this embedding of $$\\mathbb{R}$$ into $$\\mathbb{C}$$, and use the properties proven for all complex numbers to help us with real numbers.", "some quadratics ? Can we reduce the number of variables? …Try such thinking on your own… 2) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers such that $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0$ and $\\max_{1 \\leq i . Prove that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \\leq 10$. 3) Let a, b, c be positive real numbers. Prove that $\\frac{1}{2a} + \\frac{1}{2b} + \\frac{1}{2d} \\geq \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}$ More later Nalin Pithwa.", "Let $x$, $y$, and $z$ be real numbers such that $x+y+z=6$ and $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}=2$. Find the value of $\\dfrac{(x+y)}{z}+\\dfrac{(y+z)}{x}+\\dfrac{(x+z)}{y}$.", "formulas as well. Example: Prove that $\\frac{sin(x+y)}{sin(x-y)}$=$\\frac{tan x + tan y}{tan x – tan y}$ LHS=$\\frac{sin(x+y)}{sin(x-y)}$ =$\\frac{sin x cos y + cos x + sin y}{sin x cos y- cos x sin y}$ =$\\frac{tan x + tan y}{tan x – tan y}$ ———–Proved.", "+0 # pls halp?? :< 0 79 1 Given that $x$ is real and $x^3+\\frac{1}{x^3}=52$, find $x+\\frac{1}{x}$. tysm!!! Feb 23, 2021 $\\left(x+\\frac{1}{x}\\right)^3=x^3+\\frac{1}{x^3}+3x+\\frac{3}{x}=52+3\\left(x+\\frac{1}{x}\\right)$ $x+\\frac{1}{x}=k\\to k^3=52+3k$. We can guess and check or use rational root theorem to find $k=4\\implies\\boxed{x+\\frac{1}{x}=4}$.", "+0 0 53 2 Let a and b be positive real numbers such that a^b = b^a, b = 9a, Then can be expressed in the form sqrt(n,m) where m and are positive integers and n is as small as possible. Find m + n Apr 6, 2020 #1 +24972 +2 Let $$a$$ and $$b$$ be positive real numbers such that $$a^b = b^a,\\ b = 9a$$, Then $$a$$ can be expressed in the form $$sqrt(n,m)$$ where $$m$$ and $$n$$ are positive integers and $$n$$ is as small as possible. Find $$m + n$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{a^b} &=& \\mathbf{b^a} \\quad | \\quad \\mathbf{b = 9a} \\\\\\\\ a^{9a} &=& (9a)^a \\\\\\\\ a^{9a*\\frac{1}{a}} &=& \\left(9a\\right)^{a*\\frac{1}{a}} \\\\\\\\ a^{9} &=& 9a \\quad | \\quad *(a^{-1}) \\\\\\\\ a^{9}a^{-1} &=& 9a^1a^{-1} \\\\\\\\ a^{9-1} &=& 9a^{1-1} \\\\\\\\ a^{8} &=& 9a^0 \\quad | \\quad a^0 = 1 \\\\\\\\ a^{8} &=& 9 \\\\\\\\ a^{8*\\frac{1}{8}} &=& 9^{\\frac{1}{8}} \\\\\\\\ a &=& 9^{\\frac{1}{8}} \\\\\\\\ a &=& (3^2)^{\\frac{1}{8}} \\\\\\\\ a &=& 3^{2*\\frac{1}{8}} \\\\\\\\ a &=& 3^{\\frac{2}{8}} \\\\\\\\ a &=& 3^{\\frac{1}{4}} \\\\\\\\ \\mathbf{a} &=& \\mathbf{\\sqrt[4]{3}} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline m+n &=& 3+4 \\\\ \\mathbf{m+n} &=& \\mathbf{7} \\\\ \\hline \\end{array}$$ Apr 6, 2020 #2 +1 Thank you so much for explaining this to me! Apr 6, 2020", "$\\cos x = 0$. If $\\sinh y = 0$ then $y=0$ and hence $z$ is real. But every real number x satisfies $\\sin x \\in [-1,1]$. So $z$ can be any real number. Now suppose that $\\sinh y \\ne0$ and look at the other possibility, namely $\\cos x = 0$. If $\\cos x = 0$ then $\\sin x = \\pm1$; and if $\\sinh y \\ne0$ then $\\cosh y > 1$. So in this case $|\\sin{x}\\cosh{y}| > 1$ and hence $\\sin z\\notin[-1,1]$. Conclusion: $\\sin z \\in [-1,1]\\;\\Longleftrightarrow\\;z\\in\\mathbb{R}$. 3. Originally Posted by Opalg If $z = x+iy$ then $\\sin{z}=\\sin{x}\\cosh{y}+i\\cos{x}\\sinh{y}$. If that number lies in the interval [–1,1] then its imaginary part must be 0. Therefore either $\\sinh y = 0$ or $\\cos x = 0$. If $\\sinh y = 0$ then $y=0$ and hence $z$ is real. But every real number x satisfies $\\sin x \\in [-1,1]$. So $z$ can be any real number. Now suppose that $\\sinh y \\ne0$ and look at the other possibility, namely $\\cos x = 0$. If $\\cos x = 0$ then $\\sin x = \\pm1$; and if $\\sinh y \\ne0$ then $\\cosh y > 1$. So in this case $|\\sin{x}\\cosh{y}| > 1$ and hence $\\sin z\\notin[-1,1]$. Conclusion: $\\sin z \\in [-1,1]\\;\\Longleftrightarrow\\;z\\in\\mathbb{R}$. One question. By setting a = cos, I can eliminate the problem of", "# Solve this and plzz don't give any link. Q.14. Let a, b, c be positive real numbers. Let $\\theta ={\\mathrm{tan}}^{-1}\\sqrt{\\frac{a\\left(a+b+c\\right)}{bc}}+{\\mathrm{tan}}^{-1}\\sqrt{\\frac{b\\left(a+b+c\\right)}{ca}}+{\\mathrm{tan}}^{-1}\\sqrt{\\frac{c\\left(a+b+c\\right)}{ab}}$. Then • 0 What are you looking for?", "$C$ such that for any real numbers $x, y \\ge 0$, $\\pi(x+y)-\\pi(y) \\le \\frac{1}{3}x+C$. - Yea, exactly TonyK. – VVV Mar 15 '12 at 22:52 @VVV: Still not right, I'm afraid: now your statement is vacuously true. I suppose that's an improvement! – TonyK Mar 16 '12 at 8:34", "real numbers, $\\{\\rho(x,Y):y\\in Y\\}$ has a supremum. In particular, $\\sup\\{\\rho(x,Y):x\\in X\\}\\le \\delta(X)+\\rho(X,Y)+\\delta(Y).\\ \\ \\ \\ (6)$ By means of symmetry of (6), this also means that $\\sup\\{\\rho(y,X):y\\in Y\\}\\le \\delta(X)+\\rho(X,Y)+\\delta(Y),\\ \\ \\ \\ (7)$ which completes the proof.", "If $x \\neq 1$ then there exists a real number y s.t. $\\frac{y+1}{y-2} = x$ Ex-6(b) For $\\frac{y+1}{y-2} = x$ to be true $y$ = $\\frac{1 + 2x}{x - 1}$ which is defined only when $(x-1) \\neq 0$ => $x \\neq 1$ Thus, if there exists a real number y s.t. $\\frac{y+1}{y-2} = x$ then $x \\neq 1$ Ex-7 Let x be an arbitrary real number and x > 2. For a real number y = $\\frac{x + \\sqrt{x^2 - 4}}{2}$ $\\sqrt{x^2 - 4}$ is defined because $x > 2$ => $x^2 > 4$ => $x^2 - 4 > 0$ Now, $y + \\frac{1}{y}$ = $\\frac{x + \\sqrt{x^2 - 4}}{2}$ + $\\frac{2}{x + \\sqrt{x^2 - 4}}$ = $\\frac{(x + \\sqrt{x^2 - 4})(x + \\sqrt{x^2 - 4}) + 4}{2(x + \\sqrt{x^2 - 4})}$ = $\\frac{x^2 + (x^2 - 4) + 2x\\sqrt{x^2 - 4} + 4}{2(x + \\sqrt{x^2 - 4})}$ = $\\frac{2x^2 + 2x\\sqrt{x^2 - 4}}{2(x + \\sqrt{x^2 - 4})}$ = $x$ Thus, for any real number x, if x > 2, then there exists a real number y s.t. $y + \\frac{1}{y}$ = x. Ex-8 Note: $F$ in this ex means a Family of Sets. given: goal: if $A \\in F$ then $A \\subseteq \\cup F$ Assume the hypotheses given: $A \\in F$ goal: $A \\subseteq \\cup F$ $\\equiv$", "do you see some quadratics ? Can we reduce the number of variables? …Try such thinking on your own… 2) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers such that $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0$ and $\\max_{1 \\leq i . Prove that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \\leq 10$. 3) Let a, b, c be positive real numbers. Prove that $\\frac{1}{2a} + \\frac{1}{2b} + \\frac{1}{2d} \\geq \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}$ More later Nalin Pithwa. PROMYS: Boston; Program in Mathematics for Young Scientists https://promys.org For promising Indian students, the Mehta Scholarships are also available. — Shared by Nalin Pithwa for spreading awareness of math in India also…", "$$\\frac{a}{b} + \\frac{c}{d} = \\frac{ad + cb}{bd}$$ $$a,b,c,d$$ = any real number Remember to simplify the fraction (if possible) Math $$\\frac{a}{b} \\cdot \\frac{c}{d} = \\frac{ac}{bd}$$ $$a,b,c,d$$ = any real number Remember to simplify the fraction (if possible) Math $$\\frac{a}{b} \\div \\frac{c}{d} = \\frac{ad}{bc}$$ $$a,b,c,d$$ = any real number Remember to simplify the fraction (if possible) Math $$a \\frac{b}{c} = \\frac{ac+b}{c}$$ $$a,b,c$$ = any real number Remember to simplify the fraction (if possible) Math $$x^a \\cdot x^b = x^{a+b}$$ $$a,b,x$$ = any real number Math $$\\dfrac{x^a}{x^b} = x^{a-b}$$ $$a,b,x$$ = any real number Math $$(x^a)^b = x^{a\\cdot b}$$ $$a,b,x$$ = any real number Math $$(xy)^a = x^a \\cdot y^a$$ $$a,x,y$$ = any real number Math $$x^1 = x$$ $$x$$ = any real number Math $$x^0 = 1$$ $$x$$ = any real number ($$x\\ne 0$$) Math $$x^{-a} = \\frac{1}{x^a}$$ $$a,x$$ = any real number ($$x\\ne 0$$) Math $$x^{\\frac{a}{b}} = \\sqrt[b]{x^a} = (\\sqrt[b]{x})^a$$ $$a,b,x$$ = any real number Math $$a^x = b \\rightarrow \\log _a b = x$$ $$a,b,x$$ = any real number Math $$\\ln (x) = \\log _e x$$ $$x$$ = any real number $$e \\approx 2.718$$ = Euler’s number Math $$a^{\\log_a x} = x$$ $$a,x$$ = any real number Math $$\\log (a\\cdot b) = \\log (a) + \\log (b)$$ $$a,b$$ = any real number Math $$\\log (a\\div b) =", "can let $\\frac{Z_1}{Z_2} = x + yi$ where $x, y \\in \\mathbb R$ So you end up with $|x+(y+1)i| = |x+(y-1)i|$ which gives $x^2 + (y+1)^2 = x^2 + (y-1)^2 \\implies (y+1)^2 - (y-1)^2 = 0 \\implies 2y(2) = 0 \\implies y = 0$. Since its imaginary part is zero, $\\frac{Z_1}{Z_2}$ is real.", "any real number $$p,$$ there is at least one real number, usually two, that could be the value of $$y$$ in the true equation $$2 y^2 - (p+3)y + 2(p-1) = 0.$$ But now we have to remember that $$\\cos x$$ cannot be just any real number. It has to be in the interval $$[-1,1].$$ So instead of just being concerned about whether a real number exists that makes $$2 y^2 - (p+3)y + 2(p-1) = 0$$ true when we plug this number in for $$y,$$ we need to think about which real number we might have to plug in for $$y,$$ and make sure that that number is in the interval $$[-1,1].$$ The use of the variable $$p$$ makes it a little difficult to guess a factorization of the polynomial $$2 y^2 - (p+3)y + 2(p-1),$$ so let's just apply the standard formula to find solutions of any solvable quadratic equation in $$y$$: $$y = \\frac{-b \\pm \\sqrt{b^2 - 4ac} }{2a}.$$ Putting $$a = 2,$$ $$b = -(p+3),$$ $$c = 2(p-1),$$ we have $$y = \\frac{(p+3) \\pm \\sqrt{(p+3)^2 - 16(p-1)} }{4} = \\frac{(p+3) \\pm (p - 5)}{4}.$$ If we choose $$+$$ for $$\\pm$$ we get $$y = (p - 1)/2$$ and if we choose $$-$$ we get $$y = 2.$$ Obviously $$y = 2$$ is no good", "# 52. Quadratic Equations Problems Part 4 1. If $$x$$ be real, show that $$\\frac{x}{x^2 - 5x + 9}$$ lies between $$1$$ and $$-\\frac{1}{11}$$. 2. If $$p > 1$$ and $$x$$ is real, show that $$\\frac{x^2 - 2x + p^2}{x^2 + 2x + p^2}$$ lies between $$\\frac{p -1}{p + 1}$$ and $$\\frac{p + 1}{p - 1}$$ 3. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 + 2x + 1}{x^2 + 2x + 7}$$ 4. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 + 14x + 9}{x^2 + 2x + 3}$$ 5. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 -x + 1}{x^2 + x + 1}$$ 6. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 - 3x - 3}{2x^2 + 2x + 1}$$ 7. If $$x$$ be real, find the greatest and the least values of $$\\frac{2x^2 + x - 1}{x^2 + 4x + 2}$$ 8. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 - 4x + 9}{x^2 + 4x + 9}$$ 9. If $$x$$ be real, find the greatest and the least values of $$\\frac{x^2 + 7x + 16}{x^2 - 5x + 16}$$ 10. If $$x$$ be real, find the greatest and", "y$ Ex-2 To see that it exists, try x = 4 and y be arbitrary real number. LHS = xy + x - 4 = 4y + 4 - 4 = 4y = RHS To prove it is unique, let say there exists another real number z s.t. (zy + z - 4 = 4y) Then, z(y + 1) = 4y + 4 => z = $\\frac{4y + 4}{y + 1} = \\frac{4(y + 1)}{(y + 1)}$ which is clearly not defined if y+1 = 0 else z = 4 = x. Ex-3 To see that it exists, try y = $\\frac{x^2}{x - 1}$ , its defined since $x - 1 \\neq 0$ $\\frac{y}{x}$ = $\\frac{x^2}{x(x - 1)}$ = $\\frac{x}{x - 1}$ y - x = $\\frac{x^2}{x - 1} - x$ = $\\frac{x^2 - x^2 + x}{x - 1}$ = $\\frac{x}{x - 1}$ clearly $\\frac{y}{x} = y - x$ To see that its unique. Let say there exists another real number z s.t.$\\frac{z}{x} = z - x$ => $z = xz - x^2$ => $xz - z = x^2$ => z = $\\frac{x^2}{x - 1}$ = y Ex-4 To see that it exists, try y = $\\frac{1}{x}$ For any real number z, zy = $\\frac{z}{x}$ To see that it is unique, Let say there exists another real number", "Home » » If x is a real number and x + 11 < 0, evaluate $$\\frac{|x|}x$$. If x is a real number and x + 11 < 0, evaluate $$\\frac{|x|}x$$. Question If x is a real number and x + 11 < 0, evaluate $$\\frac{|x|}x$$. Options A) 0 B) -1 C) 1 D) 2 The correct answer is C. Explanation: x + 11 < 0 x < 0 - 11 x < -11 $$\\frac{|x|}{x}=\\frac{\\pm x}{x} = \\frac x x$$ or $$\\frac{-x} x$$ $$\\frac{|x|}{x}=\\frac x x$$ or $$\\frac{-x} x$$ 1 or -1 Discussion (2) • Ayoola Peace option b because if we ignore the sign for the numerator, we don't ignore it for the denominator. hence the answer will be negative • x + 11 < 0 x < 0 - 11 x < -11 $$\\frac{|x|}{x}=\\frac{\\pm x}{x} = \\frac x x$$ or $$\\frac{-x} x$$ $$\\frac{|x|}{x}=\\frac x x$$ or $$\\frac{-x} x$$ 1 or -1" ]

[ "by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42 • @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00 All you need to do now is$$\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\sin y\\cos x-\\cos y\\sin x}=\\frac{\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\cos x\\cos y}}{\\frac{\\sin y\\cos x-\\cos y\\sin x}{\\cos x\\cos y}}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}.$$ Consider fractions that are equivalent to $2/3$. $$\\frac{2}{3}\\cdot\\frac{2}{2} = \\frac{4}{6}$$ $$\\frac{2}{3}\\cdot\\frac{3}{3} = \\frac{6}{9}$$ $$etc.$$ Conclusion ... you're allowed to multiply by one. Similarly, with the expression $\\displaystyle\\frac{\\sin y\\cos x+\\cos y \\sin x}{\\sin y \\cos x−\\cos y \\sin x}$, you're allowed to multiply by $\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}}$ because $$\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}} = 1$$ And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\\displaystyle\\frac{\\frac{\\sin y\\cos x+\\cos y \\sin x}{1}}{\\frac{\\sin y \\cos x−\\cos y \\sin x}{1}}$$ before multiplying. • What does $\\cos y \\cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19 • Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17", "via analytic methods\" Here is my approach: Let $$\\arctan(x) := \\int_0^x \\frac{dt}{1+t^2}$$ and $\\tan(x)$ is defined as the inverse of $\\arctan(x)$, so that $\\tan(0) = 0$. Consider the differential equation $$\\frac{d x}{1+x^2} + \\frac{dy}{1+y^2} = 0 \\tag{DE},$$ one solution is $$\\arctan x + \\arctan y = c,$$ but the equation has also the solution $$\\frac{x + y}{1 – x y} = C.$$ Since the differential equation has but one distinct solution, the two solutions must be related to one another in a definite way. This relation is expressed by the equation $$C = f(c)$$ Now, let $$x = \\tan u, \\quad y = \\tan v,$$ then \\begin{align} u + v &= c \\\\ \\\\ \\frac{\\tan u + \\tan v}{1 – \\tan u \\tan v} &= f(c) = f(u +v) \\end{align} Let $v = 0$, then $$\\tan u = f(u)$$ and therefore $$\\color{blue}{\\frac{\\tan u + \\tan v}{1 – \\tan u \\tan v} = \\tan(u + v).}$$ Construction of the second solution Let $x = \\tan u$ and $y = \\tan v$. By definition $$\\frac{d x}{d u} = 1 + x^2 \\quad \\Longrightarrow \\quad \\frac{d^2 x}{d u^2} = 2x(1+x^2).$$ Similarly $$\\frac{d y}{d u} = -\\frac{d y}{d v} = -(1+y^2), \\mbox{ and } \\frac{d^2 y}{d u^2} = \\frac{d^2 y}{d v^2} = 2y(1+y^2)$$ from which follows that $$x \\frac{d^2 y}{d u^2} –", "# Solving differentialequation $y'=y*tan(x)+4sin(x)$ So I'm pretty new to differential equations and I currently struggle with this one. First things first. I know it's an first order linear differential equation with an undertermined coefficient. To solve it I have to find the particulate and the uniform (is this the right term?) solutions. The general solution is then $y = y_h + y_p$ $y'=y\\cdot \\tan(x)+4\\sin(x)$ 1. uniform solution $y_h'=y_h\\cdot \\tan(x)$ $\\int\\frac{y_h'}{y_h} dx = \\int \\tan(x) dx$ $\\int\\frac{1}{y_h} dy = \\int \\tan(x) dx$ $\\ln|y_h| = -\\ln(\\cos(x))+c$ $\\ln|y_h| = \\ln(cos^{-1}(x))+c$ $|y_h| = e^c\\cdot \\frac{1}{\\cos(x)}$ $y_h = C\\cdot \\frac{1}{\\cos(x)}$ 1. particulate solution I use the uniform solution. $y_p = C(x)\\cdot \\frac{1}{\\cos(x)}$ $y_p' = C'(x)\\cdot \\frac{1}{\\cos(x)}+C(x)\\cdot \\tan(x)\\cdot 1/\\cos(x)$ $y_p' = C'(x)\\cdot \\frac{1}{\\cos(x)}+y_h\\cdot \\tan(x)$ Back into y': $C'(x)\\cdot \\frac{1}{\\cos(x)}+ y\\cdot \\tan(x) = y\\cdot \\tan(x)+4\\sin(x)$ $C'(x) = 4\\sin(x)\\cos(x)$ Integrating gives: $C(x) = -2\\cos^2(x)$ Back into $y_p'$: $y_p = -2\\cos^2(x) \\cdot \\frac{1}{\\cos(x)}$ $y_p = -2\\cos(x)$ Now everything back together. $y = y_h + y_p$ $y = C\\cdot \\frac{1}{\\cos(x)} - 2\\cos(x)$ I definitly got something wrong… I'm pretty sure that we learned it this way or I might messed something up. Thanks for the help, I tried solving it now for quite some time. • \"Homogeneous\" is the word, not uniform. – Bobson Dugnutt Feb 29 '16 at 19:50 • The step where you go from $y_p'", "sin y cos (x-y) = cos x cos y + sin x sin y sin (π/2 – x) = cos x cos (π/2 – x) = sin x (II) tan (x+y) = (tan x + tan y) /(1−tan x tan y) tan (x-y) = (tan x − tan y)/(1 + tan x tan y) cot (x+y) = (cot x cot y −1)/(cot y + cot x) cot(x-y) = (cot x cot y + 1)/( cot y − cot x) (III) cos 2x = cos2 x-sin2 x = 2cos2 x-1 = 1-2sin2 x = (1-tan2 x)/(1+tan2 x) sin 2x = 2sin x cos x= 2tan x/(1+ tan2 x) tan 2x = 2 tan x/(1-tan2 x) (IV) sin 3x = 3 sin x – 4 sin3 x cos 3x = 4 cosx – 3 cos x tan 3x = [3tan x-tan3 x]/[1-3 tan2 x] (V) $cos\\ x+ cos\\ y=2\\ cos{(\\frac{x+y}{2})}\\ cos{(\\frac{x-y}{2})}\\\\ cos\\ x – cos\\ y = -2\\ sin{(\\frac{x+y}{2})}\\ sin{(\\frac{x-y}{2})}\\\\ sin\\ x + sin\\ y = 2\\ sin{(\\frac{x+y}{2})}\\ cos{(\\frac{x-y}{2})}\\\\ sin\\ x – sin\\ y = 2\\ cos{(\\frac{x+y}{2})}\\ sin{(\\frac{x-y}{2})}$ (VI) 2 cos x cos y = cos (x+y) + cos (x-y) 2 sin x sin y = cos (x-y) – cos (x+y) 2 sin x cos y= sin (x+y) + sin (x-y) 2 cos x sin y = sin (x+y) –", "\\tan b}}{{1 - \\tan a \\cdot \\tan b}}$$ $$\\tan (a - b) = \\frac{{\\tan a - \\tan b}}{{1 + \\tan a \\cdot \\tan b}}$$ $$\\cos x + \\cos y = 2\\cos \\frac{{x + y}}{2}\\cos \\frac{{x - y}}{2}$$ $$\\cos x - \\cos y = - 2\\sin \\frac{{x + y}}{2}\\sin \\frac{{x - y}}{2}$$ $$\\sin x + \\sin y = 2\\sin \\frac{{x + y}}{2}\\cos \\frac{{x - y}}{2}$$ $$\\sin x - \\sin y = 2\\cos \\frac{{x + y}}{2}\\sin \\frac{{x - y}}{2}$$ $$\\cos x \\cdot \\cos y = \\frac{1}{2}[\\cos (x - y) + \\cos (x + y)]$$ $$\\sin x\\sin y = \\frac{1}{2}[\\cos (x - y) - \\cos (x + y)]$$ $$\\sin x\\cos y = \\frac{1}{2}[\\sin (x - y) + \\sin (x + y)]$$ $$\\tan x + \\tan y = \\frac{{\\sin (x + y)}}{{\\cos x\\cos y}}$$ $$\\tan x - \\tan y = \\frac{{\\sin (x - y)}}{{\\cos x\\cos y}}$$ $$\\cot x + \\cot y = \\frac{{\\sin (x + y)}}{{\\sin x\\sin y}}$$ $$\\cot x - \\cot y = \\frac{{\\sin (x - y)}}{{\\sin x\\sin y}}$$ 2. CÔNG THỨC ĐẠO HÀM $${\\left( {{x^\\alpha }} \\right)^\\prime } = \\alpha .{x^{\\alpha - 1}}$$ $${\\left( {{u^\\alpha }} \\right)^\\prime } = \\,\\alpha .\\,{u^{\\alpha - 1}}.u'$$ $${\\left( {\\frac{1}{x}} \\right)^\\prime } = \\, - \\frac{1}{{{x^2}}}\\,\\,(x \\ne 0)$$ $${\\left( {\\frac{1}{u}} \\right)^\\prime } = \\, - \\frac{{u'}}{{{u^2}}}\\,\\,\\left( {u \\ne 0} \\right)$$ $${\\left( {\\sqrt x } \\right)^\\prime", "L.H.S= $\\frac{1 – (\\frac{3}{4})^{2}}{1 + (\\frac{3}{4})^{2}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$=$\\frac{1 – (\\frac{3}{4})^{2}}{1 + (\\frac{3}{4})^{2}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$ = $\\frac{1 – \\frac{9}{16}}{ 1+ \\frac{9}{16}}$ Take L.C.M of both numerator and denominator; $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$ = $\\frac{\\frac{16 – 9}{16}}{\\frac{16 + 9}{16}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$= $\\frac{7}{25}$ …. (6) Now we take R.H.S of equation whether $\\frac{1 – \\tan ^{2} A}{1 + \\tan ^{2 }A}= \\cos ^{2 }A – \\sin ^{2}A$ R.H.S = $\\cos ^{2}A – \\sin ^{2}A$ Putting value of sin A and cos A from equation (4) and (5); R.H.S= $(\\frac{4}{5})^{2} – {(\\frac{3}{5}^{2})}$ $\\cos ^{2}A – \\sin ^{2}A$= $(\\frac{4}{5})^{2} – {(\\frac{3}{5}^{2})}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{16}{25} – \\frac{9}{25}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{16 – 9}{25}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{7}{25}$ ….(7) On Comparing eq. (6) and (7); $\\frac{1 – \\tan ^{2} A}{1 + \\tan ^{2 }A}= \\cos ^{2 }A – \\sin ^{2}A$ Hence, proved. 9.) If tan θ =a/b , find the value of $\\frac{\\cos \\Theta + \\sin \\Theta }{\\cos \\Theta – \\sin \\Theta }$. Solution. Given here; tan θ =a/b ….…. (1) Now, we know that tan θ = sin θ/cos θ Therefore, we can write equation (1) as; sin θ/cos θ = a/b Now, by using invertendo;", "2 \\theta = 2 sin \\theta cos \\theta$ = $sin^{-1}(sin 2 \\theta)$ = $2 \\theta$ = $2 cos ^{-1} (x)$ Example 4: Write $tan^{-1} (\\frac{cos y}{1 – sin y}), \\frac{-\\pi}{2} < y < \\frac{\\pi}{2}$ in the simplest form. Solution: Given, $tan^{-1} (\\frac{cos y}{1 – sin y})$ We are going to simplify the given $tan^{-1} \\frac{cos y}{1 – sin y}$ Formula: $cos y = sin (\\frac{\\pi}{2} – y)$ and $sin y = cos (\\frac{\\pi}{2} – y)$ $tan^{-1}( \\frac{cos y}{1 – sin y}) = tan^{-1}\\frac{sin(\\frac{\\pi}{2} – y)}{1 – cos (\\frac{\\pi}{2} – y)}$ = $tan ^{-1} (tan (\\frac{\\pi}{4} + \\frac{y}{2})$) = $\\frac{\\pi}{4} + \\frac{y}{2}$", "y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c ∴ (sec x + tan x) . y = ∫(sec x tan x + tan2x) dx + c ∴ (sec x + tan x) . y = ∫(sec x tan x + sec2x – 1) dx + c ∴ (sec x + tan x) . y = sec x + tan x – x + c ∴ y(sec x + tan x) = sec x + tan x – x + c This is the general solution. (v) x $$\\frac{d y}{d x}$$ + 2y = x2 . log x Solution: (vi) (x + y) $$\\frac{d y}{d x}$$ = 1 Solution: (x + y) $$\\frac{d y}{d x}$$ = 1 ∴ $$\\frac{d x}{d y}$$ = x + y ∴ $$\\frac{d x}{d y}$$ – x = y ∴ $$\\frac{d x}{d y}$$ + (-1) x = y ……….(1) This is the linear differential equation of the form This is the general solution. (vii) (x + a) $$\\frac{d y}{d x}$$ – 3y = (x + a)5 Solution: (viii) dr + (2r cot θ + sin 2θ) dθ = 0 Solution: dr + (2r cot θ + sin 2θ) dθ = 0 ∴ $$\\frac{d r}{d \\theta}$$ + (2r cot θ + sin 2θ) = 0 ∴ $$\\frac{d r}{d", "(1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Trigonometry is a well acknowledged name in the geometric domain of mathematics, which is in relevance in this domain since ages and is also practically applied across the number of occasions. BC, The opposite site of angle B is b. i.e. They are called trigonometric functions. Trigonometric Ratio is known for the relationship between the measurement of the angles and the length of the sides of the right triangle. Trigonometry is a branch of mathematics that deals with triangles. In Trigonometry, different types of problems can be solved using trigonometry formulas. In trigonometry class 12, we study trigonometry which finds its application in the field of astronomy, engineering, architectural design, and physics.Trigonometry Formulas for class 12 contains all the essential trigonometric identities which can fetch some direct … More help with trigonometry at mathportal.org. More important identities Less important identities Truly obscure identities About the Java applet. In a right-angled triangle, we have 3 sides namely – Hypotenuse, Opposite side (Perpendicular), and Adjacent side (Height). Required", "Solve this Question: If $\\mathrm{y}=\\log \\sqrt{\\tan \\mathrm{x}}$, write $\\frac{\\mathrm{dy}}{\\mathrm{dx}}$ Solution: This particular problem is a perfect way to demonstrate how simple but powerful the Chain Rule of Differentiation is. It is important to identify and break the problem into the individual functions with respect to which successive differentiation shall be done. In this case, this is the way to break down the problem – $\\frac{d y}{d x}=\\frac{d y}{d(\\sqrt{\\tan x})} \\cdot \\frac{d(\\sqrt{\\tan x})}{d(\\tan x)} \\cdot \\frac{d(\\tan x)}{d x}$ i.e., $\\frac{d y}{d x}=\\frac{d(\\log \\sqrt{\\tan x})}{d(\\sqrt{\\tan x})} \\cdot \\frac{d(\\sqrt{\\tan x})}{d(\\tan x)} \\cdot \\frac{d(\\tan x)}{d x}$ $=\\frac{1}{\\sqrt{\\tan x}} \\cdot \\frac{1}{2 \\sqrt{\\tan x}} \\cdot \\sec ^{2} x$ $=\\frac{\\sec ^{2} x}{2 \\tan x}$ $=\\frac{1+\\tan ^{2} x}{2 \\tan x}$ $=\\frac{1}{2}(\\tan x+\\cot x)$ (Ans)", "x.$$ $\\displaystyle \\frac{1}{\\sin(x)\\cos(x)}-\\frac{1}{\\tan(x)}=\\frac{1}{\\sin(x)\\cos(x)}-\\frac{\\cos(x)}{\\sin(x)}=\\frac{1}{\\sin(x)\\cos(x)}-\\frac{\\cos^2x}{\\sin(x)\\cos(x)}=\\frac{1-\\cos^2(x)}{\\sin(x)\\cos(x)}=\\frac{\\sin^2(x)}{\\sin(x)\\cos(x)}=\\boxed{\\frac{\\sin(x)}{\\cos(x)}}$ P.S. For your first problem, $\\displaystyle \\cos(x)\\tan(x)=\\cos(x)\\frac{\\sin(x)}{\\cos(x)}=\\sin(x)$ $$\\frac{1}{\\sin x\\cos x} - \\frac{1}{\\tan x}=\\frac{\\sin^2x+\\cos^2x}{\\sin x\\cos x} - \\frac{1}{\\tan x}=\\frac{\\sin x}{\\cos x} + \\require{cancel}\\cancel{\\frac{\\cos x}{\\sin x}}- \\cancel{\\frac{1}{\\tan x}}=\\tan x.$$", "we haven’t derived that yet, so here we go: $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ $\\tan(x + y) = \\frac{\\sin(x+y)}{\\cos(x+y)}$ Using the sin & cos sum/difference identities: $\\tan(x + y) = \\frac{\\sin(x)\\cos(y) + \\cos(x)\\sin(y)}{\\cos(x)\\cos(y)-\\sin(x)\\sin(y)}$ $= \\frac{\\sin(x)\\cos(y) + \\cos(x)\\sin(y)}{\\cos(x)\\cos(y)-\\sin(x)\\sin(y)} * \\frac{\\frac{1}{\\cos(x)\\cos(y)}} {\\frac{1}{\\cos(x)\\cos(y)}}$ $= \\frac{\\frac{\\sin(x)\\cos(y)}{\\cos(x)\\cos(y)} + \\frac{\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)} } {\\frac{\\cos(x)\\cos(y)}{\\cos(x)\\cos(y)} - \\frac{\\sin(x)\\sin(y)}{\\cos(x)\\cos(y)}}$ $\\tan(x + y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$ Phew! Now, if x and y are equal, then just call them both x: $\\tan(2x) = \\frac{\\tan(x)+\\tan(x)}{1-\\tan(x)\\tan(x)}$ $\\tan(2x) = \\frac{2\\tan(x)}{1-\\tan^2(x)}$ # Trig Identities #3 Periodicity You can hopefully see from the diagram above, that if we add $2\\pi$ to our angle $\\theta$, we would come ‘full circle’ (ba-dum tish) and be back where we started from (think of it as adding 360 degrees if it makes you happier). Hence: $(\\cos(\\theta \\pm 2\\pi), \\sin(\\theta \\pm 2\\pi)) = (\\cos(\\theta), sin(\\theta))$ $\\cos(\\theta) = \\cos(\\theta \\pm 2\\pi)$ $\\sin(\\theta) = \\sin(\\theta \\pm 2\\pi)$ And since csc and sec are basically just utilisations of sin and cos, then they have the same property: $\\csc(\\theta) = \\csc(\\theta \\pm 2\\pi)$ $\\sec(\\theta) = \\sec(\\theta \\pm 2\\pi)$ Now $\\tan \\theta$ has a periodic cycle half the size of $\\sin \\theta$, which you can perhaps visualise easily by studying their respective plots side by side: and so : $\\tan(\\theta) = \\tan(\\theta \\pm \\pi)$ $\\cot(\\theta) = \\cot(\\theta \\pm \\pi)$", "and second equation by $y$ to get following equations $$x^2\\cos \\theta+xy\\sin \\theta=ax$$ And $$xy\\sin\\theta-y^2\\cos\\theta=by$$ Hence we get $$\\cos\\theta=\\frac {ax-by}{x^2+y^2}$$ By subtracting resultant $2^{nd}$ equation from the resultant $1^{st}$ equation. Using these values of $\\sin\\theta$ and $\\cos\\theta$ we get $$\\tan\\theta = \\frac {bx+ay}{ax-by}$$ Hint: Solve the two simultaneous linear equation for $\\sin\\theta,\\cos\\theta$ $$x\\cos \\theta+y\\sin\\theta=a$$ $$\\implies x+y\\tan\\theta=a\\sec\\theta$$ $$\\implies \\sec \\theta=\\frac{x+y\\tan\\theta}{a} ...(1)$$ $$x\\sin\\theta-y\\cos\\theta=b$$ $$\\implies x\\tan\\theta-y=b\\sec\\theta$$ $$\\implies \\sec\\theta=\\frac{x\\tan\\theta-y}{b}...(2)$$ From $$1$$ and $$2$$, $$\\frac{x+y\\tan\\theta}{a}=\\frac{x\\tan\\theta-y}{b}$$ $$\\implies bx+by\\tan\\theta=ax\\tan\\theta-ay$$ $$\\implies \\tan\\theta(ax-by)=bx+ay$$ $$\\tan\\theta=\\frac{bx+ay}{ax-by}$$ $$x \\cos\\theta + y\\sin\\theta = a, \\,\\, x \\sin\\theta - y\\cos\\theta = b$$. Therefore: $$\\dfrac{x \\cos\\theta + y\\sin\\theta}{x \\sin\\theta - y\\cos\\theta} = \\dfrac{a}{b}$$ $$\\implies \\dfrac{x + y\\tan\\theta}{x \\tan\\theta - y} = \\dfrac{a}{b}$$ $$\\implies bx + by\\tan\\theta = ax \\tan\\theta - ay$$ $$\\implies (ax - by) \\tan\\theta = bx + ay$$ $$\\implies \\tan\\theta = \\dfrac{bx + ay}{ax - by}$$", "x}{1 - \\tan^2 2x \\tan^2 x} = \\tan 3x \\tan x$ RHS $\\displaystyle = \\tan 3x \\tan x$ $\\displaystyle = \\tan (2x+x) \\tan (2x-x)$ $\\displaystyle = [ \\frac{\\tan 2x + \\tan x}{1- \\tan 2x \\tan x} ] [ \\frac{\\tan 2x - \\tan x}{1+ \\tan 2x \\tan x} ]$ $\\displaystyle = \\frac{\\tan^2 2x - \\tan^2 x}{1 - \\tan^2 2x \\tan^2 x}$ $\\displaystyle =$ LHS. Hence proved. $\\displaystyle \\\\$ $\\displaystyle \\text{Question 19: } \\frac{\\sin(x+y)}{\\sin(x-y)} = \\frac{a+b}{a-b}, \\text{ show that } \\frac{\\tan x}{\\tan y} = \\frac{a}{b}$ $\\displaystyle \\frac{\\sin(x+y)}{\\sin(x-y)} = \\frac{a+b}{a-b}$ $\\displaystyle \\Rightarrow \\frac{\\sin x \\cos y + \\cos x \\sin y}{\\sin x \\cos y - \\cos x \\sin y} = \\frac{a+b}{a-b}$ Using Componendo and Dividendo $\\displaystyle \\Rightarrow \\frac{\\sin x \\cos y + \\cos x \\sin y + \\sin x \\cos y - \\cos x \\sin y}{\\sin x \\cos y + \\cos x \\sin y - \\sin x \\cos y + \\cos x \\sin y} = \\frac{a+b + a - b}{a + b - a+b}$ $\\displaystyle \\Rightarrow \\frac{2\\sin x \\cos y}{2\\cos x \\sin y} = \\frac{2a}{2b}$ $\\displaystyle \\Rightarrow \\frac{\\sin x \\cos y}{\\cos x \\sin y} = \\frac{a}{b}$ $\\displaystyle \\Rightarrow \\frac{\\tan x}{\\tan y} = \\frac{a}{b}$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 20: If } \\tan A = x \\tan B , \\text{ prove that } \\frac{\\sin(A-B)}{\\sin(A+B)} = \\frac{x-1}{x+1}$ $\\displaystyle \\tan A = x \\tan B$", "{ y } =\\cos { y } +x\\sin { y } \\\\ \\\\ \\frac { dx }{ dy } \\cdot \\cos { y } =\\cos { y } +\\frac { \\sin ^{ 2 }{ y } }{ \\cos { y } } \\\\ \\\\ \\frac { 1 }{ \\cos { y } } \\cdot \\frac { dx }{ dy } \\cdot \\cos { y } =\\frac { 1 }{ \\cos { y } } \\left( \\cos { y+\\frac { \\sin ^{ 2 }{ y } }{ \\cos { y } } } \\right) \\\\ \\\\ \\frac { dx }{ dy } =1+\\tan ^{ 2 }{ y } \\\\ \\\\ \\therefore \\quad \\frac { dx }{ dy } =\\sec ^{ 2 }{ y }$ And finally: $\\frac { dy }{ dy } \\cdot \\frac { dy }{ dx } =\\frac { dy }{ dx }$ Now, using implicit differentiation: $y=\\arctan { x } \\\\ \\\\ \\tan { y } =x\\\\ \\\\ \\sec ^{ 2 }{ y } \\cdot \\frac { dy }{ dx } =1\\\\ \\\\ \\frac { dy }{ dx } =\\frac { 1 }{ \\sec ^{ 2 }{ y } } \\\\ \\\\ \\frac { dy }{ dx } =\\frac { 1 }{ \\tan ^{ 2 }{ y+1 } } \\\\ \\\\ \\therefore \\quad \\frac { dy", "= 0\\ or\\ \\frac{1}{2}$ Therefore we have $x = 0\\ or\\ x= \\frac{1}{2}$. When we have $x= \\frac{1}{2}$, we can see that : $L.H.S. = \\sin ^{-1}\\left ( 1 - \\frac{1}{2} \\right ) - 2\\sin^{-1}\\frac{1}{2} = - \\sin^{-1}\\frac{1}{2} = -\\frac{\\pi}{6}$ So, it is not equal to the R.H.S. $-\\frac{\\pi}{6} \\neq \\frac{\\pi}{2}$ Thus we have only one solution which is x = 0 Hence the correct answer is (C). (A) $\\frac{\\pi}{2}$ (B) $\\frac{\\pi}{3}$ (C) $\\frac{\\pi}{4}$ (D) $\\frac{3\\pi}{4}$ Applying formula: $\\left [ \\tan^{-1} x - \\tan^{-1}y = \\tan^{-1} \\left ( \\frac{x-y}{1+xy} \\right ) \\right ]$. We get, $\\tan^{-1}\\left (\\frac{x}{y} \\right )-\\tan^{-1} \\left ( \\frac{x-y}{x+y} \\right ) = \\tan^{-1} \\left [\\frac{ \\frac{x}{y} - \\frac{x-y}{x+y}}{1+\\left ( \\frac{x}{y} \\right ) \\left ( \\frac{x-y}{x+y} \\right ) } \\right ]$ $= \\tan^{-1} \\left [\\frac{ \\frac{x}{y} - \\frac{x-y}{x+y}}{1+\\left ( \\frac{x}{y} \\right ) \\left ( \\frac{x-y}{x+y} \\right ) } \\right ] = \\tan^{-1} \\left [ \\frac{\\frac{x(x+y)-y(x-y)}{y(x+y)}}{\\frac{y(x+y) + x(x-y)}{y(x+y)}} \\right ]$ $= \\tan^{-1}\\left ( \\frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \\right )$ $= \\tan^{-1}\\left ( \\frac{x^2 + y^2}{ y^2 + x^2 } \\right ) = \\tan^{-1} 1 = \\frac{\\pi}{4}$ Hence, the correct answer is C. ## NCERT solutions for class 12 maths chapter-wise chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 NCERT Solutions for Class", "(-\\!\\sin x) + 2x \\cos x + \\cos x\\\\&= - x^2 \\sin x + 2x \\cos x + \\cos x.$ Example 3: Find $dy/dx$ if $y = \\frac {\\cos x}{1-\\tan x}$ . What is the slope of the tangent line at $x = \\pi/3$? Solution: Using the quotient rule and the formulas above, we obtain $\\frac {dy}{dx} &= \\frac {(1-\\tan x)(-\\!\\sin x) - (\\cos x)(-\\!\\sec^2 x)} {(1-\\tan x)^2}\\\\ &= \\frac {-\\!\\sin x + \\tan x \\sin x + \\cos x \\sec^2 x}{(1- \\tan x)^2}\\\\ &= \\frac {-\\!\\sin x + \\tan x \\sin x + \\sec x}{(1- \\tan x)^2}$ To calculate the slope of the tangent line, we simply substitute $x = \\pi/3$: $\\frac {dy}{dx} \\Bigg|_{x = \\frac {\\pi}{3}} = \\frac {-\\!\\sin( \\frac{\\pi}{3} ) + \\tan ( \\frac{\\pi}{3} ) \\sin ( \\frac{\\pi}{3} ) + \\sec( \\frac{\\pi}{3} )} {(1 - \\tan( \\frac{\\pi}{3} ))^2}.$ We finally get the slope to be approximately $\\frac {dy}{dx} \\Bigg|_{x = \\frac {\\pi}{3}} = 4.9.$ Example 4: If $y = \\sec x$, find $y'' (\\pi/3)$. Solution: $y' &= \\sec x \\tan x\\\\y'' &= \\sec x (\\sec^2 x) + (\\sec x \\tan x) \\tan x\\\\&= \\sec^3 x + \\sec x \\tan^2 x.$ Substituting for $x = \\pi/3$, $y'' &= \\sec^3 \\left(\\frac{\\pi}{3}\\right) + \\sec \\left(\\frac{\\pi}{3}\\right) \\tan^2 \\left(\\frac{\\pi}{3}\\right)\\\\&= (2)^3 + (2)(\\sqrt{3})^2\\\\ &= 8 + (2)(3)\\\\&= 14.$ Thus $y'' (\\pi/3)", "Find : Question: Find $\\frac{d y}{d x}$ : $y=\\tan ^{-1}\\left(\\frac{3 x-x^{3}}{1-3 x^{2}}\\right),-\\frac{1}{\\sqrt{3}}<x<\\frac{1}{\\sqrt{3}}$ Solution: The given relationship is $y=\\tan ^{-1}\\left(\\frac{3 x-x^{3}}{1-3 x^{2}}\\right)$ $y=\\tan ^{-1}\\left(\\frac{3 x-x^{3}}{1-3 x^{2}}\\right)$ $\\Rightarrow \\tan y=\\frac{3 x-x^{3}}{1-3 x^{2}}$ …(1) It is known that, $\\tan y=\\frac{3 \\tan \\frac{y}{3}-\\tan ^{3} \\frac{y}{3}}{1-3 \\tan ^{2} \\frac{y}{3}}$ …(2) Comparing equations (1) and (2), we obtain $x=\\tan \\frac{y}{3}$ Differentiating this relationship with respect to x, we obtain $\\frac{d}{d x}(x)=\\frac{d}{d x}\\left(\\tan \\frac{y}{3}\\right)$ $\\Rightarrow 1=\\sec ^{2} \\frac{y}{3} \\cdot \\frac{d}{d x}\\left(\\frac{y}{3}\\right)$ $\\Rightarrow 1=\\sec ^{2} \\frac{y}{3} \\cdot \\frac{1}{3} \\cdot \\frac{d y}{d x}$ $\\Rightarrow \\frac{d y}{d x}=\\frac{3}{\\sec ^{2} \\frac{y}{2}}=\\frac{3}{1+\\tan ^{2} \\frac{y}{2}}$ $\\therefore \\frac{d y}{d x}=\\frac{3}{1+x^{2}}$", "# Q9. Find the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$, given that $y = 1$ when $x = 0$. G Gautam harsolia Given equation is $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ we can rewrite it as $\\frac{dy}{dx}= -\\frac{(1+y^2)e^x}{(1+e^{2x})}\\\\ \\\\ \\frac{dy}{1+y^2}= \\frac{-e^xdx}{1+e^{2x}}$ Now, integrate both the sides $\\tan^{-1}y + C' =\\int \\frac{-e^{x}dx}{1+e^{2x}}$ $\\int \\frac{-e^{x}dx}{1+e^{2x}}\\\\$ Put $e^x = t \\\\ e^xdx = dt$ $\\int \\frac{dt}{1+t^2}= \\tan^{-1}t + C''$ Put $t = e^x$ again $\\int \\frac{-e^{x}dx}{1+e^{2x}} = -\\tan ^{-1}e^x+C''$ Put this in our equation $\\tan^{-1}y = -\\tan ^{-1}e^x+C\\\\ \\tan^{-1}y +\\tan ^{-1}e^x=C$ Now, by using boundary conditions we will find the value of C It is given that y = 1 when x = 0 $\\\\ \\tan^{-1}1 +\\tan ^{-1}e^0=C\\\\ \\\\ \\frac{\\pi}{4}+\\frac{\\pi}{4}= C\\\\ C = \\frac{\\pi}{2}$ Now, put the value of C $\\tan^{-1}y +\\tan ^{-1}e^x=\\frac{\\pi}{2}$ Therefore, the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ is $\\tan^{-1}y +\\tan ^{-1}e^x=\\frac{\\pi}{2}$ Exams Articles Questions", "(2π + A) = sin A & cos (2π + A) = cos A, tan(x+y) = (tan x + tan y)/ (1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Further the formulas of Trigonometry are drafted in accordance to the various ratios used in the domain, such as sine, tangent, cosine etc. Integral Formula for Trig Identities If you are a mathmatics students then you can easily get the significance of integration formulas. Summary of Trigonometry formulas. Here below we are mentioning the list of different types of formulas of Trigonometry. This web site owner is mathematician Miloš Petrović. Thanks byjus. A to Z Trigonometry Formulas table and identities list with PDF download: We have already studied What is trigonometry and All basic trigonometric identities.Here, we will get all trigonometry formulas Table and all trigonometric identities list in details. It mainly deals with triangles and their angles. Fórmulas e identidades trigonométricas: ángulo medio, ángulo doble, suma de ángulos, paso de suma a producto, fórmulas del área, etcétera; junto con sus demostraciones. Here we provide" ]

[ "by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42 • @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00 All you need to do now is$$\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\sin y\\cos x-\\cos y\\sin x}=\\frac{\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\cos x\\cos y}}{\\frac{\\sin y\\cos x-\\cos y\\sin x}{\\cos x\\cos y}}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}.$$ Consider fractions that are equivalent to $2/3$. $$\\frac{2}{3}\\cdot\\frac{2}{2} = \\frac{4}{6}$$ $$\\frac{2}{3}\\cdot\\frac{3}{3} = \\frac{6}{9}$$ $$etc.$$ Conclusion ... you're allowed to multiply by one. Similarly, with the expression $\\displaystyle\\frac{\\sin y\\cos x+\\cos y \\sin x}{\\sin y \\cos x−\\cos y \\sin x}$, you're allowed to multiply by $\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}}$ because $$\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}} = 1$$ And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\\displaystyle\\frac{\\frac{\\sin y\\cos x+\\cos y \\sin x}{1}}{\\frac{\\sin y \\cos x−\\cos y \\sin x}{1}}$$ before multiplying. • What does $\\cos y \\cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19 • Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17", "(1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Trigonometry is a well acknowledged name in the geometric domain of mathematics, which is in relevance in this domain since ages and is also practically applied across the number of occasions. BC, The opposite site of angle B is b. i.e. They are called trigonometric functions. Trigonometric Ratio is known for the relationship between the measurement of the angles and the length of the sides of the right triangle. Trigonometry is a branch of mathematics that deals with triangles. In Trigonometry, different types of problems can be solved using trigonometry formulas. In trigonometry class 12, we study trigonometry which finds its application in the field of astronomy, engineering, architectural design, and physics.Trigonometry Formulas for class 12 contains all the essential trigonometric identities which can fetch some direct … More help with trigonometry at mathportal.org. More important identities Less important identities Truly obscure identities About the Java applet. In a right-angled triangle, we have 3 sides namely – Hypotenuse, Opposite side (Perpendicular), and Adjacent side (Height). Required", "x.$$ $\\displaystyle \\frac{1}{\\sin(x)\\cos(x)}-\\frac{1}{\\tan(x)}=\\frac{1}{\\sin(x)\\cos(x)}-\\frac{\\cos(x)}{\\sin(x)}=\\frac{1}{\\sin(x)\\cos(x)}-\\frac{\\cos^2x}{\\sin(x)\\cos(x)}=\\frac{1-\\cos^2(x)}{\\sin(x)\\cos(x)}=\\frac{\\sin^2(x)}{\\sin(x)\\cos(x)}=\\boxed{\\frac{\\sin(x)}{\\cos(x)}}$ P.S. For your first problem, $\\displaystyle \\cos(x)\\tan(x)=\\cos(x)\\frac{\\sin(x)}{\\cos(x)}=\\sin(x)$ $$\\frac{1}{\\sin x\\cos x} - \\frac{1}{\\tan x}=\\frac{\\sin^2x+\\cos^2x}{\\sin x\\cos x} - \\frac{1}{\\tan x}=\\frac{\\sin x}{\\cos x} + \\require{cancel}\\cancel{\\frac{\\cos x}{\\sin x}}- \\cancel{\\frac{1}{\\tan x}}=\\tan x.$$", "L.H.S= $\\frac{1 – (\\frac{3}{4})^{2}}{1 + (\\frac{3}{4})^{2}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$=$\\frac{1 – (\\frac{3}{4})^{2}}{1 + (\\frac{3}{4})^{2}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$ = $\\frac{1 – \\frac{9}{16}}{ 1+ \\frac{9}{16}}$ Take L.C.M of both numerator and denominator; $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$ = $\\frac{\\frac{16 – 9}{16}}{\\frac{16 + 9}{16}}$ $\\frac{1 – \\tan ^{2}A}{1 + \\tan ^{2}A}$= $\\frac{7}{25}$ …. (6) Now we take R.H.S of equation whether $\\frac{1 – \\tan ^{2} A}{1 + \\tan ^{2 }A}= \\cos ^{2 }A – \\sin ^{2}A$ R.H.S = $\\cos ^{2}A – \\sin ^{2}A$ Putting value of sin A and cos A from equation (4) and (5); R.H.S= $(\\frac{4}{5})^{2} – {(\\frac{3}{5}^{2})}$ $\\cos ^{2}A – \\sin ^{2}A$= $(\\frac{4}{5})^{2} – {(\\frac{3}{5}^{2})}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{16}{25} – \\frac{9}{25}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{16 – 9}{25}$ $\\cos ^{2}A – \\sin ^{2}A$ = $\\frac{7}{25}$ ….(7) On Comparing eq. (6) and (7); $\\frac{1 – \\tan ^{2} A}{1 + \\tan ^{2 }A}= \\cos ^{2 }A – \\sin ^{2}A$ Hence, proved. 9.) If tan θ =a/b , find the value of $\\frac{\\cos \\Theta + \\sin \\Theta }{\\cos \\Theta – \\sin \\Theta }$. Solution. Given here; tan θ =a/b ….…. (1) Now, we know that tan θ = sin θ/cos θ Therefore, we can write equation (1) as; sin θ/cos θ = a/b Now, by using invertendo;", "sin y cos (x-y) = cos x cos y + sin x sin y sin (π/2 – x) = cos x cos (π/2 – x) = sin x (II) tan (x+y) = (tan x + tan y) /(1−tan x tan y) tan (x-y) = (tan x − tan y)/(1 + tan x tan y) cot (x+y) = (cot x cot y −1)/(cot y + cot x) cot(x-y) = (cot x cot y + 1)/( cot y − cot x) (III) cos 2x = cos2 x-sin2 x = 2cos2 x-1 = 1-2sin2 x = (1-tan2 x)/(1+tan2 x) sin 2x = 2sin x cos x= 2tan x/(1+ tan2 x) tan 2x = 2 tan x/(1-tan2 x) (IV) sin 3x = 3 sin x – 4 sin3 x cos 3x = 4 cosx – 3 cos x tan 3x = [3tan x-tan3 x]/[1-3 tan2 x] (V) $cos\\ x+ cos\\ y=2\\ cos{(\\frac{x+y}{2})}\\ cos{(\\frac{x-y}{2})}\\\\ cos\\ x – cos\\ y = -2\\ sin{(\\frac{x+y}{2})}\\ sin{(\\frac{x-y}{2})}\\\\ sin\\ x + sin\\ y = 2\\ sin{(\\frac{x+y}{2})}\\ cos{(\\frac{x-y}{2})}\\\\ sin\\ x – sin\\ y = 2\\ cos{(\\frac{x+y}{2})}\\ sin{(\\frac{x-y}{2})}$ (VI) 2 cos x cos y = cos (x+y) + cos (x-y) 2 sin x sin y = cos (x-y) – cos (x+y) 2 sin x cos y= sin (x+y) + sin (x-y) 2 cos x sin y = sin (x+y) –", "2 \\theta = 2 sin \\theta cos \\theta$ = $sin^{-1}(sin 2 \\theta)$ = $2 \\theta$ = $2 cos ^{-1} (x)$ Example 4: Write $tan^{-1} (\\frac{cos y}{1 – sin y}), \\frac{-\\pi}{2} < y < \\frac{\\pi}{2}$ in the simplest form. Solution: Given, $tan^{-1} (\\frac{cos y}{1 – sin y})$ We are going to simplify the given $tan^{-1} \\frac{cos y}{1 – sin y}$ Formula: $cos y = sin (\\frac{\\pi}{2} – y)$ and $sin y = cos (\\frac{\\pi}{2} – y)$ $tan^{-1}( \\frac{cos y}{1 – sin y}) = tan^{-1}\\frac{sin(\\frac{\\pi}{2} – y)}{1 – cos (\\frac{\\pi}{2} – y)}$ = $tan ^{-1} (tan (\\frac{\\pi}{4} + \\frac{y}{2})$) = $\\frac{\\pi}{4} + \\frac{y}{2}$", "= 0\\ or\\ \\frac{1}{2}$ Therefore we have $x = 0\\ or\\ x= \\frac{1}{2}$. When we have $x= \\frac{1}{2}$, we can see that : $L.H.S. = \\sin ^{-1}\\left ( 1 - \\frac{1}{2} \\right ) - 2\\sin^{-1}\\frac{1}{2} = - \\sin^{-1}\\frac{1}{2} = -\\frac{\\pi}{6}$ So, it is not equal to the R.H.S. $-\\frac{\\pi}{6} \\neq \\frac{\\pi}{2}$ Thus we have only one solution which is x = 0 Hence the correct answer is (C). (A) $\\frac{\\pi}{2}$ (B) $\\frac{\\pi}{3}$ (C) $\\frac{\\pi}{4}$ (D) $\\frac{3\\pi}{4}$ Applying formula: $\\left [ \\tan^{-1} x - \\tan^{-1}y = \\tan^{-1} \\left ( \\frac{x-y}{1+xy} \\right ) \\right ]$. We get, $\\tan^{-1}\\left (\\frac{x}{y} \\right )-\\tan^{-1} \\left ( \\frac{x-y}{x+y} \\right ) = \\tan^{-1} \\left [\\frac{ \\frac{x}{y} - \\frac{x-y}{x+y}}{1+\\left ( \\frac{x}{y} \\right ) \\left ( \\frac{x-y}{x+y} \\right ) } \\right ]$ $= \\tan^{-1} \\left [\\frac{ \\frac{x}{y} - \\frac{x-y}{x+y}}{1+\\left ( \\frac{x}{y} \\right ) \\left ( \\frac{x-y}{x+y} \\right ) } \\right ] = \\tan^{-1} \\left [ \\frac{\\frac{x(x+y)-y(x-y)}{y(x+y)}}{\\frac{y(x+y) + x(x-y)}{y(x+y)}} \\right ]$ $= \\tan^{-1}\\left ( \\frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \\right )$ $= \\tan^{-1}\\left ( \\frac{x^2 + y^2}{ y^2 + x^2 } \\right ) = \\tan^{-1} 1 = \\frac{\\pi}{4}$ Hence, the correct answer is C. ## NCERT solutions for class 12 maths chapter-wise chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 NCERT Solutions for Class", "{ y } =\\cos { y } +x\\sin { y } \\\\ \\\\ \\frac { dx }{ dy } \\cdot \\cos { y } =\\cos { y } +\\frac { \\sin ^{ 2 }{ y } }{ \\cos { y } } \\\\ \\\\ \\frac { 1 }{ \\cos { y } } \\cdot \\frac { dx }{ dy } \\cdot \\cos { y } =\\frac { 1 }{ \\cos { y } } \\left( \\cos { y+\\frac { \\sin ^{ 2 }{ y } }{ \\cos { y } } } \\right) \\\\ \\\\ \\frac { dx }{ dy } =1+\\tan ^{ 2 }{ y } \\\\ \\\\ \\therefore \\quad \\frac { dx }{ dy } =\\sec ^{ 2 }{ y }$ And finally: $\\frac { dy }{ dy } \\cdot \\frac { dy }{ dx } =\\frac { dy }{ dx }$ Now, using implicit differentiation: $y=\\arctan { x } \\\\ \\\\ \\tan { y } =x\\\\ \\\\ \\sec ^{ 2 }{ y } \\cdot \\frac { dy }{ dx } =1\\\\ \\\\ \\frac { dy }{ dx } =\\frac { 1 }{ \\sec ^{ 2 }{ y } } \\\\ \\\\ \\frac { dy }{ dx } =\\frac { 1 }{ \\tan ^{ 2 }{ y+1 } } \\\\ \\\\ \\therefore \\quad \\frac { dy", "(2π + A) = sin A & cos (2π + A) = cos A, tan(x+y) = (tan x + tan y)/ (1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Further the formulas of Trigonometry are drafted in accordance to the various ratios used in the domain, such as sine, tangent, cosine etc. Integral Formula for Trig Identities If you are a mathmatics students then you can easily get the significance of integration formulas. Summary of Trigonometry formulas. Here below we are mentioning the list of different types of formulas of Trigonometry. This web site owner is mathematician Miloš Petrović. Thanks byjus. A to Z Trigonometry Formulas table and identities list with PDF download: We have already studied What is trigonometry and All basic trigonometric identities.Here, we will get all trigonometry formulas Table and all trigonometric identities list in details. It mainly deals with triangles and their angles. Fórmulas e identidades trigonométricas: ángulo medio, ángulo doble, suma de ángulos, paso de suma a producto, fórmulas del área, etcétera; junto con sus demostraciones. Here we provide", "$\\begin{itemize}$ • $\\frac d {dx}(a^x)\\;\\;\\text {for each real number}\\; x\\;\\text {and} \\; a>0\\; \\text {and} \\; a\\neq 1$ • $\\frac d {dx}(e^x)=e^x\\; \\text{for each real number}\\;x$ • $\\frac d {dx}(sin\\;x)=cos\\;x\\;\\text{for each real number}\\;x$ • $\\frac d {dx}(cos\\;x)=-sin\\;x\\;\\text{for each real number}\\;x$ • $\\frac d {dx}(tan\\;x)=\\frac 1 {cos^2x}\\;x\\;\\text{for each real number}\\;x$ ### Exercises Exercise 1. Compute the following derivative: $\\frac{d}{dx}(4x^{3} - 3x^{2} + 2x + 10)$ Solution. Using the sum, difference and constant multiple rules, we get: $\\frac{d}{dx}(4x^{3} - 3x^{2} + 2x + 10)=$ $= 4\\frac{d}{dx}(x^{3})- 3\\frac{d}{dx}(x^{2})+ 2\\frac{d}{dx}(x)+ 0=$ $=4\\cdot 3x^{2} - 3\\cdot 2x +2\\cdot 1= 12x^{2} - 6x + 2.$ ### Exercises Exercise 2. Compute the following derivative: $\\frac{d}{dx}(4\\sin x - 3\\cos x)$ Solution. Using the sum, difference and constant multiple rules, we get: $\\frac{d}{dx}(4\\sin x - 3\\cos x)=$ $=4\\frac{d}{dx}(\\sin x) - 3\\frac{d}{dx}(\\cos x)=$ $=4\\cdot \\cos x - 3\\cdot (-\\sin x)=4\\cos x +3\\sin x$ ### Exercises Exercise 3. Compute the following derivative: $\\frac{d}{dx}(x\\sin x + x^{2}\\cos x).$ Solution. Using the sum and product rules, we get: $\\frac{d}{dx}(x\\sin x + x^{2}\\cos x)=$ $=\\frac{d}{dx}(x\\sin x)+\\frac{d}{dx}(x^{2}\\cos x)=$ $=\\sin x + x\\cos x + 2x\\cos x - x^{2}\\sin x=$ $=\\sin x + 3x\\cos x - x^{2}\\sin x$ ### Exercises Exercise 4. Compute the following derivative: $\\frac{d}{dx}(\\frac{x^{3}+1}{x^{2}+4}).$ Solution. Using the sum and quotient rules, we get: $\\frac{d}{dx}(\\frac{x^{3}+1}{x^{2}+4})=$ $=\\frac{3x^{2}(x^{2}+4)-(x^{3}+1)2x}{(x^{2}+4)^{2}}=$ $\\frac{3x^{4}+ 12x^{2}- 2x^{4} - 2x}{(x^{2}+4)^{2}}=$ $=\\frac{x^{4}+", "x \\Big)$ RHS. Hence proved. $\\\\$ Question 8: $\\frac{\\cos x}{1- \\sin x}$ $= \\tan \\Big($ $\\frac{\\pi}{4}$ $+$ $\\frac{x}{2}$ $\\Big)$ LHS $=$ $\\frac{\\cos x}{1 - \\sin x}$ $=$ $\\frac{\\cos^2 \\frac{x}{2} - \\sin^2 \\frac{x}{2}}{\\sin^2 \\frac{x}{2}+ \\cos^2 \\frac{x}{2} - 2 \\sin \\frac{x}{2} \\cos \\frac{x}{2} }$ $=$ $\\frac{(\\cos \\frac{x}{2} - \\sin \\frac{x}{2} ) ( \\cos \\frac{x}{2} + \\sin \\frac{x}{2}) }{( \\cos \\frac{x}{2} - \\sin \\frac{x}{2})^2}$ $=$ $\\frac{\\cos \\frac{x}{2} + \\sin \\frac{x}{2} }{\\cos \\frac{x}{2} - \\sin \\frac{x}{2}}$ $=$ $\\frac{1 + \\tan \\frac{x}{2}}{1- \\tan \\frac{x}{2}}$ $=$ $\\frac{\\tan \\frac{\\pi}{4} + \\tan \\frac{x}{2}}{1- \\tan \\frac{\\pi}{4} \\tan \\frac{x}{2}}$ $= \\tan \\Big($ $\\frac{\\pi}{4}$ $+$ $\\frac{x}{2}$ $\\Big)$ RHS. Hence proved. $\\\\$ Question 9: $\\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2$ $\\frac{3\\pi}{8}$ $+\\cos^2$ $\\frac{5\\pi}{8}$ $+\\cos^2$ $\\frac{7\\pi}{8}$ $= 2$ LHS $= \\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2$ $\\frac{3\\pi}{8}$ $+\\cos^2$ $\\frac{5\\pi}{8}$ $+\\cos^2$ $\\frac{7\\pi}{8}$ $= \\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2$ $\\frac{3\\pi}{8}$ $+\\cos^2 \\big( \\pi -$ $\\frac{3\\pi}{8}$ $\\big) +\\cos^2\\big( \\pi -$ $\\frac{\\pi}{8}$ $\\big)$ $= \\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2$ $\\frac{3\\pi}{8}$ $+\\cos^2$ $\\frac{3\\pi}{8}$ $+\\cos^2$ $\\frac{\\pi}{8}$ $= 2 \\Big( \\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2$ $\\frac{3\\pi}{8}$ $\\Big)$ $= 2 \\Big( \\cos^2$ $\\frac{\\pi}{8}$ $+ \\cos^2 \\big($ $\\frac{\\pi}{2}$ $-$ $\\frac{\\pi}{8}$ $\\big) \\Big)$ $= 2 \\Big( \\cos^2$ $\\frac{\\pi}{8}$ $+ \\sin^2$ $\\frac{\\pi}{8}$ $\\Big)$ $= 2 =$ RHS. Hence proved. $\\\\$ Question 10: $\\sin^2$ $\\frac{\\pi}{8}$ $+ \\sin^2$ $\\frac{3\\pi}{8}$ $+ \\sin^2$ $\\frac{5\\pi}{8}$ $+\\sin^2$ $\\frac{7\\pi}{8}$ $= 2$ LHS $= \\sin^2$ $\\frac{\\pi}{8}$ $+ \\sin^2$ $\\frac{3\\pi}{8}$ $+ \\sin^2$ $\\frac{5\\pi}{8}$ $+\\sin^2$ $\\frac{7\\pi}{8}$ $= \\sin^2$ $\\frac{\\pi}{8}$", "# Simplifying $\\cos(2\\arctan(y/x))$ Hi I am looking for a way to calculate/simplify the following $$e^{2i \\phi}= \\cos(2\\phi)+i\\sin(2\\phi).$$ where $\\phi=\\arctan(y/x)$. I want to simplify so it is an expression with no trig functions. As done for the more simple case, $$e^{i\\phi}=\\cos \\phi+i\\sin \\phi$$ So I get $$\\cos \\phi =\\cos(\\arctan(y/x))=\\frac{x}{\\sqrt{x^2+y^2}},\\quad \\sin \\phi = \\frac{y}{\\sqrt{x^2+y^2}}$$ Thus I can simplify $e^{i\\phi}$ easily, but How can I calculate $\\cos (2\\phi), \\sin(2\\phi)$ ? (without using wolfram alpha) $$\\cos(2\\phi)=\\cos(2\\arctan(y/x))=?,\\quad \\sin(2\\phi)=\\sin(2\\arctan(y/x))=?$$ I get stuck here. Thanks a lot! $$\\cos 2\\phi = 2\\cos^2 \\phi - 1 = \\frac{2x^2}{x^2 + y^2} - 1 = \\frac{x^2 - y^2}{x^2+y^2}$$ cos(2tan$^{-1}$($\\frac{y}{x}$)). We if Z = tan$^{-1}$($\\frac{y}{x}$) then, tan(Z) = $\\frac{y}{x}$. What you want is cos(2tan$^{-1}$($\\frac{y}{x}$)) = cos(2Z) = $\\frac{1-tan^{2}Z}{1+tan^{2}Z}$ and you already have tan(Z).", "\\sin(x + y) \\sin y$. 11. Prove $\\frac{\\cos (c+d)}{\\cos (c-d)} = \\frac{1 - \\tan c \\tan d}{1 + \\tan c \\tan d}$ 12. Find all solutions to $2 \\cos^2 \\left (x+ \\frac{\\pi}{2} \\right ) = 1$, when $x$ is between $[0, 2\\pi)$. 13. Solve for all values of $x$ between $[0, 2\\pi)$ for $2 \\tan^2 \\left (x+ \\frac{\\pi}{6} \\right ) + 1 = 7$. 14. Find all solutions to $\\sin \\left (x+ \\frac{\\pi}{6} \\right ) = \\sin \\left (x- \\frac{\\pi}{4} \\right )$, when $x$ is between $[0, 2\\pi)$. 1. $\\cos \\frac{5 \\pi}{12} & = \\cos \\left (\\frac{2 \\pi}{12} + \\frac{3 \\pi}{12} \\right ) = \\cos \\left (\\frac{\\pi}{6} + \\frac{\\pi}{4} \\right ) = \\cos \\frac{\\pi}{6} \\cos \\frac{\\pi}{4} - \\sin \\frac{\\pi}{6} \\sin \\frac{\\pi}{4} \\\\& = \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2} - \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{6}}{4} - \\frac{\\sqrt{2}}{4} = \\frac{\\sqrt{6} - \\sqrt{2}}{4}$ 2. $\\cos \\frac{7 \\pi}{12} & = \\cos \\left (\\frac{4 \\pi}{12} + \\frac{3 \\pi}{12} \\right ) = \\cos \\left (\\frac{\\pi}{3} + \\frac{\\pi}{4} \\right ) = \\cos \\frac{\\pi}{3} \\cos \\frac{\\pi}{4} - \\sin \\frac{\\pi}{3} \\sin \\frac{\\pi}{4} \\\\& = \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} - \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{2}}{4} - \\frac{\\sqrt{6}}{4} = \\frac{\\sqrt{2} - \\sqrt{6}}{4}$ 3. $\\sin 345^\\circ & = \\sin (300^\\circ + 45^\\circ) = \\sin 300^\\circ \\cos 45^ \\circ + \\cos 300^\\circ \\sin 45^\\circ \\\\& = - \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2} + \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2}", "\\tan b}}{{1 - \\tan a \\cdot \\tan b}}$$ $$\\tan (a - b) = \\frac{{\\tan a - \\tan b}}{{1 + \\tan a \\cdot \\tan b}}$$ $$\\cos x + \\cos y = 2\\cos \\frac{{x + y}}{2}\\cos \\frac{{x - y}}{2}$$ $$\\cos x - \\cos y = - 2\\sin \\frac{{x + y}}{2}\\sin \\frac{{x - y}}{2}$$ $$\\sin x + \\sin y = 2\\sin \\frac{{x + y}}{2}\\cos \\frac{{x - y}}{2}$$ $$\\sin x - \\sin y = 2\\cos \\frac{{x + y}}{2}\\sin \\frac{{x - y}}{2}$$ $$\\cos x \\cdot \\cos y = \\frac{1}{2}[\\cos (x - y) + \\cos (x + y)]$$ $$\\sin x\\sin y = \\frac{1}{2}[\\cos (x - y) - \\cos (x + y)]$$ $$\\sin x\\cos y = \\frac{1}{2}[\\sin (x - y) + \\sin (x + y)]$$ $$\\tan x + \\tan y = \\frac{{\\sin (x + y)}}{{\\cos x\\cos y}}$$ $$\\tan x - \\tan y = \\frac{{\\sin (x - y)}}{{\\cos x\\cos y}}$$ $$\\cot x + \\cot y = \\frac{{\\sin (x + y)}}{{\\sin x\\sin y}}$$ $$\\cot x - \\cot y = \\frac{{\\sin (x - y)}}{{\\sin x\\sin y}}$$ 2. CÔNG THỨC ĐẠO HÀM $${\\left( {{x^\\alpha }} \\right)^\\prime } = \\alpha .{x^{\\alpha - 1}}$$ $${\\left( {{u^\\alpha }} \\right)^\\prime } = \\,\\alpha .\\,{u^{\\alpha - 1}}.u'$$ $${\\left( {\\frac{1}{x}} \\right)^\\prime } = \\, - \\frac{1}{{{x^2}}}\\,\\,(x \\ne 0)$$ $${\\left( {\\frac{1}{u}} \\right)^\\prime } = \\, - \\frac{{u'}}{{{u^2}}}\\,\\,\\left( {u \\ne 0} \\right)$$ $${\\left( {\\sqrt x } \\right)^\\prime", "– sin A, sin (3π/2 + A) = – cos A & cos (3π/2 + A) = sin A, sin (π – A) = sin A & cos (π – A) = – cos A, sin (π + A) = – sin A & cos (π + A) = – cos A, sin (2π – A) = – sin A & cos (2π – A) = cos A, sin (2π + A) = sin A & cos (2π + A) = cos A, tan(x+y) = (tan x + tan y)/ (1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Best regards from, Odhiambo Stephen Otumba. Differentiation Formula: In mathmatics differentiation is a well known term, which is generally studied in the domain of calculus portion of mathematics.We all have studied and solved its numbers of problems in our high school and +2 levels. There is an enormous number of uses of trigonometry and its formulae. It mainly deals with triangles and their angles. Trigonometry problems are very diverse and learning the below formulae help in", "= – cos A, sin (2π – A) = – sin A & cos (2π – A) = cos A, sin (2π + A) = sin A & cos (2π + A) = cos A, tan(x+y) = (tan x + tan y)/ (1−tan x •tan y), tan(x−y) = (tan x–tan y)/ (1+tan x • tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Trigonometry is a well acknowledged name in the geometric domain of mathematics, which is in relevance in this domain since ages and is also practically applied across the number of occasions. Here we provide a list of all Trigonometry formula for the students. It mainly deals with triangles and their angles. Trigonometry is also known as the study of relationships between lengths and angles of triangles. So, one thing is very important to realize here and that is: “ Focus on learning the basic identities and basic formulas ”. The formula is $a^2+b^2=c^2$. In simple language trigonometry can be defined as that branch of algebra, which is concerned with the triangle. There is an enormous number of fields where these identities of trigonometry and", "tan y), sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan, $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$, $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$, $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$, $$\\sin\\: x\\cdot \\cos\\:y=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$$, $$\\cos\\: x\\cdot \\cos\\:y=\\frac{\\cos(x+y)+\\cos(x-y)}{2}$$, $$\\sin\\: x\\cdot \\sin\\:y=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$$, $$\\sin\\: x+\\sin\\: y=2\\sin\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\sin\\: x-\\sin\\: y=2\\cos\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$, $$\\cos\\: x+\\cos\\: y=2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}$$, $$\\cos\\: x-\\cos\\: y=-2\\sin\\frac{x+y}{2}\\sin\\frac{x-y}{2}$$. Trigonometry Formulas TABLE : Now, we discuss about the value of sin, cos, tan, cot, sec and cosec at some particular given values. So, basically there are the numbers of the formulas which are generally used in Trigonometry to measure the sides of the triangle. These formulas are what simplifies the sides of triangles so that you can easily measure all its sides. It was in this context agitated that Abraham MOIVRE, a Protestant French mathematician, enabled a breakthrough. Images in Dave’s Short Trig Course are illustrated with a Java applet. Trigonometry can be divided into two sub-branches called plane trigonometry and spherical geometry. Trigonometry is an important branch of mathematics. Trigonometry Formulas Table and List . This periodicity constant is different for different trigonometric identities. Get to know some special rules for angles and various other important functions, definitions, and translations. The formulas mentioned here were very handy and my kid learns in a quick way. Periodicity Identities. Related Links. You can find Formulas for all the topics lying within the Introduction to Trigonometry Class 10 Introduction", "via analytic methods\" Here is my approach: Let $$\\arctan(x) := \\int_0^x \\frac{dt}{1+t^2}$$ and $\\tan(x)$ is defined as the inverse of $\\arctan(x)$, so that $\\tan(0) = 0$. Consider the differential equation $$\\frac{d x}{1+x^2} + \\frac{dy}{1+y^2} = 0 \\tag{DE},$$ one solution is $$\\arctan x + \\arctan y = c,$$ but the equation has also the solution $$\\frac{x + y}{1 – x y} = C.$$ Since the differential equation has but one distinct solution, the two solutions must be related to one another in a definite way. This relation is expressed by the equation $$C = f(c)$$ Now, let $$x = \\tan u, \\quad y = \\tan v,$$ then \\begin{align} u + v &= c \\\\ \\\\ \\frac{\\tan u + \\tan v}{1 – \\tan u \\tan v} &= f(c) = f(u +v) \\end{align} Let $v = 0$, then $$\\tan u = f(u)$$ and therefore $$\\color{blue}{\\frac{\\tan u + \\tan v}{1 – \\tan u \\tan v} = \\tan(u + v).}$$ Construction of the second solution Let $x = \\tan u$ and $y = \\tan v$. By definition $$\\frac{d x}{d u} = 1 + x^2 \\quad \\Longrightarrow \\quad \\frac{d^2 x}{d u^2} = 2x(1+x^2).$$ Similarly $$\\frac{d y}{d u} = -\\frac{d y}{d v} = -(1+y^2), \\mbox{ and } \\frac{d^2 y}{d u^2} = \\frac{d^2 y}{d v^2} = 2y(1+y^2)$$ from which follows that $$x \\frac{d^2 y}{d u^2} –", "we haven’t derived that yet, so here we go: $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ $\\tan(x + y) = \\frac{\\sin(x+y)}{\\cos(x+y)}$ Using the sin & cos sum/difference identities: $\\tan(x + y) = \\frac{\\sin(x)\\cos(y) + \\cos(x)\\sin(y)}{\\cos(x)\\cos(y)-\\sin(x)\\sin(y)}$ $= \\frac{\\sin(x)\\cos(y) + \\cos(x)\\sin(y)}{\\cos(x)\\cos(y)-\\sin(x)\\sin(y)} * \\frac{\\frac{1}{\\cos(x)\\cos(y)}} {\\frac{1}{\\cos(x)\\cos(y)}}$ $= \\frac{\\frac{\\sin(x)\\cos(y)}{\\cos(x)\\cos(y)} + \\frac{\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)} } {\\frac{\\cos(x)\\cos(y)}{\\cos(x)\\cos(y)} - \\frac{\\sin(x)\\sin(y)}{\\cos(x)\\cos(y)}}$ $\\tan(x + y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$ Phew! Now, if x and y are equal, then just call them both x: $\\tan(2x) = \\frac{\\tan(x)+\\tan(x)}{1-\\tan(x)\\tan(x)}$ $\\tan(2x) = \\frac{2\\tan(x)}{1-\\tan^2(x)}$ # Trig Identities #3 Periodicity You can hopefully see from the diagram above, that if we add $2\\pi$ to our angle $\\theta$, we would come ‘full circle’ (ba-dum tish) and be back where we started from (think of it as adding 360 degrees if it makes you happier). Hence: $(\\cos(\\theta \\pm 2\\pi), \\sin(\\theta \\pm 2\\pi)) = (\\cos(\\theta), sin(\\theta))$ $\\cos(\\theta) = \\cos(\\theta \\pm 2\\pi)$ $\\sin(\\theta) = \\sin(\\theta \\pm 2\\pi)$ And since csc and sec are basically just utilisations of sin and cos, then they have the same property: $\\csc(\\theta) = \\csc(\\theta \\pm 2\\pi)$ $\\sec(\\theta) = \\sec(\\theta \\pm 2\\pi)$ Now $\\tan \\theta$ has a periodic cycle half the size of $\\sin \\theta$, which you can perhaps visualise easily by studying their respective plots side by side: and so : $\\tan(\\theta) = \\tan(\\theta \\pm \\pi)$ $\\cot(\\theta) = \\cot(\\theta \\pm \\pi)$", "= \\frac{\\cos x-2}{3}$ Therefore, the answer is $\\frac{\\cos x-2}{3}$ Given function is $2 x + 3 y = \\sin y$ We can rewrite it as $\\sin y - 3y = 2x$ Now, differentiation w.r.t. x is $\\frac{dy}{dx}(\\sin y - 3y) = \\frac{d( 2x)}{dx}$ $(\\cos y\\frac{dy}{dx} - 3\\frac{dy}{dx}) = 2$ $\\frac{dy}{dx} = \\frac{2}{\\cos y -3}$ Therefore, the answer is $\\frac{2}{\\cos y -3}$ Given function is $ax + by ^2 = \\cos y$ We can rewrite it as $by^2-\\cos y = -ax$ Now, differentiation w.r.t. x is $\\frac{dy}{dx}(2by - (-\\sin y)) = \\frac{d( -ax)}{dx} = -a$ $\\frac{dy}{dx} = \\frac{-a}{2b y +\\sin y}$ Therefore, the answer is $\\frac{-a}{2b y +\\sin y}$ Question:4. Find dy/dx in the following: $xy + y^2 = \\tan x + y$ Given function is $xy + y^2 = \\tan x + y$ We can rewrite it as $xy+y^2-y= \\tan x$ Now, differentiation w.r.t. x is $y+\\frac{dy}{dx}(x+2y-1) = \\frac{d( \\tan x)}{dx} = \\sec^2 x$ $\\frac{dy}{dx} = \\frac{\\sec^2 x- y}{x+2y-1}$ Therefore, the answer is $\\frac{\\sec^2 x- y}{x+2y-1}$ Given function is $x^2 + xy + y^2 = 100$ We can rewrite it as $xy + y^2 = 100 - x^2$ Now, differentiation w.r.t. x is $y+\\frac{dy}{dx}(x+2y) = \\frac{d( 100-x^2)}{dx} = -2x$ $\\frac{dy}{dx} = \\frac{-2 x- y}{x+2y}$ Therefore, the answer is $\\frac{-2 x- y}{x+2y}$ Question:6 Find dy/dx in the following: $x" ]

[ "Thread: a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? 1. a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? 2. Re: a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? I think we need an additional assumption that $\\mathbf{a} \\ne \\mathbf{0}$; otherwise the claim is not true. So assuming $\\mathbf{a} \\ne \\mathbf{0}$, as a first step let's see if we can show that if $\\mathbf{a} \\times \\mathbf{v} = \\mathbf{0}$ and $\\mathbf{a} \\cdot \\mathbf{v} = 0$ then $\\mathbf{v} = \\mathbf{0}$. We have $\\mathbf{0} = \\mathbf{v} \\times (\\mathbf{a} \\times \\mathbf{v}) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v}) - \\mathbf{v} (\\mathbf{a} \\cdot \\mathbf{v}) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v}) - \\mathbf{v} (0) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v})$ so $\\mathbf{v} \\cdot \\mathbf{v} = 0$, hence $\\mathbf{v} = \\mathbf{0}$. Finally, to show that if $\\mathbf{a} \\cdot \\mathbf{b} = \\mathbf{a} \\cdot \\mathbf{c}$ and $\\mathbf{a} \\times \\mathbf{b} = \\mathbf{a} \\times \\mathbf{c}$ then $\\mathbf{b} = \\mathbf{c}$, let $\\mathbf{v} = \\mathbf{a} - \\mathbf{c}$ above.", "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "# a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? • August 31st 2012, 12:03 AM TaTzRE66 a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? • September 1st 2012, 08:14 AM awkward Re: a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc? I think we need an additional assumption that $\\mathbf{a} \\ne \\mathbf{0}$; otherwise the claim is not true. So assuming $\\mathbf{a} \\ne \\mathbf{0}$, as a first step let's see if we can show that if $\\mathbf{a} \\times \\mathbf{v} = \\mathbf{0}$ and $\\mathbf{a} \\cdot \\mathbf{v} = 0$ then $\\mathbf{v} = \\mathbf{0}$. We have $\\mathbf{0} = \\mathbf{v} \\times (\\mathbf{a} \\times \\mathbf{v}) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v}) - \\mathbf{v} (\\mathbf{a} \\cdot \\mathbf{v}) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v}) - \\mathbf{v} (0) = \\mathbf{a} (\\mathbf{v} \\cdot \\mathbf{v})$ so $\\mathbf{v} \\cdot \\mathbf{v} = 0$, hence $\\mathbf{v} = \\mathbf{0}$. Finally, to show that if $\\mathbf{a} \\cdot \\mathbf{b} = \\mathbf{a} \\cdot \\mathbf{c}$ and $\\mathbf{a} \\times \\mathbf{b} = \\mathbf{a} \\times \\mathbf{c}$ then $\\mathbf{b} = \\mathbf{c}$, let $\\mathbf{v} = \\mathbf{a} - \\mathbf{c}$ above.", "Vectors $\\mathbf{a}$ and $\\mathbf{b}$ must be going in the same direction.", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "A is: $\\|\\mathbf{A}\\|^2 = (A^0)^2 - (A^1)^2 - (A^2)^2 - (A^3)^2$ With the signature (+−−−), four-vectors may be classified as either spacelike if ||A|| < 0, timelike if ||A|| > 0, and null vectors if ||A|| = 0. ##### Standard basis, (−+++) signature Some authors define η with the opposite sign, in which case we have the (−+++) metric signature. Evaluating the summation with this signature: $\\mathbf{A \\cdot B} = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3$ while the matrix form is: $\\mathbf{A \\cdot B} = \\left( \\begin{matrix}A^0 & A^1 & A^2 & A^3 \\end{matrix} \\right) \\left( \\begin{matrix} -1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{matrix} \\right) \\left( \\begin{matrix}B^0 \\\\ B^1 \\\\ B^2 \\\\ B^3 \\end{matrix} \\right)$ Note that in this case, in one frame: $\\mathbf{A}\\cdot\\mathbf{B} = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3 = -C$ while in another: $\\mathbf{A}'\\cdot\\mathbf{B}' = - {A'}^0 {B'}^0 + {A'}^1 {B'}^1 + {A'}^2 {B'}^2 + {A'}^3 {B'}^3 = -C'$ so that: $-C = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3 = - {A'}^0 {B'}^0 + {A'}^1 {B'}^1 + {A'}^2 {B'}^2 + {A'}^3{B'}^3$ which", "& 1 \\end{array}} \\right| = 1\\cdot\\left| {\\begin{array}{*{20}{c}} -1 & -1\\\\ 1 & 1 \\end{array}} \\right| - 2\\cdot\\left| {\\begin{array}{*{20}{c}} 1 & -1\\\\ 1 & 1 \\end{array}} \\right| + 3\\cdot\\left| {\\begin{array}{*{20}{c}} 1 & -1\\\\ 1 & 1 \\end{array}} \\right| = 1\\cdot0 - 2\\cdot2 + 3\\cdot2 = 2.$ ### Example 2. Show that four points $$A\\left({-1,2,1}\\right),$$ $$B\\left({0,4,2}\\right),$$ $$C\\left({-1,5,5}\\right),$$ and $$D\\left({2,5,0}\\right)$$ lie in the same plane. Solution. Compute the coordinates of vectors $$\\mathbf{a} = \\mathbf{AB},$$ $$\\mathbf{b} = \\mathbf{AC},$$ and $$\\mathbf{c} = \\mathbf{AD}:$$ $\\mathbf{a} = \\mathbf{AB} = \\left({1,2,1}\\right),\\;\\mathbf{b} = \\mathbf{AC} = \\left({0,3,4}\\right),\\;\\mathbf{c} = \\mathbf{AD} = \\left({3,3,-1}\\right).$ Check that these vectors are coplanar (this will mean that they lie in the same plane). Calculate the determinant: $\\left| {\\begin{array}{*{20}{c}} 1 & 2 & 1\\\\ 0 & 3 & 4\\\\ 3 & 3 & -1 \\end{array}} \\right| = 1\\cdot\\left| {\\begin{array}{*{20}{c}} 3 & 4\\\\ 3 & -1 \\end{array}} \\right| + 3\\cdot\\left| {\\begin{array}{*{20}{c}} 2 & 1\\\\ 3 & 4 \\end{array}} \\right| = 1\\cdot\\left({-15}\\right) + 3\\cdot5 = 0.$ The determinant is zero. Hence, the vectors are coplanar. ### Example 3. Show that vectors $$\\mathbf{a} = 2\\mathbf{i} + 3\\mathbf{k},$$ $$\\mathbf{b} = \\mathbf{i} - \\mathbf{j}$$ and $$\\mathbf{c} = 2\\mathbf{j} + 3\\mathbf{k}$$ are coplanar and express vector $$\\mathbf{c}$$ in terms of $$\\mathbf{a}$$ and $$\\mathbf{b}.$$ Solution. The coordinates of the vectors $$\\mathbf{a},$$ $$\\mathbf{b},$$ and $$\\mathbf{c}$$ are equal $\\mathbf{a} = 2\\mathbf{i} + 3\\mathbf{k} =", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "is an $$1\\times m$$ row vector: $\\underset{(n\\times m)}{\\mathbf{A}}=\\left[\\begin{array}{cccc} a_{11} & a_{12} & \\ldots & a_{1m}\\\\ a_{21} & a_{22} & \\ldots & a_{2m}\\\\ \\vdots & \\vdots & \\ddots & \\vdots\\\\ a_{n1} & a_{n2} & \\ldots & a_{nm} \\end{array}\\right]=\\left[\\begin{array}{cccc} \\mathbf{a}_{1} & \\mathbf{a}_{2} & \\ldots & \\mathbf{a}_{m}\\end{array}\\right].$ Two vectors $$\\mathbf{a}_{1}$$ and $$\\mathbf{a}_{2}$$ are linearly independent if $$c_{1}\\mathbf{a}_{1}+c_{2}\\mathbf{a}_{2}=0$$ implies that $$c_{1}=c_{2}=0$$. A set of vectors $$\\mathbf{a}_{1},\\mathbf{a}_{2},\\ldots,\\mathbf{a}_{m}$$ are linearly independent if $$c_{1}\\mathbf{a}_{1}+c_{2}\\mathbf{a}_{2}+\\cdots+c_{m}\\mathbf{a}_{m}=0$$ implies that $$c_{1}=c_{2}=\\cdots=c_{m}=0$$. That is, no vector $$\\mathbf{a}_{i}$$ can be expressed as a non-trivial linear combination of the other vectors. The column rank of the $$n\\times m$$ matrix $$\\mathbf{A}$$, denoted $$\\mathrm{rank}(\\mathbf{A})$$, is equal to the maximum number of linearly independent columns. The row rank of a matrix is equal to the maximum number of linearly independent rows, and is given by $$\\mathrm{rank}(\\mathbf{A}^{\\prime})$$. It turns out that the column rank and row rank of a matrix are the same. Hence, $$\\mathrm{rank}(\\mathbf{A})$$$$=\\mathrm{rank}(\\mathbf{A}^{\\prime})\\leq\\mathrm{min}(m,n)$$. If $$\\mathrm{rank}(\\mathbf{A})=m$$ then $$\\mathbf{A}$$ is called full column rank, and if $$\\mathrm{rank}(\\mathbf{A})=n$$ then $$\\mathbf{A}$$ is called full row rank. If $$\\mathbf{A}$$ is not full rank then it is called reduced rank. Example 3.6 (Determining the rank of a matrix in R) Consider the $$2\\times3$$ matrix $\\mathbf{A}=\\left(\\begin{array}{ccc} 1 & 3 & 5\\\\ 2 & 4 & 6 \\end{array}\\right)$ Here, $$\\mathrm{rank}(\\mathbf{A})\\leq min(3,2)=2$$, the number of rows. Now, $$\\mathrm{rank}(\\mathbf{A})=2$$ since the rows of $$\\mathbf{A}$$", "and $C$ have position vectors $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ respectively.", "things enough times. So what’s a basis of $\\mathbb{R}^{3}$? A basis is a list of 3 vectors in terms of which all vectors can be written. If $\\{\\mathbf{e}_{1},\\mathbf{e}_{2},\\mathbf{e}_{3}\\}$ is a basis then it means any vector $\\mathbf{v}$ can be written as a linear combination of them, $\\mathbf{v} = A_{1}\\mathbf{e}_{1}+A_{2}\\mathbf{v}_{2}+A_{3}\\mathbf{v}_{3}$. We call $A_{i}$ the components of $\\mathbf{v}$ in the basis $\\{\\mathbf{e}_{1},\\mathbf{e}_{2},\\mathbf{e}_{3}\\}$. If we change basis to $\\{\\mathbf{f}_{1},\\mathbf{f}_{2},\\mathbf{f}_{3}\\}$ then we get new components because now we have $\\mathbf{v} = B_{1}\\mathbf{f}_{1}+B_{2}\\mathbf{f}_{2}+B_{3}\\mathbf{f}_{3}$. A simple example is that if $\\mathbf{f}_{i} = 2\\mathbf{e}_{i}$ then obviously we have $B_{i} = \\frac{1}{2}A_{i}$ but more complicated combinations are allowed. So how do we know if a set of vectors form a basis? They must be linearly independent, which means you can’t write one in terms of the others. A quick way to check this is to compute the triple vector product, $\\mathbf{e}_{1} \\cdot (\\mathbf{e}_{2} \\times \\mathbf{e}_{3})$. If it is zero then they are not a basis. If it is non-zero then you’re good to go. It’s often convenient to have the vectors be mutually orthogonal, which means they are at right angles to one another so $\\mathbf{e}_{i}\\cdot \\mathbf{e}_{j} = 0$ if $i \\neq j$. All of these things are true for the x,y,z axis directions, $\\mathbf{e}_{x},\\mathbf{e}_{y},\\mathbf{e}_{z}$. If we take $\\mathbf{e}_{1} = \\mathbf{e}_{x}$ etc then the components of", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "indicated calculations; subsequently it can be easier to use the properties instead of calculating with coordinates. Theorem 12.3.5 If $${\\bf u}$$, $${\\bf v}$$, and $${\\bf w}$$ are vectors and $$a$$ is a real number, then 1. $${\\bf u}\\cdot{\\bf u} = |{\\bf u}|^2$$ 2. $${\\bf u}\\cdot{\\bf v} = {\\bf v}\\cdot{\\bf u}$$ 3. $${\\bf u}\\cdot({\\bf v}+{\\bf w}) = {\\bf u}\\cdot{\\bf v}+{\\bf u}\\cdot{\\bf w}$$ 4. $$(a{\\bf u})\\cdot{\\bf v}=a({\\bf u}\\cdot{\\bf v}) ={\\bf u}\\cdot(a{\\bf v})$$", "all linear combinations of linear combinations of $$\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}$$ is called the linear span of S and is denoted by $$L(S)$$, namely $L(S)=\\{c_{1}\\mathbf{v}_{1}+\\cdots+c_{k}\\mathbf{v}_{k}|\\ c_{1}\\in\\mathbb{R},\\cdots,c_{k}\\in\\mathbb{R}\\}.$ Let $$S=\\{\\mathbf{v}\\}$$, and $$\\mathbf{v}\\neq O$$ is a vector in $$\\mathbb{R}^{3}$$. Then $$L(S)=\\{c\\mathbf{v}|\\ c\\in\\mathbb{R}\\}$$ is the line passing through the origin in the direction of $$\\mathbf{v}.$$ Let u and v be two non-parallel vectors in space. Then $$L(S)=\\{c_{1}\\mathbf{u}+c_{2}\\mathbf{v}|\\ c_{1}\\in\\mathbb{R},\\ c_{2}\\in\\mathbb{R}\\}$$ is the plane passing through the origin generated by u and v. Every set $$S=\\{\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}\\}$$ spans the zero vector. Or $$\\mathbf{O}\\in L(S)$$. Def: A set S = {A,, . . . ,Ak} of vectors in , is said to span X uniquely if it spans X and if Theorem: A set S spans every vector in L(S) uniquely if and only if S spans the zero vector uniquely. ## Linear Independence A set $$S=\\{\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}\\}$$ which spans the zero vector uniquely is said to be a linearly independent set of vectors. Otherwise, S is called linearly dependent. ## Change of Basis Use of the coordinate matrix of a vector $$\\mathbf{v}$$ relative to an ordered basis $$\\mathbb{\\mathcal{B}}$$ $\\mathbf{v}=\\left[\\begin{array}{c} x_{1}\\\\ \\vdots\\\\ x_{n} \\end{array}\\right],$ is often more convenient than writing it as an n–tuple $$(x_{1},\\cdots,x_{n})$$ of coordinates. We often write the coordinate matrix as $\\left[\\mathbf{v}\\right]_{\\mathbb{\\mathcal{B}}}$ to indicate its dependence on the basis. If $$\\mathbb{\\mathcal{B}}$$ is the standard basis, then we", "= 2^k$ vectors $$\\big( \\mathbf{a}^{(j)} , \\mathbf{b}^{(j)} , \\mathbf{G}^{(j)} , \\mathbf{H}^{(j)} \\big)​$$ which is initially $$\\big( \\mathbf{a} , \\mathbf{b} , \\mathbf{G} , \\mathbf{H} \\big) ​$$ However, when $j < k$, the input is updated to $$\\big(\\mathbf{a}^{(j-1)}, \\mathbf{b}^{(j-1)}, \\mathbf{G}^{(j-1)},\\mathbf{H}^{(j-1)} \\big)​$$ quadruple vectors each of size $2^{k-1}$, where $$\\mathbf{a}^{(j-1)} = \\mathbf{a}_{lo} \\cdot u_j + \\mathbf{a}_{hi} \\cdot u_j^{-1} ​\\\\ \\mathbf{b}^{(j-1)} = \\mathbf{b}_{lo} \\cdot u_j^{-1} + \\mathbf{b}_{hi} \\cdot u_j​ \\\\ \\mathbf{G}^{(j-1)} = \\mathbf{G}_{lo} \\cdot u_j^{-1} + \\mathbf{G}_{hi} \\cdot u_j​ \\\\ \\mathbf{H}^{(j-1)} = \\mathbf{H}_{lo} \\cdot u_j + \\mathbf{H}_{hi} \\cdot u_j^{-1} \\\\$$ and $u_k$ is the verifier's challenge. The vectors $$\\mathbf{a}_{lo}, \\mathbf{b}_{lo}, \\mathbf{G}_{lo}, \\mathbf{H}_{lo}​ \\\\ \\mathbf{a}_{hi}, \\mathbf{b}_{hi}, \\mathbf{G}_{hi}, \\mathbf{H}_{hi} ​$$ are the left and the right halves of the vectors $$\\mathbf{a}, \\mathbf{b}, \\mathbf{G}, \\mathbf{H}​$$ respectively. Figure 2: Inner-product Proof - Prover Side Figure 2 is included here not only to display the recursive nature of the IPP, but will also be handy and pivotal in understanding amortization strategies that will be applied to the IPP. ### Inductive Proofs As noted earlier, \"for-loops\" and \"while-loops\" are powerful in efficiently executing repetitive computations. However, they are not sufficient to reach the level of scalability needed in blockchains. The basic reason for this is that the iterative executions compute each instance of the recursive function. This is very expensive and clearly undesirable, because the aim in blockchain", "$\\mathbf{\\vec a}$ and $\\mathbf{\\vec b}$ and we subtract $\\mathbf{\\vec b}$ from $\\mathbf{\\vec a}$, then: $% %]]>$ $% %]]>$ $% %]]>$ $% = \\mathbf{\\mathbf{\\vec b}a} %]]>$ If we subtract $\\mathbf{\\vec b}​$ from $\\mathbf{\\vec a}​$, we get a different result: $% %]]>$ $% = \\mathbf{\\mathbf{\\vec a}b} %]]>$ Note that the vector $\\mathbf{\\vec b} - \\mathbf{\\vec a}$ has the same length as the vector $\\mathbf{\\vec a} - \\mathbf{\\vec b}$, but goes in the opposite direction. In general, if we have two vectors, $\\mathbf{\\vec a}$ and $\\mathbf{\\vec b}$, then $\\mathbf{\\vec a} - \\mathbf{\\vec b}$ is a vector that is calculated as follows: $% %]]>$ $% %]]>$ $% = \\mathbf{\\mathbf{\\vec b}a} %]]>$ Vector subtraction is commonly used to find vectors between points. So if we need to find a vector that goes from the tip point of the position vector $\\mathbf{\\vec b}$ to the tip point of the position vector $\\mathbf{\\vec a}$, then we use vector subtraction $(\\mathbf{\\vec a}-\\mathbf{\\vec b})$ as shown in Figure (11). ### Vector properties There are eight properties of vectors. If a, b, and c are vectors, and s and t are numbers, then: $\\mathbf{\\vec a} + \\mathbf{\\vec b} = \\mathbf{\\vec b} + \\mathbf{\\vec a}$ $\\mathbf{\\vec a} + 0 = a$ $s * (\\mathbf{\\vec a} + \\mathbf{\\vec b}) = s * a + s * \\mathbf{\\vec b}$ $s *", "# Problem: Let vectors: A = (1,0,-3), B = (-2,5,1), and C = (3,1,1)Cross product of two vectors, 2B and 3CCalculate (2B) x (3C) ###### FREE Expert Solution $\\begin{array}{rcl}\\mathbf{2}\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}& \\mathbf{=}& \\mathbf{-}\\mathbf{4}\\mathbf{+}\\mathbf{10}\\mathbf{+}\\mathbf{2}\\\\ \\mathbf{3}\\stackrel{\\mathbf{⇀}}{\\mathbf{C}}& \\mathbf{=}& \\mathbf{9}\\mathbf{+}\\mathbf{3}\\mathbf{+}\\mathbf{3}\\\\ \\mathbf{\\left(}\\mathbf{2}\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}\\mathbf{\\right)}\\mathbf{×}\\mathbf{\\left(}\\mathbf{3}\\stackrel{\\mathbf{⇀}}{\\mathbf{C}}\\mathbf{\\right)}& \\mathbf{=}& \\mathbf{|}\\begin{array}{ccc}\\stackrel{\\mathbf{^}}{\\mathbf{i}}& \\stackrel{\\mathbf{^}}{\\mathbf{j}}& \\stackrel{\\mathbf{^}}{\\mathbf{k}}\\\\ \\mathbf{-}\\mathbf{4}& \\mathbf{10}& \\mathbf{2}\\\\ \\mathbf{9}& \\mathbf{3}& \\mathbf{3}\\end{array}\\mathbf{|}\\end{array}$ ###### Problem Details Let vectors: A = (1,0,-3), B = (-2,5,1), and C = (3,1,1) Cross product of two vectors, 2B and 3C Calculate (2B) x (3C) What scientific concept do you need to know in order to solve this problem? Our tutors have indicated that to solve this problem you will need to apply the Calculating Cross Product Using Components concept. If you need more Calculating Cross Product Using Components practice, you can also practice Calculating Cross Product Using Components practice problems.", "# Tagged: imaginary number ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 316 Let $n$ be an odd positive integer. Determine whether there exists an $n \\times n$ real matrix $A$ such that $A^2+I=O,$ where $I$ is the $n \\times n$ identity matrix and $O$ is the $n \\times n$ zero matrix. If such a matrix $A$ exists, find an example. If not, prove that there is no such $A$. How about when $n$ is an even positive number?", "# Orthogonal Nonzero Vectors Are Linearly Independent ## Problem 591 Let $S=\\{\\mathbf{v}_1, \\mathbf{v}_2, \\dots, \\mathbf{v}_k\\}$ be a set of nonzero vectors in $\\R^n$. Suppose that $S$ is an orthogonal set. (a) Show that $S$ is linearly independent. (b) If $k=n$, then prove that $S$ is a basis for $\\R^n$. ## Proof. ### (a) Show that $S$ is linearly independent. Consider the linear combination $c_1\\mathbf{v}_1+c_2\\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_k=\\mathbf{0}.$ Our goal is to show that $c_1=c_2=\\cdots=c_k=0$. We compute the dot product of $\\mathbf{v}_i$ and the above linear combination for each $i=1, 2, \\dots, k$: \\begin{align*} 0&=\\mathbf{v}_i\\cdot \\mathbf{0}\\\\ &=\\mathbf{v}_i \\cdot (c_1\\mathbf{v}_1+c_2\\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_k)\\\\ &=c_1\\mathbf{v}_i \\cdot \\mathbf{v}_1+c_2\\mathbf{v}_i \\cdot \\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_i \\cdot\\mathbf{v}_k. \\end{align*} As $S$ is an orthogonal set, we have $\\mathbf{v}_i\\cdot \\mathbf{v}_j=0$ if $i\\neq j$. Hence all terms but the $i$-th one are zero, and thus we have $0=c_i\\mathbf{v}_i\\cdot \\mathbf{v}_i=c_i \\|\\mathbf{v}_i\\|^2.$ Since $\\mathbf{v}_i$ is a nonzero vector, its length $\\|\\mathbf{v}_i\\|$ is nonzero. It follows that $c_i=0$. As this computation holds for every $i=1, 2, \\dots, k$, we conclude that $c_1=c_2=\\cdots=c_k=0$. Hence the set $S$ is linearly independent. ### (b) If $k=n$, then prove that $S$ is a basis for $\\R^n$. Suppose that $k=n$. Then by part (a), the set $S$ consists of $n$ linearly independent vectors in the dimension $n$ vector space $\\R^n$. Thus, $S$ is also a spanning set of $\\R^n$," ]

[ "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "+ (-a_1 - b_1i) = (a_1 - a_1) + (b_1 - b_1)i = 0 + 0i = 0$. • FM1: $z_1 \\cdot z_2 = (a_1 + b_1i)(a_2 + b_2i) = a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i$ and so $(z_1 \\cdot z_2) \\in \\mathbb{C}$. • FM2: $z_1 \\cdot z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i = (a_2a_1 - b_2b_1) + (a_2b_1 + b_2a_1)i = z_2 \\cdot z_1$. • FM3: (1) \\begin{align} z_1 \\cdot (z_2 \\cdot z_3) \\\\ = (a_1 + b_1i) [ (a_2a_3 - b_2b_3) + (a_2b_3 + b_2a_3)i ] \\\\ = (a_1a_2a_3 - a_1b_2b_3) + (a_1a_2b_3 + a_1b_2a_3)i + (b_1a_2a_3 - b_1b_2b_3)i + (b_1a_2b_3 + b_1b_2a_3)i^2 \\\\ \\quad = [(a_1a_2a_3 - a_1b_2b_3) - (b_1a_2b_3 + b_1b_2a_3)] + [(a_1a_2b_3 + a_1b_2a_3) + (b_1a_2a_3 - b_1b_2b_3)]i \\\\ = [ (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i ] (a_3 + b_3i) \\\\ = (z_1 \\cdot z_2) \\cdot z_3 \\end{align} • FM4: The multiplicative identity is $1 = 1 + 0i$. That is $z_1 \\cdot 1 = (a_1 + b_1)(1 + 0i) = (a_1 + a_1 \\cdot 0i) + (b_1 + b_1 \\cdot 0)i = a_1 + b_1i = z_1$. • FM5: The multiplicative inverse for $z_1$ where $z_1 \\neq 0$ is $z_1^{-1} = \\frac{a_1 -b_1i}{a_1^2 + b_1^2}$. That is: (2)", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "+ \\mathbf{c}\\times \\mathbf{a} + \\mathbf{a}\\times \\mathbf{b}| }. \\,$ The circumcenter $\\mathbf{O}= \\frac {\\mathbf{a^2}(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b^2}(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c^2}(\\mathbf{a} \\times \\mathbf{b})} {2\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})}. \\,$ The Monge point $\\mathbf{M} = \\frac {\\mathbf{a} \\cdot (\\mathbf{b} + \\mathbf{c})(\\mathbf{b} \\times \\mathbf{c}) + \\mathbf{b}\\cdot (\\mathbf{c} + \\mathbf{a})(\\mathbf{c} \\times \\mathbf{a}) + \\mathbf{c} \\cdot (\\mathbf{a} + \\mathbf{b})(\\mathbf{a} \\times \\mathbf{b})} {2\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})}. \\,$ The Euler line relationships are: $\\mathbf{G} = \\mathbf{M} + \\frac{1}{2} (\\mathbf{O}-\\mathbf{M})\\,$ $\\mathbf{T} = \\mathbf{M} + \\frac{1}{3} (\\mathbf{O}-\\mathbf{M})\\,$ where $\\mathbf{T}\\,$ is twelve-point center. Also: $\\mathbf{a} \\cdot \\mathbf{O} = \\frac {\\mathbf{a^2}}{2} \\quad\\quad \\mathbf{b} \\cdot \\mathbf{O} = \\frac {\\mathbf{b^2}}{2} \\quad\\quad \\mathbf{c} \\cdot \\mathbf{O} = \\frac {\\mathbf{c^2}}{2}\\,$ and: $\\mathbf{a} \\cdot \\mathbf{M} = \\frac {\\mathbf{a} \\cdot (\\mathbf{b} + \\mathbf{c})}{2} \\quad\\quad \\mathbf{b} \\cdot \\mathbf{M} = \\frac {\\mathbf{b} \\cdot (\\mathbf{c} + \\mathbf{a})}{2} \\quad\\quad \\mathbf{c} \\cdot \\mathbf{M} = \\frac {\\mathbf{c} \\cdot (\\mathbf{a} + \\mathbf{b})}{2}.\\,$ ## Geometric relations A tetrahedron is a 3-simplex. Unlike the case of the other Platonic solids, all the vertices of a regular tetrahedron are equidistant from each other (they are the only possible arrangement of four equidistant points in 3-dimensional space). A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual. A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is", "• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a 12-18-2010, 07:33 PM Post: #1 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a (1) $(\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c} = (\\mathbf{a}\\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot \\mathbf{c}) \\mathbf{a}$ is actually equivalent to (2) $(\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d})$ Proof: By the identity $(\\mathbf{u} \\times \\mathbf{v})\\cdot \\mathbf{w} = \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})$, dot product with $\\mathbf{d}$ both sides of (1) we get (2). Conversely, from (2) we have $((\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c})\\cdot \\mathbf{d} = (\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot\\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a}\\cdot \\mathbf{d}) = ((\\mathbf{a} \\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot\\mathbf{c}) \\mathbf{a}) \\cdot \\mathbf{d}$. Since this is true for all $\\mathbf{d}$, (1) follows. Now we prove (2). Let $(a_1,a_2,a_3) = \\mathbf{a}$ and likewise for $\\mathbf{b}$ and $\\mathbf{c}$, then $(\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot \\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d}) =$ $\\displaystyle{\\sum_{1\\le i,j \\le 3} (a_i c_i b_j d_j - b_i c_i a_j d_j) = \\sum_{1\\le i<j \\le 3} ((a_i c_i b_j d_j + a_j c_j b_i d_i)-(b_i c_i a_j d_j + b_j c_j a_i d_i))=}$ $\\displaystyle{\\sum_{1\\le i<j \\le 3} (a_i b_j(c_i d_j-c_j d_i)-a_j b_i (c_i d_j - c_j", "of this is 2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3, so you normally you would form the array (\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}), which has length (2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32. However, writing the ternary form as 2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2)), the array would be (\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})], which only has length (2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most 94, and we’ve now solved the problem! The ideas above are enough to pass the problem already, but I’m sure with a couple more tricks you can reduce the length even more. ### Time Complexity: [O(N)](https://en.wikipedia.org/wiki/Output-sensitive _ algorithm) ### AUTHOR’S AND TESTER’S SOLUTIONS: #2 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then create p pairs of numbers so", "B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2 B \\right) \\\\ &\\qquad - \\mathbf{x}_2 \\left( \\mathbf{x}_{34} \\cdot \\partial_1 B \\right) - \\mathbf{x}_3 \\left( \\mathbf{x}_{42} \\cdot \\partial_1 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{23} \\cdot \\partial_1 B \\right)} \\Bigr) \\\\", "\\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2", "}}{\\partial t}}\\end{aligned}}} • {\\displaystyle {\\begin{aligned}{\\mathbf {E} }\\cdot {\\mathbf {J} }&={\\mathbf {E} }\\cdot \\left[{\\frac {1}{\\mu _{0}}}(\\nabla \\times {\\mathbf {B} })-\\epsilon _{0}{\\frac {\\partial {\\mathbf {E} }}{\\partial t}}\\right]\\\\&={\\frac {1}{\\mu _{0}}}({\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} }))-\\epsilon _{0}{\\mathbf {E} }\\cdot {\\frac {\\partial {\\mathbf {E} }}{\\partial t}}\\end{aligned}}} 5. 5 Use a vector calculus identity to write an expression with a term where both fields are in the cross product. This apparent complication is solved when we invoke Faraday’s law ${\\displaystyle \\nabla \\times {\\mathbf {E} }=-{\\frac {\\partial {\\mathbf {B} }}{\\partial t}}.}$ • ${\\displaystyle \\nabla \\cdot ({\\mathbf {E} }\\times {\\mathbf {B} })={\\mathbf {B} }\\cdot (\\nabla \\times {\\mathbf {E} })-{\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} })}$ • ${\\displaystyle {\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} })=-{\\mathbf {B} }\\cdot {\\frac {\\partial {\\mathbf {B} }}{\\partial t}}-\\nabla \\cdot ({\\mathbf {E} }\\times {\\mathbf {B} })}$ 6. 6 Simplify the expressions containing the dot product of a field with its time derivative. The extra ½ factor stems from the recognition of the product rule. Confirm this by taking the derivative of ${\\displaystyle {\\mathbf {B} }\\cdot {\\mathbf {B} }.}$ • ${\\displaystyle {\\mathbf {B} }\\cdot {\\frac {\\partial {\\mathbf {B} }}{\\partial t}}={\\frac {\\partial }{\\partial t}}\\left({\\frac {1}{2}}B^{2}\\right)}$ • ${\\displaystyle {\\mathbf {E} }\\cdot {\\frac {\\partial {\\mathbf {E} }}{\\partial t}}={\\frac {\\partial }{\\partial t}}\\left({\\frac {1}{2}}E^{2}\\right)}$ 7. 7 Substitute into the equation in step 4. Rewrite the", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "length $||\\mathbf{v}||\\neq 0$. Thus, we must have $c_1=0$. We repeat the above argument with $\\mathbf{v}_1$ replaced by $\\mathbf{v}_2$ as follows. Since $c_1=0$, the (*) becomes $c_2\\mathbf{v}_2+c_3\\mathbf{v}_3=\\mathbf{0}.$ We have \\begin{align*} \\mathbf{0}&=\\mathbf{v}_2\\cdot \\mathbf{0}=\\mathbf{v}_2\\cdot (c_2\\mathbf{v}_2+c_3\\mathbf{v}_3)\\\\ &=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2+c_3\\mathbf{v}_2\\cdot\\mathbf{v}_3. \\end{align*} Since $\\mathbf{v}_2\\cdot\\mathbf{v}_3=0$, we have $\\mathbf{0}=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2=c_2||\\mathbf{v}_2||^2.$ Since $\\mathbf{x}\\neq \\mathbf{0}$, we have $||\\mathbf{v}_2||\\neq 0$, and thus $c_2=0$. Hence (*) is now just $c_3\\mathbf{v}_3=\\mathbf{0}$, and since $\\mathbf{x}_3\\neq \\mathbf{0}$, we conclude that $c_3=0$. Therefore, we have shown that $c_1=c_2=c_3=0$, and thus the set $\\{\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3\\}$ is linearly independent. ## Comment. These are Quiz 3 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. ### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017 There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. ### 3 Responses 1. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 2. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 3. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] ##### Find a Nonsingular Matrix Satisfying Some Relation Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A,$ where $B$ is the following matrix. If such a nonsingular... Close", "= \\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}).$$ From (13) and (14), \\begin{align*} \\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}) &= \\mathbf{E}(X^{2}) - 2 \\cdot \\mu \\cdot \\mathbf{E}(X) + \\mu^{2} \\\\ &= \\mathbf{E}(X^{2}) - \\mu^{2} \\\\ &= \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}. \\end{align*} $$\\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}) = \\mathbf{E}(X^{2}) - 2 \\cdot \\mu \\cdot \\mathbf{E}(X) + \\mu^{2} = \\mathbf{E}(X^{2}) - \\mu^{2} = \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}.$$ For the discrete or continuous random variable $$X$$ and the constant $$c \\in \\real$$, $$\\text{Var}(cX) = c^{2} \\cdot \\text{Var}(X). \\tag{17}$$ The proof of (17) holds for $$X$$ discrete or continuous. Again, let $$\\mu = \\mathbf{E}(X)$$. Then, $$\\text{Var}(cX) = \\mathbf{E}((cX - c\\mu)^2) = \\mathbf{E}(c^{2}\\cdot(X - \\mu)^{2}) = c^{2} \\cdot \\mathbf{E}((X - \\mu)^2) = c^{2} \\cdot \\text{Var}(X)$$. Finally, for discrete or continuous independent random variables $$X$$ and $$Y$$, $$\\text{Var}(X + Y) = \\text{Var}(X) + \\text{Var}(Y). \\tag{18}$$ $$\\text{Var}(X + Y) = \\text{Var}(X) + \\text{Var}(Y).$$ $$(18)$$ The proof of (18) holds for $$X$$ and $$Y$$ discrete or continuous. As a shorthand notation, let $$\\mu_{X} = \\mathbf{E}(X)$$ and $$\\mu_{Y} = \\mathbf{E}(Y)$$. Then, \\begin{align*} \\text{Var}(X + Y) &= \\mathbf{E}\\Big\\{\\big((X + Y) - (\\mu_{X} + \\mu_{Y})\\big)^{2}\\Big\\} \\\\ &= \\mathbf{E}\\Big\\{\\big((X - \\mu_{X}) + (Y - \\mu_{Y})\\big)^{2}\\Big\\} \\\\ &= \\mathbf{E}\\Big\\{ (X - \\mu_{X})^{2} + 2(X - \\mu_{X}) \\cdot (Y - \\mu_{Y}) + (Y - \\mu_{Y})^{2}", "\\mathbf{c} ) + ( \\mathbf{d} \\cdot \\mathbf{a} ) ( \\mathbf{c} \\cdot \\mathbf{b} ) - ( \\mathbf{c} \\cdot \\mathbf{a} ) ( \\mathbf{d} \\cdot \\mathbf{b} ) ) \\\\ &= ( \\mathbf{a} \\cdot \\mathbf{d} ) ( \\mathbf{b} \\cdot \\mathbf{c} ) - ( \\mathbf{a} \\cdot \\mathbf{c} ) ( \\mathbf{b} \\cdot \\mathbf{d} ) \\\\ \\end{aligned} Summarizing with a comparison to the $\\mathbb{R}^{3}$ relations we have: \\begin{aligned}(\\mathbf{a} \\wedge \\mathbf{b}) \\cdot (\\mathbf{c} \\wedge \\mathbf{d}) = -(\\mathbf{a} \\times \\mathbf{b}) \\cdot (\\mathbf{c} \\times \\mathbf{d}) = ( \\mathbf{a} \\cdot \\mathbf{d} ) ( \\mathbf{b} \\cdot \\mathbf{c} ) - ( \\mathbf{a} \\cdot \\mathbf{c} ) ( \\mathbf{b} \\cdot \\mathbf{d} )\\end{aligned} \\quad\\quad\\quad(25) \\begin{aligned}(\\mathbf{a} \\wedge \\mathbf{c}) \\cdot (\\mathbf{b} \\wedge \\mathbf{c}) = -(\\mathbf{a} \\times \\mathbf{c}) \\cdot (\\mathbf{b} \\times \\mathbf{c}) = ( \\mathbf{a} \\cdot \\mathbf{c} ) ( \\mathbf{b} \\cdot \\mathbf{c} ) - \\mathbf{c}^2 ( \\mathbf{a} \\cdot \\mathbf{b} )\\end{aligned} \\quad\\quad\\quad(26) The bivector relations hold for all of $\\mathbb{R}^{N}$.", "{ \\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot" ]

[ "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "시간 제한메모리 제한제출정답맞힌 사람정답 비율 12 초 (추가 시간 없음) 1024 MB0000.000% ## 문제 This problem might be well-known in some countries, but how do other countries learn about such problems if nobody poses them. There are $n$ points on the plane, where the $i$-th point $(x_i,y_i)$ has value $\\mathbf d_i \\in D$. Two sets $D$ and $O$ are given, with the following properties: • There exists a special element $\\varepsilon_D$ in $D$. • There exists a special element $\\varepsilon_O$ in $O$. • A binary operation $+: D \\times D \\to D$ is given with the following properties: • $\\forall \\mathbf a,\\mathbf b \\in D$, $\\mathbf a+\\mathbf b = \\mathbf b+\\mathbf a$ • $\\forall \\mathbf a,\\mathbf b,\\mathbf c \\in D$, $(\\mathbf a+\\mathbf b)+\\mathbf c = \\mathbf a+(\\mathbf b+\\mathbf c)$ • $\\forall \\mathbf x \\in D$, $\\mathbf x + \\varepsilon_D = \\varepsilon_D + \\mathbf x = \\mathbf x$ • A binary operation $\\cdot: O \\times D \\to D$ is given with the following properties: • $\\forall \\mathbf a,\\mathbf b \\in O, \\mathbf x \\in D$, $(\\mathbf a \\cdot \\mathbf b) \\cdot \\mathbf x = \\mathbf a \\cdot (\\mathbf b \\cdot \\mathbf x)$ • $\\forall \\mathbf a \\in O, \\mathbf x,\\mathbf y \\in D$, $\\mathbf a \\cdot (\\mathbf x + \\mathbf y) = \\mathbf a \\cdot \\mathbf x + \\mathbf", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a 12-18-2010, 07:33 PM Post: #1 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a (1) $(\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c} = (\\mathbf{a}\\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot \\mathbf{c}) \\mathbf{a}$ is actually equivalent to (2) $(\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d})$ Proof: By the identity $(\\mathbf{u} \\times \\mathbf{v})\\cdot \\mathbf{w} = \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})$, dot product with $\\mathbf{d}$ both sides of (1) we get (2). Conversely, from (2) we have $((\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c})\\cdot \\mathbf{d} = (\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot\\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a}\\cdot \\mathbf{d}) = ((\\mathbf{a} \\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot\\mathbf{c}) \\mathbf{a}) \\cdot \\mathbf{d}$. Since this is true for all $\\mathbf{d}$, (1) follows. Now we prove (2). Let $(a_1,a_2,a_3) = \\mathbf{a}$ and likewise for $\\mathbf{b}$ and $\\mathbf{c}$, then $(\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot \\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d}) =$ $\\displaystyle{\\sum_{1\\le i,j \\le 3} (a_i c_i b_j d_j - b_i c_i a_j d_j) = \\sum_{1\\le i<j \\le 3} ((a_i c_i b_j d_j + a_j c_j b_i d_i)-(b_i c_i a_j d_j + b_j c_j a_i d_i))=}$ $\\displaystyle{\\sum_{1\\le i<j \\le 3} (a_i b_j(c_i d_j-c_j d_i)-a_j b_i (c_i d_j - c_j", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "B , \\quad \\mathbf E' = \\mathbf E + \\frac{1}{c}[\\mathbf u \\times \\mathbf B ].$$ By substitution these equations to $(.1)$ you will get $$(\\nabla \\cdot \\mathbf B)= 0 , \\quad [\\nabla \\times \\mathbf E] + \\frac{1}{c}[\\nabla \\times [\\mathbf u \\times \\mathbf B]] + \\frac{1}{c}\\partial_{t}\\mathbf B + \\frac{1}{c}(\\mathbf u \\cdot \\nabla) \\mathbf B = [\\nabla \\times \\mathbf E] + \\frac{1}{c}\\partial_{t}\\mathbf B = 0,$$ because for $\\mathbf u = const$ $$[\\nabla \\times [\\mathbf u \\times \\mathbf B]] = \\mathbf u (\\nabla \\cdot \\mathbf B) - (\\mathbf u \\cdot \\nabla )\\mathbf B = - (\\mathbf u \\cdot \\nabla )\\mathbf B .$$ So the first pair of Maxwell's equations is clearly invariant under galilean transformations. Let's look to the other pair of Maxwell's equations: $$[\\nabla \\times \\mathbf B] - \\frac{1}{c}\\partial_{t}\\mathbf E = 0 , \\quad (\\nabla \\cdot \\mathbf E ) = 0 . \\qquad (.3)$$ By using an expressions which were derived above, you can rewrite $(.3)$ as $$[\\nabla \\times \\mathbf B] - \\frac{1}{c}\\partial_{t}\\mathbf E ' - \\frac{1}{c}(\\mathbf u \\cdot \\nabla)\\mathbf E' =$$ $$=[\\nabla \\times \\mathbf B] - \\frac{1}{c}\\partial_{t}\\mathbf E - \\frac{1}{c}(\\mathbf u \\cdot \\nabla)\\mathbf E - \\frac{1}{c^{2}}\\partial_{t}[\\mathbf u \\times \\mathbf B] - \\frac{1}{c^{2}}(\\mathbf u \\cdot \\nabla)[\\mathbf u \\times \\mathbf B]= 0,$$ $$(\\nabla \\cdot \\mathbf E ) + \\frac{1}{c}(\\nabla \\cdot [\\mathbf u \\times \\mathbf B]) = (\\nabla \\cdot \\mathbf E )", "a \\to b, b \\to c, c \\to a$$ This gives $${\\bf A} = \\frac{{\\bf b}\\times {\\bf c}}{{\\bf a}\\cdot({\\bf b}\\times {\\bf c})} \\quad {\\bf B} = \\frac{{\\bf c}\\times {\\bf a}}{{\\bf b}\\cdot({\\bf c}\\times {\\bf a})} \\quad {\\bf C} = \\frac{{\\bf a}\\times {\\bf b}}{{\\bf c}\\cdot({\\bf a}\\times {\\bf b})}$$ and $${\\bf a} = \\frac{{\\bf B}\\times {\\bf C}}{{\\bf A}\\cdot({\\bf B}\\times {\\bf C})} \\quad {\\bf b} = \\frac{{\\bf C}\\times {\\bf A}}{{\\bf B}\\cdot({\\bf C}\\times {\\bf A})} \\quad {\\bf c} = \\frac{{\\bf A}\\times {\\bf B}}{{\\bf C}\\cdot({\\bf A}\\times {\\bf B})}$$ [..] the \"crystallographer's\" definition, comes from defining the reciprocal lattice to be $e^{2 \\pi i\\mathbf{K}\\cdot\\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be $\\mathbf{b_{1}}=\\frac{\\mathbf{a_{2}} \\times \\mathbf{a_{3}}}{\\mathbf{a_{1}} \\cdot (\\mathbf{a_{2}} \\times \\mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\\mathbf{b_{1}}$ is just the reciprocal magnitude of $\\mathbf{a_{1}}$ in the direction of $\\mathbf{a_{2}} \\times \\mathbf{a_{3}}$, dropping the factor of $2 \\pi$. Checking the magnitude: $$\\lVert{\\bf A}\\rVert = \\frac{\\lVert {\\bf b} \\times{\\bf c}\\rVert}{\\lVert{\\bf a}\\rVert\\lVert{\\bf b}\\times {\\bf c}\\rVert\\cos\\angle({\\bf a}, {\\bf b} \\times{\\bf c})} = \\frac{1}{\\lVert{\\bf a}\\rVert ({\\bf e_a} \\cdot {\\bf e}_{{\\bf b} \\times{\\bf c}})}$$ Solving the question: Using the \"bac-cab\" rule ${\\bf a}\\times({\\bf b}\\times{\\bf c}) = {\\bf b}({\\bf a}\\cdot {\\bf c}) - {\\bf c}({\\bf a}\\cdot {\\bf b})$ we go for the nominator of ${\\bf B}\\times{\\bf C}$: $$({\\bf c}\\times", "+ \\mathbf{b} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert ^ 2 - \\lvert \\mathbf{b} \\rvert ^2 \\end{array}$ 2. $\\begin{array}{ll} \\lvert a + b \\rvert ^ 2 &= (\\mathbf{a} + \\mathbf{b}) \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot (\\mathbf{a} + \\mathbf{b}) + \\mathbf{b} \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{b} + \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert^2 + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\lvert \\mathbf{b} \\rvert ^ 2 \\\\ \\end{array}$ Exercise 16 1. We know that $$\\lvert \\mathbf{a} \\rvert = \\lvert \\mathbf{b} \\rvert = 1$$ and $$\\mathbf{a} \\cdot \\mathbf{b} = 0$$. If the two vectors are perpendicular we must have $(2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) = 0$ We can rewrite this as $\\begin{array}{ll} 0 &= (2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot (m\\mathbf{a} + \\mathbf{b}) + 3\\mathbf{b} \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot m\\mathbf{a} + 2\\mathbf{a} \\cdot \\mathbf{b} + 3\\mathbf{b} \\cdot m\\mathbf{a} + 3\\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert 2\\mathbf{a} \\rvert \\lvert m\\mathbf{a} \\rvert + \\lvert 3\\mathbf{b} \\rvert \\lvert \\mathbf{b} \\rvert \\\\ &= 2m + 3 \\end{array}$ Therefore $$m = -\\frac{3}{2}$$", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "and get rid of the Levi-Civita symbol: $$2\\int_{\\mathbb{R}^3} x a_y \\,dxdydz = \\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_z \\,dxdydz - \\int_{\\mathbb{R}^3} xy\\nabla \\cdot \\mathbf{a}\\,dxdydz, \\\\ 2\\int_{\\mathbb{R}^3} y a_x \\,dxdydz = -\\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_z \\,dxdydz - \\int_{\\mathbb{R}^3} yx\\nabla \\cdot \\mathbf{a}\\,dxdydz, \\\\ 2\\int_{\\mathbb{R}^3} y a_z \\,dxdydz = \\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_x \\,dxdydz - \\int_{\\mathbb{R}^3} yz\\nabla \\cdot \\mathbf{a}\\,dxdydz, \\\\ 2\\int_{\\mathbb{R}^3} z a_y \\,dxdydz = -\\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_x \\,dxdydz - \\int_{\\mathbb{R}^3} zy\\nabla \\cdot \\mathbf{a}\\,dxdydz, \\\\ 2\\int_{\\mathbb{R}^3} z a_x \\,dxdydz = \\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_x \\,dxdydz - \\int_{\\mathbb{R}^3} zx\\nabla \\cdot \\mathbf{a}\\,dxdydz, \\\\ 2\\int_{\\mathbb{R}^3} x a_z \\,dxdydz = -\\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_x \\,dxdydz - \\int_{\\mathbb{R}^3} xz\\nabla \\cdot \\mathbf{a}\\,dxdydz.$$ Here I will check if the first one is correct. Others can be checked in the exactly same fashion. Now $$\\mathbf{r}\\times \\mathbf{a} = (ya_z - za_y, za_x - xa_z, xa_y-ya_x).$$ The right hand side of the first identity is: $$\\text{R.H.S}= \\int_{\\mathbb{R}^3} (\\mathbf{r}\\times \\mathbf{a})_z \\,dxdydz - \\int_{\\mathbb{R}^3} xy\\nabla \\cdot \\mathbf{a}\\,dxdydz \\\\ = \\int_{\\mathbb{R}^3} (xa_y-ya_x) \\,dxdydz + \\int_{\\mathbb{R}^3} \\nabla(xy)\\cdot \\mathbf{a}\\,dxdydz,$$ in which we assume the decaying condition of $|\\mathbf{a}| = o(r^{-4})$ where $r = |\\mathbf{r}|$, this assumption just guarantees that if we are integrating on a really large ball in $\\mathbb{R}^3$, the boundary integral term produced by Gauss-Divergence theorem vanishes when the radius of the ball goes to infinity. Expanding we have: $$\\text{R.H.S} = \\int_{\\mathbb{R}^3} (xa_y-ya_x) \\,dxdydz + \\int_{\\mathbb{R}^3} (y,x,0)\\cdot \\mathbf{a}\\,dxdydz \\\\ =\\int_{\\mathbb{R}^3} (xa_y-ya_x)", "of this is 2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3, so you normally you would form the array (\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}), which has length (2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32. However, writing the ternary form as 2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2)), the array would be (\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})], which only has length (2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most 94, and we’ve now solved the problem! The ideas above are enough to pass the problem already, but I’m sure with a couple more tricks you can reduce the length even more. ### Time Complexity: [O(N)](https://en.wikipedia.org/wiki/Output-sensitive _ algorithm) ### AUTHOR’S AND TESTER’S SOLUTIONS: #2 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then create p pairs of numbers so", "$\\textbf{Ring}$ $\\text{Let a, b, c } \\in \\mathbf{R}$ $\\text{There are two binary operations: addition and multiplication. They satisfy}$ $\\text{Associative Law}$ $a \\times b \\times c = a \\times (b \\times c)$ $\\text{Distritutive Law}$ $a \\times (b + c) = a \\times b + a \\times c$ $\\text{Additive inverse}$ $\\text{For all a in \\mathbf{R}, there exists b such that}$ $a + b = 0$ $\\text{Multiplicative identity}$ $\\forall a \\in \\mathbb{R}, \\quad \\exists 1 \\in \\mathbb{R}\\text{ such as }$ $1 \\times a = a \\times 1$ $\\textbf{Ideal}$ Let $(\\mathbf{I}, +)$ is the subgroup of $(\\mathbf{R}, +)$ Let $a \\in \\mathbf{I}$ and $r \\in \\mathbf{R}$, then $a \\cdot r, r \\cdot a \\in \\mathbf{I}$, $\\mathbf{I}$ is called ideal of $\\mathbf{R}$ For example: $2\\mathbb{Z}$ is the ideal of $\\mathbb{Z}$ Proof: $\\text{let } a \\in \\mathbb{Z}, r = 2k \\in 2\\mathbb{Z} \\quad \\text{ where } a, k \\in \\mathbb{Z} \\\\ a \\cdot r, r \\cdot a = 2a \\cdot k \\in 2\\mathbb{Z} \\\\ \\implies 2\\mathbb{Z} \\text{ is ideal of } \\mathbb{Z} \\\\$ Following screenshot can represent the ring $\\mathbf{R} = (\\mathbb{Z}, \\mathbb{Z})$ and and ideal $\\mathbf{I} = (2\\mathbb{Z}, 2\\mathbb{Z})$ blue dots Integral Domain Integral domain is commutative \\textbb{Ring} $R$, if $a, b \\in R$ and $a, b \\neq 0$ then $ab \\neq 0$ Euclidean Domain An integral domain is called Euclidean if there", "length $||\\mathbf{v}||\\neq 0$. Thus, we must have $c_1=0$. We repeat the above argument with $\\mathbf{v}_1$ replaced by $\\mathbf{v}_2$ as follows. Since $c_1=0$, the (*) becomes $c_2\\mathbf{v}_2+c_3\\mathbf{v}_3=\\mathbf{0}.$ We have \\begin{align*} \\mathbf{0}&=\\mathbf{v}_2\\cdot \\mathbf{0}=\\mathbf{v}_2\\cdot (c_2\\mathbf{v}_2+c_3\\mathbf{v}_3)\\\\ &=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2+c_3\\mathbf{v}_2\\cdot\\mathbf{v}_3. \\end{align*} Since $\\mathbf{v}_2\\cdot\\mathbf{v}_3=0$, we have $\\mathbf{0}=c_2\\mathbf{v}_2\\cdot\\mathbf{v}_2=c_2||\\mathbf{v}_2||^2.$ Since $\\mathbf{x}\\neq \\mathbf{0}$, we have $||\\mathbf{v}_2||\\neq 0$, and thus $c_2=0$. Hence (*) is now just $c_3\\mathbf{v}_3=\\mathbf{0}$, and since $\\mathbf{x}_3\\neq \\mathbf{0}$, we conclude that $c_3=0$. Therefore, we have shown that $c_1=c_2=c_3=0$, and thus the set $\\{\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3\\}$ is linearly independent. ## Comment. These are Quiz 3 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. ### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017 There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. ### 3 Responses 1. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 2. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] 3. 04/21/2017 […] Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent […] ##### Find a Nonsingular Matrix Satisfying Some Relation Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A,$ where $B$ is the following matrix. If such a nonsingular... Close", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "on del that do not apply. $$\\mathbf{A_1} \\times (\\nabla_2 \\times \\mathbf{A_2}) =\\nabla_2(\\mathbf{A_1}\\cdot\\mathbf{A_2}) - (\\mathbf{A_1} \\cdot \\nabla_2)\\mathbf{A_2}$$ is true and makes what is happening clear. A1 and A2 are equal, but nabla2 acts only on A2. The grad(A^2) part is the place where the distinction matters. In order to switch back to more standard notation we note that $$\\nabla_1(\\mathbf{A_1}\\cdot\\mathbf{A_2})=\\nabla_2(\\mathbf{A_1}\\cdot\\mathbf{A_2})$$ and $$\\nabla(\\mathbf{A}\\cdot\\mathbf{A})=\\nabla_1(\\mathbf{A_1}\\cdot\\mathbf{A_2})+\\nabla_2(\\mathbf{A_1}\\cdot\\mathbf{A_2})$$ You infact did not use B(A.C)-C(A.B), you used B(A.C)-(A.B)C so you saw that not all vector idenetities hold if a vector is replaced by del, but you still used an identity that was not true. for comparison $$(\\mathbf{A_1} \\times \\nabla_2) \\times \\mathbf{A_2} =\\nabla_2(\\mathbf{A_1}\\cdot\\mathbf{A_2}) - \\mathbf{A_1}(\\nabla_2\\cdot\\mathbf{A_2})$$ thus $$(\\mathbf{A} \\times \\nabla) \\times \\mathbf{A}= \\frac{1}{2} \\nabla(A^2) - \\mathbf{A}(\\nabla\\cdot\\mathbf{A})$$ ah! thank you very much. that is exactly what i needed. thanks a lot. found another way to do the problem (without invoking the scary kronecker deltas and epsilon tensors!). this is the proof suggested by david griffiths' intro to electrodynamics: $$\\nabla (\\mathbf{A} \\cdot \\mathbf{B}) = \\mathbf{A} \\times (\\nabla \\times \\mathbf{B}) + \\mathbf{B} \\times (\\nabla \\times \\mathbf{A}) + (\\mathbf{A} \\cdot \\nabla)\\mathbf{B} + (\\mathbf{B} \\cdot \\nabla)\\mathbf{A}.$$ when $$\\mathbf{A} = \\mathbf{B},$$ $$\\nabla (\\mathbf{A} \\cdot \\mathbf{A}) = 2\\mathbf{A} \\times (\\nabla \\times \\mathbf{A}) + 2(\\mathbf{A} \\cdot \\nabla)\\mathbf{A}.$$ note: $$\\mathbf{A} \\cdot \\mathbf{A} = A^2.$$ so $$\\mathbf{A} \\times (\\nabla \\times \\mathbf{A}) = \\frac{1}{2} \\nabla(A^2) - (\\mathbf{A} \\cdot \\nabla)\\mathbf{A}.$$ Last", "apply our vector notation then we have: $\\textbf{a}_1 = (a_{11}, a_{12}, ..., a_{1M})$ $\\textbf{a}_2 = (a_{21}, a_{22}, ..., a_{2M})$ $\\textbf{a}_N = (a_{N1}, a_{N2}, ..., a_{NM})$ $\\textbf{b}_1 = (b_{11}, b_{12}, ..., b_{1M})$ $\\textbf{b}_2 = (b_{21}, b_{22}, ..., b_{2M})$ $\\textbf{b}_N = (b_{N1}, b_{N2}, ..., b_{NM})$ $\\textbf{c}_1 = (c_{11}, c_{12}, ..., c_{1M})$ $\\textbf{c}_2 = (c_{21}, c_{22}, ..., c_{2M})$ $\\textbf{c}_N = (c_{N1}, c_{N2}, ..., c_{NM})$ $\\textbf{d}_1 = (d_{11}, d_{12}, ..., d_{1M})$ $\\textbf{d}_2 = (d_{21}, d_{22}, ..., d_{2M})$ $\\textbf{d}_N = (d_{N1}, d_{N2}, ..., d_{NM})$ And the re-written equations: $\\textbf{a}_1 = 3 \\cdot \\textbf{b}_1 + 4 \\cdot \\textbf{c}_1 + 5 \\cdot \\textbf{d}_1$ $\\textbf{a}_2 = 1 \\cdot \\textbf{b}_2 + 5 \\cdot \\textbf{c}_2 + 4 \\cdot \\textbf{d}_2$ $\\textbf{a}_N = 7 \\cdot \\textbf{b}_N + 3 \\cdot \\textbf{c}_N + 9 \\cdot \\textbf{d}_N$ So we replaced a long set of equations, with a shorter set of equations. But again if we start performing proofs on such equations, we would end up repeating everything as in the first example. We can also think about another way of representing the same set of equations. Instead of grouping all $a$‘s and all $b$‘s together, we could use following grouping: $\\textbf{p}_1 = (3, 4, 5)$ $\\textbf{p}_2 = (1, 5, 4)$ $...$ $\\textbf{p}_N = (7, 3, 9)$ $\\textbf{q}_{11} = (b_{11}, c_{11}, d_{11})$ $\\textbf{q}_{12} = (b_{12}, c_{12}, d_{12})$ $...$ $\\textbf{q}_{NM} = (b_{NM}, c_{NM}, d_{NM})$ Our", "\\textbf{b}_1 + 4 \\cdot \\textbf{c}_1 + 5 \\cdot \\textbf{d}_1$ $\\textbf{a}_2 = 1 \\cdot \\textbf{b}_2 + 5 \\cdot \\textbf{c}_2 + 4 \\cdot \\textbf{d}_2$ $\\textbf{a}_N = 7 \\cdot \\textbf{b}_N + 3 \\cdot \\textbf{c}_N + 9 \\cdot \\textbf{d}_N$ So we replaced a long set of equations, with a shorter set of equations. But again if we start performing proofs on such equations, we would end up repeating everything as in the first example. We can also think about another way of representing the same set of equations. Instead of grouping all $a$‘s and all $b$‘s together, we could use following grouping: $\\textbf{p}_1 = (3, 4, 5)$ $\\textbf{p}_2 = (1, 5, 4)$ $...$ $\\textbf{p}_N = (7, 3, 9)$ $\\textbf{q}_{11} = (b_{11}, c_{11}, d_{11})$ $\\textbf{q}_{12} = (b_{12}, c_{12}, d_{12})$ $...$ $\\textbf{q}_{NM} = (b_{NM}, c_{NM}, d_{NM})$ Our second vector form becomes then: $a_{11} = \\textbf{p}_1 \\cdot \\textbf{q}_{11}$ $a_{12} = \\textbf{p}_1 \\cdot \\textbf{q}_{12}$ $...$ $a_{1M} = \\textbf{p}_1 \\cdot \\textbf{q}_{1M}$ $a_{21} = \\textbf{p}_2 \\cdot \\textbf{q}_{21}$ $a_{22} = \\textbf{p}_2 \\cdot \\textbf{q}_{22}$ $...$ $a_{2M} = \\textbf{p}_2 \\cdot \\textbf{q}_{2M}$ $...$ $a_{N1} = \\textbf{p}_N \\cdot \\textbf{q}_{N1}$ $a_{N2} = \\textbf{p}_N \\cdot \\textbf{q}_{N2}$ $...$ $a_{NM} = \\textbf{p}_N \\cdot \\textbf{q}_{NM}$ While in this case we didn’t decrease the number of equations, we made each of them much simpler. As for the previous transformation, we could vectorize again and get yet simpler notation for", "Find $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ for $$\\textbf{u} = (1, 2, 4)$$, $$\\textbf{v} = (2, 2, 0)$$, $$\\textbf{w} = (1, 3, 0)$$. Solution Since $$\\textbf{u} \\cdot \\textbf{v} = 6$$ and $$\\textbf{u} \\cdot \\textbf{w} = 7$$, then \\begin{align*} \\textbf{u} \\times (\\textbf{v} \\times \\textbf{w}) &= (\\textbf{u} \\cdot \\textbf{w})\\textbf{v} - (\\textbf{u} \\cdot \\textbf{v})\\textbf{w}\\\\ &= 7\\,(2, 2, 0) - 6\\,(1, 3, 0) = (14, 14, 0) - (6, 18, 0)\\\\ &= (8, -4, 0) \\end{align*} Note that $$\\textbf{v}$$ and $$\\textbf{w}$$ lie in the $$xy$$-plane, and that $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ also lies in that plane. Also, $$\\textbf{u} \\times (\\textbf{v} \\times \\textbf{w})$$ is perpendicular to both $$\\textbf{u}$$ and $$\\textbf{v} \\times \\textbf{w} = (0, 0, 4)$$ (see Figure 1.4.8). Figure 1.4.8 For vectors $$\\textbf{v} = v_{1}\\textbf{i} + v_{2}\\textbf{j} + v_{3}\\textbf{k}$$ and $$\\textbf{w} = w_{1}\\textbf{i} + w_{2}\\textbf{j} + w_{3}\\textbf{k}$$ in component form, the cross product is written as: $$\\textbf{v} \\times \\textbf{w} = (v_{2}w_{3} - v_{3}w_{2})\\textbf{i} + (v_{3}w_{1} - v_{1}w_{3})\\textbf{j} + (v_{1}w_{2} - v_{2}w_{1})\\textbf{k}$$. It is often easier to use the component form for the cross product, because it can be represented as a $$\\textit{determinant}$$. We will not go too deeply into the theory of determinants; we will just cover what is essential for our purposes. A 2 $$\\times$$ 2 matrix} is an array of two rows and two columns of scalars, written as $$\\nonumber" ]

[ "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "+0 # let x and y be vectors such that and compute the ratio ​ 0 61 1 let x and y be vectors such that $$\\operatorname{proj}_{\\mathbf{x}}(\\mathbf{y})=\\dbinom{5}{3}$$and $$\\operatorname{proj}_{\\mathbf{y}}(\\mathbf{x})=\\dbinom{2}{-2}$$ compute the ratio $$\\dfrac{\\|\\mathbf{x}\\|}{\\|\\mathbf{y}\\|}$$ Feb 13, 2022", "+0 # PLZ help asap!!! (repost) 0 48 1 Let v and w be vectors such that $${proj}_{{w}}({v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {w}} (3 {v})$$. Oct 11, 2019 #1 +439 +5 The equation, $$projx⃗ y⃗ =x⃗ ⋅y⃗ |x⃗ |x⃗ |x⃗ |projx→y→=x→⋅y→|x→|x→|x→|$$ Holds for any vectors we can let those be $$πa⃗ πa→ and eb⃗ eb→$$. Where ππ and ee are some random constants like −2−2 and 33. $$projπa⃗ eb⃗ =πa⃗ ⋅eb⃗ |πa⃗ |πa⃗ |πa⃗ |projπa→eb→=πa→⋅eb→|πa→|πa→|πa→|$$ Now use$$a⃗ ⋅(cb⃗ )=ca⃗ ⋅b⃗ a→⋅(cb→)=ca→⋅b→ and |cw⃗ |=c|w⃗ ||cw→|=c|w→|.$$ To get e(pro j base a to the power of b) is the answer because $$π2eπ2a⃗ ⋅b⃗ |a⃗ |a⃗ |a⃗ |=π2eπ2a→⋅b→|a→|a→|a→| =$$ . Oct 11, 2019 edited by SVS2652 Oct 11, 2019", "+0 # help 0 91 1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ Sep 3, 2019 #1 +23357 +1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\ \\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ $$\\begin{array}{|rcll|} \\hline \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\left\\| \\dfrac{\\mathbf{v} \\cdot \\mathbf{w}}{\\left\\|\\mathbf{w}\\right\\|^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\left\\| \\dfrac{10}{7^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\| \\mathbf{w} \\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\cdot 7 \\\\\\\\ \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\mathbf{\\dfrac{10}{7}} \\\\ \\hline \\end{array}$$ Sep 4, 2019", "the transpose operation $\\mathbf{x}^{T} = \\begin{pmatrix} 1 & 3 \\end{pmatrix}$ and $\\mathbf{a}^{T} = \\begin{pmatrix} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\end{pmatrix}.$ Sometimes, we may not want to specify the dimension of a vector with an exact value. Rather, we might write $$\\mathbf{y} \\in \\mathbb{R}^{n}$$ to indicate the vector $$\\mathbf{y}$$ contains $$n$$ elements whose values are real numbers $\\mathbf{y} = \\begin{pmatrix} y_{1} \\\\ y_{2} \\\\ \\vdots \\\\ y_{n} \\end{pmatrix}.$ How do we make vectors in R? In R, there are several ways we can make vectors. Regardless of the method we choose, however, we have to specify the entries. First, we can use the combine function c() to combine several numbers into a vector. We can also use the is.vector() function to find out whether or not R views a particular object as a vector. a <- c(5, 2, 7, 1, 3) a #> [1] 5 2 7 1 3 is.vector(a) #> [1] TRUE If we want to take a sequence of numbers in increments of $$1$$ or $$-1$$ from some start value to some end value, we can do that using the colon operator : as in the example below. b <- 1:5 b #> [1] 1 2 3 4 5 is.vector(b) #> [1] TRUE Notice that if we want to increment by $$-1$$, we", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "+0 # Consider the vectors , and If the vectors v, w, and x are linearly independent, answer with 0. If they aren't, find coefficients a, 0 143 1 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors v, w, and x are linearly independent, answer with 0. If they aren't, find coefficients a, b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$and answer with (a+b)/c. Oct 6, 2019 #1 +6179 +1 3 2D vectors cannot be linearly independent. By eye a=2 b=-3 c=7 5/7 Oct 7, 2019 edited by Rom Oct 7, 2019", "+0 # halp!!! 0 98 2 Let v and w be vectors such that $${proj}_{{w}}( {v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $${proj}_{-2 {w}} (3 {v})$$. Oct 9, 2019", "## Vectors operations Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes ### Vectors operations Vectors operations - IB Maths HL If $a= \\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix}$ and $b= \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ How can we find the vectors $a-3b$ and $4a+3b$ Thanks lora Posts: 0 Joined: Wed Apr 10, 2013 7:36 pm ### Re: Vectors operations Vectors operations, addition, subtraction, scalar multiplication - IB Maths HL First of all let me remind you the properties of vector addition The vector addition (subtraction) is Commutative, Associative, there is an additive identity which is the zero vector ad there is an additive inverse which is the vector with equal magnitude and opposite direction. Also there is the scalar multiplication which is when a vector is multiplied by a scalar. $a-3b=\\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} -3 \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ $=\\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} -\\begin{pmatrix} -12 \\\\ 18 \\\\ 24 \\end{pmatrix}$ $=\\begin{pmatrix} -3+12 \\\\ 5-18 \\\\ 11-24 \\end{pmatrix}$ $=\\begin{pmatrix} 9 \\\\ -13 \\\\ -13 \\end{pmatrix}$ and $4a+3b =4 \\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} + 3 \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ $=\\begin{pmatrix} -12 \\\\ 20 \\\\", "# How to complete a set of independent vectors to a basis of a subspace? $\\newcommand{\\Span}{\\operatorname{Span}}$Let $V$ be a vector such that $\\dim V = n$, and let $v_1,\\ldots,v_k \\in V$ be independent vectors such that $1<k\\leq n$. Now Let $w_1,\\ldots,w_r\\in$ $\\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ be independent vectors such that $1\\leq r <k$. My question is this: How do I find the missing vectors $w_{r+1},\\ldots,w_k\\in \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ so that $\\Span\\left\\{w_1,\\ldots,w_r,w_{r+1},\\ldots,w_k\\right\\} = \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ ? Now I know they exist, I just don't know how I actually find them. For example: Lets look at $\\mathbb{R}^5$ and $$U=\\Span\\left\\{ \\begin{pmatrix} 5\\\\2\\\\3\\\\7\\\\3 \\end{pmatrix},\\begin{pmatrix} 2\\\\4\\\\4\\\\8\\\\1 \\end{pmatrix} ,\\begin{pmatrix} 3\\\\4\\\\7\\\\6\\\\1 \\end{pmatrix},\\begin{pmatrix} 5\\\\8\\\\6\\\\4\\\\8 \\end{pmatrix} \\right\\}$$ Those are all independent vectors. Now lets take $2$ independent vectors that are linear combination of those. lets say, $$w_1 = \\begin{pmatrix} 6\\\\2\\\\11\\\\15\\\\-3 \\end{pmatrix} \\; w_2=\\begin{pmatrix} 0\\\\4\\\\-2\\\\12\\\\1 \\end{pmatrix}$$ How can we complete those two vectors to form a basis of $U$ ? I would really like to understand the general idea of this. This is actually a general question of something I need it for, which is the process of finding a Jordan basis for matrices\\transformations. Thanks for any help Here's one method for the case of $V = \\Bbb R^n$: row-reduce the matrix $$\\pmatrix{w_1 & \\cdots & w_r & v_1 & \\cdots & v_k}$$ The columns of this matrix", "# Linear Operators • May 18th 2008, 11:33 PM steph_b23 Linear Operators Show that the range of the linear operator defined by the equations w1 = 4x1 - 2x2 w2 = 2x1 - x2 is not all of R2, and find a vector that is not in the range. Ok I can easily show the range is not all of R2 by finding the determinant to be zero... but how the vector which is not in the range? I know this is probably really easy. • May 19th 2008, 08:24 AM flyingsquirrel Hi $ w_1 =4x_1-2x_2$ and $ w_2 =2x_1-x_2 \\implies w_1=2w_2 \\implies\\begin{pmatrix}w_1\\\\w_2 \\end{pmatrix}= w_2\\begin{pmatrix}2\\\\1 \\end{pmatrix}$ All the vectors of the range can be written as $\\lambda \\begin{pmatrix}2\\\\1 \\end{pmatrix}$ hence, if you manage to find a vector of $\\mathbb{R}^2$ which direction is not that of $\\begin{pmatrix}2\\\\1 \\end{pmatrix}$, you are sure it does not belong to the range of the operator. • May 19th 2008, 09:21 AM steph_b23 Okay I get it, thanks so much :)", "# arrow_back Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. 7 views Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. If $\\mathbf{n}=(a, b, c)$ is a unit vector, use this formula to show that (perhaps surprisingly) the orthogonal projection of $\\mathbf{x}$ into the plane perpendicular to $\\mathbf{n}$ is given bya) Let $\\mathbf{v}:=(a, b, c)$ and $\\mathbf{x}:=(x, y, z)$ be any vectors in $\\mathbb{R}^{3}$. Viewed as column vectors, find a $3 \\times 3$ m ... Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\mathbf{a}=(-1,0,2,1)$ and $\\mathbf{w}_{2}$ is orthogonal to a. Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\m ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Prove the orthogonal projection theorem 1 answer 7 views Prove the orthogonal projection theoremProve the orthogonal projection theorem ... close 1 answer 8 views Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the positive$x$-axis in$R^{2}$.Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss symmetry. 0 answers 7 views Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss", "+0 # halp!!! 0 147 2 Let v and w be vectors such that $${proj}_{{w}}( {v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $${proj}_{-2 {w}} (3 {v})$$. Oct 9, 2019", "+0 # For vectors u and v, we have if and ​ find p*u 0 60 1 For vectors u and v, we have $$\\mathbf{p}=\\operatorname{proj}_{\\mathbf{v}}(\\mathbf{u})$$ if $$\\|\\mathbf{u}\\|=11$$ and $$\\|\\mathbf{p}\\|=6$$ find p*u Feb 13, 2022 #1 0 p*u = 22/3. Feb 14, 2022", "# Recent questions tagged planes If $\\mathbf{n}=(a, b, c)$ is a unit vector, use this formula to show that (perhaps surprisingly) the orthogonal projection of $\\mathbf{x}$ into the plane perpendicular to $\\mathbf{n}$ is given bya) Let $\\mathbf{v}:=(a, b, c)$ and $\\mathbf{x}:=(x, y, z)$ be any vectors in $\\mathbb{R}^{3}$. Viewed as column vectors, find a $3 \\times 3$ m ... close close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Prove the orthogonal projection theorem Prove the orthogonal projection theoremProve the orthogonal projection theorem ... close The planes $$x+2 y-2 z=3 \\text { and } 2 x+4 y-4 z=7$$ are parallel since their normals, $(1,2,-2)$ and $(2,4,-4)$, are parallel vectors. Find the distance between these planes.The planes $$x+2 y-2 z=3 \\text { and } 2 x+4 y-4 z=7$$ are parallel since their normals, $(1,2,-2)$ and $(2,4,-4)$, are parallel vectors. Find the d ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain.Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the lengths of the sides and the interior angles of the triangle in $R^{4}$ whose vertices are $$P(2,4,2,4,2), \\quad Q(6,4,4,4,6), \\quad R(5,7,5,7,2)$$ Find the", "1\\amp1\\amp1\\\\ 0\\amp0\\amp-1 \\end{bmatrix} \\end{equation*} Evaluate the following matrices: 1. $A^T$ 2. $A^{-1}$ 3. $(A^{-1})^T$ 4. $(A^T)^{-1}$ 5. $(A^{-1})^2$ 6. $(A^2)^{-1}$ Solution ##### 4 1. Find a unit vector in the direction of $\\mathbf w=(4,-3,2,1)\\text{.}$ 2. Find all values $t$ so that vectors $(2,5,-3,6)$ and $(4,t,7,1)$ are orthogonal. 3. Calculate $\\proj_{\\mathbf v} \\mathbf u\\text{,}$ the projection of $\\mathbf u$ along $\\mathbf v\\text{,}$ where $\\mathbf u=(2,-2,3)$ and $\\mathbf v=(2,1,3)\\text{.}$ Solution ##### 5 Let $\\mathbf{u}=(1,1,1,-1)\\text{,}$ $\\mathbf{v}=(1,1,-1,-1)$ and $\\mathbf{w}=(1,1,1,1)$ be three vectors in 4-space. 1. What is the length $\\|\\mathbf{u}\\|$ 2. What is the length $\\|\\mathbf{v}\\|$ 3. What is the length $\\|\\mathbf{w}\\|$ 1. What is the angle between $\\mathbf{u}$ and $\\mathbf{v}$ 2. What is the angle between $\\mathbf{u}$ and $\\mathbf{w}$ 3. What is the angle between $\\mathbf{v}$ and $\\mathbf{w}$ 1. Find a vector in 4-space that is orthogonal to $\\mathbf{u}\\text{,}$ $\\mathbf{v}$ and $\\mathbf{w}\\text{.}$ Solution ##### 6 Consider the following system of linear equations: $\\begin{array}{rrrrrr} x \\amp +2y \\amp +z \\amp +w \\amp =\\amp 2\\\\ 2x \\amp -y \\amp +z \\amp \\amp =\\amp 1\\\\ -3x \\amp -y \\amp -2z \\amp -w \\amp =\\amp -3\\\\ x \\amp +7y \\amp +2z \\amp +3w \\amp =\\amp 5 \\end{array}$ 1. Give the augmented matrix of the system. 2. Put the augmented matrix in reduced row echelon form. 3. Give all of the solutions to the system. Solution", "Vectors Satisfy $A\\mathbf{x}=0$ then Find Another Solution Suppose that the vectors \\[\\mathbf{v}_1=\\begin{bmatrix} -2 \\\\ 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\qquad \\mathbf{v}_2=\\begin{bmatrix} -4 \\\\ 0... Close", "or $\\mathbf W_2 \\subset \\mathbf W_1$. (See HW2 pp. 20-21, #19.) ## Linear combinations Definition: A vector u is a \"linear combination\" (l.c.) of vectors u1, ..., un if there exists scalars a1, ..., an such that $u = a_1 u_1 + a_2 u_2 + \\ldots + a_n u_n$ Example: $\\mathbb P_n(F) = \\{ \\mbox{Polynomials of degree at most } n \\mbox{ with coefficients in } F \\}$ $= \\left\\{ \\sum_{i=0}^n a_i x^i : a_i \\in F \\right\\}$ Definition: A subset $S \\subset \\mathbf V$ \"generates\" or \"spans\" V iff the set of linear combinations of elements of S is all of V. Example: Let $\\mathbf V = \\mathbb M_{n \\times n}(\\real)$ Let $M_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix}, M_2 = \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\end{pmatrix}, M_3 = \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}, M_4 = \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix}$ Then $S = \\{ M_1, M_2, M_3, M_4 \\}$ generates V. Proof: Given $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\in \\mathbb M_{2 \\times 2}(\\real)$, write $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4.$ Example: Let $N_1 = \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}, N_2 = \\begin{pmatrix}", "## IB Maths HL Vectors operations Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes ### IB Maths HL Vectors operations IB Mathematics HL – Vectors operations If $a= \\begin{pmatrix} 1 \\\\ 2 \\\\ 4 \\end{pmatrix}$ and $b= \\begin{pmatrix} -1 \\\\ 4 \\\\ 5 \\end{pmatrix}$ How can we find the vectors $a-2b$ and $3a+2b$ Thanks elizabeth Posts: 0 Joined: Mon Jan 28, 2013 8:09 pm ### Re: IB Maths HL Vectors operations IB Mathematics HL – Vectors operations, addition, subtraction, scalar multiplication First of all let me remind you the properties of vector addition The vector addition (subtraction) is Commutative, Associative, there is an additive identity which is the zero vector ad there is an additive inverse which is the vector with equal magnitude and opposite direction. Also there is the scalar multiplication which is when a vector is multiplied by a scalar. $a-2b=\\begin{pmatrix} 1 \\\\ 2 \\\\ 4 \\end{pmatrix}-2 \\begin{pmatrix} -1 \\\\ 4 \\\\ 5 \\end{pmatrix}$ $=\\begin{pmatrix} 1 \\\\ 2 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} -2 \\\\ 8 \\\\ 10 \\end{pmatrix}$ $=\\begin{pmatrix} 1+2 \\\\ 2-8 \\\\ 4-10 \\end{pmatrix}$ $=\\begin{pmatrix} 3 \\\\ -6 \\\\ -6 \\end{pmatrix}$ and $3a+2b=3 \\begin{pmatrix} 1 \\\\ 2 \\\\ 4 \\end{pmatrix} + 2 \\cdot \\begin{pmatrix} -1 \\\\ 4 \\\\" ]

[ "\\propto\\begin{bmatrix}b \\\\ \\omega^2 - a\\end{bmatrix},\\end{aligned} \\hspace{\\stretch{1}}(1.0.6) or \\begin{aligned}\\boxed{\\boldsymbol{\\epsilon} \\propto\\begin{bmatrix}2 b \\\\ d - a \\pm \\sqrt{ (d - a)^2 + 4 b c }\\end{bmatrix}.}\\end{aligned} \\hspace{\\stretch{1}}(1.0.6) Square lattice There is not too much to gain by expanding out the projection operators explicitly in general. However, let’s do this for the specific case of a square lattice (as on the exam problem). In that case, our projection operators are \\begin{aligned}\\hat{\\mathbf{a}} \\hat{\\mathbf{a}}^\\text{T} = \\begin{bmatrix}1 \\\\ 0\\end{bmatrix}\\begin{bmatrix}1 & 0\\end{bmatrix}=\\begin{bmatrix}1 & 0 \\\\ 0 & 0\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16a) \\begin{aligned}\\hat{\\mathbf{b}} \\hat{\\mathbf{b}}^\\text{T} = \\begin{bmatrix}0\\\\ 1 \\end{bmatrix}\\begin{bmatrix}0 &1 \\end{bmatrix}=\\begin{bmatrix}0 & 0 \\\\ 0 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16b) \\begin{aligned}\\hat{\\mathbf{r}} \\hat{\\mathbf{r}}^\\text{T} = \\frac{1}{{2}}\\begin{bmatrix}1 \\\\ 1 \\end{bmatrix}\\begin{bmatrix}1 &1 \\end{bmatrix}=\\frac{1}{{2}}\\begin{bmatrix}1 & 1 \\\\ 1 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16c) \\begin{aligned}\\hat{\\mathbf{s}} \\hat{\\mathbf{s}}^\\text{T} = \\frac{1}{{2}}\\begin{bmatrix}-1 \\\\ 1 \\end{bmatrix}\\begin{bmatrix}-1 &1 \\end{bmatrix}=\\frac{1}{{2}}\\begin{bmatrix}1 & -1 \\\\ -1 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) \\begin{aligned}\\begin{aligned}S_1 &= \\sin^2\\left( { \\mathbf{a} \\cdot \\mathbf{q} } \\right) \\\\ S_2 &= \\sin^2\\left( { \\mathbf{b} \\cdot \\mathbf{q} } \\right) \\\\ S_3 &= \\sin^2\\left( { (\\mathbf{b} + \\mathbf{a}) \\cdot \\mathbf{q} } \\right) \\\\ S_4 &= \\sin^2\\left( { (\\mathbf{b} - \\mathbf{a}) \\cdot \\mathbf{q} } \\right),\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) Our matrix is \\begin{aligned}A = \\frac{2}{m}\\begin{bmatrix}2 K_1 S_1 + K_3 S_3 + K_4 S_4 & K_3 S_3 - K_4 S_4 \\\\ K_3 S_3 - K_4 S_4 & 2 K_2 S_2 + K_3 S_3 + K_4 S_4\\end{bmatrix},\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) where, specifically, the squared sines for this geometry", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "\\begin{aligned}\\boldsymbol{\\epsilon} \\propto\\begin{bmatrix}b \\\\ \\omega^2 - a\\end{bmatrix},\\end{aligned} \\hspace{\\stretch{1}}(1.0.6) or \\begin{aligned}\\boxed{\\boldsymbol{\\epsilon} \\propto\\begin{bmatrix}2 b \\\\ d - a \\pm \\sqrt{ (d - a)^2 + 4 b c }\\end{bmatrix}.}\\end{aligned} \\hspace{\\stretch{1}}(1.0.6) Square lattice There is not too much to gain by expanding out the projection operators explicitly in general. However, let’s do this for the specific case of a square lattice (as on the exam problem). In that case, our projection operators are \\begin{aligned}\\hat{\\mathbf{a}} \\hat{\\mathbf{a}}^\\text{T} = \\begin{bmatrix}1 \\\\ 0\\end{bmatrix}\\begin{bmatrix}1 & 0\\end{bmatrix}=\\begin{bmatrix}1 & 0 \\\\ 0 & 0\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16a) \\begin{aligned}\\hat{\\mathbf{b}} \\hat{\\mathbf{b}}^\\text{T} = \\begin{bmatrix}0\\\\ 1 \\end{bmatrix}\\begin{bmatrix}0 &1 \\end{bmatrix}=\\begin{bmatrix}0 & 0 \\\\ 0 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16b) \\begin{aligned}\\hat{\\mathbf{r}} \\hat{\\mathbf{r}}^\\text{T} = \\frac{1}{{2}}\\begin{bmatrix}1 \\\\ 1 \\end{bmatrix}\\begin{bmatrix}1 &1 \\end{bmatrix}=\\frac{1}{{2}}\\begin{bmatrix}1 & 1 \\\\ 1 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16c) \\begin{aligned}\\hat{\\mathbf{s}} \\hat{\\mathbf{s}}^\\text{T} = \\frac{1}{{2}}\\begin{bmatrix}-1 \\\\ 1 \\end{bmatrix}\\begin{bmatrix}-1 &1 \\end{bmatrix}=\\frac{1}{{2}}\\begin{bmatrix}1 & -1 \\\\ -1 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) \\begin{aligned}\\begin{aligned}S_1 &= \\sin^2\\left( { \\mathbf{a} \\cdot \\mathbf{q} } \\right) \\\\ S_2 &= \\sin^2\\left( { \\mathbf{b} \\cdot \\mathbf{q} } \\right) \\\\ S_3 &= \\sin^2\\left( { (\\mathbf{b} + \\mathbf{a}) \\cdot \\mathbf{q} } \\right) \\\\ S_4 &= \\sin^2\\left( { (\\mathbf{b} - \\mathbf{a}) \\cdot \\mathbf{q} } \\right),\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) Our matrix is \\begin{aligned}A = \\frac{2}{m}\\begin{bmatrix}2 K_1 S_1 + K_3 S_3 + K_4 S_4 & K_3 S_3 - K_4 S_4 \\\\ K_3 S_3 - K_4 S_4 & 2 K_2 S_2 + K_3 S_3 + K_4 S_4\\end{bmatrix},\\end{aligned} \\hspace{\\stretch{1}}(1.0.16d) where, specifically, the squared sines for this", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "# What is the projection of < 5 , -6, 0 > onto < -1, -4 , 6 >? Aug 23, 2017 see below #### Explanation: Use the formula $p r o {j}_{a} b = \\frac{a \\cdot b}{|} a {|}^{2} a$ That is, $p r o {j}_{a} b = \\frac{< 5 , - 6 , 0 > \\cdot < - 1 , - 4 , 6 >}{\\sqrt{{5}^{2} + {\\left(- 6\\right)}^{2} + {0}^{2}}} ^ 2 < 5 , - 6 , 0 >$ $= \\frac{- 5 + 24 + 0}{25 + 36} < 5 , - 6 , 0 >$ $= \\frac{19}{61} < 5 , - 6 , 0 >$ $= < \\frac{95}{61} , - \\frac{114}{61} , 0 >$", "the formula (here, x/y is used as the projection plane; literature also may use x/z): [11] {\\displaystyle {\\begin{aligned}\\mathbf {b} _{x}&={\\frac {\\mathbf {e} _{z}}{\\mathbf {d} _{z}}}\\mathbf {d} _{x}+\\mathbf {e} _{x},\\\\[5pt]\\mathbf {b} _{y}&={\\frac {\\mathbf {e} _{z}}{\\mathbf {d} _{z}}}\\mathbf {d} _{y}+\\mathbf {e} _{y}.\\end{aligned}}} Or, in matrix form using homogeneous coordinates, the system ${\\displaystyle {\\begin{bmatrix}\\mathbf {f} _{x}\\\\\\mathbf {f} _{y}\\\\\\mathbf {f} _{w}\\end{bmatrix}}={\\begin{bmatrix}1&0&{\\frac {\\mathbf {e} _{x}}{\\mathbf {e} _{z}}}\\\\0&1&{\\frac {\\mathbf {e} _{y}}{\\mathbf {e} _{z}}}\\\\0&0&{\\frac {1}{\\mathbf {e} _{z}}}\\end{bmatrix}}{\\begin{bmatrix}\\mathbf {d} _{x}\\\\\\mathbf {d} _{y}\\\\\\mathbf {d} _{z}\\end{bmatrix}}}$ in conjunction with an argument using similar triangles, leads to division by the homogeneous coordinate, giving {\\displaystyle {\\begin{aligned}\\mathbf {b} _{x}&=\\mathbf {f} _{x}/\\mathbf {f} _{w}\\\\\\mathbf {b} _{y}&=\\mathbf {f} _{y}/\\mathbf {f} _{w}\\end{aligned}}} The distance of the viewer from the display surface, ${\\displaystyle \\mathbf {e} _{z}}$, directly relates to the field of view, where ${\\displaystyle \\alpha =2\\cdot \\arctan(1/\\mathbf {e} _{z})}$ is the viewed angle. (Note: This assumes that you map the points (-1,-1) and (1,1) to the corners of your viewing surface) The above equations can also be rewritten as: {\\displaystyle {\\begin{aligned}\\mathbf {b} _{x}&=(\\mathbf {d} _{x}\\mathbf {s} _{x})/(\\mathbf {d} _{z}\\mathbf {r} _{x})\\mathbf {r} _{z},\\\\\\mathbf {b} _{y}&=(\\mathbf {d} _{y}\\mathbf {s} _{y})/(\\mathbf {d} _{z}\\mathbf {r} _{y})\\mathbf {r} _{z}.\\end{aligned}}} In which ${\\displaystyle \\mathbf {s} _{x,y}}$ is the display size, ${\\displaystyle \\mathbf {r} _{x,y}}$ is the recording surface size (CCD or film), ${\\displaystyle \\mathbf {r} _{z}}$ is the distance", "of the camera) in the order x, y, z (reading left-to-right)\". Note that if the camera is not rotated (\\mathbf{\\theta}_{x,y,z} = \\langle 0,0,0\\rangle), then the matrices drop out (as identities), and this reduces to simply a shift: \\mathbf{d} = \\mathbf{a} - \\mathbf{c}. Alternatively, without using matrices (let's replace (ax-cx) with x and so on, and abbreviate cosθ to c and sinθ to s): \\begin{array}{lcl} \\mathbf{d}_x = c_y (s_z \\mathbf{y}+c_z \\mathbf{x})-s_y \\mathbf{z} \\\\ \\mathbf{d}_y = s_x (c_y \\mathbf{z}+s_y (s_z \\mathbf{y}+c_z \\mathbf{x}))+c_x (c_z \\mathbf{y}-s_z \\mathbf{x}) \\\\ \\mathbf{d}_z = c_x (c_y \\mathbf{z}+s_y (s_z \\mathbf{y}+c_z \\mathbf{x}))-s_x (c_z \\mathbf{y}-s_z \\mathbf{x}) \\\\ \\end{array} This transformed point can then be projected onto the 2D plane using the formula (here, x/y is used as the projection plane; literature also may use x/z):[6] \\begin{array}{lcl} \\mathbf{b}_x &= & \\frac{\\mathbf{e}_z}{\\mathbf{d}_z} \\mathbf{d}_x - \\mathbf{e}_x \\\\ \\mathbf{b}_y &= & \\frac{\\mathbf{e}_z}{\\mathbf{d}_z} \\mathbf{d}_y - \\mathbf{e}_y\\\\ \\end{array}. Or, in matrix form using homogeneous coordinates, the system \\begin{bmatrix} \\mathbf{f}_x \\\\ \\mathbf{f}_y \\\\ \\mathbf{f}_z \\\\ \\mathbf{f}_w \\\\ \\end{bmatrix}=\\begin{bmatrix} 1 & 0 & -\\frac{\\mathbf{e}_x}{\\mathbf{e}_z} & 0 \\\\ 0 & 1 & -\\frac{\\mathbf{e}_y}{\\mathbf{e}_z} & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 1/\\mathbf{e}_z & 0 \\\\ \\end{bmatrix}\\begin{bmatrix} \\mathbf{d}_x \\\\ \\mathbf{d}_y \\\\ \\mathbf{d}_z \\\\ 1 \\\\ \\end{bmatrix} in conjunction with an argument using similar triangles, leads to division by the homogeneous coordinate, giving", "### Control Keys move to next slide (also Enter or Spacebar). move to previous slide. d enable/disable drawing on slides p toggles between print and presentation view CTRL + zoom in CTRL - zoom out CTRL 0 reset zoom Slides can also be advanced by clicking on the left or right border of the slide. ### Notation Type Font Examples Variables (scalars) italics $a, b, x, y$ Functions upright $\\mathrm{f}, \\mathrm{g}(x), \\mathrm{max}(x)$ Vectors bold, elements row-wise $\\mathbf{a}, \\mathbf{b}= \\begin{pmatrix}x\\\\y\\end{pmatrix} = (x, y)^\\top,$ $\\mathbf{B}=(x, y, z)^\\top$ Matrices Typewriter $\\mathtt{A}, \\mathtt{B}= \\begin{bmatrix}a & b\\\\c & d\\end{bmatrix}$ Sets calligraphic $\\mathcal{A}, B=\\{a, b\\}, b \\in \\mathcal{B}$ Number systems, Coordinate spaces double-struck $\\mathbb{N}, \\mathbb{Z}, \\mathbb{R}^2, \\mathbb{R}^3$ ## Parallel Projection ### Parallel Projection • With parallel projection, objects with larger distance to the viewer or camera are not getting smaller • There are no vanishing points Perspective projection Parallel projection Source: 3D Model:Turbosquid (Standard Royalty Free License); Renderer: 3ds Max ### Parallel Projection projection rays image plane $x$ $y$ $z$ $\\tilde{\\mathbf{P}}$ $\\mathbf{P}$ • The projection rays are parallel, i.e., the projection center is located at infinity • The formula for mapping a 3D point $\\mathbf{P}=(p_x,p_y,p_z)^\\top$ to a point $\\tilde{\\mathbf{P}}= (\\tilde{p}_x,\\tilde{p}_y,\\tilde{p}_z)^\\top$ located at the image plane of the camera is given by: $\\tilde{\\mathbf{P}}= \\left(p_x, p_y, 0 \\right)^\\top$ ### Parallel Projection • Using homogeneous coordinates the", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "# Prove that perspective projection of circle is ellipse How to prove that perspective projection of a circle is an ellipse? I start with the parametric equation of circle and ellipse: Circle: $x = r\\cos t$ $y = r\\sin t$ Ellipse: $x = a\\cos(t)$ $y = b\\sin(t)$ Then I have perspective projection matrix: $$\\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & 1 \\\\ \\end{pmatrix}$$ And obtain projected coordinates (affine transform for simplicity): $$\\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ 0 & 0 & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} u \\\\ v \\\\ 1 \\\\ \\end{pmatrix}$$ If we convert this to equations: $u = a_{11}x+a_{12}y+a_{13}$ $v = a_{21}x+a_{22}y+a_{23}$ Substituting $x,y$ with parametric equations: $a\\cos t = a_{11}\\,r\\, \\cos t + a_{12}\\, r \\,\\sin t + a_{13}$ $b \\sin t = a_{21}\\,r \\,\\cos t + a_{22}\\, r \\,\\sin t + a_{23}$ What to do next? I can't see how these equations can be equal. • Please write a_{11} for indexed variables. You could also consider removing the multiplication operator * because in modern mathematics this operator is implicitly implied if the variables are written without an operator in the middle of them. If the spacing is to", "the projection matrix $\\mathbf{P}_{W_T}$ projects onto a 24-dimensional subspace since we have removed one dimension by subtracting out $\\mathrm{C}(\\mathbf{1}_N)$. The same goes for $\\mathbf{P}_{W_J}$ (projecting onto a three-dimensional subspace), and $\\mathbf{P}_{W_Q}$ (projecting onto a four-dimensional subspace), and $\\mathbf{P}_{W{JQ}}$ (projecting onto a 12-dimensional subspace). That means the residual projection matrix $\\mathbf{P}_\\perp$ is projecting into a $N-44=430-44=386$-dimensional subspace. All of this means that for our model: $E\\|\\mathbf{P}_0\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_0\\boldsymbol{\\tau}\\|^2 + \\sigma^2$ $E\\|\\mathbf{P}_{W_T}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_T}\\boldsymbol{\\tau}\\|^2 + 24\\sigma^2$ $E\\|\\mathbf{P}_{W_J}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_J}\\boldsymbol{\\tau}\\|^2 + 3\\sigma^2$ $E\\|\\mathbf{P}_{W_Q}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_Q}\\boldsymbol{\\tau}|^2 + 4\\sigma^2$ $E\\|\\mathbf{P}_{W_{JQ}}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_{JQ}}\\boldsymbol{\\tau}\\|^2 + 12\\sigma^2$ $E\\|\\mathbf{P}_\\perp\\mathbf{Y}\\|^2 = (N-44)\\sigma^2$ where the last one comes from the fact that $\\boldsymbol{\\tau}$ is orthogonal to the residual subspace. The left-hand side of each of these is just an expected sum of squared random values. On the right hand side, we have two terms: the first due to the deterministic effects of the levels in a factor, and the second due to iid noise in the measurements. If there is no effect at a factor, then its deterministic component will be zero. In addition, if there are no differences between the effects in a factor, then the projections will be zero. Hence, to test for significant differences between the effects in a factor, all we need to do is compare the empirical sum of squares of the projections of", "into clip-space. But in practice, the projection matrix for a perspective camera always contains four nonzero constants, and they’re always in the same place: $\\begin{split}\\begin{bmatrix} A & 0 & 0 & 0 \\\\ 0 & 0 & B & 1 \\\\ 0 & C & 0 & 0 \\\\ 0 & 0 & D & 0 \\end{bmatrix}\\end{split}$ The result is that the panda projection matrix boils down to these simple equations: \\begin{split}\\begin{aligned} x_{clip} &= x_{view} \\cdot A \\\\ y_{clip} &= z_{view} \\cdot C \\\\ z_{clip} &= y_{view} \\cdot B + D \\\\ w_{clip} &= y_{view} \\end{aligned}\\end{split} Look out, there has been a coordinate system change! In the scene graph, Z corresponds to “up”, but in clip-space, Z is the depth value (and X,Y address a pixel). After panda calculates clip-space coordinates, it divides by W. Finally, it rescales the depth-value: \\begin{split}\\begin{aligned} x_{screen} &= \\frac{x_{clip}}{w_{clip}} \\\\ y_{screen} &= \\frac{y_{clip}}{w_{clip}} \\\\ z_{screen} &= \\frac{z_{clip}}{w_{clip}} \\\\ depth &= 0.5 \\cdot z_{screen} + 0.5 \\end{aligned}\\end{split} So now we have some equations defining $$\\begin{pmatrix}x_{clip}&y_{clip}&z_{clip}&w_{clip}\\end{pmatrix}$$ in terms of $$\\begin{pmatrix}x_{view}&y_{view}&z_{view}\\end{pmatrix}$$, and $$\\begin{pmatrix}x_{screen}&y_{screen}&z_{screen}\\end{pmatrix}$$ in terms of $$\\begin{pmatrix}x_{clip}&y_{clip}&z_{clip}&w_{clip}\\end{pmatrix}$$. It’s basic algebra to solve these equations for $$\\begin{pmatrix}x_{view}&y_{view}&z_{view}\\end{pmatrix}$$ in terms of $$\\begin{pmatrix}x_{screen}&y_{screen}&z_{screen}\\end{pmatrix}$$. Here, I have shown all my algebraic steps: \\begin{split}\\begin{aligned} depth &= 0.5 \\cdot z_{screen} + 0.5 \\\\ depth &= \\frac{z_{clip}}{w_{clip}} \\cdot 0.5 + 0.5 \\\\ depth", "# Problem 1 [60 points] Projection on a set¶ Review the main ideas in Lec. 17, p. 2--4. ## (a) [20 points] Projection on standard simplex¶ Analytically compute the projection of a point $x_{0}\\in\\mathbb{R}^{n}_{\\geq 0}\\setminus\\{0\\}$ onto the standard simplex $\\mathcal{S} := \\{x\\in\\mathbb{R}^{n}_{\\geq 0}\\mid \\mathbf{1}^{\\top}x = 1\\}$ with respect to the generalized Kullback-Leibler divergence $$D_{\\text{KL}}(x\\parallel x_{0}) := \\sum_{i=1}^{n}\\left(x_{i}\\log\\frac{x_{i}}{x_{0i}} - x_{i} + x_{0i}\\right).$$ In other words, compute $\\operatorname{proj}^{D_{\\text{KL}}}_{\\mathcal{S}}\\left(x_{0}\\right)$ as a function of $x_{0}$. (Hint: Set up the relevant constrained optimization problem, argue why you can invoke statement 2 in Lec. 15, p. 2, and then solve the KKT conditions from Lec. 14, p. 18-19.) ## (b) [30 points] Projection on $\\mathbb{S}_{+}^{n}$¶ Analytically compute the projection of $X_{0}\\in\\mathbb{S}^{n}$ onto the cone $\\mathbb{S}_{+}^{n}$ with respect to the Frobenius norm (Lec. 17, p. 7, 1st row and second column). In other words, compute $\\operatorname{proj}^{\\|\\cdot\\|_{\\text{F}}}_{\\mathbb{S}^{n}_{+}}\\left(X_{0}\\right)$ as a function of $X_{0}$. (Hint: Same hints as in part (a). Review Lec. 4 for taking derivative of scalar with respect to matrix. Also, the spectral theorem for symmertic matrices should help.) ## (c) [10 points] Nonconvexity of rank ball¶ During Lec. 17, p. 9-10, Example 2, we mentioned that the $k$ rank ball $$\\mathcal{C}_{mn}^{k} := \\{X\\in\\mathbb{R}^{m\\times n} \\mid \\text{rank}(X)\\leq k\\}, \\quad\\text{for fixed}\\;k\\leq \\min\\{m,n\\},$$ is a nonconvex set. To understand this set nonconvexity, fix $m=n=3$ and $k=2$. Construct", "= \\hat{\\mathbf{y}} \\hat{\\mathbf{y}}^\\text{T}= \\begin{bmatrix}0 \\\\ 1\\end{bmatrix}\\begin{bmatrix}0 & 1\\end{bmatrix}=\\begin{bmatrix}0 & 0 \\\\ 0 & 1\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(2.0.19b) For the $\\hat{\\mathbf{r}}$ and $\\hat{\\mathbf{s}}$ projection operators, we can use half angle formulations \\begin{aligned}\\text{Proj}_{\\hat{\\mathbf{r}}} = \\hat{\\mathbf{r}} \\hat{\\mathbf{r}}^\\text{T}= \\begin{bmatrix}\\cos\\theta \\\\ \\sin\\theta\\end{bmatrix}\\begin{bmatrix}\\cos\\theta & \\sin\\theta\\end{bmatrix}=\\begin{bmatrix}\\cos^2\\theta & \\cos\\theta \\sin\\theta \\\\ \\cos\\theta \\sin\\theta & \\sin^2 \\theta\\end{bmatrix}=\\frac{1}{2}\\begin{bmatrix}1 + \\cos \\left( 2 \\theta \\right) & \\sin \\left( 2 \\theta \\right) \\\\ \\sin \\left( 2 \\theta \\right) & 1 - \\cos \\left( 2 \\theta \\right)\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(2.0.20.20) \\begin{aligned}\\text{Proj}_{\\hat{\\mathbf{s}}} = \\hat{\\mathbf{s}} \\hat{\\mathbf{s}}^\\text{T}= \\begin{bmatrix}-\\cos\\theta \\\\ \\sin\\theta\\end{bmatrix}\\begin{bmatrix}-\\cos\\theta & \\sin\\theta\\end{bmatrix}=\\begin{bmatrix}\\cos^2\\theta & -\\cos\\theta \\sin\\theta \\\\ -\\cos\\theta \\sin\\theta & \\sin^2 \\theta\\end{bmatrix}=\\frac{1}{2}\\begin{bmatrix}1 + \\cos \\left( 2 \\theta \\right) & -\\sin \\left( 2 \\theta \\right) \\\\ -\\sin \\left( 2 \\theta \\right) & 1 - \\cos \\left( 2 \\theta \\right)\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(2.0.20.20) After some manipulation, and the following helper functions \\begin{aligned}\\begin{aligned}\\alpha_\\pm &= K_3 \\sin^2 (\\mathbf{r} \\cdot \\mathbf{q}/2) \\pm K_4 \\sin^2 (\\mathbf{s} \\cdot \\mathbf{q}/2) \\\\ \\beta_\\pm &= K_1 \\cos\\left( (\\mathbf{r} - \\mathbf{s}) \\cdot \\mathbf{q}/2 \\right) \\pm K_2 \\cos\\left( (\\mathbf{r} + \\mathbf{s}) \\cdot \\mathbf{q}/2 \\right),\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(2.0.20.20) the block matrices of eq. 2.0.17.17 take the form \\begin{aligned}A = \\begin{bmatrix}K_1 + \\alpha_+ (1 + \\cos\\left( 2 \\theta \\right)) & \\alpha_- \\sin\\left( 2 \\theta \\right) \\\\ \\alpha_- \\sin\\left( 2 \\theta \\right) & K_2 + \\alpha_+ (1 - \\cos\\left( 2 \\theta \\right))\\end{bmatrix}\\end{aligned} \\hspace{\\stretch{1}}(2.0.22.22) \\begin{aligned}B = e^{ i (\\mathbf{r} + \\mathbf{s}) \\cdot \\mathbf{q}/2 }\\begin{bmatrix} \\beta_+ (1 + \\cos \\left(", "\\theta & =\\mathbf{v} \\cdot \\mathbf{w} \\end{aligned} When $v,w$ are unit vectors then, $$v \\cdot w=\\cos \\theta$$ Which is maximized when $\\theta=0$, i.e. $v=w$. Thus the dot product is often used to measure the similarity between two vectors. ##### Projections Scalar projection of $\\mathbf{a}$ onto $\\mathbf{b}$: $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction} \\cdot \\mathbf{a}$$ This essentially gives the signed length of the projection. $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\|a\\|\\cos(\\theta) = \\|\\mathbf{a}\\|\\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|\\|\\mathbf{b}\\|} = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|}$$ Vector projection of $\\mathbf{a}$ onto $\\mathbf{b}:$ $$\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}= \\underbrace{\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}}_{signed \\ length}\\quad \\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction}=\\frac{\\mathbf{b} \\cdot \\mathbf{a}}{|\\mathbf{b}|^{2}} \\mathbf{b}$$ The first equality is more meaningful, as it gives the length-direction decomposition. For any non-zero constant $c$ $$\\operatorname{proj}_{\\mathbf{cb}} \\mathbf{a}=\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}$$ In other words, the projection of $\\mathbf{a}$ onto $\\mathbf{b}$ is independent of the length of $\\mathbf{b}$, only the direction of the projection matters. #### The Cross Product ##### Motivation Given two vectors $u,v$, we want to find a vector $w$, that is perpendicular to both $u$ and $v$. This can be done by solving the following equations: $$u\\cdot w = u_1w_1+u_2w_2+u_3w_3 = 0$$ $$v\\cdot w = v_1w_1+v_2w_2+v_3w_3 = 0$$ Which implies $$(u_1v_3-u_3v_1)w_1+(u_2v_3-u_3v_2)w_2=0$$ One trivial solution is to take $$w_1=u_2v_3-u_3v_2,\\ w_2 = u_3v_1-u_1v_3, \\ w_3=u_1v_2-u_2v_1$$ The solution give the cross product of $u$ and $v$. ##### Computational Properties Often the above the formula is not very intuitive, and difficult to recall. Instead, one often use", "# Quantum Field theory, solving delta / green function I have read an equation as follow $$[-(k^2-m^2)g^{\\mu\\nu}+k^{\\mu}k^{\\nu}]D_{\\nu\\lambda}(k)=\\delta^{\\mu}_{\\lambda}$$ The solution is given as: $$D_{\\nu\\lambda}(k)=\\large{\\frac{-g_{\\nu\\lambda}+k_{\\nu}k_{\\lambda}/m^2}{k^2-m^2}}$$ I can only verify the answer but do not know how to yield the result step by step... are there any standard procedures to yield the solution? We can make use of the projection operators \\begin{align} & P^{T}_{\\mu \\nu} = g_{\\mu \\nu} - \\frac{k_{\\mu}k_{\\nu}}{k^{2}}, \\\\ & P^{L}_{\\mu \\nu} = \\frac{k_{\\mu}k_{\\nu}}{k^{2}}, \\end{align} where $$P^{T}_{\\mu \\nu}$$ and $$P^{L}_{\\mu \\nu}$$ are the transverse and longitudinal projectors. $$P^{T}_{\\mu \\nu}$$ and $$P^{L}_{\\mu \\nu}$$ are the projection operators onto vectors orthogonal and parallel to $$k^{\\mu}$$, respectively. You can verify by yourself that they satisfy the properties of projectors. We then write \\begin{align} -(k^{2}-m) g^{\\mu \\nu} + k^{\\mu}k^{\\nu} = A \\cdot P^{T,\\mu \\nu} + B \\cdot P^{L,\\mu \\nu} \\end{align} where we can find \\begin{align} A = -(k^{2} - m^{2}),\\ B = m^{2}. \\end{align} Now, to find the inverse $$D_{\\mu \\nu}(k)$$ we simply invert the $$A$$ and $$B$$ and then lower the $$\\mu$$ and $$\\nu$$ in $$A \\cdot P^{T,\\mu \\nu} + B \\cdot P^{L,\\mu \\nu}$$, where we yield \\begin{align} D_{\\mu \\nu}(k) & = \\frac{1}{A} P^{T}_{\\mu \\nu} + \\frac{1}{B} P^{L}_{\\mu \\nu} = \\frac{1}{-(k^{2}-m^{2})}P^{T}_{\\mu \\nu} + \\frac{1}{m^{2}} P^{L}_{\\mu \\nu} \\\\ & = \\frac{1}{-(k^{2}-m^{2})}\\left( g_{\\mu \\nu} - \\frac{k_{\\mu}k_{\\nu}}{k^{2}} \\right) + \\frac{1}{m^{2}} \\frac{k_{\\mu}k_{\\nu}}{k^{2}} \\\\ & = \\frac{1}{k^{2}-m^{2}}\\left(", "imaginary image plane in front of the projection center, so that the image is no longer upside-down object camera pinhole image of the object center of projection image plane ### Perspective Projection focal length center of projection image plane focal length image plane $x$ $y$ $z$ $x$ $f$ $z$ $\\tilde{\\mathbf{P}}$ $\\mathbf{P}$ $\\tilde{\\mathbf{P}}$ $\\mathbf{P}$ $f \\frac{p_x}{p_z}$ $f$ • The formula for mapping a 3D point $\\mathbf{P}=(p_x,p_y,p_z)^\\top$ to a point $\\tilde{\\mathbf{P}}= (\\tilde{p}_x,\\tilde{p}_y,\\tilde{p}_z)^\\top$ located at the image plane of the camera is given by: $\\tilde{\\mathbf{P}}= \\left( f \\frac{p_x}{p_z}, f \\frac{p_y}{p_z}, f \\right)^\\top$ • This follows immediately from the figure by application of the intercept theorem, since $\\frac{\\tilde{p}_x}{f} = \\frac{p_x}{p_z}$ and $\\frac{\\tilde{p}_y}{f} = \\frac{p_y}{p_z}$ ### Perspective Projection • Using homogeneous coordinates the perspective projection can be written as a linear mapping using a $4 \\times 4$ matrix: $\\tilde{\\mathbf{P}}= \\begin{pmatrix} \\tilde{p}_x \\\\ \\tilde{p}_y \\\\ \\tilde{p}_z \\end{pmatrix}= \\begin{pmatrix} f \\frac{p_x}{p_z}\\\\ f \\frac{p_y}{p_z}\\\\ f \\end{pmatrix} \\in \\mathbb{R}^3 \\longmapsto \\underline{\\tilde{\\mathbf{P}}}= \\begin{pmatrix}f \\, p_x \\\\f \\, p_y \\\\ f \\, p_z\\\\ p_z\\end{pmatrix} \\in \\mathbb{H}^3$ \\begin{align}\\underline{\\tilde{\\mathbf{P}}} & = \\begin{pmatrix}f \\, p_x \\\\f \\, p_y \\\\ f \\, p_z\\\\ p_z\\end{pmatrix} = \\underbrace{\\begin{bmatrix} f & 0 & 0 & 0 \\\\ 0 & f & 0 & 0 \\\\ 0 & 0 & f & 0 \\\\ 0 & 0 & 1 & 0 \\end{bmatrix}}_{\\mathtt{A}} \\begin{pmatrix}p_x \\\\p_y \\\\ p_z\\\\ 1\\end{pmatrix}\\\\", "work, we can precompute these constants: \\begin{split}\\begin{aligned} x_{proj} &= \\frac{0.5D}{A} \\\\ y_{proj} &= 0.5D \\\\ z_{proj} &= \\frac{0.5D}{C} \\\\ w_{proj} &= -0.5-0.5B \\end{aligned}\\end{split} So, here are the equations in their final form: \\begin{split}\\begin{aligned} x_{view} &= \\frac{x_{screen} \\cdot x_{proj}}{depth + w} \\\\ y_{view} &= \\frac{y_{proj}}{depth + w} \\\\ z_{view} &= \\frac{y_{screen} \\cdot z_{proj}}{depth + w} \\end{aligned}\\end{split} Back to the List of Sample Programs: Sample Programs in the Distribution", "&0 &0 &1/\\mathrm{T}_{32} \\\\ 0 &0 &1 &-\\mathrm{T}_{22}/\\mathrm{T}_{32}\\end{bmatrix} \\! . \\end{align} /** Returns the projection-to-view matrix of this perspective camera. @return The projection-to-view matrix of this perspective camera. */ virtual const XMMATRIX GetProjectionToViewMatrix() const noexcept override { const XMMATRIX view_to_projection = GetViewToProjectionMatrix(); const F32 m00 = 1.0f / XMVectorGetX(view_to_projection.r[0]); const F32 m11 = 1.0f / XMVectorGetY(view_to_projection.r[1]); const F32 m23 = 1.0f / XMVectorGetZ(view_to_projection.r[3]); const F32 m33 = -XMVectorGetZ(view_to_projection.r[2]) * m23; return XMMATRIX { m00, 0.0f, 0.0f, 0.0f, 0.0f, m11, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, m23, 0.0f, 0.0f, 1.0f, m33 }; } Orthographic Camera \\begin{align} \\mathrm{T}_{\\mathrm{view \\rightarrow projection}} &= \\begin{bmatrix} \\mathrm{T}_{00} &0 &0 &0 \\\\ 0 &\\mathrm{T}_{11} &0 &0 \\\\ 0 &0 &\\mathrm{T}_{22} &0 \\\\ 0 &0 &\\mathrm{T}_{32} &1 \\end{bmatrix} \\! , \\\\ \\mathrm{T}_{\\mathrm{projection \\rightarrow view}} &=\\begin{bmatrix} 1/\\mathrm{T}_{00} &0 &0 &0 \\\\ 0 &1/\\mathrm{T}_{11} &0 &0 \\\\ 0 &0 &1/\\mathrm{T}_{22} &0 \\\\ 0 &0 &-\\mathrm{T}_{32}/\\mathrm{T}_{22} &1\\end{bmatrix} \\! . \\end{align} /** Returns the projection-to-view matrix of this orthographic camera. @return The projection-to-view matrix of this orthographic camera. */ virtual const XMMATRIX GetProjectionToViewMatrix() const noexcept override { const XMMATRIX view_to_projection = GetViewToProjectionMatrix(); const F32 m00 = 1.0f / XMVectorGetX(view_to_projection.r[0]); const F32 m11 = 1.0f / XMVectorGetY(view_to_projection.r[1]); const F32 m22 = 1.0f / XMVectorGetZ(view_to_projection.r[2]); const F32 m32 = -XMVectorGetZ(view_to_projection.r[3]) * m22; return XMMATRIX { m00, 0.0f, 0.0f, 0.0f, 0.0f, m11, 0.0f, 0.0f," ]

[ "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "& 0 \\\\ 0 & 0 & 0 & 0 & w_{0, 0} & w_{0, 1} & w_{0, 2} & 0 & w_{1, 0} & w_{1, 1} & w_{1, 2} & 0 & w_{2, 0} & w_{2, 1} & w_{2, 2} & 0 \\\\ 0 & 0 & 0 & 0 & 0 & w_{0, 0} & w_{0, 1} & w_{0, 2} & 0 & w_{1, 0} & w_{1, 1} & w_{1, 2} & 0 & w_{2, 0} & w_{2, 1} & w_{2, 2} \\end{bmatrix} }$$ Similarly, we reshape the input $$\\mathbf{I}$$ as a 16-dimensional vector $$\\mathbf{I}' \\in \\mathbb{R}^{16}$$. $${\\scriptstyle \\mathbf{I}' = \\begin{bmatrix} i_{0, 0} & i_{0, 1} & i_{0, 2} & i_{0, 3} & i_{1, 0} & i_{1, 1} & i_{1, 2} & i_{1, 3} & i_{2, 0} & i_{2, 1} & i_{2, 2} & i_{2, 3} & i_{3, 0} & i_{3, 1} & i_{3, 2} & i_{3, 3} \\end{bmatrix}^T }$$ Then the convolution of $$\\mathbf{W}$$ and $$\\mathbf{I}$$, that is $$\\mathbf{W} \\circledast \\mathbf{I} = \\mathbf{O} \\in \\mathbb{R}^{2 \\times 2},$$ where $$\\circledast$$ is the convolution operator, is equivalently defined as $$\\mathbf{W}' \\cdot \\mathbf{I}' = \\mathbf{O}' \\in \\mathbb{R}^{4},$$ where $$\\cdot$$ is the matrix-vector multiplication operator. The produced vector $$\\mathbf{O}'$$ can then be reshaped as a $$2 \\times 2$$ feature map. You can easily verify that this representation is correct", "+0 # help 0 91 1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ Sep 3, 2019 #1 +23357 +1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\ \\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ $$\\begin{array}{|rcll|} \\hline \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\left\\| \\dfrac{\\mathbf{v} \\cdot \\mathbf{w}}{\\left\\|\\mathbf{w}\\right\\|^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\left\\| \\dfrac{10}{7^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\| \\mathbf{w} \\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\cdot 7 \\\\\\\\ \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\mathbf{\\dfrac{10}{7}} \\\\ \\hline \\end{array}$$ Sep 4, 2019", "+0 # help 0 46 1 Let $$\\mathbf{x}$$ and $$\\mathbf{y}$$ be vectors such that $$\\operatorname{proj}_{\\mathbf{x}}(\\mathbf{y})=\\dbinom{5}{3}$$ and $$\\operatorname{proj}_{\\mathbf{y}}(\\mathbf{x})=\\dbinom{2}{-2}$$. Compute the ratio $$\\dfrac{\\|\\mathbf{x}\\|}{\\|\\mathbf{y}\\|}$$. Feb 28, 2020 #1 +24366 +1 Let $$x$$ and $$y$$ be vectors such that $$\\operatorname{proj}_{\\mathbf{x}}(\\mathbf{y})=\\dbinom{5}{3}$$and $$\\operatorname{proj}_{\\mathbf{x}}(\\mathbf{y})=\\dbinom{5}{3}$$. Compute the ratio $$\\dfrac{\\|\\mathbf{x}\\|}{\\|\\mathbf{y}\\|}$$. $$\\begin{array}{|rcll|} \\hline \\cos(\\varphi) = \\dfrac{\\sqrt{8}}{\\|\\mathbf{x}\\|} &=& \\dfrac{\\sqrt{34}}{\\|\\mathbf{y}\\|} \\\\\\\\ \\dfrac{\\sqrt{8}}{\\|\\mathbf{x}\\|} &=& \\dfrac{\\sqrt{34}}{\\|\\mathbf{y}\\|} \\\\\\\\ \\dfrac{\\|\\mathbf{x}\\|}{\\|\\mathbf{y}\\|} &=& \\dfrac{\\sqrt{8}}{\\sqrt{34}} \\\\ \\hline \\end{array}$$ Feb 28, 2020", "# arrow_back Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. 7 views Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. If $\\mathbf{n}=(a, b, c)$ is a unit vector, use this formula to show that (perhaps surprisingly) the orthogonal projection of $\\mathbf{x}$ into the plane perpendicular to $\\mathbf{n}$ is given bya) Let $\\mathbf{v}:=(a, b, c)$ and $\\mathbf{x}:=(x, y, z)$ be any vectors in $\\mathbb{R}^{3}$. Viewed as column vectors, find a $3 \\times 3$ m ... Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\mathbf{a}=(-1,0,2,1)$ and $\\mathbf{w}_{2}$ is orthogonal to a. Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\m ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Prove the orthogonal projection theorem 1 answer 7 views Prove the orthogonal projection theoremProve the orthogonal projection theorem ... close 1 answer 8 views Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the positive$x$-axis in$R^{2}$.Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss symmetry. 0 answers 7 views Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss", "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "Also, let $\\mathbf y$ be the $m$-dimensionl vector with your target values: $$\\mathbf y=\\begin{pmatrix}y^{(1)}\\\\y^{(2)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}.$$ Now, since $h_w(x^{(i)})=(x^{(i)})^T\\mathbf w$, we can verify $$\\begin{array}{rcl}X\\mathbf w -\\mathbf y&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w\\end{pmatrix}-\\begin{pmatrix}y^{(1)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}\\\\&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w-y^{(1)}\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w-y^{(m)}\\end{pmatrix}.\\end{array}$$ Thus, using $\\mathbf v^T\\mathbf v = \\sum_i\\mathbf v_i^2$ for a vector $\\mathbf v$, we obtain $$\\begin{array}{rcl}E(w)&=&\\frac{1}{2}\\sum_{i=1}^m(h_w(x^{(i)}) - y^{(i)})^2\\\\&=&\\frac{1}{2}(X\\mathbf w-\\mathbf y)^T(X\\mathbf w-\\mathbf y).\\end{array}$$ Summarizing. To obtain $$\\mathbf w=\\begin{pmatrix}w_0\\\\w_1\\end{pmatrix},$$ proceed: 1. Set your collections $x$ and $y$: $$x=\\begin{pmatrix}1\\\\3\\\\5\\\\7\\\\9\\end{pmatrix}\\qquad y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2. Loop until you think that $w_j$ converge (try with $10$ iterations) { for $i=1$ to $m$ { $$w_j:=w_j+\\eta\\sum_{i=1}^m(y^{(i)}-h_w(x^{(i)}))x_j^{(i)}\\qquad(\\text{for each }j)$$ } } Insted you can use 1'. Set your vectors $X$ and $\\mathbf y$: $$X=\\begin{pmatrix}1&1\\\\1&3\\\\1&5\\\\1&7\\\\1&9\\end{pmatrix}\\qquad\\mathbf y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2'. Loop until you think that $\\mathbf w$ converge (try with $10$ iterations) { $$\\mathbf w:=\\mathbf w-\\frac{\\eta}{m}\\left((X\\mathbf w-\\mathbf y)^TX\\right)$$ } Note. With the above it can prove a analytical formula to obtain the parameters, using the normal equations to computed $\\mathbf w$ by $$\\mathbf w=(X^TX)^{-1}X^T\\mathbf y.$$ Obviously you need to use a computer software (like python or octave) or a calculator device (with matrix-vector support) to calculate the values.", "+0 # dumb precalculus 0 45 1 For $$\\bold{v} = \\begin{pmatrix} \\phantom -2 \\\\ \\phantom -3 \\\\ -1 \\end{pmatrix} and \\bold{w} = \\begin{pmatrix} \\phantom -2 \\\\ -1 \\\\ \\phantom -0 \\end{pmatrix}$$ compute the projection of $\\mathbf v$ onto $\\mathbf w.$ i literally dont know how to solve this Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ MaxWong Apr 18, 2022", "\\theta & =\\mathbf{v} \\cdot \\mathbf{w} \\end{aligned} When $v,w$ are unit vectors then, $$v \\cdot w=\\cos \\theta$$ Which is maximized when $\\theta=0$, i.e. $v=w$. Thus the dot product is often used to measure the similarity between two vectors. ##### Projections Scalar projection of $\\mathbf{a}$ onto $\\mathbf{b}$: $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction} \\cdot \\mathbf{a}$$ This essentially gives the signed length of the projection. $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\|a\\|\\cos(\\theta) = \\|\\mathbf{a}\\|\\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|\\|\\mathbf{b}\\|} = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|}$$ Vector projection of $\\mathbf{a}$ onto $\\mathbf{b}:$ $$\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}= \\underbrace{\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}}_{signed \\ length}\\quad \\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction}=\\frac{\\mathbf{b} \\cdot \\mathbf{a}}{|\\mathbf{b}|^{2}} \\mathbf{b}$$ The first equality is more meaningful, as it gives the length-direction decomposition. For any non-zero constant $c$ $$\\operatorname{proj}_{\\mathbf{cb}} \\mathbf{a}=\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}$$ In other words, the projection of $\\mathbf{a}$ onto $\\mathbf{b}$ is independent of the length of $\\mathbf{b}$, only the direction of the projection matters. #### The Cross Product ##### Motivation Given two vectors $u,v$, we want to find a vector $w$, that is perpendicular to both $u$ and $v$. This can be done by solving the following equations: $$u\\cdot w = u_1w_1+u_2w_2+u_3w_3 = 0$$ $$v\\cdot w = v_1w_1+v_2w_2+v_3w_3 = 0$$ Which implies $$(u_1v_3-u_3v_1)w_1+(u_2v_3-u_3v_2)w_2=0$$ One trivial solution is to take $$w_1=u_2v_3-u_3v_2,\\ w_2 = u_3v_1-u_1v_3, \\ w_3=u_1v_2-u_2v_1$$ The solution give the cross product of $u$ and $v$. ##### Computational Properties Often the above the formula is not very intuitive, and difficult to recall. Instead, one often use", "is, the projection a vector $$\\mathbf a$$ is given by $$\\operatorname{proj}_{\\mathbf G}(\\mathbf a) = \\frac{\\mathbf g_1\\cdot \\mathbf a}{\\|\\mathbf g_1\\|^2} \\mathbf g_1 + \\cdots + \\frac{\\mathbf g_k \\cdot \\mathbf a}{\\|\\mathbf g_k\\|^2}\\mathbf g_k \\\\ = (\\mathbf g_1 \\cdot \\mathbf a)\\mathbf g_1 + \\cdots + (\\mathbf g_k \\cdot \\mathbf a)\\mathbf g_k.$$ The coefficients of this sum are $$\\mathbf g_1\\cdot \\mathbf a, \\dots, \\mathbf g_k \\cdot \\mathbf a$$, which are the entries of the product $$\\mathbf G^T\\mathbf a.$$ To get the corresponding projected vector, take the corresponding linear combination of the columns of $$\\mathbf G$$. In other words, we have $$\\operatorname{proj}_{\\mathbf G}(\\mathbf a) = \\mathbf G(\\mathbf G^T \\mathbf a).$$ With that in mind, we can find the coefficients of the projection for each column of the matrix $$\\mathbf M = \\mathbf X - \\mathbb E(\\mathbf X)$$ by computing $$\\mathbf G^T\\mathbf M$$ and obtain the vectors corresponding to these columns of coefficients with the product $$\\mathbf G\\mathbf G^T\\mathbf M$$.", "2 \\\\ 2 \\end{bmatrix}$ and obtain \\begin{align*} \\mathbf{x}\\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} =t\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}\\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} + \\mathbf{v} \\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} \\\\ \\Leftrightarrow \\,\\,\\,\\, x_1+2x_2+2x_3=9t \\end{align*} Here $\\mathbf{v} \\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}=\\mathbf{0}$ since $\\mathbf{v}$ is perpendicular to $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$. Therefore we have $t=\\frac{1}{9}(x_1+2x_2+2x_3)$, and the formula is $T(\\mathbf{x})=\\frac{1}{9}(x_1+2x_2+2x_3)\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}.$ ### (b) Find a basis for the image subspace of $T$ Let $V$ be the subspace spanned by $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$ in $\\R^3$. Since $T$ is a projection to the subspace $W$, the image is $W$ itself. (Any vector in $W$ is mapped to itself by the projection $T$.) Since $W$ is spanned by just one vector $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$, it is one-dimensional and a basis is $\\left\\{\\, \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} \\,\\right\\}$. ### (c) Find a basis for the kernel subspace of $T$ If $\\mathbf{x}=\\begin{bmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{bmatrix} \\in \\ker T$, then by the formula in part (a), we have $x_1+2x_2+2x_3=0$. Thus the vectors in the kernel of $T$ can be written as $\\mathbf{x}=\\begin{bmatrix} -2s-2t \\\\ s \\\\ t \\end{bmatrix}= \\begin{bmatrix} -2", "2 \\\\ 2 \\end{bmatrix}$ and obtain \\begin{align*} \\mathbf{x}\\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} =t\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}\\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} + \\mathbf{v} \\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} \\\\ \\Leftrightarrow \\,\\,\\,\\, x_1+2x_2+2x_3=9t \\end{align*} Here $\\mathbf{v} \\cdot \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}=\\mathbf{0}$ since $\\mathbf{v}$ is perpendicular to $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$. Therefore we have $t=\\frac{1}{9}(x_1+2x_2+2x_3)$, and the formula is $T(\\mathbf{x})=\\frac{1}{9}(x_1+2x_2+2x_3)\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}.$ ### (b) Find a basis for the image subspace of $T$ Let $V$ be the subspace spanned by $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$ in $\\R^3$. Since $T$ is a projection to the subspace $W$, the image is $W$ itself. (Any vector in $W$ is mapped to itself by the projection $T$.) Since $W$ is spanned by just one vector $\\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix}$, it is one-dimensional and a basis is $\\left\\{\\, \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\end{bmatrix} \\,\\right\\}$. ### (c) Find a basis for the kernel subspace of $T$ If $\\mathbf{x}=\\begin{bmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{bmatrix} \\in \\ker T$, then by the formula in part (a), we have $x_1+2x_2+2x_3=0$. Thus the vectors in the kernel of $T$ can be written as $\\mathbf{x}=\\begin{bmatrix} -2s-2t \\\\ s \\\\ t \\end{bmatrix}= \\begin{bmatrix} -2", "v_2)^2$ is minimized. W.l.o.g. we can take $\\mathbf\\pi_1$ to be the $u$-$v$ plane, so that its camera matrix is the canonical $$P_1=\\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}.$$ This just picks out the $u$- and $v$-coordinates of $\\mathbf U$. The projection matrix onto $\\mathbf\\pi_2$ will have the form $$P_2=\\left[\\begin{array}{c|c}M_{2\\times 3}&\\mathbf t\\\\\\mathbf 0^T&1\\end{array}\\right].$$ The point $\\mathbf U$ then satisfies the linear equations $P_1\\mathbf U =\\mathbf u_1=(u_1,v_1)$ and $P_2\\mathbf U = \\mathbf u_2=(u_2,v_2)$. The affine fundamental matrix of this epipolar configuration is $$F_A=\\begin{bmatrix}0&0&a\\\\0&0&b\\\\c&d&e\\end{bmatrix}$$ where $$a = m_{23} \\\\ b=-m_{13} \\\\ c=m_{13}m_{21}-m_{11}m_{23} \\\\ d=m_{13}m_{22}-m_{12}m_{23} \\\\ e=m_{13}t_2-m_{23}t_1.$$ (In your case, where $P_2$ has no translation component, we would have $e=0$.) The adjusted coordinates are then given by the formula $$\\begin{bmatrix}\\hat u_2 \\\\ \\hat v_2 \\\\ \\hat u_1 \\\\ \\hat v_1\\end{bmatrix} = \\begin{bmatrix}u_2 \\\\ v_2 \\\\ u_1 \\\\ v_1\\end{bmatrix}-{au_2+bv_2+cu_1+dv_1+e\\over a^2+b^2+c^2+d^2}\\begin{bmatrix}a\\\\b\\\\c\\\\d\\end{bmatrix}.$$ (You might recognize this as the result of a standard least-squares fit, which itself is basically a projection onto a hyperplane in $\\mathbb R^4$.) With these adjusted coordinates in hand, you can then find the intersection of their back-projections in the ways described previously. • Thank you for taking the time to explain this, however, I do not understand what you mean when you say that U will lie on the intersection of those two lines (I assume lambda and mu are just arbitrary variables you chose without any", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection" ]

[ "+0 help 0 94 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times", "+0 # help 0 52 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 ### Best Answer #1 +24983 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 ### 1+0 Answers #1 +24983 +3 Best Answer Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a}", "Fluency exercise ## Solution A square is enclosed by the lines: 1. $\\mathbf{r}=\\begin{pmatrix}0\\\\2\\end{pmatrix} + a\\begin{pmatrix}1\\\\1\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}0\\\\6\\end{pmatrix} + b\\begin{pmatrix}-1\\\\1\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}2\\\\1\\end{pmatrix} + c\\begin{pmatrix}2\\\\2\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}1\\\\2\\end{pmatrix} + d\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ Can you convince yourself that the adjacent edges are perpendicular? We need to consider the direction vector of each line. Lines (1) and (2) do not have the same direction vector and so they will cross at some point. If we now think about the behaviour of each line as it moves away from the crossing point we can show that they are perpendicular using the diagram below. Can you use a similar argument to convince yourself about the other adjacent edges of the square? Find the area of the square. To do this we need to identify the positions of some of the vertices. We could aim to find the length of one side by finding two adjacent vertices, or we could find the length of the diagonal by identifying vertices opposite each other. It doesn’t matter which pair we choose to find so let’s say we look for vertices A and B shown on the diagram below. To find the position of point A, we need to find the point that is common to equations (1) and (4), (they need to have the same position vector $\\mathbf{r}$). With that", "\\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ heureka Apr 30, 2020", "Fluency exercise ## Suggestion A square is enclosed by the lines: 1. $\\mathbf{r}=\\begin{pmatrix}0\\\\2\\end{pmatrix} + a\\begin{pmatrix}1\\\\1\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}0\\\\6\\end{pmatrix} + b\\begin{pmatrix}-1\\\\1\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}2\\\\1\\end{pmatrix} + c\\begin{pmatrix}2\\\\2\\end{pmatrix}$ 1. $\\mathbf{r}=\\begin{pmatrix}1\\\\2\\end{pmatrix} + d\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ Find the area of the square. • Which line is which? • What are the positions of the vertices of the square? • Could you have predicted that the enclosed square would lie entirely within the first quadrant?", "treat the top numbers and bottom numbers completely separately. \\begin{aligned} 2\\begin{pmatrix} 3\\\\\\text{-}1 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} &= \\begin{pmatrix} 6\\\\\\text{-}2 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} \\\\ &=\\begin{pmatrix} 6-7\\\\\\text{-}2-1 \\end{pmatrix} \\\\ &= \\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}} \\end{aligned} This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down: We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down. The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour. This is the vector: $\\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}}$ ### Question 4: The Area of a Triangle The area of a triangle can be found my multiplying the base by the height and dividing by 2. However, this requires us to know the perpendicular height of the triangle. If we don’t know this measurement, we can use a slightly different formula to find the area. Consider the triangle below: Any triangle can be labelled like this. The lowercase letters denote the sides and the uppercase letters denote the angles. Each angle is", "## Algebra and Trigonometry 10th Edition $\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}=5$ Fifth row of Pascal's triangle: $1~~5~~10~~10~~5~~1$ It corresponds to: $\\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$ $\\begin{pmatrix} 5 \\\\ 2 \\end{pmatrix}$ $\\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix}$ $\\begin{pmatrix} 5 \\\\ 4 \\end{pmatrix}$ $\\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$ So, $\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}=5$", "Math Help - Altitudes of tetrahedron intersect, coordinate vectors 1. Altitudes of tetrahedron intersect, coordinate vectors A tetrahedron has vertices A, B, C and D with coordinate vectors for the points being: $\\mathbf{a}=\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix},\\ \\mathbf{b}=\\begin{pmatrix}-1\\\\4\\\\1\\end{pmatrix},$ $\\mathbf{c}= \\begin{pmatrix}1\\\\0\\\\3\\end{pmatrix},\\ and\\ \\mathbf{d}=\\begin{pmatrix}-3\\\\1\\\\2\\end{pmatrix}$ Find parametric vector equations for the two altitudes of the tetrahedron which pass through the vertices A and B, and determine whether the two altitudes intersect or not. Altitude is a line through the vertex and perpendicular to the opposite face. For my altitudes, I managed to get: $\\mathbf{x}=\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-2\\\\6\\\\14\\end{pmatrix} and\\ \\mathbf{x}=\\begin{pmatrix}-1\\\\4\\\\1\\end{pmatrix}+\\lambda\\begin{pmatrix}0\\\\-3\\\\3\\end{pmatrix}$ Not too sure if that's right 2. Originally Posted by acevipa A tetrahedron has vertices A, B, C and D with coordinate vectors for the points being: $\\mathbf{a}=\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix},\\ \\mathbf{b}=\\begin{pmatrix}-1\\\\4\\\\1\\end{pmatrix},$ $\\mathbf{c}= \\begin{pmatrix}1\\\\0\\\\3\\end{pmatrix},\\ and\\ \\mathbf{d}=\\begin{pmatrix}-3\\\\1\\\\2\\end{pmatrix}$ Find parametric vector equations for the two altitudes of the tetrahedron which pass through the vertices A and B, and determine whether the two altitudes intersect or not. Altitude is a line through the vertex and perpendicular to the opposite face. For my altitudes, I managed to get: $\\mathbf{x}=\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-2\\\\6\\\\14\\end{pmatrix} and\\ \\mathbf{x}=\\begin{pmatrix}-1\\\\4\\\\1\\end{pmatrix}+\\lambda\\begin{pmatrix}0\\\\-3\\\\3\\end{pmatrix}$ Not too sure if that's right The equations of the lines containing the altitudes are correct. BTW: The two lines don't intersect. 3. Originally Posted by earboth The equations of the lines containing the altitudes are correct. Ok great, so when it asks", "# Find area of the triangle with vertices at the point given in (1, 0), (6, 0), (4, 3) $\\begin{array}{1 1} 15/2 \\\\ 30 \\\\ 15 \\\\ 60 \\end{array}$ Toolbox: • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by • $\\bigtriangleup=\\frac{1}{2}\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 &y_3 &1\\end{vmatrix}$ • $\\bigtriangleup=\\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$ Let $(x_1,y_1)$ be=(1,0) Let $(x_2,y_2)$ be=(6,0) Let $(x_3,y_3)$ be=(4,3) Now area of a triangle is given by $\\bigtriangleup=\\frac{1}{2}\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 & y_3 & 1\\end{vmatrix}$ Substituting the respective values we get, $\\bigtriangleup=\\frac{1}{2}\\begin{vmatrix}1 & 0 & 1\\\\6 & 0 & 1\\\\4 & 3 & 1\\end{vmatrix}$ Now expanding along the row $R_1$ we get $\\frac{1}{2}\\begin{bmatrix}1\\begin{vmatrix}0 & 1\\\\3 & 1\\end{vmatrix}-0\\begin{vmatrix}6 & 1\\\\4 & 1\\end{vmatrix}+1\\begin{vmatrix}6 & 0\\\\4 & 3\\end{vmatrix}\\end{bmatrix}$ $=\\frac{1}{2}[1(-3)-0+1(18)]$ $=\\frac{1}{2}[-3+18]=15/2.$ Hence the area is 15/2sq.units. edited Feb 24, 2013", "the edge. Consider a triangle $T$ with three vertices $(\\mathbf{a},\\mathbf{b},\\mathbf{c})$. Let denote the vertices numbers by $i_{a},i_{b},i_{c}$, and define the three edge vectors $\\mathbf{e}^{0},\\mathbf{e}^{1},\\mathbf{e}^{2}$ by $sgn(i_{b}-i_{c})(\\mathbf{b}-\\mathbf{c})$, $sgn(i_{c}-i_{a})(\\mathbf{c}-\\mathbf{a})$, $sgn(i_{a}-i_{b})(\\mathbf{a}-\\mathbf{b})$. The three basis functions are: where $|T|$ is the area of the triangle $T$. So we have $N=2$, $\\mathtt{kPi}=6; \\mathtt{npPi}=3;$ and: • $P_{p} = \\left\\{\\frac{\\mathbf{b}+\\mathbf{c}}{2}, \\frac{\\mathbf{a}+\\mathbf{c}}{2}, \\frac{\\mathbf{b}+\\mathbf{a}}{2} \\right\\}$ • $\\alpha_{0}= - \\mathbf{e}^{0}_{2}, \\alpha_{1}= \\mathbf{e}^{0}_{1}$, $\\alpha_{2}= - \\mathbf{e}^{1}_{2}, \\alpha_{3}= \\mathbf{e}^{1}_{1}$, $\\alpha_{4}= - \\mathbf{e}^{2}_{2}, \\alpha_{5}= \\mathbf{e}^{2}_{1}$ (effectively, the vector $(-\\mathbf{e}^{m}_{2}, \\mathbf{e}^{m}_{1})$ is orthogonal to the edge $\\mathbf{e}^{m}= (e^m_{1},e^m_{2})$ with a length equal to the side of the edge or equal to $\\int_{e^m} 1$). • $i_{k}=\\{0,0,1,1,2,2\\}$, • $p_{k}=\\{0,0,1,1,2,2\\}$ , $j_{k}=\\{0,1,0,1,0,1,0,1\\}$. Add file FE_ADD.cpp in directory FreeFem-sources/src/femlib for example first to initialize : 1 2 3 4 5 6 7 8 9 #include \"error.hpp\" #include \"rgraph.hpp\" using namespace std; #include \"RNM.hpp\" #include \"fem.hpp\" #include \"FESpace.hpp\" #include \"AddNewFE.h\" namespace Fem2D { ... } Then add a class which derive for public TypeOfFE like: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 class TypeOfFE_RTortho : public TypeOfFE { public: static int Data[]; //some numbers TypeOfFE_RTortho(): TypeOfFE( 0+3+0, //nb degree of freedom on element 2, //dimension N of vectorial FE (1 if scalar FE) Data,", "\\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ heureka Apr 30, 2020", "Find the value of $k$ if area of triangle is $4\\; sq. units$ and vertices are: $(k, 0), (4, 0), (0, 2)$ $\\begin{array}{1 1} 8 \\\\ 4 \\\\ 0 \\\\ -2 \\end{array}$ Toolbox: • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by • $\\bigtriangleup=\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 &y_3 &1\\end{vmatrix}$ • $\\mid\\bigtriangleup\\mid=\\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$ Let $(x_1,y_1)=(k,0)$ $\\;\\;\\;(x_2,y_2)=(4,0)$ $\\;\\;\\;(x_3,y_3)=(0,2)$ Given the area of the triangle is $\\bigtriangleup=\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 &y_3 &1\\end{vmatrix}$ But the area given is 4sq.units. Substituting the respective values we get, $4=\\frac{1}{2}\\begin{vmatrix}k & 0 & 1\\\\4 & 0 & 1\\\\0 & 2 & 1\\end{vmatrix}$ Now expanding along the row $R_1$ $4=\\frac{1}{2}\\begin{bmatrix}k\\begin{vmatrix}0 & 1\\\\2 & 1\\end{vmatrix}-0\\begin{vmatrix}4 & 1\\\\0 & 1\\end{vmatrix}+1\\begin{vmatrix}4 & 0\\\\0 & 2\\end{vmatrix}\\end{bmatrix}$ $4=\\pm[\\frac{1}{2}[k(0-2)-0+1(8-0)]]$ Since the area can only be positive, the value of the determinant is modulus of | -2k + 8|. That is it will be either +(-2k+8 ) or - (2k + 8). 8=+[-2k+8] $\\Rightarrow$ then 2k=0. Hence k=0. If 8=-[-2k+8] 8=-[-2k+8] 8=2k-8$\\Rightarrow 2k=16.$ or $k=8.$ Hence the value of k is (0,8). edited Feb 24, 2013", "$OP$ be $\\mathbf{p} = \\begin{pmatrix}p_i \\\\ p_j \\\\ p_k\\end{pmatrix}$, so we have $\\mathbf{p}.\\mathbf{a} = 3p_i+4p_j+5p_k$. But $\\mathbf{p}.\\mathbf{a} = |\\mathbf{p}||\\mathbf{a}|\\cos \\alpha$, so $\\cos \\alpha = \\dfrac{\\mathbf{p}.\\mathbf{a}}{|\\mathbf{p}|\\sqrt{50}}=\\dfrac{3p_i+4p_j+5p_k}{|\\mathbf{p}|\\sqrt{50}}$. We have used the fact that $|\\mathbf{a}|=\\sqrt{3^2+4^2+5^2} = \\sqrt{50}$. Since the three triangles are congruent, they each contain the same angle $\\alpha$. Also notice that $|\\mathbf{b}|=|\\mathbf{c}|=\\sqrt{50}$. So for vector $\\mathbf{b}$ we have $\\cos \\alpha= \\dfrac{7p_i-p_k}{|\\mathbf{p}|\\sqrt{50}}$, and for $\\mathbf{c}$ we have $\\cos \\alpha = \\dfrac{5p_i+5p_j}{|\\mathbf{p}|\\sqrt{50}}$. Combining these we can write $3p_i+4p_j+5p_k = 7p_i-p_k= 5p_i+5p_j$. Using the second and third parts gives $p_k = 2p_i - 5p_j$, and substituting this into the first two parts gives $3p_i+4p_j+5(2p_i - 5p_j) = 7p_i-(2p_i - 5p_j)$. Thus $p_j = \\dfrac{4}{13} p_i$, and $p_k = \\dfrac{6}{13} p_i$, and so the vector $\\mathbf{p}$ is $t \\begin{pmatrix}13 \\\\ 4 \\\\ 6\\end{pmatrix}$ for some scalar $t$. To find the unit vector in this direction, we divide by the magnitude $\\sqrt{13^2+4^2+6^2} = \\sqrt{221}$, and so the unit vector in the direction of $\\mathbf{OP}$ is $\\hat{\\mathbf{p}} = \\frac{13}{\\sqrt{221}} \\mathbf{i} + \\frac{4}{\\sqrt{221}}\\mathbf{j} + \\frac{6}{\\sqrt{221}}\\mathbf{k}.$ Given that the line $OP$ meets the plane $ABC$ at $Q$, find $\\mathbf{OQ}$. We can define the plane $ABC$ from the three position vectors $\\mathbf{a}$, $\\mathbf{b}$ and $\\mathbf{c}$ via $\\mathbf{r} = \\mathbf{a} + \\lambda(\\mathbf{b}-\\mathbf{a}) + \\mu(\\mathbf{c}-\\mathbf{a}),$ where $\\lambda$ and $\\mu$ are scalar parameters. If we start at the origin, $\\mathbf{a}$ takes", "mere rotation and scaling of the triangle, disregarding the translation, i.e. $$\\mathbf{F}=\\mathbf{B}\\hat{\\mathbf{x}}$$. Let's compute $$\\phi_1^{(V)}$$ and $$\\phi_1^{(IV)}$$ and show that they in fact correspond to the restriction of plane $$f_1(x,y)$$ to the contiguous triangles $$\\mathcal{K}_V$$ and $$\\mathcal{K}_{IV}$$. The same operations can be done for all couples of triangles to check the glueing of the local basis functions. Triangle V $$j_g=1\\to j=1,\\quad j_g=6\\to j=2,\\quad j_g=7\\to j=3\\\\ \\mathbf{B}=\\begin{pmatrix}1 & 1 \\\\ 0 & 1\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{B}^{-1}=\\begin{pmatrix}1 & -1 \\\\ 0 & 1\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{F}^{-1}=\\mathbf{B}^{-1}\\mathbf{x}=\\begin{pmatrix}x-y\\\\y\\end{pmatrix}\\\\ \\phi_1^{(V)}=\\hat{\\varphi}_1\\circ\\mathbf{F}^{-1}=(1-\\hat{x}-\\hat{y})\\circ\\begin{pmatrix}x-y\\\\y\\end{pmatrix}=1-(x-y)-(y)=1-x\\equiv f_1(x,y)\\vert_{\\mathcal{K}_{V}}$$ Triangle IV $$j_g=1\\to j=1,\\quad j_g=5\\to j=2,\\quad j_g=7\\to j=6\\\\ \\mathbf{B}=\\begin{pmatrix}1 & 1 \\\\ -1 & 0\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{B}^{-1}=\\begin{pmatrix}0 & -1 \\\\ 1 & 1\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{F}^{-1}=\\mathbf{B}^{-1}\\mathbf{x}=\\begin{pmatrix}-y\\\\x+y\\end{pmatrix}\\\\ \\phi_1^{(IV)}=\\hat{\\varphi}_1\\circ\\mathbf{F}^{-1}=(1-\\hat{x}-\\hat{y})\\circ\\begin{pmatrix}-y\\\\x+y\\end{pmatrix}=1-(-y)-(x+y)=1-x\\equiv f_1(x,y)\\vert_{\\mathcal{K}_{IV}}$$ We can easily demonstrate that, as long as we keep consistent the ordering of the nodes it does not matter which node we do number first; for example for Triangle V let's change the first node not to be the global origin but the upper right corner. Now the global origin will coincide with the second locally numbered node according to the convention we specified. Therefore we'll have Triangle V $$j_g=7\\to j=1,\\quad j_g=1\\to j=2,\\quad j_g=6\\to j=3\\\\ \\mathbf{B}=\\begin{pmatrix}-1 & 0 \\\\ -1 & -1\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{B}^{-1}=\\begin{pmatrix}-1 & 0 \\\\ 1 & -1\\end{pmatrix}\\;\\Rightarrow\\;\\mathbf{F}^{-1}=\\mathbf{B}^{-1}(\\mathbf{x}-\\mathbf{x}_1)=\\begin{pmatrix}-(x-1)\\\\(x-1)-(y-1)\\end{pmatrix}\\\\ \\phi_1^{(V)}=\\hat{\\varphi}_1\\circ\\mathbf{F}^{-1}=(\\hat{x})\\circ\\begin{pmatrix}1-x\\\\x-y\\end{pmatrix}=1-x\\equiv f_1(x,y)\\vert_{\\mathcal{K}_{V}}$$ which reresults in the identical global basis function. For the P1-element, global continuity is enforced by choosing function values at the vertices as degrees", "a triangle of area 18 sq units, then the centroid of the triangle is (a) (8,8) (b) (4,4) (c) (3,3) (d) (2,2) 8. The locus of a point which is equidistant from (-1,1) and (4,2) is (a) 5x+3y+9=0 (b) 5x+3y-9=0 (c) 3x-5y=0 (d) 3x+5y-9=0 9. On using elementary row operation R1⟶R1-3R2 in the following matrix equation $\\begin{pmatrix} 4 & 2 \\\\ 3 & 3 \\end{pmatrix}=\\begin{pmatrix} 1 & 2 \\\\ 0 & 3 \\end{pmatrix}\\begin{pmatrix} 2 & 0 \\\\ 1 & 1 \\end{pmatrix}$ (a) $\\begin{pmatrix} -5 & -7 \\\\ 3 & 3 \\end{pmatrix}=\\begin{pmatrix} 1 & -7 \\\\ 0 & 3 \\end{pmatrix}\\begin{pmatrix} 2 & 0 \\\\ 1 & 1 \\end{pmatrix}$ (b) $\\begin{pmatrix} -5 & -7 \\\\ 3 & 3 \\end{pmatrix}=\\begin{pmatrix} 1 & 2 \\\\ 0 & 3 \\end{pmatrix}\\begin{pmatrix} -1 & -3 \\\\ 1 & 1 \\end{pmatrix}$ (c) $\\begin{pmatrix} -5 & -7 \\\\ 3 & 3 \\end{pmatrix}=\\begin{pmatrix} 1 & 2 \\\\ 1 & -7 \\end{pmatrix}\\begin{pmatrix} 2 & 0 \\\\ 1 & 1 \\end{pmatrix}$ (d) $\\begin{pmatrix} 4 & 2 \\\\ -5 & -7 \\end{pmatrix}=\\begin{pmatrix} 1 & 2 \\\\ -3 & -3 \\end{pmatrix}\\begin{pmatrix} 2 & 0 \\\\ 1 & 1 \\end{pmatrix}$ 10. Choose the correct statement (a) (b) (c) Matrix multiplication is associative (d) Matrix multiplication is commutative", "diagonal, length 3, of the enclosed square. Now that we have the position vectors of vertices A and B we could find the length of the line AB using Pythagoras. However, you might notice that the area of the square will be given by $AB^2$ and we can write this down directly, $AB^2 = 2 \\times \\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ You could also have noticed that the right-angled triangle joining points A and B is isosceles. This tells us that the square could be divided up into four of these triangles. We can now show that the area of the square is again equal to $4 \\times \\frac{1}{2}\\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ What can we now say about the positions of the other vertices and the diagonals of the square? Find a vector equation for each of the two diagonals. In order to write down a vector equation of a line we need to know the position of a point on that line and a direction vector. For the vertical diagonal of the square we know that point B, with position vector $\\begin{pmatrix}2\\\\4\\end{pmatrix}$ lies on the line and that the direction vector is $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. A vector equation of the vertical diagonal of the square is therefore $\\mathbf{r}=\\begin{pmatrix}2\\\\4\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix}.$ We could also have written $\\mathbf{r}=\\begin{pmatrix}2\\\\1\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix},$ or any vector equation", "\\end{bmatrix}$$ $$\\begin{bmatrix} -3 & 1 & 0\\\\ 2 & 5 & 3 \\end{bmatrix}$$ $$\\begin{bmatrix} -3 & 1 & 0\\\\ -2 & -5 & -3 \\end{bmatrix}$$ $$\\begin{bmatrix} 3 & -1 & 0\\\\ -2 & -5 & -3 \\end{bmatrix}$$ To use matrices to rotate the coordinates of the vertices of triangle T by $$90°$$ we will multiply the coordinates of the vertices of triangle T by $$\\begin{bmatrix} 0 & -1\\\\ 1 & 0 \\end{bmatrix}$$, which results in the following coordinates, $$\\begin{bmatrix} 3 & -1 & 0\\\\ 2 & 5 & 3 \\end{bmatrix}$$.", "# Find values of $k$ if area of triangle is $4\\; sq. units$ and vertices are $(-2, 0), (0, 4), (0, k)$ $\\begin{array}{1 1} 1 \\;and\\; -8 \\\\ 1\\; and\\; 8 \\\\ 0\\; and\\; -8 \\\\ 0\\; and\\; 8 \\end{array}$ Toolbox: • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by • $\\bigtriangleup=\\frac{1}{2}\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 &y_3 &1\\end{vmatrix}$ • $|\\bigtriangleup|=\\frac{1}{2}\\begin{bmatrix}x_1\\begin{vmatrix}y_2 & 1\\\\y_3 & 1\\end{vmatrix}-y_1\\begin{vmatrix}x_2 & 1\\\\x_3 & 1\\end{vmatrix}+1\\begin{vmatrix}x_2 & y_2\\\\x_3 & y_3\\end{vmatrix}\\end{bmatrix}$ Let $(-2,0)=(x_1,y_1);(0,4)=(x_2,y_2);(0,k)=(x_3,y_3)$ The area of the triangle is given by $\\bigtriangleup=\\frac{1}{2}\\begin{vmatrix}x_1 & y_1 & 1\\\\x_2 & y_2 & 1\\\\x_3 &y_3 &1\\end{vmatrix}$ Now substituting the respective values we get Given area =4sq.units. Hence $4=\\frac{1}{2}\\begin{vmatrix}-2 & 0 & 1\\\\0 & 4 & 1\\\\0 & k & 1\\end{vmatrix}$ Now expanding along the row $R_1$ we get $\\mid\\bigtriangleup\\mid=\\frac{1}{2}\\begin{bmatrix}-2\\begin{vmatrix}4 & 1\\\\k & 1\\end{vmatrix}-0\\begin{vmatrix}0 & 1\\\\0 & 1\\end{vmatrix}+1\\begin{vmatrix}0 & 4\\\\0 & k\\end{vmatrix}\\end{bmatrix}$ We take + or - because the determinant value is absolute value and hence it can be either + or -. 8=$\\pm[-2(4-k)]$ Therefore if 8=+(-8+2k) then 2k=16 k=8 Therefore if 8=-(-8+2k) then k=8-8 k=0. Hence the value of k is (0,8)", "# Area of the triangle formed by three vector lines I've had made several failed attempts at this question, but I'll show you my most recent one. Question: Show that the lines with equations $$\\mathbf r_1 = \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\\\ \\end{pmatrix} + \\lambda\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\\\ \\end{pmatrix}$$ $$\\mathbf r_2 = \\begin{pmatrix} -2 \\\\ 3 \\\\ -1 \\\\ \\end{pmatrix} + \\mu\\begin{pmatrix} 1 \\\\ -1 \\\\ 0 \\\\ \\end{pmatrix}$$ $$\\mathbf r_3 = \\begin{pmatrix} 2 \\\\ -1 \\\\ -1 \\\\ \\end{pmatrix} + t\\begin{pmatrix} -1 \\\\ 2 \\\\ 3 \\\\ \\end{pmatrix}$$ form a triangle, and find its area. I've 5.53 and 6.02 square units in my solution. For the sake of brevity, I'll outline my method as my actual attempt is several pages long. 1. Find co-ordinates/position vectors for the vertices of the triangle using: $$r_1 = r_2$$ $$r_1 = r_3$$ $$r_2 = r_3$$ 2. Using position vectors of vertices to find vector equations of the three lines. 3. Calculate the magnitude of two of the lines. 4. Calculate the dot product of these two lines. 5. Find the angle between these two lines using $$cos\\theta = \\frac{a \\cdot b}{\\lvert a\\rvert \\lvert b\\rvert}$$ Now the area of the triangle is $$Area = \\frac{1}{2}\\lvert a\\rvert \\lvert b\\rvert sin\\theta$$ However, this has given me 6.02 and 5.53", "the area of the paralellogram and the original square. Since the original square has area $1$, the area of the parallelogram is the determinant of $\\bc{\\A}$. Working this out requires only a small amount of basic geometry. Here’s the simplest way to do it. We first name the four elements of our matrix: $\\bc{\\A} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ We can now name the four corners of our parallelogram in terms of these four scalars, by multiplying the corner points of the unit square by $\\bc{\\A}$: $\\begin{matrix} \\times & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}1\\\\ 0\\end{pmatrix} & \\begin{pmatrix}0\\\\ 1\\end{pmatrix} & \\begin{pmatrix}1\\\\ 1\\end{pmatrix} \\\\ \\bc{\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}} & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}a\\\\ c\\end{pmatrix} & \\begin{pmatrix}b\\\\ d\\end{pmatrix} & \\begin{pmatrix}a+b\\\\ c+d\\end{pmatrix} \\\\ \\end{matrix}$ This gives us the following picture. Here we can see the area of the paralellogram clearly: there is one large rectangle of area $\\oc{ad}$. To get from this area to the area of our parallelogram, we should subtract the area of the green triangle in the bottom, which is part of the rectangle but not the parallelogram. But then, there’s a green triangle at the top, with the same size, which is (mostly) part of the parallelogram but not of the rectangle, so these cancel each other out. We" ]

[ "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "& -3 \\end{array}} \\right| = \\left({-12+21}\\right)\\mathbf{i} - \\left({8-14}\\right)\\mathbf{j} + 0\\cdot\\mathbf{k} = 9\\mathbf{i} + 6\\mathbf{j}.$ Hence, $$\\mathbf{M_A} = \\left({9,6,0}\\right).$$ ### Example 3. Calculate the area of a triangle with vertices $$A\\left({1,-2,0}\\right),$$ $$B\\left({3,1,-1}\\right)$$ and $$C\\left({2,-1,3}\\right).$$ Solution. The area of the triangle is equal to half the area of the parallelogram spanned by the vectors AB and AC, that is $A_{\\triangle} = \\frac{1}{2}\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert.$ Calculate the coordinates of vectors AB and AC: $\\mathbf{AB} = \\underbrace{\\left({3,1,-1}\\right)}_B - \\underbrace{\\left({1,-2,0}\\right)}_A = \\left({2,3,-1}\\right);$ $\\mathbf{AC} = \\underbrace{\\left({2,-1,3}\\right)}_C - \\underbrace{\\left({1,-2,0}\\right)}_A = \\left({1,1,3}\\right);$ Find the cross product: $\\mathbf{AB} \\times \\mathbf{AC} = \\left| {\\begin{array}{*{20}{c}} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 2 & 3 & -1\\\\ 1 & 1 & 3 \\end{array}} \\right| = \\mathbf{i}\\left| {\\begin{array}{*{20}{c}} 3 & -1\\\\ 1 & 3 \\end{array}} \\right| - \\mathbf{j}\\left| {\\begin{array}{*{20}{c}} 2 & -1\\\\ 1 & 3 \\end{array}} \\right| + \\mathbf{k}\\left| {\\begin{array}{*{20}{c}} 2 & 3\\\\ 1 & 1 \\end{array}} \\right| = \\left({9+1}\\right)\\mathbf{i} - \\left({6+1}\\right)\\mathbf{j} + \\left({2-3}\\right)\\mathbf{k} = 10\\mathbf{i} - 7\\mathbf{j} - \\mathbf{k}.$ The absolute value of this vector is $\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert = \\sqrt{10^2 + \\left({-7}\\right)^2 + \\left({-1}\\right)^2} = \\sqrt{150} =5\\sqrt{6}.$ Then the area of the triangle is $A_{\\triangle} = \\frac{1}{2}\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert = \\frac{5\\sqrt{6}}{2}.$ ### Example 4. Calculate the area of a parallelogram built on the vectors $$\\mathbf{a} = \\mathbf{i} + \\mathbf{j} + \\mathbf{k}$$ and $$\\mathbf{b} = \\mathbf{j} - \\mathbf{i}.$$ Solution. Determine the", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "dot(\"E\", E, NW); dot(\"F\", F, SW); dot(\"G\", G, S); dot(\"H\", H, N); dot(\"I\", I, NE); label(\"X\", X,SE); dot(\"J\", J, SE);[/asy]$ First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "i\\times\\mathbf i)\\times\\mathbf j$$ and $$\\mathbf i\\times(\\mathbf i\\times\\mathbf j)$$. Theorem. (a) Let $$\\mathbf u=\\vec{OP}$$ and $$\\mathbf v=\\vec{OQ}$$ and $$\\theta=\\measuredangle POQ$$. Then the area of the parallelogram with adjacent edges $$\\overline{OP}$$ and $$\\overline{OQ}$$ is $\\|\\mathbf u\\times\\mathbf v\\|=\\sin\\theta\\|\\mathbf u\\|\\,\\|\\mathbf v\\|.$ (b) Let in addition $$\\mathbf w=\\vec{OR}$$. Then the area of the parallelepiped with adjacent edges $$\\overline{OP},\\,\\overline{OQ}$$ and $$\\overline{OR}$$ is $|\\mathbf u(\\mathbf v\\times\\mathbf w)|.$ Corollary. Let $$P,Q,R$$ be points. Then the area of the triangle $$\\triangle PQR$$ is $$\\frac12\\|\\vec{PQ}\\times\\vec{PR}\\|$$. Remark. The scalar $$\\mathbf u(\\mathbf v\\times\\mathbf w)$$ is called a triple product. Exercise. Find the area of the triangle with vertices $$P(2,0,3),Q(-1,3,0),R(-2,1,2)$$.", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "the area of the blue region, which is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[5] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\\cdot16=\\boxed{\\textbf{(C) }96}$ ~quacker88 ~IceMatrix", "and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\\cdot16=\\boxed{\\textbf{(C) }96}$ ~quacker88 ## Solution 4 (Triangles) $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\",", "# Difference between revisions of \"2006 UNCO Math Contest II Problems/Problem 5\" ## Problem In the figure $BD$ is parallel to $AE$ and also $BF$ is parallel to $DE$. The area of the larger triangle $ACE$ is $128$. The area of the trapezoid $BDEA$ is $78$. Determine the area of triangle $ABF$. $[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP(\"A\",(4,0),SE);MP(\"C\",(1,2),N);MP(\"E\",(0,0),SW); MP(\"D\",(.5,1),W);MP(\"B\",(2.5,1),NE);MP(\"F\",(2,0),S); [/asy]$ ## Solution Since we know $BF$ is equal to the short side of both triangles and that $DB$ is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line $DF$ (as seen below), all four triangles in $\\triangle ACE$ are congruent. $[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); draw((1/2,1)--(2,0),black); MP(\"A\",(4,0),SE);MP(\"C\",(1,2),N);MP(\"E\",(0,0),SW); MP(\"D\",(.5,1),W);MP(\"B\",(2.5,1),NE);MP(\"F\",(2,0),S); [/asy]$ Then, since we know the sum of all four triangles is $128$, we can solve the easy division problem $128/4$ and see that the area of $\\triangle ABF = \\boxed{32}$", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{(B) }30}$. - fath2012 ## Solution 9 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$, we have $$\\dfrac{BF}{FC} \\cdot \\dfrac{CA}{DA} \\cdot \\dfrac{DE}{BE} = 3\\dfrac{BF}{FC} = 1 \\implies \\dfrac{BF}{FC} = \\dfrac13 \\implies \\dfrac{BF}{BC} = \\dfrac14.$$ Therefore, $$[EBF] = \\dfrac{BE}{BD}\\cdot\\dfrac{BF}{BC}\\cdot [BCD] = \\dfrac12 \\cdot \\dfrac 14 \\cdot \\left( \\dfrac23 \\cdot [ABC]\\right) = \\boxed{\\textbf{(B) }30}.$$ ## Solution 10 (Graph Paper) $[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F); draw(B--D);", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$" ]

[ "+0 # Area of Parallelogram with two adjacent 2d vectors. 0 67 1 Find the area of the parallelogram with u and v as adjacent edges. u = 5i -2j v = 6i -2j Jun 11, 2020 #1 +8340 0 The area is equal to $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right|$$. Notice that $$\\mathbf{u}\\times \\mathbf{v} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\6&-2&0\\end{bmatrix} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\1&0&0\\end{bmatrix} = 2\\mathbf{k}$$ Therefore $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right| = |2\\mathbf{k}| = 2$$ Therefore the area of the required parallelogram is 2 sq. units. Jun 11, 2020", "with two vectors: $\\mathbf{a}=\\begin{pmatrix}{a_1 \\\\a_2} \\end{pmatrix}$ and $\\mathbf{b} = \\mathbf{e}_1$, such that $\\mathbf{a}$ and $\\mathbf{b}$ are not colinear and where $\\mathbf{b}$ has been chosen for simplicity. Then a linear map, $A$, can be constructed to send these vectors to there corresponding line segments. If we get that the determinant of $A$ is such that $\\det{A}=\\frac{3}{4}$ the result will be proved. After drawing a picture of a triangle defined in such a way it is clear that we want the map $A$ such that: $$\\mathbf{a} \\mapsto \\mathbf{b} - \\frac{1}{2}\\mathbf{a} ,$$ $$\\mathbf{b} \\mapsto \\mathbf{a} - \\frac{1}{2}\\mathbf{b} .$$ This completely determines $A$ and after solving some equations we get that: $$A=\\begin{pmatrix}1-\\frac{a_1}{2} & -\\frac{1+a_1^2}{2a_2}\\\\-\\frac{a_2}{2} & 1+\\frac{a_1}{2} \\end{pmatrix}.$$ Upon computing $\\det(A)$ we get that it is $\\frac{3}{4}$ proving what is required. - I believe there is a nice visual explanation for this. Then construct congruent triangles to form a hexagon, and connect every other vertex on its perimeter. Let $A=$ (area of the original triangle) and $B=$ (area of the triangle of medians). The shaded triangle has area $4B$, because its sides are twice as long as the medians, and it is also equal to half the area of the hexagon, which is $6A/2=3A$. $3A=4B$, so $B=\\frac{3}{4}A$. - great answer! up-voted. – Mick Feb 17 '14 at 18:49 This is just awesome :)", "+0 help 0 94 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times", "also equal to the area of the parallelogram formed with two sides $\\mathbf{a}$ and $\\mathbf{b}$ . Let's see if we can verify why this is true by using Figure 2.28. Figure 2.28A parallelogram with sides $\\mathbf{a}$ and $\\mathbf{b}$ Figure 2.29Area of a parallelogram First, from planar geometry, we know that the area of the parallelogram is $bh$ , the product of the base and the height. (In Figure 2.28, the base is $b=\\parallel \\mathbf{b}\\parallel$ .) We can verify this rule by “clipping” off a triangle from one end and moving it to the other end, forming a rectangle, as shown in Figure 2.29. The area of a rectangle is given by its length and width. In this case, this area is the product $bh$ . Since the area of the rectangle is equal to the area of the parallelogram, the area of the parallelogram must also be $bh$ . Returning to Figure 2.28, let $a$ and $b$ be the lengths of $\\mathbf{a}$ and $\\mathbf{b}$ , respectively, and note that $\\mathrm{sin}\\theta =h/a$ . Then $\\begin{array}{rl}A& =bh\\\\ & =b\\left(a\\mathrm{sin}\\theta \\right)\\\\ & =\\parallel \\mathbf{a}\\parallel \\parallel \\mathbf{b}\\parallel \\mathrm{sin}\\theta \\\\ & =\\parallel \\mathbf{a}×\\mathbf{b}\\parallel .\\end{array}$ If $\\mathbf{a}$ and $\\mathbf{b}$ are parallel, or if $\\mathbf{a}$ or $\\mathbf{b}$ is the zero vector, then $\\mathbf{a}×\\mathbf{b}=\\mathbf{0}$ . So the cross product interprets the zero vector as being parallel", "& -3 \\end{array}} \\right| = \\left({-12+21}\\right)\\mathbf{i} - \\left({8-14}\\right)\\mathbf{j} + 0\\cdot\\mathbf{k} = 9\\mathbf{i} + 6\\mathbf{j}.$ Hence, $$\\mathbf{M_A} = \\left({9,6,0}\\right).$$ ### Example 3. Calculate the area of a triangle with vertices $$A\\left({1,-2,0}\\right),$$ $$B\\left({3,1,-1}\\right)$$ and $$C\\left({2,-1,3}\\right).$$ Solution. The area of the triangle is equal to half the area of the parallelogram spanned by the vectors AB and AC, that is $A_{\\triangle} = \\frac{1}{2}\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert.$ Calculate the coordinates of vectors AB and AC: $\\mathbf{AB} = \\underbrace{\\left({3,1,-1}\\right)}_B - \\underbrace{\\left({1,-2,0}\\right)}_A = \\left({2,3,-1}\\right);$ $\\mathbf{AC} = \\underbrace{\\left({2,-1,3}\\right)}_C - \\underbrace{\\left({1,-2,0}\\right)}_A = \\left({1,1,3}\\right);$ Find the cross product: $\\mathbf{AB} \\times \\mathbf{AC} = \\left| {\\begin{array}{*{20}{c}} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 2 & 3 & -1\\\\ 1 & 1 & 3 \\end{array}} \\right| = \\mathbf{i}\\left| {\\begin{array}{*{20}{c}} 3 & -1\\\\ 1 & 3 \\end{array}} \\right| - \\mathbf{j}\\left| {\\begin{array}{*{20}{c}} 2 & -1\\\\ 1 & 3 \\end{array}} \\right| + \\mathbf{k}\\left| {\\begin{array}{*{20}{c}} 2 & 3\\\\ 1 & 1 \\end{array}} \\right| = \\left({9+1}\\right)\\mathbf{i} - \\left({6+1}\\right)\\mathbf{j} + \\left({2-3}\\right)\\mathbf{k} = 10\\mathbf{i} - 7\\mathbf{j} - \\mathbf{k}.$ The absolute value of this vector is $\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert = \\sqrt{10^2 + \\left({-7}\\right)^2 + \\left({-1}\\right)^2} = \\sqrt{150} =5\\sqrt{6}.$ Then the area of the triangle is $A_{\\triangle} = \\frac{1}{2}\\Vert{\\mathbf{AB} \\times \\mathbf{AC}}\\Vert = \\frac{5\\sqrt{6}}{2}.$ ### Example 4. Calculate the area of a parallelogram built on the vectors $$\\mathbf{a} = \\mathbf{i} + \\mathbf{j} + \\mathbf{k}$$ and $$\\mathbf{b} = \\mathbf{j} - \\mathbf{i}.$$ Solution. Determine the", "+0 # help 0 52 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 ### Best Answer #1 +24983 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 ### 1+0 Answers #1 +24983 +3 Best Answer Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a}", "# Meaning of formula The exercise was: Draw the triangle with vertices $A = (2;2)$, $B = (-1;3)$, and $C = (0;0)$. By regarding it as half of a parallelogram, explain why its area equals $$\\mathrm{area}(ABC) = \\dfrac{1}{2} \\det\\begin{pmatrix} 2 & 2 \\\\ -1 & 3 \\end{pmatrix}.$$ So I did explain this, explaining the area of the parallelogram as the determinant the give. The 2nd question was: Move the third vertex to $C = (1;-4)$ and justify the formula $$\\mathrm{area}(ABC) = \\dfrac{1}{2} \\det\\begin{pmatrix} 2 & 2 & 1 \\\\ -1 & 3 & 1 \\\\ 1 & -4 & 1 \\end{pmatrix}.$$ Now I'm having trouble explaining this. The solution says that the triangle has the same area, this provoked just a translation... But I don't understand why and how because now we have a different determinant... Also, how did they get the third column $(1,1,1)$? I'm confused... Thanks The usual way to calculate the area is the cross-product in $\\mathbb R_3$. Since we have $x_3=0$, we get $[a_1,a_2,0]\\times[b_1,b_2,0]=[0,0,a_1b_2-a_2b_1]$. The length of this vector is $|a_1b_2-a_2b_1|=|det(\\pmatrix{a_1&a_2\\\\b_1&b_2})|$. Now, if $c\\ne 0$, we have $[a_1-c_1,a_2-c_2,0]\\times [b_1-c_1,b_2-c_2,0]=[0,0,(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)]$. And we have $$|det(\\pmatrix{a_1&a_2&1\\\\b_1&b_2&1\\\\c_1&c_2&1})|=|det(\\pmatrix{a_1-c_1&a_2-c_2&0\\\\b_1-c_1&b_2-c_2&0\\\\c_1&c_2&1}|$$ , which coincides with the length of the vector $[0,0,(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)]$ when three coordinates are given either we can arrange them row wise or column wise which results into an empty column", "of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[6] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ • When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D1 of (any) one diagonal, then the area can be found from Heron's formula. Specifically it is $K=2\\sqrt{S(S-B)(S-C)(S-D_1)}$ where $S=(B+C+D_1)/2$ and the leading factor 2 comes from the fact that the number of congruent triangles that the chosen diagonal divides the parallelogram into is two. ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V = \\begin{bmatrix} a_1 & a_2 & \\dots & a_n \\\\ b_1 & b_2 & \\dots & b_n \\end{bmatrix} \\in\\R^{2 \\times n}$ Then the area of the parallelogram generated by a and b is equal to $\\sqrt{\\det(V V^\\mathrm{T})}$. Let points $a,b,c\\in\\R^2$. Then the area of the parallelogram with vertices at a, b and c is equivalent to the absolute value of the determinant of", ", 6 )$ $u = AB = \\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 6 \\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1 } \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 5 & 6 \\\\ 2 & 4 \\end{bmatrix}= 20 \\: – \\: 12 = 8$ The area of the parallelogram is $8$. ## Numerical Result The area of the parallelogram is $8$. ## Example Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$ Solution A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram. $A = ( 0 , 0 )$ $B = ( 6 , 8 )$ $C = ( 5 , 4 )$ $D = ( 11 , 6 )$ $u = AB = \\begin{bmatrix} 6\\\\ 8 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 5\\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1", "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", "# Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1). Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1). • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. un4t5o4v ###### Have a similar question? content_user Consider the parallelogram A=(-3;0), B(-1,5), C(7;4) and D=(5;-1) The objective is to find the area of parallelogram Find the are as follows: The area of the parallelogram is the magnitude of the cross product of the adjacent edges. That is, $$Area=|\\overline{AB}\\times\\overline{AD}|$$ Find the adjacent edges $$\\overline{AB}$$ and $$\\overline{AD}$$ as follows $$\\overline{AB}=B-A$$ $$=(-1;5)-(-3;0)$$ $$=(-1+3;5-0)$$ $$=(2;5)$$ $$\\overline{AD}=D-A$$ $$=(5;-1)-(-3;0)$$ $$=(8;1)$$ Find $$\\overline{AB}\\times\\overline{AD}$$ as, $$\\overline{AB}\\times\\overline{AD}=\\begin{bmatrix}i&j&k\\\\2&5&0\\\\8&-1&0\\end{bmatrix}$$ $$=5(0)-0(-1)i-(2(0)-0(8))j+(2(-1)-5(8))k$$ $$=0i+0j-42k$$ The area of the parallelogram is, Area $$=|AB\\times AD|$$ $$=|(0i+0j-42k|$$ $$=\\sqrt{(0)^2+(0)^2+(-42)^2}$$ $$=\\sqrt{(42)^2}$$ = 42 square unit Hence, the area of parallelogram is 42", "and (7,0). Find the diagonals of the parallelogram. AB =| cm b: to store the breadth in decimal . $\\begingroup$ If you knew which two points are not connected on the perimeter of the parallelogram, you would use those as $\\mathbf P_1$ and $\\mathbf P_2$ $\\endgroup$ – Joffan Feb 20 '17 at 17:32 $\\begingroup$ you've been very helpful, thank you very much :) $\\endgroup$ – PhysicsMathsLove Feb 20 '17 at 17:41 This could also be stated as 2*Side A + 2*Side B or 2*(Side A + Side B). Suitable examples and sample outputs are provided for the easy comprehension of the whole scheme of things. I have to find the perimeter. Let's move everything to the right 6 and up 5; this makes the vertices become: A = (0, 0), B = (4, 9), C = (11,9), and D =(7,0) Let side AD be the base of the parallelogram; it runs along the x axis for 7 units, therefore, this is the length of the base, b. They are (2,1.5), (8,6) (16,6) and (10, 1.5). Home Contact About Subject Index. Java program to calculate the Perimeter of a parallelogram – In this specific article, we will detail in on the multiple methods to calculate the perimeter of a parallelogram in Java Programming. Therefore, the perimeter of a parallelogram", "the area of the blue region, which is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[5] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V", "Review question # Can we show that $OAQY$ is a parallelogram? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8704 ## Solution In the figure the points $X$ and $Y$ are such that $AX = \\dfrac{1}{2}XB$ and $OY = YX$, while the point $P$ is such that $OA = 3AP$. The lines $YQ$ and $PQ$ are parallel to $OA$ and $OB$ respectively. Given that $\\mathbf{OA} = \\mathbf{a}$ and $\\mathbf{OB} = \\mathbf{b}$, express $\\mathbf{OP}$ and $\\mathbf{OY}$ in terms of $\\mathbf{a}$ and $\\mathbf{b}$. Since $\\mathbf{OA} = 3\\mathbf{AP}$, we know $\\mathbf{OP}$ is $\\dfrac{4}{3}\\mathbf{a}$. We can see that the vector $\\mathbf{OX} = \\mathbf{OA} + \\mathbf{AX} = \\mathbf{a} + \\dfrac{1}{3}(\\mathbf{b}-\\mathbf{a}) = \\dfrac{2}{3}\\mathbf{a} + \\dfrac{1}{3}\\mathbf{b}.$ We also have that the vector $\\mathbf{OY}=\\dfrac{1}{2}\\mathbf{OX} = \\dfrac{1}{3}\\mathbf{a} + \\dfrac{1}{6}\\mathbf{b}.$ Given that $\\mathbf{YQ} = m\\mathbf{a}$ and $\\mathbf{PQ} = n\\mathbf{b}$, find the values of $m$ and $n$. Hence show that $OAQY$ is a parallelogram. The diagram tells us that $\\mathbf{OP} + \\mathbf{PQ} = \\mathbf{OY} + \\mathbf{YQ}$. Thus $\\dfrac{4}{3}\\mathbf{a} + n\\mathbf{b} = \\dfrac{1}{3}\\mathbf{a}+\\dfrac{1}{6}\\mathbf{b} + m\\mathbf{a}.$ Since $\\mathbf{a}$ and $\\mathbf{b}$ are linearly independent (they are not parallel) we can equate coefficients of $\\mathbf{a}$ and $\\mathbf{b}$ on either side of this equation. We are effectively saying here that if $\\alpha \\mathbf{a} + \\beta \\mathbf{b} = \\gamma \\mathbf{a} +", "is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[6] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V = \\begin{bmatrix} a_1 & a_2 & \\dots", "= \\int_0^1 x^2 (1-x) \\, dx = \\frac{1}{12}$ $\\mathbf{C}^0_{xy} = \\int_{\\Delta} xy \\, dA = \\int_{x=0}^1 x \\int_{y=0}^{1-x} y \\, dy \\, dx = \\int_0^1 x \\frac{(1-x)^2}{2} \\, dx = \\frac{1}{24}$ $\\mathbf{C}^0_{yy} = \\mathbf{C}^0_{xx}$ The rest components of $C$ are zero because the triangle is in $z=0$. As a result $\\mathbf{C}^0 = \\frac{1}{24} \\begin{bmatrix} 2 & 1 & 0 \\\\ 1 & 2 & 0 \\\\ 0 & 0 & 0 \\\\ \\end{bmatrix} = \\frac{1}{48} \\begin{bmatrix} 1 \\\\ -1 \\\\ 0 \\end{bmatrix} \\begin{bmatrix} 1 & -1 & 0 \\end{bmatrix}^{\\mathrm{T}} + \\frac{1}{16} \\begin{bmatrix} 1 \\\\ 1 \\\\ 0 \\end{bmatrix} \\begin{bmatrix} 1 & 1 & 0 \\end{bmatrix}^{\\mathrm{T}}$ Covariance of the triangle with a vertex in the origin Consider a linear operator $\\mathbf{x}' = \\mathbf{A}\\mathbf{x}^0$ that maps the canonical triangle in the triangle $\\mathbf{v}'_0 = \\mathbf{0}$, $\\mathbf{v}'_1 = \\mathbf{v}_1 - \\mathbf{v}_0$, $\\mathbf{v}'_2 = \\mathbf{v}_2 - \\mathbf{v}_0$. The first two columns of $\\mathbf{A}$ contain $\\mathbf{v}'_1$ and $\\mathbf{v}'_2$ respectively, while the third column is arbitrary. The target triangle is equal to the triangle in question (in particular their areas are equal), but shifted with its zero vertex in the origin. $\\mathbf{C}' = \\int_{\\Delta'} \\mathbf{x}'\\mathbf{x}'^{\\mathrm{T}} \\, dA' = \\int_{\\Delta^0} \\mathbf{A}\\mathbf{x}^0\\mathbf{x}^{0\\mathrm{T}}\\mathbf{A}^{\\mathrm{T}} a\\, dA^0 = a \\mathbf{A} \\mathbf{C}^0 \\mathbf{A}^{\\mathrm{T}}$ $\\mathbf{C}' = \\frac{a}{48}(\\mathbf{v}_1 - \\mathbf{v}_2)(\\mathbf{v}_1 - \\mathbf{v}_2)^{\\mathrm{T}} +\\frac{a}{16}(\\mathbf{v}_1 + \\mathbf{v}_2 - 2\\mathbf{v}_0)(\\mathbf{v}_1 + \\mathbf{v}_2 - 2\\mathbf{v}_0)^{\\mathrm{T}}$ Covariance", "# Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cristiano Sears ###### Not exactly what you’re looking for? content_user Vertices are $$A(-3,0),\\ B(-1,6),\\ C(8,5)$$ and $$D(6,-1)$$ So $$\\vec{AB}=<-1-(-3),6-0>=<2,6>$$ $$\\vec{AD}=<6-(-3),-1-0>=<9,-1>$$ Hence area of parallelogram $$=|\\vec{AB}\\times\\vec{AD}|$$ $$\\vec{AB}\\times\\vec{AD}=\\begin{bmatrix}i&j&k\\\\2&6&0\\\\9&-1&0\\end{bmatrix}$$ $$=i(0-0)+j(0-0)+k(-2-54)$$ $$=-56k$$ Hence Area $$=|\\vec{AB}\\times\\vec{AD}|$$ $$=\\sqrt{(-56)^2}=56$$", "# Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). I can't figure out what I'm doing wrong.. I let: $a = |DC| = (-2,-1) - (-6,3) = (4,-4)$ $b = |AD| = (-1,1) - (-5,5) = (4,-4)$ so: $\\begin{bmatrix} 4&-4\\\\4&-4\\end{bmatrix} \\implies$ Area $= -16 + 16 = 0$ ? The determinant is the area of the parallelogram is it not? • Not sure how you're labelling vertices, but $a$ appears to be $|CD|$, while $b$ appears to be $|AB|$. You get a result of zero because those lines are parallel. – Joey Zou Jul 6 '16 at 22:08 • You are correct that was the problem. The answer is $|-12|$ – Yusha Jul 6 '16 at 22:11", "back and think about what properties a mathematical object, describing area should have. Suppose we have two (possibly infinitesimal) vectors, forming a two-dimensional parallelogram and we need the mathematical object that tells us the area of this parallelogram: \\mathrm{area}(V,W) = \\text{ area of the parallelogram formed by V and W}. An obious property is that the number $\\mathrm{area}(V,W)$ ought to double if we double the length of one vector: \\mathrm{area}(2V,W) = 2 \\times \\text{ area of the parallelogram formed by V and W} \\mathrm{area}(V,W+Z) = \\mathrm{area}(V,W) + \\mathrm{area}(V,Z) A pictorial proof can be seen in the following figure: Together these properties are what we called linearity above, and we see that this fits perfectly with the defintion of a $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. Another important propery is that $\\mathrm{area}(V,W)$ must vanish if $V$ and $W$ are parallel, i.e. if $W=aV$ for some number $a$, we have \\mathrm{area}(V,W) =\\mathrm{area}(V,aV) \\stackrel{!}{=} 0. .Because we have $\\mathrm{area}(V,aV)=a \\times \\mathrm{area}(V,V) \\stackrel{!}{=} 0$, we start with $\\mathrm{area}(V,V) \\stackrel{!}{=} 0$ in the next step to see that $\\mathrm{area}(V,W)$ must be an antisymmetric $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. The proof is simple: From \\mathrm{area}(V,V) = 0, if we write $V= U+W$, it follows \\mathrm{area}(U+W,U+W) = 0. Using the linearity yields \\mathrm{area}(U,U) + \\mathrm{area}(U,W) + \\mathrm{area}(W,U) + \\mathrm{area}(W,W) = 0. Using now $\\mathrm{area}(V,V)", "# Area of Parallelogram from Determinant ## Theorem Let $OABC$ be a parallelogram in the Cartesian plane whose vertices are located at: $\\ds O$ $=$ $\\ds \\tuple {0, 0}$ $\\ds A$ $=$ $\\ds \\tuple {a, c}$ $\\ds B$ $=$ $\\ds \\tuple {a + b, c + d}$ $\\ds C$ $=$ $\\ds \\tuple {b, d}$ The area of $OABC$ is given by: $\\map \\Area {OABC} = \\begin {vmatrix} a & b \\\\ c & d \\end {vmatrix}$ where $\\begin {vmatrix} a & b \\\\ c & d \\end {vmatrix}$ denotes the determinant of order $2$. ## Proof Arrange for the parallelogram to be situated entirely in the first quadrant. First need we establish that $OABC$ is actually a parallelogram in the first place. Indeed: $\\ds \\vec {AB}$ $=$ $\\ds \\tuple {a + b - a, c + d - c}$ $\\ds$ $=$ $\\ds \\tuple {b, d}$ $\\ds$ $=$ $\\ds \\vec {CB}$ $\\ds \\vec {OA}$ $=$ $\\ds \\tuple {a + b - b, c + d - d}$ $\\ds$ $=$ $\\ds \\tuple {a, c}$ $\\ds$ $=$ $\\ds \\vec {OA}$ Thus: $OA = CB$ $OC = AB$ and it follows from Opposite Sides Equal implies Parallelogram that $OABC$ is indeed a parallelogram. Now we calculate the area of $OABC$ as equal to: the area occupied by the large rectangle in the" ]

[ "# Arcsine online calculator @QED arcsine calculator calculates the arcsine of given numbers in degrees and radians as follows. The asin word works as a command to calculate the arcsine values and it can take several arguments separated by spaces. In the above example, the calculator calculates the arcsine of 0.5, -0.32, 1. The results are in degrees and radians. Let's input asin and numbers in the above form!", "must be expressed in radians. Returns the arcsine (the number whose sine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arccosine (the number whose cosine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arctangent (the number whose tangent is Number) in radians.", "Express your answer in radians per second to three significant figures.", "asin Returns the arcsine of a given angle for the nth repetition. COPY/// asin(rad,n) // // Returns the arcsine of a given angle for the nth repetition. // // n repetition, integer, real //", "calculator, giving your answer in exact terms in radians. Problem Statement Evaluate $$\\sec[\\sin^{-1}(1/4)]$$ without using a calculator, giving your answer in exact terms in radians. Solution ### 3020 video Log in to rate this practice problem and to see it's current rating. Evaluate $$\\csc[\\cos^{-1}(-7/9)]$$ without using a calculator, giving your answer in exact terms in radians. Problem Statement Evaluate $$\\csc[\\cos^{-1}(-7/9)]$$ without using a calculator, giving your answer in exact terms in radians. Solution ### 3021 video Log in to rate this practice problem and to see it's current rating. Unless otherwise instructed, evaluate (simplify) $$\\cos(\\arcsin(1/2))$$ without using a calculator and give your answer in exact, simplified form. Problem Statement Unless otherwise instructed, evaluate (simplify) $$\\cos(\\arcsin(1/2))$$ without using a calculator and give your answer in exact, simplified form. Solution ### 1561 video video by PatrickJMT Log in to rate this practice problem and to see it's current rating. Unless otherwise instructed, evaluate (simplify) $$\\tan(\\arcsin(2/3))$$ without using a calculator and give your answer in exact, simplified form. Problem Statement Unless otherwise instructed, evaluate (simplify) $$\\tan(\\arcsin(2/3))$$ without using a calculator and give your answer in exact, simplified form. Solution ### 1562 video video by PatrickJMT Log in to rate this practice problem and to see it's current rating. Unless otherwise instructed, evaluate (simplify) $$\\sin(2\\arctan(\\sqrt{2}))$$ without using a calculator and give your", "radian!", "# Nested trig Calculus Level pending Let $$f\\left( x \\right) \\equiv \\arcsin { (\\cos { x } ) }$$ where $$x$$ is measured in radians and $$0<x<\\pi$$ What is the value of the following expression? $$\\left| f^{ \\prime }\\left( \\frac { 2\\pi }{ 13 } \\right) \\right|$$ ×", "Question # Use a calculator to evaluate the expression. Round your answer to the nearest hundredth. arcsin (0.25) approx ? radians Decimals Use a calculator to evaluate the expression. Round your answer to the nearest hundredth. arcsin $$\\displaystyle{\\left({0.25}\\right)}\\approx?$$ radians Step 1 Given, $$\\arcsin (0.25) \\sin^{-1}(0.25)$$ Step 2 Now using a calculator, 0.25268025514 $$\\displaystyle\\Rightarrow{0.25}$$(rounding to two decimals)", "Computes the arctangent of one value divided by another value (y/x). Computes both the sine and cosine of a specified value (x), where x is in radians.", "Version: Computes the cosine, in radians, of the input elements. ## Syntax c = cos(a) ## a Real or complex scalar or array of any dimension. ## c Cosine, in radians, of the elements in a. c is a scalar or an array of the same size as a.", "## Algebra and Trigonometry 10th Edition $$30\\ degrees$$ The arcsin(1/2) is the angle whose sine is 1/2. Thus, we see that the expression evaluates to: $$30\\ degrees$$", "at home, please don’t make the mistake I did and have your calculator in radians mode. Also, know what arctangent is on your calculator; I was getting weird numbers because mine wanted “atan” not “arctan” for it. ❦ ❦ ❦", "this post, or send us an email with a subject line such as arcotangent 0 in radians. If our calculator and the information on ctg 0 inverse have been helpful, please hit the sharing buttons to spread the word about our content, and don’t forget to bookmark us. Or, even better, install absolutely free PWA app (see menu or sidebar)! Thanks for visiting our post.", "\\pi}{2}$ radians.", "# ASIN Chandra Bergmann #### The ASIN function returns the arcsine (in radians) of a given Number. This function takes a single Number as input. It outputs another Number: the arcsine, in radians, of the given Number. ### Declaration ASIN(number) -> arcsine ### Parameters number (required, type: Number) Any number. ### Return Values arcsine (type: Number) The arcsine, in radians, of the given number. ### Examples The following example returns the arcsine of 1. Note that the arcsine of one is ninety degrees, or 0.5π radians, but the ASIN function doesn't output an angle in degrees or in radians in terms of pi. Rather, it provides a numerical estimation of the angle in radians: ASIN(1) -> 1.5707963267948966 To calculate the arctangent in degrees, the ASIN function must be used in tandem with the DEGREES function, such as in the following example: DEGREES(ASIN(1)) -> 90 ASIN can also be used in tandem with the PI function to provide insight into how the angle in radians relates to terms of pi, such as in the following example: ASIN(1)/PI() -> 0.5", "Function ArcCos(x) # ArcCos ArcCos( x ! (input) numerical expression ) ## Arguments x A scalar numerical expression in the range $$[-1,1]$$. ## Return Value The ArcCos function returns the arccosine of x in the range 0 to $$\\pi$$ radians. Note • A run-time error results if x is outside the range $$[-1,1]$$. • The function ArcCos can be used in constraints of nonlinear mathematical programs. The functions ArcSin, ArcTan, Cos. Arithmetic functions are discussed in full detail in Arithmetic Functions of the Language Reference.", "it returns results in radians. I was puzzled for a few minutes on Kalid’s example to solve for sine of angle x until I realised the answer Google was computing for arcsin(.60) [0.64] was in radians. Performing a conversion from that to degrees (by multiplying the radian value by 180/π) yielded the expected result of 36.9 degrees. Happy Math!", "# Question #2001348: Algebra: Solving Equations, Reducing Expressions and Factoring Question: Find all solutions in radians of the equation $$2{{\\cos }^{2}}x+\\cos x=1$$ that are in $$[0,2\\pi )$$ Solution: The solution consists of 61 words (1 page) Deliverables: Word Document 0", "Computes the cosecant of a specified value (x). Computes the cosine of a specified value (x), where x is in radians. Computes the cotangent of a specified value (x). Computes the inverse cosecant of a specified value (x), where x is in radians. Computes the arccosine of a specified value (x), where x is in radians. Computes the inverse cotangent of a specified value (x), where x is in radians. Computes the inverse secant of a specified value (x), where x is in radians. Computes the arcsine of a specified value (x), where x is in radians. Computes the arctangent of a specified value (x), where x is in radians. Computes the arctangent of one value divided by another value (y/x). Computes the secant of a specified value (x). Computes the sine of a specified value divided by that value (sin(x)/x), where x is in radians. Computes the sine of a specified value (x), where x is in radians. Computes both the sine and cosine of a specified value (x), where x is in radians. Computes the tangent of a specified value (x), where x is in radians.", "\\rm radians. {/eq}" ]

[ "= \\boxed{\\frac{\\pi}{4}.}$$ $$\\int_{0}^{\\frac{\\pi}{2}}\\frac{{cos}^{n}\\left(x\\right)\\:dx}{{cos}^{n}\\left(x\\right)+ {sin}^{n}\\left(x\\right)} .$$", "# How do you calculate arcsin (-1/ sqrt 2)? Jul 20, 2015 Use properties of right angled triangle with sides $\\frac{1}{\\sqrt{2}}$, $\\frac{1}{\\sqrt{2}}$ and $1$ to find $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = - \\frac{\\pi}{4}$ #### Explanation: $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = \\alpha$ where $\\alpha \\in \\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ such that $\\sin \\left(\\alpha\\right) = - \\frac{1}{\\sqrt{2}}$ $\\frac{1}{\\sqrt{2}}$ is one side of the right angled triangle with sides: $\\frac{1}{\\sqrt{2}}$, $\\frac{1}{\\sqrt{2}}$ and $1$ (Note ${\\left(\\frac{1}{\\sqrt{2}}\\right)}^{2} + {\\left(\\frac{1}{\\sqrt{2}}\\right)}^{2} = \\frac{1}{2} + \\frac{1}{2} = 1 = {1}^{2}$) This triangle has angles $\\frac{\\pi}{4}$, $\\frac{\\pi}{4}$ and $\\frac{\\pi}{2}$. So $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{1}{\\sqrt{2}}$ Now $\\sin \\left(- \\theta\\right) = - \\sin \\left(\\theta\\right)$ So $\\sin \\left(- \\frac{\\pi}{4}\\right) = - \\frac{1}{\\sqrt{2}}$ So $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = - \\frac{\\pi}{4}$", "like this: $$${ \\frac {d}{dx}} \\ arctan(\\sin({x}^{2}))=-2\\,{ \\frac { \\cos({x}^{2})x}{-2+\\left ( \\cos({x}^{2}) \\right )^{2}}}$$$ the code looks correct (to me!) but i still get <br /> tags unless i put it all in a block. ## provenshop thanks for the tutoriials", "$\\arctan \\left(1\\right)$, its value is $\\arctan \\left(1\\right) = \\frac{\\pi}{4}$", "# How do you evaluate arcsin(sin((7pi)/3))? Apr 13, 2016 $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\left(\\frac{\\pi}{3}\\right)$ #### Explanation: If $\\arcsin x = \\theta$ then $\\sin \\theta = x$. Let $\\theta = \\left(\\frac{7 \\pi}{3}\\right)$ and $\\sin \\left(\\frac{7 \\pi}{3}\\right) = x$ Then $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\arcsin x = \\theta = \\frac{7 \\pi}{3}$ However, in case of trigonometric ratios, for each $x$, their could be multiple vales of $\\theta$ (such as $2 n \\pi + \\theta$) e.g. not only $\\sin \\left(\\frac{7 \\pi}{3}\\right) = x$, but so is for $\\sin \\left(\\frac{\\pi}{3}\\right)$, $\\sin \\left(- 5 \\frac{\\pi}{3}\\right)$, $\\sin \\left(13 \\frac{\\pi}{3}\\right)$ etc. Hence, the range for $\\theta = \\arcsin x$ is limited to $\\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ Hence $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\left(\\frac{\\pi}{3}\\right)$", "$= \\frac{56}{65}$. If you want to write this without the names $\\alpha$ and $\\beta$ 9Or something like that, you could write: $\\sin \\left(\\arcsin \\left(\\frac{4}{5}\\right) + \\arctan \\left(\\frac{12}{5}\\right)\\right)$ $= \\sin \\left(\\arcsin \\left(\\frac{4}{5}\\right)\\right) \\cos \\left(\\arctan \\left(\\frac{12}{5}\\right)\\right) + \\cos \\left(\\arcsin \\left(\\frac{4}{5}\\right)\\right) \\sin \\left(\\arctan \\left(\\frac{12}{5}\\right)\\right)$ But I, personally do not think that is more clear. (I do think it is good for students to see it this way, however.)", "\\frac{x-1}{x+1}\\right)=\\frac{\\pi}{2}$ $\\Rightarrow \\arcsin\\left(\\dfrac{x-1}{x+1}\\right)=2\\arctan(\\sqrt{x})-\\dfrac{\\pi}{2}$ Status Not open for further replies.", "= 5$ $\\sin x = \\frac{5}{7}$ $x = \\arcsin \\left(\\frac{5}{7}\\right) + 2 \\pi k$ for integer $k$ or $\\left(\\pi - \\arcsin \\left(\\frac{5}{7}\\right)\\right) + 2 \\pi k$ for integer $k$ 1. Know that arccos, arcsec functions are restricted to quad 1 and 2 2. Know that arctan, arccot, arcsin, arccsc are restricted to quad 1 and 4 3. And when you go to quadrant 4 with arctan, arccot, arcsin, arccsc, use negative angles Examples 1.$\\arcsin \\left(1\\right) = \\frac{\\pi}{2}$ 2. $\\arctan \\left(- 1\\right) = - \\frac{\\pi}{4}$ 3. $a r c \\sec \\left(\\frac{1}{2}\\right) = D N E$ 4. $a r c \\csc \\left(- \\sqrt{2}\\right) = - \\frac{\\pi}{4}$", "positive on $\\left[0 , \\frac{\\pi}{4}\\right]$. $= \\ln \\left(\\sec \\left(\\frac{\\pi}{4}\\right) + \\tan \\left(\\frac{\\pi}{4}\\right)\\right) - \\ln \\left(\\sec \\left(0\\right) + \\tan \\left(0\\right)\\right)$ $= \\ln \\left(\\sqrt{2} + 1\\right) - \\ln \\left(1\\right)$ $= \\ln \\left(\\sqrt{2} + 1\\right)$ The arc has a length of $\\ln \\left(\\sqrt{2} + 1\\right) .$", "$-1\\le t\\le 1.$ and then $\\frac{-\\pi}{2}\\le\\arcsin(t)\\le\\frac{\\pi}{2}.$", "# What does arcsin(sin ((-pi)/2)) equal? Jan 20, 2016 $- \\frac{\\pi}{2}$ #### Explanation: If you want to know how please follow. $\\sin \\left(- \\frac{\\pi}{2}\\right) = - \\sin \\left(\\frac{\\pi}{2}\\right)$ since $\\sin \\left(x\\right)$ is a odd function. $\\sin \\left(- \\frac{\\pi}{2}\\right) = - 1$ since $\\sin \\left(\\frac{\\pi}{2}\\right) = 1$ $\\arcsin \\left(\\sin \\left(- \\frac{\\pi}{2}\\right)\\right) = \\arcsin \\left(- 1\\right)$ Now comes the range of $\\arcsin \\left(x\\right)$ The range of $\\arcsin \\left(x\\right)$ is $\\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ So $\\arcsin \\left(- 1\\right)$ would give $- \\frac{\\pi}{2}$ Therefore, $\\arcsin \\left(\\sin \\left(- \\frac{\\pi}{2}\\right)\\right) = - \\frac{\\pi}{2}$", "# How do you evaluate arcsin(sin3)? Aug 21, 2015 $\\arcsin \\left(\\sin \\left(3\\right)\\right) = \\pi - 3$ #### Explanation: Note that $\\arcsin$ maps $\\left[- 1 , 1\\right]$ to $\\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$. We need to find an angle $x \\in \\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ where $\\sin \\left(x\\right) = \\sin \\left(3\\right)$. Since $\\sin \\left(x\\right)$ is symmetric about $x = \\frac{\\pi}{2}$ we have $\\sin \\left(3\\right) = \\sin \\left(\\pi - 3\\right)$. $\\pi - 3$ is in $\\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$, therefore we have: $\\arcsin \\left(\\sin \\left(3\\right)\\right) = \\pi - 3$", "1 =\\tfrac{\\pi}2$$ and $$\\operatorname{arcsin} \\tfrac 12 =\\tfrac{\\pi}6$$. Thus for each natural $$n$$ we have $$f\\left(\\tfrac 1n, \\tfrac 1n\\right)= \\tfrac{\\pi}2$$ and $$f\\left(\\tfrac 1n, \\tfrac 2n\\right)= \\tfrac{\\pi}6$$, whereas $$\\left|\\left(\\tfrac 1n, \\tfrac 1n\\right)- \\left(\\tfrac 1n, \\tfrac 2n\\right)\\right|=\\tfrac 1n$$.", "- \\arcsin(\\text{-}\\tfrac{1}{2}) \\;=\\;\\frac{\\pi}{6} - \\left(\\text{-}\\frac{\\pi}{6}\\right) \\;=\\;\\frac{\\pi}{3}$", "\\right]' = 2 \\\\\\\\ \\boxed{ \\left[\\ {\\tan^{-1}(2x+1)} \\ \\right]' = \\frac{2} {1+(2x+1)^2} } }$$", "\\frac{3}{2} \\arctan \\left(\\frac{x + 1}{2}\\right) + C$", "the double angle identity yields :$anleft\\left(4 arctanleft\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\} ight\\right) ight\\right) = frac\\left\\{120\\right\\}\\left\\{119\\right\\}$ and so :$q = frac\\left\\{frac\\left\\{120\\right\\}\\left\\{119\\right\\} - 1\\right\\}\\left\\{1 +frac\\left\\{120\\right\\}\\left\\{119 = frac\\left\\{1\\right\\}\\left\\{239\\right\\}.$ Other formulas may be generated using complex numbers. For example the angle of a complex number a+bI is given by $arctanfrac\\left\\{b\\right\\}\\left\\{a\\right\\}$ and when you multiply complex numbers you add their angles. If a=b then $arctanfrac\\left\\{b\\right\\}\\left\\{a\\right\\}$ is 45 degrees or $frac\\left\\{pi\\right\\}\\left\\{4\\right\\}$. This means that if the real part and complex part are equal then the arctangent will equal $frac\\left\\{pi\\right\\}\\left\\{4\\right\\}$. Since the arctangent of one has a very slow convergence rate if we find two complex numbers that when multiplied will result in the same real and imaginary part we will have a Machin-like formula. An example is $\\left(2 + i\\right)$ and $\\left(3 + i\\right)$. If we multiply these out we will get $\\left(5 + 5i\\right)$. Therefore $arctanfrac\\left\\{1\\right\\}\\left\\{2\\right\\} + arctanfrac\\left\\{1\\right\\}\\left\\{3\\right\\} = frac\\left\\{pi\\right\\}\\left\\{4\\right\\}$. If you want to use complex numbers to show that $frac\\left\\{pi\\right\\}\\left\\{4\\right\\} = 4arctanfrac\\left\\{1\\right\\}\\left\\{5\\right\\} - arctanfrac\\left\\{1\\right\\}\\left\\{239\\right\\}$ you first must know that when multiplying angles you put the complex number to the power of the number that you are multiplying by. So $\\left(5 + i\\right)$4$\\left(-239+i\\right) = \\left(-114244 - 114244i\\right)$ since the real part and imaginary part are equal $4arctanfrac\\left\\{1\\right\\}\\left\\{5\\right\\} - arctanfrac\\left\\{1\\right\\}\\left\\{239\\right\\} = frac\\left\\{pi\\right\\}\\left\\{4\\right\\}$ Two-term formulas There are exactly three additional Machin-like formulas with two terms; these are Euler's :$frac\\left\\{pi\\right\\}\\left\\{4\\right\\}", "1$ $\\implies x = \\arcsin \\left(- 1\\right)$ $\\implies x = \\frac{3 \\pi}{2} + 2 \\pi n$ or $\\implies 2 \\sin \\left(x\\right) - 1 = 0$ $\\implies 2 \\sin \\left(x\\right) = 1$ $\\implies \\sin \\left(x\\right) = \\frac{1}{2}$ $\\implies x = \\arcsin \\left(\\frac{1}{2}\\right)$ $\\implies x = \\frac{\\pi}{6} + 2 \\pi n , \\frac{5 \\pi}{6} + 2 \\pi n$ However, the domain is given as $x \\in \\left[0 , 2 \\pi\\right]$. Therefore, the points of inflection occur at $x = \\frac{\\pi}{6}$, $x = \\frac{5 \\pi}{6}$ and $x = \\frac{3 \\pi}{2}$.", "# Machin-like formula In mathematics, Machin-like formulas are a class of identities involving &pi; = 3.14159... that generalize John Machin's formula from 1706: :$frac\\left\\{pi\\right\\}\\left\\{4\\right\\} = 4 arctanfrac\\left\\{1\\right\\}\\left\\{5\\right\\} - arctanfrac\\left\\{1\\right\\}\\left\\{239\\right\\},$ which he used along with the Taylor series expansion of arctan to compute π to 100 decimal places. Machin-like formulas have the form :$frac\\left\\{pi\\right\\}\\left\\{4\\right\\} = sum_\\left\\{n\\right\\}^N a_n arctanfrac\\left\\{1\\right\\}\\left\\{b_n\\right\\}$ with $a_n$ and $b_n$ integers. The same method is still among the most efficient known for computing a large number of digits of π with digital computers. Derivation To understand where this formula comes from, start with following basic ideas: :$frac\\left\\{pi\\right\\}\\left\\{4\\right\\} = arctan\\left(1\\right)$:$an\\left(2arctan\\left(a\\right)\\right) = frac\\left\\{2 a\\right\\} \\left\\{ 1 - a ^ 2\\right\\}$ (tangent double angle identity):$an\\left(a-arctan\\left(b\\right)\\right) = frac\\left\\{ an\\left(a\\right)-b\\right\\} \\left\\{ 1 + an\\left(a\\right) b\\right\\}$ (tangent difference identity):$frac\\left\\{pi\\right\\}\\left\\{16\\right\\} = 0.196349dots$ (approximately):$arctanleft\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\} ight\\right) = arctan\\left(0.2\\right) = 0.197395dots$ (approximately) In other words, for small numbers, arctangent is to a good approximation just the identity function. This leads to the possibility that a number $q$ can be found such that :$frac\\left\\{pi\\right\\}\\left\\{16\\right\\} = arctan\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\}\\right) - frac\\left\\{1\\right\\}\\left\\{4\\right\\} arctan\\left(q\\right).$ Using elementary algebra, we can isolate $q$: :$q = anleft\\left(4 arctanleft\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\} ight\\right) - frac\\left\\{pi\\right\\}\\left\\{4\\right\\} ight\\right)$ Using the identities above, we substitute arctan(1) for π/4 and then expand the result. :$q = frac\\left\\{ anleft\\left(4 arctanleft\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\} ight\\right) ight\\right) - 1\\right\\} \\left\\{ 1 + anleft\\left(4 arctanleft\\left(frac\\left\\{1\\right\\}\\left\\{5\\right\\} ight\\right) ight\\right)\\right\\}$ Similarly, two applications of", "\\Rightarrow 0 \\leq 2x \\leq 720^{o}$ $sin(2x) = \\frac{3}{4}$ $2x = Arcsin(\\frac{3}{4})$ $2x = 360^{o} + Arcsin(\\frac{3}{4})$ $2x = 180^{o} - Arcsin(\\frac{3}{4})$ $2x = 540^{o} - Arcsin(\\frac{3}{4})$ $\\Rightarrow$ $x = \\frac{Arcsin(\\frac{3}{4})}{2} \\approx 24.3^o$ $x = \\frac{180^{o} - Arcsin(\\frac{3}{4})}{2} \\approx 65.7^o$ $x = \\frac{360^{o} + Arcsin(\\frac{3}{4})}{2} \\approx 204.3^o$ $x = \\frac{540^{o} - Arcsin(\\frac{3}{4})}{2} \\approx 245.7^o$" ]

[ "\\frac{2}{\\sqrt{\\theta}}\\right) = 0$ $\\implies \\theta = 2$ radian Hence option (D) is correct", "# When working in decimal radians, how do you find sin^-1(-1/6)? $- 0.167448079 \\setminus \\setminus \\textrm{r a d}$ #### Explanation: For small value of angle $\\setminus {\\sin}^{- 1} \\left(- \\frac{1}{6}\\right) \\setminus \\approx - \\frac{1}{6} = - 0.167 \\setminus \\setminus \\textrm{r a d}$ Or $\\setminus {\\sin}^{- 1} \\left(- \\frac{1}{6}\\right)$ $= - \\setminus {\\sin}^{- 1} \\left(\\frac{1}{6}\\right)$ $= - 0.167448079 \\setminus \\setminus \\textrm{r a d}$", "\\dfrac{-1}{2} \\right)+\\arccos \\left( \\dfrac{-1}{2} \\right)=\\dfrac{\\pi }{2}$. Using this relation, we can directly rewrite the expression in the question as $\\arctan 1+\\dfrac{\\pi }{2}$.", "must be expressed in radians. Returns the arcsine (the number whose sine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arccosine (the number whose cosine is Number) in radians. If Number is less than -1 or greater than 1, the function yields a blank result (empty string). Returns the arctangent (the number whose tangent is Number) in radians.", "ranges over radians (or over degrees). If are the Cartesian coordinates, then we may compute from with: :$\\begin x_1 &= r \\cos\\left(\\varphi_1\\right) \\\\ x_2 &= r \\sin\\left(\\varphi_1\\right) \\cos\\left(\\varphi_2\\right) \\\\ x_3 &= r \\sin\\left(\\varphi_1\\right) \\sin\\left(\\varphi_2\\right) \\cos\\left(\\varphi_3\\right) \\\\ &\\vdots\\\\ x_ &= r \\sin\\left(\\varphi_1\\right) \\cdots \\sin\\left(\\varphi_\\right) \\cos\\left(\\varphi_\\right) \\\\ x_n &= r \\sin\\left(\\varphi_1\\right) \\cdots \\sin\\left(\\varphi_\\right) \\sin\\left(\\varphi_\\right) . \\end$ Except in the special cases described below, the inverse transformation is unique: :$\\begin r &= \\sqrt \\\\\\left[6pt\\right] \\varphi_1 &= \\arccot \\frac &&= \\arccos \\frac \\\\\\left[6pt\\right] \\varphi_2 &= \\arccot \\frac &&= \\arccos \\frac \\\\\\left[6pt\\right] &\\vdots &&\\vdots \\\\\\left[6pt\\right] \\varphi_ &= \\arccot \\frac &&= \\arccos \\frac \\\\\\left[6pt\\right] \\varphi_ &= 2\\arccot \\frac &&= \\begin \\arccos \\frac & x_n\\geq 0 \\\\\\left[6pt\\right] 2\\pi - \\arccos \\frac & x_n < 0 \\end \\,. \\end$ where if for some but all of are zero then when , and (180 degrees) when . There are some special cases where the inverse transform is not unique; for any will be ambiguous whenever all of are zero; in this case may be chosen to be zero. ## Spherical volume and area elements To express the volume element of -dimensional Euclidean space in terms of spherical coordinates, first observe that the Jacobian matrix and determinant, Jacobian matrix of the transformation is: :$J_n = \\begin \\cos\\left(\\varphi_\\right) &-r\\sin\\left(\\varphi_\\right) &0 &0&\\cdots &0 \\\\ \\sin\\left(\\varphi_\\right)\\cos\\left(\\varphi_\\right) &r\\cos\\left(\\varphi_\\right)\\cos\\left(\\varphi_\\right) &-r\\sin\\left(\\varphi_\\right)\\sin\\left(\\varphi_\\right)&0&\\cdots &0 \\\\ \\vdots", "$$1.98$$ 18) $$\\arcsin (0.23)$$ 19) $$\\arccos \\left(\\dfrac{3}{5}\\right)$$ Answer $$0.93$$ 20) $$\\cos^{-1}(-0.8)$$ 21) $$\\tan^{-1}(6)$$ Answer $$1.41$$ For the exercises 22-23, find the angle $$\\theta$$ in the given right triangle. Round answers to the nearest hundredth. 22) 23) Answer $$0.56$$ radians For the exercises 24-36, find the exact value, if possible, without a calculator. If it is not possible, explain why. 24) $$\\sin^{-1}(\\cos(\\pi))$$ 25) $$\\tan^{-1}(\\sin(\\pi))$$ Answer $$0$$ 26) $$\\cos^{-1}\\left(\\sin \\left(\\dfrac{\\pi}{3} \\right)\\right)$$ 27) $$\\tan^{-1}\\left(\\sin \\left(\\dfrac{\\pi}{3} \\right)\\right)$$ Answer $$0.71$$ 28) $$\\sin^{-1}\\left(\\cos \\left(\\dfrac{-\\pi}{2} \\right)\\right)$$ 29) $$\\tan^{-1}\\left(\\sin \\left(\\dfrac{4\\pi}{3} \\right)\\right)$$ Answer $$-0.71$$ 30) $$\\sin^{-1}\\left(\\sin \\left(\\dfrac{5\\pi}{6} \\right)\\right)$$ 31) $$\\tan^{-1}\\left(\\sin \\left(\\dfrac{-5\\pi}{2} \\right)\\right)$$ Answer $$-\\dfrac{\\pi}{4}$$ 32) $$\\cos \\left(\\sin^{-1} \\left(\\dfrac{4}{5} \\right)\\right)$$ 33) $$\\sin \\left(\\cos^{-1} \\left(\\dfrac{3}{5} \\right)\\right)$$ Answer $$0.8$$ 34) $$\\sin \\left(\\tan^{-1} \\left(\\dfrac{4}{3} \\right)\\right)$$ 35) $$\\cos \\left(\\tan^{-1} \\left(\\dfrac{12}{5} \\right)\\right)$$ Answer $$\\dfrac{5}{13}$$ 36) $$\\cos \\left(\\sin^{-1} \\left(\\dfrac{1}{2} \\right)\\right)$$ For the exercises 37-41, find the exact value of the expression in terms of $$x$$ with the help of a reference triangle. 37) $$\\tan \\left(\\sin^{-1} (x-1)\\right)$$ Answer $$\\dfrac{x-1}{\\sqrt{-x^2+2x}}$$ 38) $$\\sin \\left(\\sin^{-1} (1-x)\\right)$$ 39) $$\\cos \\left(\\sin^{-1} \\left(\\dfrac{1}{x}\\right)\\right)$$ Answer $$\\dfrac{\\sqrt{x^2-1}}{x}$$ 40) $$\\cos \\left(\\tan^{-1} (3x-1)\\right)$$ 41) $$\\tan \\left(\\sin^{-1} \\left(x+\\dfrac{1}{2}\\right)\\right)$$ Answer $$\\dfrac{x+0.5}{\\sqrt{-x^2-x+\\tfrac{3}{4}}}$$ ### Extensions For the exercise 42, evaluate the expression without using a calculator. Give the exact value. 42) $$\\dfrac{\\sin^{-1}\\left ( \\tfrac{1}{2} \\right )-\\cos^{-1}\\left ( \\tfrac{\\sqrt{2}}{2} \\right )+\\sin^{-1}\\left ( \\tfrac{\\sqrt{3}}{2} \\right )-\\cos^{-1}(1)}{\\cos^{-1}\\left ( \\tfrac{\\sqrt{3}}{2} \\right )-\\sin^{-1}\\left ( \\tfrac{\\sqrt{2}}{2} \\right )+\\cos^{-1}\\left ( \\tfrac{1}{2} \\right )-\\sin^{-1}(0)}$$ For the exercises", "# How do you calculate arc cot(tan ((79pi)/180))? $\\left(\\frac{11}{180}\\right) \\pi$, radian. $\\tan \\left(\\frac{79 \\pi}{180}\\right) = \\cot \\left(\\frac{\\pi}{2} - \\frac{79 \\pi}{180}\\right) = \\cot \\left(\\frac{11 \\pi}{180}\\right)$ Use $a r c \\cot \\left(\\cot a\\right) = a$ Here, the given expression is $a r c \\cot \\left(\\frac{11 \\pi}{180}\\right) = \\left(\\frac{11}{180}\\right) \\pi$.", "\\ldots \\ldots \\ldots . \\left(\\star\\right)$ Now, regarding the Soln. of the Problem, we note that, there is no mention about the Measure of the Angle $2 ,$ i.e., it is not clear, it is ${2}^{\\circ} ,$ or $2 \\text{ radian.}$ If it is ${2}^{\\circ} ,$then, it follows from $\\left(\\star\\right)$ that, ${\\sin}^{-} 1 \\left(\\sin {2}^{\\circ}\\right) = {2}^{\\circ} .$ In case, it is $2 \\text{ radian,}$ we note that, $\\sin 2 = \\sin \\left(\\pi - \\left(\\pi - 2\\right)\\right) = \\sin \\left(\\pi - 2\\right) ,$ where, since $\\left(\\pi - 2\\right) \\in \\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right] ,$ we have, by $\\left(\\star\\right) ,$ ${\\sin}^{-} 1 \\left(\\sin 2\\right) = \\pi - 2.$", "### Home > INT3 > Chapter 9 > Lesson 9.1.5 > Problem9-67 9-67. Colleen and Jolleen both use their calculators to determine $\\text{sin}\\left(30^\\circ\\right)$. Colleen gets $\\text{sin}\\left(30^\\circ\\right) = –0.9880316241$, but Jolleen gets $\\text{sin}\\left(30^\\circ\\right) = 0.5$. Is one of their calculators broken, or is something else going on? Why do they get different answers? Try calculating $\\text{sin}\\left(30^\\circ\\right)$ in both degree mode and radian mode on your calculator.", "# How do you evaluate arcsin(sin((7pi)/3))? Apr 13, 2016 $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\left(\\frac{\\pi}{3}\\right)$ #### Explanation: If $\\arcsin x = \\theta$ then $\\sin \\theta = x$. Let $\\theta = \\left(\\frac{7 \\pi}{3}\\right)$ and $\\sin \\left(\\frac{7 \\pi}{3}\\right) = x$ Then $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\arcsin x = \\theta = \\frac{7 \\pi}{3}$ However, in case of trigonometric ratios, for each $x$, their could be multiple vales of $\\theta$ (such as $2 n \\pi + \\theta$) e.g. not only $\\sin \\left(\\frac{7 \\pi}{3}\\right) = x$, but so is for $\\sin \\left(\\frac{\\pi}{3}\\right)$, $\\sin \\left(- 5 \\frac{\\pi}{3}\\right)$, $\\sin \\left(13 \\frac{\\pi}{3}\\right)$ etc. Hence, the range for $\\theta = \\arcsin x$ is limited to $\\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ Hence $\\arcsin \\left(\\sin \\left(\\frac{7 \\pi}{3}\\right)\\right) = \\left(\\frac{\\pi}{3}\\right)$", "# How do you use the properties of inverse trigonometric functions to evaluate tan(arcsin (0.31))? Nov 25, 2014 Let $\\theta = \\arcsin \\left(0.31\\right)$ by rewriting in terms of sine, $\\sin \\theta = 0.31 = \\frac{O}{H} \\implies$ Let $\\left\\{\\begin{matrix}O = 0.31 \\\\ H = 1\\end{matrix}\\right.$ By Pythagorean Theorem, $A = \\sqrt{{H}^{2} - {O}^{2}} = \\sqrt{{1}^{2} - {\\left(0.31\\right)}^{2}} = \\sqrt{0.9039}$ Hence, $\\tan \\left(\\arcsin \\left(0.31\\right)\\right) = \\tan \\theta = \\frac{O}{A} = \\frac{0.31}{\\sqrt{0.9039}}$ I hope that this was helpful.", "$\\phi = \\arctan \\left(\\frac{1 + \\sqrt{3}}{1 - \\sqrt{3}}\\right)$", "$\\arctan \\left(1\\right)$, its value is $\\arctan \\left(1\\right) = \\frac{\\pi}{4}$", "# How do you solve sin(arcsin(3/5) + arccos(3/5))? May 15, 2015 Picture a right angled triangle with sides 3, 4 and 5 (since ${3}^{2} + {4}^{2} = {5}^{2}$). Call the smallest angle $\\alpha$ and the next smallest $\\beta$. $\\sin \\alpha = \\frac{3}{5}$, being the length of the opposite side divided by the length of the hypotenuse. $\\cos \\beta = \\frac{3}{5}$, being the length of the adjacent side divided by the length of the hypotenuse. So $\\arcsin \\left(\\frac{3}{5}\\right) + \\arccos \\left(\\frac{3}{5}\\right) = \\alpha + \\beta = \\frac{\\pi}{2}$ (or ${90}^{o}$) since together with the right angle, the internal angles of the triangle must add up to $\\pi$ (${180}^{o}$). $\\sin \\left(\\frac{\\pi}{2}\\right) = 1$. So $\\sin \\left(\\arcsin \\left(\\frac{3}{5}\\right) + \\arccos \\left(\\frac{3}{5}\\right)\\right) = 1$.", "# What are the coordinates of -2 pi radians on the unit circle? $\\left(1 , 0\\right)$ #### Explanation: In general, the coordinates of a point lying on a circle of radius $r$ at an angle $\\setminus \\theta$ in CCW from +ve x-axis are given as follows $\\left(r \\setminus \\cos \\setminus \\theta , r \\setminus \\sin \\setminus \\theta\\right)$ hence. the coordinates of a point lying on a unit circle (radius $r = 1$) at an angle $\\setminus \\theta = - 2 \\setminus \\pi$ are given as follows $\\left(1 \\setminus \\cos \\left(- 2 \\setminus \\pi\\right) , 1 \\setminus \\sin \\left(- 2 \\setminus \\pi\\right)\\right) \\setminus \\equiv \\left(1 , 0\\right)$", "# How do you calculate arcsin 2? Nov 11, 2015 As a Real valued function $\\arcsin 2$ is undefined, since $\\sin \\left(x\\right) \\in \\left[- 1 , 1\\right]$ for all $x \\in \\mathbb{R}$. Defining $\\sin \\left(z\\right)$ for $z \\in \\mathbb{C}$ we find: $\\arcsin \\left(2\\right) = \\frac{\\pi}{2} + \\ln \\left(2 + \\sqrt{3}\\right) i$ #### Explanation: $\\sin \\left(x\\right) \\in \\left[- 1 , 1\\right]$ for all $x \\in \\mathbb{R}$. So there is no Real value of $x$ such that $\\sin \\left(x\\right) = 2$. However, it is possible to define $\\sin \\left(z\\right)$ for $z \\in \\mathbb{C}$ and hence find a workable definition and value for $\\arcsin \\left(2\\right)$. Use the following: ${e}^{i x} = \\cos \\left(x\\right) + i \\sin \\left(x\\right)$ $\\cos \\left(- x\\right) = \\cos \\left(x\\right)$ $\\sin \\left(- x\\right) = - \\sin \\left(x\\right)$ to find: $\\sin \\left(x\\right) = \\frac{{e}^{i x} - {e}^{- i x}}{2 i}$ for all $x \\in \\mathbb{R}$ Define: $\\sin \\left(z\\right) = \\frac{{e}^{i z} - {e}^{- i z}}{2 i}$ for all $z \\in \\mathbb{C}$ We want to solve $\\sin \\left(z\\right) = 2$, that is: $\\frac{{e}^{i z} - {e}^{- i z}}{2 i} = 2$ Let $t = i {e}^{i z}$ Then this equation becomes: $2 = \\frac{- i t - \\frac{i}{t}}{2 i} = - \\frac{t + \\frac{1}{t}}{2} = - \\frac{{t}^{2} + 1}{2 t}$ Multiply both ends by $- 2 t$ to get: ${t}^{2}", "### Home > PC3 > Chapter 2 > Lesson 2.3.3 > Problem2-154 2-154. Evaluate each expression (in radians) and leave your answers in fraction form when appropriate. Remember the restricted ranges for the functions. 1. $\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)$ 1. $\\sin^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(-\\frac{\\sqrt{2}}{2}\\right)$ 1. $\\cos^{-1}\\left(\\frac{\\sqrt{3}}{2}\\right)$ The domain of inverse sine is $\\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$. The domain of inverse cosine is $[0,\\pi]$.", "# How do you evaluate arcsin(sin((7*pi)/3))? Jun 8, 2018 $\\arcsin \\sin \\left(\\frac{7 \\pi}{3}\\right) = \\frac{7 \\pi}{3} + 2 \\pi k \\mathmr{and} - \\frac{\\pi}{3} + 2 \\pi k \\quad$ integer $k .$ $\\textrm{A r c} \\textrm{\\sin} \\setminus \\sin \\left(\\frac{7 \\pi}{3}\\right) = \\frac{\\pi}{3}$ #### Explanation: If we treat $\\arcsin$ as multivalued then $x = \\arcsin \\sin a$ is the same as $\\sin x = \\sin a$ which has supplementary angle solutions $x = a + 2 \\pi k \\quad \\mathmr{and} \\quad x = \\left(\\pi - a\\right) + 2 \\pi k \\quad$ for integer $k$ $x = \\arcsin \\sin \\left(\\frac{7 \\pi}{3}\\right)$ $\\sin x = \\sin \\left(\\frac{7 \\pi}{3}\\right)$ $x = \\frac{7 \\pi}{3} + 2 \\pi k \\mathmr{and} x = \\left(\\pi - \\frac{7 \\pi}{3}\\right) + 2 \\pi k$ $x = \\frac{7 \\pi}{3} + 2 \\pi k \\mathmr{and} x = - \\frac{\\pi}{3} + 2 \\pi k$ $\\arcsin \\sin \\left(\\frac{7 \\pi}{3}\\right) = \\frac{7 \\pi}{3} + 2 \\pi k \\mathmr{and} - \\frac{\\pi}{3} + 2 \\pi k \\quad$ integer $k .$ The principal value for the inverse sine is in the range $- \\frac{\\pi}{2}$ to $\\frac{\\pi}{2}$ so we take $k = - 1$ in the first clause and get $\\textrm{A r c} \\textrm{\\sin} \\setminus \\sin \\left(\\frac{7 \\pi}{3}\\right) = \\frac{\\pi}{3}$ [Had this one sitting around in my browser; I forget to hit post.]", "# How do you calculate arcsin (-1/ sqrt 2)? Jul 20, 2015 Use properties of right angled triangle with sides $\\frac{1}{\\sqrt{2}}$, $\\frac{1}{\\sqrt{2}}$ and $1$ to find $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = - \\frac{\\pi}{4}$ #### Explanation: $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = \\alpha$ where $\\alpha \\in \\left[- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$ such that $\\sin \\left(\\alpha\\right) = - \\frac{1}{\\sqrt{2}}$ $\\frac{1}{\\sqrt{2}}$ is one side of the right angled triangle with sides: $\\frac{1}{\\sqrt{2}}$, $\\frac{1}{\\sqrt{2}}$ and $1$ (Note ${\\left(\\frac{1}{\\sqrt{2}}\\right)}^{2} + {\\left(\\frac{1}{\\sqrt{2}}\\right)}^{2} = \\frac{1}{2} + \\frac{1}{2} = 1 = {1}^{2}$) This triangle has angles $\\frac{\\pi}{4}$, $\\frac{\\pi}{4}$ and $\\frac{\\pi}{2}$. So $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{1}{\\sqrt{2}}$ Now $\\sin \\left(- \\theta\\right) = - \\sin \\left(\\theta\\right)$ So $\\sin \\left(- \\frac{\\pi}{4}\\right) = - \\frac{1}{\\sqrt{2}}$ So $\\arcsin \\left(- \\frac{1}{\\sqrt{2}}\\right) = - \\frac{\\pi}{4}$", "# How do you evaluate arcsin(sqrt 2/2)? May 24, 2015 $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}$ is the length of one side of the isoceles right angled triangle with sides $\\frac{\\sqrt{2}}{2}$, $\\frac{\\sqrt{2}}{2}$ and $1$, which has internal angles $\\frac{\\pi}{4}$, $\\frac{\\pi}{4}$ and $\\frac{\\pi}{2}$. ($\\frac{\\pi}{4}$ radians = ${45}^{o}$ and $\\frac{\\pi}{2}$ radians = ${90}^{o}$ if you prefer) To show this is right angled, check with Pythagoras: ${\\left(\\frac{\\sqrt{2}}{2}\\right)}^{2} + {\\left(\\frac{\\sqrt{2}}{2}\\right)}^{2}$ $= {\\sqrt{2}}^{2} / {2}^{2} + {\\sqrt{2}}^{2} / {2}^{2}$ $= \\frac{2}{4} + \\frac{2}{4} = \\frac{1}{2} + \\frac{1}{2} = 1 = {1}^{2}$ So since $\\sin \\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}$ and $\\frac{\\pi}{4}$ is in the required range for $\\arcsin$ viz $- \\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}$, we find $\\arcsin \\left(\\frac{\\sqrt{2}}{2}\\right) = \\frac{\\pi}{4}$" ]

[ "constants) and is $\\text{deg}_{\\epsilon}(\\text{AND}_n)= \\Theta(\\sqrt{n\\log(1/\\epsilon)})$. Indeed, the degree gets higher as you ... • 13.3k • 35.7k", "22:45 • Nope, when you do that $d$ and $\\theta_2$ become your two variables. – Sean E. Lake Sep 5 '16 at 22:48 • Sure, I thought $d-d_0$ was a constant, but because $d$ is a function of the angles it can be taken as a generalized coordinate too... It is clear now I think, thank you! – Saavestro Sep 5 '16 at 22:51 There are naively three degrees of freedom: the length $d$ of the spring, and the angles $\\theta_1$ and $\\theta_2$. We also have the constraint $$d - d_0 = l(\\theta_1 - \\theta_2).$$ You can use this constraint to eliminate any one of these variables, leaving two independent degrees of freedom.", "positive constants $K_1, K_2$, there exists a list length $N_0$ such that: $\\forall N > N_0: K_1 \\Theta(N^2) > K_2 \\Theta(N \\log N)$", "constants $c_1,c_2$ such that $|\\cos(n\\theta)|\\geq c_1n^{-c_2}$, but there is a lot of work to finish, ie to find explicitely $c_1$ and $c_2$... – Kelenner Jun 26 '17 at 15:23 $$T_n (\\cos (\\theta))=\\cos (n\\theta)$$ where $T_n$ is the Tchebychev polynomial. • how do you get a lower bound for $|a_n|$ using $T_n$? – uniquesolution Jun 26 '17 at 14:09", "\\int^{t_2} dt_3 \\delta(t_3-t_1)$$ $$\\sim \\int^{t+\\Delta t}_tdt_1\\int^{t_1}_t dt_2 \\Theta(t_2-t_1)=0$$ using $\\sim$ as not keeping track of constants.", "theta = 4/3$ 2. $cos theta = 3/5$ 3. $cos theta = 4/5$ 4. $cos theta = 3/4$ If $theta$ is the angle formed between a line from the origin to the point $(4,3)$ and the $x$ axis, find $tan theta$. 1. $tan theta = 4/3$ 2. $tan theta = 3/5$ 3. $tan theta = 4/5$ 4. $tan theta = 3/4$ If $theta$ is the angle formed between a line from the origin to the point $(-1,-4)$ and the $x$ axis, find $cos theta$. 1. $cos theta = -4/sqrt(17)$ 2. $cos theta =- 1/sqrt(17)$ 3. $cos theta = 4$ 4. $cos theta = -1/4$", "I do find it a bit odd that they bother to introduce $\\theta$ though, because in the second form of the solution, the $2 \\times 2 +1$ constants in $\\vec{b}_1, \\vec{b}_2$ and $\\theta$ are not all independent, in terms of the $2 \\times 2$ initial conditions. I suppose there is a reason in the textbook for them to do this but seems a bit unnatural to me. – secavara Aug 27 '18 at 11:30", "the rotation through this line (so $1$ stays fixed). Then $\\theta^2 = e$. Now confirm that $\\rho\\theta , \\rho^2 \\theta , \\rho^3 \\theta, \\rho^4 \\theta$ are all distinct. Thus, we have created $10$ distinct symettries. And we see that each one is made out of $\\rho$, a rotation, and $\\theta$ - a reflection.", "where $r = |\\vec{r}|$ and $\\theta$ is the angle between $\\vec{r}$ and $\\vec{n}$. So, the equation says $r \\cos(\\theta) = a r,$ hence either $r = 0$ or $\\cos(\\theta) = a.$ For $a \\in (-1,1),$ this says that $\\theta =$ constant, so describes a cone.", "Is $\\Theta$ symmetric? For example if $$f(x)= \\Theta (g(x))$$ from the definition of the theta notation, there exist c1 and c2 constants such that $$c_1 g(x) \\le f(x) \\le c_2 g(x)$$ then if only we took the constants $1/c_1$ and $1/c_2$ we could say from the definition that $$g(x)= \\Theta (f(x))$$ Right? • A different way to see it would be that f=O(g) is equivalent to g=Omega(f), and f=Theta(g) is equivalent to both f=O(g) and f=Omega(g). – chirlu Sep 15 '13 at 19:18 Right, except that the constants are actually $1/c_2$ and $1/c_1$. That is, $$c_1 g(x) \\leq f(x) \\leq c_2g(x) \\Rightarrow \\frac{1}{c_2}f(x) \\leq g(x) \\leq \\frac{1}{c_1}f(x)\\,.$$ Also, remember that the inequalities only apply for large enough $x$.", "able to write that in terms of your constants. You may need to use the facts that cos(theta- theta0)= cos(theta0) cos(theta)+ sin(theta0)sin(theta) and sin(theta- theta0)= cos(theta0) sin(theta)- sin(theta0)cos(theta) to put it in the form you want.", "way, we find that up to constants, if $f_k(d,n) = \\Theta(2^{r_kd}n)$ then $r_{k+1} = \\max_{1 \\leq \\ell \\leq k} (r_\\ell + r_{k+1-\\ell})$ (except for the base cases $k \\leq 1$). For example, we have \\begin{align*} &r_0 = 0 \\\\ &r_1 = 1 \\\\ &r_2 = 2 \\\\ &r_3 = 3 \\\\ &r_4 = 4 \\end{align*} and so on. Indeed, induction shows that $f_k(d,n) = \\Theta_k(2^{kd}n)$. With more effort, we could estimate the hidden constant.", "that the length of the side opposite vertex $A$ is called $a$, and so on. Let $\\theta=\\angle C$. Then the Cosine Law says that $$c^2=a^2+b^2-2ab\\cos \\theta.$$ Since we know $a$, $b$, and $c$, we can use the above formula to calculate $\\cos\\theta$. Then we can use the $\\cos^{-1}$ button on the calculator to find $\\theta$ to excellent accuracy.", "and angles are constants, it is an integral of $$dl$$ resulting in $$uB\\frac{h}{cos(\\theta)}sin(\\theta)$$", "right path for this problem? 2. Jan 23, 2009 ### gabbagabbahey It might be more useful to write this as: $$\\left( sin^4 (\\theta ) - C^2 sin^2(\\theta ) \\right) \\phi ' ^ 2 = C^2$$ I think the point of the hint is that the above relation holds for all allowable values of $\\theta$ and that $\\theta=0$ is one such allowable value. What does $C^2$ equal when $\\theta=0$ assuming that $\\phi'$ is bounded (i.e. not infinite)? Since $C$ is a constant, it must have this value for all $\\theta$. 3. Jan 24, 2009 ### roeb Now I can see why C = 0. If I am doing my math correctly that means $$\\phi (\\theta ) = A$$ (another constant). If $$\\phi (\\theta )$$ is just some constant, does that mean this constant A will be a segment of the great circle between two locations on the sphere? As the choice of axis is arbitrary, you can choose one such that $$\\theta$$ is zero. Then varying $$\\phi$$ (which as you point out should have no $$\\theta$$ dependence) is just tracing out a great circle.", "That was the key I was missing, probably missed due to writting it so abstractly :P. Thanks! vanhees71 $$\\partial_{t_1} \\Theta(t_1-t_2)=-\\partial_{t_2} \\Theta(t_2-t_1)=\\delta(t_1-t_2)$$", "(a + \\mu \\cos\\theta)$ and $T_B = m_2a + \\mu m_2 \\cos\\theta$ e. Solve by using $d = 1/2at^2$ and substituting $h$ for $d$ 8. a. Yes, because it is static and you know the angle and $m_1$ b. Yes, $T_A$ and the angle gives you $m_1$ and the angle and $T_C$ gives you $m_2, m_1 = T_A \\cos 25/\\mathrm{g}$ and $m_2 = T_C \\cos 30/\\mathrm{g}$ 9. a. $3$ seconds d. $90 \\;\\mathrm{m}$", ", \\, {\\theta}_1 + {\\theta}_2 }$ /////// $\\, (\\cos \\alpha + i \\sin \\alpha)(\\cos \\beta + i \\sin \\beta) = \\,$ $\\, = \\, \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta + i (\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta) = \\,$ $\\, = \\, \\cos (\\alpha + \\beta) + i \\sin (\\alpha + \\beta)$ If we define $\\, e^{ \\, i \\, \\theta} \\, \\stackrel {\\mathrm{def}}{=} \\, \\cos \\theta + i \\sin \\theta$, the product above can be written $\\, e^{ \\, i \\, \\alpha} e^{ \\, i \\, \\beta} = e^{ \\, i \\, ( \\alpha + \\beta) }$ which conforms to the usual rules of exponentiation (by transforming multiplication of two unitary rotations to addition of their respective angles). This formula, discovered by Euler, is considered to be one of the most beautiful formulae in all of mathematics. It ties in with the transformation formulae from polar to cartesian coordinates: $\\, x = r \\cos \\theta \\, , \\, y = r \\sin \\theta \\,$ $\\, x + i \\, y = r \\cos \\theta + i \\, r \\sin \\theta = r (\\cos \\theta + i \\, \\sin \\theta) = r e^{ \\, i \\theta}$ ////////////////////////////////////////////////////// Geometric numbers in the Euclidean plane: $\\, [ \\, G_{eometric}N_{umber} \\, ]_{\\text{Polar} \\, \\mathbb{R}^2}", "a circle with radius $r$ and angle $\\Delta \\theta$) $\\approx (a+b\\theta)\\Delta \\theta$; if $\\Delta \\theta$ is constant, you can see that this expression will go to infinity at $\\theta$ grows. If $\\Delta \\theta$ were constant, then each step would have the same change in $\\theta$, but the line segment (that is, each piece of the spiral) drawn by the program would get longer and longer. – florence Aug 13 '16 at 8:13 • Try running the program yourself with constant dtheta and see how it affects the output. – florence Aug 13 '16 at 8:13", "you need $[x,y] = [r \\cos \\theta, r \\sin \\theta]$ which leads to the solution $$\\boxed{ r (\\theta) = \\frac{a b}{\\sqrt{a^2 + (b^2-a^2) \\cos^2 \\theta}} }$$ If you plot the above for any constants $a$ and $b$ you will get an ellipse with semi-major and semi-minor radii of $a$ and $b$ respectively." ]

[ "Suppose $Bx\\le\\Theta(x)\\le Cx$ for each $\\Theta$. \\begin{align} \\frac{1+\\Theta(\\frac{1}{2n})}{(1+\\Theta(\\frac{1}{n}))^2} &\\le\\frac{1+\\frac{C}{2n}}{(1+\\frac{B}{n})^2}\\\\ &=(1+\\frac{C}{2n})(1-2\\frac{B}{n}+O(\\frac{1}{n^2}))\\\\ &=(1+(\\frac{C}{2}-2B)\\frac{1}{n}+O(\\frac{1}{n^2}))\\\\ &=1+O(\\frac{1}{n}) \\end{align} \\begin{align} \\frac{1+\\Theta(\\frac{1}{2n})}{(1+\\Theta(\\frac{1}{n}))^2} &\\ge\\frac{1+\\frac{B}{2n}}{(1+\\frac{C}{n})^2}\\\\ &=(1+\\frac{B}{2n})(1-2\\frac{C}{n}+O(\\frac{1}{n^2}))\\\\ &=(1+(\\frac{B}{2}-2C)\\frac{1}{n}+O(\\frac{1}{n^2}))\\\\ &=1+O(\\frac{1}{n}) \\end{align} - Huh. Over an hour, and no upvotes on either of our answers. Well, +1 from me, because it's useful to see how to do this from the definition, too. – Mike Spivey Sep 9 '11 at 4:12 Yes. One way to see it is to do the long division; i.e., divide the denominator directly into the numerator. The numerator is $1 + \\Theta(\\frac{1}{2n}) = 1 + \\Theta(\\frac{1}{n})$, and the denominator is $\\left(1 + \\Theta(\\frac{1}{n})\\right)^2 = 1 + \\Theta(\\frac{1}{n}) + \\Theta(\\frac{1}{n^2}) = 1 + \\Theta(\\frac{1}{n})$. I'm not going to attempt to typeset the long division on this forum, but try it, and you'll see that you get $1$ with a remainder of $O(\\frac{1}{n})$. So you have $$\\frac {1 + \\Theta(\\frac{1}{2n})} {\\left(1 + \\Theta(\\frac{1}{n})\\right)^2} = \\frac{1 + \\Theta(\\frac{1}{n})}{1 + \\Theta(\\frac{1}{n})} = 1 + \\frac{O(\\frac{1}{n})}{1 + \\Theta(\\frac{1}{n})} = 1 + O\\left(\\frac{1}{n}\\right),$$ since the denominator is $O(1)$. (Remember that $\\Theta(\\frac{1}{2n}) = \\Theta(\\frac{1}{n})$, since the constant $\\frac{1}{2}$ doesn't affect the asymptotic order.) - I don't think you can bound the ratio away from $1$, so I don't think you can say $1+\\Theta(1/n)$. – robjohn Sep 9 '11 at 2:54 @robjohn: Right; the terms might cancel. I'll fix it. – Mike Spivey Sep", "\\theta )\\\\T_{2}(\\cos \\theta )&=&\\cos 2\\theta &=&{\\frac {1}{3}}(4P_{2}(\\cos \\theta )-P_{0}(\\cos \\theta ))\\\\T_{3}(\\cos \\theta )&=&\\cos 3\\theta &=&{\\frac {1}{5}}(8P_{3}(\\cos \\theta )-3P_{1}(\\cos \\theta ))\\\\T_{4}(\\cos \\theta )&=&\\cos 4\\theta &=&{\\frac {1}{105}}(192P_{4}(\\cos \\theta )-80P_{2}(\\cos \\theta )-7P_{0}(\\cos \\theta ))\\\\T_{5}(\\cos \\theta )&=&\\cos 5\\theta &=&{\\frac {1}{63}}(128P_{5}(\\cos \\theta )-56P_{3}(\\cos \\theta )-9P_{1}(\\cos \\theta ))\\\\T_{6}(\\cos \\theta )&=&\\cos 6\\theta &=&{\\frac {1}{1155}}(2560P_{6}(\\cos \\theta )-1152P_{4}(\\cos \\theta )-220P_{2}(\\cos \\theta )-33P_{0}(\\cos \\theta ))\\end{aligned}}} Another property is the expression for ${\\displaystyle \\sin(n+1)\\theta }$, which is ${\\displaystyle {\\frac {\\sin(n+1)\\theta }{\\sin \\theta }}=\\sum _{\\ell =0}^{n}P_{\\ell }(\\cos \\theta )P_{n-\\ell }(\\cos \\theta )}$ Legendre polynomials are symmetric or antisymmetric, that is ${\\displaystyle P_{n}(-x)=(-1)^{n}P_{n}(x).\\,}$[5] Since the differential equation and the orthogonality property are independent of scaling, the Legendre polynomials' definitions are \"standardized\" (sometimes called \"normalization\", but note that the actual norm is not unity) by being scaled so that ${\\displaystyle P_{n}(1)=1.\\,}$ The derivative at the end point is given by ${\\displaystyle P_{n}'(1)={\\frac {n(n+1)}{2}}.\\,}$ As discussed above, the Legendre polynomials obey the three term recurrence relation known as Bonnet’s recursion formula ${\\displaystyle (n+1)P_{n+1}(x)=(2n+1)xP_{n}(x)-nP_{n-1}(x)\\,}$ and ${\\displaystyle {x^{2}-1 \\over n}{d \\over dx}P_{n}(x)=xP_{n}(x)-P_{n-1}(x).}$ Useful for the integration of Legendre polynomials is ${\\displaystyle (2n+1)P_{n}(x)={d \\over dx}\\left[P_{n+1}(x)-P_{n-1}(x)\\right].}$ ${\\displaystyle {d \\over dx}P_{n+1}(x)=(2n+1)P_{n}(x)+(2(n-2)+1)P_{n-2}(x)+(2(n-4)+1)P_{n-4}(x)+\\ldots }$ or equivalently ${\\displaystyle {d \\over dx}P_{n+1}(x)={2P_{n}(x) \\over \\|P_{n}\\|^{2}}+{2P_{n-2}(x) \\over \\|P_{n-2}\\|^{2}}+\\ldots }$ where ${\\displaystyle \\|P_{n}\\|}$ is the norm over the interval −1 ≤ x ≤ 1 ${\\displaystyle \\|P_{n}\\|={\\sqrt {\\int _{-1}^{1}(P_{n}(x))^{2}\\,dx}}={\\sqrt {\\frac {2}{2n+1}}}.}$ From Bonnet’s recursion formula one", "13, \\ \\theta = \\tan^{-1} \\frac{5}{12} + 180^\\circ \\approx 202.6^\\circ.\\end{align*} 1. \\begin{align*}(x - 4)^2 + (y - 3)^2 & = 25 \\\\ x^2 - 8x + 16 + y^2 - 6y + 9 & = 25 \\\\ x^2 - 8x + y^2 - 6y + 25 & = 25 \\\\ x^2 - 8x + y^2 - 6y & = 0 \\\\ x^2 + y^2 - 8x - 6y & = 0 && \\text{From graphing}\\ r - 8 \\cos \\theta - 6 \\sin \\theta = 0,\\ \\text{we see that the} \\\\ r^2 - 8(r \\cos \\theta) - 6(r \\sin \\theta) & = 0 && \\text{additional solutions are}\\ 0\\ \\text{and}\\ 8. \\\\ r^2 - 8r \\cos \\theta - 6r \\sin \\theta & = 0 \\\\ r(r - 8 \\cos \\theta - 6 \\sin \\theta) & = 0 \\\\ r = 0\\ \\text{or}\\ r - 8 \\cos \\theta - 6 \\sin \\theta & = 0 \\\\ r = 0\\ \\text{or}\\ r & = 8 \\cos \\theta + 6 \\sin \\theta\\end{align*} 2. \\begin{align*}3x - 2y & = 1 \\\\ 3r \\cos \\theta - 2r \\sin \\theta & = 1 \\\\ r (3 \\cos \\theta - 2 \\sin \\theta) & = 1 \\\\ r & = \\frac{1}{3 \\cos \\theta - 2 \\sin \\theta}\\end{align*} 3. \\begin{align*}x^2 + y^2 - 4x + 2y & =", "\\frac{1}{z(6i-8)} \\frac{1}{1 - \\frac{3i+4}{z}} + \\frac{1}{z(8-6i)} \\frac{1}{1 - \\frac{4-3i}{z}} \\\\ &= \\frac{1}{z(6i-8)} \\sum (\\frac{3i+4}{z})^n + \\frac{1}{z(8-6i)} \\sum (\\frac{4-3i}{z})^n \\\\ &= \\frac{1}{(6i-8)} \\sum \\frac{(3i+4)^n}{z^{n+1}} + \\frac{1}{(8-6i)} \\sum \\frac{(4-3i)^n}{z^{n+1}} \\end{align*} Still, $|\\frac{3i+4}{z}| < 1$ and $|\\frac{-3i+4}{z}| < 1$ lead to $|z| > 5$. Thus $R = 5$. Use same approach as above, we can get \\begin{align*} &f(5e^{it}) = K_3 \\sum (\\frac{5 e^{i\\theta}}{5e^{it}}) ^ n \\frac{1}{z} + K_4 \\sum (\\frac{5 e^{i(-\\theta)}}{5e^{it}})^n \\frac{1}{z} \\\\ &= \\frac{K_3}{5} \\sum e^{ni(\\theta - t) - it} + \\frac{K_4}{5} \\sum e^{ni(-\\theta - t) - it} \\\\ &= \\frac{K_3}{5} \\sum e^{i(n\\theta -(n+1)t)} + \\frac{K_4}{5} \\sum e^{i(n(-\\theta) -(n+1)t)} \\\\ &= \\frac{K_3}{5} \\sum e^{i(n\\theta -(n+1)t)} + \\frac{K_4}{5} \\sum e^{i(n(-\\theta) -(n+1)t)} \\\\ &= |\\frac{K_3}{5}| \\sum e^{i(n\\theta -(n+1)t)} e^{i(\\theta - \\pi)} + |\\frac{K_4}{5}| \\sum e^{i(n(-\\theta) -(n+1)t)} e^{i\\theta} \\\\ &= |\\frac{K_3}{5}| \\sum e^{i((n+1)\\theta -(n+1)t - \\pi)} + |\\frac{K_4}{5}| \\sum e^{i((n+1)(-\\theta) -(n+1)t)} \\\\ \\end{align*} if $f(5e^{it})$ converges, then $Im\\{f(5e^{it})\\}$ converges, since \\begin{align*} Im\\{f(5e^{it})\\} = |K_5| \\sum -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sum \\sin {((n+1)(-\\theta) -(n+1)t)} \\end{align*} then $\\lim_{n \\rightarrow \\infty} |K_5| -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sin {((n+1)(-\\theta) -(n+1)t)} = 0$, since $|K_5| = |K_6|$, $\\lim_{n \\rightarrow \\infty} |K_5| -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sin {((n+1)(-\\theta) -(n+1)t)} = -2|K_5| \\lim_{n \\rightarrow \\infty} \\cos (-(n+1) t) \\sin ((n+1)\\theta) = 0$ a contradiction. Thus, on the $|z| = 5$, series does not converge. 11 MAT334--Lectures & Home Assignments /", "^{2}\\theta ={\\frac {1+\\cos(2\\theta )}{2}}}$ ${\\displaystyle \\sin ^{2}\\theta \\cos ^{2}\\theta ={\\frac {1-\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{3}\\theta ={\\frac {3\\sin \\theta -\\sin(3\\theta )}{4}}}$ ${\\displaystyle \\cos ^{3}\\theta ={\\frac {3\\cos \\theta +\\cos(3\\theta )}{4}}}$ ${\\displaystyle \\sin ^{3}\\theta \\cos ^{3}\\theta ={\\frac {3\\sin(2\\theta )-\\sin(6\\theta )}{32}}}$ ${\\displaystyle \\sin ^{4}\\theta ={\\frac {3-4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\cos ^{4}\\theta ={\\frac {3+4\\cos(2\\theta )+\\cos(4\\theta )}{8}}}$ ${\\displaystyle \\sin ^{4}\\theta \\cos ^{4}\\theta ={\\frac {3-4\\cos(4\\theta )+\\cos(8\\theta )}{128}}}$ ${\\displaystyle \\sin ^{5}\\theta ={\\frac {10\\sin \\theta -5\\sin(3\\theta )+\\sin(5\\theta )}{16}}}$ ${\\displaystyle \\cos ^{5}\\theta ={\\frac {10\\cos \\theta +5\\cos(3\\theta )+\\cos(5\\theta )}{16}}}$ ${\\displaystyle \\sin ^{5}\\theta \\cos ^{5}\\theta ={\\frac {10\\sin(2\\theta )-5\\sin(6\\theta )+\\sin(10\\theta )}{512}}}$ and in general terms of powers of ${\\displaystyle \\sin \\theta }$ or ${\\displaystyle \\cos \\theta }$ the following is true, and can be deduced using De Moivre's formula, Euler's formula and the binomial theorem[citation needed]. Cosine Sine ${\\displaystyle {\\text{if }}n{\\text{ is odd}}}$ ${\\displaystyle \\cos ^{n}\\theta ={\\frac {2}{2^{n}}}\\sum _{k=0}^{\\frac {n-1}{2}}{\\binom {n}{k}}\\cos {{\\big (}(n-2k)\\theta {\\big )}}}$ ${\\displaystyle \\sin ^{n}\\theta ={\\frac {2}{2^{n}}}\\sum _{k=0}^{\\frac {n-1}{2}}(-1)^{\\left({\\frac {n-1}{2}}-k\\right)}{\\binom {n}{k}}\\sin {{\\big (}(n-2k)\\theta {\\big )}}}$ ${\\displaystyle {\\text{if }}n{\\text{ is even}}}$ ${\\displaystyle \\cos ^{n}\\theta ={\\frac {1}{2^{n}}}{\\binom {n}{\\frac {n}{2}}}+{\\frac {2}{2^{n}}}\\sum _{k=0}^{{\\frac {n}{2}}-1}{\\binom {n}{k}}\\cos {{\\big (}(n-2k)\\theta {\\big )}}}$ ${\\displaystyle \\sin ^{n}\\theta ={\\frac {1}{2^{n}}}{\\binom {n}{\\frac {n}{2}}}+{\\frac {2}{2^{n}}}\\sum _{k=0}^{{\\frac {n}{2}}-1}(-1)^{\\left({\\frac {n}{2}}-k\\right)}{\\binom {n}{k}}\\cos {{\\big (}(n-2k)\\theta {\\big )}}}$ ## Product-to-sum and sum-to-product identities The product-to-sum identities or prosthaphaeresis formulae can be proven by expanding their right-hand sides using the angle addition theorems. See amplitude modulation", "\\\\ r^2 - 8(r \\cos \\theta) - 6(r \\sin \\theta) & = 0 \\\\ r^2 - 8r \\cos \\theta - 6r \\sin \\theta & = 0 \\\\ r(r - 8 \\cos \\theta - 6 \\sin \\theta) & = 0 \\\\ r = 0\\ \\text{or}\\ r - 8 \\cos \\theta - 6 \\sin \\theta & = 0 \\\\ r = 0\\ \\text{or}\\ r & = 8 \\cos \\theta + 6 \\sin \\theta\\end{align*} From graphing \\begin{align*}r-8\\cos \\theta -6\\sin \\theta =0\\end{align*}, we see that the additional solutions are 0 and 8. 2. \\begin{align*}3x - 2y & = 1 \\\\ 3r \\cos \\theta - 2r \\sin \\theta & = 1 \\\\ r (3 \\cos \\theta - 2 \\sin \\theta) & = 1 \\\\ r & = \\frac{1}{3 \\cos \\theta - 2 \\sin \\theta}\\end{align*} 3. \\begin{align*}x^2 + y^2 - 4x + 2y & = 0 \\\\ r^2 \\cos^2 \\theta + r^2 \\sin^2 \\theta - 4 r \\cos \\theta + 2 r \\sin \\theta & = 0 \\\\ r^2 (\\sin^2 \\theta + \\cos^2 \\theta) - 4 r \\cos \\theta + 2 r \\sin \\theta & = 0 \\\\ r (r - 4 \\cos \\theta + 2 \\sin \\theta) & = 0 \\\\ r = 0\\ \\text{or}\\ r - 4 \\cos \\theta + 2 \\sin \\theta & = 0 \\\\ r = 0\\ \\text{or}\\ r", "u={\\frac {J_{3}P_{3}^{0}(\\sin \\theta )}{r^{4}}}=J_{3}{\\frac {1}{r^{4}}}{\\frac {1}{2}}\\sin \\theta \\left(5\\sin ^{2}\\theta -3\\right)=J_{3}{\\frac {1}{r^{7}}}{\\frac {1}{2}}z\\left(5z^{2}-3r^{2}\\right)}$ are {\\displaystyle {\\begin{aligned}F_{\\theta }&=-{\\frac {1}{r}}{\\frac {\\partial u}{\\partial \\theta }}=-J_{3}{\\frac {1}{r^{5}}}{\\frac {3}{2}}\\cos \\theta \\left(5\\sin ^{2}\\theta -1\\right)\\\\F_{r}&=-{\\frac {\\partial u}{\\partial r}}=J_{3}{\\frac {1}{r^{5}}}2\\sin \\theta \\left(5\\sin ^{2}\\theta -3\\right)\\end{aligned}}} (14) and {\\displaystyle {\\begin{aligned}F_{x}&=-{\\frac {\\partial u}{\\partial x}}=J_{3}{\\frac {xz}{r^{9}}}\\left(10z^{2}-{\\frac {15}{2}}(x^{2}+y^{2})\\right)\\\\F_{y}&=-{\\frac {\\partial u}{\\partial y}}=J_{3}{\\frac {yz}{r^{9}}}\\left(10z^{2}-{\\frac {15}{2}}(x^{2}+y^{2})\\right)\\\\F_{z}&=-{\\frac {\\partial u}{\\partial z}}=J_{3}{\\frac {1}{r^{9}}}\\left(4z^{2}\\ \\left(z^{2}-3(x^{2}+y^{2})\\right)+{\\frac {3}{2}}(x^{2}+y^{2})^{2}\\right)\\end{aligned}}} (15) The exact numerical values for the coefficients deviate (somewhat) between different Earth models but for the lowest coefficients they all agree almost exactly. For the JGM-3 model (see below) the values are: μ = 398600.440 km3⋅s−2 J2 = 1.75553 × 1010 km5⋅s−2 J3 = −2.61913 × 1011 km6⋅s−2 For example, at a radius of 6600 km (about 200 km above Earth's surface) J3/(J2r) is about 0.002; i.e., the correction to the \"J2 force\" from the \"J3 term\" is in the order of 2 permille. The negative value of J3 implies that for a point mass in Earth's equatorial plane the gravitational force is tilted slightly towards the south due to the lack of symmetry for the mass distribution of Earth's \"north–south\". ### Derivation A compact derivation of the spherical harmonics used to model Earth's gravitational field. The spherical harmonics are derived from the approach of looking for harmonic functions of the form ${\\displaystyle \\phi =R(r)\\ \\Theta (\\theta )\\ \\Phi (\\varphi )}$ (16)", "# How do you convert x^2+(y-4)^2=16 to polar form? May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: If $\\left(r , \\theta\\right)$ is in polar form and $\\left(x , y\\right)$ in Cartesian form the relation between them is as follows: $x = r \\cos \\theta$, $y = r \\sin \\theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\\tan \\theta = \\frac{y}{x}$ Hence, ${x}^{2} + {\\left(y - 4\\right)}^{2} = 16$ can be written as ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ or ${x}^{2} + {y}^{2} - 8 y = 0$ or ${r}^{2} - 8 r \\sin \\theta = 0$ or $r \\left(r - 8 \\sin \\theta\\right) = 0$ dividing by $r$ $r - 8 \\sin \\theta = 0$ or $r = 8 \\sin \\theta$ May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: To convert from Cartesian to Polar coordinates use the following formulae that link them. • x=rcostheta\" and \" y=rsintheta ${x}^{2} + {y}^{2} - 8 y + 16 = 16 \\text{ (expanding bracket) }$ $\\Rightarrow {r}^{2} {\\cos}^{2} \\theta + {r}^{2} {\\sin}^{2} \\theta - 8 r \\sin \\theta + 16 - 16 = 0$ then ${r}^{2} \\left({\\cos}^{2} \\theta + {\\sin}^{2} \\theta\\right) - 8 r \\sin \\theta = 0$ using the identity: ${\\cos}^{2} \\theta + {\\sin}^{2} \\theta = 1$ $\\Rightarrow {r}^{2} = 8 r \\sin", "&+d\\theta''^2[1+2(r''-r_*)(\\frac{1}{r_*}+a_{212})+2a_{222}\\theta'']\\\\ &+d\\theta''dr''[(a_{112}+a_{211})(r''-r_*)+(a_{122}+a_{212})\\theta''] \\end{align} We want the linear terms vanish, so we get \\begin{align} a_{111}=0,a_{112}=0,a_{122}=\\frac{1}{r_*}\\\\ a_{211}=0,a_{212}=-\\frac{1}{r_*},a_{222}=0 \\end{align} Now we know that \\begin{align} dr' = dr''+\\frac{1}{r_*}\\theta'' d\\theta''\\\\ d\\theta' = -\\frac{1}{r_*}\\theta''dr''+\\biggl[1-\\frac{r''-r_*}{r_*}\\biggr]d\\theta''\\\\ r = r' = r''+\\frac{1}{2r_*}\\theta''^2 \\end{align} Now plug back to eq. (*) and retain first and second order of $$(r''-r)$$ and $$\\theta''$$, we finally have \\begin{align} ds^2 = \\biggl[1+\\frac{1}{r_*^2}\\theta''^2\\biggr]dr''^2 -\\frac{2}{r_*^2}\\theta''(r''-r_*)d\\theta''dr''+\\biggl[1+\\frac{2}{r_*^2}\\theta''^2-\\frac{2}{r_*^2}(r''-r_*)^2\\biggr]d\\theta''^2 \\end{align} Notice in new coordinates the metric has quadratic expansion form. The desired $$B_{\\mu\\nu,\\lambda\\sigma}$$ is the coefficients in the above expression, \\begin{align} B_{12,12}=-\\frac{1}{2r_*^2},B_{22,11}=-\\frac{2}{r_*^2},B_{11,22}=\\frac{1}{r_*^2} \\end{align} and the intrinsic curvature is $$2B_{12,12}-B_{22,11}-B_{11,22}=0$$ as expected because the given metric is just polar coordinates.", "\\theta)(\\sin \\theta) \\\\ &= 2r\\cos^2 \\theta + 2r \\sin^2 \\theta \\\\ &= 2r \\end{align} And also: (4) \\begin{align} \\quad \\frac{\\partial f}{\\partial \\theta} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial \\theta} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial \\theta} \\\\ &= (2x) (-r \\sin \\theta) + (2y)(r \\cos \\theta) \\\\ &= (2r \\cos \\theta)(-r \\sin \\theta) + (2r \\sin \\theta)(r \\cos \\theta) \\\\ &= -2r^2 \\cos \\theta \\sin \\theta + 2r^2 \\sin \\theta \\cos \\theta \\\\ &= 0 \\end{align} By direct substitution we have that: (5) \\begin{align} \\quad f(x, y) = x^2 + y^2 = (r \\cos \\theta)^2 + (r \\sin \\theta)^2 = r^2 \\cos^2 \\theta + r^2 \\sin^2 \\theta = r^2(\\cos^2 \\theta + \\sin^2 \\theta) = r^2 = f(r, \\theta) \\end{align} Therefore $\\displaystyle{\\frac{\\partial f}{\\partial r} = 2r}$ and $\\displaystyle{\\frac{\\partial f}{\\partial \\theta} = 0}$ as above. ## Example 2 Let $f(x, y, z) = x^2ye^z$, $x(s, t) = s^2t$, $y(s, t) = st^2$, and $z(s, t) = \\sin (st)$. Find $\\displaystyle{\\frac{\\partial f}{\\partial s}}$ by first using the chain rule and then by direct substitution. We have that: (6) \\begin{align} \\quad \\frac{\\partial f}{\\partial s} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial s} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial s} + \\frac{\\partial f}{\\partial z} \\frac{\\partial z}{\\partial s} \\\\ &= (2xye^z)(2st) + (x^2e^z)(t^2) + (x^2ye^z)(t \\cos (st)) \\\\ &= 2(s^2t)(st^2)e^{\\sin (st)}(2st) + (s^2t)^2e^{\\sin (st)} (t^2)", "# Find $\\cos(2\\alpha)$ given $\\cos(\\theta -\\alpha)$ and $\\sin(\\theta +\\alpha)$ My question is: If $\\cos(\\theta -\\alpha) = \\frac{3}{5}$ and $\\sin(\\theta +\\alpha) =\\frac{12}{13}$, find $\\cos(2\\alpha)$. Attempt I: \\begin{align*} &\\cos^2(\\theta -\\alpha)+\\sin^2(\\theta +\\alpha) = \\frac{9}{25} + \\frac{144}{169}\\\\ \\Rightarrow &\\cos^2(\\theta -\\alpha)- \\cos^2(\\theta +\\alpha) = \\frac{9}{25} + \\frac{144}{169} - 1 = \\frac{896}{4225}. \\end{align*} But I thought it won't work. Then I tried this: Attempt II: \\begin{align*} &\\begin{cases} \\cos\\theta \\cos\\alpha + \\sin\\theta \\sin\\alpha = \\cos(\\theta -\\alpha) = \\frac{3}{5},\\\\ \\sin\\theta \\cos\\alpha + \\cos\\theta \\sin\\alpha = \\sin(\\theta +\\alpha) =\\frac{12}{13}, \\end{cases}\\\\ &\\cos\\theta (\\sin\\alpha + \\cos\\alpha) + \\sin\\theta (\\sin\\alpha + \\cos\\alpha) = \\frac{99}{65},\\\\ &\\sqrt{2}\\left[\\sin\\left(\\alpha+\\frac{\\pi}{4}\\right) \\cos\\left(\\alpha -\\frac{\\pi}{4}\\right)\\right] = \\frac{99}{65}. \\end{align*} But I thought here its better to convert both parts to cosines, so I did: \\begin{align*} &\\sqrt{2}\\left[\\cos\\left(\\alpha-\\frac{\\pi}{4}\\right) \\cos\\left(\\alpha -\\frac{\\pi}{4}\\right)\\right] = \\frac{99}{65},\\\\ &\\sqrt{2} \\cos^2\\left(\\alpha-\\frac{\\pi}{4}\\right) = \\frac{99}{65}. \\end{align*} But I think it also didn't work .... - HINT: $$\\cos(2\\alpha)=\\cos\\{\\theta+\\alpha-(\\theta-\\alpha)\\}=\\cos(\\theta+\\alpha)\\cos(\\theta-\\alpha)+\\sin(\\theta+\\alpha)\\sin(\\theta-\\alpha)$$ As $\\cos(\\theta-\\alpha)=\\frac35,$ $\\sin(\\theta-\\alpha)=\\pm \\sqrt{1-\\left(\\frac35\\right)^2}=\\pm\\frac45$ Similarly, as $\\sin(\\theta+\\alpha)=\\frac{12}{13},$ $\\cos(\\theta+\\alpha)=\\pm\\sqrt{1-\\left(\\frac{12}{13}\\right)^2}=\\pm \\frac5{13}$ If we assume $0<\\theta \\pm \\alpha<\\frac\\pi2,$ we can consider the positive values only. - Nice. +1. I took five seconds to realise that the 'HONT' is actually a 'HINT', though :-D. – user1551 Jul 6 '13 at 9:19 +1 very nice solution – Ovi Jul 6 '13 at 10:30 @user1551 If your question was answered you can accept this answer by clicking the check mark below the vote count. – Ovi Jul 6 '13", "& = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence,$$15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2)$$ Taking $\\frac{(1) + (2)}{30}$, we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$. Thus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Taking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$. Hence,\\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*} Hence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$. In $\\triangle O O_1 B$,$$O_1 B = 2 r \\sin \\frac{\\alpha}{2} .$$ In $\\triangle O O_2 A$, by applying the law of sines, we get$$O_2 A = 2 r \\sin \\frac{\\beta}{2} .$$ Because circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$$A$ is on circle $\\omega_2$,\\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*} Thus, $\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} = \\frac{12}{\\sqrt{193}}$. Now, we compute $\\sin \\alpha$ and $\\sin", "answer, interested in the method. But people can have fun while solving it as well! – william122 Dec 20 '18 at 22:56 • Probably $\\tan\\theta=u$ may work – Tito Eliatron Dec 20 '18 at 22:56 HINT First of all, notice that \\begin{align*} \\frac{\\cos(2\\theta)e^{\\cot(\\theta)}}{\\sin^{3}(2\\theta)(e^{\\cot(\\theta)}-e^{\\tan(\\theta)})} = \\frac{\\cot(2\\theta)\\csc^{2}(2\\theta)}{1 - e^{\\tan(\\theta)-\\cot(\\theta)}} = \\frac{\\cot(2\\theta)\\csc^{2}(2\\theta)}{1 - e^{-2\\cot(2\\theta)}} \\end{align*} Hence, according to the substitution $$w = \\cot(2\\theta)$$, where $$\\mathrm{d}w = -2\\csc^{2}(2\\theta)\\mathrm{d}\\theta$$, we obtain \\begin{align*} \\int\\frac{\\cos(2\\theta)e^{\\cot(\\theta)}}{\\sin^{3}(2\\theta)(e^{\\cot(\\theta)}-e^{\\tan(\\theta)})}\\mathrm{d}\\theta = -\\frac{1}{2}\\int\\frac{w}{1-e^{-2w}}\\mathrm{d}w \\end{align*} Can you proceed from here? • Yes, thank you. – william122 Dec 20 '18 at 23:12", "and $z$ (for other coordinate systems this might be very difficult to obtain): \\begin{align} r &= \\sqrt{x^2+y^2+z^2} \\\\ \\theta &= \\arctan\\left(\\frac{\\sqrt{x^2+y^2}}{z}\\right) \\\\ \\phi &= \\arctan\\left(\\frac{y}{x}\\right) \\end{align} The partial derivatives are: \\begin{align} \\frac{\\partial r}{\\partial x} &= \\frac{x}{\\sqrt{x^2+y^2+z^2}} & &= \\sin\\theta\\cos\\phi \\\\ \\frac{\\partial r}{\\partial y} &= \\frac{y}{\\sqrt{x^2+y^2+z^2}} & &= \\sin\\theta\\sin\\phi \\\\ \\frac{\\partial r}{\\partial z} &= \\frac{z}{\\sqrt{x^2+y^2+z^2}} & &= \\cos\\theta \\end{align} \\begin{align} \\frac{\\partial \\theta}{\\partial x} &= \\frac{zx}{\\sqrt{x^2+y^2}\\left(x^2+y^2+z^2\\right)} & &= \\frac{\\cos\\theta\\cos\\phi}{r} \\\\ \\frac{\\partial \\theta}{\\partial y} &= \\frac{zy}{\\sqrt{x^2+y^2}\\left(x^2+y^2+z^2\\right)} & &= \\frac{\\cos\\theta\\sin\\phi}{r} \\\\ \\frac{\\partial \\theta}{\\partial z} &= -\\frac{\\sqrt{x^2+y^2}}{x^2+y^2+z^2} & &= -\\frac{\\sin\\phi}{r} \\end{align} \\begin{align} \\frac{\\partial \\phi}{\\partial x} &= -\\frac{y}{x^2+y^2} & &= -\\frac{\\sin\\phi}{r\\sin\\theta} \\\\ \\frac{\\partial \\phi}{\\partial y} &= \\frac{x}{x^2+y^2} & &= \\frac{\\cos\\phi}{r\\sin\\theta} \\\\ \\frac{\\partial \\phi}{\\partial z} &= 0 & &= 0 \\end{align} Our Jacobian is then: $$J = \\begin{bmatrix} \\sin\\theta\\cos\\phi & \\sin\\theta\\sin\\phi & \\cos\\theta \\\\ \\frac{\\cos\\theta\\cos\\phi}{r} & \\frac{\\cos\\theta\\sin\\phi}{r} & -\\frac{\\sin\\phi}{r} \\\\ -\\frac{\\sin\\phi}{r\\sin\\theta} & \\frac{\\cos\\phi}{r\\sin\\theta} & 0 \\end{bmatrix}$$ Alternatively, we could have calculated the inverse Jacobian (which is straightforward) and then inverted it (which is a nightmare). We can use Wolfram Alpha to confirm that it gives the same result: Finally, we use the dot product to find the coefficients $r'$, $\\theta'$, and $\\phi'$: $$r' = \\vec\\nabla f\\cdot\\mathbf{\\hat e}_r = \\begin{bmatrix}\\frac{\\partial f}{\\partial r} & \\frac{\\partial f}{\\partial \\theta} & \\frac{\\partial f}{\\partial \\phi}\\end{bmatrix} J \\begin{bmatrix}\\sin\\theta\\cos\\phi \\\\ \\sin\\theta\\sin\\phi \\\\ \\cos\\theta\\end{bmatrix} = \\begin{bmatrix}\\frac{\\partial f}{\\partial r} & \\frac{\\partial f}{\\partial \\theta} & \\frac{\\partial", "\\sin \\left( \\theta + \\frac{\\alpha + \\beta}{2} \\right) \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence, $$15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2)$$ Taking $\\frac{(1) + (2)}{30}$, we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$. Thus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Taking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$. Hence, \\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*} Hence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$. In $\\triangle O O_1 B$, $$O_1 B = 2 r \\sin \\frac{\\alpha}{2} .$$ In $\\triangle O O_2 A$, by applying the law of sines, we get $$O_2 A = 2 r \\sin \\frac{\\beta}{2} .$$ Because circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$, $A$ is on circle $\\omega_2$, \\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*}", "\\left( {E_x'}^2 - \\frac{1}{{2}} ({\\mathbf{E}'}^2 + {\\mathbf{B}'}^2)\\right) \\\\ &= \\frac{1}{{4 \\pi}} \\left( {E'}^2 \\sin^2 \\theta - \\frac{1}{{2}} 2 {E'}^2 \\right) \\\\ &= -\\frac{1}{{4 \\pi}} {E'}^2 \\cos^2\\theta\\end{aligned} So by comparison we see that we have \\begin{aligned}\\sigma_{1 1} = {\\sigma}_{x x}^{\\text{incident}} +{\\sigma}_{x x}^{\\text{reflected}} \\end{aligned} \\hspace{\\stretch{1}}(2.30) Moving on, for our other component on the $x,y$ place $\\sigma_{12}$ we have \\begin{aligned}\\sigma_{12} &= \\frac{1}{{4 \\pi}} \\mathcal{E}^1 \\mathcal{E}^2 \\\\ &= \\frac{1}{{4 \\pi}} (E + E') \\sin\\theta (-E + E') \\cos\\theta \\\\ &= \\frac{1}{{4 \\pi}} ((E')^2 - E^2) \\sin\\theta \\cos\\theta \\end{aligned} Again we can compare to the sums of the reflected and incident tensors for this $x,y$ component. Those are \\begin{aligned}\\sigma_{12}^{\\text{incident}} &= \\frac{1}{{4 \\pi}} ( E^1 E^2 ) \\\\ &= -\\frac{1}{{4 \\pi}} E^2 \\sin\\theta \\cos\\theta,\\end{aligned} and \\begin{aligned}\\sigma_{12}^{\\text{reflected}} &= \\frac{1}{{4 \\pi}} ( {E'}^1 {E'}^2 ) \\\\ &= \\frac{1}{{4 \\pi}} {E'}^2 \\sin\\theta \\cos\\theta\\end{aligned} Which demonstrates that we have \\begin{aligned}\\sigma_{12} = \\sigma_{12}^{\\text{incident}} + \\sigma_{12}^{\\text{reflected}} \\end{aligned} \\hspace{\\stretch{1}}(2.31) Summarizing, for the components in the $x,y$ plane we have found that we have \\begin{aligned}\\sigma_{\\alpha\\beta}^{\\text{total}} n_\\beta = \\sigma_{\\alpha 1 }^{\\text{total}} = \\sigma_{\\alpha 1} + {\\sigma'}_{\\alpha 1}\\end{aligned} \\hspace{\\stretch{1}}(2.32) (where $n_\\beta = \\delta^{\\beta 1}$) This result, assumed in the text, was non-trivial to derive. It is also not generally true. We have \\begin{aligned}\\sigma_{2 2} &= \\frac{1}{{4 \\pi}} \\left( (\\mathcal{E}^y)^2 - \\frac{1}{{2}} ( \\vec{\\mathcal{E}}^2 + \\vec{\\mathcal{B}}^2 ) \\right) \\\\ &= \\frac{1}{{8 \\pi}} \\left( (\\mathcal{E}^y)^2 -", "parts, $$\\int \\sec^3 \\Theta\\, d\\Theta$$ would become $$\\frac{\\sec\\Theta tan\\Theta + \\ln\\left | \\sec\\Theta + \\tan\\Theta \\right |}{2}$$. Finally, $$\\int \\sec^2 \\Theta\\, d\\Theta$$ would simply become $$\\tan \\Theta$$ and $$\\int \\sec \\Theta$$ would be $$ln\\left | \\sec\\Theta + \\tan\\Theta \\right |$$ . Since $$\\sec \\Theta$$ is equal to $$\\frac{x+5}{3}$$ and $$\\tan\\Theta$$ is equal to $$\\frac{\\sqrt{x^2+10x+16}}{3}$$, the whole integral would be (I had added all like terms before this): $$9\\left [ \\frac{\\sqrt{x^2+10x+16}}{3} \\right ]^3 + 252\\left ( \\frac{\\sqrt{x^2+10x+16}}{3}\\right ) -\\frac{135}{2} \\left ( \\frac{x+5}{3}\\right ) \\left ( \\frac{\\sqrt{x^2+10x+16}}{3}\\right ) - \\frac{385}{2} ln \\left | \\frac{x+5}{3} + \\frac{\\sqrt{x^2+10x+16}}{3}\\right | + C$$ Lastly, I simplified the integral: $$\\frac{1}{3} \\left ( x^2 + 10x+16 \\right )^\\frac{3}{2} + \\frac{(93-15x)\\sqrt{x^2+10x+16}}{2} - \\frac{385}{2} ln \\left | \\frac{x+5+\\sqrt{x^2+10x+16}}{3}\\right | + C$$ Have I integrated the integrand appropriately? Thanks in advanced. • No. How did you eliminate the square root in the first step? By the way, you should get: $$\\frac{1}{6} \\sqrt{x^2+10 x+16} \\left(2 x^2-25 x+311\\right)-\\frac{385}{2} \\log \\left(\\sqrt{x^2+10 x+16}+x+5\\right))$$ – David G. Stork Dec 24 '19 at 1:59 • It's a typo. I edited it. – romeoPH Dec 24 '19 at 2:20 • @DavidG.Stork, he is getting this only, if you combine first two terms. – Martund Dec 24 '19 at 2:38 It is correct. Here is another method without trigonometric substitution: $$\\int\\frac{x^3}{\\sqrt{x^2+10x+16}}dx=\\int\\frac{x^3+10x^2+16x-10x^2-100x-160+84x+160}{\\sqrt{x^2+10x+16}}dx$$ $$= \\int\\frac{(x-10)(x^2+10x+16)+84x+160}{\\sqrt{x^2+10x+16}}dx$$ $$= \\int(x-10)\\sqrt{x^2+10x+16}\\ dx+\\int\\frac{84x+420}{\\sqrt{x^2+10x+16}}dx-260\\int\\frac{1}{\\sqrt{x^2+10x+16}}dx$$ The", "# How do you convert r = 8 sin theta into cartesian form? Dec 29, 2016 The answer is ${x}^{2} + {\\left(y - 4\\right)}^{2} = {4}^{2}$ #### Explanation: To convert from polar coordinates $\\left(r , \\theta\\right)$ to cartesian coordinates $\\left(x , y\\right)$, we use the following equations $x = r \\cos \\theta$ $y = r \\sin \\theta$ ${x}^{2} + {y}^{2} = {r}^{2}$ Here, we have $r = 8 \\sin \\theta$ $r = 8 \\cdot \\frac{y}{r}$ ${r}^{2} = 8 y$ ${x}^{2} + {y}^{2} = 8 y$ ${x}^{2} + {y}^{2} - 8 y = 0$ Completing the squares ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ ${x}^{2} + {\\left(y - 4\\right)}^{2} = {4}^{2}$ This is the equation of a circle, center $\\left(0 , 4\\right)$ and radius $= 4$ graph{x^2+(y-4)^2=16 [-13.88, 14.6, -2.85, 11.39]} Dec 29, 2016 ${x}^{2} + {y}^{2} = 8 y$ #### Explanation: we understand that; $x = r \\cos \\theta$, $y = r \\sin \\theta$ and ${x}^{2} + {y}^{2} = {r}^{2}$ --@ from the question, $r = 8 \\sin \\theta$ we can multiply with r, then $r \\cdot r = r \\cdot 8 \\sin \\theta$ ${r}^{2} = 8 \\left(r \\sin \\theta\\right) w h e r e , y = r \\sin \\theta$ ${r}^{2} = 8 y$ therefore from @ the answer is ${x}^{2} + {y}^{2} =", "## Precalculus (6th Edition) Blitzer The required polar equation is $r=12\\cos \\theta$. To find the polar equation, substitute $x=r\\cos \\theta \\ \\text{ and }\\ y=r\\sin \\theta$ in the provided equation, where r is the distance of that point from the origin and $\\theta$ is the respective angle. So, \\begin{align} & {{\\left( x-6 \\right)}^{2}}+{{y}^{2}}=36 \\\\ & {{\\left( r\\cos \\theta -6 \\right)}^{2}}+{{\\left( r\\sin \\theta \\right)}^{2}}=36 \\\\ & {{r}^{2}}{{\\cos }^{2}}\\theta +36-12r\\cos \\theta +{{r}^{2}}{{\\sin }^{2}}\\theta =36 \\\\ & {{r}^{2}}\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)-12r\\cos \\theta =0 \\end{align} Since, ${{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta =1$ so, \\begin{align} & {{r}^{2}}\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)-12r\\cos \\theta =0 \\\\ & {{r}^{2}}-12r\\cos \\theta =0 \\\\ & r\\left( r-12\\cos \\theta \\right)=0 \\\\ & r=0,r=12\\cos \\theta \\end{align} Because, $r=0$ is a single point. Therefore, the polar form of the equation ${{\\left( x-6 \\right)}^{2}}+{{y}^{2}}=36$ is $r=12\\cos \\theta$. Hence, the required polar equation is $r=12\\cos \\theta$.", "given by $$\\omega = \\sqrt{\\frac{U''(x)}{m}}$$. Perhaps a similar technique can be applied to angles? $$U(\\theta) = \\frac{1}{2}k x_1^2 + \\frac{1}{2}k x_2^2 + mg\\Delta h$$ $$x_1 = L sin(\\theta)$$ $$x_2 = \\frac{L}{2} sin(\\theta)$$ $$\\Delta h = h (1-cos(\\theta))$$ Then using small angle approximations, $$U(\\theta) = \\frac{1}{2}k (L \\theta)^2 + \\frac{1}{2}k (\\frac{L}{2}\\theta)^2$$ $$U(\\theta) = \\frac{k}{2} L^2 \\theta^2 + \\frac{k}{2} \\frac{L^2}{4}\\theta^2$$ $$U(\\theta) = \\frac{k}{2} (L^2 \\theta^2 + \\frac{L^2}{4}\\theta^2)$$ $$U(\\theta) = \\frac{k}{2} (\\frac{5}{4}L^2 \\theta^2)$$ $$U(\\theta) = \\frac{5k}{8}L^2 \\theta^2$$ $$U''(\\theta) = \\frac{5k}{4}L^2$$ Then $$\\omega = \\sqrt{\\frac{\\frac{5k}{4}L^2}{m}}$$ $$=\\frac{L}{2}\\sqrt{\\frac{5k}{m}}$$ So, I'm not sure which technique is valid or how to continue from here. Any help would be greatly appreciated. Thank you. #### Attached Files: • ###### phys.bmp File size: 270.9 KB Views: 171 2. Nov 23, 2009 ### Redbelly98 Staff Emeritus Yes, though the middle term on the LHS should not have an L in the denominator. Perhaps just a typo? From here, you can get ω simply by inspection of the differential equation. 3. Nov 23, 2009 ### Math Jeans Your period is contained in the term before your $$\\theta$$. You should make the quick correction suggested, and see if you can get equivalent answers. If you can, then you're fine for either. If you cannot, try using Lagrange's equations to check a third method. 4. Nov 23, 2009 ### Redbelly98 Staff Emeritus Just" ]

[ "In order to find these, we will need to use a product rule (more on the product rule). Doing these one at a time we find, \\begin{align*} \\frac{d}{d\\theta}3\\sin(4\\theta)\\sin(\\theta) &=SF’+FS’ \\end{align*} where $$S=\\sin(\\theta)$$, $$F=3\\sin(4\\theta)$$, $$S’=\\cos(\\theta)$$ and for $$F’$$ we need to use a chain rule (more on chain rule). For this one, we have $$i=4 \\theta$$, $$o=3\\sin(i)$$, so $$o’=3\\cos(i)$$ and $$i’=4$$, thus giving $$3\\cos(i)*4=12\\cos(4\\theta)$$. Combining these, we find, \\begin{align*} \\frac{d}{d\\theta}&3\\sin(4\\theta)\\sin(\\theta)\\\\ &=SF’+FS’ \\\\ &=\\sin(\\theta)*12\\cos(4\\theta)+3\\sin(4\\theta)\\cos(\\theta). \\end{align*} If you follow a similar process for $$\\frac{dx}{d\\theta}$$, you will find that \\begin{align*} \\frac{d}{d\\theta}&(3\\sin(4\\theta)\\cos(\\theta) \\\\ &=\\cos(\\theta)*12\\cos(4\\theta)-3\\sin(4\\theta)\\sin(\\theta). \\end{align*} We can now find that \\begin{align*} \\frac{dy}{dx}=\\frac{\\sin(\\theta)*12\\cos(4\\theta)+3\\sin(4\\theta)\\cos(\\theta)}{\\cos(\\theta)*12\\cos(4\\theta)-3\\sin(4\\theta)\\sin(\\theta)}. \\end{align*} This is indeed what we were looking for. If we were going to use it further, I would suggest simplifying what you could. However, the question only asked to find $$\\frac{dy}{dx}$$, so I would leave the problem as is. ### Areas of the leaves We now move on to finding the area enclosed by the function $$r=3\\sin(4\\theta)$$ on each of its leaves and as a whole. In order to do this, we will be using the techniques we used in Finding Area using Integrals. In this case; however, we will be cutting triangles instead of a rectangles. I have drawn an example of such a triangle below. In this case, we see that area of the triangle will be $$\\frac{1}{2}bh$$,", "holds for all real numbers $$n$$. Therefore, it makes no difference whether one considers the case $$n = 0$$ in the solution; either way, the same constants will be obtained. To see why this is the case, we note that if $$\\sin \\theta = \\frac{7}{25}$$, with $$0 < \\theta < \\frac{\\pi}{2}$$, then $$\\cos \\theta = \\frac{24}{25}$$. Then \\begin{align*} \\sin 2\\theta &= 2 \\sin \\theta \\cos \\theta = \\frac{336}{625}, \\\\ \\cos 2\\theta &= 1 - 2\\sin^2 \\theta = \\frac{527}{625}. \\end{align*} Hence \\begin{align*} a_{n+2} &= \\sin n \\theta \\cos 2\\theta + \\cos n \\theta \\sin 2\\theta \\\\ &= \\frac{527}{625} \\sin n\\theta + \\frac{336}{625} \\cos n\\theta. \\end{align*} But we also have \\begin{align*} a_{n+2} &= k_1 a_{n+1} + k_0 a_n \\\\ &= k_1 (\\sin n \\theta \\cos \\theta + \\cos n \\theta \\sin \\theta) + k_0 \\sin n \\theta \\\\ &= \\left( k_0 + \\frac{24k_1}{25} \\right) \\sin n\\theta + \\frac{7k_1}{25} \\cos n\\theta. \\end{align*} So if this recurrence is to be satisfied for positive integers $$n$$, then it is immediately obvious that it is also satisfied for arbitrary real $$n$$, since the equality of the coefficients gives identical equality. In particular, we must have \\begin{align*} k_0 + \\frac{24}{25} k_1 &= \\frac{527}{625}, \\\\ \\frac{7}{25} k_1 &= \\frac{336}{625}, \\end{align*} the solution of which is $$k_0 = -1$$, $$k_1 = \\frac{48}{25}$$. Therefore, $$k_0 + k_1 =", "the graph of a polar equation? For the following exercises, test the equation for symmetry. 6. $r=5\\cos 3\\theta$ 7. $r=3 - 3\\cos \\theta$ 8. $r=3+2\\sin \\theta$ 9. $r=3\\sin 2\\theta$ 10. $r=4$ 11. $r=2\\theta$ 12. $r=4\\cos \\frac{\\theta }{2}$ 13. $r=\\frac{2}{\\theta }$ 14. $r=3\\sqrt{1-{\\cos }^{2}\\theta }$ 15. $r=\\sqrt{5\\sin 2\\theta }$ For the following exercises, graph the polar equation. Identify the name of the shape. 16. $r=3\\cos \\theta$ 17. $r=4\\sin \\theta$ 18. $r=2+2\\cos \\theta$ 19. $r=2 - 2\\cos \\theta$ 20. $r=5 - 5\\sin \\theta$ 21. $r=3+3\\sin \\theta$ 22. $r=3+2\\sin \\theta$ 23. $r=7+4\\sin \\theta$ 24. $r=4+3\\cos \\theta$ 25. $r=5+4\\cos \\theta$ 26. $r=10+9\\cos \\theta$ 27. $r=1+3\\sin \\theta$ 28. $r=2+5\\sin \\theta$ 29. $r=5+7\\sin \\theta$ 30. $r=2+4\\cos \\theta$ 31. $r=5+6\\cos \\theta$ 32. ${r}^{2}=36\\cos \\left(2\\theta \\right)$ 33. ${r}^{2}=10\\cos \\left(2\\theta \\right)$ 34. ${r}^{2}=4\\sin \\left(2\\theta \\right)$ 35. ${r}^{2}=10\\sin \\left(2\\theta \\right)$ 36. $r=3\\text{sin}\\left(2\\theta \\right)$ 37. $r=3\\text{cos}\\left(2\\theta \\right)$ 38. $r=5\\text{sin}\\left(3\\theta \\right)$ 39. $r=4\\text{sin}\\left(4\\theta \\right)$ 40. $r=4\\text{sin}\\left(5\\theta \\right)$ 41. $r=-\\theta$ 42. $r=2\\theta$ 43. $r=-3\\theta$ For the following exercises, use a graphing calculator to sketch the graph of the polar equation. 44. $r=\\frac{1}{\\theta }$ 45. $r=\\frac{1}{\\sqrt{\\theta }}$ 46. $r=2\\sin \\theta \\tan \\theta$, a cissoid 47. $r=2\\sqrt{1-{\\sin }^{2}\\theta }$ , a hippopede 48. $r=5+\\cos \\left(4\\theta \\right)$ 49. $r=2-\\sin \\left(2\\theta \\right)$ 50. $r={\\theta }^{2}$ 51. $r=\\theta +1$ 52. $r=\\theta \\sin \\theta$ 53. $r=\\theta \\cos \\theta$ For the following exercises, use a graphing utility to", "\\frac{1}{z(6i-8)} \\frac{1}{1 - \\frac{3i+4}{z}} + \\frac{1}{z(8-6i)} \\frac{1}{1 - \\frac{4-3i}{z}} \\\\ &= \\frac{1}{z(6i-8)} \\sum (\\frac{3i+4}{z})^n + \\frac{1}{z(8-6i)} \\sum (\\frac{4-3i}{z})^n \\\\ &= \\frac{1}{(6i-8)} \\sum \\frac{(3i+4)^n}{z^{n+1}} + \\frac{1}{(8-6i)} \\sum \\frac{(4-3i)^n}{z^{n+1}} \\end{align*} Still, $|\\frac{3i+4}{z}| < 1$ and $|\\frac{-3i+4}{z}| < 1$ lead to $|z| > 5$. Thus $R = 5$. Use same approach as above, we can get \\begin{align*} &f(5e^{it}) = K_3 \\sum (\\frac{5 e^{i\\theta}}{5e^{it}}) ^ n \\frac{1}{z} + K_4 \\sum (\\frac{5 e^{i(-\\theta)}}{5e^{it}})^n \\frac{1}{z} \\\\ &= \\frac{K_3}{5} \\sum e^{ni(\\theta - t) - it} + \\frac{K_4}{5} \\sum e^{ni(-\\theta - t) - it} \\\\ &= \\frac{K_3}{5} \\sum e^{i(n\\theta -(n+1)t)} + \\frac{K_4}{5} \\sum e^{i(n(-\\theta) -(n+1)t)} \\\\ &= \\frac{K_3}{5} \\sum e^{i(n\\theta -(n+1)t)} + \\frac{K_4}{5} \\sum e^{i(n(-\\theta) -(n+1)t)} \\\\ &= |\\frac{K_3}{5}| \\sum e^{i(n\\theta -(n+1)t)} e^{i(\\theta - \\pi)} + |\\frac{K_4}{5}| \\sum e^{i(n(-\\theta) -(n+1)t)} e^{i\\theta} \\\\ &= |\\frac{K_3}{5}| \\sum e^{i((n+1)\\theta -(n+1)t - \\pi)} + |\\frac{K_4}{5}| \\sum e^{i((n+1)(-\\theta) -(n+1)t)} \\\\ \\end{align*} if $f(5e^{it})$ converges, then $Im\\{f(5e^{it})\\}$ converges, since \\begin{align*} Im\\{f(5e^{it})\\} = |K_5| \\sum -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sum \\sin {((n+1)(-\\theta) -(n+1)t)} \\end{align*} then $\\lim_{n \\rightarrow \\infty} |K_5| -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sin {((n+1)(-\\theta) -(n+1)t)} = 0$, since $|K_5| = |K_6|$, $\\lim_{n \\rightarrow \\infty} |K_5| -\\sin{((n+1)\\theta -(n+1)t)} + |K_6| \\sin {((n+1)(-\\theta) -(n+1)t)} = -2|K_5| \\lim_{n \\rightarrow \\infty} \\cos (-(n+1) t) \\sin ((n+1)\\theta) = 0$ a contradiction. Thus, on the $|z| = 5$, series does not converge. 11 MAT334--Lectures & Home Assignments /", "Suppose $Bx\\le\\Theta(x)\\le Cx$ for each $\\Theta$. \\begin{align} \\frac{1+\\Theta(\\frac{1}{2n})}{(1+\\Theta(\\frac{1}{n}))^2} &\\le\\frac{1+\\frac{C}{2n}}{(1+\\frac{B}{n})^2}\\\\ &=(1+\\frac{C}{2n})(1-2\\frac{B}{n}+O(\\frac{1}{n^2}))\\\\ &=(1+(\\frac{C}{2}-2B)\\frac{1}{n}+O(\\frac{1}{n^2}))\\\\ &=1+O(\\frac{1}{n}) \\end{align} \\begin{align} \\frac{1+\\Theta(\\frac{1}{2n})}{(1+\\Theta(\\frac{1}{n}))^2} &\\ge\\frac{1+\\frac{B}{2n}}{(1+\\frac{C}{n})^2}\\\\ &=(1+\\frac{B}{2n})(1-2\\frac{C}{n}+O(\\frac{1}{n^2}))\\\\ &=(1+(\\frac{B}{2}-2C)\\frac{1}{n}+O(\\frac{1}{n^2}))\\\\ &=1+O(\\frac{1}{n}) \\end{align} - Huh. Over an hour, and no upvotes on either of our answers. Well, +1 from me, because it's useful to see how to do this from the definition, too. – Mike Spivey Sep 9 '11 at 4:12 Yes. One way to see it is to do the long division; i.e., divide the denominator directly into the numerator. The numerator is $1 + \\Theta(\\frac{1}{2n}) = 1 + \\Theta(\\frac{1}{n})$, and the denominator is $\\left(1 + \\Theta(\\frac{1}{n})\\right)^2 = 1 + \\Theta(\\frac{1}{n}) + \\Theta(\\frac{1}{n^2}) = 1 + \\Theta(\\frac{1}{n})$. I'm not going to attempt to typeset the long division on this forum, but try it, and you'll see that you get $1$ with a remainder of $O(\\frac{1}{n})$. So you have $$\\frac {1 + \\Theta(\\frac{1}{2n})} {\\left(1 + \\Theta(\\frac{1}{n})\\right)^2} = \\frac{1 + \\Theta(\\frac{1}{n})}{1 + \\Theta(\\frac{1}{n})} = 1 + \\frac{O(\\frac{1}{n})}{1 + \\Theta(\\frac{1}{n})} = 1 + O\\left(\\frac{1}{n}\\right),$$ since the denominator is $O(1)$. (Remember that $\\Theta(\\frac{1}{2n}) = \\Theta(\\frac{1}{n})$, since the constant $\\frac{1}{2}$ doesn't affect the asymptotic order.) - I don't think you can bound the ratio away from $1$, so I don't think you can say $1+\\Theta(1/n)$. – robjohn Sep 9 '11 at 2:54 @robjohn: Right; the terms might cancel. I'll fix it. – Mike Spivey Sep", "\\theta )\\\\T_{2}(\\cos \\theta )&=&\\cos 2\\theta &=&{\\frac {1}{3}}(4P_{2}(\\cos \\theta )-P_{0}(\\cos \\theta ))\\\\T_{3}(\\cos \\theta )&=&\\cos 3\\theta &=&{\\frac {1}{5}}(8P_{3}(\\cos \\theta )-3P_{1}(\\cos \\theta ))\\\\T_{4}(\\cos \\theta )&=&\\cos 4\\theta &=&{\\frac {1}{105}}(192P_{4}(\\cos \\theta )-80P_{2}(\\cos \\theta )-7P_{0}(\\cos \\theta ))\\\\T_{5}(\\cos \\theta )&=&\\cos 5\\theta &=&{\\frac {1}{63}}(128P_{5}(\\cos \\theta )-56P_{3}(\\cos \\theta )-9P_{1}(\\cos \\theta ))\\\\T_{6}(\\cos \\theta )&=&\\cos 6\\theta &=&{\\frac {1}{1155}}(2560P_{6}(\\cos \\theta )-1152P_{4}(\\cos \\theta )-220P_{2}(\\cos \\theta )-33P_{0}(\\cos \\theta ))\\end{aligned}}} Another property is the expression for ${\\displaystyle \\sin(n+1)\\theta }$, which is ${\\displaystyle {\\frac {\\sin(n+1)\\theta }{\\sin \\theta }}=\\sum _{\\ell =0}^{n}P_{\\ell }(\\cos \\theta )P_{n-\\ell }(\\cos \\theta )}$ Legendre polynomials are symmetric or antisymmetric, that is ${\\displaystyle P_{n}(-x)=(-1)^{n}P_{n}(x).\\,}$[5] Since the differential equation and the orthogonality property are independent of scaling, the Legendre polynomials' definitions are \"standardized\" (sometimes called \"normalization\", but note that the actual norm is not unity) by being scaled so that ${\\displaystyle P_{n}(1)=1.\\,}$ The derivative at the end point is given by ${\\displaystyle P_{n}'(1)={\\frac {n(n+1)}{2}}.\\,}$ As discussed above, the Legendre polynomials obey the three term recurrence relation known as Bonnet’s recursion formula ${\\displaystyle (n+1)P_{n+1}(x)=(2n+1)xP_{n}(x)-nP_{n-1}(x)\\,}$ and ${\\displaystyle {x^{2}-1 \\over n}{d \\over dx}P_{n}(x)=xP_{n}(x)-P_{n-1}(x).}$ Useful for the integration of Legendre polynomials is ${\\displaystyle (2n+1)P_{n}(x)={d \\over dx}\\left[P_{n+1}(x)-P_{n-1}(x)\\right].}$ ${\\displaystyle {d \\over dx}P_{n+1}(x)=(2n+1)P_{n}(x)+(2(n-2)+1)P_{n-2}(x)+(2(n-4)+1)P_{n-4}(x)+\\ldots }$ or equivalently ${\\displaystyle {d \\over dx}P_{n+1}(x)={2P_{n}(x) \\over \\|P_{n}\\|^{2}}+{2P_{n-2}(x) \\over \\|P_{n-2}\\|^{2}}+\\ldots }$ where ${\\displaystyle \\|P_{n}\\|}$ is the norm over the interval −1 ≤ x ≤ 1 ${\\displaystyle \\|P_{n}\\|={\\sqrt {\\int _{-1}^{1}(P_{n}(x))^{2}\\,dx}}={\\sqrt {\\frac {2}{2n+1}}}.}$ From Bonnet’s recursion formula one", "\\theta)(\\sin \\theta) \\\\ &= 2r\\cos^2 \\theta + 2r \\sin^2 \\theta \\\\ &= 2r \\end{align} And also: (4) \\begin{align} \\quad \\frac{\\partial f}{\\partial \\theta} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial \\theta} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial \\theta} \\\\ &= (2x) (-r \\sin \\theta) + (2y)(r \\cos \\theta) \\\\ &= (2r \\cos \\theta)(-r \\sin \\theta) + (2r \\sin \\theta)(r \\cos \\theta) \\\\ &= -2r^2 \\cos \\theta \\sin \\theta + 2r^2 \\sin \\theta \\cos \\theta \\\\ &= 0 \\end{align} By direct substitution we have that: (5) \\begin{align} \\quad f(x, y) = x^2 + y^2 = (r \\cos \\theta)^2 + (r \\sin \\theta)^2 = r^2 \\cos^2 \\theta + r^2 \\sin^2 \\theta = r^2(\\cos^2 \\theta + \\sin^2 \\theta) = r^2 = f(r, \\theta) \\end{align} Therefore $\\displaystyle{\\frac{\\partial f}{\\partial r} = 2r}$ and $\\displaystyle{\\frac{\\partial f}{\\partial \\theta} = 0}$ as above. ## Example 2 Let $f(x, y, z) = x^2ye^z$, $x(s, t) = s^2t$, $y(s, t) = st^2$, and $z(s, t) = \\sin (st)$. Find $\\displaystyle{\\frac{\\partial f}{\\partial s}}$ by first using the chain rule and then by direct substitution. We have that: (6) \\begin{align} \\quad \\frac{\\partial f}{\\partial s} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial s} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial s} + \\frac{\\partial f}{\\partial z} \\frac{\\partial z}{\\partial s} \\\\ &= (2xye^z)(2st) + (x^2e^z)(t^2) + (x^2ye^z)(t \\cos (st)) \\\\ &= 2(s^2t)(st^2)e^{\\sin (st)}(2st) + (s^2t)^2e^{\\sin (st)} (t^2)", "From the previous question we have that $$x = 5$$, $$y = 8$$ and $$r = \\sqrt{89}$$. \\begin{align*} \\cot \\theta & = \\frac{x}{y} \\\\ & = \\frac{5}{8} \\end{align*} $$\\sin^{2} \\theta + \\cos^{2} \\theta$$ From the previous question we have that $$x = 5$$, $$y = 8$$ and $$r = \\sqrt{89}$$. \\begin{align*} \\sin^{2} \\theta + \\cos^{2} \\theta & = \\left(\\frac{y}{r}\\right)^{2} + \\left(\\frac{x}{r}\\right)^{2} \\\\ & = \\frac{y^{2}}{r^{2}} + \\frac{x^{2}}{r^{2}} \\\\ & = \\frac{y^{2} + x^{2}}{r^{2}} \\\\ & = \\frac{64 + 25}{89} \\\\ & = 1 \\end{align*} Given the following diagram and that $$\\cos \\theta = -\\frac{24}{25}$$. State two sets of possible values of $$a$$ and $$b$$. We first need to use the given information to find a possible set of values for $$a$$ and $$b$$. Using $$\\cos \\theta = -\\frac{24}{25}$$ and the fact that $$\\cos \\theta = \\dfrac{x}{r}$$ we can determine that $$x = -24$$ and $$r = 25$$. Now we can find $$y$$: \\begin{align*} y^{2} & = r^{2} - x^{2} \\\\ & = (25)^2 - (24)^2 \\\\ & = 625 - 576 \\\\ & = 49 \\\\ y & = \\pm 7 \\end{align*} From the diagram we see that $$y$$ must be negative. This gives us one possible set of values for $$a$$ and $$b$$: $$a = 24$$ and $$b = -7$$. Now we note that we can simply", "r \\cos \\theta+2 r \\sin \\theta && \\text{isolate the squared terms} \\\\ r^{2}\\left(\\cos ^{2} \\theta+\\sin ^{2} \\theta\\right)&=-2 r(4 \\cos \\theta-\\sin \\theta) && \\text{ factor } r^{2} \\text{-a common factor} \\\\ r^{2}&=-2 r(4 \\cos \\theta-\\sin \\theta) && \\text{ Pythagorean Identity} \\\\ r&=-2(4 \\cos \\theta-\\sin \\theta) && \\text{ Divide each side by } r \\end{aligned} Example $$\\PageIndex{1}$$ Earlier, you were asked to convert an equation to polar form. Solution The original equation to convert is: $$(x+1)^2−(y+2)^2=7$$ You can substitute $$x=r\\cos \\theta$$ and $$y=r\\sin \\theta$$ into the equation, and then simplify: \\begin{aligned} (r \\cos \\theta+1)^{2}-(r \\sin \\theta+2)^{2}&=7 \\\\ \\left(r^{2} \\cos ^{2} \\theta+2 r \\cos \\theta+1\\right)-\\left(r^{2} \\sin ^{2} \\theta+4 r \\sin \\theta+4\\right)&=7 \\\\ r^{2}\\left(\\cos ^{2} \\theta-\\sin ^{2} \\theta\\right)+2 r(\\cos \\theta-2 \\sin \\theta)-3&=7 \\\\ r^{2}\\left(\\cos ^{2} \\theta-\\sin ^{2} \\theta\\right)+2 r(\\cos \\theta-2 \\sin \\theta)&=10 \\end{aligned} Example $$\\PageIndex{2}$$ Write the rectangular equation $$(x−4)^2+(y−3)^2=25$$ in polar form. Solution \\begin{aligned} (x-4)^{2}+(y-3)^{2} &=25 \\\\ x^{2}-8 x+16+y^{2}-6 y+9 &=25 \\\\ x^{2}-8 x+y^{2}-6 y+25 &=25 \\\\ x^{2}-8 x+y^{2}-6 y &=0 \\\\ x^{2}+y^{2}-8 x-6 y &=0 \\\\ r^{2}-8(r \\cos \\theta)-6(r \\sin \\theta) &=0 \\\\ r^{2}-8 r \\cos \\theta-6 r \\sin \\theta &=0 \\\\ r(r-8 \\cos \\theta-6 \\sin \\theta) &=0 \\\\ r=0 \\text { or } r-8 \\cos \\theta-6 \\sin \\theta &=0 \\\\ r=0 \\text { or } r &=8 \\cos \\theta+6 \\sin \\theta \\end{aligned} From graphing r−8\\cos \\theta−6\\sin \\theta=0, we see", "a}{3\\tan^2 a + 4} = \\frac{\\sin 2a}{7 + \\cos 2a}$. Starting from the left side: $\\frac{\\tan a}{3\\tan^2 a + 4}$ \\begin{aligned} &= \\frac{\\tan a}{3\\tan^2 a + 4} \\cdot \\frac{2\\cos^2 a}{2\\cos^2 a} \\\\ &= \\frac{2\\sin a \\cos a}{6\\sin^2 a + 8\\cos^2 a} \\\\ &= \\frac{\\sin 2a}{6\\sin^2 a + 6\\cos^2 a + 2\\cos^2 a} \\end{aligned} \\begin{aligned} &= \\frac{\\sin 2a}{6 + 2\\cos^2 a} \\\\ &= \\frac{\\sin 2a}{7 + 2\\cos^2 a - 1} \\\\ &= \\frac{\\sin 2a}{7 + \\cos 2a} \\end{aligned} 01 • Jul 25th 2009, 07:48 AM Soroban Hello, flyinhigh123! Quote: Prove: . $\\tan A \\:=\\: \\sqrt{\\frac{1-\\cos2A}{1+\\cos2A}}$ We need these two identities: . $\\begin{array}{ccccccc}\\sin^2\\theta &=& \\dfrac{1-\\cos2\\theta}{2} & \\Rightarrow & \\sin\\theta &=& \\sqrt{\\dfrac{1-\\cos2\\theta}{2}} \\\\ \\\\[-3mm] \\cos^2\\theta &=& \\dfrac{1+\\cos2\\theta}{2} & \\Rightarrow & \\cos\\theta &=& \\sqrt{\\dfrac{1+\\cos2\\theta}{2}} \\end{array}$ We have: . $\\tan A \\;=\\;\\frac{\\sin A}{\\cos A} \\;=\\;\\frac{\\sqrt{\\frac{1-\\cos2A}{2}} }{\\sqrt{\\frac{1+\\cos2A}{2}}} \\;=\\; \\sqrt{\\frac{\\frac{1-\\cos2A}{2}} {\\frac{1+\\cos2A}{2}}}$ . . Therefore: . $\\boxed{\\tan A \\;=\\;\\sqrt{\\frac{1-\\cos2A}{1+\\cos2A}}}$ Quote: If $4\\tan(A-B) \\:=\\: 3\\tan A$, prove that: . $\\tan B \\:=\\:\\frac{\\sin2A}{7+\\cos2A}$ We are given: . $4\\tan(A-B) \\:=\\:3\\tan A$ . . $4\\left(\\frac{\\tan A - \\tan B}{1 + \\tan A\\tan B}\\right) \\;=\\;3\\tan A$ . . $4(\\tan A - \\tan B) \\;=\\;3\\tan A(1 + \\tan A\\tan B)$ . . $4\\tan A - 4\\tan B \\;=\\;3\\tan A + 3\\tan^2\\!A\\tan B$ . . $3\\tan^2\\!A\\tan B + 4\\tan B \\;=\\;\\tan A$ . . $\\tan B(3\\tan^2\\!A + 4) \\;=\\;\\tan A$ . . $\\tan", "\\\\ r^2 - 8(r \\cos \\theta) - 6(r \\sin \\theta) & = 0 \\\\ r^2 - 8r \\cos \\theta - 6r \\sin \\theta & = 0 \\\\ r(r - 8 \\cos \\theta - 6 \\sin \\theta) & = 0 \\\\ r = 0\\ \\text{or}\\ r - 8 \\cos \\theta - 6 \\sin \\theta & = 0 \\\\ r = 0\\ \\text{or}\\ r & = 8 \\cos \\theta + 6 \\sin \\theta\\end{align*} From graphing \\begin{align*}r-8\\cos \\theta -6\\sin \\theta =0\\end{align*}, we see that the additional solutions are 0 and 8. 2. \\begin{align*}3x - 2y & = 1 \\\\ 3r \\cos \\theta - 2r \\sin \\theta & = 1 \\\\ r (3 \\cos \\theta - 2 \\sin \\theta) & = 1 \\\\ r & = \\frac{1}{3 \\cos \\theta - 2 \\sin \\theta}\\end{align*} 3. \\begin{align*}x^2 + y^2 - 4x + 2y & = 0 \\\\ r^2 \\cos^2 \\theta + r^2 \\sin^2 \\theta - 4 r \\cos \\theta + 2 r \\sin \\theta & = 0 \\\\ r^2 (\\sin^2 \\theta + \\cos^2 \\theta) - 4 r \\cos \\theta + 2 r \\sin \\theta & = 0 \\\\ r (r - 4 \\cos \\theta + 2 \\sin \\theta) & = 0 \\\\ r = 0\\ \\text{or}\\ r - 4 \\cos \\theta + 2 \\sin \\theta & = 0 \\\\ r = 0\\ \\text{or}\\ r", "# How do you convert x^2+(y-4)^2=16 to polar form? May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: If $\\left(r , \\theta\\right)$ is in polar form and $\\left(x , y\\right)$ in Cartesian form the relation between them is as follows: $x = r \\cos \\theta$, $y = r \\sin \\theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\\tan \\theta = \\frac{y}{x}$ Hence, ${x}^{2} + {\\left(y - 4\\right)}^{2} = 16$ can be written as ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ or ${x}^{2} + {y}^{2} - 8 y = 0$ or ${r}^{2} - 8 r \\sin \\theta = 0$ or $r \\left(r - 8 \\sin \\theta\\right) = 0$ dividing by $r$ $r - 8 \\sin \\theta = 0$ or $r = 8 \\sin \\theta$ May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: To convert from Cartesian to Polar coordinates use the following formulae that link them. • x=rcostheta\" and \" y=rsintheta ${x}^{2} + {y}^{2} - 8 y + 16 = 16 \\text{ (expanding bracket) }$ $\\Rightarrow {r}^{2} {\\cos}^{2} \\theta + {r}^{2} {\\sin}^{2} \\theta - 8 r \\sin \\theta + 16 - 16 = 0$ then ${r}^{2} \\left({\\cos}^{2} \\theta + {\\sin}^{2} \\theta\\right) - 8 r \\sin \\theta = 0$ using the identity: ${\\cos}^{2} \\theta + {\\sin}^{2} \\theta = 1$ $\\Rightarrow {r}^{2} = 8 r \\sin", "& = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence,$$15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2)$$ Taking $\\frac{(1) + (2)}{30}$, we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$. Thus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Taking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$. Hence,\\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*} Hence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$. In $\\triangle O O_1 B$,$$O_1 B = 2 r \\sin \\frac{\\alpha}{2} .$$ In $\\triangle O O_2 A$, by applying the law of sines, we get$$O_2 A = 2 r \\sin \\frac{\\beta}{2} .$$ Because circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$$A$ is on circle $\\omega_2$,\\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*} Thus, $\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} = \\frac{12}{\\sqrt{193}}$. Now, we compute $\\sin \\alpha$ and $\\sin", "determine an expression for $$\\cos \\theta$$. \\begin{align*} B\\hat{A}E &= \\text{90} ° \\quad (\\angle \\text{ in semi circle}) \\\\ \\therefore \\cos \\theta &= \\frac{AE}{BE} \\end{align*} In $$\\triangle ACE$$, determine an expression for $$\\sin \\theta$$. \\begin{align*} A\\hat{C}E &= \\text{90} ° \\quad (\\text{given}) \\\\ \\therefore \\sin \\theta &= \\frac{CA}{AE} \\end{align*} In $$\\triangle ACO$$, determine an expression for $$\\sin 2 \\theta$$. \\begin{align*} A\\hat{C}O &= \\text{90} ° \\quad (\\text{given}) \\\\ A\\hat{O}C &= 2 \\theta \\quad (\\text{proven}) \\\\ \\therefore \\sin 2 \\theta &= \\frac{CA}{AO} \\end{align*} Use the results from the previous questions to show that $$\\sin2\\theta =2\\sin\\theta\\cos \\theta$$. $\\begin{array}{rll} \\cos \\theta &= \\frac{AE}{BE} & \\\\ \\therefore AE &= BE \\cos \\theta & \\\\ \\sin \\theta &= \\frac{CA}{AE} & \\\\ \\therefore CA &= AE \\sin \\theta & \\\\ &= (BE \\cos \\theta) \\sin \\theta & \\\\ BE &= 2(OE) & \\text{ (radii)}\\\\ \\therefore CA &= 2(OE) \\sin \\theta \\cos \\theta & \\\\ \\therefore \\frac{CA}{OE} &= 2 \\sin \\theta \\cos \\theta & \\\\ \\text{but }\\frac{CA}{OE} &= \\frac{CA}{OA} = \\sin 2\\theta & \\text{ (in }\\triangle ACO \\text{)}\\\\ \\therefore \\sin 2\\theta &= 2 \\sin \\theta \\cos \\theta & \\end{array}$ $$DC$$ is a diameter of the circle with centre $$O$$ and radius $$r$$. $$CA=r$$, $$AE=2DE$$ and $$D\\hat{O}E=\\theta$$. Show that $$\\cos\\theta =\\frac{1}{4}$$. \\begin{align*} \\text{In }\\triangle DOE \\text{, let }DE &= k \\\\ k^{2} &= r^{2} + r^{2} - 2r^{2} \\cos", "parts, $$\\int \\sec^3 \\Theta\\, d\\Theta$$ would become $$\\frac{\\sec\\Theta tan\\Theta + \\ln\\left | \\sec\\Theta + \\tan\\Theta \\right |}{2}$$. Finally, $$\\int \\sec^2 \\Theta\\, d\\Theta$$ would simply become $$\\tan \\Theta$$ and $$\\int \\sec \\Theta$$ would be $$ln\\left | \\sec\\Theta + \\tan\\Theta \\right |$$ . Since $$\\sec \\Theta$$ is equal to $$\\frac{x+5}{3}$$ and $$\\tan\\Theta$$ is equal to $$\\frac{\\sqrt{x^2+10x+16}}{3}$$, the whole integral would be (I had added all like terms before this): $$9\\left [ \\frac{\\sqrt{x^2+10x+16}}{3} \\right ]^3 + 252\\left ( \\frac{\\sqrt{x^2+10x+16}}{3}\\right ) -\\frac{135}{2} \\left ( \\frac{x+5}{3}\\right ) \\left ( \\frac{\\sqrt{x^2+10x+16}}{3}\\right ) - \\frac{385}{2} ln \\left | \\frac{x+5}{3} + \\frac{\\sqrt{x^2+10x+16}}{3}\\right | + C$$ Lastly, I simplified the integral: $$\\frac{1}{3} \\left ( x^2 + 10x+16 \\right )^\\frac{3}{2} + \\frac{(93-15x)\\sqrt{x^2+10x+16}}{2} - \\frac{385}{2} ln \\left | \\frac{x+5+\\sqrt{x^2+10x+16}}{3}\\right | + C$$ Have I integrated the integrand appropriately? Thanks in advanced. • No. How did you eliminate the square root in the first step? By the way, you should get: $$\\frac{1}{6} \\sqrt{x^2+10 x+16} \\left(2 x^2-25 x+311\\right)-\\frac{385}{2} \\log \\left(\\sqrt{x^2+10 x+16}+x+5\\right))$$ – David G. Stork Dec 24 '19 at 1:59 • It's a typo. I edited it. – romeoPH Dec 24 '19 at 2:20 • @DavidG.Stork, he is getting this only, if you combine first two terms. – Martund Dec 24 '19 at 2:38 It is correct. Here is another method without trigonometric substitution: $$\\int\\frac{x^3}{\\sqrt{x^2+10x+16}}dx=\\int\\frac{x^3+10x^2+16x-10x^2-100x-160+84x+160}{\\sqrt{x^2+10x+16}}dx$$ $$= \\int\\frac{(x-10)(x^2+10x+16)+84x+160}{\\sqrt{x^2+10x+16}}dx$$ $$= \\int(x-10)\\sqrt{x^2+10x+16}\\ dx+\\int\\frac{84x+420}{\\sqrt{x^2+10x+16}}dx-260\\int\\frac{1}{\\sqrt{x^2+10x+16}}dx$$ The", "## Precalculus (6th Edition) Blitzer The required polar equation is $r=12\\cos \\theta$. To find the polar equation, substitute $x=r\\cos \\theta \\ \\text{ and }\\ y=r\\sin \\theta$ in the provided equation, where r is the distance of that point from the origin and $\\theta$ is the respective angle. So, \\begin{align} & {{\\left( x-6 \\right)}^{2}}+{{y}^{2}}=36 \\\\ & {{\\left( r\\cos \\theta -6 \\right)}^{2}}+{{\\left( r\\sin \\theta \\right)}^{2}}=36 \\\\ & {{r}^{2}}{{\\cos }^{2}}\\theta +36-12r\\cos \\theta +{{r}^{2}}{{\\sin }^{2}}\\theta =36 \\\\ & {{r}^{2}}\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)-12r\\cos \\theta =0 \\end{align} Since, ${{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta =1$ so, \\begin{align} & {{r}^{2}}\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)-12r\\cos \\theta =0 \\\\ & {{r}^{2}}-12r\\cos \\theta =0 \\\\ & r\\left( r-12\\cos \\theta \\right)=0 \\\\ & r=0,r=12\\cos \\theta \\end{align} Because, $r=0$ is a single point. Therefore, the polar form of the equation ${{\\left( x-6 \\right)}^{2}}+{{y}^{2}}=36$ is $r=12\\cos \\theta$. Hence, the required polar equation is $r=12\\cos \\theta$.", "Comment Share Q) # A line makes angle $\\theta$ with both $x\\:axis$ and $z\\:axis$. If it makes angle $\\alpha$ with $y\\:axis$ and $\\sin^2\\alpha=2 \\sin^2\\theta$ then $\\cot^2\\theta=?$ $(a)\\;2/3\\:\\:\\:\\qquad\\:\\:(b)\\:\\:1\\:\\:\\:\\qquad\\:\\:(c)\\:\\:3/5\\:\\:\\:\\qquad\\:\\:(d)\\:\\:5/3$ Since the line makes angle $\\theta,\\alpha\\:and\\:\\theta$ with the coordinate axes, $cos^2\\theta+cos^2\\alpha+cos^2\\theta=1$ $\\Rightarrow\\:2cos^2\\theta+cos^2\\alpha=1$....(i) Given: $sin^2\\alpha=2sin^2\\theta$ $\\Rightarrow\\:1-cos^2\\alpha=2(1-cos^2\\theta)$ $\\Rightarrow\\:cos^2 \\alpha=2cos^2\\theta-1$.....(ii) Solving (i) and (ii), $cos^2\\theta=\\large\\frac{1}{2}$ $\\Rightarrow\\:cot^2\\theta=1$", "the formulation in the OP was not really clear to me because the reason for this expression to be constant is already in there. In any case, taking derivatives easily shows that it must be constant. Feb 5 '19 at 14:38 I'm going to use $$y$$ instead of $$x$$. Let $$P(\\theta) = (1, y)$$. Then $$\\theta = \\arctan(y).$$ Then $$\\sin \\theta = \\dfrac{y}{\\sqrt{1+y^2}} \\qquad \\text{and} \\qquad \\cos \\theta = \\dfrac{1}{\\sqrt{1+y^2}}$$ Then \\begin{align} \\tan \\bigg( \\frac 12 \\theta - \\frac{\\pi}{4} \\bigg) &= \\dfrac{\\tan \\frac 12 \\theta - \\tan \\frac{\\pi}{4}} {1 + \\tan(\\frac 12 \\theta) \\tan(\\frac{\\pi}{4})}\\\\ &=\\dfrac{\\left(\\dfrac{\\sin \\theta}{1+\\cos \\theta}-1 \\right)} {\\left( 1+\\dfrac{\\sin \\theta}{1+\\cos \\theta} \\right)} \\\\ &= \\frac{\\sin \\theta - 1 - \\cos \\theta} {1 + \\cos \\theta + \\sin \\theta} \\\\ &= \\frac{\\left( \\dfrac{y}{\\sqrt{1+y^2}} - 1 - \\dfrac{1}{\\sqrt{1+y^2}} \\right)} {\\left( 1 + \\dfrac{1}{\\sqrt{1+y^2}} + \\dfrac{y}{\\sqrt{1+y^2}} \\right)} \\\\ &= \\frac{y - \\sqrt{1+y^2} - 1}{\\sqrt{1+y^2} + 1 + y} \\\\ &= \\frac{(y - 1) - \\sqrt{1+y^2}}{(y + 1) + \\sqrt{1+y^2}} \\cdot \\frac{(y + 1) - \\sqrt{1+y^2}}{(y + 1) - \\sqrt{1+y^2}} \\\\ &= \\dfrac{2y^2 - 2y\\sqrt{1+y^2}}{2y} \\\\ &= y - \\sqrt{1+y^2} \\end{align} Since $$\\tan \\bigg( \\dfrac 12 \\theta - \\dfrac{\\pi}{4} \\bigg) = y - \\sqrt{1+y^2}$$, then $$\\dfrac 12 \\theta - \\dfrac{\\pi}{4} = \\arctan\\left(y - \\sqrt{1+y^2}\\right).$$ In other words $$\\dfrac 12 \\arctan(y) = \\arctan\\left(y - \\sqrt{1+y^2}\\right) + \\dfrac{\\pi}{4}.$$ Let $$\\arctan x-\\dfrac\\pi2=2y$$ Using Principal", "\\sin^4\\theta \\end{equation*} To express this as a polynomial in $$\\cos \\theta$$ only, we use the identity $$\\sin^2\\theta = 1- \\cos^2\\theta$$ to obtain \\begin{align*} \\cos5\\theta &= \\cos^5\\theta- 10\\cos^3\\theta \\sin^2\\theta+5 \\cos\\theta \\sin^4\\theta \\\\ &= \\cos^5\\theta- 10\\cos^3\\theta (1-\\cos^2\\theta)+5 \\cos\\theta (1-\\cos^2\\theta)^2 \\\\ &= \\cos^5\\theta -10\\cos^3\\theta+10\\cos^5\\theta +5\\cos\\theta(1-2\\cos^2\\theta+\\cos^4\\theta) \\\\ &= 16\\cos^5\\theta – 20 \\cos^3\\theta +5 \\cos\\theta \\end{align*} as required. ii) Note that $$\\text{Re}(e^{i\\frac{\\pi}{10}}) = \\text{R}e(\\cos\\frac{\\pi}{10}+i\\sin\\frac{\\pi}{10})=\\cos\\frac{\\pi}{10}$$. Let $$\\theta = \\frac{\\pi}{10}$$; then, using (i): \\begin{align*} 16\\cos^5{\\theta} – 20\\cos^3{\\theta} + 5\\cos{\\theta} &= \\cos\\left(\\frac{5\\pi}{10}\\right)=0. \\end{align*} If we now let $$x=\\cos{\\theta}$$, we can write: \\begin{align*} 16x^5 – 20x^3 + 5x &= 0 \\\\ 16x^4 – 20x^2 + 5 &= 0 \\end{align*} where we note that $$x\\neq0$$, making it valid to divide both sides by $$x$$. Treating this as a quadratic equation in $$x^2$$ we obtain \\begin{align*} x^2 &= \\frac{-(-20)\\pm\\sqrt{(20)^2 – 4\\times 16\\times 5}}{2\\times16} \\\\ &= \\frac{20\\pm\\sqrt{80}}{32} \\\\ &= \\frac{5\\pm\\sqrt{5}}{8} \\\\ x &= \\pm \\sqrt{\\frac{5\\pm\\sqrt{5}}{8}}. \\end{align*} Now, $$\\cos\\frac{\\pi}{10}>0$$ as it is in the first quadrant; so we have \\begin{align*} x &= \\sqrt{\\frac{5\\pm\\sqrt{5}}{8}} = 0.951 \\textrm{ or } 0.5877. \\end{align*} where 0.951 is taking $$+\\sqrt{5}$$ and 0.5877 is taking $$-\\sqrt{5}$$. For the inner sign, we note that $$0<\\frac{\\pi}{10}<\\frac{\\pi}{4}$$, and so \\begin{equation*} 0.707\\approx \\cos \\frac{\\pi}{4} < \\cos \\frac{\\pi}{10} < \\cos 0 = 1 \\end{equation*} and hence we must choose the inner $$+$$ sign to obtain $$x \\approx 0.951$$ since $$0.5877$$ is too", "\\sin \\left( \\theta + \\frac{\\alpha + \\beta}{2} \\right) \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence, $$15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2)$$ Taking $\\frac{(1) + (2)}{30}$, we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$. Thus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Taking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$. Hence, \\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*} Hence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$. In $\\triangle O O_1 B$, $$O_1 B = 2 r \\sin \\frac{\\alpha}{2} .$$ In $\\triangle O O_2 A$, by applying the law of sines, we get $$O_2 A = 2 r \\sin \\frac{\\beta}{2} .$$ Because circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$, $A$ is on circle $\\omega_2$, \\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*}" ]

[ "pair of complex numbers.", "of ordered pairs $$\\{(a,1), (b,3), (c,2)\\}.$$", "Find the number of ordered pairs of real numbers $$(a,b)$$ such that whenever $$\\gamma$$ is a root of $$x^2+ax+b=0$$, $$\\gamma^{2}-2$$ is also a root of the same equation.", "# Number of ordered pairs Find the number of ordered pairs of whole numbers $$(x,y)$$ such that $$(xy-7)^2 = x^2 +y^2$$. ×", "# Complex System of Equations Algebra Level 5 For how many ordered pairs $$(a,b)$$ of complex numbers does the system of equations $\\begin{cases} x^2y-x=a\\\\ xy^2+y=b \\end{cases}$ have exactly one complex number solution? Details and assumptions A complex number solution is an ordered pair of complex numbers $$(x,y)$$ that satisfies both equations. Clarification: $$a, b, x$$ and $$y$$ are all complex numbers. As an explicit example, $$(x,y) = (i, i)$$ is a solution to $$(a,b) = (-2i, 0 )$$. ×", "Five digit pair The ordered pair of four-digit numbers (2025; 3136) has the property that each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number. Suppose $$(a,b)$$ is the sole ordered pair of five-digit number with the same property. Find $$a+b$$. ×", "# Quiet complex -4 Algebra Level 5 Number of ordered pairs $$(a,b)$$ of real numbers such that $$(a + ib)^{2008}=(a - ib)$$ holds is: ×", "# $$3 + 4 = 7$$!! Find the number of pairs $$(x,y) \\in \\mathbb{Z}$$ such that $x^3+y^4=7$ For more problems try the sets My Problems and THRILLER. ×", "For certain ordered pairs $$(a,b)\\,$$ of real numbers, the system of equations $$ax+by=1\\,$$ $$x^2+y^2=50\\,$$ has at least one solution, and each solution is an ordered pair $$(x,y)\\,$$ of integers. How many such ordered pairs $$(a,b)\\,$$ are there? (第十二届AIME1994 第7题)", "Determine the number of ordered pairs $$(a,b)$$ of integers such that $$\\log_a b + 6\\log_b a=5, 2 \\leq a \\leq 2005,$$ and $$2 \\leq b \\leq 2005.$$ (第二十三届AIME2 2005 第5题)", "# Equality of Complex Numbers ## Theorem Let $z_1 := a_1 + i b_1$ and $z_2 := a_2 + i b_2$ be complex numbers. Then $z_1 = z_2$ if and only if $a_1 = a_2$ and $b_1 = b_2$. ## Proof By definition of a complex number, $z_1$ and $z_2$ can be expressed in the form: $z_1 = \\tuple {a_1, b_1}$ $z_2 = \\tuple {a_2, b_2}$ where $\\tuple {a, b}$ denotes an ordered pair. The result follows from Equality of Ordered Pairs.", "+0 0 148 1 The equation of the line joining the complex numbers $-5 + 4i$ and $7 + 2i$ can be expressed in the form $az + b \\overline{z} = 38$for some complex numbers $a$ and $b$. Find the ordered pair $(a, b)$. Oct 6, 2019", "# RMO 2011 Find the number of ordered pairs $$(x,y)$$ of real numbers such that $$\\large 16^{x^2 + y} + 16^{x+y^2} = 1$$. ×", "# The ordered pair (1, –3) is represented by which number in the complex plane? $1 - 3 i$. The ordered pair $\\left(x , y\\right)$ in the complex plane represents the complex no. $z = x + i y$. So, the given pair corresponds to the complex no. $1 - 3 i$.", "# Such A Long Number! Find the number of ordered pairs of digits $$(A,B)$$ such that $A3640548981270644B$ is divisible by 99. ×", "12$ and a pair of Complex zeros.", "# A number theory problem by Kushal Bose Find the number of ordered pairs of $(x,y)$, where $x,y \\in \\mathbb{N} \\leq 100$ and such that $\\large 5 \\mid 3^x+3^y$ ×", "+0 0 64 1 +95 Let $$(a_1,b_1), (a_2,b_2), \\dots, (a_n,b_n)$$ be all the ordered pairs $$(a,b)$$ of complex numbers with $$a^2+b^2\\neq 0, a+\\frac{10b}{a^2+b^2}=5, \\text{ and } b+\\frac{10a}{a^2+b^2}=4.$$ Find $$a_1 + b_1 + a_2 + b_2 + \\dots + a_n + b_n.$$ Sep 17, 2022", "Find the number of ordered quadruples of complex numbers $$(a,b,c,d)$$ such that $\\begin{cases} ab=b-c\\\\ bc=c-d\\\\ cd=d-a\\\\ ad=a-b \\end{cases}.$", "## anonymous 4 years ago Given the ordered pair (x, 2), what value of x would make the ordered pair a solution of the equation 5x − 3y = 19? A. 8 B. 24 C. 21 D. 5 1. anonymous d 2. anonymous replace y by 2 and get $5x-3\\times 2=19$ then solve for x" ]

[ "primitive eighth root of unity satisfying ${\\displaystyle \\omega _{3}'^{8}=\\left(e^{\\frac {2\\pi i}{2^{3}}}\\right)^{8}=1}$ (since ${\\displaystyle N=2^{3}=8}$). For short, setting ${\\displaystyle \\omega =\\omega _{3}'=\\omega _{8}}$, the matrix representation of this transformation on 3 qubits is: ${\\displaystyle F_{2^{3}}={\\frac {1}{\\sqrt {2^{3}}}}{\\begin{bmatrix}1&1&1&1&1&1&1&1\\\\1&\\omega &\\omega ^{2}&\\omega ^{3}&\\omega ^{4}&\\omega ^{5}&\\omega ^{6}&\\omega ^{7}\\\\1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}&\\omega ^{8}&\\omega ^{10}&\\omega ^{12}&\\omega ^{14}\\\\1&\\omega ^{3}&\\omega ^{6}&\\omega ^{9}&\\omega ^{12}&\\omega ^{15}&\\omega ^{18}&\\omega ^{21}\\\\1&\\omega ^{4}&\\omega ^{8}&\\omega ^{12}&\\omega ^{16}&\\omega ^{20}&\\omega ^{24}&\\omega ^{28}\\\\1&\\omega ^{5}&\\omega ^{10}&\\omega ^{15}&\\omega ^{20}&\\omega ^{25}&\\omega ^{30}&\\omega ^{35}\\\\1&\\omega ^{6}&\\omega ^{12}&\\omega ^{18}&\\omega ^{24}&\\omega ^{30}&\\omega ^{36}&\\omega ^{42}\\\\1&\\omega ^{7}&\\omega ^{14}&\\omega ^{21}&\\omega ^{28}&\\omega ^{35}&\\omega ^{42}&\\omega ^{49}\\\\\\end{bmatrix}}={\\frac {1}{\\sqrt {2^{3}}}}{\\begin{bmatrix}1&1&1&1&1&1&1&1\\\\1&\\omega &\\omega ^{2}&\\omega ^{3}&\\omega ^{4}&\\omega ^{5}&\\omega ^{6}&\\omega ^{7}\\\\1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}&1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}\\\\1&\\omega ^{3}&\\omega ^{6}&\\omega &\\omega ^{4}&\\omega ^{7}&\\omega ^{2}&\\omega ^{5}\\\\1&\\omega ^{4}&1&\\omega ^{4}&1&\\omega ^{4}&1&\\omega ^{4}\\\\1&\\omega ^{5}&\\omega ^{2}&\\omega ^{7}&\\omega ^{4}&\\omega &\\omega ^{6}&\\omega ^{3}\\\\1&\\omega ^{6}&\\omega ^{4}&\\omega ^{2}&1&\\omega ^{6}&\\omega ^{4}&\\omega ^{2}\\\\1&\\omega ^{7}&\\omega ^{6}&\\omega ^{5}&\\omega ^{4}&\\omega ^{3}&\\omega ^{2}&\\omega \\\\\\end{bmatrix}}.}$ It could be simplified further by using ${\\displaystyle \\omega ^{4}=-1}$, ${\\displaystyle \\omega ^{2}=i}$ and ${\\displaystyle \\omega ^{6}=-i}$ and then even more given that ${\\displaystyle \\omega ^{5}=-\\omega }$, ${\\displaystyle \\omega ^{3}=i\\omega }$ and ${\\displaystyle \\omega ^{7}=-i\\omega }$. The 3-qubit quantum Fourier transform can be rewritten as: ${\\displaystyle {\\text{QFT}}(|x_{1},x_{2},x_{3}\\rangle )={\\frac {1}{\\sqrt {2^{3}}}}\\ \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{2}x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{1}x_{2}x_{3}]}|1\\rangle \\right)}$ or ${\\displaystyle {\\text{QFT}}(|x_{1},x_{2},x_{3}\\rangle )={\\frac {1}{\\sqrt {2^{3}}}}\\ \\left(|0\\rangle +\\omega _{1}'^{[x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +\\omega _{2}'^{[x_{2}x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +\\omega _{3}'^{[x_{1}x_{2}x_{3}]}|1\\rangle \\right).}$ In case that we use the circuit", "& 0 & \\frac{11}{12}Mb^2 \\end{pmatrix}\\notag$ The identical roots $$I_{22} = I_{33} = \\frac{11}{12} Mb^2$$ imply that the principal axis associated with $$I_{11}$$ must be a symmetry axis. The orientation can be found by substituting $$I_{11}$$ into the above equation $(\\{\\mathbf{I}\\} − I \\{\\mathbb{I}\\}) \\cdot \\boldsymbol{\\omega} = \\frac{1}{12} Mb^2 \\begin{pmatrix} 6 & −3 & −3\\\\ −3& 6 & −3\\\\ −3 &−3 & 6 \\end{pmatrix} \\begin{pmatrix} \\omega_{11} \\\\ \\omega_{21} \\\\ \\omega_{31} \\end{pmatrix} = 0 \\notag$ where the second subscript 1 attached to $$\\omega_i$$ signifies that this solution corresponds to $$I_{11}$$. This gives \\begin{aligned} 2\\omega_{11} − \\omega_{21} − \\omega_{31} = 0 \\\\ −\\omega_{11} + 2\\omega_{21} − \\omega_{31} = 0 \\\\ −\\omega_{11} − \\omega_{21} + 2\\omega_{31} = 0 \\end{aligned} Solving these three equations gives the unit vector for the first principal axis for which $$I_{11} = \\frac{1}{6} Mb^2$$ to be $$\\mathbf{\\hat{e}}_1= \\frac{1}{\\sqrt{3}} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}$$. This can be repeated to find the other two principal axes by substituting $$I_{22} = \\frac{11}{12} Mb^2$$. This gives for the second principal moment $$I_{22}$$ $(\\{\\mathbf{I}\\} − I \\{\\mathbb{I}\\}) \\cdot \\boldsymbol{\\omega} = \\frac{1}{12} Mb^2 \\begin{pmatrix} −3 & −3 & −3\\\\ −3& −3& −3\\\\ −3 &−3 &−3 \\end{pmatrix} \\begin{pmatrix} \\omega_{12} \\\\ \\omega_{22} \\\\ \\omega_{32} \\end{pmatrix} = 0 \\notag$ This results in three identical equations for the components of $$\\omega$$ but all three equations are", "fifth root of unity. Now $$\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5=x$$ so \\begin{align} \\omega_5x&=\\omega_5(\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5)\\\\ &=\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5+\\omega_5^6\\\\ &=\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5+\\omega_5\\\\ &=x \\end{align} so $\\omega_5x=x$, and so $x\\neq 0$ would imply $\\omega_5=1$ which is false. So $x=0$. Now use $$e^{i\\theta}=\\cos(\\theta)+i\\sin(\\theta)$$ Now take $\\theta=\\frac{2k\\pi}5$ for $k=1,2,3,4,5$ to get $\\omega_5^k$. So we get: \\begin{align} 0&=\\Re(0)\\\\ &=\\Re(\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5)\\\\ &=\\Re(e^{2i\\pi/5}+e^{4i\\pi/5}+e^{6i\\pi/5}+e^{8i\\pi/5}+e^{10i\\pi/5})\\\\ &=\\cos(\\frac{2\\pi}5)+\\cos(\\frac{4\\pi}5)+\\cos(\\frac{6\\pi}5)+\\cos(\\frac{8\\pi}5)+\\cos(\\frac{10\\pi}5)\\\\ &=\\cos(\\frac{2\\pi}5)+\\cos(\\frac{4\\pi}5)+\\cos(\\frac{6\\pi}5)+\\cos(\\frac{8\\pi}5)+1 \\end{align} - I presume lines 3 and 4 are meant to be understood as equivalent since \\omega_5^6\\ = \\omega_5\\ by periodic formula. – CandiedOrange Feb 18 at 10:01 @CandiedOrange, yes, that's the step I took. We use $\\omega_5^5=1$ – vrugtehagel Feb 18 at 10:05 Let's try this - Using $$2\\cos a \\sin b = \\sin(a+b) - \\sin(a- b)$$ we get: $$2\\cos(2π/5)\\sin(2π/5)=\\sin(4π/5 )- \\sin(0)$$ $$2\\cos(4π/5)\\sin(2π/5)=\\sin(6π/5)-\\sin(2π/5)$$ $$2\\cos(6π/5)\\sin(2π/5)=\\sin(8π/5) - \\sin(4π/5)$$ $$2\\cos(8π/5)\\sin(2π/5)=\\sin(10π/5) -\\sin(6π/5)$$2\\cos(10π/5)\\sin(2π/5)=\\sin(12π/5) - \\sin(8π/5) Adding: \\begin{align} &2\\sin(2π/5)\\{\\cos(10π/5)+\\cos(2π/5)+\\cos(4π/5)+\\cos(6π/5)+\\cos(8π/5)\\}\\\\ &=2\\sin(2π/5)\\{1+\\cos(2π/5)+\\cos(4π/5)+\\cos(6π/5)+\\cos(8π/5)\\} \\\\ &=\\sin(12π/5) - \\sin(2π/5)\\\\ &=2\\sin(10π/5)\\cos(14π/5)\\\\&=0\\qquad [\\;\\because \\sin a - \\sin b = 2\\sin(a-b)\\cos(a+b)] \\end{align}1+\\cos(2π/5)+\\cos(4π/5)+ \\cos(6π/5)+\\cos(8π/5)=0$$- I used it by grouping the last term with the second and the onea in the middle. I disnt manage to do something – Razvan Paraschiv Feb 17 at 16:28 @RazvanParaschiv Check my method. I edited my answer. – Win Vineeth Feb 18 at 0:02 Draw a regular pentagon ABCDE with all sides one unit long. Treat these sides as vectors from A to B, then B to C, etc. From the \"head to tail\" rule for resultants the resultant of", "only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \\frac{d - b}{2} = 51$. Therefore, the perimeter of $\\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \\boxed{\\textbf{(459) }}$. ~Steven Chen (www.professorchenedu.com) ## Solution (Number Theory Part) We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \\cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for. Notice that $r^2+r+1=(r-\\omega)(r-{\\omega}^2)$, where $\\omega=e^{i\\frac{2\\pi}{3}}$. Thus, $$x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\\frac{x}{y}-\\omega)(\\frac{x}{y}-{\\omega}^2) = (x-y\\omega)(x-y{\\omega}^2).$$ Note that $8^2+1^2+8 \\cdot 1=73$. Thus, $(8-\\omega)(8-{\\omega}^2)=73$. Squaring both sides yields \\begin{align} (8-\\omega)^2(8-{\\omega}^2)^2&=73^2\\\\ (63-17\\omega)(63-17{\\omega}^2)&=73^2. \\end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=\\boxed{459}$. ~ Leo.Euler ## Solution(Visual geometry) We look at upper and middle diagrams and get $\\angle BAC = 120^\\circ$. Next we use only the lower Diagram. Let $I$ be incenter $\\triangle ABC$, E be midpoint of biggest arc $\\overset{\\Large\\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\\triangle ABC$ at point $E$. Quadrilateral $ABEC$ is cyclic, so $$\\angle BEC = 180^\\circ - \\angle ABC = 60^\\circ \\implies BE = CE = IE = BC.$$", "out yo find $$215^2$$. \\begin{align} 215^2 &= (21 \\times (21 + 1)) \\omega 25\\\\ &= (21 \\times 22) \\omega 25\\\\ &= 462\\omega25\\\\ &= 46225\\tag{Answer} \\end{align} This always seemed to work out very well. The problem was, can I trust this trick? Will this always hold true? I didn’t have answers to those questions, until I proved it myself using Digit Math. Good news is that this trick will always hold true. # The proof Let the number be $$x = a\\omega 5$$. $$a$$ can have any number of digits. ### Case 1: $$a$$ has exactly one digit. So, $$]a[ = ]5[ = 1$$. \\begin{align} \\therefore (a\\omega 5)^2 &= aa\\omega (5a+5a) \\omega 5.5\\tag{Using Bimultiplication formula}\\\\ &= a^2 \\omega 10a \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega0) \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega2) \\omega 5\\\\ &= (a^2 + a) \\omega 2 \\omega 5\\\\ &= \\big(a(a+1)\\big) \\omega 25\\tag{Proved} \\end{align} ### Case 2: $$a$$ has more than one digits. So, $$]a[ > (]5[ = 1)$$. But to apply Bimultiplication formula $$a$$ must have the same number of digits in $$5$$, which is obviously not the case here. So, we will use one trick. We will pad $$5$$ with some number of zeroes on the right, so that, $$]a[ = ]5\\omega c[$$, where, $$c$$ is all zeroes and $$]c[ = ]a[ – 1$$. So,", "for other digit groups which are at higher place order. The reason is simple, they may have to be added with carry-overs from the groups (terms) on the right. ## A corollary of the General Formula of Transformation (for two digits) ### With to Plus form $$a\\omega b = 10a+b$$ The above equation is true only when $$b$$ has one digit. The more general equation is:- $$\\boxed{a\\omega b = a.10^d+b}\\hspace1em\\text{where d}=]b[\\hspace1em\\ldots\\text{Corollary 1}$$ ### Plus to With form $$a\\omega b = 10a +b \\Rightarrow a\\omega b = 9a + a + b \\Rightarrow \\boxed{a+b = a\\omega b -9a}\\hspace1em\\ldots\\text{Corollary 2}$$ ### Negative digits We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this. \\begin{align} a\\omega(-b) &= 10a + (-b)\\\\ &= 10a -b\\\\ &= 10a + b -2b\\\\ &= a\\omega b -2b \\end{align} Altire, \\begin{align} a\\omega(-b) &= -(-10a + b)\\\\ &= -(10a+b -20a)\\\\ &= -(a\\omega b – 20a)\\\\ &= -(a\\omega b) +20a \\end{align} $$\\therefore \\boxed{a\\omega(-b) = a\\omega b-2b = -(a\\omega b)+20a}$$ Remember, $$(-a)\\omega b \\neq -(a\\omega b)$$. The former means that only the digit $$a$$ is negative. The later means that the complete number is negative. Similarly, $$\\boxed{(-a)\\omega b = a\\omega b-20a = -(a\\omega b)+2b}$$", "= 2$ and $(a,b) = (0.5,3.0625)\\in)A,B($ The solutions $u$ and $v$ separately for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.9511,1.2)\\in)A,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.1,1.2976)\\in)I,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.5,1.1371)\\in)I,B($ The solutions $u$ and $v$ separately for $(a,b) = (0.1,1.2976)\\in)I,B($ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (0.2,0.4)$ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (2,4)$ The chaotic behavior for $p = 2.5$, $w = 3$ and $(a,b) = (0.2,0.14)$ The chaotic behavior for $p = 2.5$, $w = 2$ and $(a,b) = (2,4)$ The partition of the plane according to $\\Gamma_1$, $\\Gamma_2$ and $\\Lambda$ for $w = 0$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.5,1)\\in R_1$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (2.5,3.5)\\in R_1$ The portrait $(u',u)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The portrait $(v',v)$ for $p = 1.5$, $w = 0$ and $(a,b) =", "(terms) on the right. ## A corollary of the General Formula of Transformation (for two digits) ### With to Plus form $$a\\omega b = 10a+b$$ The above equation is true only when $$b$$ has one digit. The more general equation is:- $$\\boxed{a\\omega b = a.10^d+b}\\hspace1em\\text{where d}=]b[\\hspace1em\\ldots\\text{Corollary 1}$$ ### Plus to With form $$a\\omega b = 10a +b \\Rightarrow a\\omega b = 9a + a + b \\Rightarrow \\boxed{a+b = a\\omega b -9a}\\hspace1em\\ldots\\text{Corollary 2}$$ ### Negative digits We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this. \\begin{align} a\\omega(-b) &= 10a + (-b)\\\\ &= 10a -b\\\\ &= 10a + b -2b\\\\ &= a\\omega b -2b \\end{align} Altire, \\begin{align} a\\omega(-b) &= -(-10a + b)\\\\ &= -(10a+b -20a)\\\\ &= -(a\\omega b – 20a)\\\\ &= -(a\\omega b) +20a \\end{align} $$\\therefore \\boxed{a\\omega(-b) = a\\omega b-2b = -(a\\omega b)+20a}$$ Remember, $$(-a)\\omega b \\neq -(a\\omega b)$$. The former means that only the digit $$a$$ is negative. The later means that the complete number is negative. Similarly, $$\\boxed{(-a)\\omega b = a\\omega b-20a = -(a\\omega b)+2b}$$ Also similarly evaluating, we get, $$\\boxed{(-a)\\omega(-b) = -(a\\omega b)}\\hspace1em\\ldots\\text{Corollary 3}$$ So, we see that in a negative number all the digits too are negative. Not", "$H(s) = Y(s)/X(s)$, of a linear time invariant (LTI) system. – Robert L. Sep 5 '18 at 22:05 • Hi. Thanks for clearing that up. There's a lot to learn indeed. – antwoord Sep 8 '18 at 12:13 Here is a way that you can approach this mathematically. It's not the only way. Another way to approach this is with Tchebyshev polynomials and I will leave that to someone else. This approach requires understanding two ingredients. 1. an identity that is a consequence of Euler's formula: $$\\cos(\\omega t) = \\tfrac12(e^{j\\omega t}+e^{-j\\omega t})$$ or $$\\sin(\\omega t) = \\tfrac1{2j}(e^{j\\omega t}-e^{-j\\omega t})$$ 1. the binomial theorem: \\begin{align} (a+b)^N &= \\sum_{n=0}^N {N \\choose n} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N {N \\choose n} \\, a^n \\, b^{N-n} \\\\ \\end{align} or \\begin{align} (a+b)^N &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^n \\, b^{N-n} \\\\ \\end{align} Now you might see where this is going. So using the cosine identity (so i don't have to mess around with powers of $j$ in the sine identity): \\begin{align} \\cos^k(\\omega t) &= \\frac{1}{2^k}(e^{j\\omega t}+e^{-j\\omega t})^k \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, (e^{j\\omega t})^{k-n} \\, (e^{-j\\omega t})^n \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, e^{j(k-n)\\omega t} \\, e^{-jn\\omega t} \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k", "n_e/n_e$$ becomes7,37: \\begin{aligned} \\partial ^2 b /\\partial t^2 = -\\Gamma _{sc}^2 a_0 a_1, \\end{aligned} (5) with $$\\Gamma _{sc}^3 = (v_g/c)^2 \\omega _{pi}^2 \\omega _0/(2g) = \\omega _0^3 (Zm_e/m_i) (n_0/n_{cr}) (1-n_0/n_{cr})/(2g) = 2\\omega _s \\Gamma _B^2$$, $$\\alpha _{sc} = \\omega _{pe}^2/(4\\omega _0 \\Gamma _{sc})$$. For $$a_1(t) > a_{00}$$, and in similar fashion to our treatment of Raman amplification, we consider a quasi-$$\\pi$$-pulse “attractor” solution for (1) and (5) that scales as $$||a_1||(t) = A (a_{00}^2 \\Gamma _{sc} t)^{3/4}$$ and $$\\Gamma _{sc} \\tau _1(t) = \\Delta \\xi /(a_{00}^2 \\Gamma _{sc} t)^{1/2}$$, with $$\\Delta \\xi \\approx 3.3$$ and $$A \\approx 0.62$$7,27,38. Again using $$\\tau _0 = 2t$$, we find the following non-linear matching conditions27: \\begin{aligned} \\Gamma _{sc}^3 a_{00}^2 \\tau _0 \\tau _1^2&= 2(\\Delta \\xi )^2 \\approx 22, \\end{aligned} (6) \\begin{aligned} \\Gamma _{sc} ||a_1||^{2/3} \\tau _1&= A^{2/3} \\Delta \\xi \\approx (13.8)^{1/3} \\approx 2.40. \\end{aligned} (7) The asymptotic efficiency for this case is $$\\eta = a_1^2(t)\\tau _1(t)/(2a_{00}^2 t) = A^2 \\Delta \\xi /2 \\approx 0.63$$. The role of (6) and (7) matches that of (3) and (4) for Raman amplification. Like (3) and (4), (6) and (7) require $$a_1(0) > a_{00}$$27. While the notion that $$a_1^2 \\Gamma _{sc}^3 \\tau _1^3 = \\text {const.}$$ for a Brillouin-amplified seed pulse (or $$a_1 \\Gamma _R \\tau _1 = \\text {const.}$$ for Raman) was advanced in our", "- 2abcd = 0$ w and $${\\omega ^2}$$ are roots of this equation. Let a and b be the other two roots. \\begin{align} \\Rightarrow \\omega + {\\omega ^2} + \\alpha + \\beta = - \\frac{{(a + b + c + d)}}{2}\\end{align} Since $$\\omega + {\\omega ^2} = - 1,$$ this reduces to $\\alpha + \\beta = 1 - \\frac{{(a + b + c + d)}}{2} \\cdots (1)$ Also, the coefficient of $${x^2}$$ is 0 $\\begin{array}{l} \\Rightarrow \\alpha \\beta + \\alpha \\omega + \\alpha {\\omega ^2} + \\beta \\omega + \\beta {\\omega ^2} + {\\omega ^3} = 0\\\\ \\Rightarrow \\alpha \\beta - \\alpha - \\beta + 1 = 0\\\\ \\Rightarrow (\\alpha - 1)(\\beta - 1) = 0\\\\ \\Rightarrow \\alpha = 1 {\\rm{or}} \\beta = 1\\end{array}$ If $$\\alpha = 1$$ , b can be determined from (1) and vice-versa Thus, x = 1 is a root of the given equation Example- 12 If $$|{a_n}| < 2$$ for $$1 \\le n \\le k,$$ show that the equation $1 + {a_1}z + {a_2}{z^2} + ...... + {a_k}{z^k} = 0$ cannot be satisfied by any z whose modulus is less than $$\\frac{1}{3}$$ Solution: Assume that $$|z|\\, < \\frac{1}{3}.$$ We will use this assumption to arrive at a contradiction, proving the assumption wrong. \\begin{align}&p = {a_1}z + {a_2}{z^2} + .... + {a_k}{z^k}\\\\\\\\\\Rightarrow\\quad &|p|\\, =", "Hand Side (LHS) = ${{a}^{2}}+{{\\omega }^{3}}{{b}^{2}}+{{\\omega }^{3}}{{c}^{2}}+{{\\omega }^{2}}ab+\\omega ab+{{\\omega }^{2}}bc+{{\\omega }^{4}}bc+{{\\omega }^{2}}ac+\\omega ac$ Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{\\omega }^{3}}{{b}^{2}}+{{\\omega }^{3}}{{c}^{2}}+{{\\omega }^{2}}ab+\\omega ab+{{\\omega }^{2}}bc+{{\\omega }^{3+1}}bc+{{\\omega }^{2}}ac+\\omega ac$ As we know that we can write ${{a}^{m+n}}$ as ${{a}^{m}}{{a}^{n}}$ therefore in the above equation we can replace ${{\\omega }^{3+1}}$ by ${{\\omega }^{3}}\\omega$ therefore we will get, Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{\\omega }^{3}}{{b}^{2}}+{{\\omega }^{3}}{{c}^{2}}+{{\\omega }^{2}}ab+\\omega ab+{{\\omega }^{2}}bc+{{\\omega }^{3}}\\omega bc+{{\\omega }^{2}}ac+\\omega ac$ As we know that in the given equation $1,\\omega ,{{\\omega }^{2}}$ are the cube roots of unity and to proceed further in the solution we should know the property of cube roots of unity given below, Property: ${{\\omega }^{3}}=1$ If we use the above property in LHS we will get, Therefore, Left Hand Side (LHS) = ${{a}^{2}}+1\\times {{b}^{2}}+1\\times {{c}^{2}}+{{\\omega }^{2}}ab+\\omega ab+{{\\omega }^{2}}bc+1\\times \\omega bc+{{\\omega }^{2}}ac+\\omega ac$ Further simplification in the above equation will give, Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{\\omega }^{2}}ab+\\omega ab+{{\\omega }^{2}}bc+\\omega bc+{{\\omega }^{2}}ac+\\omega ac$ If we take ‘ab’ common from the above equation we will get, Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\\left( {{\\omega }^{2}}+\\omega \\right)ab+{{\\omega }^{2}}bc+\\omega bc+{{\\omega }^{2}}ac+\\omega ac$ Similarly if we take ‘bc’ and ‘ac’ common from the above equation we will get, Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\\left( {{\\omega }^{2}}+\\omega \\right)ab+\\left( {{\\omega }^{2}}+\\omega \\right)bc+\\left( {{\\omega }^{2}}+\\omega \\right)ac$ Now before we", "(twelve)» can also occur as a result of the following six combinations: $$(6,5,1),(6,4,2),(6,3,3),(5,5,2),(5,4,3),(4,4,4) \\text {. }$$ Consequently, these events are equally probable. Here, the mistake of de Mere is that the possible outcomes that he considered are not equally probable. For example, the event $(6,4,1)$ can occur in $3 !=6$ cases: $(6,4,1),(6,1,4), \\ldots$, $(1,4,6)$. At the same time, for example, a combination $(4,4,4)$ can occur only in one case. In modern language, de Mere incorrectly constructed the space of elementary events corresponding to the given problem. The solution of the problem. We define $\\Omega$ as $$\\Omega={(i, j, k): i, j, k=\\overline{1,6}}=\\Omega_{0} \\times \\Omega_{0} \\times \\Omega_{0},$$ where $\\Omega_{0}={1,2,3,4,5,6}$. Let’s introduce the events: $A_{11}={$ the sum of points is equal to 11$}, A_{12}={$ the sum of points is equal to 12$}$. Hence \\begin{aligned} &A_{11}={(i, j, k) \\in \\Omega: i+j+k=11}, \\ &A_{12}={(i, j, k) \\in \\Omega: i+j+k=12}, \\end{aligned} We have $$\\left|A_{11}\\right|=27, \\quad\\left|A_{12}\\right|=25, \\quad|\\Omega|=6^{3}=216,$$ Therefore $$P\\left(A_{11}\\right)=\\frac{27}{216}>\\frac{25}{216}=P\\left(A_{12}\\right) .$$ 1. From the general population $\\Omega_{0}=\\left{a_{1}, a_{2}, \\ldots, a_{n}\\right}$ (for example, from an urn with numbered balls) of size $n$, a random sample with replacement of size $r$ is extracted. a) Find the probability that the extracted sample is a sample without replacement (that is, all the extracted balls have different numbers). b) Find the probability that the first sample element is the first element", "(7) yields (bc-1)(c-b)=0, but we have seen that c\\ne b, so we must have cb=1 or equivalently$$c =\\frac{1}{b}=\\overline{b}\\,.\\tag{9}$$Replacing back in (7) we get b^2+c+1+b+c^2=0 or equivalently$$b^4+b^3+b^2+b+1=0\\,.$$Thus, b is a fifth root of 1 different from 1, that is$$b\\in\\{\\omega,\\omega^2,\\omega^3,\\omega^4\\}\\quad\\hbox{ where\\omega=\\exp \\left(\\frac{2\\pi i}{5}\\right)\\,.}$$It follows that c=b^{-1}=b^4 and c^2=b^8=b^3. From (5) we conclude that$$\\eqalign{ a&\\,=\\,\\frac{b+b^3+2b^4+1}{b-b^4}\\,=\\,\\frac{b^4\\!-b^2}{b-b^4}\\,=\\,b^3\\,,\\cr d&\\,=\\,b^2\\,,}$$where we used the identities 1+b+b^2+ b^3+b^4=0 and b^5=1. We get the solution (z_0,z_1,z_2,z_3,z_4) with$$ z_0=b^3z_4\\,,~z_1=b z_4\\,,~z_2=b^4 z_4\\,,~z_3=b^2 z_4\\,,$which are the vertices of a regular pentagon in some order, (A regular pentagon, or a regular pentagonal star). • Nice solution, I wonder if it can be simplified. The \"which is equivalent to\" between (7) and (8) seems to be pure luck. – Ewan Delanoy May 22 '14 at 16:35 • @Ewan Delanoy. It is not pure luck: the \"which is equivalent to\" between$(7)$and$(8)$is$(b\\leftrightarrow c)$-symmetric to the \"which is equivalent to\" between$(6)$and$(7)\\$. – chizhek Aug 27 '14 at 16:59", "$$J_2(1.38) = f_3^2(6)$$ $$J_2(1.39) = f_3^2(7)$$ $$J_2(1.405) = f_3^3(4)$$ $$J_2(1.705) = f_4^3(5)$$ $$J_2(2.305) = f_6^3(7)$$ $$J_2(2.355) = f_6^4(7)$$ $$J_2(2.405) = f_6^5(7)$$ $$J_2(2.455) = f_6^6(7)$$ $$J_2(2.5) = f_{\\omega}(7)$$ $$J_2(2.505) = f_{\\omega}^2(6)$$ $$J_2(2.51) = f_{\\omega}^4(6)$$ $$J_2(2.52) = f_{\\omega+1}(6)$$ $$J_2(2.53) = f_{\\omega+1}^5(6)$$ $$J_2(2.54) = f_{\\omega+2}^2(5)$$ $$J_2(2.55) = f_{\\omega+2}^5(7)$$ $$J_2(2.56) = f_{\\omega+3}^3(4)$$ $$J_2(2.57) = f_{\\omega+3}^6(7)$$ $$J_2(2.58) = f_{\\omega+4}^3(7)$$ $$J_2(2.6) = f_{\\omega+5}^4(7)$$ $$J_2(2.65) = f_{\\omega.2+1}^3(4)$$ $$J_2(2.7) = f_{\\omega.2+4}^2(5)$$ $$J_2(2.72) = f_{\\omega.2+5}^2(7)$$ $$J_2(2.725) = f_{\\omega.2+5}^4(7)$$ $$J_2(2.75) = f_{\\omega.2+6}^6(7)$$ $$J_2(2.8) = f_{\\omega.3+2}^5(7)$$ $$J_2(2.9) = f_{\\omega.4+1}^3(6)$$ $$J_2(3) = f_{\\omega.4+6}^6(7)$$ $$J_2(3.1) = f_{\\omega.5+5}^4(7)$$ $$J_2(3.2) = f_{\\omega.6+4}^2(7)$$ $$J_2(3.21) = f_{\\omega.6+4}^5(7)$$ $$J_2(3.22) = f_{\\omega.6+5}^2(7)$$ $$J_2(3.23) = f_{\\omega.6+5}^6(7)$$ $$J_2(3.24) = f_{\\omega.6+6}^3(7)$$ $$J_2(3.245) = f_{\\omega.6+6}^5(7)$$ $$J_2(3.25) = f_{\\omega^2}(2)$$ $$J_2(3.252) = f_{\\omega^2}(3)$$ $$J_2(3.255) = f_{\\omega^2}(4)$$ $$J_2(3.26) = f_{\\omega^2}^2(2)$$ $$J_2(3.27) = f_{\\omega^2}^3(4)$$ $$J_2(3.28) = f_{\\omega^2}^4(5)$$ $$J_2(3.29) = f_{\\omega^2}^5(6)$$ $$J_2(3.3) = f_{\\omega^2}^6(7)$$ $$J_2(3.32) = f_{\\omega^2+1}^2(6)$$ $$J_2(3.34) = f_{\\omega^2+1}^5(6)$$ $$J_2(3.36) = f_{\\omega^2+2}(3)$$ $$J_2(3.38) = f_{\\omega^2+2}(6)$$ $$J_2(3.4) = f_{\\omega^2+2}^2(5)$$ $$J_2(3.45) = f_{\\omega^2+3}^5(6)$$ $$J_2(3.5) = f_{\\omega^2+4}^4(6)$$ $$J_2(3.55) = f_{\\omega^2+5}^4(6)$$ $$J_2(3.6) = f_{\\omega^2+6}^4(7)$$ $$J_2(3.65) = f_{\\omega^2.2}^2(3)$$ $$J_2(3.7) = f_{\\omega^2.2}^4(6)$$ $$J_2(3.75) = f_{\\omega^2.2}^6(7)$$ $$J_2(3.8) = f_{\\omega^2.2+1}^3(4)$$ $$J_2(3.85) = f_{\\omega^2.2+1}^5(6)$$ $$J_2(3.9) = f_{\\omega^2.2+2}(4)$$ $$J_2(3.95) = f_{\\omega^2.2+2}^2(3)$$ $$J_2(4) = f_{\\omega^2.2+2}^3(3)$$ Work in progress", "A)\\mathsf{P}(M). \\end{align} 3. Taking $M=\\Omega$ and using the equality between the first and the third terms, \\begin{align} \\mathsf{P}((B_{t_1}^{*},\\ldots,B_{t_k}^{*})\\in A)&=\\mathsf{P}(\\{(B_{t_1}^{*},\\ldots,B_{t_k}^{*})\\in A\\}\\cap \\Omega) \\\\ &=\\mathsf{P}((B_{t_1},\\ldots,B_{t_k})\\in A). \\end{align} • 1. Okay so each member of the union is a measurable set of $\\Omega \\times T$? 2. I'm still a little confused about this. Lets pick two elements $a$ and $b$ in $V$. Then we will have $B^*_{t_i}=B_{t_i+a} - B_{a}$. Also there's another $b$: $B'^*_{t_i}=B_{t_i+b} - B_{b}$. Are the $t_i$'s chosen in $B^*$ are different from $B'^*$? 3. This calculation is much clearer to me except for second to last line - I have a question but it depends on the answer 2. 4. Ah - okay I see what's happening. Jul 10 '18 at 3:40 • 1. If $(\\Omega,\\mathcal{F},\\mathsf{P})$ is the underlying prob. space, then $\\{\\tau=t_0\\}\\in\\mathcal{F}$ and $\\{B_t'\\in A\\}\\in \\mathcal{F}$ and so $\\{\\tau=t_0\\}\\cap\\{B_t'\\in A\\}\\in \\mathcal{F}$. Jul 10 '18 at 3:49 • 2. Look at these notes Jul 10 '18 at 4:21 • okay, BM is translation invariant. 3. So for the calculation, in the second to last line the term $\\mathsf{P}((B_{t_1},\\ldots,B_{t_k})\\in A)$ pulls out because of translation invariance then right? Technically, for each $t_0 \\in V$, $(B_{t_1},\\ldots,B_{t_k})$ represent different time points Jul 10 '18 at 12:37 • $(B_{t_1}',\\ldots,B_{t_k}')$ and $(B_{t_1},\\ldots,B_{t_k})$ have the same distribution and the latter does not depend on", "is the Golden Ratio. - I love solutions invoking the golden ratio. It'd be even better with a pentagram. (+1) – Ron Gordon Feb 11 '13 at 9:05 We have $\\cos 36 = \\frac{\\omega + \\omega^{-1}}{2}$ where $\\omega = \\text{exp} \\left ( \\frac{2 \\pi}{10} \\right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\\omega^4 - \\omega^3 + \\omega^2 - \\omega + 1 = 0$. Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \\frac{1}{x} + \\frac{1}{x^2}) = x^2\\left ( \\left (x + \\frac{1}{x} \\right )^2 - \\left (x + \\frac{1}{x} \\right ) -1 \\right )$. Thus $0 = \\omega^2 \\cdot ((2 \\cos 36)^2 - (2 \\cos 36) - 1)$. Therefore $4 \\cos^3 36 - 2 \\cos 36 - 1 = 0$. Using the quadratic formula we then arrive at $$\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$$", "$$\\omega$$ equal in number of digits. So, \\begin{align} &x\\omega (5 \\times 10^{n-1}) \\times 0 \\omega (2\\times 10^{n-1})\\\\ &= 0 \\omega (2x\\times 10^{n-1})\\omega (10\\times 10^{n-1})\\\\ &= (2x\\times 10^{n-1})\\omega 10^n\\\\ &= (2x\\times 10^{n-1} + 1) \\omega ]0[^n\\tag{Moving extra 1 to left} \\end{align} In the above equation it is easy to see that the result ends with $$10$$ after we divide the $$10^{n-1}$$. Therefore, this is not going to leave any remainder. (Proved) Case 3: $$\\frac 5 5 = 1$$. (Proved) Yeah, this is a trivial case, but for the sake of completeness. Check out other applications of Digit Math Link to list of other applications of Digit Math. The problem Someday, somewhere I came to know that any number which ends with the digit 5 can be easily squared. The trick can be easily demonstrate using an example. Suppose we want to find the square of 25. Trick is to take the number before 5 (which will be 2 here), add one to it (2 + 1 = 3) and then multiply them together (2 x 3 = 6). Now the final answer would be the product followed by the number 25, i.e. 625 in this case. Now let’s try it out yo find $$215^2$$. \\begin{align} 215^2 &= (21 \\times (21 + 1)) \\omega 25\\\\ &= (21 \\times 22) \\omega", "# Fourier coordinates If the given directed graph with boundary is rotation invariant then its hitting matrix is diagonal in Fourier coordinates. Let $\\omega = e^{2 \\pi i/N}$ be the N'th root of unity or any other root of unity not equal to 1. $\\omega^N = 1,\\quad \\omega \\ne 1$. We consider the following symmetric Vandermonde matrix: $\\mathbf{F}_N = \\begin{bmatrix} 1 & 1 & 1 & \\ldots & 1 \\\\ 1 & \\omega & \\omega^2 & \\ldots & \\omega^{(N-1)} \\\\ 1 & \\omega^2 & \\vdots & \\ldots & \\omega^{2(N-1)} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 1 & \\omega^{(N-1) } & \\omega^{2(N-1)} & \\ldots & \\omega^{(N-1)^2} \\\\ \\end{bmatrix}$ For example, $\\mathbf{F}_5 = \\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\\\ 1 & \\omega & \\omega^2 & \\omega^3 & \\omega^4 \\\\ 1 & \\omega^2 & \\omega^4 & \\omega^6 & \\omega^8 \\\\ 1 & \\omega^3 & \\omega^6 & \\omega^9 & \\omega^{12} \\\\ 1 & \\omega^4 & \\omega^8 & \\omega^{12} & \\omega^{16} \\\\ \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\\\ 1 & \\omega & \\omega^2 & \\omega^3 & \\omega^4 \\\\ 1 & \\omega^2 & \\omega^4 & \\omega & \\omega^3 \\\\ 1 & \\omega^3 & \\omega & \\omega^4 & \\omega^2 \\\\ 1 & \\omega^4 & \\omega^3 & \\omega^2 &", "& \\omega^{n-1} & \\omega^{2(n-1)} & ... & \\omega^{(n-1)(n-1)} \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 1 &1 \\\\ 1 &\\omega & \\omega^2 & \\omega^3\\\\ 1 &\\omega^2 & \\omega^4 & \\omega^6\\\\ 1 &\\omega^3 & \\omega^6 & \\omega^9 \\end{bmatrix}$$ Given the $$\\omega_4 = 4^{th}$$ roots of unity, we have the ordered set equality: $$\\{\\omega^{0},\\omega^1, \\omega^2, \\omega^3, \\omega^4, \\omega^5, ... \\} = \\{1, i, -1, -i, 1, i, ...\\}$$ This can be visualized as iterating thru roots of the unit circle in the counter-clockwise direction. Also, notice the mod n nature, i.e. $$\\omega^6 = \\omega^{6 \\mod n} = \\omega^{2} = -1$$ and $$-i = \\omega^{3} = \\omega^{3+n}$$ To complete step 1 (evaluation) we find $$A', B'$$ by performing $$A' = M * \\vec{a} = \\begin{bmatrix} 1 & 1 & 1 &1 \\\\ 1 &\\omega & \\omega^2 & \\omega^3\\\\ 1 &\\omega^2 & \\omega^4 & \\omega^6\\\\ 1 &\\omega^3 & \\omega^6 & \\omega^9 \\end{bmatrix} \\begin{bmatrix} 3 \\\\ 1 \\\\ 0 \\\\ 0 \\end{bmatrix} = \\begin{bmatrix} 3 + 1 \\\\ 3 + 1 \\omega \\\\ 3 + \\omega^2 \\\\ 3 + \\omega^3 \\end{bmatrix} = \\begin{bmatrix} 4 \\\\ 3 + i \\\\ 2 \\\\ 3 - i \\end{bmatrix} \\\\ B' = M * \\vec{b} = \\begin{bmatrix} 1 & 1 & 1 &1 \\\\ 1 &\\omega & \\omega^2 & \\omega^3\\\\ 1 &\\omega^2 & \\omega^4 & \\omega^6\\\\ 1" ]

[ "= 2$ and $(a,b) = (0.5,3.0625)\\in)A,B($ The solutions $u$ and $v$ separately for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.9511,1.2)\\in)A,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.1,1.2976)\\in)I,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.5,1.1371)\\in)I,B($ The solutions $u$ and $v$ separately for $(a,b) = (0.1,1.2976)\\in)I,B($ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (0.2,0.4)$ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (2,4)$ The chaotic behavior for $p = 2.5$, $w = 3$ and $(a,b) = (0.2,0.14)$ The chaotic behavior for $p = 2.5$, $w = 2$ and $(a,b) = (2,4)$ The partition of the plane according to $\\Gamma_1$, $\\Gamma_2$ and $\\Lambda$ for $w = 0$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.5,1)\\in R_1$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (2.5,3.5)\\in R_1$ The portrait $(u',u)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The portrait $(v',v)$ for $p = 1.5$, $w = 0$ and $(a,b) =", "primitive eighth root of unity satisfying ${\\displaystyle \\omega _{3}'^{8}=\\left(e^{\\frac {2\\pi i}{2^{3}}}\\right)^{8}=1}$ (since ${\\displaystyle N=2^{3}=8}$). For short, setting ${\\displaystyle \\omega =\\omega _{3}'=\\omega _{8}}$, the matrix representation of this transformation on 3 qubits is: ${\\displaystyle F_{2^{3}}={\\frac {1}{\\sqrt {2^{3}}}}{\\begin{bmatrix}1&1&1&1&1&1&1&1\\\\1&\\omega &\\omega ^{2}&\\omega ^{3}&\\omega ^{4}&\\omega ^{5}&\\omega ^{6}&\\omega ^{7}\\\\1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}&\\omega ^{8}&\\omega ^{10}&\\omega ^{12}&\\omega ^{14}\\\\1&\\omega ^{3}&\\omega ^{6}&\\omega ^{9}&\\omega ^{12}&\\omega ^{15}&\\omega ^{18}&\\omega ^{21}\\\\1&\\omega ^{4}&\\omega ^{8}&\\omega ^{12}&\\omega ^{16}&\\omega ^{20}&\\omega ^{24}&\\omega ^{28}\\\\1&\\omega ^{5}&\\omega ^{10}&\\omega ^{15}&\\omega ^{20}&\\omega ^{25}&\\omega ^{30}&\\omega ^{35}\\\\1&\\omega ^{6}&\\omega ^{12}&\\omega ^{18}&\\omega ^{24}&\\omega ^{30}&\\omega ^{36}&\\omega ^{42}\\\\1&\\omega ^{7}&\\omega ^{14}&\\omega ^{21}&\\omega ^{28}&\\omega ^{35}&\\omega ^{42}&\\omega ^{49}\\\\\\end{bmatrix}}={\\frac {1}{\\sqrt {2^{3}}}}{\\begin{bmatrix}1&1&1&1&1&1&1&1\\\\1&\\omega &\\omega ^{2}&\\omega ^{3}&\\omega ^{4}&\\omega ^{5}&\\omega ^{6}&\\omega ^{7}\\\\1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}&1&\\omega ^{2}&\\omega ^{4}&\\omega ^{6}\\\\1&\\omega ^{3}&\\omega ^{6}&\\omega &\\omega ^{4}&\\omega ^{7}&\\omega ^{2}&\\omega ^{5}\\\\1&\\omega ^{4}&1&\\omega ^{4}&1&\\omega ^{4}&1&\\omega ^{4}\\\\1&\\omega ^{5}&\\omega ^{2}&\\omega ^{7}&\\omega ^{4}&\\omega &\\omega ^{6}&\\omega ^{3}\\\\1&\\omega ^{6}&\\omega ^{4}&\\omega ^{2}&1&\\omega ^{6}&\\omega ^{4}&\\omega ^{2}\\\\1&\\omega ^{7}&\\omega ^{6}&\\omega ^{5}&\\omega ^{4}&\\omega ^{3}&\\omega ^{2}&\\omega \\\\\\end{bmatrix}}.}$ It could be simplified further by using ${\\displaystyle \\omega ^{4}=-1}$, ${\\displaystyle \\omega ^{2}=i}$ and ${\\displaystyle \\omega ^{6}=-i}$ and then even more given that ${\\displaystyle \\omega ^{5}=-\\omega }$, ${\\displaystyle \\omega ^{3}=i\\omega }$ and ${\\displaystyle \\omega ^{7}=-i\\omega }$. The 3-qubit quantum Fourier transform can be rewritten as: ${\\displaystyle {\\text{QFT}}(|x_{1},x_{2},x_{3}\\rangle )={\\frac {1}{\\sqrt {2^{3}}}}\\ \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{2}x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +e^{2\\pi i\\,[0.x_{1}x_{2}x_{3}]}|1\\rangle \\right)}$ or ${\\displaystyle {\\text{QFT}}(|x_{1},x_{2},x_{3}\\rangle )={\\frac {1}{\\sqrt {2^{3}}}}\\ \\left(|0\\rangle +\\omega _{1}'^{[x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +\\omega _{2}'^{[x_{2}x_{3}]}|1\\rangle \\right)\\otimes \\left(|0\\rangle +\\omega _{3}'^{[x_{1}x_{2}x_{3}]}|1\\rangle \\right).}$ In case that we use the circuit", "set of solutions but this is also included in $a+b+c=0$. – mathsdiscussion.com Jan 19 '20 at 12:32 • For complex numbers $a$, $b$, and $c$, $a^3+b^3+c^3=3abc$ if and only if the (possibly degenerate) triangle formed by complex coordinates $a$, $b$, and $c$ is an equilateral, or has its centroid at the origin $0$ (or equivalently, $a+b+c=0$, $a+b\\omega+c\\omega^2=0$, or $a+b\\omega^2+c\\omega=0$, where $\\omega$ is a complex root of $x^2+x+1=0$). From this observation, it follows immediately that when $a$, $b$, and $c$ are real, $a^3+b^3+c^3=3abc$ iff $a+b+c=0$ or $a=b=c$. – Batominovski Jan 19 '20 at 14:45 By your work: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ Thus, $$a^3+b^3+c^3-3abc=0$$ for $$a+b+c=0$$ or for $$a^2+b^2+c^2-ab-ac-bc=0,$$ which is $$(a-b)^2+(a-c)^2+(b-c)^2=0,$$ which gives $$a=b=c.$$ Id est, we have no another cases for equality occurring for real values. • There are comments involving complex solutions. I added a qualification that the proof holds in the real domain. – Oscar Lanzi Jan 19 '20 at 12:38 • Oscar, if in the given of the problem we don't say about complex numbers so we say about real numbers. It's a known rule. – Michael Rozenberg Jan 19 '20 at 13:26 Without loss of generality, suppose that $$a\\ge b$$ and $$a\\ge c$$. Then $$a^2+b^2+c^2-ab-bc-ca=(a-b)(a-c)+(b-c)^2\\ge 0.$$ Equality can only occur if $$a=b=c$$ and so there are no other solutions. The equation can also be factorized as follows-", "so $a^2+b^2+44a+76b=0$, which can be rewritten as$(a+22)^2+(b+38)^2=22^2+38^2$. In order to maximize $\\Im(z)$ we want to maximize $b$, and in order to maximize $b$ we want $a+22=0$ and $a=-22$, so $\\Re(z)=a+78=-22+78=\\boxed{056}$. (Note: $\\Im(\\omega)$ is the imaginary part of $\\omega$, and $\\Re(\\omega)$ is the real part of $\\omega$) ~Stormersyle ## Solution 5 We will just bash. Let $z=a+bi$ where $a,b\\in\\mathbb{R}$. We see that $\\frac{z_3-z_1}{z_2-z_1}=\\frac{-4+15i}{11}$ after doing some calculations. We also see that $\\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\\text{some real stuff}}.$ We note that $[(a-18)+(b-39)i][(a-78)-(b-99)i]$ is a multiple of $-4-15i$ because the numerator has to be real. Thus, expanding it out, we see that $(a-18)(a-78)+(b-39)(b-99)=-4k \\\\ (a-78)(b-39)-(a-18)(b-99)=-15k.$ Hence, $(a-18)(a-78)+(b-39)(b-99)=\\frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)] \\implies a^2-96a+b^2-138b+5625=16a-16b+336 \\\\ (a-56)^2+(b-61)^2=1568.$ To maximize the imaginary part, $(a-56)^2$ must equal $0$ so hence, $a=\\boxed{56}$.", "fifth root of unity. Now $$\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5=x$$ so \\begin{align} \\omega_5x&=\\omega_5(\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5)\\\\ &=\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5+\\omega_5^6\\\\ &=\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5+\\omega_5\\\\ &=x \\end{align} so $\\omega_5x=x$, and so $x\\neq 0$ would imply $\\omega_5=1$ which is false. So $x=0$. Now use $$e^{i\\theta}=\\cos(\\theta)+i\\sin(\\theta)$$ Now take $\\theta=\\frac{2k\\pi}5$ for $k=1,2,3,4,5$ to get $\\omega_5^k$. So we get: \\begin{align} 0&=\\Re(0)\\\\ &=\\Re(\\omega_5+\\omega_5^2+\\omega_5^3+\\omega_5^4+\\omega_5^5)\\\\ &=\\Re(e^{2i\\pi/5}+e^{4i\\pi/5}+e^{6i\\pi/5}+e^{8i\\pi/5}+e^{10i\\pi/5})\\\\ &=\\cos(\\frac{2\\pi}5)+\\cos(\\frac{4\\pi}5)+\\cos(\\frac{6\\pi}5)+\\cos(\\frac{8\\pi}5)+\\cos(\\frac{10\\pi}5)\\\\ &=\\cos(\\frac{2\\pi}5)+\\cos(\\frac{4\\pi}5)+\\cos(\\frac{6\\pi}5)+\\cos(\\frac{8\\pi}5)+1 \\end{align} - I presume lines 3 and 4 are meant to be understood as equivalent since \\omega_5^6\\ = \\omega_5\\ by periodic formula. – CandiedOrange Feb 18 at 10:01 @CandiedOrange, yes, that's the step I took. We use $\\omega_5^5=1$ – vrugtehagel Feb 18 at 10:05 Let's try this - Using $$2\\cos a \\sin b = \\sin(a+b) - \\sin(a- b)$$ we get: $$2\\cos(2π/5)\\sin(2π/5)=\\sin(4π/5 )- \\sin(0)$$ $$2\\cos(4π/5)\\sin(2π/5)=\\sin(6π/5)-\\sin(2π/5)$$ $$2\\cos(6π/5)\\sin(2π/5)=\\sin(8π/5) - \\sin(4π/5)$$ $$2\\cos(8π/5)\\sin(2π/5)=\\sin(10π/5) -\\sin(6π/5)$$2\\cos(10π/5)\\sin(2π/5)=\\sin(12π/5) - \\sin(8π/5) Adding: \\begin{align} &2\\sin(2π/5)\\{\\cos(10π/5)+\\cos(2π/5)+\\cos(4π/5)+\\cos(6π/5)+\\cos(8π/5)\\}\\\\ &=2\\sin(2π/5)\\{1+\\cos(2π/5)+\\cos(4π/5)+\\cos(6π/5)+\\cos(8π/5)\\} \\\\ &=\\sin(12π/5) - \\sin(2π/5)\\\\ &=2\\sin(10π/5)\\cos(14π/5)\\\\&=0\\qquad [\\;\\because \\sin a - \\sin b = 2\\sin(a-b)\\cos(a+b)] \\end{align}1+\\cos(2π/5)+\\cos(4π/5)+ \\cos(6π/5)+\\cos(8π/5)=0$$- I used it by grouping the last term with the second and the onea in the middle. I disnt manage to do something – Razvan Paraschiv Feb 17 at 16:28 @RazvanParaschiv Check my method. I edited my answer. – Win Vineeth Feb 18 at 0:02 Draw a regular pentagon ABCDE with all sides one unit long. Treat these sides as vectors from A to B, then B to C, etc. From the \"head to tail\" rule for resultants the resultant of", "for other digit groups which are at higher place order. The reason is simple, they may have to be added with carry-overs from the groups (terms) on the right. ## A corollary of the General Formula of Transformation (for two digits) ### With to Plus form $$a\\omega b = 10a+b$$ The above equation is true only when $$b$$ has one digit. The more general equation is:- $$\\boxed{a\\omega b = a.10^d+b}\\hspace1em\\text{where d}=]b[\\hspace1em\\ldots\\text{Corollary 1}$$ ### Plus to With form $$a\\omega b = 10a +b \\Rightarrow a\\omega b = 9a + a + b \\Rightarrow \\boxed{a+b = a\\omega b -9a}\\hspace1em\\ldots\\text{Corollary 2}$$ ### Negative digits We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this. \\begin{align} a\\omega(-b) &= 10a + (-b)\\\\ &= 10a -b\\\\ &= 10a + b -2b\\\\ &= a\\omega b -2b \\end{align} Altire, \\begin{align} a\\omega(-b) &= -(-10a + b)\\\\ &= -(10a+b -20a)\\\\ &= -(a\\omega b – 20a)\\\\ &= -(a\\omega b) +20a \\end{align} $$\\therefore \\boxed{a\\omega(-b) = a\\omega b-2b = -(a\\omega b)+20a}$$ Remember, $$(-a)\\omega b \\neq -(a\\omega b)$$. The former means that only the digit $$a$$ is negative. The later means that the complete number is negative. Similarly, $$\\boxed{(-a)\\omega b = a\\omega b-20a = -(a\\omega b)+2b}$$", "The solutions $u$ and $v$ separately for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.9511,1.2)\\in)A,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.1,1.2976)\\in)I,B($ The solution $(u,v)$ for $p = 1.5$, $\\omega = 2$ and $(a,b) = (0.5,1.1371)\\in)I,B($ The solutions $u$ and $v$ separately for $(a,b) = (0.1,1.2976)\\in)I,B($ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (0.2,0.4)$ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (2,4)$ The chaotic behavior for $p = 2.5$, $w = 3$ and $(a,b) = (0.2,0.14)$ The chaotic behavior for $p = 2.5$, $w = 2$ and $(a,b) = (2,4)$ The partition of the plane according to $\\Gamma_1$, $\\Gamma_2$ and $\\Lambda$ for $w = 0$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.5,1)\\in R_1$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (2.5,3.5)\\in R_1$ The portrait $(u',u)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The portrait $(v',v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solutions $u$ (in", "(7) yields (bc-1)(c-b)=0, but we have seen that c\\ne b, so we must have cb=1 or equivalently$$c =\\frac{1}{b}=\\overline{b}\\,.\\tag{9}$$Replacing back in (7) we get b^2+c+1+b+c^2=0 or equivalently$$b^4+b^3+b^2+b+1=0\\,.$$Thus, b is a fifth root of 1 different from 1, that is$$b\\in\\{\\omega,\\omega^2,\\omega^3,\\omega^4\\}\\quad\\hbox{ where\\omega=\\exp \\left(\\frac{2\\pi i}{5}\\right)\\,.}$$It follows that c=b^{-1}=b^4 and c^2=b^8=b^3. From (5) we conclude that$$\\eqalign{ a&\\,=\\,\\frac{b+b^3+2b^4+1}{b-b^4}\\,=\\,\\frac{b^4\\!-b^2}{b-b^4}\\,=\\,b^3\\,,\\cr d&\\,=\\,b^2\\,,}$$where we used the identities 1+b+b^2+ b^3+b^4=0 and b^5=1. We get the solution (z_0,z_1,z_2,z_3,z_4) with$$ z_0=b^3z_4\\,,~z_1=b z_4\\,,~z_2=b^4 z_4\\,,~z_3=b^2 z_4\\,,$which are the vertices of a regular pentagon in some order, (A regular pentagon, or a regular pentagonal star). • Nice solution, I wonder if it can be simplified. The \"which is equivalent to\" between (7) and (8) seems to be pure luck. – Ewan Delanoy May 22 '14 at 16:35 • @Ewan Delanoy. It is not pure luck: the \"which is equivalent to\" between$(7)$and$(8)$is$(b\\leftrightarrow c)$-symmetric to the \"which is equivalent to\" between$(6)$and$(7)\\$. – chizhek Aug 27 '14 at 16:59", "L.H.S. =$$\\frac{\\left(\\mathbf{a}+\\mathbf{b} \\omega+\\mathbf{c} \\omega^{2}\\right)}{\\mathbf{c}+\\mathbf{a} \\omega+\\mathbf{b} \\omega^{2}}$$ = $$\\frac{a \\omega^{3}+b \\omega^{4}+c \\omega^{2}}{c+a \\omega+b \\omega^{2}}$$ ……[∵ ω3 = 1, ω4 = ω] = $$\\frac{\\omega^{2}\\left(c+a \\omega+b \\omega^{2}\\right)}{c+a \\omega+b \\omega^{2}}$$ = ω2 = R.H.S. Question 2. If ω is a complex cube root of unity, find the value of (i) ω + $$\\frac{1}{\\omega}$$ (ii) ω2 + ω3 + ω4 (iii) (1 + ω2)3 (iv) (1 – ω – ω2)3 + (1 – ω + ω2)3 (v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) Solution: ω is the complex cube root of unity. ∴ ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1 (i) ω + $$\\frac{1}{\\omega}$$ = $$\\frac{\\omega^{2}+1}{\\omega}$$ = $$\\frac{-\\omega}{\\omega}$$ = -1 (ii) ω2 + ω3 + ω4 = ω2 (1 + ω + ω2) = ω2 (0) = 0 (iii) (1 + ω2)3 = (-ω)3 = -ω3 = -1 (iv) (1 – ω – ω2)3 + (1 – ω + ω2)3 = [1 – (ω + ω2)]3 + [(1 + ω2) – ω]3 = [1 – (-1)]3 + (-ω – ω)3 = 23 + (-2ω)3 = 8 – 8ω3 = 8 – 8(1) = 0 (v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) = (1 + ω)(1 +", "(twelve)» can also occur as a result of the following six combinations: $$(6,5,1),(6,4,2),(6,3,3),(5,5,2),(5,4,3),(4,4,4) \\text {. }$$ Consequently, these events are equally probable. Here, the mistake of de Mere is that the possible outcomes that he considered are not equally probable. For example, the event $(6,4,1)$ can occur in $3 !=6$ cases: $(6,4,1),(6,1,4), \\ldots$, $(1,4,6)$. At the same time, for example, a combination $(4,4,4)$ can occur only in one case. In modern language, de Mere incorrectly constructed the space of elementary events corresponding to the given problem. The solution of the problem. We define $\\Omega$ as $$\\Omega={(i, j, k): i, j, k=\\overline{1,6}}=\\Omega_{0} \\times \\Omega_{0} \\times \\Omega_{0},$$ where $\\Omega_{0}={1,2,3,4,5,6}$. Let’s introduce the events: $A_{11}={$ the sum of points is equal to 11$}, A_{12}={$ the sum of points is equal to 12$}$. Hence \\begin{aligned} &A_{11}={(i, j, k) \\in \\Omega: i+j+k=11}, \\ &A_{12}={(i, j, k) \\in \\Omega: i+j+k=12}, \\end{aligned} We have $$\\left|A_{11}\\right|=27, \\quad\\left|A_{12}\\right|=25, \\quad|\\Omega|=6^{3}=216,$$ Therefore $$P\\left(A_{11}\\right)=\\frac{27}{216}>\\frac{25}{216}=P\\left(A_{12}\\right) .$$ 1. From the general population $\\Omega_{0}=\\left{a_{1}, a_{2}, \\ldots, a_{n}\\right}$ (for example, from an urn with numbered balls) of size $n$, a random sample with replacement of size $r$ is extracted. a) Find the probability that the extracted sample is a sample without replacement (that is, all the extracted balls have different numbers). b) Find the probability that the first sample element is the first element", "$p = 1.5$, $\\omega = 2$ and $(a,b) = (0.5,1.1371)\\in)I,B($ The solutions $u$ and $v$ separately for $(a,b) = (0.1,1.2976)\\in)I,B($ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (0.2,0.4)$ The chaotic behavior for $p = 1.5$, $w = 2$ and $(a,b) = (2,4)$ The chaotic behavior for $p = 2.5$, $w = 3$ and $(a,b) = (0.2,0.14)$ The chaotic behavior for $p = 2.5$, $w = 2$ and $(a,b) = (2,4)$ The partition of the plane according to $\\Gamma_1$, $\\Gamma_2$ and $\\Lambda$ for $w = 0$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.5,1)\\in R_1$ The solution $(u,v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (2.5,3.5)\\in R_1$ The portrait $(u',u)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.5538)\\in\\Lambda$ The portrait $(v',v)$ for $p = 1.5$, $w = 0$ and $(a,b) = (0.25,0.35)\\in R_2$ The solutions $u$ (in blue) and $v$ (in pink) for $p = 1.5$, $w = 0$ and $(a,b) = (0.5,1)\\in R_1$ The energy $E(u,v)(x)$ for $p = 1.5$, $(a,b) = (0.15,25)\\in R_2$, $w = 0$ at the top and", "is the Golden Ratio. - I love solutions invoking the golden ratio. It'd be even better with a pentagram. (+1) – Ron Gordon Feb 11 '13 at 9:05 We have $\\cos 36 = \\frac{\\omega + \\omega^{-1}}{2}$ where $\\omega = \\text{exp} \\left ( \\frac{2 \\pi}{10} \\right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\\omega^4 - \\omega^3 + \\omega^2 - \\omega + 1 = 0$. Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \\frac{1}{x} + \\frac{1}{x^2}) = x^2\\left ( \\left (x + \\frac{1}{x} \\right )^2 - \\left (x + \\frac{1}{x} \\right ) -1 \\right )$. Thus $0 = \\omega^2 \\cdot ((2 \\cos 36)^2 - (2 \\cos 36) - 1)$. Therefore $4 \\cos^3 36 - 2 \\cos 36 - 1 = 0$. Using the quadratic formula we then arrive at $$\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$$", "# Homework Help: Find roots of complex equation (1-x)^5 = x^5 1. Sep 6, 2011 ### DryRun 1. The problem statement, all variables and given/known data Find roots of complex equation (1-x)^5 = x^5 2. Relevant equations Probably Euler and/or De Moivre. 3. The attempt at a solution I know i need to have z^5 on one side and all the rest on the other side. But i need 5 roots, and i'm only getting one. (1-x)^5 = x^5 Use fifth root on both sides. (1-x) = x 2x= 1 x = 1/2 (only one solution, which i know is wrong, but no idea how to proceed). 2. Sep 6, 2011 ### micromass You have the right idea here. But (like you thought), it isn't true that $a^5=b^5$ implies a=b. However, we can fix this: If $a^5=b^5$ (and they are nonzero), then $\\left(\\frac{a}{b}\\right)^5=1$. So a/b are fifth roots of unity. There are 5 fifth roots of unity. So let $\\alpha_i$ be one of the fifth roots of unity, then we have $$\\frac{a}{b}=\\alpha_i$$ And thus $$a=\\alpha_i b$$ So, if $a^5=b^5$, then $a=\\alpha_i b$ where we take $\\alpha_i$ the fifth roots of unity. This gives us 5 equations. Try that on your equation. 3. Sep 6, 2011 ### Dickfore Take the complex root, to get: $$1 - x = \\omega", "out yo find $$215^2$$. \\begin{align} 215^2 &= (21 \\times (21 + 1)) \\omega 25\\\\ &= (21 \\times 22) \\omega 25\\\\ &= 462\\omega25\\\\ &= 46225\\tag{Answer} \\end{align} This always seemed to work out very well. The problem was, can I trust this trick? Will this always hold true? I didn’t have answers to those questions, until I proved it myself using Digit Math. Good news is that this trick will always hold true. # The proof Let the number be $$x = a\\omega 5$$. $$a$$ can have any number of digits. ### Case 1: $$a$$ has exactly one digit. So, $$]a[ = ]5[ = 1$$. \\begin{align} \\therefore (a\\omega 5)^2 &= aa\\omega (5a+5a) \\omega 5.5\\tag{Using Bimultiplication formula}\\\\ &= a^2 \\omega 10a \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega0) \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega2) \\omega 5\\\\ &= (a^2 + a) \\omega 2 \\omega 5\\\\ &= \\big(a(a+1)\\big) \\omega 25\\tag{Proved} \\end{align} ### Case 2: $$a$$ has more than one digits. So, $$]a[ > (]5[ = 1)$$. But to apply Bimultiplication formula $$a$$ must have the same number of digits in $$5$$, which is obviously not the case here. So, we will use one trick. We will pad $$5$$ with some number of zeroes on the right, so that, $$]a[ = ]5\\omega c[$$, where, $$c$$ is all zeroes and $$]c[ = ]a[ – 1$$. So,", "only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \\frac{d - b}{2} = 51$. Therefore, the perimeter of $\\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \\boxed{\\textbf{(459) }}$. ~Steven Chen (www.professorchenedu.com) ## Solution (Number Theory Part) We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \\cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for. Notice that $r^2+r+1=(r-\\omega)(r-{\\omega}^2)$, where $\\omega=e^{i\\frac{2\\pi}{3}}$. Thus, $$x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\\frac{x}{y}-\\omega)(\\frac{x}{y}-{\\omega}^2) = (x-y\\omega)(x-y{\\omega}^2).$$ Note that $8^2+1^2+8 \\cdot 1=73$. Thus, $(8-\\omega)(8-{\\omega}^2)=73$. Squaring both sides yields \\begin{align} (8-\\omega)^2(8-{\\omega}^2)^2&=73^2\\\\ (63-17\\omega)(63-17{\\omega}^2)&=73^2. \\end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=\\boxed{459}$. ~ Leo.Euler ## Solution(Visual geometry) We look at upper and middle diagrams and get $\\angle BAC = 120^\\circ$. Next we use only the lower Diagram. Let $I$ be incenter $\\triangle ABC$, E be midpoint of biggest arc $\\overset{\\Large\\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\\triangle ABC$ at point $E$. Quadrilateral $ABEC$ is cyclic, so $$\\angle BEC = 180^\\circ - \\angle ABC = 60^\\circ \\implies BE = CE = IE = BC.$$", "(terms) on the right. ## A corollary of the General Formula of Transformation (for two digits) ### With to Plus form $$a\\omega b = 10a+b$$ The above equation is true only when $$b$$ has one digit. The more general equation is:- $$\\boxed{a\\omega b = a.10^d+b}\\hspace1em\\text{where d}=]b[\\hspace1em\\ldots\\text{Corollary 1}$$ ### Plus to With form $$a\\omega b = 10a +b \\Rightarrow a\\omega b = 9a + a + b \\Rightarrow \\boxed{a+b = a\\omega b -9a}\\hspace1em\\ldots\\text{Corollary 2}$$ ### Negative digits We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this. \\begin{align} a\\omega(-b) &= 10a + (-b)\\\\ &= 10a -b\\\\ &= 10a + b -2b\\\\ &= a\\omega b -2b \\end{align} Altire, \\begin{align} a\\omega(-b) &= -(-10a + b)\\\\ &= -(10a+b -20a)\\\\ &= -(a\\omega b – 20a)\\\\ &= -(a\\omega b) +20a \\end{align} $$\\therefore \\boxed{a\\omega(-b) = a\\omega b-2b = -(a\\omega b)+20a}$$ Remember, $$(-a)\\omega b \\neq -(a\\omega b)$$. The former means that only the digit $$a$$ is negative. The later means that the complete number is negative. Similarly, $$\\boxed{(-a)\\omega b = a\\omega b-20a = -(a\\omega b)+2b}$$ Also similarly evaluating, we get, $$\\boxed{(-a)\\omega(-b) = -(a\\omega b)}\\hspace1em\\ldots\\text{Corollary 3}$$ So, we see that in a negative number all the digits too are negative. Not", "0,8 2,12 2 0,1 0,9 3,9 6,4 3,9 0,9 0,1 3 3 0,1 1,0 5,5 14,9 20,8 14,9 5,5 1,0 0,1 3,67 4 1,3 7,6 26,5 56,3 72,2 56,3 26,5 7,6 1,3 4,24 The most probable states of the $$x_i$$ variable system are the equilibrium states $$(X_N=0)$$. Denoting the number of equilibrium states by $$\\Omega_N$$, the value $$\\ln\\Omega_N$$ is called here the information capacity or the system entropy. First three values of $$\\Omega_N$$ are close to the Fibonacci sequence sums: $$N$$ $$\\sum_{i=0}^{2N–1}{F_i}$$ $$\\Omega_N$$ $$\\ln\\Omega_N$$ 1 (1+1) = 2 2 0,69 2 (3+2) + (1+1) = 7 6 1,79 3 (8+5) + (3+2) + (1+1) = 20 20 3 4 (21+13) + (8+5) + (3+2) + (1+1) = 54 70 4,25 The especially remarkable case is $$N=3$$, when the capacity or entropy $$\\ln\\Omega_3=2,996\\approx 3$$. The state space for a system of three pairs of binary values could be represented as a simple 3D-matrix (a cube) and also a stellated octahedron: Number of the zero pairs $$X_N$$ Number of combinations Geometry 3 0 8 Faces of internal octahedron 2 ±1 24 Сonvex edges 1 0 12 Faces ±2 12 0 ±1 6 Vertices ±3 2 The system of three pairs of binary values takes place in the nature and the culture. For example:", "### Complex Numbers goals The number of common roots of the 15th and of 25th roots of unity are The number of common roots of the 15th and of 25th roots of unity are If $\\omega$ is the cube root of unity then $(1 - \\omega + \\omega^2)^6 =$ If $\\omega$ is the cube root of unity then $(1 - \\omega + \\omega^2)^6 =$ The closest distance of the origin from a curve given as $a\\bar z + \\bar az + a\\bar a = 0$ ($a$ is a complex number) is.", "is, for every $$\\omega\\in\\Omega$$ find the smallest member of $$\\sigma({\\mathcal U})$$ that contains $$\\omega$$. Check that there is no way to separate $$1$$ and $$2$$ using unions and/or intersections from $$\\mathcal U$$, and similarly you cannot find a way to separate $$4$$ from $$5$$. On the other hand, the element $$3$$ is covered by the singleton set $$\\{3\\} = \\{1,2,3\\}\\cap\\{3,4,5\\}$$. The three sets $$\\{1,2\\}$$, $$\\{3\\}$$, $$\\{4,5\\}$$ are disjoint and their union is $$\\Omega$$, so that's your partition. • Does the set $\\{3\\}$ correspond to any of my terms given by $\\sigma_{\\mathcal{U}}=\\{\\cdots\\}$? – M B Mar 10 at 6:20 • Yes, $\\{3\\} = A\\cap B = (A^c\\cup B^c)^c$. – grand_chat Mar 10 at 6:23", "creation operators \\begin{aligned} A=(A^{\\dag})^{\\dag}=\\sqrt{\\mu {\\omega}_{A}/(2\\hbar )}x_{1}+i\\sqrt{1/(2\\mu \\hbar{\\omega}_{A})}p_{1}, \\quad B=(B^{\\dag})^{\\dag}=\\sqrt{\\mu {\\omega}_{B}/(2\\hbar )}x_{2}+i\\sqrt{1/(2\\mu \\hbar{\\omega}_{B})}p_{2}, \\end{aligned} (12) Hamiltonian (10) can be expressed as \\begin{aligned} \\tilde{H}=\\hbar{\\omega}_{A}A^{\\dag}A+\\hbar{\\omega}_{B}B^{\\dag}B+\\frac{1}{2}\\hbar ({\\omega}_{A}+{\\omega}_{B}). \\end{aligned} (13) The relations between the operators A ($$A^{\\dag}$$), B ($$B^{\\dag}$$), a ($$a^{\\dag}$$), and b ($$b^{\\dag}$$) can be obtained as \\begin{aligned} UaU^{\\dag}=f_{1}A+f_{2}A^{\\dag}+f_{3}B+f_{4}B^{\\dag},\\quad UbU^{\\dag}=-f_{5}A-f_{6}A^{\\dag}+f_{7}B+f_{8}B^{\\dag}. \\end{aligned} (14) Here the concrete forms of coefficients $$f_{i}$$ ($$i=1,2,\\ldots ,7,8$$) have been given by \\begin{aligned} f_{1,2}=&\\frac{1}{2}\\frac{\\cos (\\theta )}{\\sqrt{{\\omega}_{a}{\\omega}_{A}}}({\\omega}_{a}\\pm {\\omega}_{A}),\\quad f_{3,4}=\\frac{1}{2}\\frac{\\sin (\\theta )}{\\sqrt{{\\omega}_{a}{\\omega}_{B}}}({\\omega}_{a}\\pm {\\omega}_{B}), \\end{aligned} (15a) \\begin{aligned} f_{5,6}=&\\frac{1}{2}\\frac{\\sin (\\theta )}{\\sqrt{{\\omega}_{b}{\\omega}_{A}}}({\\omega}_{b}\\pm {\\omega}_{A}),\\quad f_{7,8}=\\frac{1}{2}\\frac{\\cos (\\theta )}{\\sqrt{{\\omega}_{b}{\\omega}_{B}}}({\\omega}_{b}\\pm {\\omega}_{B}). \\end{aligned} (15b) Based on Eq. (13), we know the eigenstates of the system in the representation associated with $$A^{\\dag}A$$ and $$B^{\\dag}B$$ as \\begin{aligned} {\\tilde{H}}\\vert m{\\rangle}_{A}\\vert n{\\rangle}_{B}=E_{m,n}\\vert m{\\rangle}_{A}\\vert n{\\rangle}_{B},\\quad m,n=0,1,2,\\ldots , \\end{aligned} (16) where the eigenvalues are given by \\begin{aligned} E_{m,n}=\\hbar{\\omega}_{A}m+\\hbar{\\omega}_{B}n+\\frac{1}{2}\\hbar ({\\omega}_{A}+{\\omega}_{B}). \\end{aligned} (17) It is obvious that the ground state of the two-oscillator system is $$\\vert 0{\\rangle}_{A}\\vert 0{\\rangle}_{B}$$. To study the virtual excitations in the system, we need to know the eigenstates which are expressed in the representation associated with $$a^{\\dag}a$$ and $$b^{\\dag} b$$. It implies that we need to diagonalize the Hamiltonian $${\\tilde{H}}$$ in the representation of a and b. To this end, we express Hamiltonian (10) with the bosonic creation (annihilation) operators $$a^{\\dag}$$ (a) and $$b^{\\dag}$$ (b) as \\begin{aligned} {\\tilde{H}}=\\frac{1}{2}\\hbar ({\\omega}_{A}^{2}/{\\omega}_{a}+{\\omega}_{a})({a}^{\\dagger}{a}+1/2)+\\hbar g_{a}({a}^{\\dagger 2}+{a}^{2})+\\frac{1}{2}\\hbar ({\\omega}_{B}^{2}/{\\omega}_{b}+{\\omega}_{b})({b}^{\\dagger}{b}+1/2)+\\hbar g_{b}({b}^{\\dagger 2}+{b}^{2}), \\end{aligned} (18) where we introduce the coupling strengths \\begin{aligned}" ]

[ "$\\bf{u}\\times \\bf{v} = A \\bf{n}$, i.e. $\\bf{u}\\times \\bf{v}$ is perpendicular to u,v and of length $A$. QED!", "$$p'_w = - w\\mathbf{v}.\\mathbf{p} + w\\mathbf{p}.\\mathbf{v} - (\\mathbf{v} \\times \\mathbf{p}).\\mathbf{v}$$ We know $$\\mathbf{v} \\times \\mathbf{p}$$ is orthogonal to $$\\mathbf{v}$$, so $$(\\mathbf{v} \\times \\mathbf{p}).\\mathbf{v} = 0$$ Therefore $$p'_w = 0$$ Giving a resulting vector with zero w coordinate like we expected. Continuing with the vector part of $$p'$$: $$\\mathbf{p'} = (\\mathbf{p}w + \\mathbf{v} \\times \\mathbf{p})w + (-\\mathbf{v})(- \\mathbf{v}.\\mathbf{p}) + (\\mathbf{p}w \\times -\\mathbf{v}) + ((\\mathbf{v} \\times \\mathbf{p}) \\times -\\mathbf{v})$$ $$\\mathbf{p'} = (\\mathbf{p}w + \\mathbf{v} \\times \\mathbf{p})w + \\mathbf{v}( \\mathbf{v}.\\mathbf{p}) + (\\mathbf{v} \\times \\mathbf{p})w + (\\mathbf{v} \\times (\\mathbf{v} \\times \\mathbf{p}))$$ $$\\mathbf{p'} = \\mathbf{p}w^2 + (\\mathbf{v} \\times \\mathbf{p})w + \\mathbf{v}( \\mathbf{v}.\\mathbf{p}) + (\\mathbf{v} \\times \\mathbf{p})w + (\\mathbf{v} \\times (\\mathbf{v} \\times \\mathbf{p}))$$ $$\\mathbf{p'} = \\mathbf{p}w^2 + 2(\\mathbf{v} \\times \\mathbf{p})w + \\mathbf{v}( \\mathbf{v}.\\mathbf{p}) + (\\mathbf{v} \\times (\\mathbf{v} \\times \\mathbf{p}))$$ Remember that for a unit quaternion, $$\\mathbf{v}$$ is a (non-unit) vector around which the quaternion gives a rotation, in particular $$||\\mathbf{v}|| = \\sin(\\Omega)$$ Where $$\\Omega$$ is half the angle of the quaternion rotation, e.g. $$||\\mathbf{v}|| = \\sin(\\Omega) = \\sin(\\theta/2)$$ So now we decompose the vector $$\\mathbf{p}$$ into the vector projection and vector rejection (see wikipedia) components with respect to the unit vector $$\\hat{\\mathbf{v}} = \\mathbf{v}/||\\mathbf{v}||$$, giving $$\\mathbf{p} = \\hat{\\mathbf{v}}(\\hat{\\mathbf{v}}.\\mathbf{p}) + (-\\hat{\\mathbf{v}} \\times (\\hat{\\mathbf{v}} \\times \\mathbf{p}))$$ Substituting in $$\\hat{\\mathbf{v}} = \\mathbf{v}/||\\mathbf{v}|| = { \\mathbf{v} \\over \\sin(\\theta/2) }$$ gives $$\\mathbf{p} = ({ \\mathbf{v} \\over \\sin(\\theta/2) })(({ \\mathbf{v}", "# Vectors-Scalar Triple product Show that $$(a \\times b) \\times (c \\times d)= [a,b,d]c - [a,b,c]d.$$ I understand $[a,b,d] = a \\cdot (b \\times d)$ and $[a,b,c] = a \\cdot (b \\times c)$ but not really sure where else to go with this. - YES! Thank you that works. – Apeman Aug 11 '12 at 22:09 As Rahul mentioned: $$n \\times (m \\times o) = (n \\cdot o)m - (n \\cdot m)o$$ Then it follows: $$(a \\times b) \\times (c \\times d) = (a \\times b \\cdot d)c -(a \\times b \\cdot c)d$$ Using the expressions you mentioned above this can be further simplified to:$$[a,b,d]c - [a,b,c]d$$", "\\times \\mathbf{j} = \\mathbf{k},\\;$$ $$\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i},\\;$$ $$\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}.$$", "scalar product of the two vectors $$\\overrightarrow{\\mathbf{v}}_{1}$$ = $$\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+\\mathbf{3} \\hat{\\mathbf{k}}$$ and $$\\overrightarrow{\\mathbf{v}}_{2}$$ = $$3 \\hat{\\mathbf{i}}+4 \\hat{\\mathbf{j}}-\\mathbf{5} \\hat{\\mathbf{k}}$$ Solution: Answer: Scalar product of two given vectors is – 4. Question 51. A force $$\\overrightarrow{\\mathbf{F}}$$ = $$4 \\hat{\\mathbf{i}}+6 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$$ acting on a particle produces a displacement of $$\\overrightarrow{\\mathbf{S}}$$ = $$\\overrightarrow{\\mathrm{s}}=2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+\\mathbf{5} \\hat{\\mathbf{k}}$$ where F is expressed in newton and s in metre. Find the work done by the force. Solution: Answer: The work done by the force is 41 J. Question 52. Find ‘a’ if $$\\overrightarrow{\\mathbf{A}}$$ = $$3 \\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}+4 \\hat{\\mathbf{k}}$$ and $$\\overrightarrow{\\mathbf{B}}$$ = $$\\mathbf{a} \\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}$$ are perpendicular to one another. Solution: Question 53. If $$\\overrightarrow{\\mathbf{A}}$$ = $$5 \\hat{\\mathbf{i}}+6 \\hat{\\mathbf{j}}+4 \\hat{\\mathbf{k}}$$ and $$\\overrightarrow{\\mathbf{B}}$$ = $$2 \\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$$ determine the angle between and . Solution: Solution: Question 54. Find the angle between the vectors $$\\overrightarrow{\\mathbf{A}}$$ = $$\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}$$ and $$\\overrightarrow{\\mathbf{B}}$$ = $$-\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}-\\mathbf{2} \\hat{\\mathbf{k}}$$. Solution: Let angle between the vectors be θ Answer: The angle between the vectors is 60°. Question 55. If $$\\overrightarrow{\\mathbf{A}}$$ = $$2 \\hat{\\mathbf{i}}+7 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$$ and $$\\vec{B}$$ = $$3 \\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+\\mathbf{5} \\hat{\\mathbf{k}}$$, find the component of $$\\overrightarrow{\\mathbf{A}}$$ along $$\\overrightarrow{\\mathbf{B}}$$. Solution: Question 56. $$\\hat{\\mathbf{i}}$$ and $$\\hat{\\mathbf{j}}$$ are unit vectors along X-axis and Y-axis respectively. What is the magnitude and direction of the vector $$\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}$$ and $$\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}$$? What are the components of a vector $$\\overrightarrow{\\mathbf{A}}=2", "\\times \\mathbf{A} = -\\mathbf{A} \\times [-(\\mathbf{B} \\times \\mathbf{A})] = \\mathbf{A} \\times (\\mathbf{B} \\times \\mathbf{A})$ So my example is testing a case that is not valid for the inequality $(\\mathbf{A} \\times \\mathbf{B}) \\times \\mathbf{C} \\ne \\mathbf{A} \\times (\\mathbf{B} \\times \\mathbf{C})$, which assumes the three vectors are unique. In regards to Mariano's valid answer, the fact that $(c\\times a) \\times b = 0$ is true in my case because $c = a$. - A longuish comment The cross product satisfies the Jacobi identity $$(a\\times b)\\times c+(b\\times c)\\times a+(c\\times a)\\times b=0.$$ Using this and the fact that it is antisymmetric, you can easily see that $$(a\\times b)\\times c=a\\times(b\\times c)\\iff(c\\times a)\\times b=0.$$ This immediately explains your example where the equality holds and the one where it doesn't. - After thinking about Mariano's answer, I realized I've tricked myself and there is no significance in the above result. The inequality is false when $\\mathbf{A} = \\mathbf{C}$, which in reality is my first case. $(\\mathbf{A} \\times \\mathbf{B}) \\times \\mathbf{A} = [-(\\mathbf{B} \\times \\mathbf{A})] \\times \\mathbf{A} = -\\mathbf{A} \\times [-(\\mathbf{B} \\times \\mathbf{A})] = \\mathbf{A} \\times (\\mathbf{B} \\times \\mathbf{A})$ So my example is testing a case that is not valid for the inequality $(\\mathbf{A} \\times \\mathbf{B}) \\times \\mathbf{C} \\ne \\mathbf{A} \\times (\\mathbf{B} \\times \\mathbf{C})$, which assumes the three vectors are unique. In regards to Mariano's valid", "+q(\\mathbf {v} \\times \\mathbf {B} )}$", "For any vector $\\bar r,\\bar {i} \\times (\\bar {r} \\times \\bar {i}) \\div \\bar {j} \\times (\\bar {r} \\times \\bar {j})+ \\bar k \\times (\\bar {r} \\times {\\bar k})=$ $(a)\\;\\bar {0} \\qquad(b)\\;2 \\bar {r} \\qquad(c)\\;3 \\bar {r} \\qquad (d)\\;4 \\bar {r}$ $(b)\\;2 \\bar {r}$", "+ w\\mathbf{t} + \\mathbf{v}\\times\\mathbf{t}) \\end{array}$ with $\\mathbf{t}=2\\mathbf{v}\\times\\mathbf{u}$ Which is a pretty nice finding! We can then rotate a vector $$\\mathbf{u}$$ by $\\begin{array}{rl} \\mathbf{t}&=2\\mathbf{v}\\times\\mathbf{u}\\\\ R(\\mathbf{u}) &= \\mathbf{u} + w\\mathbf{t} + \\mathbf{v}\\times\\mathbf{t} \\end{array}$", "Math Help - [SOLVED] Find scalar ||kv||=4 1. [SOLVED] Find scalar ||kv||=4 v = (-1,2,5) find scalar k such that ||kv||n = 4 First, what does ||k|| mean? Isn't that the distance between the norm of a vector? or the distance between two points? I've been able to set this up as square root (k-1)^2 + (k2)^2+ (k5)^2 I cannot solve this however. 2. Originally Posted by thekrown v = (-1,2,5) find scalar k such that ||kv||n = 4 First, what does ||k|| mean? Isn't that the distance between the norm of a vector? or the distance between two points? I've been able to set this up as square root (k-1)^2 + (k2)^2+ (k5)^2 I cannot solve this however. What is $n$? 3. Originally Posted by Prove It What is $n$? My mistake, there is no n. It's: v = (-1,2,5) find scalar k such that ||kv|| = 4 4. Originally Posted by thekrown My mistake, there is no n. It's: v = (-1,2,5) find scalar k such that ||kv|| = 4 Ok, this is saying that there is some vector that has magnitude $4$ and goes in the direction of $\\mathbf{v}$. Therefore it is a scalar multiple of $\\mathbf{v}$. So you need to find the magnitude of $\\mathbf{v}$ and then figure out what the scale factor $k$", "be an element in $W$. Then we have $f(x)=f(1-x).$ It follows from this that $cf(x)=cf(1-x),$ and this shows that the scalar product $cf(x)$ is in $W$. Therefore condition 3 holds, and we have proved the subspace criteria for $W$. Thus $W$ is a subspace of the vector space $V$. ##### Linearly Independent vectors $\\mathbf{v}_1, \\mathbf{v}_2$ and Linearly Independent Vectors $A\\mathbf{v}_1, A\\mathbf{v}_2$ for a Nonsingular Matrix Let $\\mathbf{v}_1$ and $\\mathbf{v}_2$ be $2$-dimensional vectors and let $A$ be a $2\\times 2$ matrix. (a) Show that if \\$\\mathbf{v}_1,...", "\\mathbb{Z}_{49}$ 6. $\\mathbb{Z}_4 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_9 \\times \\mathbb{Z}_7 \\times \\mathbb{Z}_7$ 7. $\\mathbb{Z}_4 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_{49}$ 8. $\\mathbb{Z}_4 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_7 \\times \\mathbb{Z}_7$ 9. $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_9 \\times \\mathbb{Z}_{49}$ 10. $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_9 \\times \\mathbb{Z}_7 \\times \\mathbb{Z}_7$ 11. $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_{49}$ 12. $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_3 \\times \\mathbb{Z}_7 \\times \\mathbb{Z}_7$ $\\Box$ 3. Let $V$ be a vector space over a field $F$ and let $T:V\\to V$ be a linear operator such that $T^2=T$. Prove that $V=\\text{ker}(T)\\oplus\\text{im}(T)$. Proof. 😦 4. Let $A$, $B$ be $n\\times n$ matrices with entries in a field $F$ and suppose that $AB=BA$. Prove that if $B$ has an eigenspace of dimension one, then $A$ and $B$ share a common eigenvector. Proof. Since $B$ has an eigenspace of dimension one, let $\\lambda$ be an eigenvalue of $B$ and let $v$ be the one eigenvector corresponding to $\\lambda$. Then we have that $ABv=BAv$ and $ABv=A\\lambda v=\\lambda Av$. Thus, $BAv=\\lambda Av$ which means that $Av$ is an eigenvector of $B$ with corresponding eigenvalue $\\lambda$. But $v$ is the only eigenvector corresponding to $\\lambda$ so $Av=v$. This shows that $v$ is an eigenvector of $A$ with corresponding", "Study guide and 4 practice problems on: ## Length of a scalar vector multiplication For any scalar $c$ and vector $\\mathbf{x}$, $|c \\mathbf{x} | = |c|\\ |\\mathbf{x} |$, where $|c|$ is the absolute value of $c$.", "$\\mathbf j \\times \\mathbf k=\\mathbf i$. Applying it again gives $\\mathbf k \\times \\mathbf i=\\mathbf j$. There's also a rotation (of $180^\\circ$ about $\\mathbf i+\\mathbf j$) that takes $\\mathbf i$ to $\\mathbf j$, $\\mathbf j$ to $\\mathbf i$, and $\\mathbf k$ to $-\\mathbf k$. Applying invariance under this matrix to our equation $\\mathbf i \\times \\mathbf j=\\mathbf k$ gives us $\\mathbf j \\times \\mathbf i=-\\mathbf k$. Similarly we have $\\mathbf k \\times \\mathbf j=-\\mathbf i$ and $\\mathbf i \\times \\mathbf k=-\\mathbf j$. Finally we want to know what $\\mathbf i \\times \\mathbf i$ is. Let $M$ be the $180^\\circ$ rotation about the $\\mathbf k$-axis, as before. Then $$M(\\mathbf i \\times \\mathbf i)=(M\\mathbf i)\\times(M\\mathbf i)=(-\\mathbf i)\\times(-\\mathbf i)=\\mathbf i\\times\\mathbf i$$ so $\\mathbf i \\times \\mathbf i$ is fixed by $M$ and therefore lies on the $\\mathbf k$-axis. But the same argument applied with a rotation about the $\\mathbf j$-axis shows that $\\mathbf i \\times \\mathbf i$ lies on the $\\mathbf j$-axis too! These two axes only intersect at $\\mathbf 0$. So $\\mathbf i \\times \\mathbf i=\\mathbf 0$ and by the same argument $\\mathbf j \\times \\mathbf j=\\mathbf 0$ and $\\mathbf k \\times \\mathbf k=\\mathbf 0$. Now since we know how to cross product any two basis vectors we can calculate the cross product of any two vectors by multiplying out", "## Calculus 8th Edition $(a-b) \\times (a+b)= 2(a \\times b)$ $(a-b) \\times (a+b)=a \\times (a+b)-b \\times (a+b)$ $(a-b) \\times (a+b)=a \\times (a+b)-b \\times (a+b)= a \\times a+ a \\times b - b \\times a- b \\times b$ As we know the cross product of a vector with itself is $0$. $(a-b) \\times (a+b)=a \\times (a+b)-b \\times (a+b)= a \\times a+ a \\times b - b \\times a- b \\times b= 0+ a \\times b - b \\times a- 0$ Note: $-b \\times a =a \\times b$ $=a \\times b + a \\times b$ $= 2(a \\times b)$ Hence, $(a-b) \\times (a+b)= 2(a \\times b)$", "# Scalar Multiplication of Vectors is Distributive over Vector Addition ## Theorem Let $\\mathbf a, \\mathbf b$ be a vector quantities. Let $m$ be a scalar quantity. Then: $m \\paren {\\mathbf a + \\mathbf b} = m \\mathbf a + m \\mathbf b$ ## Proof Let $\\mathbf a = \\vec {OP}$ and $\\mathbf b = \\vec {PQ}$. Then: $\\vec {OQ} = \\mathbf a + \\mathbf b$ Let $P'$ and $Q'$ be points on $OP$ and $OQ$ respectively so that: $OP' : OP = OQ' : OQ = m$ Then $P'Q'$ is parallel to $PQ$ and $m$ times it in length. Thus: $\\vec {P'Q'} = m \\mathbf b$ which shows that: $\\ds m \\paren {\\mathbf a + \\mathbf b}$ $=$ $\\ds \\vec {OQ'}$ $\\ds$ $=$ $\\ds \\vec {OP} + \\vec {OP'}$ $\\ds$ $=$ $\\ds m \\mathbf a + m \\mathbf b$ $\\blacksquare$", "B) \\times C \\quad \\ldots$ (iv) From (iii) and (iv), we get: (A ∩ B) × C = (A × C) ∩ (B × C)", "(B + C) = A \\otimes B + A \\otimes C$ whenever $B$ and $C$ are matrices of the same size. c) If $k$ is a scalar then $k(A \\otimes B) = (kA) \\otimes B = A \\otimes (kB)$. Note that in general we cannot expect that $A \\otimes B$ will equal $B \\otimes A$.", "complex number, or you can use the $$2\\times 2$$ inverse formula (if that is in the scope of the discussion). Try to handle and solve for the $$x$$'s conceptually this system, $$A\\mathbf x = \\mathbf b$$ where $$A$$ is a $$K\\times K$$ matrix, and $$\\mathbf x, \\mathbf b$$ are $$K \\times 1$$ vectors, $$K=10^6$$. Hmm, why not set $$K$$ equal to one trillion? Or a hundred thousand trillion.", "b = np.array([[4, 5, 6], [7, 8, 9]]) np.cross(a, b, axis=1) Out[18]: array([[-3, 6, -3], [ 1, -2, 1]]) ## Problems¶ 1. Given the vectors, $\\mathbf{a}$=(1, 0, 0) and $\\mathbf{b}$=(1, 1, 1), calculate the dot product. 2. Calculate two unit vectors for (2, −2, 3) and (3, −3, 2) and determine an orthogonal vector for the two. 3. Given the vectors $\\mathbf{a}$=(1, 0, 0) and $\\mathbf{b}$=(1, 1, 1), calculate $\\mathbf{a} \\times \\mathbf{b}$ and verify that this vector is orthogonal to vectors $\\mathbf{a}$ and $\\mathbf{b}$. Also, calculate $\\mathbf{b} \\times \\mathbf{a}$ and compare it with $\\mathbf{a} \\times \\mathbf{b}$. If you are new to scalars and vectors, you should solve these problems first by hand and then use Python to check the answers." ]

[ "# 12.3: Vector Identities ## Algebraic Identities \\begin{align} \\mathbf { A } \\cdot ( \\mathbf { B } \\times \\mathbf { C } ) &= \\mathbf { B } \\cdot ( \\mathbf { C } \\times \\mathbf { A } ) = \\mathbf { C } \\cdot ( \\mathbf { A } \\times \\mathbf { B } ) \\\\[5pt] \\mathbf { A } \\times ( \\mathbf { B } \\times \\mathbf { C } ) &= \\mathbf { B } ( \\mathbf { A } \\cdot \\mathbf { C } ) - \\mathbf { C } ( \\mathbf { A } \\cdot \\mathbf { B } ) \\end{align} \\nonumber ## Identities Involving Differential Operators \\begin{align} \\nabla \\cdot ( \\nabla \\times \\mathbf { A } ) &= 0\\\\[5pt] \\nabla \\times ( \\nabla f ) &= 0\\\\[5pt] \\nabla \\times ( f \\mathbf { A } ) &= f ( \\nabla \\times \\mathbf { A } ) + ( \\nabla f ) \\times \\mathbf { A }\\\\[5pt] \\nabla \\cdot ( \\mathbf { A } \\times \\mathbf { B } ) &= \\mathbf { B } \\cdot ( \\nabla \\times \\mathbf { A } ) - \\mathbf { A } \\cdot ( \\nabla \\times \\mathbf { B } )\\\\[5pt] \\nabla \\cdot ( \\nabla f ) &= \\nabla ^ { 2 } f\\\\[5pt]", "# The curl of a special cross product When given two vectors $\\mathbf{A}$ and $\\mathbf{B}$, the curl of the cross product of these two is given by $$\\nabla\\times(\\mathbf{A}\\times\\mathbf{B})=(\\mathbf{B}\\cdot\\nabla)\\mathbf{A}-\\mathbf{B}(\\nabla\\cdot\\mathbf{A})-(\\mathbf{A}\\cdot\\nabla)\\mathbf{B}+\\mathbf{A}(\\nabla\\cdot\\mathbf{B}).$$ Using this relation, we can write $$\\nabla\\times(\\boldsymbol{\\mu}_I\\times\\mathbf{r})=(\\mathbf{r}\\cdot\\nabla)\\boldsymbol{\\mu}_I-\\mathbf{r}(\\nabla\\cdot\\boldsymbol{\\mu}_I)-(\\boldsymbol{\\mu}_I\\cdot\\nabla)\\mathbf{r}+\\boldsymbol{\\mu}_I(\\nabla\\cdot\\mathbf{r}).$$ In a lecture course that I'm reading, it is stated that this can actually be rewritten as $$\\nabla\\times(\\boldsymbol{\\mu}_I\\times\\mathbf{r})=-(\\boldsymbol{\\mu}_I\\cdot\\nabla)\\mathbf{r}+\\boldsymbol{\\mu}_I(\\nabla\\cdot\\mathbf{r}),$$ which means that the first two terms on the right hand side cancel. These can be written out, which results in $$\\begin{array}{l@{\\;}l} (\\mathbf{r}\\cdot\\nabla)\\boldsymbol{\\mu}_I-\\mathbf{r}(\\nabla\\cdot\\boldsymbol{\\mu}_I)&=\\left(y\\partial_y\\mu_x-x\\partial_y\\mu_y+z\\partial_z\\mu_x-x\\partial_z\\mu_z\\right)\\boldsymbol{\\hat{\\textbf{i}}}\\\\ &+\\left(x\\partial_x\\mu_y-y\\partial_x\\mu_x+z\\partial_z\\mu_y-y\\partial_z\\mu_z\\right)\\boldsymbol{\\hat{\\textbf{j}}}\\\\ &+\\left(x\\partial_x\\mu_z+y\\partial_y\\mu_z-z\\partial_x\\mu_x-z\\partial_y\\mu_y\\right)\\boldsymbol{\\hat{\\textbf{k}}}. \\end{array}$$ (Sorry for the ugly ihat, SE doesn't support \\imath for some reason.) In order for this to be zero, each term should be zero. Taking for example the $\\boldsymbol{\\hat{\\textbf{i}}}$ component, this means that $$y\\partial_y\\mu_x-x\\partial_y\\mu_y+z\\partial_z\\mu_x-x\\partial_z\\mu_z=0.$$ In this case, $\\boldsymbol{\\mu}_I$ is proportional to an angular momentum operator. Is this a special case, or is the above equation always true? And if so, why? • Which lecture course are you reading? Author, title, page, etc. Jan 4, 2014 at 17:38 • As another interpretation, it could mean that $\\boldsymbol{\\mu}_I$ is constant in space. Jan 4, 2014 at 18:08 • @Qmechanic Atomic Physics 1 & 2, written by our teacher at the University of Amsterdam for the two courses with the same name. Jan 4, 2014 at 18:30 • @KyleKanos Well, looking at the (probably correct) answer by anecdote, this does seem to", "try taking its dot and cross product with other vectors, and with itself. Product and chain rules Just as with ordinary differentiation, there are product rules for grad, div and curl. • If g is a scalar and v is a vector, then the divergence of gv is ${\\displaystyle \\nabla \\cdot (g\\mathbf {v} )=g\\nabla \\cdot \\mathbf {v} +(\\mathbf {v} \\cdot \\nabla )g}$ the curl of gv is ${\\displaystyle \\nabla \\times (g\\mathbf {v} )=g(\\nabla \\times \\mathbf {v} )+(\\nabla g)\\times \\mathbf {v} }$ • If u and v are both vectors then the gradient of their dot product is ${\\displaystyle \\nabla (\\mathbf {u} \\cdot \\mathbf {v} )=\\mathbf {u} \\times (\\nabla \\times \\mathbf {v} )+\\mathbf {v} \\times (\\nabla \\times \\mathbf {u} )+(\\mathbf {u} \\cdot \\nabla )\\mathbf {v} +(\\mathbf {v} \\cdot \\nabla )\\mathbf {u} }$ the divergence of their cross product is ${\\displaystyle \\nabla \\cdot (\\mathbf {u} \\times \\mathbf {v} )=\\mathbf {v} \\cdot (\\nabla \\times \\mathbf {u} )-\\mathbf {u} \\cdot (\\nabla \\times \\mathbf {v} )}$ the curl of their cross product is ${\\displaystyle \\nabla \\times (\\mathbf {u} \\times \\mathbf {v} )=(\\mathbf {v} \\cdot \\nabla )\\mathbf {u} -(\\mathbf {u} \\cdot \\nabla )\\mathbf {v} +\\mathbf {u} (\\nabla \\cdot \\mathbf {v} )-\\mathbf {v} (\\nabla \\cdot \\mathbf {u} )}$ We can also write chain rules. In the general case, when both functions are vectors and the", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "a \\to b, b \\to c, c \\to a$$ This gives $${\\bf A} = \\frac{{\\bf b}\\times {\\bf c}}{{\\bf a}\\cdot({\\bf b}\\times {\\bf c})} \\quad {\\bf B} = \\frac{{\\bf c}\\times {\\bf a}}{{\\bf b}\\cdot({\\bf c}\\times {\\bf a})} \\quad {\\bf C} = \\frac{{\\bf a}\\times {\\bf b}}{{\\bf c}\\cdot({\\bf a}\\times {\\bf b})}$$ and $${\\bf a} = \\frac{{\\bf B}\\times {\\bf C}}{{\\bf A}\\cdot({\\bf B}\\times {\\bf C})} \\quad {\\bf b} = \\frac{{\\bf C}\\times {\\bf A}}{{\\bf B}\\cdot({\\bf C}\\times {\\bf A})} \\quad {\\bf c} = \\frac{{\\bf A}\\times {\\bf B}}{{\\bf C}\\cdot({\\bf A}\\times {\\bf B})}$$ [..] the \"crystallographer's\" definition, comes from defining the reciprocal lattice to be $e^{2 \\pi i\\mathbf{K}\\cdot\\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be $\\mathbf{b_{1}}=\\frac{\\mathbf{a_{2}} \\times \\mathbf{a_{3}}}{\\mathbf{a_{1}} \\cdot (\\mathbf{a_{2}} \\times \\mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\\mathbf{b_{1}}$ is just the reciprocal magnitude of $\\mathbf{a_{1}}$ in the direction of $\\mathbf{a_{2}} \\times \\mathbf{a_{3}}$, dropping the factor of $2 \\pi$. Checking the magnitude: $$\\lVert{\\bf A}\\rVert = \\frac{\\lVert {\\bf b} \\times{\\bf c}\\rVert}{\\lVert{\\bf a}\\rVert\\lVert{\\bf b}\\times {\\bf c}\\rVert\\cos\\angle({\\bf a}, {\\bf b} \\times{\\bf c})} = \\frac{1}{\\lVert{\\bf a}\\rVert ({\\bf e_a} \\cdot {\\bf e}_{{\\bf b} \\times{\\bf c}})}$$ Solving the question: Using the \"bac-cab\" rule ${\\bf a}\\times({\\bf b}\\times{\\bf c}) = {\\bf b}({\\bf a}\\cdot {\\bf c}) - {\\bf c}({\\bf a}\\cdot {\\bf b})$ we go for the nominator of ${\\bf B}\\times{\\bf C}$: $$({\\bf c}\\times", "\\cdot \\mathbf{A}_\\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum. Momentum Now move on to (16). For the factors other than $\\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a \\mathbf{a} (\\mathbf{k} \\wedge \\mathbf{c}) }}\\right\\rangle&=\\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) - \\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) \\\\ &=\\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}) - \\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}),\\end{aligned} and the other \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a (\\mathbf{k} \\wedge \\mathbf{c}) \\mathbf{a} }}\\right\\rangle&=\\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) - \\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) \\\\ &=\\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}).\\end{aligned} The momentum $\\mathbf{P} = \\int T(\\gamma^0) \\wedge \\gamma_0$ in this volume follows by computation of \\begin{aligned}&\\sigma_a \\left\\langle{{ \\sigma_a\\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} - i \\mathbf{k} \\wedge {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\right) \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} + i \\mathbf{k} \\wedge \\mathbf{A}_\\mathbf{k} \\right) }}\\right\\rangle \\\\ &= i \\mathbf{A}_\\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{k} \\right) - i \\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{A}_\\mathbf{k} \\right) \\\\ &- i \\mathbf{k} \\left( \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} \\right) \\cdot {{\\mathbf{A}_\\mathbf{k}}}^{*} \\right) + i {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\left( \\left(", "unnecessary and unnecessarily requires n vectors $\\mathbf{c}_1, \\ldots, \\mathbf{c}_{k}, \\ldots, \\mathbf{c}_{n}$ at a time. Solving each equation is independent in this case. This has been shown to clarify the usage, as far as what not to do, unless one has an unusual need to solve a particular vector $\\mathbf{a}_{k}$. Instead, the following can be done in the case of projecting vectors $\\mathbf{c}_{k}$ onto a new arbitrary basis $\\mathbf{x}_{k}$. ### Projecting a vector onto an arbitrary basis. Projecting any vector $\\mathbf{c}$ onto a new arbitrary basis $\\mathbf{x}_1, \\ldots, \\mathbf{x}_k, \\ldots, \\mathbf{x}_n$ as $\\begin{array}{rcl} \\mathbf{c} & = & c_1 \\mathbf{e}_1 + \\cdots + c_k \\mathbf{e}_k + \\cdots + c_n \\mathbf{e}_n\\\\ & = & a_1 \\mathbf{x}_1 + \\cdots + a_k \\mathbf{x}_k + \\cdots + a_n \\mathbf{x}_n\\end{array}$ where each $\\mathbf{x}_k$ is written in the form $\\mathbf{x}_k = x_{k 1} \\mathbf{e}_1 + x_{k 2} \\mathbf{e}_2 + \\cdots + x_{k k} \\mathbf{e}_k + \\cdots + x_{k n} \\mathbf{e}_n$ is a system of n scalar equations in n unknown coordinates $a_k$ $\\begin{array}{rcl} a_1 x_{11} + \\cdots + a_k x_{k 1} + \\cdots + a_n x_{n 1} & = & c_1\\\\ & \\vdots & \\\\ a_1 x_{1 k} + \\cdots + a_k x_{k k} + \\cdots + a_n x_{n k} & = & c_k\\\\ & \\vdots & \\\\ a_1 x_{1 n} + \\cdots + a_k x_{k", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "right order. So let’s look at our matrix product again: we wan’t the i-th column of the matrix product $\\mathbf{B}\\mathbf{A}$, so we look at $(\\mathbf{B}\\mathbf{A})\\mathbf{e}_i = \\mathbf{B}(\\mathbf{A}\\mathbf{e}_i) = \\mathbf{B} \\mathbf{a}_i$ exactly as claimed. As always, there’s a corresponding construction using row vectors. Using the dual basis of linear functionals (note no transpose this time!) $\\mathbf{e}^1 = \\begin{pmatrix} 1 & 0 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^2 = \\begin{pmatrix} 0 & 1 & \\cdots & 0 \\end{pmatrix}$ $\\mathbf{e}^m = \\begin{pmatrix} 0 & 0 & \\cdots & 1 \\end{pmatrix}$ we can disassemble a matrix into its rows: $\\mathbf{A} = \\begin{pmatrix} \\mathbf{a}^1 \\\\ \\mathbf{a}^2 \\\\ \\vdots \\\\ \\mathbf{a}^m \\end{pmatrix} = \\begin{pmatrix} \\mathbf{e}^1 \\mathbf{A} \\\\ \\mathbf{e}^2 \\mathbf{A} \\\\ \\vdots \\\\ \\mathbf{e}^m \\mathbf{A} \\end{pmatrix}$ and since $\\mathbf{e}^i (\\mathbf{A} \\mathbf{B}) = (\\mathbf{e}^i \\mathbf{A}) \\mathbf{B} = \\mathbf{a}^i \\mathbf{B}$, we can write matrix products in terms of what happens to the row vectors too (though this time, we’re expanding in terms of the row vectors of the first not second factor): $\\mathbf{A} \\mathbf{B} = \\begin{pmatrix} \\mathbf{a}^1 \\mathbf{B} \\\\ \\mathbf{a}^2 \\mathbf{B} \\\\ \\vdots \\\\ \\mathbf{a}^m \\mathbf{B} \\end{pmatrix}$. ### Gluing it back together Above, the trick was to write the “slicing” step as a matrix product with one of the basis vectors – $\\mathbf{A} \\mathbf{e}_i$ and so forth. We’ll soon need to deal with the gluing step too,", "### Vector Indentities Triple Products $$\\vec{A}\\cdot(\\vec{B}\\times\\vec{C})=\\vec{B}\\cdot(\\vec{C}\\times\\vec{A})=\\vec{C}\\cdot(\\vec{A}\\times\\vec{B})$$$$\\vec{A}\\times(\\vec{B}\\times\\vec{C})=\\vec{B}(\\vec{A}\\cdot\\vec{C})-\\vec{C}(\\vec{A}\\cdot\\vec{B})$$ Product Rules $$\\nabla(fg) = f(\\nabla g) + g (\\nabla f)$$ $$\\nabla(\\vec{A}\\cdot\\vec{B})=\\vec{A}\\times(\\nabla\\times\\vec{B})+\\vec{B}\\times(\\nabla\\times\\vec{A})+(\\vec{A}\\cdot\\nabla)\\vec{B}+(\\vec{B}\\cdot\\nabla)\\vec{A}$$ $$\\nabla\\cdot(f\\vec{A}) = f(\\nabla\\cdot\\vec{A}) + \\vec{A} \\cdot (\\nabla f)$$ $$\\nabla\\cdot(\\vec{A}\\times\\vec{B})=\\vec{B}\\cdot(\\nabla\\times\\vec{A})-\\vec{A}\\cdot(\\nabla\\times\\vec{B})$$ $$\\nabla\\times(f\\vec{A}) = f(\\nabla\\times\\vec{A}) - \\vec{A} \\times(\\nabla f)$$ $$\\nabla\\times(\\vec{A}\\times\\vec{B})=(\\vec{B}\\cdot\\nabla)\\vec{A}-(\\vec{A}\\cdot\\nabla)\\vec{B}+\\vec{A}(\\nabla\\cdot\\vec{B})-\\vec{B}(\\nabla\\cdot\\vec{A})$$ Second Derivatives $$\\nabla\\cdot(\\nabla\\times\\vec{A})=0$$ $$\\nabla\\times(\\nabla f)=0$$ $$\\nabla\\times(\\nabla\\times\\vec{A})=\\nabla(\\nabla\\cdot\\vec{A})-\\nabla^{2}\\vec{A}$$ Fundamental Theorems Gradient Theorem: $$\\displaystyle \\int_a^b(\\nabla f)\\cdot d\\vec{l} = f(b) - f(a)$$ Divergence Theorem: $$\\int_V(\\nabla\\cdot\\vec{A})d\\vec{\\tau}=\\oint_{S} \\! \\vec{A}\\cdot d\\vec{a}$$ Curl Theorem: $$\\int_S(\\nabla\\times\\vec{A})\\cdot d\\vec{a}=\\oint_{L} \\! \\vec{A}\\cdot d\\vec{l}$$", "to the boundary and therefore any cross product involving $$\\hat{\\bf n}$$ will be perpendicular to $$\\hat{\\bf n}$$. The corresponding component of the current density is $$\\hat{\\bf n} \\times \\left( \\hat{\\bf n} \\times {\\bf J}_s \\right)$$, so Equation \\ref{m0023_eBC1} may be equivalently written as follows: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s \\label{m0023_eBC2}$ Applying a vector identity ($$\\mathbf{A} \\times(\\mathbf{B} \\times \\mathbf{C})=\\mathbf{B}(\\mathbf{A} \\cdot \\mathbf{C})-\\mathbf{C}(\\mathbf{A} \\cdot \\mathbf{B})$$ of Section B3) to the right side of Equation \\ref{m0023_eBC2} we obtain: \\begin{aligned} \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s &= \\hat{\\bf n} \\left(\\hat{\\bf n}\\cdot{\\bf J}_s\\right) - {\\bf J}_s\\left(\\hat{\\bf n}\\cdot\\hat{\\bf n}\\right) \\nonumber \\\\ &= \\hat{\\bf n} \\left(0\\right) - {\\bf J}_s\\left(1\\right) \\nonumber \\\\ &= -{\\bf J}_s\\end{aligned} Therefore: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = -{\\bf J}_s$ The minus sign on the right can be eliminated by swapping $${\\bf H}_2$$ and $${\\bf H}_1$$ on the left, yielding $\\boxed{ \\hat{\\bf n} \\times \\left( {\\bf H}_1 -{\\bf H}_2 \\right) = {\\bf J}_s }$ This is the form in which the boundary condition is most commonly expressed. It is worth noting what this means for the magnetic field intensity $${\\bf B}$$. Since $${\\bf B}=\\mu{\\bf H}$$: In the absence of surface current, the tangential component of $${\\bf B}$$ across the boundary between two material regions", "\\mathbf{A}_\\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum. ## Momentum Now move on to (16). For the factors other than $\\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a \\mathbf{a} (\\mathbf{k} \\wedge \\mathbf{c}) }}\\right\\rangle&=\\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) - \\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) \\\\ &=\\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}) - \\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}),\\end{aligned} and the other \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a (\\mathbf{k} \\wedge \\mathbf{c}) \\mathbf{a} }}\\right\\rangle&=\\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) - \\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) \\\\ &=\\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}).\\end{aligned} The momentum $\\mathbf{P} = \\int T(\\gamma^0) \\wedge \\gamma_0$ in this volume follows by computation of \\begin{aligned}&\\sigma_a \\left\\langle{{ \\sigma_a\\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} - i \\mathbf{k} \\wedge {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\right) \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} + i \\mathbf{k} \\wedge \\mathbf{A}_\\mathbf{k} \\right) }}\\right\\rangle \\\\ &= i \\mathbf{A}_\\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{k} \\right) - i \\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{A}_\\mathbf{k} \\right) \\\\ &- i \\mathbf{k} \\left( \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} \\right) \\cdot {{\\mathbf{A}_\\mathbf{k}}}^{*} \\right) + i {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\left( \\left(", "v})^2$$ This, along with ${\\bf u}\\cdot{\\bf v}=|{\\bf u}||{\\bf v}|\\cos\\theta$, will lead to \\eqref{eq:crossprod3}. From \\eqref{eq:crossprod3}, we can easily see that two nonzero vectors ${\\bf u}$ and ${\\bf v}$ are parallel if and only if ${\\bf u}\\times{\\bf v}=0$. The standard basis vectors ${\\bf i}$, ${\\bf j}$, ${\\bf k}$ satisfy the following cross products: $${\\bf i}\\times{\\bf j}={\\bf k},\\ {\\bf j}\\times{\\bf k}={\\bf i},\\ {\\bf k}\\times{\\bf i}={\\bf j}$$ The following theorem summarizes the properties of the cross product. Theorem. Let ${\\bf u}$, ${\\bf v}$, and ${\\bf w}$ be vectors and $c$ a scalar. Then 1. ${\\bf u}\\times{\\bf v}=-{\\bf v}\\times{\\bf u}$ 2. $(c{\\bf u})\\times{\\bf v}=c({\\bf u}\\times{\\bf v})={\\bf u}\\times(c{\\bf v})$ 3. ${\\bf u}\\times({\\bf v}+{\\bf w})={\\bf u}\\times{\\bf v}+{\\bf u}\\times{\\bf w}$ 4. $({\\bf u}+{\\bf v})\\times{\\bf w}={\\bf u}\\times{\\bf w}+{\\bf v}\\times{\\bf w}$ 5. ${\\bf u}\\cdot({\\bf v}\\times{\\bf w})=({\\bf u}\\times{\\bf v})\\cdot{\\bf w}$ 6. ${\\bf u}\\times({\\bf v}\\times{\\bf w})=({\\bf u}\\cdot{\\bf w}){\\bf v}-({\\bf u}\\cdot{\\bf v}){\\bf w}$ The products in 5 and 6 are called, respectively, a scalar triple product and a vector triple product. From Figure 3, we see that $$\\label{eq:areaparallelogram}|{\\bf u}\\times{\\bf v}|$$ is equal to the area of the parallelogram determined by ${\\bf u}$ and ${\\bf v}$. Example. Find a vector perpendicular to the plane that passes through the points $P(1,4,6)$, $Q(-2,5,-1)$, and $R(1,-1,1)$. Solution. The vectors $\\overrightarrow{PQ}=(-3,1,-7)$ and $\\overrightarrow{PR}=(0,-5,-5)$ lie in the plane through $P,Q,R$. So the cross product $$\\overrightarrow{PQ}\\times\\overrightarrow{PR}=(-40,-15,15)$$ is", "## Identity for Product of a Function With a Vector Field Satisfying Poisson's Equations Theorem If $f$ is a function and $\\mathbf{v}$ is a vector field satisfying $\\nabla^2 (f \\mathbf{v}) =0$ and $\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))=0$ then $\\mathbf{\\nabla} ((\\mathbf{\\nabla} f) \\cdot \\mathbf{v} + f \\mathbf{\\nabla} \\cdot (\\mathbf{v})) =0$ Proof $\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))= \\mathbf{\\nabla} (\\mathbf{\\nabla} \\cdot (f \\mathbf{v}))- \\nabla^2 (f \\mathbf{v})$ $\\nabla^2 (f \\mathbf{v}) =\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))=0$ so $\\mathbf{\\nabla} (\\mathbf{\\nabla} \\cdot (f \\mathbf{v} ))=0$ Now use the identity $\\mathbf{\\nabla} \\cdot (f \\mathbf{v}) =(\\mathbf{\\nabla} f ) \\cdot \\mathbf{v} + f ( \\mathbf{\\nabla} \\cdot \\mathbf{v})$ to obtain $\\mathbf{\\nabla} ((\\mathbf{\\nabla} f ) \\cdot \\mathbf{v} + f \\mathbf{\\nabla} \\cdot \\mathbf{v}) =0$", "}}{\\partial t}}\\end{aligned}}} • {\\displaystyle {\\begin{aligned}{\\mathbf {E} }\\cdot {\\mathbf {J} }&={\\mathbf {E} }\\cdot \\left[{\\frac {1}{\\mu _{0}}}(\\nabla \\times {\\mathbf {B} })-\\epsilon _{0}{\\frac {\\partial {\\mathbf {E} }}{\\partial t}}\\right]\\\\&={\\frac {1}{\\mu _{0}}}({\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} }))-\\epsilon _{0}{\\mathbf {E} }\\cdot {\\frac {\\partial {\\mathbf {E} }}{\\partial t}}\\end{aligned}}} 5. 5 Use a vector calculus identity to write an expression with a term where both fields are in the cross product. This apparent complication is solved when we invoke Faraday’s law ${\\displaystyle \\nabla \\times {\\mathbf {E} }=-{\\frac {\\partial {\\mathbf {B} }}{\\partial t}}.}$ • ${\\displaystyle \\nabla \\cdot ({\\mathbf {E} }\\times {\\mathbf {B} })={\\mathbf {B} }\\cdot (\\nabla \\times {\\mathbf {E} })-{\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} })}$ • ${\\displaystyle {\\mathbf {E} }\\cdot (\\nabla \\times {\\mathbf {B} })=-{\\mathbf {B} }\\cdot {\\frac {\\partial {\\mathbf {B} }}{\\partial t}}-\\nabla \\cdot ({\\mathbf {E} }\\times {\\mathbf {B} })}$ 6. 6 Simplify the expressions containing the dot product of a field with its time derivative. The extra ½ factor stems from the recognition of the product rule. Confirm this by taking the derivative of ${\\displaystyle {\\mathbf {B} }\\cdot {\\mathbf {B} }.}$ • ${\\displaystyle {\\mathbf {B} }\\cdot {\\frac {\\partial {\\mathbf {B} }}{\\partial t}}={\\frac {\\partial }{\\partial t}}\\left({\\frac {1}{2}}B^{2}\\right)}$ • ${\\displaystyle {\\mathbf {E} }\\cdot {\\frac {\\partial {\\mathbf {E} }}{\\partial t}}={\\frac {\\partial }{\\partial t}}\\left({\\frac {1}{2}}E^{2}\\right)}$ 7. 7 Substitute into the equation in step 4. Rewrite the", "#### Solutions Collecting From Web of \"prove $\\frac{1}{2}\\mathbf{\\nabla (u \\cdot u) = u \\times (\\nabla \\times u ) + (u \\cdot \\nabla)u}$ using index notation\" So, your first step is indeed correct, but the one you are not sure about is wrong. You are probably confused by your own notation, I prefer to write $\\partial_i=\\frac{\\partial}{\\partial x_i}$ instead of $\\partial x_i$. And remember that repeated indices are being summed over, so e.g. $u_ju_j=\\mathbf{u}\\cdot\\mathbf{u}$. Given that, you have u_j\\partial _i u_j-u_j\\partial_j u_i =(1/2)\\partial_i(u_j u_j)-(\\mathbf{u}\\cdot \\nabla)u_i =(1/2) \\partial_i (\\mathbf{u}\\cdot\\mathbf{u})-(\\mathbf{u}\\cdot \\nabla)u_i, which (taking into account your earlier work) is the $i$-th component of the equation \\mathbf{u}\\times(\\nabla\\times \\mathbf{u})=\\frac{1}{2}\\nabla(\\mathbf{u}\\cdot\\mathbf{u})-(\\mathbf{u}\\cdot \\nabla)\\mathbf{u} Hint: Let ${\\bf e}_i$ be orthonormal unit vectors then in index notation $${\\bf u} = u_i {\\bf e}_i$$ The dot product between two vectors can then be written $${\\bf a}\\cdot {\\bf b} = a_ib_i$$ in particular when ${\\bf a} = {\\bf \\nabla}$ we get ${\\bf \\nabla}\\cdot {\\bf b} = \\partial_ib_i$. The curl can be written $${\\bf a}\\times {\\bf b} = \\epsilon_{kij}a_i b_j {\\bf e}_k$$ where $\\epsilon_{ijk}$ is the Levi-Civita symbol. Using this then the first term on the right hand side in your equation becomes $${\\bf u}\\times(\\nabla\\times {\\bf u}) = \\epsilon_{kij}u_i(\\nabla\\times {\\bf u})_j{\\bf e}_k = \\epsilon_{kij}\\epsilon_{jlm}u_i\\partial_lu_m{\\bf e}_k$$ Now repeat this for the other terms and compare. You will have to use the identity (the", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "Now, let us look at the squared length of this vector, \\begin{align} \\ln{\\vc{b}}^2 &= (\\vc{u}\\cdot \\vc{u})^2 (\\vc{v}\\cdot\\vc{v}) -2(\\vc{u}\\cdot \\vc{u})(\\vc{u}\\cdot \\vc{v})^2 + (\\vc{u}\\cdot \\vc{v})^2(\\vc{u} \\cdot \\vc{u}) \\\\ &=(\\vc{u}\\cdot \\vc{u})^2 (\\vc{v}\\cdot\\vc{v}) -(\\vc{u}\\cdot \\vc{u})(\\vc{u}\\cdot \\vc{v})^2 \\\\ &= \\ln{\\vc{u}}^4\\ln{\\vc{v}}^2 - \\ln{\\vc{u}}^2(\\vc{u}\\cdot \\vc{v})^2, \\end{align} (4.32) which is the same as Equation (4.29), which proves the theorem. $\\square$ Next, the full vector triple product, sometimes called Lagrange's formula, or triple product expansion, is presented in theorem below. Theorem 4.5: Vector Triple Product The vector triple product of $\\vc{u}$, $\\vc{v}$, and $\\vc{w}$ is $$\\vc{u} \\times (\\vc{v} \\times \\vc{w}) = (\\vc{u} \\cdot \\vc{w})\\vc{v} - (\\vc{u} \\cdot \\vc{v})\\vc{w}.$$ (4.33) We assume that the vectors $\\vc{v}$ and $\\vc{w}$ not are parallel, because otherwise the product will be the zero vector. Hence, $\\vc{u}$ can be expressed in terms of $\\vc{v}$ and $\\vc{w}$: $$\\vc{u} = a\\vc{v} + b\\vc{w} + c(\\vc{v} \\times \\vc{w}),$$ (4.34) for some values, $a$, $b$, and $c$. Let us simplify the expression for the dot products, $\\vc{u}\\cdot \\vc{v}$ and $\\vc{u}\\cdot \\vc{w}$, \\begin{align} \\vc{u}\\cdot \\vc{v} & = (a\\vc{v} + b\\vc{w} + c(\\vc{v} \\times \\vc{w}))\\cdot \\vc{v} \\\\ & = a(\\vc{v}\\cdot\\vc{v}) + b(\\vc{w}\\cdot \\vc{v}) + c\\underbrace{(\\vc{v} \\times \\vc{w})\\cdot \\vc{v}}_{=0}\\\\ & = a(\\vc{v}\\cdot\\vc{v}) + b(\\vc{w}\\cdot \\vc{v}) . \\end{align} (4.35) In the same way, we get $$\\vc{u}\\cdot \\vc{w} = a(\\vc{v}\\cdot\\vc{w}) + b(\\vc{w}\\cdot \\vc{w}).$$ (4.36) Now, let us use Equation (4.34) and Theorem", "A \\\\ &= (\\mathbf{A} \\cdot \\nabla) \\mathbf{A} + \\mathbf{A} \\times (\\nabla \\times \\mathbf{A}) \\ . \\end{align} Vector cross product $\\nabla \\cdot (\\mathbf{A} \\times \\mathbf{B}) = (\\nabla \\times \\mathbf{A}) \\cdot \\mathbf{B} - \\mathbf{A} \\cdot (\\nabla \\times \\mathbf{B})$ \\begin{align}\\nabla \\times (\\mathbf{A} \\times \\mathbf{B}) &= \\mathbf{A} (\\nabla \\cdot \\mathbf{B}) - \\mathbf{B} (\\nabla \\cdot \\mathbf{A}) + (\\mathbf{B} \\cdot \\nabla) \\mathbf{A} - (\\mathbf{A} \\cdot \\nabla) \\mathbf{B} \\\\ &= (\\nabla \\cdot \\mathbf{B} + \\mathbf{B} \\cdot \\nabla)\\mathbf{A} -(\\nabla \\cdot \\mathbf{A} + \\mathbf{A} \\cdot \\nabla )\\mathbf{B} \\\\ &= \\nabla \\cdot (\\mathbf{B} \\mathbf{A}^\\mathrm{T}) - \\nabla \\cdot (\\mathbf{A} \\mathbf{B}^\\mathrm{T}) \\\\ &= \\nabla \\cdot (\\mathbf{B} \\mathbf{A}^\\mathrm{T} - \\mathbf{A} \\mathbf{B}^\\mathrm{T}) \\end{align} Second derivatives Curl of the gradient The curl of the gradient of any scalar field $\\ \\phi$ is always the zero vector: $\\nabla \\times ( \\nabla \\phi ) = \\mathbf{0}$ Divergence of the curl The divergence of the curl of any vector field A is always zero: $\\nabla \\cdot ( \\nabla \\times \\mathbf{A} ) = 0$ Divergence of the gradient The Laplacian of a scalar field is defined as the divergence of the gradient: $\\nabla^2 \\psi = \\nabla \\cdot (\\nabla \\psi)$ Note that the result is a scalar quantity. Curl of the curl $\\nabla \\times \\left( \\nabla \\times \\mathbf{A} \\right) = \\nabla(\\nabla \\cdot \\mathbf{A}) - \\nabla^{2}\\mathbf{A}$ Here,∇2 is the vector Laplacian operating on the vector field A. Summary of" ]

[ "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "# Problem Set 2 1. Easy Let $$\\mathbf{u},\\mathbf{v}$$ be two vectors in $$\\mathbb{R}^{3}$$. Compute $$\\mathbf{u}\\times\\mathbf{v}$$, $$\\mathbf{u}\\cdot\\mathbf{v}$$ for the following choices of $$\\mathbf{u}$$ and $$\\mathbf{v}$$, and state whether or not $$\\mathbf{u}$$ and $$\\mathbf{v}$$ are either orthogonal or colinear. 1. $$\\mathbf{u} = (6,0,-2)$$, $$\\mathbf{v} = (0,8,0)$$ 2. $$\\mathbf{u} = (1,1,-1)$$, $$\\mathbf{v} = (2,4,6)$$ 2. Easy Let $$\\vec{a},\\vec{b},\\vec{c},\\vec{d} \\in \\mathbb{R}^{3}$$ be vectors. Determine which of the following expressions are meaningless: 1. $$(\\vec{a}\\cdot\\vec{b})\\cdot\\vec{c}$$ 2. $$\\vert\\vec{a}\\vert(\\vec{b}\\cdot\\vec{c})$$ 3. $$\\vec{a}\\cdot(\\vec{b}+\\vec{c})$$ 4. $$(\\vec{a}\\cdot\\vec{b})\\vec{c}$$ 5. $$\\vec{a}\\cdot\\vec{b}+c$$ 6. $$\\vec{a}\\cdot(\\vec{b}\\times\\vec{c})$$ 7. $$(\\vec{a}\\cdot\\vec{b})\\cdot(\\vec{c}\\cdot\\vec{d})$$ 8. $$(\\vec{a}\\times\\vec{b})\\cdot(\\vec{c}\\times\\vec{d})$$ 3. Easy If $$S$$ is not closed, then there exists $$x \\in \\overline{S}$$ but $$x \\notin S$$ 4. Medium Prove that for any $$x,y \\in \\mathbb{R}^{n}$$, $$\\vert x - y \\vert \\geq \\big\\vert \\vert x \\vert - \\vert y \\vert \\big\\vert$$. This is commonly called, the reverse triangle inequality''. It is extremely useful when you want to prove inequalities like $$\\vert (x-a)^{-1} \\vert \\leq M$$ for $$x$$ in some fixed closed set. 5. Medium If $$U$$ and $$V$$ are open (resp. closed) then $$U\\cup V$$ is open (resp. $$U\\cap V$$ is closed). If $$\\left\\{U_{i}\\right\\}_{i\\in I}$$ is a countable collection of open sets, must $$\\bigcap_{i\\in I} U_{i}$$ be open? Provide a proof or counterexample. Similarly, if $$\\left\\{A_{i}\\right\\}_{i\\in I}$$ is an infinite collection of closed sets, must $$\\bigcap_{i\\in I} A_{i}$$ be closed? 6. Medium Prove that the", "try taking its dot and cross product with other vectors, and with itself. Product and chain rules Just as with ordinary differentiation, there are product rules for grad, div and curl. • If g is a scalar and v is a vector, then the divergence of gv is ${\\displaystyle \\nabla \\cdot (g\\mathbf {v} )=g\\nabla \\cdot \\mathbf {v} +(\\mathbf {v} \\cdot \\nabla )g}$ the curl of gv is ${\\displaystyle \\nabla \\times (g\\mathbf {v} )=g(\\nabla \\times \\mathbf {v} )+(\\nabla g)\\times \\mathbf {v} }$ • If u and v are both vectors then the gradient of their dot product is ${\\displaystyle \\nabla (\\mathbf {u} \\cdot \\mathbf {v} )=\\mathbf {u} \\times (\\nabla \\times \\mathbf {v} )+\\mathbf {v} \\times (\\nabla \\times \\mathbf {u} )+(\\mathbf {u} \\cdot \\nabla )\\mathbf {v} +(\\mathbf {v} \\cdot \\nabla )\\mathbf {u} }$ the divergence of their cross product is ${\\displaystyle \\nabla \\cdot (\\mathbf {u} \\times \\mathbf {v} )=\\mathbf {v} \\cdot (\\nabla \\times \\mathbf {u} )-\\mathbf {u} \\cdot (\\nabla \\times \\mathbf {v} )}$ the curl of their cross product is ${\\displaystyle \\nabla \\times (\\mathbf {u} \\times \\mathbf {v} )=(\\mathbf {v} \\cdot \\nabla )\\mathbf {u} -(\\mathbf {u} \\cdot \\nabla )\\mathbf {v} +\\mathbf {u} (\\nabla \\cdot \\mathbf {v} )-\\mathbf {v} (\\nabla \\cdot \\mathbf {u} )}$ We can also write chain rules. In the general case, when both functions are vectors and the", "a \\to b, b \\to c, c \\to a$$ This gives $${\\bf A} = \\frac{{\\bf b}\\times {\\bf c}}{{\\bf a}\\cdot({\\bf b}\\times {\\bf c})} \\quad {\\bf B} = \\frac{{\\bf c}\\times {\\bf a}}{{\\bf b}\\cdot({\\bf c}\\times {\\bf a})} \\quad {\\bf C} = \\frac{{\\bf a}\\times {\\bf b}}{{\\bf c}\\cdot({\\bf a}\\times {\\bf b})}$$ and $${\\bf a} = \\frac{{\\bf B}\\times {\\bf C}}{{\\bf A}\\cdot({\\bf B}\\times {\\bf C})} \\quad {\\bf b} = \\frac{{\\bf C}\\times {\\bf A}}{{\\bf B}\\cdot({\\bf C}\\times {\\bf A})} \\quad {\\bf c} = \\frac{{\\bf A}\\times {\\bf B}}{{\\bf C}\\cdot({\\bf A}\\times {\\bf B})}$$ [..] the \"crystallographer's\" definition, comes from defining the reciprocal lattice to be $e^{2 \\pi i\\mathbf{K}\\cdot\\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be $\\mathbf{b_{1}}=\\frac{\\mathbf{a_{2}} \\times \\mathbf{a_{3}}}{\\mathbf{a_{1}} \\cdot (\\mathbf{a_{2}} \\times \\mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\\mathbf{b_{1}}$ is just the reciprocal magnitude of $\\mathbf{a_{1}}$ in the direction of $\\mathbf{a_{2}} \\times \\mathbf{a_{3}}$, dropping the factor of $2 \\pi$. Checking the magnitude: $$\\lVert{\\bf A}\\rVert = \\frac{\\lVert {\\bf b} \\times{\\bf c}\\rVert}{\\lVert{\\bf a}\\rVert\\lVert{\\bf b}\\times {\\bf c}\\rVert\\cos\\angle({\\bf a}, {\\bf b} \\times{\\bf c})} = \\frac{1}{\\lVert{\\bf a}\\rVert ({\\bf e_a} \\cdot {\\bf e}_{{\\bf b} \\times{\\bf c}})}$$ Solving the question: Using the \"bac-cab\" rule ${\\bf a}\\times({\\bf b}\\times{\\bf c}) = {\\bf b}({\\bf a}\\cdot {\\bf c}) - {\\bf c}({\\bf a}\\cdot {\\bf b})$ we go for the nominator of ${\\bf B}\\times{\\bf C}$: $$({\\bf c}\\times", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "unnecessary and unnecessarily requires n vectors $\\mathbf{c}_1, \\ldots, \\mathbf{c}_{k}, \\ldots, \\mathbf{c}_{n}$ at a time. Solving each equation is independent in this case. This has been shown to clarify the usage, as far as what not to do, unless one has an unusual need to solve a particular vector $\\mathbf{a}_{k}$. Instead, the following can be done in the case of projecting vectors $\\mathbf{c}_{k}$ onto a new arbitrary basis $\\mathbf{x}_{k}$. ### Projecting a vector onto an arbitrary basis. Projecting any vector $\\mathbf{c}$ onto a new arbitrary basis $\\mathbf{x}_1, \\ldots, \\mathbf{x}_k, \\ldots, \\mathbf{x}_n$ as $\\begin{array}{rcl} \\mathbf{c} & = & c_1 \\mathbf{e}_1 + \\cdots + c_k \\mathbf{e}_k + \\cdots + c_n \\mathbf{e}_n\\\\ & = & a_1 \\mathbf{x}_1 + \\cdots + a_k \\mathbf{x}_k + \\cdots + a_n \\mathbf{x}_n\\end{array}$ where each $\\mathbf{x}_k$ is written in the form $\\mathbf{x}_k = x_{k 1} \\mathbf{e}_1 + x_{k 2} \\mathbf{e}_2 + \\cdots + x_{k k} \\mathbf{e}_k + \\cdots + x_{k n} \\mathbf{e}_n$ is a system of n scalar equations in n unknown coordinates $a_k$ $\\begin{array}{rcl} a_1 x_{11} + \\cdots + a_k x_{k 1} + \\cdots + a_n x_{n 1} & = & c_1\\\\ & \\vdots & \\\\ a_1 x_{1 k} + \\cdots + a_k x_{k k} + \\cdots + a_n x_{n k} & = & c_k\\\\ & \\vdots & \\\\ a_1 x_{1 n} + \\cdots + a_k x_{k", "# 12.3: Vector Identities ## Algebraic Identities \\begin{align} \\mathbf { A } \\cdot ( \\mathbf { B } \\times \\mathbf { C } ) &= \\mathbf { B } \\cdot ( \\mathbf { C } \\times \\mathbf { A } ) = \\mathbf { C } \\cdot ( \\mathbf { A } \\times \\mathbf { B } ) \\\\[5pt] \\mathbf { A } \\times ( \\mathbf { B } \\times \\mathbf { C } ) &= \\mathbf { B } ( \\mathbf { A } \\cdot \\mathbf { C } ) - \\mathbf { C } ( \\mathbf { A } \\cdot \\mathbf { B } ) \\end{align} \\nonumber ## Identities Involving Differential Operators \\begin{align} \\nabla \\cdot ( \\nabla \\times \\mathbf { A } ) &= 0\\\\[5pt] \\nabla \\times ( \\nabla f ) &= 0\\\\[5pt] \\nabla \\times ( f \\mathbf { A } ) &= f ( \\nabla \\times \\mathbf { A } ) + ( \\nabla f ) \\times \\mathbf { A }\\\\[5pt] \\nabla \\cdot ( \\mathbf { A } \\times \\mathbf { B } ) &= \\mathbf { B } \\cdot ( \\nabla \\times \\mathbf { A } ) - \\mathbf { A } \\cdot ( \\nabla \\times \\mathbf { B } )\\\\[5pt] \\nabla \\cdot ( \\nabla f ) &= \\nabla ^ { 2 } f\\\\[5pt]", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "f_1(\\mathbf{v}) + f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}} + \\frac{\\partial f_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $f(\\mathbf{v}) = f_1(\\mathbf{v})~ f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)~f_2(\\mathbf{v}) + f_1(\\mathbf{v})~\\left(\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $f(\\mathbf{v}) = f_1(f_2(\\mathbf{v}))$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial f_1}{\\partial f_2}~\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}$ ### Derivatives of vector valued functions of vectors Let $\\mathbf{f}(\\mathbf{v})$ be a vector valued function of the vector $\\mathbf{v}$. Then the derivative of $\\mathbf{f}(\\mathbf{v})$ with respect to $\\mathbf{v}$ (or at $\\mathbf{v}$) in the direction $\\mathbf{u}$ is the second-order tensor defined as $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = D\\mathbf{f}(\\mathbf{v})[\\mathbf{u}] = \\left[\\frac{d }{d \\alpha}~\\mathbf{f}(\\mathbf{v} + \\alpha~\\mathbf{u})\\right]_{\\alpha = 0}$ for all vectors $\\mathbf{u}$. Properties: 1. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v}) + \\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}} + \\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v})\\times\\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)\\times\\mathbf{f}_2(\\mathbf{v}) + \\mathbf{f}_1(\\mathbf{v})\\times\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{f}_2(\\mathbf{v}))$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{f}_2}\\cdot\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ ### Derivatives of scalar valued functions of second-order tensors Let $f(\\boldsymbol{S})$ be a real valued function of the second order tensor $\\boldsymbol{S}$. Then the derivative of $f(\\boldsymbol{S})$ with respect to $\\boldsymbol{S}$ (or at $\\boldsymbol{S}$) in the direction $\\boldsymbol{T}$ is the second order tensor defined as $\\frac{\\partial f}{\\partial \\boldsymbol{S}}:\\boldsymbol{T} = Df(\\boldsymbol{S})[\\boldsymbol{T}] = \\left[\\frac{d }{d \\alpha}~f(\\boldsymbol{S} + \\alpha\\boldsymbol{T})\\right]_{\\alpha =", "## Finding a Vector With a Given Curl Suppose we have a vector field $\\mathbf{v}$ satisfying $\\mathbf{\\nabla} \\cdot \\mathbf{v} =0$ for which we want to find a vector field $\\mathbf{w}$ such that $\\mathbf{\\nabla} \\times \\mathbf{w} = \\mathbf{v}$ . (1) For any vector $\\mathbf{u}$ , $\\mathbf{\\nabla} \\cdot (\\mathbf{\\nabla}\\times \\mathbf{u}) =0$ so all solutions of (1) are given by $\\mathbf{u} = \\mathbf{u_o} + \\mathbf{\\nabla} f$ where $f=f(x,y,z)$ is an arbitrary scalar and $\\mathbf{w_0}$ is any vector satisfying nbsp; $\\mathbf{\\nabla} \\times \\mathbf{w} = \\mathbf{v}$ . Example $\\mathbf{v}=3x \\mathbf{i} +2y \\mathbf{j} - 5z \\mathbf{k}$ $\\mathbf{\\nabla} \\cdot \\mathbf{v} = \\frac{\\partial (3x) }{\\partial x} + \\frac{\\partial (2y) }{\\partial y} + \\frac{\\partial (-5x) }{\\partial z}=0$ Because $\\mathbf{w_0}$ is arbitrary we can choose $\\mathbf{w_0} = w_1 \\mathbf{i} + w_2 \\mathbf{j}$ Then \\begin{aligned} \\mathbf{\\nabla} \\times \\nabla{w_0} &= (\\frac{\\partial}{\\partial x} \\mathbf{i} + \\frac{\\partial}{\\partial y} \\mathbf{j} + \\frac{\\partial}{\\partial z} \\mathbf{k}) \\times (w_1 \\mathbf{i} + w_2 \\mathbf{j}) \\\\ &=- \\frac{\\partial w_2}{\\partial z} \\mathbf{i} + \\frac{\\partial w_1}{\\partial z} \\mathbf{j} + (\\frac{\\partial w_2}{\\partial x} - \\frac{\\partial w_1}{\\partial y})\\mathbf{k} = 3x \\mathbf{i} + 2y \\mathbf{j} -5z \\mathbf{k} \\end{aligned} Equating components gives $- \\frac{\\partial w_2}{\\partial z} = 3x$ , $- \\frac{\\partial w_1}{\\partial z} = 2y$ , $\\frac{\\partial w_2}{\\partial x} - \\frac{\\partial w_1}{\\partial y} = -5z$ Integration gives $w_2 = -3xz+ g(x,y)$ and $w_1 = 2yz+ h(x,y),$ Substituting thse into $\\frac{\\partial w_2}{\\partial x} - \\frac{\\partial", "## Identity for Product of a Function With a Vector Field Satisfying Poisson's Equations Theorem If $f$ is a function and $\\mathbf{v}$ is a vector field satisfying $\\nabla^2 (f \\mathbf{v}) =0$ and $\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))=0$ then $\\mathbf{\\nabla} ((\\mathbf{\\nabla} f) \\cdot \\mathbf{v} + f \\mathbf{\\nabla} \\cdot (\\mathbf{v})) =0$ Proof $\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))= \\mathbf{\\nabla} (\\mathbf{\\nabla} \\cdot (f \\mathbf{v}))- \\nabla^2 (f \\mathbf{v})$ $\\nabla^2 (f \\mathbf{v}) =\\mathbf{\\nabla} \\times (\\mathbf{\\nabla} \\times (f \\mathbf{v}))=0$ so $\\mathbf{\\nabla} (\\mathbf{\\nabla} \\cdot (f \\mathbf{v} ))=0$ Now use the identity $\\mathbf{\\nabla} \\cdot (f \\mathbf{v}) =(\\mathbf{\\nabla} f ) \\cdot \\mathbf{v} + f ( \\mathbf{\\nabla} \\cdot \\mathbf{v})$ to obtain $\\mathbf{\\nabla} ((\\mathbf{\\nabla} f ) \\cdot \\mathbf{v} + f \\mathbf{\\nabla} \\cdot \\mathbf{v}) =0$", "\\mathbf{z} \\cdot (\\mathbf{x} \\times \\mathbf{y})$ 3. Malcev identity:[8] $(\\mathbf{x} \\times \\mathbf{y}) \\times (\\mathbf{x} \\times \\mathbf{z}) = ((\\mathbf{x} \\times \\mathbf{y}) \\times \\mathbf{z}) \\times \\mathbf{x} + ((\\mathbf{y} \\times \\mathbf{z}) \\times \\mathbf{x}) \\times \\mathbf{x} + ((\\mathbf{z} \\times \\mathbf{x}) \\times \\mathbf{x}) \\times \\mathbf{y}$ $\\mathbf{x} \\times (\\mathbf{x} \\times \\mathbf{y}) = -|\\mathbf{x}|^2 \\mathbf{y} + (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{x}.$ Other properties follow only in the three-dimensional case, and are not satisfied by the seven-dimensional cross product, notably, 1. Vector triple product: $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) = (\\mathbf{x} \\cdot \\mathbf{z}) \\mathbf{y} - (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{z}$ 2. Jacobi identity:[8] $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) + \\mathbf{y} \\times (\\mathbf{z} \\times \\mathbf{x}) + \\mathbf{z} \\times (\\mathbf{x} \\times \\mathbf{y}) = 0$ ## Coordinate expressions To define a particular cross product, an orthonormal basis {ej} may be selected and a multiplication table provided that determines all the products {ei × ej}. One possible multiplication table is described in the Example section, but it is not unique.[5] Unlike three dimensions, there are many tables because every pair of unit vectors is perpendicular to five other unit vectors, allowing many choices for each cross product. Once we have established a multiplication table, it is then applied to general vectors x and y by expressing x and y in terms of the basis and expanding x × y through bilinearity. × e1", "to the boundary and therefore any cross product involving $$\\hat{\\bf n}$$ will be perpendicular to $$\\hat{\\bf n}$$. The corresponding component of the current density is $$\\hat{\\bf n} \\times \\left( \\hat{\\bf n} \\times {\\bf J}_s \\right)$$, so Equation \\ref{m0023_eBC1} may be equivalently written as follows: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s \\label{m0023_eBC2}$ Applying a vector identity ($$\\mathbf{A} \\times(\\mathbf{B} \\times \\mathbf{C})=\\mathbf{B}(\\mathbf{A} \\cdot \\mathbf{C})-\\mathbf{C}(\\mathbf{A} \\cdot \\mathbf{B})$$ of Section B3) to the right side of Equation \\ref{m0023_eBC2} we obtain: \\begin{aligned} \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s &= \\hat{\\bf n} \\left(\\hat{\\bf n}\\cdot{\\bf J}_s\\right) - {\\bf J}_s\\left(\\hat{\\bf n}\\cdot\\hat{\\bf n}\\right) \\nonumber \\\\ &= \\hat{\\bf n} \\left(0\\right) - {\\bf J}_s\\left(1\\right) \\nonumber \\\\ &= -{\\bf J}_s\\end{aligned} Therefore: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = -{\\bf J}_s$ The minus sign on the right can be eliminated by swapping $${\\bf H}_2$$ and $${\\bf H}_1$$ on the left, yielding $\\boxed{ \\hat{\\bf n} \\times \\left( {\\bf H}_1 -{\\bf H}_2 \\right) = {\\bf J}_s }$ This is the form in which the boundary condition is most commonly expressed. It is worth noting what this means for the magnetic field intensity $${\\bf B}$$. Since $${\\bf B}=\\mu{\\bf H}$$: In the absence of surface current, the tangential component of $${\\bf B}$$ across the boundary between two material regions", "Why is it that for $\\vec{a}$ and $\\vec{b}$ the vectors represented by the corss products were used? In other words, why did they cross $\\vec{g}_{1}\\times\\vec{g}_{2} and \\vec{g}_{1} \\times \\vec{v}$ to get to $g_{1}$ and $v_{1}?$ If I can figure that out then I feel like I will understand this problem a lot better. I have looked up projections, but that hasn't proven beneficial as of yet. Maybe I'm looking in the wrong topics/source? Thanks Again! I'm not sure exactly what you mean here, but I can guess that you are assuming something like if $\\mathbf{v} =v_1\\mathbf{g}_1 + v_2\\mathbf{g}_2$, then $\\mathbf{v} \\cdot \\mathbf{g_1} = v_1$, but this is only true if $\\mathbf{g}_1$ and $\\mathbf{g}_2$ are orthonormal (like $\\mathbf{i}$ and $\\mathbf{j}$. In general, $\\mathbf{g}_1$ and $\\mathbf{g}_2$ won't be orthogonal, or normalised (have a magnitude of 1), so all taking the dot product gives you is $\\mathbf{v} \\cdot \\mathbf{g_1} = v_1 |\\mathbf{g}_1|^2 +v_2 \\mathbf{g}_1 \\cdot \\mathbf{g}_2$, which isn't all that useful here. Instead, do exactly what they say to in the Hint. Start by taking the vector (cross) product of $\\mathbf{g}_i$ with $\\mathbf{v}$ from the right. That is, start by computing $\\mathbf{v} \\times \\mathbf{g}_i$... what do you get when you do that? I'm sorry, I had a typo in that first line. This is really my 2nd time using LaTEX. What", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "$\\begin{split}\\mathbf{a} \\times \\mathbf{b} &\\neq \\mathbf{b} \\times \\mathbf{a} \\\\ \\mathbf{a} \\times \\mathbf{b} &= - \\mathbf{b} \\times \\mathbf{a}\\end{split}$ and not associative: $(\\mathbf{a} \\times \\mathbf{b} ) \\times \\mathbf{c} \\neq \\mathbf{a} \\times (\\mathbf{b} \\times \\mathbf{c})$ Two parallel vectors will have a zero cross product. The outer product between two vectors will not be not be discussed here, but instead in the inertia section (that is where it is used). Other useful vector properties and relationships are: $\\begin{split}\\alpha (\\mathbf{a} + \\mathbf{b}) &= \\alpha \\mathbf{a} + \\alpha \\mathbf{b}\\\\ \\mathbf{a} \\cdot (\\mathbf{b} + \\mathbf{c}) &= \\mathbf{a} \\cdot \\mathbf{b} + \\mathbf{a} \\cdot \\mathbf{c}\\\\ \\mathbf{a} \\times (\\mathbf{b} + \\mathbf{c}) &= \\mathbf{a} \\times \\mathbf{b} + \\mathbf{a} \\times \\mathbf{b}\\\\ (\\mathbf{a} \\times \\mathbf{b}) \\cdot \\mathbf{c} & \\textrm{ gives the scalar triple product.}\\\\ \\mathbf{a} \\times (\\mathbf{b} \\cdot \\mathbf{c}) & \\textrm{ does not work, as you cannot cross a vector and a scalar.}\\\\ (\\mathbf{a} \\times \\mathbf{b}) \\cdot \\mathbf{c} &= \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})\\\\ (\\mathbf{a} \\times \\mathbf{b}) \\cdot \\mathbf{c} &= (\\mathbf{b} \\times \\mathbf{c}) \\cdot \\mathbf{a} = (\\mathbf{c} \\times \\mathbf{a}) \\cdot \\mathbf{b}\\\\ (\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c} &= \\mathbf{b}(\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{a}(\\mathbf{b} \\cdot \\mathbf{c})\\\\ \\mathbf{a} \\times (\\mathbf{b} \\times \\mathbf{c}) &= \\mathbf{b}(\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c}(\\mathbf{a} \\cdot \\mathbf{b})\\\\\\end{split}$ ### Alternative Representation¶ If we have three non-coplanar unit vectors $$\\mathbf{\\hat{n}_x},\\mathbf{\\hat{n}_y},\\mathbf{\\hat{n}_z}$$, we can represent any vector $$\\mathbf{a}$$ as $$\\mathbf{a} = a_x \\mathbf{\\hat{n}_x} + a_y \\mathbf{\\hat{n}_y}", "# The curl of a special cross product When given two vectors $\\mathbf{A}$ and $\\mathbf{B}$, the curl of the cross product of these two is given by $$\\nabla\\times(\\mathbf{A}\\times\\mathbf{B})=(\\mathbf{B}\\cdot\\nabla)\\mathbf{A}-\\mathbf{B}(\\nabla\\cdot\\mathbf{A})-(\\mathbf{A}\\cdot\\nabla)\\mathbf{B}+\\mathbf{A}(\\nabla\\cdot\\mathbf{B}).$$ Using this relation, we can write $$\\nabla\\times(\\boldsymbol{\\mu}_I\\times\\mathbf{r})=(\\mathbf{r}\\cdot\\nabla)\\boldsymbol{\\mu}_I-\\mathbf{r}(\\nabla\\cdot\\boldsymbol{\\mu}_I)-(\\boldsymbol{\\mu}_I\\cdot\\nabla)\\mathbf{r}+\\boldsymbol{\\mu}_I(\\nabla\\cdot\\mathbf{r}).$$ In a lecture course that I'm reading, it is stated that this can actually be rewritten as $$\\nabla\\times(\\boldsymbol{\\mu}_I\\times\\mathbf{r})=-(\\boldsymbol{\\mu}_I\\cdot\\nabla)\\mathbf{r}+\\boldsymbol{\\mu}_I(\\nabla\\cdot\\mathbf{r}),$$ which means that the first two terms on the right hand side cancel. These can be written out, which results in $$\\begin{array}{l@{\\;}l} (\\mathbf{r}\\cdot\\nabla)\\boldsymbol{\\mu}_I-\\mathbf{r}(\\nabla\\cdot\\boldsymbol{\\mu}_I)&=\\left(y\\partial_y\\mu_x-x\\partial_y\\mu_y+z\\partial_z\\mu_x-x\\partial_z\\mu_z\\right)\\boldsymbol{\\hat{\\textbf{i}}}\\\\ &+\\left(x\\partial_x\\mu_y-y\\partial_x\\mu_x+z\\partial_z\\mu_y-y\\partial_z\\mu_z\\right)\\boldsymbol{\\hat{\\textbf{j}}}\\\\ &+\\left(x\\partial_x\\mu_z+y\\partial_y\\mu_z-z\\partial_x\\mu_x-z\\partial_y\\mu_y\\right)\\boldsymbol{\\hat{\\textbf{k}}}. \\end{array}$$ (Sorry for the ugly ihat, SE doesn't support \\imath for some reason.) In order for this to be zero, each term should be zero. Taking for example the $\\boldsymbol{\\hat{\\textbf{i}}}$ component, this means that $$y\\partial_y\\mu_x-x\\partial_y\\mu_y+z\\partial_z\\mu_x-x\\partial_z\\mu_z=0.$$ In this case, $\\boldsymbol{\\mu}_I$ is proportional to an angular momentum operator. Is this a special case, or is the above equation always true? And if so, why? • Which lecture course are you reading? Author, title, page, etc. Jan 4, 2014 at 17:38 • As another interpretation, it could mean that $\\boldsymbol{\\mu}_I$ is constant in space. Jan 4, 2014 at 18:08 • @Qmechanic Atomic Physics 1 & 2, written by our teacher at the University of Amsterdam for the two courses with the same name. Jan 4, 2014 at 18:30 • @KyleKanos Well, looking at the (probably correct) answer by anecdote, this does seem to", "Is there a counterexample to the claim that if $\\mathbf y\\cdot\\mathbf y=1$ and $\\mathbf x\\cdot\\mathbf y=c$ then $\\mathbf x=c\\mathbf y$? Let $x,y$ be arbitrary vectors where $\\mathbf{y} \\cdot \\mathbf{y} = 1$ and $c$ be a real valued scalar. If $\\mathbf{x} \\cdot \\mathbf{y} = c = c (\\mathbf{y} \\cdot \\mathbf{y} ) = (c \\mathbf{y} ) \\cdot \\mathbf{y}$, then $\\mathbf{x} = c \\mathbf{y}$ Is this true? Feels like a sloppy conclusion no? - Now that we have counterexamples, let's look where the fault in the reasoning is. When you say $x \\cdot y = c y \\cdot y$, you can't \"eliminate\" the common $y$ because that would require invertibility of the dot product. By itself, the dot product is not invertible, and you cannot reconstruct $x$ without additional information: in particular, information from the cross product. Given a vector $x$, and a unit vector $y$, you can decompose $x$ as $$x = (x \\cdot y) y + y \\times (x \\times y) = x_\\parallel + x_\\perp$$ With the information you have, you know that $x_\\parallel = cy$, but you don't know $x \\times y$, and because of that, you don't know $x_\\perp$. If you knew that $x \\times y =0$, then $x = x_\\parallel$, and you'd know you're done. - So in Algebra sense, the vector space with the operation", "Finally in the second line we see things like $$\\mathbf x \\mathbf x^T$$. This a matrix. For example $$M=\\mathbf a \\mathbf b^T$$ is a matrix whose components are given by products of the components of the vectors: $$M_{ij}=a_i b_j$$. There's no conflict with having both $$\\mathbf x \\mathbf x^T$$ and $$\\mathbf x^T \\mathbf x$$ at the same time. They just mean different things. Note that dot product multiplication commutes so $$\\mathbf x\\cdot \\mathbf{\\hat n} = \\mathbf{\\hat n}\\cdot \\mathbf x$$. With that in mind you can write $$(\\mathbf x \\cdot \\mathbf{\\hat n})^2 = (\\mathbf x \\cdot \\mathbf{\\hat n})(\\mathbf x \\cdot \\mathbf{\\hat n}) = (\\mathbf{\\hat n}\\cdot \\mathbf x)(\\mathbf x \\cdot \\mathbf{\\hat n})=(\\mathbf{\\hat n}^T \\mathbf x)(\\mathbf x^T \\mathbf{\\hat n})$$, having switched between the various notational conventions ;). But the parentheses don't really matter here, and you could just as well imagine forming the matrix $$\\mathbf x \\mathbf x^T$$ first, and then multiplying it on both sides by $$\\mathbf{\\hat n}$$ later, instead of taking the dot products first and multiplying the resulting scalars second. The notation in this text sample was really bad, so don't feel discouraged. $$\\vec u=(\\vec x^T\\,\\vec n)\\,\\vec n$$ \\begin{align*} & \\text{\\vec u~ is also}\\\\ &\\vec{u}=\\vec{n}\\,\\vec{n}^T\\,\\vec x= \\left[ \\begin {array}{ccc} {n_{{x}}}^{2}&n_{{x}}n_{{y}}&n_{{x}}n_{{z} }\\\\ n_{{x}}n_{{y}}&{n_{{y}}}^{2}&n_{{y}}n_{{z}} \\\\ n_{{x}}n_{{z}}&n_{{y}}n_{{z}}&{n_{{z}}}^{2} \\end {array} \\right]\\vec x =\\boldsymbol A\\,\\vec{x}\\\\ &\\text{with}\\\\ &\\boldsymbol A^T=A\\\\ &\\boldsymbol A^T\\,A=A\\,A=\\vec{n}\\,\\vec{n}^T\\,\\vec{n}\\,\\vec{n}^T=\\boldsymbol A \\end{align*} where", "\\cdot \\textbf{c} ) \\textbf{b} + ( \\textbf{a} \\cdot \\textbf{b} ) \\textbf{$$ Let r(t) be the position vector for a particle moving in $\\mathbb{R^3}.$ How to show that $$\\frac{d}{dt}(r \\times (v\\times r))=||r||^2 *a+ (r\\cdot v)*v-(||v||^2+ r\\cdot a)*r \\tag{1}$$ Where r(t) is a position vector (x(t),y(t),z(t)), $v(t)=\\frac{dr}{dt}=(x'(t),y'(t),z'(t)), a(t)=\\frac{dv}{dt}=\\frac{d^2r}{dt^2}=(x''(t),y''(t),z''(t))$ Note: v=velocity, a= acceleration My attempt:- I know $\\frac{d}{dt}(f\\times g)=\\frac{df}{dt}\\times g +f\\times \\frac{dg}{dt}$. I also know for any vectors u, v, w in $\\mathbb{R^3}, u\\times (v\\times w)=(u\\cdot w)*v-(u\\cdot v)*w$ But I don't understand how to use these formulas to prove equation (1)? Is this a part of a derivation? The $$\\displaystyle \\textbf{r} \\times ( \\textbf{v} \\times \\textbf{r} )$$ looks familiar but I can't seem to find it." ]

[ "+0 # Help! 0 49 1 +11 Consider the line parametrized by \\begin{align*} x&= 2t + 1,\\\\ y& = -3t +2. \\end{align*} Find a vector $\\begin{pmatrix}a\\\\b\\end{pmatrix}$ that's parallel to this line and satisfies $a+b = 3$. Jul 27, 2020 edited by FearlessIsland3 Jul 27, 2020", "$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$ and the second has normal vector $\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\\wedge v_2$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\wedge \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $y=0$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $x=0, z=1$, thus the line of intersection is $\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$ -", "+0 # Consider the line parametrized by \\begin{align*} x&= 4t + 2,\\\\ y& = t+2. \\end 0 207 1 Consider the line parametrized by$$x= 4t + 2, y = t+2$$ Find vector $$\\begin{pmatrix}a\\\\b\\end{pmatrix}$$ that's parallel to this line and satisfies $$a+b=10$$. Feb 2, 2019 #1 +103947 +1 Here's my best effort....although there might be an easier method !!! Wnen = 0 we have the point (2, 2) When t = 1 we have the point (6, 3) So...the slope of the given line is (3 - 2) / (6 - 2) = 1/4 If we let the parallel line go through the origin....we have the line y = (1/4)x This means that every point on this line has the coordinates ( x, 1/4x) This implies that x + (1/4)x = 10 (5/4)x = 10 x = 8 and y = (1/4)x = 2 So....the required vector is < 8, 2 > Feb 2, 2019", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "# Math Help - Unit vectors parallel 1. ## Unit vectors parallel Hi there, I need to find two unit vectors parallel to line L, where L is given by x = (z-1)/2, y = 1 Not sure how to approach the problem, could anybody help? Thanks! 2. ## Re: Unit vectors parallel Start by putting the line into its vector form. You need a parameter, say \\displaystyle \\begin{align*} t \\end{align*}. So you can set \\displaystyle \\begin{align*} x = t \\end{align*}, which means \\displaystyle \\begin{align*} \\frac{z - 1}{2} = t \\implies z = 2t + 1 \\end{align*}. So the line can be written as \\displaystyle \\begin{align*} L = ( t, 1, 2t + 1 ) = t (1 , 0 , 2 ) + ( 0, 1, 1 ) \\end{align*}. So the direction vector of the line is \\displaystyle \\begin{align*} (1, 0, 2) \\end{align*}. What is the unit vector going in that direction? What is another unit vector that is parallel to this one? 3. ## Re: Unit vectors parallel Originally Posted by Prove It Start by putting the line into its vector form. You need a parameter, say \\displaystyle \\begin{align*} t \\end{align*}. So you can set \\displaystyle \\begin{align*} x = t \\end{align*}, which means \\displaystyle \\begin{align*} \\frac{z - 1}{2} = t \\implies z = 2t + 1", "\\end{align*}. So the line can be written as \\displaystyle \\begin{align*} L = ( t, 1, 2t + 1 ) = t (1 , 0 , 2 ) + ( 0, 1, 1 ) \\end{align*}. So the direction vector of the line is \\displaystyle \\begin{align*} (1, 0, 2) \\end{align*}. What is the unit vector going in that direction? What is another unit vector that is parallel to this one? Ahh I get it now. So the unit vectors parallel would be +(1/sqrt5)(1, 0, 2) and -(1/sqrt5)(1, 0, 2)? 4. ## Re: Unit vectors parallel They're fine, you could also move them around if you want...", "## Vector field example 3 Given a vector field \\begin{align*} F &= \\begin{pmatrix} y \\\\ x \\end{pmatrix} \\end{align*} an object move from $(1/\\sqrt{2}, 1/\\sqrt{2})$ to the origin, how much work is done? Solution 1: \\begin{align*} F\\cdot dr &= F \\cdot t \\; ds \\\\ &= -|F|ds \\\\ &= -s \\; ds \\\\ \\end{align*} where $t$ is the tangent vector and $s$ is arc length of movement. You only need to integrate from 0 to 1 with respect to $s$, the answer should be $-1/2$ Solution 2: the normal way. Make use of the relation $y = x$.", "## Directional and orthogonal form of a line Directional and orthogonal form of a line Directional form: \\begin{align*} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix} + \\alpha \\begin{pmatrix} d_1 \\\\ d_2 \\end{pmatrix} \\end{align*} or, concisely \\begin{align} x = a + \\alpha d \\end{align} Let $v$ be a vector perpendicular to $d$. Transpose both sides and multiply by $v$: \\begin{align} x'v &= a'v + \\alpha d'v = a'v = c \\end{align} Where c is a constant scalar. This is called the orthogonal form. If you see it as $f(x) = x'v$, then $v$ is the gradient of $f$.", "intersection, simply plug $$t$$ or $$s$$ into their corresponding vector line equations. Plugging $$t=1$$ into the first equation will get: \\begin{align} \\vec{r_1} & = \\begin{bmatrix} 2 \\\\ -4 \\\\ 3 \\end{bmatrix} + 1 \\cdot \\begin{bmatrix} 3 \\\\ 11 \\\\ -2 \\end{bmatrix} \\\\ & = \\begin{bmatrix} 5 \\\\ 7 \\\\ 1 \\end{bmatrix} .\\end{align} So the lines intersect at the point $$(5, 7, 1).$$ Let's look at one final example of a question like this, this time with the line in parametric form. Determine whether the following lines intersect. If they intersect, find the point of intersection. If they do not intersect, determine whether they are skewed or parallel. Line 1: \\begin{align} x & = 11 - 10t \\\\ y & = 2 - 7t \\\\ z & = 3. \\end{align} Line 2: \\begin{align} x & = 1 + 2s \\\\ y & = 1 + 6s \\\\ z & = 1 - 2 s. \\end{align} Solution First, determine if the lines are parallel. Since this isn't in vector form, you need to create the direction vectors to be able to check this. The direction vectors here will be made using the coefficient of the scalar terms $$t$$ and $$s$$. The direction vector for Line 1 is: $\\begin{bmatrix} -10 \\\\ -7 \\\\ 0 \\end{bmatrix},$ and the direction vector for Line", "# Lines and Planes A line starts at an initial point and proceeds straight in a constant direction. Thus, we can write the equation of a line as \\begin{align*} x(t)=mt+b \\end{align*} where $b = \\left< b_1, b_2, \\ldots, b_N \\right>$ is the initial point, $m = \\left< m_1, m_2, \\ldots, m_N \\right>$ is the constant direction in which the line travels, and $x(t) = \\left< x_1(t), x_2(t), \\ldots, x_N(t) \\right>$ is the point reached by traveling $t$ units away from $b$ in the direction of $m$. (Though $x(t)$ is actually a vector, we can also refer to it as the point where the vector lands when the vector is placed at the origin.) For example, to compute the line between the points $(1,2,3,4)$ and $(5,-2,3,7)$ in 4-dimensional space, we can start by computing the direction $m$ as the difference between the two points: \\begin{align*} m &= \\left< 5,-2,3,7 \\right> - \\left< 1,2,3,4 \\right> \\\\ &= \\left< 4,-4,0,3 \\right> \\end{align*} Taking $b=\\left< 1,2,3,4 \\right>$ as our initial point, then, we can express the line as \\begin{align*} x(t) = \\left< 4,-4,0,3 \\right> t + \\left< 1,2,3,4 \\right>. \\end{align*} If we wanted to find another point on the line, we could substitute another value for $t$, say, $t=-2$. \\begin{align*} x(2) &= \\left< 4,−4,0,3 \\right>(−2) + \\left< 1,2,3,4 \\right> \\\\ &= \\left<", "+0 # Halp 0 94 1 The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations \\begin{align*} x &= at + b, \\\\ y &= ct + d, \\end{align*} where $0 \\le t \\le 1,$ and $t = 0$ corresponds to the point $(-2,7).$ Find $a^2 + b^2 + c^2 + d^2.$ Jan 6, 2022", "equation. \\begin{align*} Ax &= b \\\\ A^{-1} A x &= A^{-1} b \\\\ x &= A^{-1} b \\end{align*} When you have calculated $A^{-1} b$, you have the point of intersection. You can plot both lines and check visually that this point is in the right neighborhood, and you should take both lines, plug in the coordinates, and check that the answers are the same. Do you understand the basic process that's going on here?", "## Section1.19Lines and Planes Two points $$A$$ and $$B$$ determine a line. But there are also other ways to describe a line. Rather than specifying two points, we can specify just one ($$A$$), then give a vector $$\\vv$$ along the line. So let $$\\AA$$ be the vector from the origin to the point $$A\\text{,}$$ and $$\\vv$$ be a vector from $$A$$ that points along the line. Then any other point $$P$$ on the line can be reached by going to $$A$$ along $$\\AA\\text{,}$$ then going along $$\\vv\\text{.}$$ Thus, the vector $$\\rr(u) = \\AA + \\vv\\,u\\tag{1.19.1}$$ from the origin to $$P$$ provides a vector parameterization of the line. Equivalently, if \\begin{align*} \\AA \\amp = a_x\\,\\xhat + a_y\\,\\yhat + a_z\\,\\zhat ,\\\\ \\vv \\amp = v_x\\,\\xhat + v_y\\,\\yhat + v_z\\,\\zhat , \\end{align*} then the line is given by the parametric equations \\begin{align*} x \\amp = a_x + v_x\\,u ,\\\\ y \\amp = a_y + v_y\\,u ,\\\\ z \\amp = a_z + v_z\\,u , \\end{align*} which gives the coordinates $$(x,y,z)$$ of the point $$P$$ in terms of the parameter $$u\\text{.}$$ A given line can have many parameterizations, depending not only on the choice of the point $$A$$ on the line and vector $$\\vv$$ along the line, but also on the choice of parameter $$u\\text{.}$$ How do two points determine a line? Using", "- \\overline{OP} \\\\ &= <1, 0, 4> - <1, 1, 2> \\\\ &= <0, -1, 2> \\end{align*} To make it infinitely long, we multiply by some parameter, \\displaystyle \\begin{align*} t \\end{align*}, to give \\displaystyle \\begin{align*} t<0, -1, 2> \\end{align*}. Then we need to position this infinitely long vector at a point. Adding either of the points you know the line goes through will do. So \\displaystyle \\begin{align*} l = <1, 1, 2> + t<0, -1, 2> \\end{align*}. This is the vector form of the line. To get the parametric equations, expand it out... \\displaystyle \\begin{align*} l &= <1, 1 - t, 2 + 2t> \\\\ \\\\ x &= 1 \\\\ \\\\ y &= 1 - t \\\\ \\\\ z &= 2 + 2t \\end{align*} To get the Cartesian equations, try to write each of them as t in terms of the other variable and set them equal.", "Grassmannian Of Lines Consider the Grassmannian $\\G(1,3)$ of lines in $\\P^3$ (a.k.a. $\\Gr(2,4)$, the Grassmanian of 2-planes passing through the origin in $\\bC^4$). Think of its elements being represented by matrices: (1) \\begin{align} \\ell = \\begin{bmatrix} a_1 & b_1\\\\ a_2 & b_2\\\\ a_3 & b_3\\\\ a_4 & b_4 \\end{bmatrix} \\in \\G(1,3). \\end{align} 1. Let $\\x = (x_1,x_2,x_3,x_4)$ be a point. What polynomial equation(s) correspond(s) to the statement “$\\ell$ contains $x$”? 2. Let $\\ell'\\in\\G(1,3)$ be another line. What polynomial equation(s) correspond(s) to the statement $\\ell'$ meets $\\ell''$? 3. Show that any pair of lines $\\ell_1$ and $\\ell_2$ in general position can represented (in suitable coordinates) as (2) \\begin{align} \\ell_1 = \\begin{bmatrix} 1& 0\\\\ 0 & 1\\\\ 0& 0\\\\ 0 & 0 \\end{bmatrix} \\text{ and } \\ell_2 = \\begin{bmatrix} 0& 0\\\\ 0 & 0\\\\ 1& 0\\\\ 0 & 1 \\end{bmatrix} \\end{align} 1. Verify that $\\ell_1$ and $\\ell_2$ meet (3) \\begin{align} X = \\begin{bmatrix} 1& 0\\\\ x & 0\\\\ 0& 1\\\\ 0 & y \\end{bmatrix} \\end{align} 1. Given (4) \\begin{align} \\ell_3 = \\begin{bmatrix} 1& 0\\\\ 0 & 1\\\\ 1& 0\\\\ 0 & 1 \\end{bmatrix} \\text{ and } \\ell_4 = \\begin{bmatrix} 1& 0\\\\ 0 & -1\\\\ 0& 1\\\\ 1 & 0 \\end{bmatrix} \\end{align} find the solutions to the problem “$X$ meets $\\ell_1,\\ell_2,\\ell_3,\\ell_4$”. 1. Find a regular homotopy in the parameter space $\\G(1,3)^4$", "# Math Help - line equation in vectors 1. ## line equation in vectors \"find the equation of the line passing through B, with positiion vector b, which is perpendicular to the line r=a + $\\lambda$c, c=/=0, given that B is not a point on the line\". hi, i dont even get this at all to be honest, how is r=a + $\\lambda$c a line? in terms of normal line equations, which is the gradient, and the intercept? because r and c are not axis they are vectors, i dont know what this is asking. 2. ## Re: line equation in vectors A line is essentially a vector (which determines the direction) that is of infinite length, which is then positioned somewhere. So basically, what has happened to get the line \\displaystyle \\begin{align*} \\mathbf{r} \\end{align*} is that \\displaystyle \\begin{align*} \\mathbf{c} \\end{align*} determines the direction of the line, it is multiplied by some parameter \\displaystyle \\begin{align*} \\lambda \\end{align*} to make it of infinite length, and then the point \\displaystyle \\begin{align*} a \\end{align*} has been added (or rather, each component of \\displaystyle \\begin{align*} \\mathbf{c} \\end{align*} is adjusted numerically by each component of \\displaystyle \\begin{align*} a \\end{align*}) to position it in the correct spot. Do you think you can get a vector-form equation for the line passing through B in the", "coordinate system, and $$oxyz$$ is the camera coordinate system attached to the camera. $$O$$ and $$o$$ represent the origin of each. The vector is expressed as being independent of the coordinate system. I will write $$\\vec{Oo}=\\vec{t}$$. Then $$P$$ can be expressed as \\begin{align*} \\vec{OP}=\\vec{Oo}+\\vec{oP}=\\vec{t}+\\vec{oP}=\\vec{t}+x\\vec{e_1}+y\\vec{e_2}+z\\vec{e_3} \\end{align*} as a point in space. If $$\\vec{e_1}$$, $$\\vec{e_2}$$, and $$\\vec{e_3}$$ are the basis vectors of the coordinate system $$oxyz$$, the coordinates of the $$P$$ point in the coordinate system $$oxyz$$ are $$(x, y, z)$$. On the other hand, the coordinates of the $$P$$ point in the coordinate system $$OXYZ$$ are assumed to be $$(X, Y, Z)$$. Here, if all the vectors are displayed as components using the coordinate system $$OXYZ$$, the result is \\begin{align*} \\vec{OP}=\\begin{pmatrix} X \\\\ Y \\\\ Z \\end{pmatrix} .\\end{align*} Also assume that it is \\begin{align*} \\vec{t}=\\begin{pmatrix} t_1 \\\\ t_2 \\\\ t_3 \\end{pmatrix}, \\vec{e_1}=\\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\end{pmatrix}, \\vec{e_2}=\\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}, \\vec{e_3}=\\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix} .\\end{align*} Then \\begin{align*} \\vec{OP}=\\vec{t}+x\\vec{e_1}+y\\vec{e_2}+z\\vec{e_3} \\end{align*} becomes \\begin{align*} \\begin{pmatrix} X \\\\ Y \\\\ Z \\end{pmatrix}= \\begin{pmatrix} t_1 \\\\ t_2 \\\\ t_3 \\end{pmatrix}+ \\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\end{pmatrix}x+ \\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}y+ \\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix}z= \\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} &", "vector {eq}\\begin{align*} \\left< 0, 1, 2 \\right> - \\left< 1, 1, 1 \\right> &= \\left< -1, 0, 1 \\right> \\end{align*} {/eq} and {eq}\\begin{align*} \\left< 3, 0, 4 \\right> - \\left< 1, 1, 1 \\right> &= \\left< 2, -1, 3 \\right> \\end{align*} {/eq} The plane also passes through the point we used as our tail: {eq}(1,1,1) {/eq}. So a parameterization is {eq}\\begin{align*} \\vec r (u,v) &= \\left< 1,1,1 \\right> + u \\left< -1, 0, 1 \\right> + v \\left< 2, -1, 3 \\right> \\\\ &= \\left< 1-u+2v, 1-v, 1+u+3v \\right> \\end{align*} {/eq} In general, we create a couple vectors that must lie in the plane using one point as the tail and the other two as the heads, then we build the plane by treating these vectors as spanning vectors and adding them to the position vector described by the point we used as the tail. We can visualize this by picturing ourselves as standing at the point we chose to be the tail, then heading directly to either point we used as heads, and sweeping out the floor created by the vectors.", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •" ]

[ "input. – Surya Nov 26 '13 at 7:29 @Surya: Yes, that's a general method. You need to make sure to assign variable names for your independent variables that aren't x and y. I've edited my answer. – Jake Nov 26 '13 at 7:49 Okay. That was the mistake then. Thanks. – Surya Nov 26 '13 at 8:11 The code: \\documentclass[margin=10pt]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.render=4; settings.prc=false; size3(8cm); defaultpen(fontsize(10pt)); import graph3; currentprojection=orthographic(5,2,3); real ymax = 3; real ymin = -ymax; real h(real y) { return sqrt(1 + (y/2)^2); } path hyperbola = graph(h, ymin, ymax, operator ..); path3 hyperbola3 = path3(hyperbola, plane=YZplane); surface hyperboloid = surface(hyperbola3, c=O, axis=Y); real umax = hyperboloid.index.length - .001; real vmax = hyperboloid.index[0].length - .001; /* Parametrize by the square [0,2pi] x [ymin,ymax] */ triple point(real theta, real y) { return hyperboloid.point(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } triple normal(real theta, real y) { return hyperboloid.normal(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } int n = 10; int m = 10; real shrinkfactor=0.8; for (int i = 0; i < n; ++i) { real theta = i/n*2pi; triple f(real t) { return point(theta, t); } for (int j = 0; j < m; ++j) { real ylow = j/m*(ymax-ymin) + ymin; real yhigh = ylow + shrinkfactor*(ymax-ymin)/m; path3 arrowpath = graph(f, ylow, yhigh, n=10, operator..); }", "May 1 '15 at 22:10 It works when replacing Cf and Cg by Cf[0][0] and Cg[0][0], which allows to obtain the desired (single) guides. path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); Note that I've also slightly modified D1 and D2: in your definitions, D1, D2 and pp did not intersect anyway. path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); For proper explanations about how to use contour, see the very well-made (alas not complete) tutorial on Asymptote by Charles Staats, p. 32. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); fill(pp,.8white); draw(pp); xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); • Great, it works fine ! thank a lot fpast, for the answer and for the suggestions. Apr 30 '15 at 18:41 Just for comparison, here is a Metapost version. There's no support for implicit equations or anything so clever (or complex) as Asymptote's contour facility, so I could not avoid some algebra to get the \"long expression\" that the OP was happy to avoid. And to avoid overflows when x is near zero, I'm exploiting the facts that these two curves have", "# Asymptote fill area between two contours I use Asymptote to make the graphics. I have two curves defined as contours C_1 and C_2. I want to fill the area between the two curves. The problem is that contour returns a guide structure in Asymptote, and this is not accepted to to buildcycle command. Can anyone tell me how to solve this problem ? Thanks. This is the code I use and which produces the two curves, but fails to fill area between these. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax)--(xmax+eps,ymax); path D2=(xmax,0)--(xmax,ymax+eps); draw(Cf); draw(Cg); //path pp=buildcycle(Cf,D1,Cg,D2); // the preceding line produces and error if commented xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); end; • Is there some reason why you are using contour to produce the lines? It seems a rather complicated way to produce a single curve. May 1 '15 at 9:26 • Using f to write y = function (x) lead to a long expression. Around zero it requires large precision for fine tuned draw. Contour allows easier control over the drawing box.", "graph3; currentprojection=orthographic(2,1.4,.6,zoom=.95); //currentprojection=obliqueX(40); // the yellow paraboloid z=f(x,y)= x^2+y^2 real a=2.25; // below z=a^2 // to take domain is a disk (instead of a rectangle) triple f(pair M) { real x = a*(M.x)*cos(M.y); real y = a*(M.x)*sin(M.y); return (x,y,x^2+y^2); } real rmax=1, rmin=0; real phimax=2*pi, phimin=0; surface sf=surface(f, (rmin,phimin), (rmax,phimax), Spline); //draw(sf, surfacepen=pink,meshpen=brown+thick()); draw(sf,surfacepen=yellow+opacity(1)); triple h(pair M) { if (1+2M.y^2>a^2) return (M.x,M.y,a^2); else return (M.x,M.y,1+2M.y^2); } real xmax=4, xmin=-4; real ymax=1.5, ymin=-1.5; surface sh=surface(h, (xmin,ymin), (xmax,ymax), Spline); //draw(sh, surfacepen=pink,meshpen=brown+thick()); draw(sh,surfacepen=blue+opacity(.8)); triple A=(xmin,ymin,0), B=(xmax,ymax,a^2); draw(box(A,B),red+opacity(.5)); // the red circle x^2+y^2=b^2 is the intersection of // the blue plane z=b^2 and the paraboloid real b=1.6; real x(real t) {return b*cos(t);} real y(real t) {return b*sin(t);} real z(real t) {return b^2;} path3 g=graph(x,y,z,phimin,phimax,operator..); //draw(g,red+1pt); //draw(shift(-3,-3,b^2)*scale(6,6,0)*unitplane,blue+opacity(1)); draw(Label(\"$x$\",EndPoint),-5*X--5*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--5*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),-Z--7*Z,Arrow3()); • This is quite beautiful. Oct 19 '21 at 19:10 • I updated the 2nd picture. Oct 20 '21 at 19:15 • @CaptainGiraffe beautiful is as beautiful does! Oct 21 '21 at 15:10 I offer a TikZ solution. I think that for this figure it's easier to draw in 2d, so I did it this way. I'm not posting the maths, because it will be tedious without MathJax or something similar. But the key here is to find the envelope of the family of ellipses (circles in 3d) that are obtained cutting the", "that the graph origin is left-right centered. I've added the black border in the plot below just to illustrate the white border to the left of the plot. // arctan.asy import settings; settings.outformat=\"pdf\"; settings.render=0; unitsize(1cm); import graph; real f(real x) { return (atan(x)+(pi/2.0))/pi; } real xmin = -3.5; real xmax = 3.5; real ymin = -0.35; real ymax = 1.5; draw((xmin,0) -- (xmax,0), blue, arrow=None); draw((0,ymin) -- (0,ymax), blue, arrow = None); path g = graph(f,xmin,xmax); draw(g,red); label(\"$\\scriptstyle (\\arctan(x)+(\\pi/2))/\\pi$\", (xmax,0.8), S); void centerAtOrigin() { pair origMinPoint = min(currentpicture)/72*2.54; pair origMaxPoint = max(currentpicture)/72*2.54; pair origSize = size(currentpicture)/72*2.54; real xmin = origMinPoint.x; real xmax = origMaxPoint.x; real ymin = origMinPoint.y; if (xmax > fabs(xmin)) { xmin = -xmax; } if (fabs(xmin) > xmax) { xmax = -xmin; } fill(shift(xmin,ymin)*scale(xmax-xmin,origSize.y)* unitsquare, opacity(0)+white); } centerAtOrigin(); Edit: Instead of showing the label even though it's outside of the bounding box, this method expands the bounding box on the other side to keep your vertical axis centered.", "Compute the t values double txmin; double txmax; double tymin; double tymax; double tzmin; double tzmax; txmin = (minPt.x - rayIn.origin.x) / rayIn.direction.x; txmax = (maxPt.x - rayIn.origin.x) / rayIn.direction.x; x=1; if (txmin > txmax) { double tmp1 = txmin; txmin = txmax; txmax = tmp1; x=-1; } if (txmin > tNear) tNear = txmin; if (txmax < tFar) tFar = txmax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray y direction is parallel to plane tymin = (minPt.y - rayIn.origin.y) / rayIn.direction.y; tymax = (maxPt.y - rayIn.origin.y) / rayIn.direction.y; y=1; if (tymin > tymax) { double tmp2 = tymin; tymin = tymax; tymax = tmp2; y=-1; } if (tymin > tNear) tNear = tymin; if (tymax < tFar) tFar = tymax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray z direction is parallel to plane tzmin = (minPt.z - rayIn.origin.z) / rayIn.direction.z; tzmax = (maxPt.z - rayIn.origin.z) / rayIn.direction.z; z=1; if (tzmin > tzmax) { double tmp3 = tzmin; tzmin = tzmax; tzmax = tmp3; z=-1; } if (tzmin > tNear) tNear = tzmin; if (tzmax < tFar) tFar = tzmax; if (tNear > tFar) return false; if (tFar < 0) return false; double t = Math.min(tNear, tFar); // Set the location point. rayIn.evaluate(outRecord.location, tNear); //rayIn.start=0; rayIn.end=t;", "the direction of the line. else{ if(dy > 0){ t0 = (top - point1.y)/dy; t1 = (bottom - point1.y)/dy; } else{ t1 = (top - point1.y)/dy; t0 = (bottom - point1.y)/dy; } if(t0 > min) min = t0; if(t1 < max) max = t1; if(min > max || max < 0) return false; } // The point of intersection ix = point1.x + dx * min; iy = point1.y + dy * min; return true; • this is 2d, yes? – SirYakalot Dec 9 '11 at 12:41 • This is only 2D, yes. Also, the code does is not as well thought as zacharmarz's, which takes care of reducing the number of divisions and tests. – sam hocevar Dec 9 '11 at 14:09 I'm surprised to see that no one has mentioned the branchless slab method by Tavian bool intersection(box b, ray r) { double tx1 = (b.min.x - r.x0.x)*r.n_inv.x; double tx2 = (b.max.x - r.x0.x)*r.n_inv.x; double tmin = min(tx1, tx2); double tmax = max(tx1, tx2); double ty1 = (b.min.y - r.x0.y)*r.n_inv.y; double ty2 = (b.max.y - r.x0.y)*r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >= tmin; } I have added to @zacharmarz answer to handle when the ray origin is inside of the AABB. In this case tmin will be negative and", "real x=pt.x, y=pt.y, z=pt.z; int toreturn=0; real xmin=a.x, xmax=b.x, ymin = a.y, ymax=b.y, zmin=a.z, zmax=b.z; if (xmin > xmax) { xmin = b.x; xmax = a.x; } if (ymin > ymax) { ymin = b.y; ymax = a.y; } if (zmin > zmax) { zmin = b.z; zmax = a.z; } if (x < xmin) --toreturn; else if (x > xmax) ++toreturn; toreturn *= 2; if (y < ymin) --toreturn; else if (y > ymax) ++toreturn; toreturn *= 2; if (z < zmin) --toreturn; else if (z > zmax) ++toreturn; } bool keepregion(int region) { return (region == 0); } // Crop the surface to box(a,b). surface crop(surface s, triple a, triple b) { int region(triple pt) { return region(pt, a, b); } return divide(s, region=region, numregions=1, keepregion=keepregion)[0]; } Then save the following in, say, foo.asy in the same directory: settings.outformat=\"png\"; settings.render=16; import crop3D; import graph3; size(390pt/2, IgnoreAspect); //390pt is the default text width for the article class currentprojection=perspective(4,2,3); real f(pair z) {return sqrt(4*(z.x)^2+(z.y)^2);} surface s = surface(f,(-1,-1),(1,1),nx=30,Spline); s = crop(s, (-1,-1,-1),(1,1,1)); draw(s, green+opacity(0.4), render(merge=true)); xaxis3(\"$x$\",Bounds,InTicks); yaxis3(\"$y$\",Bounds,InTicks(beginlabel=false)); zaxis3(\"$z$\",Bounds,InTicks); Then compile it by typing asy foo at the command line. You should end up with a file foo.png that looks like this: • You should just add this as an update to your other answer. – Svend Tveskæg Feb", "-1}}; (* M.{a,b,c,d,x,y}=0 generates all 3 constraints *) bounds = {{0, 1}, {0, 1}, {0, 1}, {0, 1}, {-Infinity, Infinity}, {-Infinity, Infinity}} (* Now get the 4 solutions to min/max x/y: *) sol = LinearProgramming[#, M, {{1, 0}, {0, 0}, {0, 0}}, bounds] & /@ obj (* and keep the {x,y}'s sol[[1 ;; 4, 5 ;; 6]] (* all downhill from here *) xmin = Min[points[[1 ;; 4, 1]]] xmax = Max[points[[1 ;; 4, 1]]] ymin = Min[points[[1 ;; 4, 2]]] ymax = Max[points[[1 ;; 4, 2]]] ListPlot[points, AxesOrigin -> {xmin - 1, ymin - 1}, Epilog -> {Orange, Line[{{xmin, ymin}, {xmax, ymin}, {xmax, ymax}, {xmin, ymax}, {xmin, ymin}}]}] - Hi A.G., I guess you missed an image by carelessness. I added it back, hope you don't mind. – Silvia Dec 22 '13 at 3:10 You have 3 free parameters, since the imposed condition over their sum takes one away. So, let's plot in 3D the region for those three free parms: RegionPlot3D[(0 < p1 < 1 && 0 < p2 < 1 && 0 < p4 < 1 && p1 + p2 + p4 < 1), {p1, 0, 1}, {p2, 0, 1}, {p4, 0, 1}, PlotPoints -> 30] Which is equivalent to: Graphics3D[gc = GraphicsComplex[{{0, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, Polygon@{{1,", "x,y) = zsqrt((x,0) + (1,1) zscaled (y,0) ) + zsqrt((x,0) + (2,+2) zscaled (y,0)) enddef; vardef g(expr x,y) = zsqrt((x,0) - (1,1) zscaled (y,0) ) + zsqrt((x,0) + (2,-2) zscaled (y,0)) enddef; vardef edge_path(suffix FUN)(expr xmin, xmax, ymin, ymax, s) = for x=xmin step s until xmax: FUN(x,ymin) .. endfor for y=ymin step s until ymax: FUN(xmax,y) .. endfor for x=xmax step -s until xmin: FUN(x,ymax) .. endfor for y=ymax step -s until ymin: FUN(xmin,y) .. endfor cycle enddef; path a, b; a = edge_path(f,0,12,1,3,1/4) scaled 1cm; b = edge_path(g,0,12,1,3,1/4) scaled 1cm; fill a withcolor .8[blue,white]; draw a; fill b withcolor .8[red+1/2green,white]; draw b; path xx, yy; xx = (1/2 left -- 9 right) scaled 1cm; yy = (7/2 down -- 3 up) scaled 1cm; drawarrow xx; drawarrow yy; endfig; \\end{mplibcode} \\end{document} Metapost has complex numbers weakly built-in. You can basically treat a pair type as a complex number. The built-in operations abs and angle work as you might expect, and the operation zscaled does complex multiplication. ## Edit I have updated the complex sqrt function with a more correct, rather simpler, and more efficient version. The original one was: vardef sqrtz(expr zz) = save c, r, w; numeric r, c; pair w; r = abs(zz); w = zz + (r,0); c = sqrt(r)/abs(w); w scaled c enddef;", "z){return 2*(z-a1)*(z-a2)*(z-a3);},(0,0),(1,1),20,30); draw(pic2, mp.uGrid,deepgreen+0.4bp); draw(pic2, mp.vGrid,blue+0.4bp); mp=zGridMap(new pair(pair z){return sin(z);},(0,0),(2*pi/3,pi/2),20,30); draw(pic3, mp.uGrid,deepgreen+0.4bp); draw(pic3, mp.vGrid,blue+0.4bp); with the help of the zGridMap.asy module: // zGridMap.asy // import graph; typedef pair pairFpair(pair); struct zGridMap{ pairFpair fmap; pair pMin, pMax; int n, m; guide[] uGrid; guide[] vGrid; guide mapLine(pair a, pair b){ pair mapPoint(real t){return fmap(a*(1-t)+b*t);} return graph(mapPoint,0,1); } void makeWgrid(){ pair p=pMin,q=pMin; real dx=(pMax.x-pMin.x)/(m-1); real dy=(pMax.y-pMin.y)/(n-1); guide g; for(int i=0;i<n;++i){ uGrid.push(mapLine(p,(p.x+pMax.x-pMin.x,p.y))); p+=(0,dy); } for(int i=0;i<m;++i){ vGrid.push(mapLine(q,(q.x,q.y+pMax.y-pMin.y))); q+=(dx,0); } } void operator init(pairFpair fmap, pair pMin, pair pMax, int n, int m=n){ assert(n>1 && m>1); this.fmap=fmap; this.pMin=pMin; this.pMax=pMax; this.n =n ; this.m =m ; makeWgrid(); } } • I have deleted an answer of mine in similar spirit (based on an Octave few-liner) because the question seems to be asking not for the how, but the on-line where. – ccorn Oct 18 '19 at 22:52 • @ccorn: This answer responds to the lines: \"ideally, one would define a function (an elementary one) and then specify a grid (say an x-range and y-range and number of horizontal and vertical lines) and it would draw the picture of the grid transformed by the function.\" The example just illustrates how this can be done with the free tool Asymptote. Perhaps, the Octave example could be useful too? – g.kov Oct 18 '19 at 23:07 •", "out tMin, out tMax ); // tMin = min( tri[0].x, ... , tri[2].x ) // tMax = max( tri[0].x, ... , tri[2].x ) if ( tMax < m_MinMax[0] || m_MinMax[1] < tMin ) return false; ProjectTri( BOX_DIR[1], tri, out tMin, out tMax ); // tMin = min( tri[0].y, ... , tri[2].y ) // tMax = max( tri[0].y, ... , tri[2].y ) if ( tMax < m_MinMax[2] || m_MinMax[3] < tMin ) return false; ProjectTri( BOX_DIR[2], tri, out tMin, out tMax ); // tMin = min( tri[0].z, ... , tri[2].z ) // tMax = max( tri[0].z, ... , tri[2].z ) if ( tMax < m_MinMax[4] || m_MinMax[5] < tMin ) return false; // cross-products of tri edges and box axis... for ( int i = 0, j = 3; i < 3; j = i++ ) { Vector3 edge = tri[i] - tri[j]; for ( int k = 0; k < 3; ++k ) { dir = Vector3.Cross( edge, BOX_DIR[k] ); ProjectBox( dir, out bMin, out bMax ); ProjectTri( dir, tri, out tMin, out tMax ); if ( tMax < bMin || bMax < tMin ) return false; } } return true; } //----------------------------------------------------------------------------------------------------------------- private void ProjectBox( Vector3 dir, out float projMin, out float projMax ) { float c = ( dir.x * m_Center.x ) + ( dir.y *", "for plotting later xminn, yminn = minimum(hcat(minimum.(trajj, dims = 2)...), dims = 2) xmaxx, ymaxx = maximum(hcat(maximum.(trajj, dims = 2)...), dims = 2) if xminn < xmin xmin = xminn end if yminn < ymin ymin = yminn end if xmaxx > xmax xmax = xmaxx end if ymaxx > ymax ymax = ymaxx end #this should always evaluate to true, just a sanity check if size(trajj[2]) == (2,args.batchsize) #we concatenate the two feature vectors for easier plotting for i in 1:length(traj) traj[i] = cat(traj[i], trajj[i], dims=2) end Y = Flux.onecold(Y) |> cpu color = cat(color, Y, dims=1) end end end return traj, color, xmin, xmax, ymin, ymax end traj, color, xmin, xmax, ymin, ymax = loop() Now let's see what we got. We'll do a visualization of the values that come out of the neural network. The neural network acts on a very high dimensional space, so we will need to project that to a visualization in a two dimensional space. If the neural network was successfully trained to be a classifier, then we should see relatively distinct clusters for the various digits, noting that they will not be fully separated in two dimensions due to potential distance warping in the projection. # A collection of all the allowed shapes shape = [:circle, :rect, :star5, :diamond, :hexagon,", "ray to alleviate 0 = -0 issues Vec3f inverse_direction(1 / r.direction.x, 1 / r.direction.y, 1 / r.direction.z); // Solving box_min/box_max0 = O + Dt txmin = (b.box_min.x - r.origin.x) * inverse_direction.x; txmax = (b.box_max.x - r.origin.x) * inverse_direction.x; tymin = (b.box_min.y - r.origin.y) * inverse_direction.y; tymax = (b.box_max.y - r.origin.y) * inverse_direction.y; tzmin = (b.box_min.z - r.origin.z) * inverse_direction.z; tzmax = (b.box_max.z - r.origin.z) * inverse_direction.z; // Depending on direction of ray tmin may > tmax, so we may need to swap if (txmin > txmax) std::swap(txmin, txmax); if (tymin > tymax) std::swap(tymin, tymax); if (tzmin > tzmax) std::swap(tzmin, tzmax); t_min = txmin; t_max = txmax; // If t-value of a min is greater than t-value of max, // we missed the object in that plane. if ((t_min > tymax) || (tymin > t_max)) return false; if (tymin > t_min) t_min = tymin; if (tymax < t_max) t_max = tymax; if ((t_min > tzmax) || (tzmin > t_max)) return false; if (tzmin > t_min) t_min = tzmin; if (tzmax < t_max) t_max = tzmax; if (t_min > 0) t = t_min; else if (t_max > 0) t = t_max; else return false; intersect_point.x = r.origin.x + r.direction.x * t; intersect_point.y = r.origin.y + r.direction.y * t; intersect_point.z = r.origin.z + r.direction.z * t; return true; } And my", "real[]{4,3})+extendcap; arrowbar arr=Arrow(HookHead,size=3); real arrowW=0.9bp; pen arrowPen0=red+arrowW; pen arrowPen1=blue+arrowW; // real xmin=0,xmax=1; real ymin=0,ymax=1; xaxis(xmin,xmax,axisPen); yaxis(ymin,ymax,axisPen); typedef pair pairFuncReal(real); pairFuncReal CubicBezier(pair A, pair B, pair C, pair D){ return new pair(real t){return A*(1-t)^3+3*B*(1-t)^2*t+3*C*(1-t)*t^2+D*t^3;}; } pair[] p; p.append(new pair[]{(0,0),(0,1),(0.25,0),(0,0.47),(-0.2,0.07),(0.25,0.26),(0.25,0.67),(0.62,0.89)}); p.push(0.6p[4]+0.4p[5]-(0,0.05)); // control points for f0 p.push(0.4p[4]+0.6p[5]-(0,0.05)); // p.push(0.6p[6]+0.4p[7]+(0,0.05)); // control points for f1 p.push(0.4p[6]+0.6p[7]+(0,0.05)); // pairFuncReal f0=CubicBezier(p[4],p[8],p[9],p[5]); pairFuncReal f1=CubicBezier(p[6],p[10],p[11],p[7]); real f0xmin=0, f0xmax=1; real f1xmin=0, f1xmax=1; guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); pair arrowTail0=relpoint(g0,0.3)+(0,0.05); pair arrowHead0=relpoint(g0,0.8)+(0,0.06); pair arrowTail1=relpoint(g1,0.6)+(0,0.08); pair arrowHead1=relpoint(g1,0.01)+(0,0.06); p.append(new pair[]{ arrowTail0,arrowHead0,arrowTail1,arrowHead1 }); draw(g0,linePen); draw(g1,linePen); draw(p[2]--p[6],dashPen); draw(p[3]--(p[2].x,p[3].y),dashPen); real r=0.04; filldraw(circle(p[5],r),white,linePen); fill(circle(p[6],r),linePen); draw(p[12]--p[13],arrowPen0,arr); draw(p[14]--p[15],arrowPen1,arr); string[] plabels; plabels[0]=\"0\"; plabels[1]=\"1\"; plabels[2]=\"X^\\pm(\\omega)\"; plabels[3]=\"\\omega\"; pair[] ppos={ plain.W, plain.W, plain.S, plain.W, plain.S, plain.E, // 5 plain.SE, plain.SE, plain.E, plain.E, plain.SE, // 10 plain.SE, plain.N, plain.N, plain.N, plain.NW, // 15 plain.SE, plain.NE, plain.NW, plain.S, }; bool[] showDot=array(p.length,true); //bool[] showDot=array(p.length,false); showDot[5]=false; showDot[6]=false; if(showDots){ for(int i=0;i<p.length;++i){ if(showDot[i]) dot(p[i],UnFill); } } if(draft){ for(int i;i<p.length;++i){ if(p.initialized(i) && showDot[i]){ label(\"$p_{\"+string(i)+\"}$\",p[i],ppos[i]); } } } if(!draft){ for(int i;i<plabels.length;++i){ if(plabels.initialized(i) && showDot[i]) label(\"$\"+plabels[i]+\"$\",p[i],ppos[i]); } } shipout(bbox(Fill(paleyellow))); Run asy diag to get asy.pdf. The code shown is a working example for the draft mode; to get the final diagram, change bool draft=true; to bool draft=false; and bool showDots=true; to bool showDots=false; . Comment 1. Functions. The two curve segments are constructed with the standard graph function for 2d-drawing: guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); Functions f0, f1 take a", "the function, with respect to which the mesh is adapted, is rescaled to be between 0 and 1. true is the default. • keepbackvertices= if true, tries to keep as many vertices from the original mesh as possible. true is the default. • IsMetric= if true, the metric is defined explicitly. false is the default. If the 3 functions $m_{11}, m_{12}, m_{22}$ are given, they directly define a symmetric matrix field whose Hessian is computed to define a metric. If only one function is given, then it represents the isotropic mesh size at every point. For example, if the partial derivatives fxx ($=\\p^2 f/\\p x^2$), fxy ($=\\p^2 f/\\p x\\p y$), fyy ($=\\p^2 f/\\p y^2$) are given, we can set Th = adaptmesh(Th, fxx, fxy, fyy, IsMetric=1, nbvx=10000, hmin=hmin); • power= exponent power of the Hessian used to compute the metric. 1 is the default. • thetamax= minimum corner angle in degrees. Default is $10^\\circ$ where the corner is $ABC$ and the angle is the angle of the two vectors ${AB}, {BC}$, ($0$ imply no corner, $90$ imply perpendicular corner, ...). • splitin2= boolean value. If true, splits all triangles of the final mesh into 4 sub-triangles. • metric= an array of 3 real arrays to set or get metric data information. The size of these three arrays must", "files and insert the line usepackage(\"siunitx\"); at the end of it. Actually, the code in the plain.asy suggests that if there is a command specified in settings.command, the Asymptote executes it and exits ... 5 My solution is a more automated version of the answer from @chishimotoji. My code breaks all the paths up into subpaths and then automatically determines which should be plotted using inside(path p, pair z) functions. I created the isOutside and getOuterSubpaths functions as defined below. Using these functions, you will only need to define the paths, ... 7 This is raw code! Clean code should be written by yourself. unitsize(1cm); guide U = circle( (0,0), 1), E = ellipse( (0,0), 1.3, 0.6 ), B = box( (-1.2, -0.5), (1.2,0.5) ), Bo = box( (-0.4, -1.2), (0.4,1.2) ), all[] = U ^^ E ^^ B ^^ Bo; pair[] Int=intersectionpoints(U,Bo); pair[] Intt=intersectionpoints(U,B); pair[] IntT=... 5 One can plot this very easily if one knows the polar coordinate representations of the rectangle and ellipse. Here is the asymptote code: \\documentclass[varwidth,border=3mm]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.outformat=\"pdf\"; import graph; size(8cm,0); real rrect(real a,real b,real t) { return 1/max(abs(cos(t)/a),... Top 50 recent answers are included", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "grid3(ZYgrid,Step=2,gridpen); // No-go zone draw((xmax,1,zmin)--(xmin,1,zmin)--(xmin,1,zmax),black+1bp); xaxis3(Label(\"$x_1$\",MidPoint,align=Y-Z),Bounds(Both,Min),InTicks(Step=1),p=black); yaxis3(Label(\"$x_2$\",MidPoint,align=X-Z),Bounds(Both,Min),InTicks(Step=.5),p=black); zaxis3(Label(\"$\\dot x_2$\",MidPoint,align=X-Y),Bounds(Both,Min),InTicks(Step=2),p=black); The code is not optimized. You obtain the picture. I do not manage the color palette. It is possible to have a very short solution (but not very readable). The \"Bézier parametrized\" process (50 lines) is replaced by 5 lines. // Global Asymptote definitions can be put here. import three; import grid3; import palette; usepackage(\"mathptmx\"); // One can globally override the default toolbar settings here: // settings.toolbar=true; import graph3; real xmin=-2, xmax=2; real ymin=-2, ymax=1.5; real zmin=-2.5, zmax=2.5; limits((xmin,ymin,zmin),(xmax,ymax,zmax)); currentprojection=perspective(camera=(1.5,2,2.5)); unitsize(3cm,3cm,2cm); real linewidth=1.1; real linewidthprojections=.15; string filebasename=\"./data/ForASY_mnl2ippGridBranche3T1_\"; bool renderPRC = false; if(renderPRC) { // PRC TRUE settings.prc=true; settings.embed=true; } else { // RASTERIZE settings.outformat=\"png\"; settings.prc=false; settings.render=3; } /////// ORBITS IN 3D, SMOOTH //////// for (int i=1; i<=50; i=i+2) { string filename1 = filebasename + string(i) + \"PartI.dat\"; file in1=input(filename1).line().csv(); real[][] a1=in1.dimension(0,0); // 0 pour dire jusqu'à la fin du fichier a1=transpose(a1); real[] x1=a1[0]; real[] y1=a1[1]; real[] z1=a1[3]; pen orbitpen=.7bp+red*(1-sqrt(1-i/50))+rgb(0,62/255,91/255)*(sqrt(1-i/50)); pen projpen=.3bp+red*(1-sqrt(1-i/50))+rgb(0,62/255,91/255)*(sqrt(1-i/50))+opacity(0.3); path3 thepath1 = graph(x1, y1, z1, operator..); draw(thepath1,orbitpen,currentlight); } //////// SURFACE PART I //////// triple[][] v=new triple[50][21]; int increment=1; for (int i=1; i<52-increment; i=i+increment) { string filename1 = filebasename + string(i) + \"PartI.dat\"; file in1=input(filename1).line().csv(); real[][] a1=in1.dimension(0,0); // 0 pour dire jusqu'à la fin du fichier a1=transpose(a1); real[] x=a1[0]; real[] y=a1[1]; real[] z=a1[3]; int step=1; for (int j=0;", "are dotted with blue. Notice that in this critical (in any sense) case, we can not use the built-in path routines mintimes, maxtimes, times. // http://asymptote.ualberta.ca/ // Find all local maximal.minimal points unitsize(1cm); import graph; real f(real x){return (.6*cos(4.5*(x-4)+2.1)-1.2*sin(x-4)+.1*x+.2);} //real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); int n=200; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } //write(x,y); for (int k=2; k<n-2; ++k){ if ( (y[k] > y[k-1]) & (y[k] > y[k+1]) ) dot( (x[k],y[k]),red); if ( (y[k] < y[k-1]) & (y[k] < y[k+1]) ) dot( (x[k],y[k]),blue); } shipout(bbox(5mm,invisible)); For f(x)=.5+sin(3x), we get Even more, this procedure can be used to find all intersection point of the graph with the horizontal line through a given point A. In other words, more or less does the job for the following routines // http://asymptote.ualberta.ca/ // Find all intersection points with horizontal line through given point A unitsize(1cm); import graph; real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); pair A=(1,.6); draw((xmin,A.y)--(xmax,A.y),blue); int n=5000; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } for (int k=2; k<n-2; ++k) if ( (y[k]-A.y)*(y[k+1]-A.y)<0) dot( ((x[k]+x[k])/2,A.y),red); shipout(bbox(5mm,invisible)); There's a quick and easy approach using the sagetex package. It gives you access to a CAS, called Sage" ]

[ "input. – Surya Nov 26 '13 at 7:29 @Surya: Yes, that's a general method. You need to make sure to assign variable names for your independent variables that aren't x and y. I've edited my answer. – Jake Nov 26 '13 at 7:49 Okay. That was the mistake then. Thanks. – Surya Nov 26 '13 at 8:11 The code: \\documentclass[margin=10pt]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.render=4; settings.prc=false; size3(8cm); defaultpen(fontsize(10pt)); import graph3; currentprojection=orthographic(5,2,3); real ymax = 3; real ymin = -ymax; real h(real y) { return sqrt(1 + (y/2)^2); } path hyperbola = graph(h, ymin, ymax, operator ..); path3 hyperbola3 = path3(hyperbola, plane=YZplane); surface hyperboloid = surface(hyperbola3, c=O, axis=Y); real umax = hyperboloid.index.length - .001; real vmax = hyperboloid.index[0].length - .001; /* Parametrize by the square [0,2pi] x [ymin,ymax] */ triple point(real theta, real y) { return hyperboloid.point(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } triple normal(real theta, real y) { return hyperboloid.normal(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } int n = 10; int m = 10; real shrinkfactor=0.8; for (int i = 0; i < n; ++i) { real theta = i/n*2pi; triple f(real t) { return point(theta, t); } for (int j = 0; j < m; ++j) { real ylow = j/m*(ymax-ymin) + ymin; real yhigh = ylow + shrinkfactor*(ymax-ymin)/m; path3 arrowpath = graph(f, ylow, yhigh, n=10, operator..); }", "that the graph origin is left-right centered. I've added the black border in the plot below just to illustrate the white border to the left of the plot. // arctan.asy import settings; settings.outformat=\"pdf\"; settings.render=0; unitsize(1cm); import graph; real f(real x) { return (atan(x)+(pi/2.0))/pi; } real xmin = -3.5; real xmax = 3.5; real ymin = -0.35; real ymax = 1.5; draw((xmin,0) -- (xmax,0), blue, arrow=None); draw((0,ymin) -- (0,ymax), blue, arrow = None); path g = graph(f,xmin,xmax); draw(g,red); label(\"$\\scriptstyle (\\arctan(x)+(\\pi/2))/\\pi$\", (xmax,0.8), S); void centerAtOrigin() { pair origMinPoint = min(currentpicture)/72*2.54; pair origMaxPoint = max(currentpicture)/72*2.54; pair origSize = size(currentpicture)/72*2.54; real xmin = origMinPoint.x; real xmax = origMaxPoint.x; real ymin = origMinPoint.y; if (xmax > fabs(xmin)) { xmin = -xmax; } if (fabs(xmin) > xmax) { xmax = -xmin; } fill(shift(xmin,ymin)*scale(xmax-xmin,origSize.y)* unitsquare, opacity(0)+white); } centerAtOrigin(); Edit: Instead of showing the label even though it's outside of the bounding box, this method expands the bounding box on the other side to keep your vertical axis centered.", "# Asymptote fill area between two contours I use Asymptote to make the graphics. I have two curves defined as contours C_1 and C_2. I want to fill the area between the two curves. The problem is that contour returns a guide structure in Asymptote, and this is not accepted to to buildcycle command. Can anyone tell me how to solve this problem ? Thanks. This is the code I use and which produces the two curves, but fails to fill area between these. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax)--(xmax+eps,ymax); path D2=(xmax,0)--(xmax,ymax+eps); draw(Cf); draw(Cg); //path pp=buildcycle(Cf,D1,Cg,D2); // the preceding line produces and error if commented xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); end; • Is there some reason why you are using contour to produce the lines? It seems a rather complicated way to produce a single curve. May 1 '15 at 9:26 • Using f to write y = function (x) lead to a long expression. Around zero it requires large precision for fine tuned draw. Contour allows easier control over the drawing box.", "May 1 '15 at 22:10 It works when replacing Cf and Cg by Cf[0][0] and Cg[0][0], which allows to obtain the desired (single) guides. path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); Note that I've also slightly modified D1 and D2: in your definitions, D1, D2 and pp did not intersect anyway. path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); For proper explanations about how to use contour, see the very well-made (alas not complete) tutorial on Asymptote by Charles Staats, p. 32. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); fill(pp,.8white); draw(pp); xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); • Great, it works fine ! thank a lot fpast, for the answer and for the suggestions. Apr 30 '15 at 18:41 Just for comparison, here is a Metapost version. There's no support for implicit equations or anything so clever (or complex) as Asymptote's contour facility, so I could not avoid some algebra to get the \"long expression\" that the OP was happy to avoid. And to avoid overflows when x is near zero, I'm exploiting the facts that these two curves have", "false; // Update the ray range t_enter = max(u0, t_enter); t_exit = min(u1, t_exit); // Ah. we missed the interval return (t_exit >= t_enter);}bool RayIntersectAxisAlignedBox(Vector& BoxMin, Vector& BoxMax, Vector& RayOrigin, Vector& RayDirection, float& t_enter/*=0.0f*/, float& t_exit/*=ray_length*/){ // x axis if(!RayIntersectInterval(BoxMin.x, BoxMax.x, RayOrigin.x, RayDirection.x, t_enter, t_exit)) return false; // y axis if(!RayIntersectInterval(BoxMin.y, BoxMax.y, RayOrigin.y, RayDirection.y, t_enter, t_exit)) return false; // z axis if(!RayIntersectInterval(BoxMin.z, BoxMax.z, RayOrigin.z, RayDirection.z, t_enter, t_exit)) return false; return true;} That's the first step anyway. the '... / (ray_direction)', that can be pre-computed (for each axis, x, y and z) prior to testing the octree, because that will never change (the box axes will always be along the global x, y and z axis). Then you save quite a few divisions. Secondly, there are loads more you can deduce when using an octree. You can save the parent's enter and exit point (t_enter and t_exit), and only calculate the intersection of the ray and the planes originating from the parent's centre position. Then you can deduce which of the child boxes actually intersects the ray (given the order of the intersection params). So you don't actually end up doing a ray-AABox intersect in the end, more like a slimlined, specific test. So in the end, a LOT faster than building a box around the ray and culling,", "the direction of the line. else{ if(dy > 0){ t0 = (top - point1.y)/dy; t1 = (bottom - point1.y)/dy; } else{ t1 = (top - point1.y)/dy; t0 = (bottom - point1.y)/dy; } if(t0 > min) min = t0; if(t1 < max) max = t1; if(min > max || max < 0) return false; } // The point of intersection ix = point1.x + dx * min; iy = point1.y + dy * min; return true; • this is 2d, yes? – SirYakalot Dec 9 '11 at 12:41 • This is only 2D, yes. Also, the code does is not as well thought as zacharmarz's, which takes care of reducing the number of divisions and tests. – sam hocevar Dec 9 '11 at 14:09 I'm surprised to see that no one has mentioned the branchless slab method by Tavian bool intersection(box b, ray r) { double tx1 = (b.min.x - r.x0.x)*r.n_inv.x; double tx2 = (b.max.x - r.x0.x)*r.n_inv.x; double tmin = min(tx1, tx2); double tmax = max(tx1, tx2); double ty1 = (b.min.y - r.x0.y)*r.n_inv.y; double ty2 = (b.max.y - r.x0.y)*r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >= tmin; } I have added to @zacharmarz answer to handle when the ray origin is inside of the AABB. In this case tmin will be negative and", "1. If you're not concerned with efficiency too much, just take the \"naïve\" code from part 2 and swap <=/>= with </>: bool intersection(box b, ray r) { double tmin = -INFINITY, tmax = INFINITY; for (int i = 0; i < 3; ++i) { if (ray.dir[i] != 0.0) { double t1 = (b.min[i] - r.origin[i])/r.dir[i]; double t2 = (b.max[i] - r.origin[i])/r.dir[i]; tmin = max(tmin, min(t1, t2)); tmax = min(tmax, max(t1, t2)); } else if (ray.origin[i] < b.min[i] || ray.origin[i] > b.max[i]) { return false; } } return tmax >= tmin && tmax >= 0.0; } 17. Amomum says: Can you please explain what is this 'double t' parameter is? static inline bool dmnsn_ray_box_intersection(dmnsn_optimized_ray optray, dmnsn_aabb box, double t) It hurts me to say this but I believe that some of the functions do require comments about their input parameters. Or more obvious names. 1. A common way of defining lines (see dmnsn_ray) is by the parametric formula origin + t*direction. Commonly we restrict t >= 0 to get a half-line starting from the given origin. So generally t specifies a position along a line. In this case, t is the closest object intersection found so far, so if the bounding box is farther away than t, we can ignore it. I agree that should be better documented!", "$(x_d, y_d, z_d)$ is the ray's direction. $t > 0$ is considered “in front” of the origin, while $t < 0$ is “behind” it. We can find where the ray intersects an axis-aligned plane by solving (for example) \\ell_x(t) = x_0 + t x_d = x for t = \\frac{x - x_0}{x_d}. The segment of the ray that intersects the $x$ planes of the bounding box is then \\begin{aligned} t_{\\min} & = \\min \\{ t_1, t_2 \\} \\\\ t_{\\max} & = \\max \\{ t_1, t_2 \\}. \\\\ \\end{aligned} The intersection of two segments $(u_{\\min}, u_{\\max})$ and $(v_{\\min}, v_{\\max})$ is just \\begin{aligned} t_{\\min} & = \\max \\{ u_{\\min}, v_{\\min} \\} \\\\ t_{\\max} & = \\min \\{ u_{\\max}, v_{\\max} \\}. \\\\ \\end{aligned} If we end up with $t_{\\min} > t_{\\max}$, the intersection is empty. Putting all this together leads to a naturally branch-free ray/AABB intersection test (assuming min() and max() are branch-free and loops are unrolled): bool intersection(ray r, box b) { float tmin = 0.0, tmax = INFINITY; for (int i = 0; i < 3; ++i) { float t1 = (b.min[i] - r.origin[i]) * r.dir_inv[i]; float t2 = (b.max[i] - r.origin[i]) * r.dir_inv[i]; tmin = max(tmin, min(t1, t2)); tmax = min(tmax, max(t1, t2)); } return tmin < tmax; } We precompute r.dir_inv[i] = 1.0 / r.dir[i] to replace", "copy-pasted straight out of the Williams et al. 2005 paper, albeit with some minor changes to use our previously defined data structures: bool rayBBoxIntersectScalar(const Ray& ray, const BBox& bbox, float& tMin, float& tMax) { FVec4 rdir = 1.0f / ray.direction; int sign[3]; sign[0] = (rdir.x < 0); sign[1] = (rdir.y < 0); sign[2] = (rdir.z < 0); float tyMin, tyMax, tzMin, tzMax; tMin = (bbox.cornersAlt[sign[0]][0] - ray.origin.x) * rdir.x; tMax = (bbox.cornersAlt[1 - sign[0]][0] - ray.origin.x) * rdir.x; tyMin = (bbox.cornersAlt[sign[1]][1] - ray.origin.y) * rdir.y; tyMax = (bbox.cornersAlt[1 - sign[1]][1] - ray.origin.y) * rdir.y; if ((tMin > tyMax) || (tyMin > tMax)) { return false; } if (tyMin > tMin) { tMin = tyMin; } if (tyMax < tMax) { tMax = tyMax; } tzMin = (bbox.cornersAlt[sign[2]][2] - ray.origin.z) * rdir.z; tzMax = (bbox.cornersAlt[1 - sign[2]][2] - ray.origin.z) * rdir.z; if ((tMin > tzMax) || (tzMin > tMax)) { return false; } if (tzMin > tMin) { tMin = tzMin; } if (tzMax < tMax) { tMax = tzMax; } return ((tMin < ray.tMax) && (tMax > ray.tMin)); } For our test, we want to intersect a ray against four boxes, so we just write a wrapper function that calls rayBBoxIntersectScalar() four times in sequence. In the wrapper function, hits is a reference to a IVec4 where each", "graph3; currentprojection=orthographic(2,1.4,.6,zoom=.95); //currentprojection=obliqueX(40); // the yellow paraboloid z=f(x,y)= x^2+y^2 real a=2.25; // below z=a^2 // to take domain is a disk (instead of a rectangle) triple f(pair M) { real x = a*(M.x)*cos(M.y); real y = a*(M.x)*sin(M.y); return (x,y,x^2+y^2); } real rmax=1, rmin=0; real phimax=2*pi, phimin=0; surface sf=surface(f, (rmin,phimin), (rmax,phimax), Spline); //draw(sf, surfacepen=pink,meshpen=brown+thick()); draw(sf,surfacepen=yellow+opacity(1)); triple h(pair M) { if (1+2M.y^2>a^2) return (M.x,M.y,a^2); else return (M.x,M.y,1+2M.y^2); } real xmax=4, xmin=-4; real ymax=1.5, ymin=-1.5; surface sh=surface(h, (xmin,ymin), (xmax,ymax), Spline); //draw(sh, surfacepen=pink,meshpen=brown+thick()); draw(sh,surfacepen=blue+opacity(.8)); triple A=(xmin,ymin,0), B=(xmax,ymax,a^2); draw(box(A,B),red+opacity(.5)); // the red circle x^2+y^2=b^2 is the intersection of // the blue plane z=b^2 and the paraboloid real b=1.6; real x(real t) {return b*cos(t);} real y(real t) {return b*sin(t);} real z(real t) {return b^2;} path3 g=graph(x,y,z,phimin,phimax,operator..); //draw(g,red+1pt); //draw(shift(-3,-3,b^2)*scale(6,6,0)*unitplane,blue+opacity(1)); draw(Label(\"$x$\",EndPoint),-5*X--5*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--5*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),-Z--7*Z,Arrow3()); • This is quite beautiful. Oct 19 '21 at 19:10 • I updated the 2nd picture. Oct 20 '21 at 19:15 • @CaptainGiraffe beautiful is as beautiful does! Oct 21 '21 at 15:10 I offer a TikZ solution. I think that for this figure it's easier to draw in 2d, so I did it this way. I'm not posting the maths, because it will be tedious without MathJax or something similar. But the key here is to find the envelope of the family of ellipses (circles in 3d) that are obtained cutting the", ">= 0. Then we also know tmin is the closest intersection and tmax is the furthest. I implemented this in C++ and it works perfectly for any ray direction. However, in GLSL it seems to have problems with negative directions. Jon olick 2015-01-22 Something to consider here is that 0 * inf =nan which occurs when the ray starts exactly on the edge of a box Tavian Barnes 2015-01-31 True, if you want consistent handling of that case while staying branch-free, you have to do a little more work. This is worth another post actually, I'll write one up. Josh 2017-08-14 Did you do another post? Tavian Barnes 2017-08-16 Chris 2015-03-08 Thanks for posting this (so long ago)! I used this in my own code, and wondered if it is possible to save 2 of the subtractions by taking advantage of the fact that min(x+a, y+a) = min(x,y)+a (for some constant a): r.offset = (r.n_inv.y * r.x0.y) - (r.n_inv.x * r.x0.x); bool intersection(box b, ray r) { double tx1 = b.min.x * r.n_inv.x; double tx2 = b.max.x * r.n_inv.x; double tmin = min(tx1, tx2) + r.offset; double tmax = max(tx1, tx2) + r.offset; double ty1 = b.min.y * r.n_inv.y; double ty2 = b.max.y * r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >=", "Compute the t values double txmin; double txmax; double tymin; double tymax; double tzmin; double tzmax; txmin = (minPt.x - rayIn.origin.x) / rayIn.direction.x; txmax = (maxPt.x - rayIn.origin.x) / rayIn.direction.x; x=1; if (txmin > txmax) { double tmp1 = txmin; txmin = txmax; txmax = tmp1; x=-1; } if (txmin > tNear) tNear = txmin; if (txmax < tFar) tFar = txmax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray y direction is parallel to plane tymin = (minPt.y - rayIn.origin.y) / rayIn.direction.y; tymax = (maxPt.y - rayIn.origin.y) / rayIn.direction.y; y=1; if (tymin > tymax) { double tmp2 = tymin; tymin = tymax; tymax = tmp2; y=-1; } if (tymin > tNear) tNear = tymin; if (tymax < tFar) tFar = tymax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray z direction is parallel to plane tzmin = (minPt.z - rayIn.origin.z) / rayIn.direction.z; tzmax = (maxPt.z - rayIn.origin.z) / rayIn.direction.z; z=1; if (tzmin > tzmax) { double tmp3 = tzmin; tzmin = tzmax; tzmax = tmp3; z=-1; } if (tzmin > tNear) tNear = tzmin; if (tzmax < tFar) tFar = tzmax; if (tNear > tFar) return false; if (tFar < 0) return false; double t = Math.min(tNear, tFar); // Set the location point. rayIn.evaluate(outRecord.location, tNear); //rayIn.start=0; rayIn.end=t;", "can use built-in function times t=times(gf,x)[0]; to get a value of the time parameter t, which corresponds to the intersection of the function curve and a vertical line at x=1. This t value allows to: first, get a missing 'y' coordinate of the tangent point P P=point(gf,t); And second, the direction of the tangent line at this point as dir(gf,t). Function drawline (part of the basic Asymptote module math.asy), allows to draw the visible portion of the (infinite) line going through two points: drawline(P,P+dir(gf,t),tanLinePen); Another useful function for this drawing is relpoint, which returns the point on curve at the relative fraction of its arclength. This is a complete MWE: // tan.asy // // run // asy tan.asy // // to get tan.pdf // settings.tex=\"pdflatex\"; import graph; import math; size(6cm); import fontsize;defaultpen(fontsize(10pt)); texpreamble(\"\\usepackage{lmodern}\" +\"\\usepackage{amsmath}\" +\"\\usepackage{amsfonts}\" +\"\\usepackage{amssymb}\" ); pen funcLinePen=darkblue+0.9bp; pen tanLinePen=orange+0.9bp; pen grayPen=gray(0.3)+0.8bp; pen dashPen=gray(0.3)+0.8bp+linetype(new real[]{5,5})+linecap(0); real xmin=-2,xmax=-xmin; real ymin=0,ymax=4; real dxmin=0.2; real dxmax=dxmin; real dymin=dxmin; real dymax=dxmax; xaxis(\"$x$\",xmin-dxmin,xmax+dxmax,RightTicks(Step=1,step=0.5),arr,above=true); yaxis(\"$y$\",ymin-dymin,ymax+dymax,LeftTicks (Step=1,step=0.5,OmitTick(0)),arr,above=true); real f(real x){return x^2;} guide gf=graph(f,xmin,xmax,operator..); real x=1, t=times(gf,x)[0]; pair P=point(gf,t), Q=relpoint(gf,6/7); draw(gf,funcLinePen); draw((P.x,0)--P--(0,P.y),dashPen); drawline(P,P+dir(gf,t),tanLinePen); dot((P.x,0)^^P^^(0,P.y)^^Q,UnFill); label(\"$y=x^2$\",relpoint(gf,1/4),UnFill); label(\"$P(1,\"+string(round(P.y))+\")$\",P,plain.SE); label(\"$Q(x,x^2)$\",Q,plain.W); label(\"$T$\",Q,3*plain.E); I'm surprised that the original question is tagged but one answer (the accepted one) uses and the other uses ! So I decide to post a TikZ answer, but this is mainly a mathematics answer. \\documentclass[tikz]{standalone} \\usetikzlibrary{through}", "for plotting later xminn, yminn = minimum(hcat(minimum.(trajj, dims = 2)...), dims = 2) xmaxx, ymaxx = maximum(hcat(maximum.(trajj, dims = 2)...), dims = 2) if xminn < xmin xmin = xminn end if yminn < ymin ymin = yminn end if xmaxx > xmax xmax = xmaxx end if ymaxx > ymax ymax = ymaxx end #this should always evaluate to true, just a sanity check if size(trajj[2]) == (2,args.batchsize) #we concatenate the two feature vectors for easier plotting for i in 1:length(traj) traj[i] = cat(traj[i], trajj[i], dims=2) end Y = Flux.onecold(Y) |> cpu color = cat(color, Y, dims=1) end end end return traj, color, xmin, xmax, ymin, ymax end traj, color, xmin, xmax, ymin, ymax = loop() Now let's see what we got. We'll do a visualization of the values that come out of the neural network. The neural network acts on a very high dimensional space, so we will need to project that to a visualization in a two dimensional space. If the neural network was successfully trained to be a classifier, then we should see relatively distinct clusters for the various digits, noting that they will not be fully separated in two dimensions due to potential distance warping in the projection. # A collection of all the allowed shapes shape = [:circle, :rect, :star5, :diamond, :hexagon,", "* | |\\ tymin --------------+-*----+----------------- | \\ | | \\ tymax --------------+----*-+----------------- | \\| | *txmax | |\\ | | \\ | | \\ | | *P1 | | in code, you start off with a ray where tfirst = 0.0f, and tlast = |P1 - P0|- first, you test intersection with the two X planes=> will give you [txmin, txmax]=> also, you notice that txmin > tfirst, and txmax < tlast, so then tfirst = txmin, tlast = txmax- second, you test with the Y planes=> will give you [tymin, tymax]=> here also, (tymin > tfirst), so tfirst = tymin=> (tymax < tlast), so tlast = tymax- finally, (tlast > tfirst). this is consistent. so we have an intersection.in case of no intersection-------------------------- first, you test intersection with the two X planes=> will give you [txmin, txmax]=> also, you notice that txmin > tfirst, and txmax < tlast, so then tfirst = txmin, tlast = txmax- second, you test with the Y planes=> will give you [tymin, tymax]=> but here, tymin is not > tfirst=> (tymax < tlast), so tlast = tymax- finally, now that we have tfirst = txmin, tlast = tymaxbut, (tfirst > tlast). discrepency, the interseciton is invalid.- bottom line-------------you keep testing intersections with pairs of planes along axes. -> you keep reducing the", "z){return 2*(z-a1)*(z-a2)*(z-a3);},(0,0),(1,1),20,30); draw(pic2, mp.uGrid,deepgreen+0.4bp); draw(pic2, mp.vGrid,blue+0.4bp); mp=zGridMap(new pair(pair z){return sin(z);},(0,0),(2*pi/3,pi/2),20,30); draw(pic3, mp.uGrid,deepgreen+0.4bp); draw(pic3, mp.vGrid,blue+0.4bp); with the help of the zGridMap.asy module: // zGridMap.asy // import graph; typedef pair pairFpair(pair); struct zGridMap{ pairFpair fmap; pair pMin, pMax; int n, m; guide[] uGrid; guide[] vGrid; guide mapLine(pair a, pair b){ pair mapPoint(real t){return fmap(a*(1-t)+b*t);} return graph(mapPoint,0,1); } void makeWgrid(){ pair p=pMin,q=pMin; real dx=(pMax.x-pMin.x)/(m-1); real dy=(pMax.y-pMin.y)/(n-1); guide g; for(int i=0;i<n;++i){ uGrid.push(mapLine(p,(p.x+pMax.x-pMin.x,p.y))); p+=(0,dy); } for(int i=0;i<m;++i){ vGrid.push(mapLine(q,(q.x,q.y+pMax.y-pMin.y))); q+=(dx,0); } } void operator init(pairFpair fmap, pair pMin, pair pMax, int n, int m=n){ assert(n>1 && m>1); this.fmap=fmap; this.pMin=pMin; this.pMax=pMax; this.n =n ; this.m =m ; makeWgrid(); } } • I have deleted an answer of mine in similar spirit (based on an Octave few-liner) because the question seems to be asking not for the how, but the on-line where. – ccorn Oct 18 '19 at 22:52 • @ccorn: This answer responds to the lines: \"ideally, one would define a function (an elementary one) and then specify a grid (say an x-range and y-range and number of horizontal and vertical lines) and it would draw the picture of the grid transformed by the function.\" The example just illustrates how this can be done with the free tool Asymptote. Perhaps, the Octave example could be useful too? – g.kov Oct 18 '19 at 23:07 •", "are dotted with blue. Notice that in this critical (in any sense) case, we can not use the built-in path routines mintimes, maxtimes, times. // http://asymptote.ualberta.ca/ // Find all local maximal.minimal points unitsize(1cm); import graph; real f(real x){return (.6*cos(4.5*(x-4)+2.1)-1.2*sin(x-4)+.1*x+.2);} //real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); int n=200; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } //write(x,y); for (int k=2; k<n-2; ++k){ if ( (y[k] > y[k-1]) & (y[k] > y[k+1]) ) dot( (x[k],y[k]),red); if ( (y[k] < y[k-1]) & (y[k] < y[k+1]) ) dot( (x[k],y[k]),blue); } shipout(bbox(5mm,invisible)); For f(x)=.5+sin(3x), we get Even more, this procedure can be used to find all intersection point of the graph with the horizontal line through a given point A. In other words, more or less does the job for the following routines // http://asymptote.ualberta.ca/ // Find all intersection points with horizontal line through given point A unitsize(1cm); import graph; real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); pair A=(1,.6); draw((xmin,A.y)--(xmax,A.y),blue); int n=5000; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } for (int k=2; k<n-2; ++k) if ( (y[k]-A.y)*(y[k+1]-A.y)<0) dot( ((x[k]+x[k])/2,A.y),red); shipout(bbox(5mm,invisible)); There's a quick and easy approach using the sagetex package. It gives you access to a CAS, called Sage", "such that: or Thus we can compute the four t values: And we can apply the test using the following pseudo-code: if max(t_x_min, t_y_min) > min(t_x_max, t_y_max) then return false # no hit else return true Note that if the ray is traveling the opposite direction we need to swap the mins and maxes. In general we are concerned with the order in which the rays reach the two lines. Thus we can generalize our approach (which also makes it easier to add a 3rd dimension later) tgmin = -infinity tgmax = infinity # for x t1 = (x_min - P_0_x) / dx t2 = (x_max - P_0_x) / dx if t1 > t2 then swap(t1, t2) # Because we want the closest t1 from the perspective of the ray if t1 > tgmin then tgmin = t1 # not a typo - remember we are finding the largest of the mins if t2 < tgmax then tgmax = t2 # smallest of the maxes # same for y, z if tgmin > tgmax then return false if tgmax < 0 then return false # box behind # If we're still here, it's an intersect, and tgmin is the intersction point if (tgmin > 0) return (true, tgmin) else return (true, tgmax) # Though we could be inside", "+ (y_i-b)^2 \\le r^2, \\> \\> r \\ge 0. \\\\ \\end{array}$ This problem can be modeled in GAMS as follows. set i 'points' /p1*p10/; parameters x(i) 'x coordinates', y(i) 'y coordinates'; * fill with random data x(i) = uniform(1,10); y(i) = uniform(1,10); variables a 'x coordinate of center of circle' b 'y coordinate of center of circle' r 'radius'; equations e(i) 'points must be inside circle'; e(i).. sqr(x(i)-a) + sqr(y(i)-b) =l= sqr(r); r.lo = 0; model m /all/; option nlp=minos; solve m using nlp minimizing r; Without help, MINOS will not be able to find an optimal solution. The problem will be declared infeasible. In this case, providing a good starting point is very easy. If we define \\begin{eqnarray*} x_{\\min} &=& \\min_i x_i,\\\\ y_{\\min} &=& \\min_i y_i,\\\\ x_{\\max} &=& \\max_i x_i,\\\\ y_{\\max} &=& \\max_i y_i, \\end{eqnarray*} then good estimates are \\begin{eqnarray*} a &=& (x_{\\min}+x_{\\max})/2, \\\\ b &=& (y_{\\min}+y_{\\max})/2, \\\\ r &=& \\sqrt{ (a-x_{\\min})^2 + (b-y_{\\min})^2}. \\end{eqnarray*} Thus we include in our model: parameters xmin,ymin,xmax,ymax; xmin = smin(i, x(i)); ymin = smin(i, y(i)); xmax = smax(i, x(i)); ymax = smax(i, y(i)); * set starting point a.l = (xmin+xmax)/2; b.l = (ymin+ymax)/2; r.l = sqrt( sqr(a.l-xmin) + sqr(b.l-ymin) ); and now the model solves very easily. Level values can also be set away from zero implicitly as a result of", "real[]{4,3})+extendcap; arrowbar arr=Arrow(HookHead,size=3); real arrowW=0.9bp; pen arrowPen0=red+arrowW; pen arrowPen1=blue+arrowW; // real xmin=0,xmax=1; real ymin=0,ymax=1; xaxis(xmin,xmax,axisPen); yaxis(ymin,ymax,axisPen); typedef pair pairFuncReal(real); pairFuncReal CubicBezier(pair A, pair B, pair C, pair D){ return new pair(real t){return A*(1-t)^3+3*B*(1-t)^2*t+3*C*(1-t)*t^2+D*t^3;}; } pair[] p; p.append(new pair[]{(0,0),(0,1),(0.25,0),(0,0.47),(-0.2,0.07),(0.25,0.26),(0.25,0.67),(0.62,0.89)}); p.push(0.6p[4]+0.4p[5]-(0,0.05)); // control points for f0 p.push(0.4p[4]+0.6p[5]-(0,0.05)); // p.push(0.6p[6]+0.4p[7]+(0,0.05)); // control points for f1 p.push(0.4p[6]+0.6p[7]+(0,0.05)); // pairFuncReal f0=CubicBezier(p[4],p[8],p[9],p[5]); pairFuncReal f1=CubicBezier(p[6],p[10],p[11],p[7]); real f0xmin=0, f0xmax=1; real f1xmin=0, f1xmax=1; guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); pair arrowTail0=relpoint(g0,0.3)+(0,0.05); pair arrowHead0=relpoint(g0,0.8)+(0,0.06); pair arrowTail1=relpoint(g1,0.6)+(0,0.08); pair arrowHead1=relpoint(g1,0.01)+(0,0.06); p.append(new pair[]{ arrowTail0,arrowHead0,arrowTail1,arrowHead1 }); draw(g0,linePen); draw(g1,linePen); draw(p[2]--p[6],dashPen); draw(p[3]--(p[2].x,p[3].y),dashPen); real r=0.04; filldraw(circle(p[5],r),white,linePen); fill(circle(p[6],r),linePen); draw(p[12]--p[13],arrowPen0,arr); draw(p[14]--p[15],arrowPen1,arr); string[] plabels; plabels[0]=\"0\"; plabels[1]=\"1\"; plabels[2]=\"X^\\pm(\\omega)\"; plabels[3]=\"\\omega\"; pair[] ppos={ plain.W, plain.W, plain.S, plain.W, plain.S, plain.E, // 5 plain.SE, plain.SE, plain.E, plain.E, plain.SE, // 10 plain.SE, plain.N, plain.N, plain.N, plain.NW, // 15 plain.SE, plain.NE, plain.NW, plain.S, }; bool[] showDot=array(p.length,true); //bool[] showDot=array(p.length,false); showDot[5]=false; showDot[6]=false; if(showDots){ for(int i=0;i<p.length;++i){ if(showDot[i]) dot(p[i],UnFill); } } if(draft){ for(int i;i<p.length;++i){ if(p.initialized(i) && showDot[i]){ label(\"$p_{\"+string(i)+\"}$\",p[i],ppos[i]); } } } if(!draft){ for(int i;i<plabels.length;++i){ if(plabels.initialized(i) && showDot[i]) label(\"$\"+plabels[i]+\"$\",p[i],ppos[i]); } } shipout(bbox(Fill(paleyellow))); Run asy diag to get asy.pdf. The code shown is a working example for the draft mode; to get the final diagram, change bool draft=true; to bool draft=false; and bool showDots=true; to bool showDots=false; . Comment 1. Functions. The two curve segments are constructed with the standard graph function for 2d-drawing: guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); Functions f0, f1 take a" ]

[ "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "line $2 x+y=k$ passes through the point which divides the line segment joining the points $(1,1)$ and $(2,4)$ in the ratio $3: 2$, then $k$ equals (A) $frac{29}{5}$ (B) 5 (C) 6 (D) $frac{11}{5}$... -- 0.085330--", "Geometry # Collinearity If three points $$A=(k,k+2),$$ $$B=(0,k-6)$$ and $$C=(k-4,k)$$ all lie on the same line, what is the value of $$k?$$ The $$x$$- and $$y$$-intercepts of a line have the same value. If the line passes through $$P=(7,-6)$$ and $$Q=(-6,k),$$ what is the value of $$k?$$ Consider the two points $$(-1, -1)$$ and $$(3, 7).$$ If a circle passes through these two points, then which of the following cannot also be a point on the circle? What is the sum of all the values of $$k$$ that make three points $$A=(1,k),$$ $$B=(k,1)$$ and $$C=(10,10)$$ lie on the same line? Consider the line passing through two points $$P=(1, 10)$$ and $$Q=(-1, 3)$$. If the line also passes through point $$R=(17, a)$$, what is the value of $$a$$? ×", "A line passes through the points with position vectors are hat i+ hat j-2 hat k and hat i-3 hat j+ hat k . The posit # Ask a Question ### Question Asked by a Student from EXXAMM.com Team Q 1387101987. A line passes through the points with position vectors are hat i+ hat j-2 hat k and hat i-3 hat j+ hat k . The position vector of a point on the line which is at a unit distance from the first point is ? A frac{1}{5}(hat i-4hat j+3hat k) B frac{1}{5}(6 hat i+ hat j-7 hat k) C none of these D frac{1}{5}(4 hat i+9 hat j-13 hat k) #### HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) #### Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis", "Vectors $\\mathbf{a}$ and $\\mathbf{b}$ must be going in the same direction.", "# A line passes through the points with position vectors are $\\hat i+\\hat j-2\\hat k$ and $\\hat i-3\\hat j+\\hat k$. The position vector of a point on the line which is at a unit distance from the first point is ? $\\begin{array}{1 1} (a)\\:\\frac{1}{5}(6\\hat i+\\hat j-7\\hat k)\\:\\qquad\\:(b)\\:\\frac{1}{5}(4\\hat i+9\\hat j-13\\hat k)\\:\\qquad\\:(c)\\:\\frac{1}{5}(\\hat i-4\\hat j+3\\hat k)\\:\\qquad\\:(d)\\:None\\:of\\:these. \\end{array}$", "values of the Brauer character of $M$ will lie in the ring of Witt-vectors over $K$.", "vectors lie on the $y$-$z$ plane perpendicular to the vector ${\\bf a}$.", "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! SB # Let $L_1$ be the line through the origin and the point $(2, 0 , -1)$. Let $L_2$ be the line through the points $(1, -1 , 1)$ and $(4, 1 , 3)$. Find the distance between $L_1$ and $L_2$. ## $\\frac{13}{\\sqrt{69}}$ Vectors ### Discussion You must be signed in to discuss. Lectures Join Bootcamp ### Video Transcript the question there asking to find The distance between the lines. L one Line passes through original and the point A 20-1. Underline L two. Mhm Passes through the Point B 1 -1 one and C41 three. So these two lines are having these coordinates and we have to find the distance between these. So in order to find first we have to find the equation of these two lines. So the line one has origin, that is the coordinates Or the original 000. And and the another point that passes through this line is 20 minus one. So therefore the direction vector poor line L one B one vector is equal to a vector that can be found using these two points Uh to zero is 20 minus 00 and minus one minus zero minus ones. Therefore the", "the last equation, k= b. Then 3a- k= 0 so a= k/3 and $a- bk= a- k^2= 3$ so that $a= k/3= 3- k^2$. 4. http://www.mathhelpforum.com/math-he...dbea0df917.png How did you get that matrix from the initial three vectors?", "Let's view it mathematically Consider a line $L$ in a parametric form described as a set of all vectors $k\\cdot\\mathbf{u}+\\mathbf{v}$ parameterized by $k\\in \\mathbb{R}$, where $\\mathbf{v}$ is a vector orthogonal to a normalized vector $\\mathbf{u}$. Below is the graphical equivalent of the statement:", "check that the planes intersect in a common line? – alex.jordan May 11 '13 at 0:32 Suppose there are $m (m\\geq 4)$ planes, for plane number $k$ we know a point $\\mathbf{x}_k = (x_{k,1},x_{k,2},x_{k,3})$ on the plane, and the unit normal vector for plane number $k$ is $\\mathbf{n}_k$. Here I am not sure whether you mean by \"a point for each plane\", or \"a common point for all planes\", both are discussed. If we assume that all planes pass through point $\\mathbf{x}_0$. Then this line can be represented using this point $\\mathbf{x}_0$ and a direction $\\mathbf{b} = (b_1,b_2,b_3)$ $$\\mathbf{x} = \\mathbf{x}_0 + t\\mathbf{b} \\quad\\text{ for } t\\in \\mathbb{R}$$ Then the problem does not need least-square treatment at all, as we are not solving an overdetermined system, $\\mathbf{b}$ can be obtained simply by cross product any two normals: $$\\mathbf{b} = \\mathbf{n}_1\\times\\mathbf{n}_2, \\tag{1}$$ where $\\times$ is the cross product of vectors. Because the normal vectors of all the planes are coplanar (thus the planes intersect at exactly one line), $\\mathbf{b}\\perp \\mathbf{n}_1, \\mathbf{n}_2,\\ldots, \\mathbf{n}_m$. And we are done. If we only know a point for each plane, and we do not know if they all passing through one common line. Things get more complicated. The least square problem may not be well-posed. We want this line being perpendicular to every plane", "• # question_answer The position vector of a point lying on the line joining the points whose position vectors are $\\hat{i}+\\hat{j}-\\hat{k}$and $\\hat{i}-\\hat{j}+\\hat{k}$is A) $\\hat{j}$ B) $\\hat{i}$ C) $\\hat{k}$ D) $\\vec{o}$ The position vector of mid point of line joining the points whose position vectors are $\\hat{i}+\\hat{j}-\\hat{k}$and $\\hat{i}-\\hat{j}+\\hat{k}$ $=\\frac{\\hat{i}+\\hat{j}-\\hat{k}+\\hat{i}-\\hat{j}+\\hat{k}}{2}=\\frac{2\\hat{i}}{2}=\\hat{i}$", "# If a,b and c are in A.P thin the straight line $ax+by+c=0$ will always pass through line $ax+by+c=0$ will always pass through a fixed point whose coordinates are $\\begin{array}{1 1}(A)\\;(1,2) \\\\(B)\\;(1,-2)\\\\(C)\\;(1,1)\\\\(D)\\;(1,3)\\end{array}$ If a,b,c are in A.P then $a+c=2b$ => $a-2b+c=0$ => $ax+by+c=0$ it passes through $(1,-2)$ Hence B is the correct answer.", "# How to find normal vector of line given point normal passes through Given a line L in three-dimensional space and a point P, how can we find the normal vector of L under the constraint that the normal passes through P? - Let the line and point have position vectors $\\vec r=\\vec a+\\lambda \\vec b$ ($\\lambda$ is real) and $\\vec p$ respectively. Set $(\\vec r-\\vec p).\\vec b=0$ and solve for $\\lambda$ to obtain $\\lambda_0$. The normal vector is simply $\\vec a+\\lambda_0 \\vec b-\\vec p$.", "+0 Let a, b, c be three vectors that lie on the same line. Find ​ 0 46 2 Let a, b, c be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}.$$ Apr 9, 2022 #1 0 The vectors is $(1,0,0)$. Apr 9, 2022 #2 0 If two vectors are parallel, their cross product (vector product ) is zero. So, all three products are zero in which case their sum will be zero. Apr 11, 2022", "to get the two endpoints of the line segment. This works because $$\\mathbf{x}(t)=(1-t)\\mathbf{a}+t\\mathbf{b}=\\mathbf{a}+(\\mathbf{b}-\\mathbf{a})t$$ starts at $\\mathbf{a}$ when $t=0$ and will go in a straight $\\mathbf{b}-\\mathbf{a}$ direction towards $\\mathbf{b}$ (when $t=1$). -", "${\\mathbb{R}}^{3}$, they have to be linearly independent; i.e. The second question. No. First because a vector with 4 entries lives in ${\\mathbb{R}}^{4}$. But if you consider ${\\mathbb{R}}^{3}$ embedded in ${\\mathbb{R}}^{4}$. Again you need 3 vectors ${v}_{1},{v}_{2},{v}_{3}\\in {\\mathbb{R}}^{4}$ that are linearly independent to span ${\\mathbb{R}}^{3}$, but here the question is a little bit different. Think of $\\mathbb{R}$ embedded in ${\\mathbb{R}}^{2}$. You can embed it as any line that passes through the origin, so you have many representations of $\\mathbb{R}$ in ${\\mathbb{R}}^{2}$, same thing happens with ${\\mathbb{R}}^{3}$ embedded in ${\\mathbb{R}}^{4}$", "+0 # help 0 139 1 Let a and b be three-dimensional vectors. Then $$(5a + 7 b) \\times (-a+ 3 b) = ka \\times b$$ for some scalar k. Find k. Apr 29, 2020 #1 +31498 +3 The cross-product of a vector with itself is zero, so the LHS reduces to: 15(a x b) - 7(b x a) = 15(a x b) + 7(a x b) = 22(a x b) So k = 22 . Apr 29, 2020", "# Math Help - Finding a vector 1. ## Finding a vector Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? I don't understand how they say the vector is (4,-3) 2. Originally Posted by nyasha Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? $3x + 4y = -2$ $4y = -3x - 2$ $y = -\\frac{3}{4}x - \\frac{1}{2}$. So for every $4$ units you have gone to the right, you have gone $3$ units down. This would suggest that this line would have direction the same as $4\\mathbf{i} - 3\\mathbf{j}$. But we don't know the length of this vector. So multiply the vector by some parameter $t$. Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as $t(4\\mathbf{i} - 3\\mathbf{j})$. Now we just need a point that lies on the line so we know where to place this vector. The $y$ intercept will suffice. Therefore, the vector form of this line is $\\left(0\\mathbf{i} - \\frac{1}{2}\\mathbf{j}\\right) + t(4\\mathbf{i} - 3\\mathbf{j})$ or if you like $4t\\mathbf{i} - \\left(\\frac{1}{2} + 3t\\right)\\mathbf{j}$. 3. Originally Posted by Prove It $3x + 4y = -2$ $4y = -3x - 2$" ]

[ "${\\mathbf A}$ having passed each other, • ${\\mathbf B}$ and ${\\mathbf F}$ determined that ${\\mathbf B}$'s indication of the passage of ${\\mathbf J}$ was simultaneous to ${\\mathbf F}$'s indication of the passage of ${\\mathbf Q}$, and • ${\\mathbf K}$ and ${\\mathbf Q}$ determined that K's indication of the passage of ${\\mathbf A}$ was simultaneous to ${\\mathbf Q}$'s indication of the passage of ${\\mathbf F}$. Question: Is thereby guaranteed that for these distance ratios obtains (1) $\\frac{{\\mathbf A}{\\mathbf B}}{{\\mathbf A}{\\mathbf F}} = \\frac{{\\mathbf J}{\\mathbf K}}{{\\mathbf J}{\\mathbf Q}}$?, and (moreover) (2) $\\left( \\left(\\frac{{\\mathbf B}{\\mathbf F}}{{\\mathbf A}{\\mathbf F}}\\right)^2 + 1 - \\left(\\frac{{\\mathbf A}{\\mathbf B}}{{\\mathbf A}{\\mathbf F}}\\right)^2 \\right) \\left( \\left(\\frac{{\\mathbf K}{\\mathbf Q}}{{\\mathbf J}{\\mathbf Q}}\\right)^2 + 1 - \\left(\\frac{{\\mathbf J}{\\mathbf K}}{{\\mathbf J}{\\mathbf Q}}\\right)^2 \\right) = 4 \\left( 1 - \\left( \\frac{{\\mathbf A}{\\mathbf B}}{{\\mathbf A}{\\mathbf F}} \\right) \\left( \\frac{{\\mathbf J}{\\mathbf K}}{{\\mathbf J}{\\mathbf Q}} \\right) \\right)$? Or otherwise: What could be concluded if (1) and/or (2) were not found satisfied? This question was deleted from Physics Stack Exchange and restored from an archive. Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the \"link\" button. You can then enter your link URL. Please consult the FAQ for as to how", "take $n-2$ more points, evenly spaced in [a,b] $u_i = k / {n-1} \\forall k \\in [1 .. n-1]$ $\\mathbf{B}_p(u (b - a) + a) = \\mathbf{B}_q(u_i)$ for each $u_i$ For a 4 control points bezier curve If we take $u$ in $(0, {1 \\over 3}, {2 \\over 3}, 1)$, hence $t_0$, $t_1$, $t_2$, $t_3$ respectively equal to $a$, $\\frac{b - 2 a}{3}$, $\\frac{2b - a}{3}$, $b$ $\\mathbf{Q}_0 = \\mathbf{B}_p(t_0)$ $\\mathbf{Q}_1 = {-15 \\mathbf{B}_p(t_0) + 54 \\mathbf{B}_p(t_1) -27 \\mathbf{B}_p(t_2) + 6 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_2 = {6 \\mathbf{B}_p(t_0) -27 \\mathbf{B}_p(t_1) + 54 \\mathbf{B}_p(t_2) -15 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_3 = \\mathbf{B}_p(b)$", "the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=(\\sin t) \\mathbf{i}+\\left(t^{2}-\\cos t\\right) \\mathbf{j}+e^{z} \\mathbf{k}, \\quad t_{0}=0 Check back soon! Problem 20 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=t^{2} \\mathbf{i}+(2 t-1) \\mathbf{j}+t^{3} \\mathbf{k}, \\quad t_{0}=2 Check back soon! Problem 21 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=\\ln t \\mathbf{i}+\\frac{t-1}{t+2} \\mathbf{j}+t \\ln t \\mathbf{k}, \\quad t_{0}=1 Check back soon! Problem 22 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$", "# Problem: Is the line through (4,6,3) and (2,0, -1) parallel to the line through (6,22,12) and (1,7,2)? ###### FREE Expert Solution The equation of a line with the direction vector d = (l,m,n) passing through a point (x1,y1,z1) is: $\\overline{)\\frac{\\mathbf{x}\\mathbf{-}{\\mathbf{x}}_{\\mathbf{1}}}{\\mathbf{L}}{\\mathbf{=}}\\frac{\\mathbf{y}\\mathbf{-}{\\mathbf{y}}_{\\mathbf{1}}}{\\mathbf{m}}{\\mathbf{=}}\\frac{\\mathbf{z}\\mathbf{-}{\\mathbf{z}}_{\\mathbf{1}}}{\\mathbf{n}}}$ For the first line: $\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{4}\\mathbf{-}\\mathbf{2}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{6}\\mathbf{-}\\mathbf{0}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{3}\\mathbf{-}\\mathbf{\\left(}\\mathbf{-}\\mathbf{1}\\mathbf{\\right)}}\\phantom{\\rule{0ex}{0ex}}\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{2}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{6}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{4}}\\phantom{\\rule{0ex}{0ex}}\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{1}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{3}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{2}}$ ###### Problem Details Is the line through (4,6,3) and (2,0, -1) parallel to the line through (6,22,12) and (1,7,2)?", "the line connecting $$\\mathbf x$$ and $$\\mathbf a$$.", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "14 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{728}$. This slightly improves upon the lower bound of 1/2124 coming from Corollary 10. Proof Combine Corollary 4 and Lemma 13. $\\Box$ Lemma 15 One has ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{2}{3}$ and $2{\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3}) ) + {\\bf P} ({\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq 1$. Proof As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length $\\sqrt{3}$ and three of length $2$. If we let a denote the number of monochromatic $\\sqrt{3}$ edges and b the number of monochromatic $2$-edges, we see from inspection of all four colorings that $(a,b)$ is either $(6, 0), (4,0), (2, 1)$, or $(0,3)$. In particular, one always has $\\frac{a}{6} + \\frac{b}{3} \\geq \\frac{2}{3}$ and $2\\frac{a}{6} + \\frac{b}{3} \\geq 1$. Taking expectations, we obtain the claim. $\\Box$ Corollary 16 We have ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(2) ) \\geq \\frac{1}{6}$, ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\sqrt{3} ) ) \\geq \\frac{1}{4}$ and ${\\bf P}( {\\mathbf c}(0) = {\\mathbf c}(\\frac{1}{\\sqrt{3}}) ) \\geq \\frac{1}{28}$. Proof Combine Lemma 2, Lemma 15, and Lemma 13. $\\Box$ Lemma 17 If $a,b,c \\gt 0$ are the lengths", "# Code for the figure below, showing a parametric cubic Hermite curve↔ Specifically, suppose the segment travels from milestone $\\mathbf{m}_0$ to $\\mathbf{m}_1$ as the interpolation parameter travels from 0 to 1. We wish to find a cubic polynomial segment that respects the initial tangent $\\mathbf{t}_0$ and the terminal tangent $\\mathbf{t}_1$. In other words, the terminal conditions are $$\\begin{split} \\mathbf{m}_0 &= poly(C_i;0) \\\\ \\mathbf{m}_1 &= poly(C_i;1) \\\\ \\mathbf{t}_0 &= \\left. \\frac{d}{du} poly(C_i;u) \\right|_{u=0} \\\\ \\mathbf{t}_1 &= \\left. \\frac{d}{du} poly(C_i;u) \\right|_{u=1} \\end{split}$$ Letting the coefficients $C_i =(\\mathbf{a},\\mathbf{b},\\mathbf{c},\\mathbf{t})$, we have $$poly(C_i;u) = \\mathbf{a} + u \\mathbf{b} + u^2 \\mathbf{c} + u^3 \\mathbf{t}$$$$\\frac{d}{du} poly(C_i;u) = \\mathbf{b} + 2 u \\mathbf{c} + 3 u^2 \\mathbf{t}$$ so immediately, we see that $\\mathbf{a} = \\mathbf{m}_0$ and $\\mathbf{b} = \\mathbf{t}_0$. We now need to solve for $\\mathbf{c}$ and $\\mathbf{t}$ such that the linear equations $$\\mathbf{m}_1 = \\mathbf{m}_0 + \\mathbf{t}_0 + \\mathbf{c} + \\mathbf{t}$$ and $$\\mathbf{t}_1 = \\mathbf{t}_0 + 2 \\mathbf{c} + 3 \\mathbf{t}$$ hold. To do so we rearrange the above into the matrix equation $$\\begin{bmatrix}\\mathbf{m}_1 - \\mathbf{m}_0 - \\mathbf{t}_0 \\\\ \\mathbf{t}_1 - \\mathbf{t}_{0} \\end{bmatrix} = \\begin{bmatrix} 1 & 1 \\\\ 2 & 3 \\end{bmatrix} \\begin{bmatrix} \\mathbf{c} \\\\ \\mathbf{t} \\end{bmatrix}$$ Which has the solution via 2x2 matrix inversion $$\\begin{bmatrix} \\mathbf{c} \\\\ \\mathbf{t} \\end{bmatrix} = \\begin{bmatrix} 1 & 1 \\\\ 2 & 3 \\end{bmatrix}^{-1} \\begin{bmatrix}\\mathbf{m}_1 - \\mathbf{m}_0", "1\\\\ 2}$$ • p127, 4th Line after Eq (4.22): \"initial $$n_0$$ edges\" should be \"initial $$n_0$$ nodes\" Chapter 5: Kernel Methods • p138, Example 5.4: $$\\mathbf{\\mu}_\\phi = \\sum_{i=1}^5 \\phi(\\mathbf{x}_i) = \\sum_{i=1}^5 \\mathbf{x}_i$$ should be $$\\mathbf{\\mu}_\\phi = \\frac{1}{5}\\sum_{i=1}^5 \\phi(\\mathbf{x}_i) = \\frac{1}{5} \\sum_{i=1}^5 \\mathbf{x}_i$$ • p140, 7th Line after Eq (5.3): \"$$\\sum_{i=1}^{m_a} \\sum_{j=1}^{m_a} \\alpha_i \\alpha_{\\!j} K(\\mathbf{x}_i, \\mathbf{x})$$\" should be \"$$\\sum_{i=1}^{m_a} \\sum_{j=1}^{m_a} \\alpha_i \\alpha_{\\!j} K(\\mathbf{x}_i, \\mathbf{x}_j)$$\" • p141, 3rd line and 10th Line before Sec 5.1.2: There is an extra left bracket in definition of $$\\phi(\\mathbf{x})$$, that is, \"$$\\big( ( K(\\mathbf{x}_1, \\mathbf{x}), ...$$\" should be \"$$\\big( K(\\mathbf{x}_1, \\mathbf{x}), ...$$\" • p144, 2nd line: \"$$\\int a(\\mathbf{x})^2\\; d\\mathbf{x} < 0$$\" should be \"$$\\int a(\\mathbf{x})^2\\; d\\mathbf{x} < \\infty$$\" • p144, last line: \"$$\\sum_{k=1}^q$$\" should be \"$$\\sum_{k=0}^q$$\" • p156, Section 5.4.2: all occurrences of \"path/paths\" should be \"walk/walks\" Chapter 6: High-dimensional Data • p164: In the definitions of the hyperball and and hypersphere \"$$\\mathbf{x} = (x_1, x_2, \\ldots, x_d)$$\" should be \"$$\\mathbf{x} = (x_1, x_2, \\ldots, x_d)^T$$\" • p171: \"$$\\mathbf{0}_d = (0_1,0_2,\\ldots,0_d)$$\" should be \"$$\\mathbf{0}_d = (0_1,0_2,\\ldots,0_d)^T$$\" • p172, Section 6.6, 1st Line after Eq. (6.11): $$\\mu$$ in equation \"$$\\mu=\\mathbf{0}_d$$\" should be in bold. • p178, section \"Volume in d dimensions\": \"$$x_1 = r \\cos\\theta_1\\cos\\theta_2 \\cos\\theta_3 = r c_2 c_2 c_3$$\" should be \"$$x_1 = r \\cos\\theta_1\\cos\\theta_2 \\cos\\theta_3 = r c_1 c_2 c_3$$\" \"$$x_3 =", "function can be written as $$F(k,\\mathbf {M})=F_f(k,\\mathbf {M}_f)+F_d(k,\\mathbf {M}_d)$$, where $$F_f(k,\\mathbf {M}_f)$$ is curvature function at fixed point and $$F_d(k,\\mathbf {M}_d)$$ is deviation from the fixed point. Applying Eq. (11) iteratively makes $$F_d(k,\\mathbf {M}_d) \\rightarrow 0$$, implying gradual decrease in the deviation of curvature function from the fixed point configuration. Hence $$F(k,\\mathbf {M})\\rightarrow F_f(k,\\mathbf {M}_f)$$. Finding the map from $$\\mathbf {M}$$ to $$\\mathbf {M}^{\\prime }$$ iteratively, broadens the curvature function $$F(k_0, \\mathbf {M})$$ until it reaches fixed point. This iterative procedure yields RG flow in parameter space indicating critical points of the system. Generic RG equation of parameters $$\\mathbf {M}$$ can be obtained by expanding Eq. (11) to leading order and writing $$d\\mathbf {M}=\\mathbf {M}^{\\prime }-\\mathbf {M}$$ and $$\\delta k^2=dl$$, as25,26 \\begin{aligned} \\frac{d\\mathbf {M}}{dl}= \\frac{1}{2} \\frac{\\partial _k^2 F(k,\\mathbf {M})|_{k=k_0}}{\\partial _{\\mathbf {M}}F(k_0,\\mathbf {M})}. \\end{aligned} (12) The critical point can be defined by the condition $$|\\frac{d\\mathbf {M}}{dl}|=\\infty$$, and fixed point can be defined by the condition $$|\\frac{d\\mathbf {M}}{dl}|=0$$. As we approach critical point, along with the divergence of the curvature function [Eq. (10)], characteristic length $$\\xi$$ in Eq. (9) also diverges \\begin{aligned} \\lim _{\\mathbf {M}\\rightarrow \\mathbf {M}_c}\\xi =\\infty . \\end{aligned} These divergences in $$F(k_0,\\mathbf {M})$$ and $$\\xi$$ give rise to divergent behavior characterized by the critical exponents \\begin{aligned} F(k_0,\\mathbf {M}) \\propto |\\mathbf {M}-\\mathbf {M}_c|^{-\\gamma } \\;\\;\\;,\\;\\;\\;\\;\\;\\; \\xi \\propto |\\mathbf {M}-\\mathbf", "with the line, and thus with $\\mathbf{l}$, then the distance from the point $\\mathbf{p}$ to the line is $d=|\\mathbf{r}|\\sin\\theta$. Now, recall that the magnitude of the cross product of two vectors $\\mathbf{a}$ and $\\mathbf{b}$ is given by $|\\mathbf{a}\\times\\mathbf{b}|=|\\mathbf{a}||\\mathbf{ b}|\\sin\\theta$. Thus, $d=\\frac{|\\mathbf{r}\\times\\mathbf{l}|}{|\\mathbf{l}| }$ $d=\\frac{|(\\mathbf{p}-\\mathbf{q})\\times\\mathbf{l}|}{|\\mathbf{l}|}$ Our point is $\\mathbf{p}=\\mathbf{\\hat{i}}-2\\mathbf{\\hat{j}}+4\\mathbf{\\hat{k}}$, and our line is that which passes through the point $\\mathbf{q}=-3\\mathbf{\\hat{j}}+2\\mathbf{\\hat{k}}$ parallel to the vector $\\mathbf{l}=2\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\ma thbf{\\hat{k}}$. Thus $\\mathbf{r}=\\mathbf{p}-\\mathbf{q}=\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\mat hbf{\\hat{k}}$, and $d=\\frac{|\\mathbf{r}\\times\\mathbf{l}|}{|\\mathbf{l}| }$ $=\\frac{|(\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\mathb f{\\hat{k}})\\times(2\\mathbf{\\hat{i}}+\\mathbf{\\hat{j }}+2\\mathbf{\\hat{k}})|}{|2\\mathbf{\\hat{i}}+\\mathbf {\\hat{j}}+2\\mathbf{\\hat{k}}|}$ $=\\frac{|2\\mathbf{\\hat{j}}-\\mathbf{\\hat{k}}|}{\\sqrt{2^2+1^2+2^2}}$ $=\\frac{\\sqrt{2^2+(-1)^2}}{3}$ $d=\\frac{\\sqrt{5}}{3}$ --Kevin C. 5. THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote \"OH NO\" when i said we have to minimize the square of d. Why? Then he gave me 0 points. 6. Originally Posted by Intsecxtanx THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote \"OH NO\" when i said we have to minimize the square of d. Why? Then he gave me 0 points. Only that instructor can really answer that question. Prehaps he wanted you to use only vector methods. But as you can see the answers are the same.", "## Elementary Linear Algebra 7th Edition $\\mathbf{z}=(\\frac{7}{2},3,\\frac{5}{2})$ $\\mathbf{u}=(1,2,3)\\quad\\mathbf{v}=(2,2-1)\\quad\\mathbf{w}=(4,0,-4)$ Find $\\mathbf{z}$ if $2\\mathbf{z}-3\\mathbf{u}=\\mathbf{w}$ We can solve for $\\mathbf{z}$: $2\\mathbf{z}=\\mathbf{w}+3\\mathbf{u}$ $\\mathbf{z}=\\frac{1}{2}\\mathbf{w}+\\frac{3}{2}\\mathbf{u}$. All we need to do now is plug in the vectors $\\mathbf{w}$ and $\\mathbf{u}$: $\\mathbf{z}=\\frac{1}{2}(4,0,-4)+\\frac{3}{2}(1,2,3)$ $\\mathbf{z}=(2,0,-2)+(\\frac{3}{2},3,\\frac{9}{2})=(\\frac{7}{2},3,\\frac{5}{2})$", "F}$ is conservative. Indeed, testing each of our conditions, we find that ${\\mathbf F}$ is conservative. So, we can use the fundamental theorem of line integrals to evaluate our answer quite easily. We’ll do it the long way too, for practice. • Using the Fundamental Theorem of Line Integrals (FTLI) To use the FTLI, we have to first find the potential function for ${\\mathbf F}$. In this case, it’s pretty straightforward. We want to find a scalar function, say $\\phi$, such that $\\nabla \\phi={\\mathbf F}$. By inspection, we see that $\\phi=\\frac{x^2}{2}+\\frac{y^2}{2}+\\frac{z^2}{2}$ will work ($\\nabla[\\frac{x^2}{2}+\\frac{y^2}{2}+\\frac{z^2}{2}] = \\langle \\frac{\\partial}{\\partial x}, \\frac{\\partial}{\\partial y}, \\frac{\\partial}{\\partial z} \\rangle[\\frac{x^2}{2}+\\frac{y^2}{2}+\\frac{z^2}{2}] = \\langle x,y,z\\rangle$). So, to finish evaluating, we simply use the formula $\\phi({\\mathbf c}(2))-\\phi({\\mathbf c}(0)) = \\phi(2,e^2,4)-\\phi(0,1,0) = \\frac{19+e^4}{2}$. • Using Parameterization Following the flowchart the wrong way leads to the formula $\\int_c {\\mathbf F}\\cdot d{\\mathbf s} = \\int_a^b {\\mathbf F}(\\mathbf{c}(t))\\cdot {\\mathbf c}^\\prime(t) dt$. Like in Example 1, we follow a three step process here. First, we find ${\\mathbf c}^{\\prime} (t) = \\langle 1, e^t,2t \\rangle$. Next, we find ${\\mathbf F}({\\mathbf c}(t))\\cdot{\\mathbf c}^{\\prime}(t) = 2t^3+t+e^{2t}$. Finally, we integrate this function, $\\int_0^2 2t^3+t+e^{2t} dt = \\frac{19+e^4}{2}$. These are the same answers (which is nice). Example 3 : Surface integral of a scalar function Let’s consider $\\int_S xy+z ds$ over the surface $x+2y+z=2$ in the first octant", "B)}{P(B)} = \\frac{\\frac{1}{3}}{\\frac{19}{45}}=\\boxed{\\frac{15}{19} \\ \\mathbf{(D)}}$.", "1)x + (k – 1)y = 2k + 1 For no, solution, $\\frac{\\mathbf{a}_{\\mathbf{1}}}{\\mathbf{a}_{\\mathbf{2}}}=\\frac{\\mathbf{b}_{\\mathbf{1}}}{\\mathbf{b}_{\\mathbf{2}}} \\neq \\frac{\\mathbf{c}_{\\mathbf{1}}}{\\mathbf{c}_{\\mathbf{2}}}$ $\\frac{\\mathbf{3}}{2 \\mathbf{k}-\\mathbf{1}}=\\frac{1}{\\mathbf{k}-\\mathbf{1}} \\neq \\frac{1}{2 \\mathbf{k}+1}$ So, $\\frac{\\mathbf{3}}{\\mathbf{2 k - 1}}=\\frac{\\mathbf{1}}{\\mathbf{k}-\\mathbf{1}} \\quad \\& \\frac{\\mathbf{1}}{\\mathbf{k}-\\mathbf{1}} \\neq \\frac{\\mathbf{1}}{\\mathbf{2 k}+\\mathbf{1}}$ 3(k – 1) = 2k – 1 2k + 1 = k – 1 k = 2 k=–2", "$\\mathbf r = \\mathbf a + t\\frac{\\mathbf b - \\mathbf a}{\\lVert \\mathbf b - \\mathbf a \\rVert}$. – Théophile May 10 '17 at 14:06 • Okay, so I'm taking the unit vector of this line, and the unit vector, when scaled to ||$b-a$||, is at $a$, so then is the scalar for this? If this is true, how do I now make it so it's 3 units away? – sangstar May 10 '17 at 18:56 ## 1 Answer \"A direction\" of the line $\\overleftrightarrow{AB}$ is $B-A=(0,-3,0)$. (Another direction is $A-B$. But we won't use that.) So every point on the line $\\overleftrightarrow{AB}$ can be expressed as $$P(t)=A +t(B-A) = (1,2,-2)-t(0,3,0)$$ The distance between the point $P(t)$ and $A$ is $||P(t)-A||=||t(0,3,0)|| = 3|t|.$ So the points on the line that are a distance of $3$ from $A$ satisfy the equation $3|t|=3$. Hence $t\\in \\{1, -1\\}$. The points are $(1,2,-2) \\pm (0,3,0)$", "$\\mathbf{E} = k_\\text{e} \\sum_{i=1}^N \\frac{Q_i}{r_i^2} \\mathbf{\\hat{r}}_i$.", "(b_3 - a_3)$ This is the parametric form of the straight line, where coordinates are expressed in terms of a parameter $$s$$. Exercise 11 1. $\\mathbf{r} = x\\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ where $x = 1 + s \\quad , \\quad y = 1 + 2s \\quad , \\quad z = 2 - s$ 2. The point cuts the $$(x, z)$$-plane when $$y = 0$$. That is, $0 = 1 + 2s$ Thus $$s = -\\frac{1}{2}$$. At this parameter the coordinates are $\\mathbf{r} = \\frac{1}{2}\\mathbf{i} + 0\\mathbf{j} + \\frac{5}{2}\\mathbf{k}$ Exercise 12 1. $\\lvert \\mathbf{a} \\rvert = \\sqrt{2^2 + (-1)^2} = \\sqrt{5}$ $\\lvert \\mathbf{b} \\rvert = \\sqrt{1^2 + 3^2 + 5^2} = \\sqrt{35}$ 2. $\\mathbf{a} + \\mathbf{b} = 2\\mathbf{i} - \\mathbf{j} + \\mathbf{i} + 3\\mathbf{j} + 5\\mathbf{k} = 3\\mathbf{i} + 2 \\mathbf{j} + 5\\mathbf{k}$ $\\begin{array}{ll} \\mathbf{2a} - \\mathbf{b} &= 2(2\\mathbf{i} - \\mathbf{j}) - (\\mathbf{i} + 3\\mathbf{j} + 5\\mathbf{k}) \\\\ &= 4\\mathbf{i} - 2\\mathbf{j} - \\mathbf{i} - 3\\mathbf{j} - 5\\mathbf{k} \\\\ & = 3\\mathbf{i} - 5 \\mathbf{j} - 5\\mathbf{k} \\end{array}$ $\\begin{array}{ll} \\mathbf{c} + \\mathbf{2b} - 3\\mathbf{a} &= (\\mathbf{i} + \\mathbf{j} - 2\\mathbf{k}) + 2(\\mathbf{i} + 3\\mathbf{j} + 5\\mathbf{k}) - 3(2\\mathbf{i} - \\mathbf{j})\\\\ &= \\mathbf{i} + \\mathbf{j} - 2\\mathbf{k} + 2\\mathbf{i} + 6\\mathbf{j} + 10\\mathbf{k} - 6\\mathbf{i} + 3\\mathbf{j}\\\\ &= -3\\mathbf{i} + 10\\mathbf{j} + 8\\mathbf{k} \\end{array}$ 3. Let $$\\mathbf{p}", "from left-to-right $$\\vecs r(t)=\\tan t\\,\\hat{\\mathbf{i}}+6\\sec t\\,\\hat{\\mathbf{j}}$$ 40) The hyperbola, $$\\frac{x^2}{49} - \\frac{y^2}{64} = 1$$, right piece is oriented from bottom-to-top $$\\vecs r(t)=7\\sec t\\,\\hat{\\mathbf{i}}+8\\tan t\\,\\hat{\\mathbf{j}}$$ ### Parameterizing a Piecewise Path For questions 41 - 44, provide a parameterization for each piecewise path. Try to write a parameterization that starts with $$t = 0$$ and progresses on through values of $$t$$ as you move from one piece to another. 41) a. $$\\vecs r_1(t)= t\\,\\hat{\\mathbf{i}} + t^4 \\,\\hat{\\mathbf{j}}$$ for $$0 \\le t \\le 1$$ $$\\vecs r_2(t)= -t\\,\\hat{\\mathbf{i}} + \\sqrt[3]{-t} \\,\\hat{\\mathbf{j}}$$ for $$-1 \\le t \\le 0$$ So a piecewise parameterization of this path is: $$\\vecs r(t) = \\begin{cases} t\\,\\hat{\\mathbf{i}} + t^4 \\,\\hat{\\mathbf{j}}, & 0 \\le t \\le 1 \\\\ \\left(2-t\\right)\\,\\hat{\\mathbf{i}} + \\sqrt[3]{2-t} \\,\\hat{\\mathbf{j}}, & 1 \\lt t\\le 2 \\end{cases}$$ b. $$\\vecs r_1(t)= t\\,\\hat{\\mathbf{i}} + \\sqrt[3]{t} \\,\\hat{\\mathbf{j}}$$ for $$0 \\le t \\le 1$$ $$\\vecs r_2(t)= -t\\,\\hat{\\mathbf{i}} + (-t)^4 \\,\\hat{\\mathbf{j}}$$ for $$-1 \\le t \\le 0$$ So a piecewise parameterization of this path is: $$\\vecs r(t) = \\begin{cases} t\\,\\hat{\\mathbf{i}} + \\sqrt[3]{t} \\,\\hat{\\mathbf{j}}, & 0 \\le t \\le 1 \\\\ \\left(2-t\\right)\\,\\hat{\\mathbf{i}} + \\left(2-t\\right)^4 \\,\\hat{\\mathbf{j}}, & 1 \\lt t\\le 2 \\end{cases}$$ 42) 43) a. b. a. $$\\vecs r_1(t)= t\\,\\hat{\\mathbf{i}} +0 \\,\\hat{\\mathbf{j}}$$ for $$0 \\le t \\le 2$$ $$\\vecs r_2(t)= -t\\,\\hat{\\mathbf{i}} + \\left(2 + t\\right) \\,\\hat{\\mathbf{j}}$$ for $$-2 \\le t \\le -1$$ $$\\vecs r_3(t)= -t\\,\\hat{\\mathbf{i}} +", "point to find the single parameter determining the line $\\frac{a}{b} = -\\frac{y_1}{x_1} = -\\frac{y_2}{x_2} \\ .$ Last edited by a moderator: May 5, 2017" ]

[ "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "# Problem: Is the line through (4,6,3) and (2,0, -1) parallel to the line through (6,22,12) and (1,7,2)? ###### FREE Expert Solution The equation of a line with the direction vector d = (l,m,n) passing through a point (x1,y1,z1) is: $\\overline{)\\frac{\\mathbf{x}\\mathbf{-}{\\mathbf{x}}_{\\mathbf{1}}}{\\mathbf{L}}{\\mathbf{=}}\\frac{\\mathbf{y}\\mathbf{-}{\\mathbf{y}}_{\\mathbf{1}}}{\\mathbf{m}}{\\mathbf{=}}\\frac{\\mathbf{z}\\mathbf{-}{\\mathbf{z}}_{\\mathbf{1}}}{\\mathbf{n}}}$ For the first line: $\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{4}\\mathbf{-}\\mathbf{2}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{6}\\mathbf{-}\\mathbf{0}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{3}\\mathbf{-}\\mathbf{\\left(}\\mathbf{-}\\mathbf{1}\\mathbf{\\right)}}\\phantom{\\rule{0ex}{0ex}}\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{2}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{6}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{4}}\\phantom{\\rule{0ex}{0ex}}\\frac{\\mathbf{x}\\mathbf{-}\\mathbf{4}}{\\mathbf{1}}\\mathbf{=}\\frac{\\mathbf{y}\\mathbf{-}\\mathbf{6}}{\\mathbf{3}}\\mathbf{=}\\frac{\\mathbf{z}\\mathbf{-}\\mathbf{3}}{\\mathbf{2}}$ ###### Problem Details Is the line through (4,6,3) and (2,0, -1) parallel to the line through (6,22,12) and (1,7,2)?", "## Elementary Linear Algebra 7th Edition $\\mathbf{z}=(\\frac{7}{2},3,\\frac{5}{2})$ $\\mathbf{u}=(1,2,3)\\quad\\mathbf{v}=(2,2-1)\\quad\\mathbf{w}=(4,0,-4)$ Find $\\mathbf{z}$ if $2\\mathbf{z}-3\\mathbf{u}=\\mathbf{w}$ We can solve for $\\mathbf{z}$: $2\\mathbf{z}=\\mathbf{w}+3\\mathbf{u}$ $\\mathbf{z}=\\frac{1}{2}\\mathbf{w}+\\frac{3}{2}\\mathbf{u}$. All we need to do now is plug in the vectors $\\mathbf{w}$ and $\\mathbf{u}$: $\\mathbf{z}=\\frac{1}{2}(4,0,-4)+\\frac{3}{2}(1,2,3)$ $\\mathbf{z}=(2,0,-2)+(\\frac{3}{2},3,\\frac{9}{2})=(\\frac{7}{2},3,\\frac{5}{2})$", "# Math Help - scalar products 1. ## scalar products The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively. Find the unit vectors in the directions of CA and CB. < Calculate the ange ACB in degrees. 2. Originally Posted by Oasis1993 The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively. Find the unit vectors in the directions of CA and CB. < Calculate the ange ACB in degrees. $CA = CO + OA$ $= -OC + OA$ $= OA - OC$ $= 2\\mathbf{i} - 2\\mathbf{j} + \\mathbf{k} - (4\\mathbf{i} + 2\\mathbf{j} + \\mathbf{k})$ $= -2\\mathbf{i} - 4\\mathbf{j} + 0\\mathbf{k}$. To find the unit vector in this direction, divide it by its length. $|CA| = \\sqrt{(-2)^2 + (-4)^2 + 0^2}$ $= \\sqrt{4 + 16 + 0}$ $= \\sqrt{20}$ $= 2\\sqrt{5}$. So $\\frac{OC}{|OC|} = \\frac{-2\\mathbf{i} - 4\\mathbf{j} + 0\\mathbf{k}}{2\\sqrt{5}}$ $= \\frac{-\\mathbf{i} - 2\\mathbf{j} + 0\\mathbf{k}}{\\sqrt{5}}$ $= \\frac{-\\sqrt{5}\\mathbf{i} - 2\\sqrt{5}\\mathbf{j} + 0\\mathbf{k}}{5}$ $= -\\frac{\\sqrt{5}}{5}\\mathbf{i} - \\frac{2\\sqrt{5}}{5}\\mathbf{j} + 0\\mathbf{k}$. Can you do the same to find the", "Griffiths 1.13 Let r be the separation vector from a fixed point (x', y', z') to the point (x,y,z), and let r be its length # Griffiths 1.13 Let r be the separation vector from a fixed point (x', y', z') to the point (x,y,z), and let r be its length P 945 points Griffiths 1.13 Let r be the separation vector from a fixed point (x', y', z') to the point (x,y,z), and let r be its length. Show that (a) $\\nabla (r^2) = 2\\mathbf{r}$ (b) $\\nabla (1/r) = -\\mathbf{r}/r^2$ (c) What is the general formula for $\\nabla (r^n)$? Griffiths 1 Pairoaj P 945 points define $$\\mathbf{\\hat{r}} = \\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}}$$ $$\\mathbf{r} = (x-x^\\prime)\\hat{\\mathbf{x}}+(y-y^\\prime)\\hat{\\mathbf{y}}+(z-z^\\prime)\\hat{\\mathbf{z}}$$ $$r=\\sqrt{(x-x^\\prime)^2+(y-y^\\prime)^2+(z-z^\\prime)^2}$$ part (a) $$\\nabla (r^2) = \\frac{\\partial}{\\partial x}(r^2)+\\frac{\\partial}{\\partial y}(r^2)+\\frac{\\partial}{\\partial z}(r^2)$$ $$\\nabla (r^2) = 2(x-x^\\prime)\\hat{\\mathbf{x}}+2(y-y^\\prime)\\hat{\\mathbf{y}}+2(z-z^\\prime)\\hat{\\mathbf{z}}$$ $$\\nabla (r^2) = 2[(x-x^\\prime)\\hat{\\mathbf{x}}+(y-y^\\prime)\\hat{\\mathbf{y}}+(z-z^\\prime)\\hat{\\mathbf{z}}]$$ $$\\nabla (r^2) = 2\\mathbf{r}$$ part (b) $$\\nabla (1/r) = \\nabla (r^{-1}) = \\frac{\\partial}{\\partial x}(r^{-1})+\\frac{\\partial}{\\partial y}(r^{-1})+\\frac{\\partial}{\\partial z}(r^{-1})$$ I will just show the partial for the x-component -- it is the same for y and z components $$\\nabla (r^{-1})=\\frac{\\partial}{\\partial x}[(x-x^\\prime)^2+(y-y^\\prime)^2+(z-z^\\prime)^2]^{-1/2}+\\ldots$$ $$\\nabla (r^{-1}) = -\\frac{1}{2}(r^2)^{-\\frac{3}{2}}2(x-x^\\prime)\\hat{\\mathbf{x}}+\\ldots$$ $$\\nabla (r^{-1}) = -(r^2)^{-\\frac{3}{2}}(x-x^\\prime)\\hat{\\mathbf{x}}+\\ldots$$ $$\\nabla (r^{-1}) = -r^{-3}(x-x^\\prime)\\hat{\\mathbf{x}}+\\ldots$$ put y and z components back in $$\\nabla (r^{-1}) = -r^{-3}(x-x^\\prime)\\hat{\\mathbf{x}}+-r^{-3}(y-y^\\prime)\\hat{\\mathbf{y}}+-r^{-3}(z-z^\\prime)\\hat{\\mathbf{z}}$$ $$\\nabla (r^{-1}) = -r^{-3}[(x-x^\\prime)\\mathbf{\\hat{x}}+(y-y^\\prime)\\mathbf{\\hat{y}}+(z-z^\\prime)\\mathbf{\\hat{z}}]$$ $$\\nabla (r^{-1}) = -\\frac{1}{r^3}\\mathbf{r}=-\\frac{1}{r^2}\\mathbf{\\hat{r}}$$ part (c) in general (for the x component -- again its the same for the y and z),", "v and w. Sci Advisor PF Gold P: 1,352 Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. P: 101 Quote by DrClaude Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? Sci Advisor PF Gold P: 1,352 I'm", "this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt},\\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. 8. Feb 14, 2013 ### strokebow Thanks for your help. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? 9. Feb 14, 2013 ### Staff: Mentor I'm not sure I follow you. The \"first step\" is the calculation of $\\mathbf{k}_1$. The first element of $\\mathbf{f}(\\mathbf{a}_n,t_n)$ is $$\\left.\\frac{du}{dt}\\right|_{t_n} = K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n )$$ so the first element of $\\mathbf{k}_1$ is $$h \\times \\left[ K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n ) \\right]$$ and so on. For instance, the 4th element of $\\mathbf{k}_1$ is $$u_n$$ Once you have calculated all six elements of $\\mathbf{k}_1$, the \"second step\" is do", "with the line, and thus with $\\mathbf{l}$, then the distance from the point $\\mathbf{p}$ to the line is $d=|\\mathbf{r}|\\sin\\theta$. Now, recall that the magnitude of the cross product of two vectors $\\mathbf{a}$ and $\\mathbf{b}$ is given by $|\\mathbf{a}\\times\\mathbf{b}|=|\\mathbf{a}||\\mathbf{ b}|\\sin\\theta$. Thus, $d=\\frac{|\\mathbf{r}\\times\\mathbf{l}|}{|\\mathbf{l}| }$ $d=\\frac{|(\\mathbf{p}-\\mathbf{q})\\times\\mathbf{l}|}{|\\mathbf{l}|}$ Our point is $\\mathbf{p}=\\mathbf{\\hat{i}}-2\\mathbf{\\hat{j}}+4\\mathbf{\\hat{k}}$, and our line is that which passes through the point $\\mathbf{q}=-3\\mathbf{\\hat{j}}+2\\mathbf{\\hat{k}}$ parallel to the vector $\\mathbf{l}=2\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\ma thbf{\\hat{k}}$. Thus $\\mathbf{r}=\\mathbf{p}-\\mathbf{q}=\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\mat hbf{\\hat{k}}$, and $d=\\frac{|\\mathbf{r}\\times\\mathbf{l}|}{|\\mathbf{l}| }$ $=\\frac{|(\\mathbf{\\hat{i}}+\\mathbf{\\hat{j}}+2\\mathb f{\\hat{k}})\\times(2\\mathbf{\\hat{i}}+\\mathbf{\\hat{j }}+2\\mathbf{\\hat{k}})|}{|2\\mathbf{\\hat{i}}+\\mathbf {\\hat{j}}+2\\mathbf{\\hat{k}}|}$ $=\\frac{|2\\mathbf{\\hat{j}}-\\mathbf{\\hat{k}}|}{\\sqrt{2^2+1^2+2^2}}$ $=\\frac{\\sqrt{2^2+(-1)^2}}{3}$ $d=\\frac{\\sqrt{5}}{3}$ --Kevin C. 5. THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote \"OH NO\" when i said we have to minimize the square of d. Why? Then he gave me 0 points. 6. Originally Posted by Intsecxtanx THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote \"OH NO\" when i said we have to minimize the square of d. Why? Then he gave me 0 points. Only that instructor can really answer that question. Prehaps he wanted you to use only vector methods. But as you can see the answers are the same.", "$\\mathbf r = \\mathbf a + t\\frac{\\mathbf b - \\mathbf a}{\\lVert \\mathbf b - \\mathbf a \\rVert}$. – Théophile May 10 '17 at 14:06 • Okay, so I'm taking the unit vector of this line, and the unit vector, when scaled to ||$b-a$||, is at $a$, so then is the scalar for this? If this is true, how do I now make it so it's 3 units away? – sangstar May 10 '17 at 18:56 ## 1 Answer \"A direction\" of the line $\\overleftrightarrow{AB}$ is $B-A=(0,-3,0)$. (Another direction is $A-B$. But we won't use that.) So every point on the line $\\overleftrightarrow{AB}$ can be expressed as $$P(t)=A +t(B-A) = (1,2,-2)-t(0,3,0)$$ The distance between the point $P(t)$ and $A$ is $||P(t)-A||=||t(0,3,0)|| = 3|t|.$ So the points on the line that are a distance of $3$ from $A$ satisfy the equation $3|t|=3$. Hence $t\\in \\{1, -1\\}$. The points are $(1,2,-2) \\pm (0,3,0)$", "$i$ for the moment and rewrite this term in the form $$y_k = \\mathbf u_i^\\prime \\mathbf m_k,\\tag{2}$$ for now we see that $c_i$ minimizes the sum $\\sum_{k=1}^N (y_k - c_i)^2$. This is the very simplest version of an Ordinary Least Squares problem, involving just one parameter $c_i$. It's easy to see that $$c_i = \\bar y_k = \\frac{1}{N}\\sum_{k=1}^N y_k\\tag{3}$$ is the unique solution, because for any number $c$, the sum of squares $$\\sum_{k=1}^N (y_k - c)^2 = \\sum_{k=1}^N (y_k - \\bar y_k)^2 + N(c - \\bar y_k)^2$$ depends on $c$ only through the second term, which obviously is minimized uniquely at $c = \\bar y_k$. Consider now the mean vector (or \"barycenter\") $$\\mathbf{\\bar m} = \\frac{1}{N}\\sum_{k=1}^N \\mathbf{m}_k.$$ To see whether it lies on $\\mathcal L$, plug it in for $\\mathbf x$ on the left hand side of $(1)$ and, for each $i$, use $(2)$ and $(3)$ to compute $$\\mathbf u_i^\\prime \\mathbf{\\bar m} = \\mathbf u_i^\\prime \\frac{1}{N}\\sum_{k=1}^N \\mathbf{m}_k = \\frac{1}{N}\\sum_{k=1}^N\\mathbf u_i^\\prime\\mathbf{m}_k = \\frac{1}{N}\\sum_{k=1}^Ny_k = \\bar y_k = c_i.$$ Thus $(1)$ holds, proving that the mean vector must lie on $\\mathcal L$. I believe you're talking about the least square regression line. Basically, this is the line where the sum of the residuals of each data point to the line is minimised. A residual is the distance from the point", "# Math Help - Finding a vector 1. ## Finding a vector Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? I don't understand how they say the vector is (4,-3) 2. Originally Posted by nyasha Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? $3x + 4y = -2$ $4y = -3x - 2$ $y = -\\frac{3}{4}x - \\frac{1}{2}$. So for every $4$ units you have gone to the right, you have gone $3$ units down. This would suggest that this line would have direction the same as $4\\mathbf{i} - 3\\mathbf{j}$. But we don't know the length of this vector. So multiply the vector by some parameter $t$. Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as $t(4\\mathbf{i} - 3\\mathbf{j})$. Now we just need a point that lies on the line so we know where to place this vector. The $y$ intercept will suffice. Therefore, the vector form of this line is $\\left(0\\mathbf{i} - \\frac{1}{2}\\mathbf{j}\\right) + t(4\\mathbf{i} - 3\\mathbf{j})$ or if you like $4t\\mathbf{i} - \\left(\\frac{1}{2} + 3t\\right)\\mathbf{j}$. 3. Originally Posted by Prove It $3x + 4y = -2$ $4y = -3x - 2$", "-5,000 8000 oooo w... ##### Evaluate the line integral (r+ 2)ds, where € is given by X=( Y=tm, 12 =47 , 0 <t<2 Evaluate the line integral (r+ 2)ds, where € is given by X=( Y=tm, 12 =47 , 0 <t<2... ##### Show that the vector-valued function $\\mathbf{r}(t)=(2 \\mathbf{i}+2 \\mathbf{j}+\\mathbf{k})$ $+\\cos t\\left(\\frac{1}{\\sqrt{2}} \\mathbf{i}-\\frac{1}{\\sqrt{2}} \\mathbf{j}\\right)+\\sin t\\left(\\frac{1}{\\sqrt{3}} \\mathbf{i}+\\frac{1}{\\sqrt{3}} \\mathbf{j}+\\frac{1}{\\sqrt{3}} \\mathbf{k}\\right)$describes the motion of a particle moving in the circle of radius 1 centered at the point (2,2,1) and lying in the plane $x+y-2 z=2$. Show that the vector-valued function $\\mathbf{r}(t)=(2 \\mathbf{i}+2 \\mathbf{j}+\\mathbf{k})$ $+\\cos t\\left(\\frac{1}{\\sqrt{2}} \\mathbf{i}-\\frac{1}{\\sqrt{2}} \\mathbf{j}\\right)+\\sin t\\left(\\frac{1}{\\sqrt{3}} \\mathbf{i}+\\frac{1}{\\sqrt{3}} \\mathbf{j}+\\frac{1}{\\sqrt{3}} \\mathbf{k}\\right)$describes the mo... ##### In Campbell v. S.S. Kresge Co., the court found that a department store security guard had... In Campbell v. S.S. Kresge Co., the court found that a department store security guard had falsely imprisoned a shopper who was attempting to leave the store where she had been shopping. If an identical situation were to occur today, on what grounds could the employer (the department store) be held ... ##### Question 4 2 pts In the circuit shown below, the emf of the battery E=34 volts,... Question 4 2 pts In the circuit shown below, the emf of the battery E=34 volts, and the resistors have values R1 = 1902, R2 = 1362, and R3 = 127 12.", "2)$ and $Q(4, 6)$ that define a line $L1$, what are one set of values of $a$ and $b$ such that the vector $n = (a, b)$ is perpendicular to $L1$? We first note that $\\vec{PQ} = (3, 4)$, and thus it follows we need to construct a dot product $\\vec{n} \\cdot \\vec{PQ} = 0$. It thus follows that we need a linear combination of $a$ and $b$ that satisfies the equation $3a + 4b = 0$. Let $a = t$. Therefore $b = -\\frac{3t}{4}$. Thus it follows that $n = (t, -\\frac{3t}{4})$ where $t \\in \\mathbb{R}$. One set of values occurs when $t = 4$ and thus $a = 4$ and $b = -3$.", "take $n-2$ more points, evenly spaced in [a,b] $u_i = k / {n-1} \\forall k \\in [1 .. n-1]$ $\\mathbf{B}_p(u (b - a) + a) = \\mathbf{B}_q(u_i)$ for each $u_i$ For a 4 control points bezier curve If we take $u$ in $(0, {1 \\over 3}, {2 \\over 3}, 1)$, hence $t_0$, $t_1$, $t_2$, $t_3$ respectively equal to $a$, $\\frac{b - 2 a}{3}$, $\\frac{2b - a}{3}$, $b$ $\\mathbf{Q}_0 = \\mathbf{B}_p(t_0)$ $\\mathbf{Q}_1 = {-15 \\mathbf{B}_p(t_0) + 54 \\mathbf{B}_p(t_1) -27 \\mathbf{B}_p(t_2) + 6 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_2 = {6 \\mathbf{B}_p(t_0) -27 \\mathbf{B}_p(t_1) + 54 \\mathbf{B}_p(t_2) -15 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_3 = \\mathbf{B}_p(b)$", "Find the components of Let $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components ofLet $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components of (a) $\\mathbf{u}+\\mathbf{w}$ (b) $\\mathbf{v}-3 \\mathbf{u}$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vectorShow that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector ... Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components ofLet $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of (a) $\\mathbf{v}-\\mathbf{w}$ (b) $6 \\mathrm{u}+2 \\ma ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$0 answers 58 views Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$0 answers 88 views Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)\\$ ...", "## Parallel Lines Vector Question In the diagram P splits AB in the ratio 1:2. BQ=k(BC) and PQD is a straight line. If $\\mathbf{AB}=\\mathbf{a}, \\; \\mathbf{AC}=\\mathbf{c}, \\; \\mathbf{CD}=3\\mathbf{a}$ find the value of $k$ . Since P splits AB in the ratio 1:2, $\\mathbf{AP}=\\frac{1}{3} \\mathbf{a}, \\; \\mathbf{PB}=\\frac{2}{3} \\mathbf{a}$ . $\\mathbf{BC}=- \\mathbf{a}+ \\mathbf{b}$ so $\\mathbf{BQ}=k(- \\mathbf{a}+ \\mathbf{b})$ . Then $\\mathbf{PQ}= \\mathbf{PB}+ \\mathbf{BQ} =\\frac{2}{3} \\mathbf{a}+k(- \\mathbf{a}+ \\mathbf{b})=(\\frac{2}{3}-k) \\mathbf{a} + k \\mathbf{b}$ $\\mathbf{PD}= - \\frac{1}{3} \\mathbf{a} ++4 \\mathbf{b}$ . $PQD$ is a straight line so $\\mathbf{PQ}, \\; \\mathbf{PD}$ are parallel and the coefficients of $\\mathbf{a}, \\; \\mathbf{b}$ are in the same ratio. Hence $\\frac{2/3-k}{-1/3}= \\frac{k}{4} \\rightarrow\\frac{8}{3}-4k=- \\frac{k}{3} \\rightarrow \\frac{8}{3}= \\frac{11k}{3} \\rightarrow k =\\frac{\\frac{8}{3}}{\\frac{11}{3}}=\\frac{8}{11}$", "About vector form of a line passing through 2 points. According to my book: Equation of line passing through 2 points with position vectors $a$ and $b$ is $$r = a + K(b - a)$$ My question: If we are given 2 points how do we determine which point is to be taken as b and which as a? For example: Here's a question from my book: Find the equation of line passing through A (5, 1, 6), B (3, 4, 1). I can get 2 equations depending on the points that I select as $a$ and $b$ respectively. If I take $a = 5i + 1j + 6k$ and $b = 3i + 4j + 1k$ I get: $$b - a = -2i + 3j - 5k$$ $$r = 5i + 1j + 6k + K(-2i + 3j - 5k)$$ $$r = (5-2K)i + (1+3K)j + (6 - 5K)k$$ In cartesian form: $$\\frac{5 - x}{2} = \\frac{y - 1}{3} = \\frac{6 - z}{5}$$ If I take $a = 3i + 4j + 1k$ and $b = 5i + 1j + 6k$ I get: $$b - a = 2i - 3j + 5k$$ $$r = 3i + 4j + 1k + K(2i - 3j + 5k)$$ $$r = (3 + 2K)i + (4 - 3K)j + (1 +", "the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=(\\sin t) \\mathbf{i}+\\left(t^{2}-\\cos t\\right) \\mathbf{j}+e^{z} \\mathbf{k}, \\quad t_{0}=0 Check back soon! Problem 20 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=t^{2} \\mathbf{i}+(2 t-1) \\mathbf{j}+t^{3} \\mathbf{k}, \\quad t_{0}=2 Check back soon! Problem 21 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$ . \\mathbf{r}(t)=\\ln t \\mathbf{i}+\\frac{t-1}{t+2} \\mathbf{j}+t \\ln t \\mathbf{k}, \\quad t_{0}=1 Check back soon! Problem 22 As mentioned in the text, the tangent line to a smooth curve $\\mathbf{r}(t)=f(t) \\mathbf{i}+g(t) \\mathbf{j}+h(t) \\mathbf{k}$ at $t=t_{0}$ is the line that passes through the point $\\left(f\\left(t_{0}\\right), g\\left(t_{0}\\right), h\\left(t_{0}\\right)\\right)$ parallel to $\\mathbf{v}\\left(t_{0}\\right),$ the curve's velocity vector at $t_{0}$ Find parametric equations for the line that is tangent to the given curve at the given parameter value $t=t_{0}$", "paper. Hölder's inequality states that $\\sum_{i=1}^n a_i b_i \\leq \\| \\mathbf{ a}\\|_{ p} \\cdot \\| \\mathbf{ b}\\|_{ q}$ for any vector $\\mathbf{ a},\\mathbf{ b}$ so long as $\\frac1p + \\frac1q = 1$. If we take $\\mathbf{ a}=\\mathbf{ b}=\\mathbf{ v}-\\mathbf{ v}_k$, $p=1$, and $q=1$, we find that $\\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 2}^2 \\leq \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1} \\ \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ \\infty}$. Further, note that by our construction of $\\mathbf{ v}_k$, $\\| \\mathbf{ v}\\|_{ 1} = \\| \\mathbf{ v}_k\\|_{ 1} + \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1}$. Then, $\\frac{\\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 2}}{\\| \\mathbf{ v}\\|_{ 1}} \\leq \\frac{\\sqrt{\\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1} \\ \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ \\infty}}}{\\| \\mathbf{ v}_k\\|_{ 1} + \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1}}$ Note that $\\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ \\infty} = \\lambda_{k+1}$ and $\\| \\mathbf{ v}_k\\|_{ 1} = \\sum_{i=1}^k \\lambda_i \\geq k\\lambda_{k+1}$. Then, if we let $\\gamma \\;{\\vcentcolon=}\\; \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1}$, we find $\\frac{\\sqrt{\\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1} \\ \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ \\infty}}}{\\| \\mathbf{ v}_k\\|_{ 1} + \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1}} \\leq \\frac{\\sqrt{\\gamma \\ \\lambda_{k+1}}}{k\\lambda_{k+1} + \\gamma}$ We don't have a good way to bound $\\gamma = \\| \\mathbf{ v}-\\mathbf{ v}_k\\|_{ 1}$, so we instead notice that the function $\\gamma \\mapsto \\frac{\\sqrt{a\\gamma}}{b+\\gamma}$ is maximized at $\\gamma=b$ (check with calculus). So, $\\frac{\\sqrt{a\\gamma}}{b+\\gamma} \\leq \\frac{\\sqrt{ab}}{2b}$ for all $\\gamma$: $\\frac{\\sqrt{\\gamma \\ \\lambda_{k+1}}}{k\\lambda_{k+1} +", "For the given position vectors $\\mathbf{r}(t)$, compute the (tangent) velocity vector $\\mathbf{r}'(t)$ for the given value of $t$ . A) $\\displaystyle \\textrm{Let } \\mathbf{r}(t)= (\\cos 5t,\\, \\sin 5t )$. Then $\\mathbf{r}'(\\frac{\\pi}{4})$= ( , )? B) $\\displaystyle \\textrm{Let } {\\mathbf{r}}(t)= (t^2,t^3)$. Then ${\\mathbf{r}}'(4)$= ( , )? C) $\\displaystyle \\textrm{Let } \\mathbf{r}(t)= e^{5t}\\mathbf{i}+ e^{-4t}\\mathbf{j}+ t\\mathbf{k}$. Then $\\mathbf{r}'(1)$= $\\mathbf{i}+$ $\\mathbf{j}+$ $\\mathbf{k}$ ? Your overall score for this problem is" ]

[ "Note that, S is the first octant part of the plane, then you will have the region in xy-plane bounded by the $x$-axis, $y$-axis and the $y=1-x$, $x\\geq 0, y\\geq 0.$ -", "# Math Help - Finding an area using matrices & determinants 1. ## Finding an area using matrices & determinants Find the area of the triangle with vertices, $(1,3)$, $(2,7)$ and $(4,1)$. I went about this by drawing a diagram and then forming a rectangle containing the area described above, and 3 other triangles, calculated the area of these and subtracted it to find the area in the question, 7. However looking at the answers it appears as if we were meant to do this using matrices and determinants. Here's what the solutions say: Area $= \\frac{1}{2}det(A)$ where $A = \\begin{pmatrix} 1&1&3 \\\\ 1&2&7 \\\\ 1&4&1 \\end{pmatrix}$ . Now I gather that the last two columns are the coordinates of the vertices, just unsure where the 1s in the first column come from.", "\\widehat{\\boldsymbol{k}}$ and ${}S$ is the surface of the region ${}T$ bounded by $z=1-x^2$ and the planes $z=0,$ $y=0$ and $y+z=2.$ $\\Box$", "The area of region in $$xy$$-plane consisting of all points $$(a,b)$$ such that quadratic equation $$ax^2 + 2(a + b – 7)x + 2b = 0$$ has fewer than $$2$$ real solutions is?", "? Free Version Easy # Plane Area with Curly Bounds II MVCALC-KKPEEC Find the area of the plane region $R$ in figure below by setting up a correct iterated integral: $$R=\\{(x,y): g(y)\\leq x\\leq h(y), c\\leq y\\leq d.\\}$$ A $\\int_c^d \\int_{g(y)}^{h(y)} 1dxdy$ B $\\int_c^d \\int_{h(y)}^{g(y)} 1dxdy$ C $\\int_c^d \\int_{g(y)}^{h(y)} 1dydx$ D $\\int_{g(y)}^{h(y)} \\int_c^d 1dydx$ E $\\int_{h(y)}^{g(y)} \\int_c^d 1dxdy$", "# Sky Region Geometry Level 4 Find the area of the sky region if radius of circle is $$\\frac{1}{4}$$ and each side of the square is 1. The area of sky region is looks like as $$\\frac{a-\\sqrt{b}}{c}$$, where $$a$$, $$b$$, $$c$$ are positive integers and $$b$$ is square free then find $$a+b-c$$. ×", "great explanation. How can I determine the region of the $(x,y)$-plane in which $z(x,y)$ is uniquely determined? – Mathmo Jan 13 '12 at 10:39 • @MiamiMaths: All $\\mathbb{R}^2$ (as both $x(s)$ and $y(s)$ are invertible. – Fabian Jan 13 '12 at 16:33", "asking for the area of a region on the given plane.) 5. It's a linear transformation $T(\\begin{bmatrix}u\\\\v \\end{bmatrix})=\\begin{bmatrix}u-v\\\\u+v\\\\u \\end{bmatrix}\\rightarrow T(\\vec{e}_1)=\\begin{bmatrix}1\\\\1\\\\1 \\end{bmatrix}~,~ T(\\vec{e}_2)=\\begin{bmatrix}-1\\\\1\\\\0 \\end{bmatrix}$ $A=\\begin{bmatrix}1&-1\\\\1&1\\\\1&0 \\end{bmatrix}~,~S_{\\Phi}=\\sqrt{\\det (A^TA)}\\cdot S_{D}=\\sqrt{6}~\\pi r^2=\\sqrt{6}~\\pi$ 6. Thank You much appreciation.", "to the axes. For each projection, the area is $1$. This makes a sum of $3$. Now rotate the cube by $45^\\circ$ along the $z$-axis. When projecting onto the $xy$-plane, you still get an area of $1$. For the other projections, you get $\\sqrt{2}$ each. This makes a total of $1 + 2\\sqrt{2} \\neq 3$.", "region is on one side of $$P$$ or the other, so it either passes through $$P^+$$ or $$P^-$$. The plane $$P^+$$ is $$(n-1)$$ dimensional, so it is divided by the other $$m-1$$ planes into $$\\sum_{j=0}^{n-1}\\binom{m-1}j$$ regions, by the previous result. These regions of $$P^+$$ correspond exactly to the original regions which are on the $$P^+$$ side. You then multiply by $$2$$ to also account for the $$P^-$$ regions.", "b, and B transform the $$xy$$-plane. Points in the plane can be written as column vectors and I want you to focus on these four: $p_1 = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix}, \\quad p_2 = \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}, \\quad p_3 = \\begin{pmatrix}0 \\\\ 1\\end{pmatrix}, \\quad \\text{and} \\quad p_4 = \\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}$ Here's a picture of the $$xy$$-plane, with the diagonal lines and the four points included: Fig. 1. The $$xy$$-plane. Let's see what happens when you multiply some of these points with a: $a \\cdot p_1 = \\begin{pmatrix}1 & 2 \\\\ 0 & 1\\end{pmatrix} \\begin{pmatrix}1 \\\\ 0\\end{pmatrix} = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix} = p_1 \\qquad \\text{and} \\qquad a \\cdot p_4 = \\begin{pmatrix}1 & 2 \\\\ 0 & 1\\end{pmatrix} \\begin{pmatrix}-1 \\\\ 0\\end{pmatrix} = \\begin{pmatrix}1 \\\\ 1\\end{pmatrix} = p_2$ So a fixes $$p_1$$, sends $$p_4$$ to $$p_2$$, and (you should check this) $$p_2$$ to somewhere to the right between the diagonal and the $$x$$-axis. You can do the same with with b; the next picture is an illustration of the two transformations: Fig. 2. The left and right figures are the before and after pictures of the $$xy$$-plane. Notice how the shaded regions $$\\mathcal A^+$$ and $$\\mathcal B^+$$ in the two after pictures don't overlap? Do you also notice how the blank regions $$\\mathcal A^-$$ and $$\\mathcal B^-$$ in the before pictures", "from the given region $R$ in the $z$ -plane onto the target region $R^{prime}$ in the $w$ -plane, and find the image of the given boundary curve....", "that it's fundamental in Data Analysis and Probability and Statistics) $e$ is like matrices. Or sets. When you first learn them, you think, \"Err...so what?\" As you learn more mathematics, you find yourself using these things more and more, until you can't imagine doing mathematics without them. For $u>0$ and $f(u)=\\int_1^u (1/x)dx$ , a geometric representation of $e$ is that $f(e)=1$. The region bounded by the $x-$axis, the line $y=1$,the graph of $y=f(x)$, and the line $y=u$, has area $1$ when $u=e$. This can be taken as a definition of $e$, analogous to the definition of $\\pi$ as the area of a semi-circle of radius $1$.", "+0 0 336 2 +115 What is the area of the region of the complex plane defined by $$|z|<5$$ ? Oct 13, 2018 #1 +115 +2 Shouldn't it just be $$\\boxed{25\\pi}$$ Oct 13, 2018 #2 +110149 +2 What is the area of the region of the complex plane defined by |z|<5 ? $$let\\\\ z=a+bi\\\\ where\\\\ a^2+b^2<25$$ I think it is that simple. Oct 13, 2018", "# Definition:Region/Plane ## Definition A point set $R$ in the plane is a region if and only if: $(1): \\quad$ Each point of $R$ is the center of a circle all of whose elements consist of points of $R$ $(2): \\quad$ Each point of $R$ can be joined by a curve consisting entirely of points of $R$.", "# Determinants and Areas. The intention here is to prove that application of a 2 x 2 matrix to regions of the plane multiplies the area by the absolute. ## Presentation on theme: \"Determinants and Areas. The intention here is to prove that application of a 2 x 2 matrix to regions of the plane multiplies the area by the absolute.\"— Presentation transcript: Determinants and Areas The intention here is to prove that application of a 2 x 2 matrix to regions of the plane multiplies the area by the absolute value of the determinant of the matrix. Begin with a unit square: (1,0) (0,1) Transform this by a matrix (1,0) (0,1) Transform this by a matrix (1,0) (0,1) (a,c) Transform this by a matrix (1,0) (0,1) (b,d) (1,0) (0,1) (b,d) (a,c) So the unit square is transformed into a parallelogram (a+b,c+d) (1,0) (0,1) (b,d) (a,c) (a+b,c+d) We need to show that the area of the parallelogram is |ad-bc| (1,0) (0,1) (b,d) (a,c) (a+b,c+d) The base of the parallelogram has length (a 2 +c 2 ) 1/2 (a 2 +c 2 ) 1/2 (1,0) (0,1) (b,d) (a,c) (a+b,c+d) We need the altitude (a 2 +c 2 ) 1/2 ? (1,0) (0,1) (b,d) (a,c) (a+b,c+d) The residual projection of (b,d) onto (a,c) is an altitude (1,0) (0,1) (b,d) (a,c)", "region in the z=0 plane is a just a rectangle which it is not. So the region is actually: $E=( (x,y) | 1 \\le x \\le 2 , 1+\\frac{x}{2} \\le y \\le 1+x )$ ? The given order is to write the specific integral as shown in the theorem...No calculation is needed... Does my representation above is now correct? Or am I still wrong at something? Hope you'll be able to help me understand my mistakes Thanks", "1. Draw a labeled picture of $T$ in the $st$-plane. 2. Find the image of the $st$-vertex $(0,1)$ in the $xy$-plane. Likewise, find the respective images of the other three vertices of the rectangle $T\\text{:}$ $(0,4)\\text{,}$ $(1,1)\\text{,}$ and $(1,4)\\text{.}$ 3. In the $xy$-plane, draw a labeled picture of the image, $T'\\text{,}$ of the original $st$-rectangle $T\\text{.}$ What appears to be the shape of the image, $T'\\text{?}$ 4. To transform an integral with a change of variables, we need to determine the area element $dA$ for image of the transformed rectangle. Note that $T'$ is not exactly a parallelogram since the equations that define the transformation are not lienar. But we can approximate the area of $T'$ with the area of a parallelogram. How would we find the area of a parallelogram that approximates the area of the $xy$-figure $T'\\text{?}$ (Hint: Remember what the cross product of two vectors tells us.) Activity 11.9.2 presents the general idea of how a change of variables works. We partition a rectangular domain in the $st$ system into subrectangles. Let $T = [a, b] \\times [a+\\Delta s, b+\\Delta t]$ be one of these subrectangles. Then we transform this into a region $T'$ in the standard $xy$ Cartesian coordinate system. The region $T'$ is called the image of $T\\text{;}$ the region $T$ is the", "# Definition:Region/Plane (Redirected from Definition:Region of Plane) A point set $R$ in the plane is a region if and only if: $(1): \\quad$ Each point of $R$ is the center of a circle all of whose elements consist of points of $R$ $(2): \\quad$ Each point of $R$ can be joined by a curve consisting entirely of points of $R$.", "find the distance PQ and the image of the point P treating this plane as a mirror Concept: Plane - Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point Chapter: [4.01] Three - Dimensional Geometry [6] 25 [6] 25.1 Using elementary row transformation, find the inverse of the matrix [(2,-3,5),(3,2,-4),(1,1,-2)] Concept: Inverse of a Matrix - Inverse of a Nonsingular Matrix by Elementary Transformation Chapter: [2.02] Matrices OR [6] 25.2 Using matrices, solve the following system of linear equations : x + 2y − 3z = −4 2x + 3y + 2z = 2 3x − 3y − 4z = 11 Concept: Determinant of a Matrix of Order 3 × 3 Chapter: [2.01] Determinants [6] 26 [6] 26.1 Using integration, find the area of the region bounded by the parabola y= 4x and the circle 4x2 + 4y2 = 9. Concept: Area of the Region Bounded by a Curve and a Line Chapter: [3.03] Applications of the Integrals OR [6] 26.2 Using the method of integration, find the area of the region bounded by the lines 3x − 2y + 1 = 0, 2x + 3y − 21 = 0 and x − 5y + 9 = 0 Concept: Area of the Region Bounded by a Curve and a Line Chapter: [3.03]" ]

[ "3Blue1Brown # Chapter 6The determinant \"The purpose of computation is insight, not numbers.\" $\\qquad$— Richard Hamming ## Scaling Area One thing that turns out to be surprisingly useful for understanding a given transformation is to measure exactly how much a given transformation stretches and squishes things. More specifically, to measure the factor by which the area of a given region increases or decreases. For example, the matrix $\\begin{bmatrix}3&0\\\\0&2\\end{bmatrix}$ scales $\\hat i$ by a factor of $3$, and scales $\\hat j$ by a factor of $2$. How can we tell the factor by which space gets stretched? Focus your attention on the $1\\times 1$ square whose bottom sits on $\\hat i$, and whose left side sits on $\\hat j$. After the transformation, this turns into a $2\\times 3$ rectangle. Since this region started with area $1$, and ended up with area $2\\times 3=6$, we can say the linear transformation has scaled its area by a factor of $6$. So even though this transformation smooshes things about, it seems to leave areas unchanged. Actually, though, if you know how much the area of that one unit square changes, it tells you how the area of any region in space changes. For starters, notice that whatever happens to one square in the grid happens to any other square in the grid,", "# Scale Factor • Nov 8th 2009, 05:42 AM stratovarius Scale Factor hello, can someone explain me what a volume scale factor is and if I have a liner Transformation T(v1) = av1 + bv2 T(v2) = cv1 + dv2 how can I find the volume scale factor of it? Thank you • Nov 8th 2009, 06:29 AM Tuufless You mean area factor? (Volume factors are for 3D...) Think of it this way: if you have an enlargement matrix, say scale factor 2, then your matrix is $\\begin{bmatrix}2 & 0\\\\0 & 2\\end{bmatrix}$. So, if we define a unit square (1x1 length) using four points, say $A = \\begin{bmatrix}0\\\\0\\end{bmatrix}, B = \\begin{bmatrix}0\\\\1\\end{bmatrix}, C = \\begin{bmatrix}1\\\\1\\end{bmatrix}$ and $D = \\begin{bmatrix}1\\\\0\\end{bmatrix}$. Then if we put all those points into the matrix, the transformed points are $A' = \\begin{bmatrix}0\\\\0\\end{bmatrix}, B' = \\begin{bmatrix}0\\\\2\\end{bmatrix}, C' = \\begin{bmatrix}2\\\\2\\end{bmatrix}$ and $D' = \\begin{bmatrix}2\\\\0\\end{bmatrix}$. These points define a matrix with an area of 4 square units, so the original unit square got enlarged by an area factor of 4. That is to say, if you used points to define a polygon with area n square units, then the transformed shape will have an area of 4n square units. For any generic transformation matrix, the area factor is equal to the determinant. i.e: For any matrix $M =", "$CDP$. Therefore the area of $CDP$ is $\\frac{1}{2} \\cdot 13 \\cdot 84=\\boxed{546}$.", "quantities If you have measured values for the quantities X, Y, and Z, with uncertainties dX, dY, and dZ, and your final University Science Books, 327 pp. The area $$area = length \\cdot width$$ can be computed from each replicate. Or in matrix notation, f ≈ f 0 + J x {\\displaystyle \\mathrm σ 6 \\approx \\mathrm σ 5 ^ σ 4+\\mathrm σ 3 \\mathrm σ 2 \\,} where J is", "3 & 3 \\end{vmatrix}$ Area = ½ (6 – 6) Area = 0 Hence, the points P, Q and R are collinear. For more maths articles, visit byjus.com and download BYJU’S – The Learning App for exciting learning videos.", "\\begin{bmatrix}a & b\\\\c & d\\end{bmatrix}$, the area factor is $det(M) = ad-bc$ (for a 2x2 matrix). • Nov 8th 2009, 07:01 AM stratovarius Thanks so much for the reply, i think I understood what you say , but no I meant volume scale factor or metric scale factor, that's how it was given to me. Regards • Nov 8th 2009, 07:42 AM Tuufless Oh, oops- my bad, I thought you were dealing with 2x2 matrices only since your original post had a system of two vectors. In general, the volume factor is just the determinant of the matrix. (We say \"volume factor\" for anything above 2D.)", "\\cdot 0.4 = 0.04$. Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \\cdot 4 = 0.16$. By periodicity, this probability is the same for all $0 \\le x \\le 2012$ and $0 \\le y \\le 2012$. Therefore, the answer is $0.16 = \\boxed{4/25 (C) }$", "$V_2 ?$ 1. 108 units squared 2. 72 units squared 3. 54 units squared 4. 36 units squared C. What is the absolute value of the determinant of the transformation matrix, $T ?$ 1. 6 2. 7 3. 12 4. 3 D. How does the absolute value of the determinant of the transformation matrix relate to the area of the two quadrilaterals (the square, $S_1$ and the transformed shape, $S_2$)? Choose the equation which correctly describes this relationship. 1. $|det(T)| = \"Area\"(S_1) \\ \"Area\"(S_2)$ 2. $\"Area\"(S_2) = |det(T)| \\ \"Area\"(S_1)$ 3. $|det(T)| = |\"Area\"(S_2) - \"Area\"(S_1)|$ 4. There is no relationship between the absolute value of the determinant of T and the area of the shapes. You need to be a HelpTeaching.com member to access free printables.", "have a combined area of $$\\mathtt{-v_{1}v_{2}}$$. Thus, distributing and subtracting, we get $\\mathtt{u_{1}u_{2} + u_{1}v_{2} – v_{1}u_{2} – v_{1}v_{2} – u_{1}u_{2} – (-v_{1}v_{2})}$ Then, after simplifying, we have $$\\mathtt{u_{1}v_{2} – u_{2}v_{1}}$$. If the two vectors u and v represented a linear transformation and were written as column vectors in a matrix, then we could say that there is a determinant of the matrix and show the determinant of the matrix in the way it is usually presented: $\\begin{vmatrix}\\mathtt{u_1} & \\mathtt{v_1}\\\\\\mathtt{u_2} & \\mathtt{v_2}\\end{vmatrix} = \\mathtt{u_{1}v_{2} – u_{2}v_{1}}$ One thing to note is that this is a signed area. The sign records a change in orientation that we won’t go into at the moment. In fact, describing the determinant as an area is a little misleading. When you look at transformations, the determinant tells you the scale factor of the change in area. A determinant of 1 would mean that areas did not change, etc. Also, if we have vectors that are simply scaled versions of one another—the components of one vector are scaled versions of the other—then the determinant will be zero, which is pretty much what we want, since the area will be zero. Let’s use lambda ($$\\mathtt{\\lambda}$$) as our scalar to be cool. $\\,\\,\\,\\,\\,\\,\\quad\\,\\,\\,\\,\\,\\begin{vmatrix}\\mathtt{u_1} & \\mathtt{\\lambda u_1}\\\\\\mathtt{u_2} & \\mathtt{\\lambda u_2}\\end{vmatrix} = \\mathtt{\\lambda u_{1}u_{2} – \\lambda u_{1}u_{2} =", "$w\\cdot v'=3$. So we get the matrix \\begin{equation*}\\begin{pmatrix}0 & 1 & 0 \\\\ 0 & 0 & -1 \\\\ -1 & 0 & 0\\end{pmatrix}\\end{equation*} Yep. All correct.", "triangles have a combined area of $$\\mathtt{-v_{1}v_{2}}$$. Thus, distributing and subtracting, we get $\\mathtt{u_{1}u_{2} + u_{1}v_{2} – v_{1}u_{2} – v_{1}v_{2} – u_{1}u_{2} – (-v_{1}v_{2})}$ Then, after simplifying, we have $$\\mathtt{u_{1}v_{2} – u_{2}v_{1}}$$. If the two vectors u and v represented a linear transformation and were written as column vectors in a matrix, then we could say that there is a determinant of the matrix and show the determinant of the matrix in the way it is usually presented: $\\begin{vmatrix}\\mathtt{u_1} & \\mathtt{v_1}\\\\\\mathtt{u_2} & \\mathtt{v_2}\\end{vmatrix} = \\mathtt{u_{1}v_{2} – u_{2}v_{1}}$ One thing to note is that this is a signed area. The sign records a change in orientation that we won’t go into at the moment. In fact, describing the determinant as an area is a little misleading. When you look at transformations, the determinant tells you the scale factor of the change in area. A determinant of 1 would mean that areas did not change, etc. Also, if we have vectors that are simply scaled versions of one another—the components of one vector are scaled versions of the other—then the determinant will be zero, which is pretty much what we want, since the area will be zero. Let’s use lambda ($$\\mathtt{\\lambda}$$) as our scalar to be cool. $\\,\\,\\,\\,\\,\\,\\quad\\,\\,\\,\\,\\,\\begin{vmatrix}\\mathtt{u_1} & \\mathtt{\\lambda u_1}\\\\\\mathtt{u_2} & \\mathtt{\\lambda u_2}\\end{vmatrix} = \\mathtt{\\lambda u_{1}u_{2} – \\lambda u_{1}u_{2}", "asking for the area of a region on the given plane.) 5. It's a linear transformation $T(\\begin{bmatrix}u\\\\v \\end{bmatrix})=\\begin{bmatrix}u-v\\\\u+v\\\\u \\end{bmatrix}\\rightarrow T(\\vec{e}_1)=\\begin{bmatrix}1\\\\1\\\\1 \\end{bmatrix}~,~ T(\\vec{e}_2)=\\begin{bmatrix}-1\\\\1\\\\0 \\end{bmatrix}$ $A=\\begin{bmatrix}1&-1\\\\1&1\\\\1&0 \\end{bmatrix}~,~S_{\\Phi}=\\sqrt{\\det (A^TA)}\\cdot S_{D}=\\sqrt{6}~\\pi r^2=\\sqrt{6}~\\pi$ 6. Thank You much appreciation.", "matrix. To scale an object by a vector v = (vx, vy, vz), each point p = (px, py, pz) would need to be multiplied with this scaling matrix: ${\\displaystyle S_{v}={\\begin{bmatrix}v_{x}&0&0\\\\0&v_{y}&0\\\\0&0&v_{z}\\\\\\end{bmatrix}}.}$ As shown below, the multiplication will give the expected result: ${\\displaystyle S_{v}p={\\begin{bmatrix}v_{x}&0&0\\\\0&v_{y}&0\\\\0&0&v_{z}\\\\\\end{bmatrix}}{\\begin{bmatrix}p_{x}\\\\p_{y}\\\\p_{z}\\end{bmatrix}}={\\begin{bmatrix}v_{x}p_{x}\\\\v_{y}p_{y}\\\\v_{z}p_{z}\\end{bmatrix}}.}$ Such a scaling changes the diameter of an object by a factor between the scale factors, the area by a factor between the smallest and the largest product of two scale factors, and the volume by the product of all three. The scaling is uniform if and only if the scaling factors are equal (vx = vy = vz). If all except one of the scale factors are equal to 1, we have directional scaling. In the case where vx = vy = vz = k, the scaling is also called an enlargement or dilation by a factor k, increasing the area by a factor of k2 and the volume by a factor of k3. A scaling in the most general sense is any affine transformation with a diagonalizable matrix. It includes the case that the three directions of scaling are not perpendicular. It includes also the case that one or more scale factors are equal to zero (projection), and the case of one or more negative scale factors. The latter corresponds to a combination", "with a scale factor for volume of one half. The basic equation for it is image over preimage. In the field of measurements, the scale factor of an instrument is sometimes referred to as sensitivity. The ratio of any two corresponding lengths in two similar geometric figures is also called a scale. ## Matrix representation A scaling can be represented by a scaling matrix. To scale an object by a vector v = (vx, vy, vz), each point p = (px, py, pz) would need to be multiplied with this scaling matrix: ${\\displaystyle S_{v}={\\begin{bmatrix}v_{x}&0&0\\\\0&v_{y}&0\\\\0&0&v_{z}\\\\\\end{bmatrix}}.}$ As shown below, the multiplication will give the expected result: ${\\displaystyle S_{v}p={\\begin{bmatrix}v_{x}&0&0\\\\0&v_{y}&0\\\\0&0&v_{z}\\\\\\end{bmatrix}}{\\begin{bmatrix}p_{x}\\\\p_{y}\\\\p_{z}\\end{bmatrix}}={\\begin{bmatrix}v_{x}p_{x}\\\\v_{y}p_{y}\\\\v_{z}p_{z}\\end{bmatrix}}.}$ Such a scaling changes the diameter of an object by a factor between the scale factors, the area by a factor between the smallest and the largest product of two scale factors, and the volume by the product of all three. The scaling is uniform if and only if the scaling factors are equal (vx = vy = vz). If all except one of the scale factors are equal to 1, we have directional scaling. In the case where vx = vy = vz = k, scaling increases the area of any surface by a factor of k2 and the volume of any solid object by a factor of k3. ### Scaling", "How to compute regions of matrix Let's say we have the following matrix: 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 I want to calculate the size of every connected region of 1s and also get the corresponding matrix elements of each region. The size should be computed as (width + heigth) / 2. In the example the bottom left region would have size (2 + 3) / 2 and the bottom right region would have size (3 + 2) / 2. Additional to the size of the regions I also want the corresponding entries of the regions, i.e. region 1 has entries (2,2), (3,1), (3,2) and so on. How can this be done in Matlab? I'm looking forward for the answers and wish everybody all the best. • I don't see any submatrix f the size 2,3 could you be more specific? – Royi May 13 '14 at 18:43 • What's your definition of connectedness? Path connected? If yes, what paths? Are diagonal steps allowed? – Jazzmaniac May 14 '14 at 16:00 Lets take the given matrix", "\\cdot$ A. $24~\\text{cm}^2$ D. $36~\\text{cm}^2$ B. $28~\\text{cm}^2$ E. $40~\\text{cm}^2$ C. $34~\\text{cm}^2$ Solution Given: $T_1 = \\begin{pmatrix} \\dfrac14 & \\dfrac58 \\\\ \\frac12 & 2 \\end{pmatrix}~~~~T_2 = \\begin{pmatrix} 4 & 1 \\\\ 8 & 1 \\end{pmatrix}$ The transformation by the two matrices is expressed by \\begin{aligned} T_2 \\cdot T_1 & = \\begin{pmatrix} 4 & 1 \\\\ 8 & 1 \\end{pmatrix}\\begin{pmatrix} \\dfrac14 & \\dfrac58 \\\\ \\frac12 & 2 \\end{pmatrix} \\\\ & = \\begin{pmatrix} \\dfrac32 & \\dfrac92 \\\\ \\dfrac52 & 7 \\end{pmatrix} \\end{aligned} The area of the captured object is as follows. \\begin{aligned} L & = \\left|\\begin{vmatrix} \\dfrac32 & \\dfrac92 \\\\ \\dfrac52 & 7 \\end{vmatrix}\\right| \\times~\\text{The original area} \\\\ & = \\left|\\dfrac{21}{2}- \\dfrac{45}{4}\\right| \\times 32~\\text{cm}^2 \\\\ & = \\left|-\\dfrac34\\right| \\times 32~\\text{cm}^2 = 24~\\text{cm}^2 \\end{aligned} Thus, the area of the captured object is $\\boxed{24~\\text{cm}^2}$ [collapse] Problem Number 23 If the rectangle $ABCD$ is dilated to $A’B’C’D’$ as shown, then the appropriate scale factor is $\\cdots \\cdot$ A. $2$ B. $3$ C. $4$ D. $9$ Solution As shown in the figure that the process of dilation takes center at the bottom leftmost point. Assume this as point $(0, 0)$, so it can be assumed that $A (0, 1)$, $B(3, 1)$, $C(3, 3)$, and $D(1, 3)$. The coordinates of the dilated points are $A'(3, 0)$, $B'(9, 3)$, $C'(9, 9)$, and $D'(3, 9)$. From this, we", "Browse Questions # If $\\begin{vmatrix} 3x & 7 \\\\ 2 & 4 \\end{vmatrix}$= 10, find x. Toolbox: • The determinant value of a $2\\times 2$ matrix is $|A|=a_{11}\\times a_{22}-a_{21}\\times a_{12}$ $\\begin{vmatrix}3x & 7\\\\2 & 4\\end{vmatrix}$=10 Now let us evaluate the value of the determinant $3x\\times 4-7\\times 2=10.$ 12x-14=10. 12x=10+14 12x=24 x=$\\frac{24}{12}=2$ Hence the value of x=2.", "n-space? If the matrix is order two, then the determinant of this matrix is equivalent to the area of the parallelogram the vectors of this matrix comprise on the plane. Scaling a sinle dimension will scale the area – Dodgie Apr 25 '15 at 1:34 • Don't mind the question mark. Impossible to edit on the phone :( – Dodgie Apr 25 '15 at 1:41 Multiplying a matrix by a scalar $c$ amounts to multiplying each entry by $c$, not just the first row. • Ohh damn stupid I am, I should delete this question, but I read below property somewhere, which implies so, is it wrong then? (Unfortunately I noted down this in my book, have not noted its source. $$\\begin{vmatrix} a & kb & c \\\\ x & ky & z \\\\ p & kq & r \\\\ \\end{vmatrix} = k\\begin{vmatrix} a & b & c \\\\ x & y & z \\\\ p & q & r \\\\ \\end{vmatrix} = \\begin{vmatrix} ka & kb & kc \\\\ x & y & z \\\\ p & q & r \\\\ \\end{vmatrix}$$ – anir123 Apr 24 '15 at 17:34 • @Mahesha999 The vertical bars represent the determinant. It is not a matrix and is probably the cause of your confusion. – JessicaK Apr 24 '15 at 17:42", "\\end{vmatrix} = 1 \\cdot \\left( 1 \\cdot \\left( 8 \\right) - 3 \\cdot \\left(4 \\right) \\right) - 2 \\cdot \\left( 3 \\cdot \\left( 8 \\right) - 2 \\cdot \\left(4 \\right) \\right) + 3 \\cdot \\left( 3 \\cdot \\left( 3 \\right) - 2 \\cdot \\left(1 \\right) \\right)$ $= 1 \\cdot \\left( -4 \\right) - 2 \\cdot \\left( 16 \\right) + 3 \\cdot \\left( 7 \\right) = -15$ Conclusion: Based on the calculations shown above, it is found that the determinant of the matrix is $$\\det A = \\displaystyle -15$$. ### Other useful matrix calculators that you can use Matrix calculations done by hand are labor-intensive, so you can take advantage of our linear algebra solvers. First, you can use this inverse matrix calculator to compute the inverse of a matrix showing steps, and you can do it either by adjoint method, or by using RREF reduction.", "sub-matrix of size $h \\times w$ in $\\mathcal{O} (n^2)$ time. Note that have to swap out all the 0's from this matrix with outside 1's. So total swaps taken will be equal to number of zeros in this sub-matrix. # EXPLANATION Find the area of rectangular region in which all the 1's will go finally As we know that the rectangular region should contain only 1's and all the 1's of the matrix. So its area will be equal to number of 1's in the entire matrix. It can be computed in $\\mathcal{O} (n^2)$ time. Fix height of the rectangular region Now we will iterate from $h = 1$ to $n$ where $h$ denotes the height of rectangular region. We can find $w$ from $h$ easily as $h$ will be $\\frac{area}{w}$. Note that we don't have to consider all $h$'s from $1$ to $n$. We only need to consider the $h$'s which divide $n$ completely, otherwise width $w$ won't be integral. Number of swaps required are related to number of 1's and 0's Let us consider a rectangle having height $h$ and width $w$, then if want this to contain 1's only, then we have to swap all the 0's from this rectangle with 1's from the remaining region of matrix. Hence total number of swaps needed will" ]

[ "asking for the area of a region on the given plane.) 5. It's a linear transformation $T(\\begin{bmatrix}u\\\\v \\end{bmatrix})=\\begin{bmatrix}u-v\\\\u+v\\\\u \\end{bmatrix}\\rightarrow T(\\vec{e}_1)=\\begin{bmatrix}1\\\\1\\\\1 \\end{bmatrix}~,~ T(\\vec{e}_2)=\\begin{bmatrix}-1\\\\1\\\\0 \\end{bmatrix}$ $A=\\begin{bmatrix}1&-1\\\\1&1\\\\1&0 \\end{bmatrix}~,~S_{\\Phi}=\\sqrt{\\det (A^TA)}\\cdot S_{D}=\\sqrt{6}~\\pi r^2=\\sqrt{6}~\\pi$ 6. Thank You much appreciation.", "# Math Help - Finding an area using matrices & determinants 1. ## Finding an area using matrices & determinants Find the area of the triangle with vertices, $(1,3)$, $(2,7)$ and $(4,1)$. I went about this by drawing a diagram and then forming a rectangle containing the area described above, and 3 other triangles, calculated the area of these and subtracted it to find the area in the question, 7. However looking at the answers it appears as if we were meant to do this using matrices and determinants. Here's what the solutions say: Area $= \\frac{1}{2}det(A)$ where $A = \\begin{pmatrix} 1&1&3 \\\\ 1&2&7 \\\\ 1&4&1 \\end{pmatrix}$ . Now I gather that the last two columns are the coordinates of the vertices, just unsure where the 1s in the first column come from.", "$CDP$. Therefore the area of $CDP$ is $\\frac{1}{2} \\cdot 13 \\cdot 84=\\boxed{546}$.", "dimensions. Here $sum[i][j]$ denotes sum of binary matrix where top left coordinate is $(0, 0)$ and bottom right coordinate is $(i, j)$. We can create this array in $\\mathcal{O} (n^2)$ time easily. Now assume we have to find number of 1's (i.e. sum of sub-matrix) of a sub-matrix whose top left coordinate is $(x1, y1)$ and bottom right coordinate is $(x2, y2)$. Then it will be equal to $sum[x2][y2] - sum[x2][y1 - 1] - sum[x1 - 1][y2] + sum[x1 - 1][y1 - 1]$. In the diagram above, we need to the area of green region (i.e. sum of subarray from $a[x1][y1]$ to $a[x2][y2]$). So for finding that, we need to subtract regions of violet, orange and blue regions. Here $sum[x2][y2]$ denotes the entire region. $sum[x2][y1 - 1]$ denotes violet and green region. $sum[x1 - 1][y2]$ denotes orange and green region. As we have subtracted green region twice, We had to subtract it only one times, so we need to add it one more times. So we also add $sum[x1 - 1][y1 - 1]$ which denotes green region. Overall time complexity So we can notice that we are taking $\\mathcal{O} (n^2)$ time for each fixed height from $1$ to $n$ which divides area. So in the worst case, as area could be up to $10^6$. Total number of divisors", "\\begin{pmatrix} 2 &1 \\\\ 0 &1 \\end{pmatrix}}$ area is $2$ determinant is $2$ area is $5$ This exercise is recommended for all readers. Problem 10 Does $\\left|TS\\right|=\\left|ST\\right|$? $\\left|T(SP)\\right|=\\left|(TS)P\\right|$? Problem 11 1. Suppose that $\\left|A\\right|=3$ and that $\\left|B\\right|=2$. Find $\\left|A^2\\cdot {{B}^{\\rm trans}}\\cdot B^{-2}\\cdot {{A}^{\\rm trans}} \\right|$. 2. Assume that $\\left|A\\right|=0$. Prove that $\\left|6A^3+5A^2+2A\\right|=0$. This exercise is recommended for all readers. Problem 12 Let $T$ be the matrix representing (with respect to the standard bases) the map that rotates plane vectors counterclockwise thru $\\theta$ radians. By what factor does $T$ change sizes? This exercise is recommended for all readers. Problem 13 Must a transformation $t:\\mathbb{R}^2\\to \\mathbb{R}^2$ that preserves areas also preserve lengths? This exercise is recommended for all readers. Problem 14 What is the volume of a parallelepiped in $\\mathbb{R}^3$ bounded by a linearly dependent set? This exercise is recommended for all readers. Problem 15 Find the area of the triangle in $\\mathbb{R}^3$ with endpoints $(1,2,1)$, $(3,-1,4)$, and $(2,2,2)$. (Area, not volume. The triangle defines a plane— what is the area of the triangle in that plane?) This exercise is recommended for all readers. Problem 16 An alternate proof of Theorem 1.5 uses the definition of determinant functions. 1. Note that the vectors forming $S$ make a linearly dependent set if and only if $\\left|S\\right|=0$, and check that the result", "+0 # Application of matrices 0 108 2 After applying $$\\mathbf M = \\begin{pmatrix} -1 & 0 \\\\ -\\sqrt 2 & 3 \\end{pmatrix}$$to the circle of radius 3 centered at (2,0), what is the area of the resulting region? Here's what I've done so far: I found two points - (2,0) and (-1,0), which I turned into vectors of $$\\ \\begin{pmatrix} 2\\\\ 0 \\end{pmatrix}$$and $$\\ \\begin{pmatrix} -1\\\\ 0 \\end{pmatrix}$$ . I then multiplied these by the original matrix, to get $$\\ \\begin{pmatrix} -2\\\\ -2\\sqrt2 \\end{pmatrix}$$ and $$\\ \\begin{pmatrix} 1\\\\ \\sqrt2 \\end{pmatrix}$$ . I then calculated the distance between these two using the distance formula, which I got to equal $$\\sqrt11$$. I then plugged in $$\\sqrt11$$ to the circle area formula, ending up with a circle with area $$​​11\\pi$$. However, this answer is incorrect. Any tips on how I should solve this? Thanks! Mar 20, 2021 #1 +1 I got it, nevermind! I calculated the distance incorrectly. The answer was $$27\\pi$$. Mar 20, 2021 #2 +121003 +1 distance between ( -2 , -2sqrt (2) ) and ( 1 , sqrt (2) ) = sqrt [ ( 1 - -2)^2 + ( sqrt 2 - - 2 sqrt 2)^2 ] = sqrt [ 3^2 + ( 3sqrt (2)) ^2 ] = sqrt [ 9 + 18 ] = sqrt (27)", "sub-matrix of size $h \\times w$ in $\\mathcal{O} (n^2)$ time. Note that have to swap out all the 0's from this matrix with outside 1's. So total swaps taken will be equal to number of zeros in this sub-matrix. # EXPLANATION Find the area of rectangular region in which all the 1's will go finally As we know that the rectangular region should contain only 1's and all the 1's of the matrix. So its area will be equal to number of 1's in the entire matrix. It can be computed in $\\mathcal{O} (n^2)$ time. Fix height of the rectangular region Now we will iterate from $h = 1$ to $n$ where $h$ denotes the height of rectangular region. We can find $w$ from $h$ easily as $h$ will be $\\frac{area}{w}$. Note that we don't have to consider all $h$'s from $1$ to $n$. We only need to consider the $h$'s which divide $n$ completely, otherwise width $w$ won't be integral. Number of swaps required are related to number of 1's and 0's Let us consider a rectangle having height $h$ and width $w$, then if want this to contain 1's only, then we have to swap all the 0's from this rectangle with 1's from the remaining region of matrix. Hence total number of swaps needed will", "$A$ is an area-preserving transformation of the plane. However if we imagine traveling counter-clockwise from the point $(1,0)$ to the point $(0,1)$ we see that the transformation $A$ has us traveling clockwise from $A\\cdot(1,0) = (0,1)$ to $A\\cdot(0,1) = (1,1)$: In other words, the matrix $A$ is an orientation-reversing transformation of the plane. #### Appearance of the Fibonacci numbers What happens when we apply the matrix over and over to a point? For instance, if we apply the matrix $A$ to the point $(x,y)$Â we get the point $A\\cdot (x,y)=(y,x+y)$, and if we then apply $A$ to this new point we get the point $A\\cdot (y,x+y) =(x+y,x+2y)$. This is the same result as if we had applied the square $A^2=\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot \\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix} =\\begin{pmatrix} 1 & 1 \\\\1 & 2 \\end{pmatrix}$ to the point $(x,y)$. This is true more generally: to figure out what applying the matrix $A$ over and over does to a point $(x,y)$, it is the same as working out the appropriate power of the matrix $A$ and applying that matrix to $(x,y)$. So can we characterize, or identify, the powers of the matrix $A=\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}$ ? Here are the first few powers: $A^1=\\begin{pmatrix} 0 & 1", "# Math Help - Area of Region 1. ## Area of Region What is the area of the region in the first quadrant bounded by the x-axis, the line $y=2x$, and the circles $x^2+y^2=20$ and $x^2+y^2=30$? 2. ## Re: Area of Region Hi, Mrdavid445. The area of a region R in the plane is given by $\\int\\int_{R}1\\cdot dA$, where dA is the area element. Since the region we're interested in is a portion of an annular region, it's best to use the polar form for dA, which is $dA=rdrd\\theta$. What remains are to determine the limits of integration and then compute the double integral. Does this get everything going in the right direction? Good luck! 3. ## Re: Area of Region Originally Posted by Mrdavid445 What is the area of the region in the first quadrant bounded by the x-axis, the line $y=2x$, and the circles $x^2+y^2=20$ and $x^2+y^2=30$? Since you posted your question in the Geometry forum you probably wanted to use a more geometrical way to solve the problem(?). 1. The area in question is the difference of 2 sectors of 2 concentric circles: $r_1 = \\sqrt{20} \\approx 4.472$ $r_2 = \\sqrt{30} \\approx 5.477$ $|\\theta| = \\arctan(2) \\approx 63.435^\\circ$ Draw a sketch! 2. The area a is calculated by: $a = \\frac{|\\theta|}{360^\\circ} \\cdot \\pi r_2^2 -", "+0 0 336 2 +115 What is the area of the region of the complex plane defined by $$|z|<5$$ ? Oct 13, 2018 #1 +115 +2 Shouldn't it just be $$\\boxed{25\\pi}$$ Oct 13, 2018 #2 +110149 +2 What is the area of the region of the complex plane defined by |z|<5 ? $$let\\\\ z=a+bi\\\\ where\\\\ a^2+b^2<25$$ I think it is that simple. Oct 13, 2018", "flightwingsd2 2022-06-26 I had a question about a problem that I was working on for my pre-calculus class. Here's the problem statement: The area of the parallelogram with vertices 0, $м$, $w$, and $v+w$ is 34. Find the area of the parallelogram with vertices 0, $Av$, $Aw$, and $Av+Aw$, where $A=\\left(\\begin{array}{cc}3& -5\\\\ -1& -3\\end{array}\\right).$ I got the answer by doing something very tedious. I set $v=\\left(\\genfrac{}{}{0}{}{17}{0}\\right)$ and $w=\\left(\\genfrac{}{}{0}{}{0}{2}\\right)$, and did some really crazy matrix multiplication and a lot of plotting points of GeoGebra to get the answer of: $\\overline{)476}$. Now, I'm 100% sure that was not the fastest way, can someone tell me the non-bash way to do the problem? Angelo Murray Expert Hint: If $v=\\left(\\begin{array}{c}a\\\\ b\\end{array}\\right)$ and $w=\\left(\\begin{array}{c}c\\\\ d\\end{array}\\right)$ then the area of the original parallelogram is $|det\\left(\\begin{array}{cc}a& b\\\\ c& d\\end{array}\\right)|=|ad-bc|.$ Provide a similar expression for the area of the second parallelogram, and relate it to the above expression using properties of determinants. sedeln5w Expert You're probably supposed to know that the absolute value of the determinant of $A$ is the area scale factor of the linear transformation represented by the matrix $A$. With this knowledge, you can easily solve the problem in your head.", "to convert the worksheet to a matrix. 4. Activate the matrix, then run xyzarea -d; in Command Window to bring up the marea dialog box. 5. Check the check box behind the Interpolate for Missing Value. 6. Click the OK button, then the area of the matrix is put to Results Log Window. ## Algorithm In three dimensions Cartesian space, the area of a general triangle $A = (x_A, y_A, z_A), B = (x_B, y_B, z_B), C = (x_C, y_C, z_C)$ is $S=\\frac{1}{2} \\sqrt{ \\left( \\det\\begin{pmatrix} x_A & x_B & x_C \\\\ y_A & y_B & y_C \\\\ 1 & 1 & 1 \\end{pmatrix} \\right)^2 + \\left( \\det\\begin{pmatrix} y_A & y_B & y_C \\\\ z_A & z_B & z_C \\\\ 1 & 1 & 1 \\end{pmatrix} \\right)^2 + \\left( \\det\\begin{pmatrix} z_A & z_B & z_C \\\\ x_A & x_B & x_C \\\\ 1 & 1 & 1 \\end{pmatrix} \\right)^2 }$. To calculate the area of a matrix surface, the matrix surface is divided into many sub-triangles like below. The the area of the matrix surface is the sum of all the area of its sub-triangles. • Matrix Surface with Boundary The steps as follow are for calculating the area of matrix surface with boundary. 1. Find the intersection. 2. Pick up all the points inside or on the", "\\widehat{\\boldsymbol{k}}$ and ${}S$ is the surface of the region ${}T$ bounded by $z=1-x^2$ and the planes $z=0,$ $y=0$ and $y+z=2.$ $\\Box$", "b, and B transform the $$xy$$-plane. Points in the plane can be written as column vectors and I want you to focus on these four: $p_1 = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix}, \\quad p_2 = \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}, \\quad p_3 = \\begin{pmatrix}0 \\\\ 1\\end{pmatrix}, \\quad \\text{and} \\quad p_4 = \\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}$ Here's a picture of the $$xy$$-plane, with the diagonal lines and the four points included: Fig. 1. The $$xy$$-plane. Let's see what happens when you multiply some of these points with a: $a \\cdot p_1 = \\begin{pmatrix}1 & 2 \\\\ 0 & 1\\end{pmatrix} \\begin{pmatrix}1 \\\\ 0\\end{pmatrix} = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix} = p_1 \\qquad \\text{and} \\qquad a \\cdot p_4 = \\begin{pmatrix}1 & 2 \\\\ 0 & 1\\end{pmatrix} \\begin{pmatrix}-1 \\\\ 0\\end{pmatrix} = \\begin{pmatrix}1 \\\\ 1\\end{pmatrix} = p_2$ So a fixes $$p_1$$, sends $$p_4$$ to $$p_2$$, and (you should check this) $$p_2$$ to somewhere to the right between the diagonal and the $$x$$-axis. You can do the same with with b; the next picture is an illustration of the two transformations: Fig. 2. The left and right figures are the before and after pictures of the $$xy$$-plane. Notice how the shaded regions $$\\mathcal A^+$$ and $$\\mathcal B^+$$ in the two after pictures don't overlap? Do you also notice how the blank regions $$\\mathcal A^-$$ and $$\\mathcal B^-$$ in the before pictures", "1. Draw a labeled picture of $T$ in the $st$-plane. 2. Find the image of the $st$-vertex $(0,1)$ in the $xy$-plane. Likewise, find the respective images of the other three vertices of the rectangle $T\\text{:}$ $(0,4)\\text{,}$ $(1,1)\\text{,}$ and $(1,4)\\text{.}$ 3. In the $xy$-plane, draw a labeled picture of the image, $T'\\text{,}$ of the original $st$-rectangle $T\\text{.}$ What appears to be the shape of the image, $T'\\text{?}$ 4. To transform an integral with a change of variables, we need to determine the area element $dA$ for image of the transformed rectangle. Note that $T'$ is not exactly a parallelogram since the equations that define the transformation are not lienar. But we can approximate the area of $T'$ with the area of a parallelogram. How would we find the area of a parallelogram that approximates the area of the $xy$-figure $T'\\text{?}$ (Hint: Remember what the cross product of two vectors tells us.) Activity 11.9.2 presents the general idea of how a change of variables works. We partition a rectangular domain in the $st$ system into subrectangles. Let $T = [a, b] \\times [a+\\Delta s, b+\\Delta t]$ be one of these subrectangles. Then we transform this into a region $T'$ in the standard $xy$ Cartesian coordinate system. The region $T'$ is called the image of $T\\text{;}$ the region $T$ is the", "$w\\cdot v'=3$. So we get the matrix \\begin{equation*}\\begin{pmatrix}0 & 1 & 0 \\\\ 0 & 0 & -1 \\\\ -1 & 0 & 0\\end{pmatrix}\\end{equation*} Yep. All correct.", "Example Find the area of an triangle with its vertices located at (-2,2), (1,3) and (3,0) (this example is also shown in our video lesson). We plug our coordinates for the vertices into our area formula $A=\\frac{1}{2}\\begin{vmatrix} -2 & 2 & 1\\\\ 1 & 3 & 1\\\\ 3 & 0 & 1 \\end{vmatrix}$ and continues with $=\\frac{1}{2}(-2\\begin{vmatrix} 3 & 1\\\\ 0 & 1 \\end{vmatrix}-2\\begin{vmatrix} 1 & 1\\\\ 3 & 1 \\end{vmatrix}+1\\begin{vmatrix} 1 & 3\\\\ 3 & 0 \\end{vmatrix})=\\\\ \\\\ =\\frac{1}{2}(-2(3\\cdot1-1\\cdot 0 )-2(1\\cdot 1-1\\cdot 3)+1(1\\cdot 0-3\\cdot 3))=\\\\ \\\\ =\\frac{1}{2}(-6+4-9)=\\frac{-11}{2}=-5.5$ We received a negative value for A and an area cannot be negative, therefore we must take the absolute value for A: $\\mid A\\mid =\\mid -5.5\\mid =5.5\\;$ So the triangles area is 5.5 square units. ## Video lesson Find the area from the example above.", "area, i.e., the amount that the map stretches or shrinks $\\dlr^*$ as it transforms it into $\\dlr$. This stretching is captured by the area expansion factor $| \\det \\jacm{\\cvarf}(\\cvarfv,\\cvarsv)|$ from equation (3) of the introductory page. For out change of variables function, the matrix of partial derivatives is \\begin{align*} \\jacm{\\cvarf}(\\cvarfv,\\cvarsv) &= \\left(\\begin{array}{rr} 2\\cvarfv &-2\\cvarsv \\\\ 2\\cvarsv & 2\\cvarfv \\end{array}\\right) \\end{align*} so that the area expansion factor is \\begin{align*} |\\det \\jacm{\\cvarf}(\\cvarfv,\\cvarsv) | &= |(2\\cvarfv)(2\\cvarfv) - (-2\\cvarsv)(2\\cvarsv)| = 4\\cvarfv^2 + 4\\cvarsv^2. \\end{align*} The amount of stretching by $\\cvarf$ increases with distance from the origin of the $\\cvarfv\\cvarsv$-plane. The following applet allows you to explore how $\\cvarf$ is changing area in the irregular region $\\dlr$ compared to area in the rectangle $\\dlr^*$. See if you can convince yourself that the above formula seems correct. Area transformation of nonlinear 2D change of variables map. When using the map $(x,y)=\\cvarf(\\cvarfv,\\cvarsv) = (\\cvarfv^2-\\cvarsv^2,2\\cvarfv \\cvarsv)$ to change variables in a double integral, one needs to calculate a region $\\dlr^*$ in the $\\cvarfv\\cvarsv$-plane (rectangle in left panel) that is mapped onto the given region $\\dlr$ in the $xy$-plane (such as the region in right panel). The transformation $\\cvarf$ also maps each small rectangle in $\\dlr^*$ to a “curvy rectangle” in $\\dlr$. Although the small rectangles in $\\dlr^*$ are the same size, the corresponding “curvy rectangles” vary", "Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the \"About\" page. # One Dimension: U-Substitution How do you solve the integral? $$\\int_{0}^{5}2xe^{x^2}\\, \\mathrm{d}x$$ You let $u = x^2$, $du = 2x \\mathrm{d}x$ $$\\int_{x = 0}^{x = 5}e^{u}\\, \\mathrm{d}u \\to \\int_{0}^{25}e^{u}\\, \\mathrm{d}u = e^{25} - 1$$ if $X$ is our region $[0,5]$, $g(X)$ is our new region $[0,25]$. $$\\int_{g(X)}f(u)\\, \\mathrm{d}u = \\int_{X}f(g(x))g'(x)\\, \\mathrm{d}x$$ Now, how can we turn this logic into one that works in general? # Determinants Represent Scaling Area $$\\to \\begin{bmatrix}2& 0 \\\\ 0 & 3\\end{bmatrix} \\to$$ The area of the first region is $4 * 4 = 16$. The area of the second region, which is the first region multiplied by the matrix above, is $8 * 12 = 96$. The determinant of the matrix = 6, and 6* the 16 = 96, which begs the question, Does $vol_n T(X) = |det(T)|vol_n(X)$? # Proof It turns out that it does. The way we show it's true is to verify that it works for the 3 elementary row operations, because if those are true, then you can create any linear transformation from the product of those. We will do a special case of it for now, which is using a rectangle, however the intuition is the", "immediately (and exactly) plug it into the integral expression! :) – Deepak Sep 11 at 23:43 The effect of replacing $$x$$ in $$\\cos x$$ with $$x + a y - a$$ is a shear. In particular, this shear can be represented by the (homogeneous matrix) $$M = \\begin{pmatrix} 1 & a & -a \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ since $$\\begin{pmatrix} \\tilde{x} \\\\ \\tilde{y} \\\\ 1 \\end{pmatrix} = M \\cdot \\begin{pmatrix} x \\\\ y \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} x + ay - a \\\\ y \\\\ 1\\end{pmatrix}$$ That is, given coordinates $$(x,y)$$ of a point of the graph of $$y = \\cos x$$, its image under the shear gives the coordinates $$(\\tilde{x}, \\tilde{y})$$ of a point on the graph $$\\tilde{y} = \\cos(\\tilde{x}+a\\tilde{y}-a)$$. This matrix applies a linear map. The effect of a linear map on area is given by its determinant -- if the determinant is, say, $$2$$, the map doubles areas. So we compute the determinant of our map. Using he first column for expansion by minors should minimize computation. We find $$\\det M = 1 \\begin{vmatrix} 1 & 0 \\\\ 0 & 1 \\end{vmatrix} + 0 \\cdot | \\dots | + 0 \\cdot | \\dots | = 1 \\text{.}$$ This says that areas are unchanged when we apply" ]

[ "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "$m$. Let $A$ be an $m$ by $n$ matrix whose entries are denoted $a_{ij}, 1 \\leq i \\leq m, 1 \\leq j \\leq n$. To each column vector $\\mathbf{v} \\in V$ of length $n$, we have an associated column vector $A\\mathbf{v} \\in W$ of length $m$, which is quite literally the product of the $m$ by $n$ matrix $A$ with the $n$ by $1$ matrix $\\mathbf{v}$: $$A \\mathbf{v} = \\begin{pmatrix} a_{11} & a_{12} & \\cdots &a_{1n} \\\\ \\vdots & & \\vdots\\\\ a_{m1} & a_{m2} &\\cdots & a_{mn} \\end{pmatrix} \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\x_n \\end{pmatrix}$$ This is the $m$ by $1$ matrix (that is, the column vector of length $m$): $$\\begin{pmatrix} a_{11}x_1 + \\cdots + a_{1n}x_n \\\\ a_{21}x_1 + \\cdots + a_{2n}x_n \\\\ \\vdots \\\\ a_{m1}x_1 + \\cdots + a_{mn}x_n \\end{pmatrix}$$ Anyway, we have just defined a function, which we will call $T$, from $V$ to $W$, namely the function which sends each column vector $\\mathbf{v}$ to the the column vector $A \\mathbf{v}$. And you can check that it is linear, by which we mean $$T(\\mathbf{v} + \\mathbf{w}) = T(\\mathbf{v}) + T(\\mathbf{w})$$ (the addition on the left is addition of column vectors of length $n$, the addition on the right is addition of column vectors of length $m$) and $$T(\\lambda \\mathbf{v}) = \\lambda T(\\mathbf{v})$$ for any $\\mathbf{v},", "+0 # math help 0 89 1 Let P be the matrix that projects onto j that is, we want P to satisfy $$\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ for all vectors v. Use the pictures below to calculate $$\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$$. Calculate the matrix P that projects onto j. Aug 15, 2019 $$\\text{v projected onto (0,1,0) is (v \\cdot (0,1,0))(0,1,0)}\\\\ v \\cdot (0,1,0) = v_y\\\\ v \\perp (0,1,0) = (0,v_y,0)\\\\ \\text{The matrix that will perform this is}\\\\ P=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$", "\\text{L}}{\\mathbf{a}^{[1]}} = \\mathbf{W}^{[2]}.T \\frac{d \\text{L}}{d \\mathbf{z}^{[2]}}$$ Now we only missing $\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}}$ and $\\frac{d \\text{L}}{db^{[1]}}$. Let's look how we calculate $\\mathbf{z}^{[1]}$: $$\\begin{matrix} (\\mathbf{W}^{[1]}) \\\\ \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\\\ 0 & 0 \\\\ \\end{bmatrix} \\end{matrix} \\begin{matrix} (\\mathbf{x}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} + \\begin{matrix} (\\mathbf{b}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} = \\begin{matrix} (\\mathbf{z}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix}$$ Now, let's look at: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1](1,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](1)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](2,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](2)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](3,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](3)}} \\mathbf{x}.\\text{T}$$ or equivalently: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}} \\mathbf{x}.\\text{T}$$ and as expected: $$\\frac{d \\text{L}}{d \\mathbf{b}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}}$$ Let's apply the backprop altogether! ALPHA = 0.4 # learning rate def get_loss(y, a): return -1 * (y * np.log(a) + (1-y) * np.log(1-a)) def get_loss_numerically_stable(y, z): return -1 * (y * -1 * np.log(1 + np.exp(-z)) + (1-y) * (-z - np.log(1 + np.exp(-z)))) def get_gradients_loops(z1, a1, z2, a2, z3, a3, x, y, W1, b1, W2, b2, W3, b3): dz3 = a3 - y db3 = dz3 dW3 = dz3 * a2.T da2 = dz3 * W3.T dz2 = da2 * (a2 * (1-a2)) db2 = dz2 dW2 = np.matmul(dz2, a1.T)", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "+0 # Help hard quick 0 68 0 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ Use the picture below to find vectors with integer components and magnitude no greater than 8 that satisfy $$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "+0 # three dimensional vectors 0 107 1 Let $$u$$ and $$v$$ be three-dimensional vectors such that the length of $$u$$ is $$2$$, the length of $$v$$ is $$4$$, and the angle between $$u$$ and $$v$$ when placed tail to tail is $$120^{\\circ}$$. Let $$\\mathbf{A}$$ be a $$3\\times 3$$ matrix such that $$\\mathbf{row}_1(\\mathbf{A}) = \\mathbf{u}, \\mathbf{row}_2(\\mathbf{A}) = \\mathbf{v}, \\mathbf{row}_3(\\mathbf{A}) =3\\mathbf{u} + 2\\mathbf{v}.$$ Then what are $$\\mathbf{Au}$$ and $$\\mathbf{Av}$$ in that order? (Your answers should be numerical.) Oct 24, 2021 $\\mathbf{A} \\mathbf{u} = \\begin{pmatrix} 4 \\\\ -4 \\\\ 8 \\end{pmatrix}$ $\\mathbf{A} \\mathbf{v} = \\begin{pmatrix} -8 \\\\ 12 \\\\ -16 \\end{pmatrix}$", "+0 Let $\\mathbf{A}$ be a matrix, and let $\\mathbf{x}$ and $\\mathbf{y}$ be linearly independent vectors such that \\[\\mathbf{A} \\mathbf{x} = \\mat +1 273 1 Let A be a matrix, and let x and y be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ for some scalars a and b. Find the ordered pair (a,b). Mar 1, 2019 #1 +23071 +2 Let $$\\mathbf{A}$$ be a matrix, and let $$\\mathbf{x}$$ and $$\\mathbf{y}$$ be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ or some scalars a and b. Find the ordered pair (a,b). $$\\begin{array}{|rcll|} \\hline Ax &=& y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ax &=& A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ x &=& A^{-1}y \\\\ \\mathbf{A^{-1}y} &\\mathbf{=}&\\mathbf{x} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline Ay &=& x + 2y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ay &=& A^{-1}x + 2A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ y &=& A^{-1}x + 2A^{-1}y \\quad | \\quad \\mathbf{A^{-1}y=x} \\\\ y &=& A^{-1}x + 2x \\\\ \\mathbf{A^{-1}x} &\\mathbf{=}&\\mathbf{-2x+y } \\\\ \\hline \\end{array}$$ $$\\mathbf{(a,b) = (-2,1)}$$ Mar 1, 2019", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "+0 # Long solution matrix part 2 0 237 1 Let A be a $$2 \\times 2$$ matrix. Suppose that for every two-dimensional vector v, there exists a two-dimensional vector w such that $$\\mathbf{A} \\mathbf{w} = \\mathbf{v}$$ Show that we can find a matrix B such that $$\\mathbf{A} \\mathbf{B} = \\mathbf{I}$$. Mar 13, 2019 $$\\exists w_1 \\ni A w_1 = \\begin{pmatrix}1\\\\0\\end{pmatrix}\\\\ \\exists w_2 \\ni A w_2 = \\begin{pmatrix}0\\\\1\\end{pmatrix}\\\\ B = \\begin{pmatrix}w_1 &w_2\\end{pmatrix}$$", "to the basis $$\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}\\text{.}$$ The vectors $$(1, 1)$$ and $$(-1, -1)$$ do not form a basis for $${\\mathbb R}^2$$ since these two vectors lie along the same line. If $$2 \\times 2$$ matrices and the rest of what we have described above make you nervous, you should work through the exercises at the end of this section. Activity3.1.1.Matrix Operations. Given the matrices and vectors \\begin{equation*} A = \\begin{pmatrix} 5 & 3 \\\\ -6 & 4 \\end{pmatrix}, B = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix}, \\mathbf x = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\mathbf y = \\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix} \\end{equation*} compute each of the following expressions. (a) $$AB\\text{,}$$ $$BA$$ (b) $$A^{-1}\\text{,}$$ $$B^{-1}\\text{,}$$ $$(AB)^{-1}\\text{,}$$ $$B^{-1}A^{-1}$$ (c) $$\\det(A), \\det(B), \\det(AB), \\det(A^{-1})$$ (d) $$A \\mathbf x\\text{,}$$ $$A \\mathbf y\\text{,}$$ $$\\mathbf y^T \\mathbf x\\text{,}$$ $$\\mathbf x \\mathbf y^T\\text{,}$$ where $$\\mathbf y^T = (2, -1)$$ Subsection3.1.3Finding Eigenvalues and Eigenvectors A nonzero vector $${\\mathbf v}$$ is an eigenvector of $$A$$ if $$A {\\mathbf v} = \\lambda {\\mathbf v}$$ for some $$\\lambda \\in {\\mathbf R}\\text{.}$$ The constant $$\\lambda$$ is called an eigenvalue of $$A\\text{.}$$ Letting \\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf v = \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\neq \\mathbf 0, \\end{equation*} we have $$A \\mathbf", "+0 # Help Quick +1 38 2 Let $\\mathbf{A}$ be a matrix and $\\mathbf{x}$ be a non-zero vector such that $\\mathbf{A} \\mathbf{x} = 2 \\mathbf{x}.$Then we have that $\\mathbf{A}^7\\mathbf{x} = a \\mathbf{x}$some value of $a$. What is $a$ equal to? Nov 12, 2020 ### 1+0 Answers #2 +31296 +2 Think about $$\\mathbf{A^7}\\mathbf{x}= \\mathbf{A^6}(\\mathbf{A}\\mathbf{x})$$ Nov 13, 2020", "## Problem 639 Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v}^\\trans \\mathbf{v} = 0$ if and only if $\\mathbf{v}$ is the zero vector $\\mathbf{0}$. ## Problem 638 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. ## Problem 637 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a) $A B^\\trans + \\mathbf{v} \\mathbf{v}^\\trans$. (b) $A \\mathbf{v} – 2 \\mathbf{v}$. (c) $\\mathbf{v}^{\\trans} B$. (d) $\\mathbf{v}^\\trans \\mathbf{v} + \\mathbf{v}^\\trans B A^\\trans \\mathbf{v}$. ## Problem 635 Let $A$ and $B$ be $n \\times n$ matrices, and $\\mathbf{v}$ an $n \\times 1$ column vector. Use the matrix components to prove that $(A + B) \\mathbf{v} = A\\mathbf{v} + B\\mathbf{v}$. ## Problem 634 Let $A$ and $B$ be $n \\times n$ matrices. Is it always true that $\\tr (A B) = \\tr (A) \\tr (B)$? If it is", "Main # MIT 18.06SC R1 Vectors and Linear Combinations \"\" #### Contents Section 1.1 [SG] (folio 2, page 13) The usage of vectors is considered as having the ability to add apples and oranges, a thing the folklore of arithmetic forbids. We consider column vectors in exploring the linear operations. (Of course things will work for row vectors as well.) Notation: A boldface lowercase letter represents a vector and the vector is displayed with its components in a verticle column (for column vectors). {$$\\textbf{ Column Vector }\\quad\\quad \\mathbf v = \\left[\\begin{matrix}{v_1}\\\\{v_2}\\end{matrix}\\right] \\quad \\begin{matrix}{v_1} =&\\text{first component}\\\\{v_2} =&\\text{second component}\\end{matrix}$$} Use Maxima or wxMaxim for symbolic operations (letters). The super lab programs like Matlab, FreeMat, and Sage are horrible on symbolic stuff. Tested wxMaxima commands are in blue. If you have the jsMath fonts, the results are beautiful and can be saved in a variety of beautiful ways by right clicking. v:matrix([v_1],[v_2]); (underscore shows as raised dot in wxMaxima) We define another vector so we will have something to talk about. {$$\\mathbf w = \\left[\\begin{matrix}{w_1}\\\\{w_2}\\end{matrix}\\right]$$} w:matrix([w_1],[w_2]); {$$\\mathbf v = \\left[\\begin{matrix} {v_1}\\cr{v_2} \\end{matrix} \\right] \\text{ and }\\mathbf w = \\left[\\begin{matrix}{w_1}\\\\{w_2}\\end{matrix}\\right]\\text{ add to }\\mathbf v + \\mathbf w = \\left[\\begin{matrix}{w_1}+{v_1}\\\\ {w_2}+{v_2}\\end{matrix}\\right]$$} v+w; Mention is made of subtraction, which works the same way, but subtraction can also be viewed as multiplying the subtrahend by", "+0 # HELP ASAP PLEEASSE!!! 0 89 1 Let $$\\mathbf{A}$$ be a matrix such that $$\\mathbf{A}^{-1} = \\begin{pmatrix} 1 & 1 \\\\ 2 & x \\end{pmatrix}$$ for some value of $$x$$. What is the vector that $$\\mathbf{A}$$ maps to $$\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}?$$ Oct 13, 2019 The answer is $$\\begin{pmatrix} 3 \\\\ -7 \\end{pmatrix}$$.", "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "Also, let $\\mathbf y$ be the $m$-dimensionl vector with your target values: $$\\mathbf y=\\begin{pmatrix}y^{(1)}\\\\y^{(2)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}.$$ Now, since $h_w(x^{(i)})=(x^{(i)})^T\\mathbf w$, we can verify $$\\begin{array}{rcl}X\\mathbf w -\\mathbf y&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w\\end{pmatrix}-\\begin{pmatrix}y^{(1)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}\\\\&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w-y^{(1)}\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w-y^{(m)}\\end{pmatrix}.\\end{array}$$ Thus, using $\\mathbf v^T\\mathbf v = \\sum_i\\mathbf v_i^2$ for a vector $\\mathbf v$, we obtain $$\\begin{array}{rcl}E(w)&=&\\frac{1}{2}\\sum_{i=1}^m(h_w(x^{(i)}) - y^{(i)})^2\\\\&=&\\frac{1}{2}(X\\mathbf w-\\mathbf y)^T(X\\mathbf w-\\mathbf y).\\end{array}$$ Summarizing. To obtain $$\\mathbf w=\\begin{pmatrix}w_0\\\\w_1\\end{pmatrix},$$ proceed: 1. Set your collections $x$ and $y$: $$x=\\begin{pmatrix}1\\\\3\\\\5\\\\7\\\\9\\end{pmatrix}\\qquad y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2. Loop until you think that $w_j$ converge (try with $10$ iterations) { for $i=1$ to $m$ { $$w_j:=w_j+\\eta\\sum_{i=1}^m(y^{(i)}-h_w(x^{(i)}))x_j^{(i)}\\qquad(\\text{for each }j)$$ } } Insted you can use 1'. Set your vectors $X$ and $\\mathbf y$: $$X=\\begin{pmatrix}1&1\\\\1&3\\\\1&5\\\\1&7\\\\1&9\\end{pmatrix}\\qquad\\mathbf y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2'. Loop until you think that $\\mathbf w$ converge (try with $10$ iterations) { $$\\mathbf w:=\\mathbf w-\\frac{\\eta}{m}\\left((X\\mathbf w-\\mathbf y)^TX\\right)$$ } Note. With the above it can prove a analytical formula to obtain the parameters, using the normal equations to computed $\\mathbf w$ by $$\\mathbf w=(X^TX)^{-1}X^T\\mathbf y.$$ Obviously you need to use a computer software (like python or octave) or a calculator device (with matrix-vector support) to calculate the values.", "Matrix determinant lemma In mathematics, in particular linear algebra, the matrix determinant lemma computes the determinant of the sum of an invertible matrix A and the dyadic product, uvT, of a column vector u and a row vector vT.[1][2] Statement Suppose A is an invertible square matrix and u, v are column vectors. Then the matrix determinant lemma states that ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\left(1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {A} ^{-1}\\mathbf {u} \\right)\\,\\det \\left(\\mathbf {A} \\right)\\,.}$ Here, uvT is the outer product of two vectors u and v. The theorem can also be stated in terms of the adjugate matrix of A: ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\det \\left(\\mathbf {A} \\right)+\\mathbf {v} ^{\\textsf {T}}\\mathrm {adj} \\left(\\mathbf {A} \\right)\\mathbf {u} \\,,}$ in which case it applies whether or not the square matrix A is invertible. Proof First the proof of the special case A = I follows from the equality:[3] ${\\displaystyle {\\begin{pmatrix}\\mathbf {I} &0\\\\\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} +\\mathbf {uv} ^{\\textsf {T}}&\\mathbf {u} \\\\0&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} &0\\\\-\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {I} &\\mathbf {u} \\\\0&1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {u} \\end{pmatrix}}.}$ The determinant of the left hand side is the product of the determinants of the three matrices. Since the first and third matrix are triangular matrices with unit diagonal, their determinants are just 1. The determinant of" ]

[ "(6), where {\\displaystyle {\\begin{aligned}&\\mathbf {D} ={\\begin{pmatrix}6&0&0\\\\0&4&0\\\\0&0&5\\end{pmatrix}},\\quad \\mathbf {U} ={\\begin{pmatrix}0&2&3\\\\0&0&2\\\\0&0&0\\end{pmatrix}},\\quad \\mathbf {L} ={\\begin{pmatrix}0&0&0\\\\1&0&0\\\\3&1&0\\end{pmatrix}}.\\quad (14)\\end{aligned}}} We then have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}} ={\\frac {1}{120}}{\\begin{pmatrix}20&0&0\\\\5&30&0\\\\13&6&24\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-L)^{-1}U} ={\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}},\\quad \\mathbf {(D-L)^{-1}k} ={\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}}.\\end{aligned}}} The Gauss-Seidel method (8) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{120}}{\\begin{pmatrix}0&40&60\\\\0&10&75\\\\0&26&51\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{120}}{\\begin{pmatrix}100\\\\-335\\\\233\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (15)} The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), somewhat faster than the Jacobi method described above. ### Successive over-relaxation method Let ω = 1.1. Using the splitting (14) of the matrix A in problem (10) for the successive over-relaxation method, we have {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}} ={\\frac {1}{12}}{\\begin{pmatrix}2&0&0\\\\0.55&3&0\\\\1.441&0.66&2.4\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {(D-\\omega L)^{-1}[(1-\\omega )D+\\omega U]} ={\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}},\\end{aligned}}} {\\displaystyle {\\begin{aligned}&\\mathbf {\\omega (D-\\omega L)^{-1}k} ={\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}}.\\end{aligned}}} The successive over-relaxation method (9) applied to the problem (10) takes the form {\\displaystyle {\\mathbf {x} }^{(m+1)}={\\begin{aligned}&{\\frac {1}{12}}{\\begin{pmatrix}-1.2&4.4&6.6\\\\-0.33&0.01&8.415\\\\-0.8646&2.9062&5.0073\\end{pmatrix}}{\\mathbf {x} }^{(m)}+{\\frac {1}{12}}{\\begin{pmatrix}11\\\\-36.575\\\\25.6135\\end{pmatrix}},\\end{aligned}}\\quad m=0,1,2,\\ldots \\quad (16)} The first few iterates for equation (16) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From the table one can see that the method is evidently converging to the solution (13), slightly faster than the Gauss-Seidel method described above. ## References • {{#invoke:citation/CS1|citation |CitationClass=citation }}.", "e_{12} \\\\ e_{13} \\end{pmatrix}, \\vec{e_2}=\\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}, \\vec{e_3}=\\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix} \\end{align*} is an orthonormal vector, it is \\begin{align*} \\begin{pmatrix} (\\vec{e_1})^T \\\\ (\\vec{e_2})^T \\\\ (\\vec{e_3})^T \\end{pmatrix} \\begin{pmatrix} \\vec{e_1} & \\vec{e_2} & \\vec{e_3} \\end{pmatrix} =\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\ e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix} \\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} & e_{23} & e_{33} \\end{pmatrix} =\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} .\\end{align*} That is, \\begin{align*} R=\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\ e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix} =(R^{-1})^T \\end{align*}\\begin{align*} R^{-1}=\\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} & e_{23} & e_{33} \\end{pmatrix} =R^T .\\end{align*} Camera and perspective transformation Consider a pinhole camera in which light from an object always passes through the pinhole. Align the origin $$o$$ of the 3D coordinate $$oxyz$$ with this pinhole. Next, consider the plane that intersects with the $$oz$$ axis at $$f$$ point $$(0, 0, f)$$ and calls it the imaging surface. However, $$f$$ is positive and the imaging surface is between the pinhole and the object. $$f$$ is also called the focus. In", "\\vdots\\\\ Y_n \\end{pmatrix}_{n\\times 1}. \\end{align*} Then we can re-express (4.10) into a weighted least squares problem, which has an exact solution: \\begin{align} \\hat{\\boldsymbol{\\beta}}_h&=\\arg\\min_{\\boldsymbol{\\beta}\\in\\mathbb{R}^{p+1}} (\\mathbf{Y}-\\mathbf{X}\\boldsymbol{\\beta})'\\mathbf{W}(\\mathbf{Y}-\\mathbf{X}\\boldsymbol{\\beta})\\tag{4.11}\\\\ &=(\\mathbf{X}'\\mathbf{W}\\mathbf{X})^{-1}\\mathbf{X}'\\mathbf{W}\\mathbf{Y}.\\tag{4.12} \\end{align} The estimate125 of $$m(x)$$ is therefore computed as126 \\begin{align} \\hat{m}(x;p,h):=&\\,\\hat{\\beta}_{h,0}\\nonumber\\\\ =&\\,\\mathbf{e}_1'(\\mathbf{X}'\\mathbf{W}\\mathbf{X})^{-1}\\mathbf{X}'\\mathbf{W}\\mathbf{Y}\\nonumber\\\\ =&\\,\\sum_{i=1}^nW^p_{i}(x)Y_i,\\tag{4.13} \\end{align} where \\begin{align*} W^p_{i}(x):=\\mathbf{e}_1'(\\mathbf{X}'\\mathbf{W}\\mathbf{X})^{-1}\\mathbf{X}'\\mathbf{W}\\mathbf{e}_i \\end{align*} and $$\\mathbf{e}_i$$ is the $$i$$-th canonical vector.127 Just as the Nadaraya–Watson was, the local polynomial estimator is a weighted linear combination of the responses.128 Two cases on (4.13) deserve special attention: • $$p=0$$ is the local constant estimator, previously referred to as the Nadaraya–Watson estimator. In this situation, the estimator has explicit weights, as we saw before: \\begin{align} W_i^0(x)=\\frac{K_h(x-X_i)}{\\sum_{j=1}^nK_h(x-X_j)}.\\tag{4.14} \\end{align} • $$p=1$$ is the local linear estimator, which has weights equal to \\begin{align} W_i^1(x)=\\frac{1}{n}\\frac{\\hat{s}_2(x;h)-\\hat{s}_1(x;h)(X_i-x)}{\\hat{s}_2(x;h)\\hat{s}_0(x;h)-\\hat{s}_1(x;h)^2}K_h(x-X_i),\\tag{4.15} \\end{align} where $$\\hat{s}_r(x;h):=\\frac{1}{n}\\sum_{i=1}^n(X_i-x)^rK_h(x-X_i).$$ Exercise 4.4 Show that (4.12) is the solution of (4.11), $$\\hat{\\boldsymbol{\\beta}}_h.$$ To do so, first prove Exercise B.1. Exercise 4.5 Show that the local polynomial estimator yields the Nadaraya–Watson estimator when $$p=0.$$ Use (4.12) to obtain (4.5). Exercise 4.6 Prove that $$\\sum_{i=1}^nW_i^0(x)=1$$ and $$\\sum_{i=1}^nW_i^1(x)\\neq1.$$ Exercise 4.7 Obtain the weight expressions (4.15) of the local linear estimator using the matrix inversion formula for $$2\\times2$$ matrices: $$\\begin{pmatrix}a & b\\\\ c & d\\end{pmatrix}^{-1}=(ad-bc)^{-1}\\begin{pmatrix}d & -b\\\\ -c & a\\end{pmatrix}.$$ Remark. The local polynomial fit is computationally more expensive than the local constant fit: $$\\hat{m}(x;p,h)$$ is obtained as the solution of a weighted", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "\\\\ 2 & 3 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\ 6 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{5} \\begin{pmatrix} 9 \\\\ 4 \\end{pmatrix} \\end{align*} \\begin{align*} 12) \\hspace{.5cm} \\begin{pmatrix} −1 & 5 \\\\ 1 & 1 \\end{pmatrix} x = \\begin{pmatrix} 2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{6} \\begin{pmatrix} −7 \\\\ 1 \\end{pmatrix} \\end{align*} \\begin{align*} 13) \\hspace{.5cm} \\begin{pmatrix} 3 & −7 \\\\ −9 & 21 \\end{pmatrix} x = \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 14) \\hspace{.5cm} \\begin{pmatrix} 2 & 7 \\\\ 5 & −1 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{37} \\begin{pmatrix} 31 \\\\ 7 \\end{pmatrix} \\end{align*} \\begin{align*} 15) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 1 \\\\ 0 & −1 & 1 \\\\ 1 & 2 & 2 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ −2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\begin{pmatrix} 3 \\\\ 0 \\\\ −2 \\end{pmatrix}\\end{align*} \\begin{align*} 16) \\hspace{.5cm} \\begin{pmatrix} 7 & 3 & 4 \\\\ 1 & 2 & 3 \\\\ 4 & −3 & −5 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 2 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 17) \\hspace{.5cm} \\begin{pmatrix} 3 & 4 & 1 \\\\ −2 & 3 & 1 \\\\ 0 & 3 & 2 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\", "\\mathbf {u} =\\mathbf {x} +\\alpha \\mathbf {e} _{1},}$ and ${\\displaystyle \\mathbf {v} ={\\mathbf {u} \\over \\|\\mathbf {u} \\|}.}$ Here, ${\\displaystyle \\alpha =-14}$ and ${\\displaystyle \\mathbf {x} =\\mathbf {a} _{1}=(12,6,-4)^{T}}$ Therefore ${\\displaystyle {\\mathbf {u} }=(-2,6,-4)^{T}=({2})(-1,3,-2)^{T}}$ and ${\\displaystyle \\mathbf {v} ={1 \\over {\\sqrt {14}}}(-1,3,-2)^{T}}$, and then ${\\displaystyle Q_{1}=I-{2 \\over {\\sqrt {14}}{\\sqrt {14}}}{\\begin{pmatrix}-1\\\\3\\\\-2\\end{pmatrix}}{\\begin{pmatrix}-1&3&-2\\end{pmatrix}}}$ ${\\displaystyle =I-{1 \\over 7}{\\begin{pmatrix}1&-3&2\\\\-3&9&-6\\\\2&-6&4\\end{pmatrix}}}$ ${\\displaystyle ={\\begin{pmatrix}6/7&3/7&-2/7\\\\3/7&-2/7&6/7\\\\-2/7&6/7&3/7\\\\\\end{pmatrix}}.}$ Now observe: ${\\displaystyle Q_{1}A={\\begin{pmatrix}14&21&-14\\\\0&-49&-14\\\\0&168&-77\\end{pmatrix}},}$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to ${\\displaystyle A'=M_{11}={\\begin{pmatrix}-49&-14\\\\168&-77\\end{pmatrix}}.}$ By the same method as above, we obtain the matrix of the Householder transformation ${\\displaystyle Q_{2}={\\begin{pmatrix}1&0&0\\\\0&-7/25&24/25\\\\0&24/25&7/25\\end{pmatrix}}}$ after performing a direct sum with 1 to make sure the next step in the process works properly. Now, we find ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}6/7&-69/175&58/175\\\\3/7&158/175&-6/175\\\\-2/7&6/35&33/35\\end{pmatrix}}}$ Then ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}0.8571&-0.3943&0.3314\\\\0.4286&0.9029&-0.0343\\\\-0.2857&0.1714&0.9429\\end{pmatrix}}}$ ${\\displaystyle R=Q_{2}Q_{1}A=Q^{T}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&-35\\end{pmatrix}}.}$ The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition. Using Givens rotations QR decompositions can also be computed with a series of Givens rotations. Each rotation zeros an element in the subdiagonal of the matrix, forming the R matrix. The concatenation of all the Givens rotations forms the orthogonal Q matrix. In practice, Givens rotations are not actually performed by building a whole matrix and doing a matrix multiplication. A Givens rotation procedure is", "& \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} <br> \\begin{aligned} \\quad \\mathbf u &= \\alpha \\mathbf a + \\beta \\mathbf b \\qquad \\text{Norm:}|| \\mathbf u|| \\end{aligned} <br> \\begin{align} \\hat{\\mathbf u} = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} <br> 2. **Matrices** \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} <br> 3. \\begin{aligned} a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5 \\end{aligned} <br> 4. **Transformations:** \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} The syntax is very close to what one writes in a .tex file. Switch html_document in the YAML header with pdf_document and you can get the corresponding pdf. This is the HTML output: To have our HTML is written with mathml, we add --mathml to the arguments passed to pandoc. This is what we see at the top of the HTML file: In this workflow, knitr produces a .tex file. As we can see, it is unremarkable, and very closely resembles the OP's original latex code, just adding elements like the enumerate environment. \\begin{document} \\centerline{\\textbf{\\large{Mathematical expressions}}} \\begin{enumerate} \\item", "A={\\begin{pmatrix}12&-51&4\\\\6&167&-68\\\\-4&24&-41\\end{pmatrix}}.}$ Recall that an orthonormal matrix ${\\displaystyle Q}$ has the property ${\\displaystyle Q^{\\textsf {T}}\\,Q=I.}$ Then, we can calculate ${\\displaystyle Q}$ by means of Gram–Schmidt as follows: {\\displaystyle {\\begin{aligned}U={\\begin{pmatrix}\\mathbf {u} _{1}&\\mathbf {u} _{2}&\\mathbf {u} _{3}\\end{pmatrix}}&={\\begin{pmatrix}12&-69&-58/5\\\\6&158&6/5\\\\-4&30&-33\\end{pmatrix}};\\\\Q={\\begin{pmatrix}{\\frac {\\mathbf {u} _{1}}{\\|\\mathbf {u} _{1}\\|}}&{\\frac {\\mathbf {u} _{2}}{\\|\\mathbf {u} _{2}\\|}}&{\\frac {\\mathbf {u} _{3}}{\\|\\mathbf {u} _{3}\\|}}\\end{pmatrix}}&={\\begin{pmatrix}6/7&-69/175&-58/175\\\\3/7&158/175&6/175\\\\-2/7&6/35&-33/35\\end{pmatrix}}.\\end{aligned}}} Thus, we have {\\displaystyle {\\begin{aligned}Q^{\\textsf {T}}A&=Q^{\\textsf {T}}Q\\,R=R;\\\\R&=Q^{\\textsf {T}}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&35\\end{pmatrix}}.\\end{aligned}}} #### Relation to RQ decomposition The RQ decomposition transforms a matrix A into the product of an upper triangular matrix R (also known as right-triangular) and an orthogonal matrix Q. The only difference from QR decomposition is the order of these matrices. QR decomposition is Gram–Schmidt orthogonalization of columns of A, started from the first column. RQ decomposition is Gram–Schmidt orthogonalization of rows of A, started from the last row. The Gram-Schmidt process is inherently numerically unstable. While the application of the projections has an appealing geometric analogy to orthogonalization, the orthogonalization itself is prone to numerical error. A significant advantage however is the ease of implementation, which makes this a useful algorithm to use for prototyping if a pre-built linear algebra library is unavailable. ### Using Householder reflections A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane or hyperplane. We can use this operation to calculate the QR factorization of", "\\mathbf{S} ) \\times \\mathcal{IW}(\\mathbf{S} | \\upsilon_0, \\mathbf{S_0}) \\end{align} I'm doing the first case (please correct me if I am wrong): \\begin{align} \\mathcal{W}(\\mathbf{W} | \\upsilon, \\mathbf{S^{-1}} ) \\times \\mathcal{W}(\\mathbf{S} | \\upsilon_0, \\mathbf{S_0}) &\\propto |\\mathbf{S}|^{\\upsilon/2} \\exp\\{-\\frac{1}{2} tr(\\mathbf{SW}) \\}\\\\ &\\times |\\mathbf{S}|^{\\frac{\\upsilon_0 - D -1 }{2}} \\exp\\{-\\frac{1}{2} tr (\\mathbf{S_0^{-1} S})\\}\\\\ &\\propto |\\mathbf{S}|^{\\frac{\\upsilon + \\upsilon_0 - D -1 }{2}} \\exp\\{-\\frac{1}{2} tr ( (\\mathbf{W} + \\mathbf{S_0^{-1}) S})\\} \\end{align} where we used the fact that $tr({\\mathbf{SW}}) = tr({\\mathbf{WS}})$. By inspection, we see that this is a Wishart distribution: \\begin{align} p(\\mathbf{S} | \\cdot) = \\mathcal{W}(\\upsilon+ \\upsilon_0, \\mathbf{(W+S_0^{-1})^{-1}}) \\end{align} Extension for $N$ draws $\\mathbf{W_1...W_N}$: For the case when we have $N$ precision matrices then the likelihood becomes a product of $N$ likelihoods and we get: \\begin{align} p(\\mathbf{S} | \\cdot) = \\mathcal{W}(N \\upsilon+ \\upsilon_0, (\\sum_{i=1}^N \\mathbf{W_i+S_0^{-1})^{-1}}) \\end{align}", "amsmath. You should consider reading the user's guide -- texdoc amsldoc at a command line if you are using a TeX Live installation. – barbara beeton Feb 12 at 17:15 You should give & symbol, where you want to align: \\documentclass{book} \\usepackage{amsmath} \\begin{document} \\begin{align} \\mathbf{i}_{t} &= \\sigma(\\mathbf{W}_{xi}\\mathbf{x}_{t} + \\mathbf{W}_{hi}\\mathbf{h}_{t-1} + \\mathbf{W}_{ci}\\circ\\mathbf{c}_{t-1} + \\mathbf{b}_{i})\\\\ \\mathbf{f}_{t} &= \\sigma(\\mathbf{W}_{xf}\\mathbf{x}_{t} + \\mathbf{W}_{hf}\\mathbf{h}_{t-1} + \\mathbf{W}_{cf}\\circ\\mathbf{c}_{t-1} + \\mathbf{b}_{f})\\\\ \\mathbf{c}_{t} &= \\mathbf{f}_{t}\\circ\\mathbf{c}_{t-1} + \\mathbf{i}_{t}\\circ \\tanh(\\mathbf{W}_{xc}\\mathbf{x}_{t} + \\mathbf{W}_{hc}\\mathbf{h}_{t-1} + \\mathbf{b}_{c}) \\end{align} \\end{document} • @Raaja done :-) – MadyYuvi Feb 12 at 15:09", "1 \\\\ 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{19} \\begin{pmatrix} 12 \\\\ 16 \\\\ −5 \\end{pmatrix} \\end{align*} \\begin{align*} 18) \\hspace{.5cm} \\begin{pmatrix} 3 & 1 & 1 \\\\ −1 & 1 & 2 \\\\ 4 & 3 & 5 \\end{pmatrix} x = \\begin{pmatrix} −3 \\\\ 3 \\\\ 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{3} \\begin{pmatrix} 1 \\\\ −34 \\\\ 22 \\end{pmatrix} \\end{align*} \\begin{align*} 19) \\hspace{.5cm} \\begin{pmatrix} 1 & 2 & −1 & −2 \\\\ 3 & 0 & 4 & 1 \\\\ 1 & 5 & 0 & 0 \\\\ 0 & 1 & 0 & 1 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{10} \\begin{pmatrix} 45 \\\\ −9 \\\\ −41 \\\\ 29 \\end{pmatrix} \\end{align*} \\begin{align*} 20) \\hspace{.5cm} \\begin{pmatrix} 0 & 7 & 1 & 0 \\\\ 1 & 1 & 1 & 0 \\\\ 1 & −1 & 1 & −1 \\\\ 0 & 1 & 2 & 3 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ 0 \\\\ 2 \\\\ −2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x &=\\frac{1}{19} \\begin{pmatrix} −31 \\\\ −2 \\\\ 33 \\\\ −34 \\end{pmatrix} \\end{align*} Tags:", "= (\\mathbf{\\mu}_{1}-\\mathbf{\\mu})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})^{T}-(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})^{T}-(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})^{T}+(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})^{T} \\end{align} \\begin{align} & \\mathbf{S}_{B} = n_{1}(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})^{T} + n_{2}(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})^{T} \\end{align} Apparently, they are very similar. Now, we are trying to find the optimal transformation. Basically, we have \\begin{align} \\mathbf{z}_{i} = \\mathbf{W}^{T}\\mathbf{x}_{i}, i=1,2,...,k-1 \\end{align} where $\\mathbf{z}_{i}$ is a $(k-1)\\times 1$ vector, $\\mathbf{W}$ is a $d\\times (k-1)$ transformation matrix, i.e. $\\mathbf{W} = [\\mathbf{w}_{1}, \\mathbf{w}_{2},..., \\mathbf{w}_{k-1}]$, and $\\mathbf{x}_{i}$ is a $d\\times 1$ column vector. Thus we obtain \\begin{align} & \\mathbf{S}_{W}^{\\ast} = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{W}^{T}\\mathbf{x}_{j}-\\mathbf{W}^{T}\\mathbf{\\mu}_{i})(\\mathbf{W}^{T}\\mathbf{x}_{j}-\\mathbf{W}^{T}\\mathbf{\\mu}_{i})^{T} \\\\ & = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{W}^{T}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i}))(\\mathbf{W}^{T}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i}))^{T} \\\\ & = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{W}^{T}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i}))((\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})^{T}\\mathbf{W}) \\\\ & = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}\\mathbf{W}^{T}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})^{T}\\mathbf{W} \\\\ & = \\mathbf{W}^{T}\\left[\\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})^{T}\\right]\\mathbf{W} \\\\ & = \\mathbf{W}^{T}\\mathbf{S}_{W}\\mathbf{W} \\end{align} Similarly, we obtain \\begin{align} & \\mathbf{S}_{B}^{\\ast} = \\sum_{i=1}^{k}n_{i}(\\mathbf{W}^{T}\\mathbf{\\mu}_{i}-\\mathbf{W}^{T}\\mathbf{\\mu})(\\mathbf{W}^{T}\\mathbf{\\mu}_{i}-\\mathbf{W}^{T}\\mathbf{\\mu})^{T} \\\\ & = \\sum_{i=1}^{k}n_{i}\\mathbf{W}^{T}(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})^{T}\\mathbf{W} \\\\ & = \\mathbf{W}^{T}\\left[ \\sum_{i=1}^{k}n_{i}(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})^{T}\\right]\\mathbf{W} \\\\ & = \\mathbf{W}^{T}\\mathbf{S}_{B}\\mathbf{W} \\end{align} Now, we use the following as our measure: \\begin{align} \\sum_{i=1}^{k}n_{i}\\|(\\mathbf{W}^{T}\\mathbf{\\mu}_{i}-\\mathbf{W}^{T}\\mathbf{\\mu})^{T}\\|^{2} \\end{align} The solution for this question is that the columns of the transformation matrix $\\mathbf{W}$ are exactly the eigenvectors that correspond to the largest $k-1$ eigenvalues with respect to {{ Template:namespace detect | type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. Please improve this article if you can. (What if we encounter complex eigenvalues? Then concept of being large does not dense. What is the solution in that case?) |", "object. To answer the next part of the question we will also use vectors, this part seems almost tailor made to do so given that we are looking for when two lines are perpendicular. \\begin{align*}\\ \\ \\overrightarrow{OH} = \\mathbf{h} = \\begin{bmatrix} p_1 + p_2 + p_3 \\\\ q_1 + q_2 + q_3 \\end{bmatrix} = \\mathbf{a}+\\mathbf{b}+\\mathbf{c}\\end{align*} \\begin{align*}\\mathbf{r_{AH}} = \\mathbf{a} + \\lambda (\\mathbf{h} - \\mathbf{a}) = \\mathbf{a} + \\lambda (\\mathbf{b} + \\mathbf{c}) \\end{align*} \\begin{align*}\\mathbf{r_{BC}} = \\mathbf{b} + \\mu (\\mathbf{c} - \\mathbf{b}) \\end{align*} If $AH \\perp BC,\\ (\\mathbf{b} + \\mathbf{c}) \\cdot (\\mathbf{c} - \\mathbf{b}) = 0 \\Rightarrow \\mathbf{b} \\cdot \\mathbf{b} = \\mathbf{c} \\cdot \\mathbf{c} \\Rightarrow |\\mathbf{b}| = |\\mathbf{c}| \\therefore p_2^2+q_2^2 = p_3^2+q_3^2 (1)$ as required. A virtually identical method is taken for $BH \\perp AC$. \\begin{align*}\\mathbf{r_{BH}} = \\mathbf{b} + \\lambda (\\mathbf{a} + \\mathbf{c}) \\end{align*} \\begin{align*}\\mathbf{r_{AC}} = \\mathbf{a} + \\mu (\\mathbf{c} - \\mathbf{a}) \\end{align*} If $BH \\perp AC,\\ (\\mathbf{a} + \\mathbf{c}) \\cdot (\\mathbf{c} - \\mathbf{a}) = 0 \\Rightarrow \\mathbf{a} \\cdot \\mathbf{a} = \\mathbf{c} \\cdot \\mathbf{c} \\Rightarrow |\\mathbf{a}| = |\\mathbf{c}| \\therefore p_1^2+q_1^2 = p_3^2+q_3^2 (2)$. The final part can be easily deduced if we subtract the two statements we have already shown to be true, eliminating $p_3,\\ q_3$. \\begin{align*}\\ \\ (1) - (2) \\Rightarrow p_1^2 + q_1^2 = p_2^2 + q_2^2 \\Rightarrow(\\mathbf{a} + \\mathbf{b}) \\cdot (\\mathbf{b} - \\mathbf{a}) = 0 \\therefore AC \\perp", "+ P(\\mathbf{x} \\mid \\mathbf{a}, \\mathbf{\\phi}) = \\frac{1}{Z} \\exp(- \\frac{(\\mathbf{x}-\\sum^{k}_{i=1} a_i \\mathbf{\\phi}_{i})^2}{2\\sigma^2}) + \\end{align} + In order to determine the distribution $P(\\mathbf{x}\\mid\\mathbf{\\phi})$, we also need to specify the prior distribution $P(\\mathbf{a})$. Assuming the independence of our source features, we can factorize our prior probability as + :\\begin{align} + P(\\mathbf{x} \\mid \\mathbf{a}, \\mathbf{\\phi}) = \\frac{1}{Z} \\exp(- \\frac{(\\mathbf{x}-\\sum^{k}_{i=1} a_i \\mathbf{\\phi}_{i})^2}{2\\sigma^2}) \\end{align} \\end{align}[/itex] - - - In order to define the distribution $P(\\mathbf{x}\\mid\\mathbf{\\phi})$, we must first specify a prior distribution over the amplitudes $a_i$", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "e&f\\\\ \\end{pmatrix}$ $\\begin{matrix} a&b\\\\ c&d\\\\ e&f\\\\ \\end{matrix}$ $\\begin{vmatrix} a&b\\\\ c&d\\\\ e&f\\\\ \\end{vmatrix}$ \\end{document} main.tex main.pdf ### Aligned mathematics (Keywords: aligned, label, mathematics.) Video demo It is possible to create aligned mathematics using: \\begin{align} (x+h)^2-x^2 & =x^2+2xh+h^2-x^2 \\nonumber\\\\ & =2xh+h^2 \\nonumber\\\\ & =h(2x+h) \\nonumber \\end{align} Note that this requires the amsmath package. Annotated text can also be added: \\begin{align} (x+h)^2-x^2 & = x^2+2xh+h^2-x^2 && \\text{(by distributivity)}\\\\ & = 2xh+h^2 && \\text{(by subtraction)}\\\\ & = h(2x+h) && \\text{(by factorisation)} \\end{align} A full document showing these is: \\documentclass{article} \\usepackage{amsmath} \\title{Aligned mathematics} \\author{V Knight} \\date{\\today} \\begin{document} \\maketitle Some examples of aligned mathematics: \\begin{align} (x+h)^2-x^2 & =x^2+2xh+h^2-x^2 \\nonumber\\\\ & =2xh+h^2 \\nonumber\\\\ & =h(2x+h) \\nonumber \\end{align} Some explanatory text: \\begin{align} (x+h)^2-x^2 & = x^2+2xh+h^2-x^2 && \\text{(by distributivity)}\\\\ & = 2xh+h^2 && \\text{(by subtraction)}\\\\ & = h(2x+h) && \\text{(by factorisation)} \\end{align} \\end{document} main.tex main.pdf ### Partitioned statements (Keywords: partitioned, mathematics.) Video demo It is possible to create partitioned statements for case dependent expressions: $1 + (-1) ^ n = \\begin{cases} 0, & \\text{if $$n$$ odd}\\\\ 2, & \\text{if $$n$$ even} \\end{cases}$ Note that this requires the amsmath package. A full document showing this is: \\documentclass{article} \\usepackage{amsmath} \\title{Partitioned statements} \\author{V Knight} \\date{\\today} \\begin{document} \\maketitle $1 + (-1) ^ n = \\begin{cases} 0, & \\text{if $$n$$ odd}\\\\ 2, & \\text{if $$n$$ even} \\end{cases}$ \\end{document} main.tex main.pdf ###", "Similar to the two class case problem, we have: max $Tr[\\mathbf{W}^{T}\\mathbf{S}_{B}\\mathbf{W}]$ subject to $Tr[\\mathbf{W}^{T}\\mathbf{S}_{W}\\mathbf{W}]=1$ To solve this optimization problem a Lagrange multiplier $\\Lambda$, which actually is a $d \\times d$ diagonal matrix, is introduced: \\begin{align} L(\\mathbf{W},\\Lambda) = Tr[\\mathbf{W}^{T}\\mathbf{S}_{W}\\mathbf{B}] - \\Lambda\\left\\{ Tr[\\mathbf{W}^{T}\\mathbf{S}_{W}\\mathbf{W}] - 1 \\right\\} \\end{align} Differentiating with respect to $\\mathbf{W}$ we obtain: \\begin{align} \\frac{\\partial L}{\\partial \\mathbf{W}} = (\\mathbf{S}_{B} + \\mathbf{S}_{B}^{T})\\mathbf{W} - \\Lambda (\\mathbf{S}_{W} + \\mathbf{S}_{W}^{T})\\mathbf{W} \\end{align} Note that the $\\mathbf{S}_{B}$ and $\\mathbf{S}_{W}$ are both symmetric matrices, thus set the first derivative to zero, we obtain: \\begin{align} \\mathbf{S}_{B}\\mathbf{W} - \\Lambda\\mathbf{S}_{W}\\mathbf{W}=0 \\end{align} Thus, \\begin{align} \\mathbf{S}_{B}\\mathbf{W} = \\Lambda\\mathbf{S}_{W}\\mathbf{W} \\end{align} where $\\mathbf{\\Lambda} = \\begin{pmatrix} \\lambda_{1} & & 0\\\\ &\\ddots&\\\\ 0 & &\\lambda_{d} \\end{pmatrix}$ and $\\mathbf{W} = [\\mathbf{w}_{1}, \\mathbf{w}_{2},..., \\mathbf{w}_{k-1}]$. As a matter of fact, $\\mathbf{\\Lambda}$ must have $\\mathbf{k-1}$ nonzero eigenvalues, because $rank({S}_{W}^{-1}\\mathbf{S}_{B})=k-1$. Therefore, the solution for this question is as same as the previous case. The columns of the transformation matrix $\\mathbf{W}$ are exactly the eigenvectors that correspond to largest $k-1$ eigenvalues with respect to \\begin{align} \\mathbf{S}_{W}^{-1}\\mathbf{S}_{B}\\mathbf{w}_{i} = \\lambda_{i}\\mathbf{w}_{i} \\end{align} ## Linear Regression Models - October 14, 2009 Regression analysis is a general statistical technique for modelling and analyzing how a dependent variable changes according to changes in independent variables. In classification, we are interested in how a label, $\\,y$, changes according to changes in $\\,X$. We will start by considering a", "with \\\\, that will give you unwanted vertical space after the math display (and an extra number for numbered equations). \\documentclass{article} \\usepackage{amsmath} \\begin{document} \\begin{alignat}{2} y_{\\mathrm{destr}} &= nL\\lambda/2d, \\text{ where } & n &= \\pm 1, \\pm 3, \\pm 5, \\dots \\\\ y_{\\mathrm{constr}} &= nL\\lambda/d, \\text{ where} & n &= 0, \\pm1, \\pm2, \\pm3, \\dots \\end{alignat} \\end{document} As Sam Carter suggested, it might be better to align at where instead of the second =. Alternatively, if the second n was a wider symbol/longer expression than the first n, then you can align at both where and =. It wont make a difference in this case though. \\documentclass{article} \\usepackage{amsmath} \\begin{document} Align at where: \\begin{alignat}{2} y_{\\mathrm{destr}} &= nL\\lambda/2d, &&\\text{ where } n = \\pm 1, \\pm 3, \\pm 5, \\dots \\\\ y_{\\mathrm{constr}} &= nL\\lambda/d, &&\\text{ where } n = 0, \\pm1, \\pm2, \\pm3, \\dots \\end{alignat} Align at where and =: \\begin{alignat}{3} y_{\\mathrm{destr}} &= nL\\lambda/2d, &&\\text{ where } & n &= \\pm 1, \\pm 3, \\pm 5, \\dots \\\\ y_{\\mathrm{constr}} &= nL\\lambda/d, &&\\text{ where } & n &= 0, \\pm1, \\pm2, \\pm3, \\dots \\end{alignat} \\end{document} • May I suggest a second version with the \"where\" aligned? (You bet my by 20 seconds :) – samcarter_is_at_topanswers.xyz Jan 23 '17 at 15:38 • @samcarter Working on it. – Torbjørn T. Jan 23 '17 at", "roll strategy we get the following PMF \\begin{align} p(\\min(x, y) = b) &= \\frac{2}{N}\\,\\sum_{i=b+1}^{N} \\frac{1}{N} + \\frac{1}{N^2} \\\\ &= \\frac{2 (N - b) + 1}{N^2} \\\\ \\end{align} From PMF we can see that advantages and disadvantages have the same shape, expect that one distributions is a flipped version of the other distribution For this problem general PMF formula for disadvantages could be applied here as well \\begin{align} p(\\min(\\max(x, y), \\max(u, v)) = b) = 2\\,p(b)\\,\\sum_{i=b+1}^{N} p(i) + p(b)^2 \\\\ \\end{align} The only difference is that now we need to use PMF from advanages rather than from a single roll distribution \\begin{align} p(\\min(\\max(x, y), \\max(u, v)) = b) &= \\left(\\frac{2b - 1}{N^2}\\right)^2 + 2 \\frac{2b - 1}{N^2} \\, \\sum_{i=b+1}^{N^2} \\frac{2i - 1}{N^2} \\\\ &= \\frac{1}{N^4}(-4\\,b^3 + 6\\,b^2 + 4\\,b\\,(N^2-1) - 2\\,N^2+1) \\end{align} And from PMF it’s easy to find CMF \\begin{align} p(\\min(\\max(x, y), \\max(u, v)) \\leq b) = \\frac{1}{N^4}(2\\,b^2\\,N^2 - b^4) \\end{align} For this problem we can apply general formula for advantages \\begin{align} p(\\max(\\min(x, y), \\min(u, v)) = b) = 2\\,p(b)\\,\\sum_{i=1}^{b-1} p(i) + p(b)^2 \\\\ \\end{align} And similarly we can derive PMF \\begin{align} p(\\max(\\min(x, y), \\min(u, v)) = b) = \\frac{1}{N^4}( &4\\,b^3 - 6\\,b^2\\,(2\\,N + 1)\\,+ \\\\ &+4\\,b\\,(2\\,N^2 + 3\\,N + 1) - (2\\,N+1)^2) \\end{align} And as before we can calculate CMF \\begin{align} p(\\max(\\min(x, y), \\min(u, v)) \\leq b)", "not constrain the values $$P_{ij}$$ well. Therefore, we will use several functions $$\\{f_k\\}$$ in our minimization, \\begin{aligned} &\\min_{P_{ij}} \\sum_k w_k \\| f^{F,k}_i - \\sum_j P_{ij} f^{C,k}_j \\|_2 \\\\ = &\\min_{P_{ij}} \\sum_k \\| \\sqrt{w_k} f^{F,k}_i - \\sqrt{w_k} \\sum_j P_{ij} f^{C,k}_j \\|_2 \\\\ = &\\min_{P_{ij}} \\| W^{1/2} \\mathbf{f}^{F}_i - W^{1/2} \\mathbf{f}^{C} p_i \\|_2 \\end{aligned} where \\begin{aligned} W &= \\begin{pmatrix} w_0 & & \\\\ & \\ddots & \\\\ & & w_K \\end{pmatrix} \\\\ \\mathbf{f}^{F}_i &= \\begin{pmatrix} f^{F,0}_i \\\\ \\vdots \\\\ f^{F,K}_i \\end{pmatrix} \\\\ \\mathbf{f}^{C} &= \\begin{pmatrix} f^{C,0}_0 & \\cdots & f^{C,0}_n \\\\ \\vdots & \\ddots & \\vdots \\\\ f^{C,K}_0 & \\cdots & f^{C,K}_n \\end{pmatrix} \\\\ p_i &= \\begin{pmatrix} P_{i0} \\\\ \\vdots \\\\ P_{in} \\end{pmatrix} \\end{aligned} or alternatively \\begin{aligned} [W]_{kk} &= w_k \\\\ [f^{F}_i]_k &= f^{F,k}_i \\\\ [f^{C}]_{kj} &= f^{C,k}_j \\\\ [p_i]_j &= P_{ij} \\end{aligned} We thus have a standard least-squares problem $\\min_{P_{ij}} \\| b - A x \\|_2$ where \\begin{aligned} A &= W^{1/2} f^{C} \\\\ b &= W^{1/2} f^{F}_i \\\\ x &= p_i \\end{aligned} which can be solved using LAPACK. We will typically perform this optimization on a multigrid level $$l$$ when the change in eigenvalue from level $$l+1$$ is relatively large, meaning $\\frac{|\\lambda_l - \\lambda_{l+1}|}{|\\lambda_l|}.$ This indicates that the generalized eigenvector associated with that eigenvalue was not adequately represented by $$P^l_{l+1}$$, and the interpolator should be recomputed. ### Balancing Domain" ]

[ "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "Also, let $\\mathbf y$ be the $m$-dimensionl vector with your target values: $$\\mathbf y=\\begin{pmatrix}y^{(1)}\\\\y^{(2)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}.$$ Now, since $h_w(x^{(i)})=(x^{(i)})^T\\mathbf w$, we can verify $$\\begin{array}{rcl}X\\mathbf w -\\mathbf y&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w\\end{pmatrix}-\\begin{pmatrix}y^{(1)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}\\\\&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w-y^{(1)}\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w-y^{(m)}\\end{pmatrix}.\\end{array}$$ Thus, using $\\mathbf v^T\\mathbf v = \\sum_i\\mathbf v_i^2$ for a vector $\\mathbf v$, we obtain $$\\begin{array}{rcl}E(w)&=&\\frac{1}{2}\\sum_{i=1}^m(h_w(x^{(i)}) - y^{(i)})^2\\\\&=&\\frac{1}{2}(X\\mathbf w-\\mathbf y)^T(X\\mathbf w-\\mathbf y).\\end{array}$$ Summarizing. To obtain $$\\mathbf w=\\begin{pmatrix}w_0\\\\w_1\\end{pmatrix},$$ proceed: 1. Set your collections $x$ and $y$: $$x=\\begin{pmatrix}1\\\\3\\\\5\\\\7\\\\9\\end{pmatrix}\\qquad y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2. Loop until you think that $w_j$ converge (try with $10$ iterations) { for $i=1$ to $m$ { $$w_j:=w_j+\\eta\\sum_{i=1}^m(y^{(i)}-h_w(x^{(i)}))x_j^{(i)}\\qquad(\\text{for each }j)$$ } } Insted you can use 1'. Set your vectors $X$ and $\\mathbf y$: $$X=\\begin{pmatrix}1&1\\\\1&3\\\\1&5\\\\1&7\\\\1&9\\end{pmatrix}\\qquad\\mathbf y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2'. Loop until you think that $\\mathbf w$ converge (try with $10$ iterations) { $$\\mathbf w:=\\mathbf w-\\frac{\\eta}{m}\\left((X\\mathbf w-\\mathbf y)^TX\\right)$$ } Note. With the above it can prove a analytical formula to obtain the parameters, using the normal equations to computed $\\mathbf w$ by $$\\mathbf w=(X^TX)^{-1}X^T\\mathbf y.$$ Obviously you need to use a computer software (like python or octave) or a calculator device (with matrix-vector support) to calculate the values.", "+0 # Help 0 35 2 The vectors $$\\dbinom{3}{2}$$ and $$\\dbinom{-4}{1}$$ can be written as linear combinations of $$\\mathbf{u}$$ and $$\\mathbf{w}$$\\begin{align*} \\dbinom{3}{2} &= 5\\mathbf{u}+8\\mathbf{w} \\\\ \\dbinom{-4}{1} &= -3\\mathbf{u}+\\mathbf{w} . \\end{align*} The vector $$\\dbinom{5}{-2}$$ can be written as the linear combination $$a\\mathbf{u}+b\\mathbf{w}$$. Find the ordered pair $$(a,b)$$. Feb 7, 2020 #1 0 (a,b) = (17,-4). Feb 7, 2020 #2 +24065 +1 The vectors $$\\dbinom{3}{2}$$ and $$\\dbinom{-4}{1}$$ can be written as linear combinations of $$u$$ and $$w$$: \\begin{align*} \\dbinom{3}{2} = 5\\mathbf{u}+8\\mathbf{w} ,\\\\\\\\ \\dbinom{-4}{1} = -3\\mathbf{u}+\\mathbf{w} . \\end{align*} The vector $$\\dbinom{5}{-2}$$ can be written as the linear combination $$a\\mathbf{u}+b\\mathbf{w}$$. Find the ordered pair $$(a,b)$$. $$\\text{Let \\dbinom{3}{2} = v_1 } \\\\ \\text{Let \\dbinom{-4}{1} = v_2 }$$ $$\\begin{array}{|lrcll|} \\hline (1): & \\mathbf{5\\mathbf{u}+8\\mathbf{w}} &=& \\mathbf{v_1} \\\\\\\\ & -3\\mathbf{u}+\\mathbf{w} &=& v_2 \\quad | \\quad \\times 8 \\\\ (2): & -24 u + 8w &=& 8v_2 \\\\ \\hline (1)-(2): & 29u &=& v_1-8v_2 \\quad | \\quad v_1=\\dbinom{3}{2},\\ v_2 = \\dbinom{-4}{1} \\\\ & 29u &=& \\dbinom{3}{2}-8\\dbinom{-4}{1} \\\\ & 29u &=& \\dbinom{3}{2}+\\dbinom{32}{-8} \\\\ & 29u &=& \\dbinom{35}{-6} \\\\ & \\mathbf{u} &=& \\mathbf{\\dfrac{1}{29} \\dbinom{35}{-6}} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline -3\\mathbf{u}+\\mathbf{w} &=& v_2 \\\\ w &=& v_2+3u \\quad | \\quad v_2 = \\dbinom{-4}{1},\\ \\mathbf{u=\\dfrac{1}{29} \\dbinom{35}{-6}} \\\\ w &=& \\dbinom{-4}{1}+\\dfrac{3}{29} \\dbinom{35}{-6} \\\\ w &=& \\dfrac{29}{29}\\dbinom{-4}{1}+\\dfrac{3}{29} \\dbinom{35}{-6} \\\\ w &=& \\dfrac{1}{29} \\left( 29\\dbinom{-4}{1}+3\\dbinom{35}{-6} \\right) \\\\ w &=& \\dfrac{1}{29} \\dbinom{-4*29+3*35}{29-18} \\\\ \\mathbf{w} &=&", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "v_2)^2$ is minimized. W.l.o.g. we can take $\\mathbf\\pi_1$ to be the $u$-$v$ plane, so that its camera matrix is the canonical $$P_1=\\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}.$$ This just picks out the $u$- and $v$-coordinates of $\\mathbf U$. The projection matrix onto $\\mathbf\\pi_2$ will have the form $$P_2=\\left[\\begin{array}{c|c}M_{2\\times 3}&\\mathbf t\\\\\\mathbf 0^T&1\\end{array}\\right].$$ The point $\\mathbf U$ then satisfies the linear equations $P_1\\mathbf U =\\mathbf u_1=(u_1,v_1)$ and $P_2\\mathbf U = \\mathbf u_2=(u_2,v_2)$. The affine fundamental matrix of this epipolar configuration is $$F_A=\\begin{bmatrix}0&0&a\\\\0&0&b\\\\c&d&e\\end{bmatrix}$$ where $$a = m_{23} \\\\ b=-m_{13} \\\\ c=m_{13}m_{21}-m_{11}m_{23} \\\\ d=m_{13}m_{22}-m_{12}m_{23} \\\\ e=m_{13}t_2-m_{23}t_1.$$ (In your case, where $P_2$ has no translation component, we would have $e=0$.) The adjusted coordinates are then given by the formula $$\\begin{bmatrix}\\hat u_2 \\\\ \\hat v_2 \\\\ \\hat u_1 \\\\ \\hat v_1\\end{bmatrix} = \\begin{bmatrix}u_2 \\\\ v_2 \\\\ u_1 \\\\ v_1\\end{bmatrix}-{au_2+bv_2+cu_1+dv_1+e\\over a^2+b^2+c^2+d^2}\\begin{bmatrix}a\\\\b\\\\c\\\\d\\end{bmatrix}.$$ (You might recognize this as the result of a standard least-squares fit, which itself is basically a projection onto a hyperplane in $\\mathbb R^4$.) With these adjusted coordinates in hand, you can then find the intersection of their back-projections in the ways described previously. • Thank you for taking the time to explain this, however, I do not understand what you mean when you say that U will lie on the intersection of those two lines (I assume lambda and mu are just arbitrary variables you chose without any", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "x_i + b \\,, \\end{align} with $$w_i$$ denoting the $$i$$-th so-called weight, i.e. parameter, of this block and $$b$$ denoting the so-called bias (also a parameter), i.e., an additive term not depending on the input. When one input is simultaneously processed by $$K$$ linear blocks, i.e., \\begin{align} y_k = \\sum\\limits^{N}_{i=1} w^k_i \\cdot x_i + b^k = \\underbrace{\\left( w^k_1, w^k_2, \\ldots, w^k_N, b^k \\right)}_{\\mathbf{w}\\transp_k} \\cdot \\underbrace{\\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_N \\\\ 1 \\end{pmatrix}}_{\\mathbf{x}} \\,, \\end{align} for block $$k$$, this can be expressed compactly via matrix-vector multiplications: \\begin{align} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_K \\end{pmatrix} &= \\begin{pmatrix} \\qquad \\mathbf{w}\\transp_1 \\qquad \\\\ \\mathbf{w}\\transp_2 \\\\ \\vdots \\\\ \\mathbf{w}\\transp_K \\end{pmatrix} \\cdot \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_N \\\\ 1 \\end{pmatrix} \\\\ &= \\qquad \\quad \\mathbf{W} \\quad \\qquad \\cdot \\quad \\mathbf{x} \\,, \\end{align} with $$\\mathbf{W} \\in \\mathbb{R}^{K \\times (N+1)}$$ and $$\\mathbf{x} \\in \\mathbb{R}^{(N+1) \\times 1}$$. This expression can be further extended for the case when multiple input vectors $$\\mathbf{x}_1, \\mathbf{x}_2, \\ldots, \\mathbf{x}_B$$, i.e., a so-called batch of size $$B$$, have to be processed simultaneously: \\begin{align} \\begin{pmatrix} \\qquad \\mathbf{w}\\transp_1 \\qquad \\\\ \\mathbf{w}\\transp_2 \\\\ \\vdots \\\\ \\mathbf{w}\\transp_K \\end{pmatrix} \\cdot \\begin{pmatrix} \\\\ \\\\ \\mathbf{x}_1 & \\mathbf{x}_2 & \\cdots & \\mathbf{x}_B \\\\ \\\\ \\\\ \\end{pmatrix} &= \\begin{pmatrix} \\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_1 & \\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_2 &\\cdots &\\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_B \\\\ \\mathbf{w}\\transp_2 \\cdot", "x}+{\\mathbf y}\\neq {\\mathbf 0} \\}\\\\ & = \\min\\{ w({\\mathbf z}) : {\\mathbf z} \\neq {\\mathbf 0} \\}\\text{.} \\end{align*} ### SubsectionLinear Codes From Example 8.16, it is now easy to check that the minimum nonzero weight is $$3\\text{;}$$ hence, the code does indeed detect and correct all single errors. We have now reduced the problem of finding “good” codes to that of generating group codes. One easy way to generate group codes is to employ a bit of matrix theory. Define the inner product of two binary $$n$$-tuples to be \\begin{equation*} {\\mathbf x} \\cdot {\\mathbf y} = x_1 y_1 + \\cdots + x_n y_n\\text{,} \\end{equation*} where $${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^\\transpose$$ and $${\\mathbf y} = (y_1, y_2, \\ldots, y_n)^\\transpose$$ are column vectors. 5 For example, if $${\\mathbf x} = (011001)^\\transpose$$ and $${\\mathbf y} = (110101)^\\transpose\\text{,}$$ then $${\\mathbf x} \\cdot {\\mathbf y} = 0\\text{.}$$ We can also look at an inner product as the product of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^\\transpose {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}\\text{.} \\end{align*} Since we will be working with matrices, we will write binary", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "2, 0) ... The given coordinates are in object space and the problem asks us to compute their representations in frames $$\\{\\mathbf{e}_i,\\mathbf{P}\\}$$ and $$\\{\\mathbf{e}_i,\\mathbf{O}\\}$$ respectively. Take $$\\mathbf{x}=(-1,2,0)$$ as an example, we want to find the coordinates $$\\mathbf{l}$$ in upright space: \\begin{align} x_i \\mathbf{e}'_i + \\mathbf{P} &= l_i \\mathbf{e}_i + \\mathbf{P}\\\\ x_j m_{ji} \\mathbf{e}_i &= l_i \\mathbf{e}_i \\end{align} Written in matrix form: $$\\mathbf{M}^T \\mathbf{x}=\\mathbf{l}$$. Next we find the coordinates $$\\mathbf{w}$$ in world space: \\begin{align} x_i \\mathbf{e}'_i + \\mathbf{P} &= w_i \\mathbf{e}_i + \\mathbf{O} \\\\ x_i \\mathbf{e}'_i + p_i\\mathbf{e}_i &= w_i \\mathbf{e}_i \\\\ x_j m_{ji} \\mathbf{e}_i + p_i\\mathbf{e}_i &= w_i \\mathbf{e}_i \\\\ x_j m_{ji} + p_i &= w_i \\end{align} Written in vector form: $$\\mathbf{M}^T \\mathbf{x} + \\mathbf{p} = \\mathbf{w}$$ Your idea is generally correct. And the \"upright\" matrix can be constructed by considering the relation ($$m_{ij}$$) between the two basis, as shown above.", "$$\\mathbf{x}=(-1,2,0)$$ as an example, we want to find the coordinates $$\\mathbf{l}$$ in upright space: \\begin{align} x_i \\mathbf{e}'_i + \\mathbf{P} &= l_i \\mathbf{e}_i + \\mathbf{P}\\\\ x_j m_{ji} \\mathbf{e}_i &= l_i \\mathbf{e}_i \\end{align} Written in matrix form: $$\\mathbf{M}^T \\mathbf{x}=\\mathbf{l}$$. Next we find the coordinates $$\\mathbf{w}$$ in world space: \\begin{align} x_i \\mathbf{e}'_i + \\mathbf{P} &= w_i \\mathbf{e}_i + \\mathbf{O} \\\\ x_i \\mathbf{e}'_i + p_i\\mathbf{e}_i &= w_i \\mathbf{e}_i \\\\ x_j m_{ji} \\mathbf{e}_i + p_i\\mathbf{e}_i &= w_i \\mathbf{e}_i \\\\ x_j m_{ji} + p_i &= w_i \\end{align} Written in vector form: $$\\mathbf{M}^T \\mathbf{x} + \\mathbf{p} = \\mathbf{w}$$ Your idea is generally correct. And the \"upright\" matrix can be constructed by considering the relation ($$m_{ij}$$) between the two basis, as shown above.", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "### $( 2 \\implies 1)$ Next suppose that the statement 2 is true. That is, there exists an invertible matrix $S$ such that $S^{-1}AS=C$. Then we have $AS=CS$. Define $\\mathbf{v}:=S\\mathbf{e}_1$ to be the first column vector of $S$. We prove that with this choice of $\\mathbf{v}$, the vectors $\\mathbf{v}, A\\mathbf{v}, A^2\\mathbf{v}, \\dots, A^{n-1}\\mathbf{v}$ form a basis of $\\C^n$. We have \\begin{align*} AS&=A[S\\mathbf{e}_1, S\\mathbf{e}_2, \\dots, S\\mathbf{e}_n]\\\\ &=[AS\\mathbf{e}_1, AS\\mathbf{e}_2, \\dots, AS\\mathbf{e}_n] \\end{align*} and \\begin{align*} SC&=S [\\mathbf{e}_2, \\mathbf{e}_3, \\dots, \\mathbf{e}_n, -\\mathbf{a}]\\\\ &=[S \\mathbf{e}_2, S\\mathbf{e}_3, \\dots, S\\mathbf{e}_n, -S\\mathbf{a}]. \\end{align*} Since $AS=SC$, comparing column vectors, we obtain \\begin{align*} AS\\mathbf{e}_1&=S\\mathbf{e}_2\\\\ AS\\mathbf{e}_2&=S\\mathbf{e}_3\\\\ \\vdots\\\\ AS\\mathbf{e}_{n-1}&=S\\mathbf{e}_n\\\\ AS\\mathbf{e}_n&=-S\\mathbf{a}.\\\\ \\end{align*} It follows that we have inductively \\begin{align*} A\\mathbf{v}&=S\\mathbf{e}_2\\\\ A^2\\mathbf{v}&=A(A\\mathbf{v})=AS\\mathbf{e}_2=S\\mathbf{e}_3\\\\ \\vdots\\\\ A^{n-1}\\mathbf{v}&=A(A^{n-2}\\mathbf{v})=AS\\mathbf{e}_{n-1}=S\\mathbf{e}_n. \\end{align*} Namely we obtain $A^i\\mathbf{v}=S\\mathbf{e}_{i+1}$ for $i=0, 1, \\dots, n-1$. Now, suppose that we have the linear combinaiton $c_1\\mathbf{v}+c_2A\\mathbf{v}+ \\dots c_n A^{n-1}\\mathbf{v}=\\mathbf{0},$ for some coefficient scalars $c_1, \\dots, c_n \\in \\C$. Then we have using the above relations \\begin{align*} c_1S\\mathbf{e}_1+c_2S\\mathbf{e}_2+\\cdots +c_nS\\mathbf{e}_n=\\mathbf{0}. \\end{align*} Since $S$ is invertible, the column vectors $S\\mathbf{e}_i$ are linearly independent. Thus, we must have $c_1=c_2=\\cdots=c_n=0$. Therefore the vectors $\\mathbf{v}, A\\mathbf{v}, A^2\\mathbf{v}, \\dots, A^{n-1}\\mathbf{v}$ are linearly independent, and they form a basis of $\\C^n$ since we have $n$ linearly independent vectors in the $n$-dimensional vector space $\\C^n$. ## Related Question. For a basic question about the companion matrix of a polynomial, check out the post “Companion", "2-i\\\\ -i\\\\ -1+2i\\end{array}\\right)$$\\mathbf{x}=\\begin{pmatrix} 1 \\\\ 2-i \\\\ -i \\\\ -1+2i \\end{pmatrix}$ We need to find the column vector $\\mathbf{X}$$\\mathbf{X}$. The general method is already shown on the Wikipedia page. But we will develop a matrix notation for the same. $\\mathbf{X}$$\\mathbf{X}$ can be easily obtained by pre multiplying $\\mathbf{x}$$\\mathbf{x}$ by the matrix: $M=\\frac{1}{\\sqrt{N}}\\left(\\begin{array}{cccc}1& 1& 1& 1\\\\ 1& w& {w}^{2}& {w}^{3}\\\\ 1& {w}^{2}& {w}^{4}& {w}^{6}\\\\ 1& {w}^{3}& {w}^{6}& {w}^{9}\\end{array}\\right)$ where $w$$w$ is ${e}^{\\frac{-2\\pi i}{N}}$$e^{\\frac{-2\\pi i}{N}}$. Each element of the matrix is basically ${w}^{ij}$$w^{ij}$. $\\frac{1}{\\sqrt{N}}$$\\frac{1}{\\sqrt{N}}$ is simply a normalization constant. Finally, $\\mathbf{X}$$\\mathbf{X}$ turns out to be: $\\frac{1}{2}\\left(\\begin{array}{c}2\\\\ -2-2i\\\\ -2i\\\\ 4+4i\\end{array}\\right)$$\\frac{1}{2}\\begin{pmatrix} 2 \\\\ -2-2i \\\\ -2i \\\\ 4+4i \\end{pmatrix}$. Now, sit back for a while and notice a few important properties: • All the columns of the matrix $M$$M$ are orthogonal to each other. • All the columns of $M$$M$ have magnitude $1$$1$. • If you post multiply $M$$M$ with a column vector having lots of zeroes (large spread) you'll end up with a column vector with only a few zeroes (narrow spread). The converse also holds true. (Check!) It can be very simply noticed that the classical DFT has a time complexity $\\mathcal{O}\\left({N}^{2}\\right)$$\\mathcal O(N^2)$. That is because for obtaining every row of $\\mathbf{X}$$\\mathbf{X}$, $N$$N$ operations need to be performed. And there are $N$$N$ rows in $\\mathbf{X}$$\\mathbf{X}$. ## The Fast fourier transform:" ]

[ "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "example of how to do this in R! a <- 5 x <- c(1, 4, 7, 9, 10) a*x #> [1] 5 20 35 45 50 is.vector(a*x) #> [1] TRUE We can also perform addition with vectors of the same size, or dimension. For vectors, the addition operation is also performed element-wise. $\\mathbf{x} + \\mathbf{y} = \\begin{pmatrix} x_{1} \\\\ x_{2} \\\\ \\vdots \\\\ x_{n} \\end{pmatrix} + \\begin{pmatrix} y_{1} \\\\ y_{2} \\\\ \\vdots \\\\ y_{n} \\end{pmatrix} = \\begin{pmatrix} x_{1} + y_{1} \\\\ x_{2} + y_{2} \\\\ \\vdots \\\\ x_{n} + y_{n} \\end{pmatrix}.$ Here’s a picture of what this looks like in two dimensions! Let’s say we want to add the two vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ below. What does this look like in two dimensions? We could start first with $$\\mathbf{a}$$. The vector $$\\mathbf{a}$$ says that we move to the right two squares and then up three squares. Once we’re there, the vector $$\\mathbf{b}$$ says that we move to the left three squares and then up one square. Alternatively, we would get the same answer if we started first with $$\\mathbf{b}$$. The vector $$\\mathbf{b}$$ says that we move to the left three squares and then up one square. Once we’re there, the vector $$\\mathbf{a}$$ says that we move to the right 2 squares and then up three squares. Remember that", "Study guide and 3 practice problems on: ## Midpoint between two vectors The midpoint between the two vectors $\\mathbf{x}$ and $\\mathbf{y}$ is $\\frac{\\mathbf{x} + \\mathbf{y}}{2}$.", "\\quad \\mathbf{b} = \\begin{pmatrix} x_b \\\\y_b\\end{pmatrix}$$ If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that $$N\\begin{pmatrix} x_a \\\\ y_a\\end{pmatrix} +M\\begin{pmatrix} x_b\\\\ y_b\\end{pmatrix}\\mathbf{b} = \\begin{pmatrix} 2p \\\\ q\\end{pmatrix}$$ Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations. Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'. So, look at some more particular cases. Look at a base vector with an $x$-component of $4$. $$\\begin{pmatrix} 4 \\\\ 0\\end{pmatrix}$$ Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\\begin{pmatrix} \\pm 2 \\\\ 0\\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components. A suitable pair would be $$\\mathbf{a}= \\begin{pmatrix} 4 \\\\ 1\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}2 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $\\mathbf{a}-\\mathbf{b}$ and $2\\mathbf{b}-\\mathbf{a}$ take us to the", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "must solve the equations \\begin{align*} z_1 & = \\alpha x_1 + \\beta y_1\\\\ z_2 & = \\alpha x_2 + \\beta y_2 \\end{align*} for $$\\alpha$$ and $$\\beta\\text{.}$$ However, this system has a unique solution since \\begin{equation*} \\det \\begin{pmatrix} x_1 & y_1 \\\\ x_2 & y_2 \\end{pmatrix} \\neq 0. \\end{equation*} Two vectors are said to be a basis for $${\\mathbb R}^2$$ if we can write any vector in $${\\mathbb R}^2$$ as a linear combination of these two vectors. By our arguments above, any two linearly independent vectors will form a basis for $${\\mathbb R}^2\\text{.}$$ The vectors $${\\mathbf e}_1 = (1, 0)$$ and $${\\mathbf e}_2 = (0, 1)$$ form a basis for $${\\mathbb R}^2\\text{.}$$ Indeed, if $${\\mathbf z} = (z_1, z_2)\\text{,}$$ then we can write \\begin{equation*} {\\mathbf z} = z_1 {\\mathbf e}_1 + z_2 {\\mathbf e}_2. \\end{equation*} The vectors $${\\mathbf e}_1$$ and $${\\mathbf e}_2$$ are called the standard basis for $${\\mathbb R}^2\\text{.}$$ Let $${\\mathbf v}_1 = (2,1)$$ and $${\\mathbf v}_2 = (3, 2)\\text{.}$$ Since \\begin{equation*} \\det \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\neq 0, \\end{equation*} these vectors form a basis for $${\\mathbb R}^2\\text{.}$$ If $${\\mathbf z} = (-5, -4)\\text{,}$$ then we can write \\begin{equation*} {\\mathbf z} = 2 {\\mathbf v}_1 - 3 {\\mathbf v}_2. \\end{equation*} We say that the coordinates of $${\\mathbf z}$$ are $$(2, -3)$$ with respect", "+0 # Let $\\mathbf{p} = \\begin{pmatrix} 2\\\\ -1\\\\ 2 \\end{pmatrix}$ and $\\mathbf{d} =\\begin{pmatrix} 1\\\\ -1\\\\ 4 \\end{pmatrix}$ and consider the par 0 71 1 a) Let $\\mathbf{p} = \\begin{pmatrix} 2\\\\ -1\\\\ 2 \\end{pmatrix}$ and $\\mathbf{d} =\\begin{pmatrix} 1\\\\ -1\\\\ 4 \\end{pmatrix}$ and consider the particle that visits the point $(x,y,z)$ at time $t$ if $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\mathbf{p} + t\\mathbf{d}.$The vectors $\\mathbf{p}$ and $\\mathbf{d}$ and the graph of the parametrization are shown below: If the intersection between this line and the $xy$-plane is $(a,b,c),$ enter $a, b, c$ in this order below. b) Let $\\mathbf{p} = \\begin{pmatrix}-1\\\\ -2 \\\\ 1\\end{pmatrix}$ and $\\mathbf{d} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ and consider the particle that visits $(x,y,z)$ at time $t$ if $\\begin{pmatrix}x \\\\ y \\\\ z \\end{pmatrix} = \\mathbf{p} + t\\mathbf{d}.$The vectors $\\mathbf{p}$ and $\\mathbf{d}$ and the graph of the parametrization are shown below: Then if $R = (-1, 3, 5)$, calculate the point $(a,b,c)$ on the line that is closest to $R$. Enter $a, b, c$ in that order below. help asap Feb 10, 2021 edited by Guest Feb 10, 2021", "# Spanning Sets for $\\R^2$ or its Subspaces ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Solution. By definition, the subspace $\\Span(S)$ spanned by $S$ is the set of all linear combinations of vectors in $S$. Thus, $\\Span(S)$ is a subset in $\\R^2$. The question is whether all of the vectors in $\\R^2$ are linear combinations of vectors in $S$ or not. ### (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ Let $\\begin{bmatrix} a \\\\ b \\end{bmatrix}$ be an arbitrary vector in $\\R^2$. We determine whether it is a linear combination of $\\mathbf{a}$ and $\\mathbf{b}$: $\\begin{bmatrix} a \\\\ b \\end{bmatrix}=c_1\\mathbf{a}+c_2\\mathbf{b}.$ The augmented matrix is \\begin{align*} \\left[\\begin{array}{rr|r} 1 & 1 & a \\\\ 0 & 1 & b \\end{array} \\right] \\xrightarrow{R_1-R_2} \\left[\\begin{array}{rr|r} 1 & 0 & a-b \\\\ 0 & 1 & b \\end{array} \\right]. \\end{align*} Hence, the solution is $c_1=a-b$, $c_2=b$. Thus, we", "# Spanning Sets for $\\R^2$ or its Subspaces ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Solution. By definition, the subspace $\\Span(S)$ spanned by $S$ is the set of all linear combinations of vectors in $S$. Thus, $\\Span(S)$ is a subset in $\\R^2$. The question is whether all of the vectors in $\\R^2$ are linear combinations of vectors in $S$ or not. ### (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ Let $\\begin{bmatrix} a \\\\ b \\end{bmatrix}$ be an arbitrary vector in $\\R^2$. We determine whether it is a linear combination of $\\mathbf{a}$ and $\\mathbf{b}$: $\\begin{bmatrix} a \\\\ b \\end{bmatrix}=c_1\\mathbf{a}+c_2\\mathbf{b}.$ The augmented matrix is \\begin{align*} \\left[\\begin{array}{rr|r} 1 & 1 & a \\\\ 0 & 1 & b \\end{array} \\right] \\xrightarrow{R_1-R_2} \\left[\\begin{array}{rr|r} 1 & 0 & a-b \\\\ 0 & 1 & b \\end{array} \\right]. \\end{align*} Hence, the solution is $c_1=a-b$, $c_2=b$. Thus, we", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "equal to ${\\displaystyle \\mathbf {a} ={\\begin{pmatrix}a_{1}\\\\a_{2}\\\\a_{3}\\\\a_{4}\\end{pmatrix}}.}$ This can be written in terms of the four standard basis vectors (e1, e2, e3, e4), given by ${\\displaystyle \\mathbf {e} _{1}={\\begin{pmatrix}1\\\\0\\\\0\\\\0\\end{pmatrix}};\\mathbf {e} _{2}={\\begin{pmatrix}0\\\\1\\\\0\\\\0\\end{pmatrix}};\\mathbf {e} _{3}={\\begin{pmatrix}0\\\\0\\\\1\\\\0\\end{pmatrix}};\\mathbf {e} _{4}={\\begin{pmatrix}0\\\\0\\\\0\\\\1\\end{pmatrix}},}$ so the general vector a is ${\\displaystyle \\mathbf {a} =a_{1}\\mathbf {e} _{1}+a_{2}\\mathbf {e} _{2}+a_{3}\\mathbf {e} _{3}+a_{4}\\mathbf {e} _{4}.}$ Vectors add, subtract and scale as in three dimensions. The dot product of Euclidean three-dimensional space generalizes to four dimensions as ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}+a_{4}b_{4}.}$ It can be used to calculate the norm or length of a vector, ${\\displaystyle \\left|\\mathbf {a} \\right|={\\sqrt {\\mathbf {a} \\cdot \\mathbf {a} }}={\\sqrt {{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}+{a_{4}}^{2}}},}$ and calculate or define the angle between two vectors as ${\\displaystyle \\theta =\\arccos {\\frac {\\mathbf {a} \\cdot \\mathbf {b} }{\\left|\\mathbf {a} \\right|\\left|\\mathbf {b} \\right|}}.}$ Minkowski spacetime is four-dimensional space with geometry defined by a nondegenerate pairing different from the dot product: ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}-a_{4}b_{4}.}$ As an example, the distance squared between the points (0,0,0,0) and (1,1,1,0) is 3 in both the Euclidean and Minkowskian 4-spaces, while the distance squared between (0,0,0,0) and (1,1,1,1) is 4 in Euclidean space and 2 in Minkowski space; increasing ${\\displaystyle b_{4}}$ actually decreases the metric distance. This leads to many of the well known apparent \"paradoxes\" of relativity. The cross product is not defined in", "\\left(1, 1\\right)$ and $\\mathbf_2 = \\left(-3, 2\\right).$ Thus, the three vectors are linearly dependent. Two vectors: Now consider the linear dependence of the two vectors $\\mathbf_1 = \\left(1, 1\\right)$ and $\\mathbf_2 = \\left(-3, 2\\right),$ and check, :$a_1 \\begin 1\\\\1\\end + a_2 \\begin -3\\\\2\\end =\\begin 0\\\\0\\end,$ or :$\\begin 1 & -3 \\\\ 1 & 2 \\end\\begin a_1\\\\ a_2 \\end= \\begin 0\\\\0\\end.$ The same row reduction presented above yields, :$\\begin 1 & 0 \\\\ 0 & 1 \\end\\begin a_1\\\\ a_2 \\end= \\begin 0\\\\0\\end.$ This shows that $a_i = 0,$ which means that the vectors ''v''1 = (1, 1) and ''v''2 = (−3, 2) are linearly independent. ## Vectors in R4 In order to determine if the three vectors in $\\mathbb^4,$ :$\\mathbf_1= \\begin1\\\\4\\\\2\\\\-3\\end, \\mathbf_2=\\begin7\\\\10\\\\-4\\\\-1\\end, \\mathbf_3=\\begin-2\\\\1\\\\5\\\\-4\\end.$ are linearly dependent, form the matrix equation, :$\\begin1&7&-2\\\\4& 10& 1\\\\2&-4&5\\\\-3&-1&-4\\end\\begin a_1\\\\ a_2 \\\\ a_3 \\end = \\begin0\\\\0\\\\0\\\\0\\end.$ Row reduce this equation to obtain, :$\\begin 1& 7 & -2 \\\\ 0& -18& 9\\\\ 0 & 0 & 0\\\\ 0& 0& 0\\end \\begin a_1\\\\ a_2 \\\\ a_3 \\end = \\begin0\\\\0\\\\0\\\\0\\end.$ Rearrange to solve for v3 and obtain, :$\\begin 1& 7 \\\\ 0& -18 \\end \\begin a_1\\\\ a_2 \\end = -a_3\\begin-2\\\\9\\end.$ This equation is easily solved to define non-zero ''a''i, :$a_1 = -3 a_3 /2, a_2 = a_3/2,$ where $a_3$ can be chosen arbitrarily. Thus, the vectors $\\mathbf_1, \\mathbf_2,$", "vector with terminal point $$(−2,−4)$$. Let $$\\vec{\\mathbf b}$$ be a vector with initial point $$(1,2)$$ and terminal point $$(−1,4)$$. Find the magnitude of vector $$−3\\vec{\\mathbf a}+\\vec{\\mathbf b}−4\\hat{\\mathbf i}+\\hat{\\mathbf j}.$$ $$15$$ 18) Let $$\\vec{\\mathbf a}$$ be a standard-position vector with terminal point at $$(2,5)$$. Let $$\\vec{\\mathbf b}$$ be a vector with initial point $$(−1,3)$$ and terminal point $$(1,0)$$. Find the magnitude of vector $$\\vec{\\mathbf a}−3\\vec{\\mathbf b}+14\\hat{\\mathbf i}−14\\hat{\\mathbf j}.$$ 19) Let $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$ be two nonzero vectors that are nonequivalent. Consider the vectors $$\\vec{\\mathbf a}=4\\vec{\\mathbf u}+5\\vec{\\mathbf v}$$ and $$\\vec{\\mathbf b}=\\vec{\\mathbf u}+2\\vec{\\mathbf v}$$ defined in terms of $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$. Find the scalar $$λ$$ such that vectors $$\\vec{\\mathbf a}+λ\\vec{\\mathbf b}$$ and $$\\vec{\\mathbf u}−\\vec{\\mathbf v}$$ are equivalent. $$λ=−3$$ 20) Let $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$ be two nonzero vectors that are nonequivalent. Consider the vectors $$\\vec{\\mathbf a}=2\\vec{\\mathbf u}−4\\vec{\\mathbf v}$$ and $$\\vec{\\mathbf b}=3\\vec{\\mathbf u}−7\\vec{\\mathbf v}$$ defined in terms of $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$. Find the scalars $$α$$ and $$β$$ such that vectors $$α\\vec{\\mathbf a}+β\\vec{\\mathbf b}$$ and $$\\vec{\\mathbf u}−\\vec{\\mathbf v}$$ are equivalent. 21) Consider the vector $$\\vec{\\mathbf a}(t)=⟨\\cos t, \\sin t⟩$$ with components that depend on a real number $$t$$. As the number $$t$$ varies, the components of $$\\vec{\\mathbf a}(t)$$ change as well, depending on the functions that define them. a. Write the vectors $$\\vec{\\mathbf a}(0)$$ and $$\\vec{\\mathbf", "we are given a vector $\\mathbf t$ in the vector space (but it might not be in $\\mathcal{L}(\\mathbf B)$) and we want to find the closest vector in $\\mathcal{L}(\\mathbf B)$. There is a simple reduction from $\\textsf{CVP}$ to $\\textsf{SVP}$ by setting $\\mathbf t = \\mathbf 0$. There is also a reduction in the other direction, which involves extending the basis of $\\mathcal{L}(\\mathbf B)$ to also include $\\mathbf t$. This is called embedding. Let us return to the example of DSA and different-moduli equations. The equations we got from before $\\begin{array}{rclc} s_1 \\cdot k_1 - r_1\\cdot x & = & H(m_1) &\\pmod q \\\\ s_2 \\cdot k_2 - r_2\\cdot x & = & H(m_1) &\\pmod q \\\\ -a\\cdot k_1 + 1\\cdot k_2 & = & c & \\pmod M \\\\ \\end{array}$ can be formulated as basis vectors. In matrix form, we have $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 \\\\ -r_2 & 0 & s_2 & 0 & q & 0 \\\\ 0 & -a & 1 & 0 & 0 & M\\end{pmatrix} \\cdot \\mathbf y = \\begin{pmatrix}H(m_1) \\\\ H(m_2) \\\\ c\\end{pmatrix}$ or (by embedding technique): $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 & H(m_1) \\\\ -r_2 & 0 & s_2 & 0 & q & 0 & H(m_2)\\\\" ]

[ "${\\displaystyle w=p\\mathbf {e} _{1}\\cdot \\mathbf {n} _{3}+q\\mathbf {e} _{2}\\cdot \\mathbf {n} _{3}+r\\mathbf {e} _{3}\\cdot \\mathbf {n} _{3}}$. ${\\displaystyle u=c_{11}p+c_{12}q+c_{13}r}$, ${\\displaystyle v=c_{21}p+c_{22}q+c_{23}r}$, ${\\displaystyle w=c_{31}p+c_{32}q+c_{33}r}$, ${\\displaystyle {\\begin{bmatrix}u\\\\v\\\\w\\\\\\end{bmatrix}}={\\begin{bmatrix}c_{11}&c_{12}&c_{13}\\\\c_{21}&c_{22}&c_{23}\\\\c_{31}&c_{32}&c_{33}\\end{bmatrix}}{\\begin{bmatrix}p\\\\q\\\\r\\end{bmatrix}}}$. ${\\displaystyle c_{11}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{12}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{13}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle c_{21}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{22}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{23}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle c_{31}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{32}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{33}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle {\\mathbf {x} }=x_{1}{\\mathbf {e} }_{1}+x_{2}{\\mathbf {e} }_{2}+x_{3}{\\mathbf {e} }_{3}.}$ ${\\displaystyle {\\mathbf {y} }-{\\mathbf {x} }=(y_{1}-x_{1}){\\mathbf {e} }_{1}+(y_{2}-x_{2}){\\mathbf {e} }_{2}+(y_{3}-x_{3}){\\mathbf {e} }_{3}.}$ ${\\displaystyle {\\mathbf {x} }_{t}=t{\\mathbf {v} }+{\\mathbf {x} }_{0},}$ ${\\displaystyle {\\mathbf {F} }=m{\\mathbf {a} }}$ ${\\displaystyle E={\\mathbf {F} }\\cdot ({\\mathbf {x} }_{2}-{\\mathbf {x} }_{1}).}$ ${\\displaystyle {\\frac {df}{d\\tau }}=\\sum _{\\alpha =1}^{n}{\\frac {dx^{\\alpha }}{d\\tau }}{\\frac {\\partial f}{\\partial x^{\\alpha }}}.}$ ${\\displaystyle t^{\\alpha }={\\frac {dx^{\\alpha }}{d\\tau }}.}$ ${\\displaystyle {\\frac {d}{d\\tau }}=\\sum _{\\alpha }t^{\\alpha }{\\frac {\\partial }{\\partial x^{\\alpha }}}.}$ ${\\displaystyle \\mathbf {a} \\equiv a^{\\alpha }{\\frac {\\partial }{\\partial x^{\\alpha }}}.}$ Wikipedia", "of $$X$$, $$f_{X}(x)$$, is then \\begin{align*} f_{X}(0) &= \\mathbf{P}(X = 0) \\\\ &= \\mathbf{P}(\\mathrm{Hit\\:E})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 81/625 \\\\ f_{X}(1) &= \\mathbf{P}(X = 1) \\\\ &= 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 126/625 \\\\ f_{X}(2) &= \\mathbf{P}(X = 2) \\\\ &= \\begin{align*}& 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:C})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &+ \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:D}) \\end{align*} \\\\ &= 139/625 \\end{align*} and so on. From (2), the expected value of $$X$$ is $$\\mathbf{E}(X) = 0.81/625 + 1 \\cdot 126/625 + 2 \\cdot 139/625 + 3 \\cdot 124/625 + ... = 2.4$$. From (4), the variance of $$X$$ is $$\\mathrm{Var}(X) = 2.4^{2} \\cdot 81/625 + 1.4^{2} \\cdot 126/625 + 0.4^{2} \\cdot 139/625 + 0.6^{2} \\cdot 124/625 + ... = 2.72$$. The distribution function of $$X$$, $$F_{X}(x)$$, is then \\begin{align*} F_{X}(0) &= \\mathbf{P}(X \\le 0) \\\\ &= f_{X}(0) \\\\ F_{X}(1) &= \\mathbf{P}(X \\le 1) \\\\ &= f_{X}(1) + f_{X}(0) \\\\ F_{X}(2) &= \\mathbf{P}(X \\le 2) \\\\ &= f_{X}(2) + f_{X}(1) + f_{X}(0) \\end{align*} and so on. The distribution function $$F_{X}(x)$$ is shown in Figure 4. Figure 4: The distribution function $$F_{X}$$ of $$X$$ for the archery target.", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2 B \\right) \\\\ &\\qquad - \\mathbf{x}_2 \\left( \\mathbf{x}_{34} \\cdot \\partial_1 B \\right) - \\mathbf{x}_3 \\left( \\mathbf{x}_{42} \\cdot \\partial_1 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{23} \\cdot \\partial_1 B \\right)} \\Bigr) \\\\", "+ (-a_1 - b_1i) = (a_1 - a_1) + (b_1 - b_1)i = 0 + 0i = 0$. • FM1: $z_1 \\cdot z_2 = (a_1 + b_1i)(a_2 + b_2i) = a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i$ and so $(z_1 \\cdot z_2) \\in \\mathbb{C}$. • FM2: $z_1 \\cdot z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i = (a_2a_1 - b_2b_1) + (a_2b_1 + b_2a_1)i = z_2 \\cdot z_1$. • FM3: (1) \\begin{align} z_1 \\cdot (z_2 \\cdot z_3) \\\\ = (a_1 + b_1i) [ (a_2a_3 - b_2b_3) + (a_2b_3 + b_2a_3)i ] \\\\ = (a_1a_2a_3 - a_1b_2b_3) + (a_1a_2b_3 + a_1b_2a_3)i + (b_1a_2a_3 - b_1b_2b_3)i + (b_1a_2b_3 + b_1b_2a_3)i^2 \\\\ \\quad = [(a_1a_2a_3 - a_1b_2b_3) - (b_1a_2b_3 + b_1b_2a_3)] + [(a_1a_2b_3 + a_1b_2a_3) + (b_1a_2a_3 - b_1b_2b_3)]i \\\\ = [ (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i ] (a_3 + b_3i) \\\\ = (z_1 \\cdot z_2) \\cdot z_3 \\end{align} • FM4: The multiplicative identity is $1 = 1 + 0i$. That is $z_1 \\cdot 1 = (a_1 + b_1)(1 + 0i) = (a_1 + a_1 \\cdot 0i) + (b_1 + b_1 \\cdot 0)i = a_1 + b_1i = z_1$. • FM5: The multiplicative inverse for $z_1$ where $z_1 \\neq 0$ is $z_1^{-1} = \\frac{a_1 -b_1i}{a_1^2 + b_1^2}$. That is: (2)", "\\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{a} = 1,\\end{aligned} \\hspace{\\stretch{1}}(1.174) and \\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{b} = 0.\\end{aligned} \\hspace{\\stretch{1}}(1.175) \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}.\\end{aligned} \\hspace{\\stretch{1}}(1.176) Dotting with $\\mathbf{a}$ we have \\begin{aligned}\\mathbf{a} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right)=\\mathbf{a} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right)^2 - \\mathbf{b}^2 \\mathbf{a}^2,\\end{aligned} \\hspace{\\stretch{1}}(1.177) but dotting with $\\mathbf{b}$ yields zero \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right) &= \\mathbf{b} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right) \\\\ &= \\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}^2 - \\mathbf{b}^2 \\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right) \\\\ &= 0.\\end{aligned} \\hspace{\\stretch{1}}(1.178) To complete the proof, we note that the product in eq. 1.177 is just the wedge squared \\begin{aligned}\\left( { \\mathbf{a} \\wedge \\mathbf{b}} \\right)^2 &= \\left\\langle{{\\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)^2}}\\right\\rangle \\\\ &= \\left\\langle{{\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)}}\\right\\rangle \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\mathbf{a} \\mathbf{b} - 2 \\left( {\\mathbf{a} \\cdot \\mathbf{b}} \\right) \\mathbf{a} \\mathbf{b}}}\\right\\rangle+\\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right)^2 \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\left( { -\\mathbf{b} \\mathbf{a} + 2 \\mathbf{a} \\cdot \\mathbf{b} }", "\\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2", "= \\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}).$$ From (13) and (14), \\begin{align*} \\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}) &= \\mathbf{E}(X^{2}) - 2 \\cdot \\mu \\cdot \\mathbf{E}(X) + \\mu^{2} \\\\ &= \\mathbf{E}(X^{2}) - \\mu^{2} \\\\ &= \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}. \\end{align*} $$\\mathbf{E}(X^{2} - 2 \\cdot \\mu \\cdot X + \\mu^{2}) = \\mathbf{E}(X^{2}) - 2 \\cdot \\mu \\cdot \\mathbf{E}(X) + \\mu^{2} = \\mathbf{E}(X^{2}) - \\mu^{2} = \\mathbf{E}(X^{2}) - \\left(\\mathbf{E}(X)\\right)^{2}.$$ For the discrete or continuous random variable $$X$$ and the constant $$c \\in \\real$$, $$\\text{Var}(cX) = c^{2} \\cdot \\text{Var}(X). \\tag{17}$$ The proof of (17) holds for $$X$$ discrete or continuous. Again, let $$\\mu = \\mathbf{E}(X)$$. Then, $$\\text{Var}(cX) = \\mathbf{E}((cX - c\\mu)^2) = \\mathbf{E}(c^{2}\\cdot(X - \\mu)^{2}) = c^{2} \\cdot \\mathbf{E}((X - \\mu)^2) = c^{2} \\cdot \\text{Var}(X)$$. Finally, for discrete or continuous independent random variables $$X$$ and $$Y$$, $$\\text{Var}(X + Y) = \\text{Var}(X) + \\text{Var}(Y). \\tag{18}$$ $$\\text{Var}(X + Y) = \\text{Var}(X) + \\text{Var}(Y).$$ $$(18)$$ The proof of (18) holds for $$X$$ and $$Y$$ discrete or continuous. As a shorthand notation, let $$\\mu_{X} = \\mathbf{E}(X)$$ and $$\\mu_{Y} = \\mathbf{E}(Y)$$. Then, \\begin{align*} \\text{Var}(X + Y) &= \\mathbf{E}\\Big\\{\\big((X + Y) - (\\mu_{X} + \\mu_{Y})\\big)^{2}\\Big\\} \\\\ &= \\mathbf{E}\\Big\\{\\big((X - \\mu_{X}) + (Y - \\mu_{Y})\\big)^{2}\\Big\\} \\\\ &= \\mathbf{E}\\Big\\{ (X - \\mu_{X})^{2} + 2(X - \\mu_{X}) \\cdot (Y - \\mu_{Y}) + (Y - \\mu_{Y})^{2}", "generates $(\\mathbb{Z}/7\\mathbb{Z})^\\times$ and since \\begin{align*} 2^4 &\\equiv 16 \\equiv 3 \\pmod{13}\\\\ 2^6 &\\equiv 3 \\cdot 4 \\equiv 12 \\pmod{13} \\end{align*} then $2$ generates $(\\mathbb{Z}/13\\mathbb{Z})^\\times$. This gives us the following isomorphism. \\begin{align*} \\mathbb{Z}/6\\mathbb{Z} \\times \\mathbb{Z}/12\\mathbb{Z} &\\overset{\\sim}{\\to} (\\mathbb{Z}/7\\mathbb{Z})^\\times \\times (\\mathbb{Z}/13\\mathbb{Z})^\\times\\\\ (a,b) &\\mapsto (3^a, 2^b) \\end{align*} Next we need to find the isomorphism $(\\mathbb{Z}/7\\mathbb{Z})^\\times \\times (\\mathbb{Z}/13\\mathbb{Z})^\\times \\overset{\\sim}{\\longrightarrow} (\\mathbb{Z}/91\\mathbb{Z})^\\times$. Since $1 = 14 - 13 = 2 \\cdot 7 + (-1) 13$, then we have an isomorphism \\begin{align*} (\\mathbb{Z}/7\\mathbb{Z})^\\times \\times (\\mathbb{Z}/13\\mathbb{Z})^\\times &\\overset{\\sim}{\\longrightarrow} (\\mathbb{Z}/91\\mathbb{Z})^\\times\\\\ (r, s) &\\longmapsto 14s - 13r \\, . \\end{align*} (See here for more on this isomorphism.) Lastly, we have an isomorphism \\begin{align*} (\\mathbb{Z}/91\\mathbb{Z})^\\times &\\overset{\\sim}{\\to} \\text{Aut}(\\mathbb{Z}/91\\mathbb{Z})\\\\ k &\\mapsto \\varphi_k \\end{align*} where $\\varphi_k: \\mathbb{Z}/91\\mathbb{Z} \\to \\mathbb{Z}/91\\mathbb{Z}$ is given by $\\varphi_k(x) = kx$. Taking the elements of order $3$ you found in $\\mathbb{Z}/6\\mathbb{Z} \\times \\mathbb{Z}/12\\mathbb{Z}$ and using these isomorphisms, you should be able to find the automorphisms of $\\mathbb{Z}/91\\mathbb{Z}$ of order $3$. Note SpamIAm's method yields: $(2,0) \\mapsto \\phi_{79}$ and $(0,4) \\mapsto \\phi_{29}$. Explicitly calculating $\\langle \\phi_{79},\\phi_{29}\\rangle$, we find the subgroup of $\\text{Aut}(\\Bbb Z_{91})$ isomorphic to $\\Bbb Z_3 \\times \\Bbb Z_3$ is: $\\{\\phi_1,\\phi_{79},\\phi_{53},\\phi_{29},\\phi_{22},\\phi_{16},\\phi_9,\\phi_{81},\\phi_{74}\\}$ So your $\\phi_9$ and $\\phi_{22}$ are indeed generators-your error was that $572 = 26$ (mod $91$), and clearly $26$ has order $\\dfrac{91}{\\gcd(26,91)} = 7$. • Lol, yup, I did. Fixed. Jul 25 '15 at 7:16", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "\\begin{pmatrix} 1 \\\\1\\\\1 \\end{pmatrix} }$$ $$\\begin{array}{|rcll|} \\hline \\langle \\mathbf {u} ,\\mathbf {v} \\rangle &=& \\begin{pmatrix} x \\\\y\\\\z \\end{pmatrix}\\begin{pmatrix} 1 \\\\1\\\\1 \\end{pmatrix} \\\\ &=& x+y+z \\\\\\\\ \\langle \\mathbf {u} ,\\mathbf {u} \\rangle &=& \\begin{pmatrix} x \\\\y\\\\z \\end{pmatrix}\\begin{pmatrix} x \\\\y\\\\z \\end{pmatrix} \\\\ &=& x^2+y^2+z^2 \\\\\\\\ \\langle \\mathbf {v} ,\\mathbf {v} \\rangle &=& \\begin{pmatrix} 1 \\\\1\\\\1 \\end{pmatrix}\\begin{pmatrix} 1 \\\\1\\\\1 \\end{pmatrix} \\\\ &=& 1^2+1^2+1^2 \\\\ &=& 3 \\\\\\\\ \\hline |\\langle \\mathbf {u} ,\\mathbf {v} \\rangle |^{2} &\\leq& \\langle \\mathbf {u} ,\\mathbf {u} \\rangle \\cdot \\langle \\mathbf {v} ,\\mathbf {v} \\rangle \\\\ (x+y+z)^2 &\\le& (x^2+y^2+z^2)\\cdot 3 \\\\ x+y+z &\\le& \\sqrt{x^2+y^2+z^2}\\cdot \\sqrt{3} \\quad | \\quad x^2+y^2+z^2 = 1 \\\\ x+y+z &\\le& 1\\cdot \\sqrt{3} \\\\ \\mathbf{ x+y+z } & \\mathbf{\\le} & \\mathbf{\\sqrt{3}} \\\\ \\hline \\end{array}$$ The maximum value of $$x + y + z$$ is $$\\mathbf{\\sqrt{3}}$$ Jul 12, 2019 #4 +106533 +2 Thanks, heureka.....that's a pretty neat method for solving this one...... CPhill Jul 12, 2019 #6 +23846 +2 Thank you CPhill ! heureka Jul 14, 2019 #2 +23846 +4 3. The sequence $$(a_n)$$ is defined recursively by $$a_0=1,\\ a_1=\\sqrt[19]{2}$$, and $$a_n=a_{n-1}a_{n-2}^2$$ for $$n\\geq 2$$. What is the smallest positive integer k such that the product $$a_1a_2\\cdots a_k$$ is an integer? $$\\begin{array}{|l|r|rcl|l|} \\hline & k & \\text{product } a_1a_2\\cdots a_k \\\\ \\hline a_2 = a_1a_0^2=a_1^1 & 2 & a_1a_2 &=& a_1^2 & =2^{\\frac{2}{19}} & =", "\\mathbf{R} \\, \\vec{u} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\mathbf{R} \\, \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -x \\\\ \\end{bmatrix} \\\\ {} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\begin{bmatrix} \\mathbf{R}_{1,1} - \\mathbf{R}_{1,4}\\,x \\\\ \\mathbf{R}_{2,1} - \\mathbf{R}_{2,4}\\,x \\\\ \\mathbf{R}_{3,1} - \\mathbf{R}_{3,4}\\,x \\\\ \\mathbf{R}_{4,1} - \\mathbf{R}_{4,4}\\,x \\\\ \\end{bmatrix} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - (\\mathbf{R}_{1,4} + \\mathbf{R}_{4,1})\\,x + \\mathbf{R}_{1,1} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - 2\\,\\mathbf{R}_{1,4}\\,x + \\mathbf{R}_{1,1} \\\\ {} & = 0 \\\\ x & = \\frac{\\mathbf{R}_{1,4} \\pm \\sqrt{\\mathbf{R}_{1,4}^2 - \\mathbf{R}_{1,1}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} Similarly, the \\hat{y}$ and $\\hat{z} planes are given by \\begin{aligned} y & = \\frac{\\mathbf{R}_{2,4} \\pm \\sqrt{\\mathbf{R}_{2,4}^2 - \\mathbf{R}_{2,2}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}} \\\\ z & = \\frac{\\mathbf{R}_{3,4} \\pm \\sqrt{\\mathbf{R}_{3,4}^2 - \\mathbf{R}_{3,3}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} This is where the above linked posts stop, but a lot more simplification can be done. ## Implementation Details Several details of the problem can make computing the planes more efficient. The first is that \\mathbf{S}$ is involutory, meaning $\\mathbf{S}^{-1} = \\mathbf{S}$. This means that the product $\\mathbf{M}\\,\\mathbf{S}^{-1}$ can be computed implicitly: it is simply $\\mathbf{M}$ with its last column negated. The last column of $\\mathbf{R} = \\mathbf{M} \\, \\mathbf{S}^{-1} \\, \\mathbf{M}^\\top$ is the same, because the last column of $\\mathbf{M}^\\top$ is $\\begin{bmatrix}0 & 0 & 0 & 1\\end{bmatrix}^\\top$. In particular, $\\mathbf{R}_{4,4} = -1$.", "of this is 2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3, so you normally you would form the array (\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}), which has length (2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32. However, writing the ternary form as 2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2)), the array would be (\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})], which only has length (2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most 94, and we’ve now solved the problem! The ideas above are enough to pass the problem already, but I’m sure with a couple more tricks you can reduce the length even more. ### Time Complexity: [O(N)](https://en.wikipedia.org/wiki/Output-sensitive _ algorithm) ### AUTHOR’S AND TESTER’S SOLUTIONS: #2 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then create p pairs of numbers so", "{ \\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "so we have \\begin{aligned} \\vec{u}^\\top \\, \\mathbf{R} \\, \\vec{u} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\mathbf{R} \\, \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -x \\\\ \\end{bmatrix} \\\\ {} & = \\begin{bmatrix} 1 & 0 & 0 & -x \\end{bmatrix} \\, \\begin{bmatrix} \\mathbf{R}_{1,1} - \\mathbf{R}_{1,4}\\,x \\\\ \\mathbf{R}_{2,1} - \\mathbf{R}_{2,4}\\,x \\\\ \\mathbf{R}_{3,1} - \\mathbf{R}_{3,4}\\,x \\\\ \\mathbf{R}_{4,1} - \\mathbf{R}_{4,4}\\,x \\\\ \\end{bmatrix} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - (\\mathbf{R}_{1,4} + \\mathbf{R}_{4,1})\\,x + \\mathbf{R}_{1,1} \\\\ {} & = \\mathbf{R}_{4,4}\\,x^2 - 2\\,\\mathbf{R}_{1,4}\\,x + \\mathbf{R}_{1,1} \\\\ {} & = 0 \\\\ x & = \\frac{\\mathbf{R}_{1,4} \\pm \\sqrt{\\mathbf{R}_{1,4}^2 - \\mathbf{R}_{1,1}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} Similarly, the $\\hat{y}$ and $\\hat{z}$ planes are given by \\begin{aligned} y & = \\frac{\\mathbf{R}_{2,4} \\pm \\sqrt{\\mathbf{R}_{2,4}^2 - \\mathbf{R}_{2,2}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}} \\\\ z & = \\frac{\\mathbf{R}_{3,4} \\pm \\sqrt{\\mathbf{R}_{3,4}^2 - \\mathbf{R}_{3,3}\\,\\mathbf{R}_{4,4}}}{\\mathbf{R}_{4,4}}. \\\\ \\end{aligned} This is where the above linked posts stop, but a lot more simplification can be done. ## Implementation Details Several details of the problem can make computing the planes more efficient. The first is that $\\mathbf{S}$ is involutory, meaning $\\mathbf{S}^{-1} = \\mathbf{S}$. This means that the product $\\mathbf{M}\\,\\mathbf{S}^{-1}$ can be computed implicitly: it is simply $\\mathbf{M}$ with its last column negated. The last column of $\\mathbf{R} = \\mathbf{M} \\, \\mathbf{S}^{-1} \\, \\mathbf{M}^\\top$ is the same, because the last column of $\\mathbf{M}^\\top$ is $\\begin{bmatrix}0 & 0 & 0 &", "# What is the number of $2n\\times 2n$-matrices $g=\\begin{pmatrix} A & B \\\\\\ C & -A^{t} \\end{pmatrix}$, $B$ and $C$ symmetric, over the finite field $\\mathbb{F}\\_{q}$ with $\\mathrm{rank}(g)=k$? It is not difficult to get a formula for the number of $n\\times n$-matrices $g\\in\\mathrm{Mat}_{n}(\\mathbb{F}_{q})$ with $\\mathrm{rank}(g)=k$. Namely we have got \\begin{align*} \\mid\\{g\\in\\mathrm{Mat}_{n}(\\mathbb{F}\\_{q})\\mid \\mathrm{rank}(g)=k\\}\\mid = \\binom{n}{k}\\_{q}\\cdot(q^{n}-1)\\cdots(q^{n}-q^{k-1}) \\end{align*} with $\\binom{n}{k}_{q}=\\frac{(q^{n}-1)\\cdots(q^{n}-q^{k-1})}{(q^{k}-1)\\cdots(q^{k}-q^{k-1})}$. My question is now, if it is possible to get a formula in the same fashion for the following number \\begin{align*} f(n,q,k):=\\mid\\{g\\in\\mathrm{Mat}_{2n}(\\mathbb{F}\\_{q})\\mid g=\\begin{pmatrix} A & B \\\\\\ C & -A^{t} \\end{pmatrix}, B=B^{t}, C=C^{t} \\\\text{ and }\\mathrm{rank}(g)=k\\}\\mid \\end{align*}. My conjecture (based on computer data for small $n$ and $k$) is the following, but I have no idea for the proof. \\begin{align*} f(2n,q,2k)=\\binom{2n}{2k}\\_{q}\\cdot q^{k(k+1)}\\displaystyle\\prod_{i=1}^{k}(q^{2i-1}-1), \\end{align*} \\begin{align*} f(2n,q,2k+1)=\\binom{2n}{2k+1}\\_{q}\\cdot q^{k(k-1)}\\displaystyle\\prod_{i=1}^{k}(q^{2i-1}-1) \\end{align*} - If k is odd, I think that will give very strict conditions on A. You might consider asking the question for k=1 or k=3, and perhaps doing a computer enumeration leveraged by whatever thery you can derive. Gerhard \"Ask Me About System Design\" Paseman, 2012.10.24 – Gerhard Paseman Oct 24 '12 at 13:37 I need the above as a lemma to deduce some deaper results about distribution of special matrices over finite rings. – micha Nov 6 '12 at 15:05 Try the following: look at the map which takes a matrix of your", "+0 # problem 0 176 1 Given that $\\displaystyle {{\\left((3!)!\\right)!}\\over{3!}}= k\\cdot n!$, where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k+n$. Jun 25, 2018 #1 +20831 0 Given that $\\displaystyle {{\\left((3!)!\\right)!}\\over{3!}}= k\\cdot n!$, where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k+n$. $$\\begin{array}{|rcll|} \\hline \\displaystyle {{\\left((3!)!\\right)!}\\over{3!}} &=& k\\cdot n! \\quad & | \\quad 3! = 6 \\\\ \\displaystyle {{\\left(6!\\right)!}\\over{6}} &=& k\\cdot n! \\quad & | \\quad 6! = 720 \\\\ \\displaystyle {{720!}\\over{6}} &=& k\\cdot n! \\\\ 720! &=& 6k\\cdot n! \\quad & | \\quad 720! = \\underbrace{720}_{=6k}\\cdot \\underbrace{719}_{=n}! \\\\\\\\ 6k &=& 720 \\quad & | \\quad : 6 \\\\ \\mathbf{k} & \\mathbf{=} & \\mathbf{120} \\\\ \\mathbf{n} & \\mathbf{=} & \\mathbf{719} \\\\ \\mathbf{k+n} & \\mathbf{=}&\\mathbf{ 839 } \\\\\\\\ \\boxed { \\dfrac{ \\left((3!)!\\right)!}{3!} = 120\\cdot 719! } \\\\ \\hline \\end{array}$$ Jun 26, 2018", "& \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} <br> \\begin{aligned} \\quad \\mathbf u &= \\alpha \\mathbf a + \\beta \\mathbf b \\qquad \\text{Norm:}|| \\mathbf u|| \\end{aligned} <br> \\begin{align} \\hat{\\mathbf u} = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} <br> 2. **Matrices** \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} <br> 3. \\begin{aligned} a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5 \\end{aligned} <br> 4. **Transformations:** \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} The syntax is very close to what one writes in a .tex file. Switch html_document in the YAML header with pdf_document and you can get the corresponding pdf. This is the HTML output: To have our HTML is written with mathml, we add --mathml to the arguments passed to pandoc. This is what we see at the top of the HTML file: In this workflow, knitr produces a .tex file. As we can see, it is unremarkable, and very closely resembles the OP's original latex code, just adding elements like the enumerate environment. \\begin{document} \\centerline{\\textbf{\\large{Mathematical expressions}}} \\begin{enumerate} \\item" ]

[ "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "the linear transformation $T$ is nonzero. Hence the range is $\\calR(T)=\\Span(\\mathbf{u}),$ and a basis for $\\calR(T)$ is the set $\\{\\mathbf{u}\\}$. The rank of $T$ is the dimension of the range $\\calR(T)$ by definition. Thus, the rank of $T$ is $1$. ### (d) Calculate the matrix $A$ representing $T$ with respect to the standard basis for $\\R^3$. Let $\\mathbf{e}_1=\\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_2=\\begin{bmatrix} 0 \\\\ 1 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_3=\\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}$ be the standard basis vectors of $\\R^3$. We calculate \\begin{align*} T(\\mathbf{e}_1)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_1}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{1}{2}\\mathbf{u}=\\begin{bmatrix} 1/2 \\\\ 1/2 \\\\ 0 \\end{bmatrix}\\6pt] T(\\mathbf{e}_2)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_2}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{1}{2}\\mathbf{u}=\\begin{bmatrix} 1/2 \\\\ 1/2 \\\\ 0 \\end{bmatrix}\\\\[6pt] T(\\mathbf{e}_3)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_3}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{0}{2}\\mathbf{u}=\\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}. \\end{align*} Thus, the matrix representation A of T is \\begin{align*} A&=\\begin{bmatrix} T(\\mathbf{e}_1) & T(\\mathbf{e}_2) & T(\\mathbf{e}_3) \\\\ \\end{bmatrix}\\\\ &=\\begin{bmatrix} 1/2 & 1/2 & 0 \\\\ 1/2 &1/2 &0 \\\\ 0 & 0 & 0 \\end{bmatrix}. \\end{align*} ### (e) Calculate the coordinate vector Let \\mathbf{v}=\\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} and let \\[ [\\mathbf{v}]_B=\\begin{bmatrix} a \\\\ b \\\\ c \\end{bmatrix} be the coordinate vector of $\\mathbf{v}$ with respect to the basis $B$. This means that we have \\begin{align*} \\mathbf{v}=a \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\end{bmatrix}+b \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\end{bmatrix}+c\\begin{bmatrix} 0", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "m}$$: (2.3.7)$\\begin{split}\\mathbf{A}=\\begin{bmatrix} a_{11} & a_{12} & \\cdots & a_{1k} \\\\ a_{21} & a_{22} & \\cdots & a_{2k} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots & a_{nk} \\\\ \\end{bmatrix},\\quad \\mathbf{B}=\\begin{bmatrix} b_{11} & b_{12} & \\cdots & b_{1m} \\\\ b_{21} & b_{22} & \\cdots & b_{2m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b_{k1} & b_{k2} & \\cdots & b_{km} \\\\ \\end{bmatrix}.\\end{split}$ Denote by $$\\mathbf{a}^\\top_{i} \\in \\mathbb{R}^k$$ the row vector representing the $$i^\\mathrm{th}$$ row of the matrix $$\\mathbf{A}$$, and let $$\\mathbf{b}_{j} \\in \\mathbb{R}^k$$ be the column vector from the $$j^\\mathrm{th}$$ column of the matrix $$\\mathbf{B}$$. To produce the matrix product $$\\mathbf{C} = \\mathbf{A}\\mathbf{B}$$, it is easiest to think of $$\\mathbf{A}$$ in terms of its row vectors and $$\\mathbf{B}$$ in terms of its column vectors: (2.3.8)$\\begin{split}\\mathbf{A}= \\begin{bmatrix} \\mathbf{a}^\\top_{1} \\\\ \\mathbf{a}^\\top_{2} \\\\ \\vdots \\\\ \\mathbf{a}^\\top_n \\\\ \\end{bmatrix}, \\quad \\mathbf{B}=\\begin{bmatrix} \\mathbf{b}_{1} & \\mathbf{b}_{2} & \\cdots & \\mathbf{b}_{m} \\\\ \\end{bmatrix}.\\end{split}$ Then the matrix product $$\\mathbf{C} \\in \\mathbb{R}^{n \\times m}$$ is produced as we simply compute each element $$c_{ij}$$ as the dot product $$\\mathbf{a}^\\top_i \\mathbf{b}_j$$: (2.3.9)$\\begin{split}\\mathbf{C} = \\mathbf{AB} = \\begin{bmatrix} \\mathbf{a}^\\top_{1} \\\\ \\mathbf{a}^\\top_{2} \\\\ \\vdots \\\\ \\mathbf{a}^\\top_n \\\\ \\end{bmatrix} \\begin{bmatrix} \\mathbf{b}_{1} & \\mathbf{b}_{2} & \\cdots & \\mathbf{b}_{m} \\\\ \\end{bmatrix} = \\begin{bmatrix} \\mathbf{a}^\\top_{1} \\mathbf{b}_1 & \\mathbf{a}^\\top_{1}\\mathbf{b}_2& \\cdots & \\mathbf{a}^\\top_{1} \\mathbf{b}_m \\\\ \\mathbf{a}^\\top_{2}\\mathbf{b}_1 & \\mathbf{a}^\\top_{2} \\mathbf{b}_2 & \\cdots", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "_{m}=m_{1}\\mathbf {b} _{1}+m_{2}\\mathbf {b} _{2}}$ Where, {\\displaystyle {\\begin{aligned}\\mathbf {b} _{1}&=2\\pi {\\frac {-\\mathbf {Q} \\,\\mathbf {a} _{2}}{-\\mathbf {a} _{1}\\cdot \\mathbf {Q} \\,\\mathbf {a} _{2}}}=2\\pi {\\frac {\\mathbf {Q} \\,\\mathbf {a} _{2}}{\\mathbf {a} _{1}\\cdot \\mathbf {Q} \\,\\mathbf {a} _{2}}}\\\\\\mathbf {b} _{2}&=2\\pi {\\frac {\\mathbf {Q} \\,\\mathbf {a} _{1}}{\\mathbf {a} _{2}\\cdot \\mathbf {Q} \\,\\mathbf {a} _{1}}}\\end{aligned}}} Here ${\\displaystyle \\mathbf {Q} }$ represents a 90 degree rotation matrix, i.e. a quarter turn. The anti-clockwise rotation and the clockwise rotation can both be used to determine the reciprocal lattice: If ${\\displaystyle \\mathbf {Q} }$ is the anti-clockwise rotation and ${\\displaystyle \\mathbf {Q'} }$ is the clockwise rotation, ${\\displaystyle \\mathbf {Q} \\,\\mathbf {v} =-\\mathbf {Q'} \\,\\mathbf {v} }$ for all vectors ${\\displaystyle \\mathbf {v} }$. Thus, using the permutation ${\\displaystyle \\sigma ={\\begin{pmatrix}1&2\\\\2&1\\end{pmatrix}}}$ we obtain {\\displaystyle {\\begin{aligned}\\mathbf {b} _{n}=2\\pi {\\frac {\\mathbf {Q} \\,\\mathbf {a} _{\\sigma (n)}}{\\mathbf {a} _{n}\\cdot \\mathbf {Q} \\,\\mathbf {a} _{\\sigma (n)}}}=2\\pi {\\frac {\\mathbf {Q} '\\,\\mathbf {a} _{\\sigma (n)}}{\\mathbf {a} _{n}\\cdot \\mathbf {Q} '\\,\\mathbf {a} _{\\sigma (n)}}}.\\end{aligned}}} ### Three dimensions For an infinite three-dimensional lattice, defined by its primitive vectors ${\\displaystyle \\left(\\mathbf {a_{1}} ,\\mathbf {a} _{2},\\mathbf {a} _{3}\\right)}$, its reciprocal lattice can be determined by generating its three reciprocal primitive vectors, through the formulae ${\\displaystyle \\mathbf {G} _{m}=m_{1}\\mathbf {b} _{1}+m_{2}\\mathbf {b} _{2}+m_{3}\\mathbf {b} _{3}}$ where the subscript ${\\displaystyle m=(m_{1},m_{2},m_{3})}$ in three dimensions, and for the", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} =A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}}$ while in matrix form: ${\\displaystyle \\mathbf {A\\cdot B} ={\\begin{pmatrix}A^{0}&A^{1}&A^{2}&A^{3}\\end{pmatrix}}{\\begin{pmatrix}1&0&0&0\\\\0&-1&0&0\\\\0&0&-1&0\\\\0&0&0&-1\\end{pmatrix}}{\\begin{pmatrix}B^{0}\\\\B^{1}\\\\B^{2}\\\\B^{3}\\end{pmatrix}}}$ It is a recurring theme in special relativity to take the expression ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} =A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}=C}$ in one reference frame, where C is the value of the inner product in this frame, and: ${\\displaystyle \\mathbf {A} '\\cdot \\mathbf {B} '={A'}^{0}{B'}^{0}-{A'}^{1}{B'}^{1}-{A'}^{2}{B'}^{2}-{A'}^{3}{B'}^{3}=C'}$ in another frame, in which C′ is the value of the inner product in this frame. Then since the inner product is an invariant, these must be equal: ${\\displaystyle \\mathbf {A} \\cdot \\mathbf {B} =\\mathbf {A} '\\cdot \\mathbf {B} '}$ that is: ${\\displaystyle C=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}={A'}^{0}{B'}^{0}-{A'}^{1}{B'}^{1}-{A'}^{2}{B'}^{2}-{A'}^{3}{B'}^{3}}$ Considering that physical quantities in relativity are four-vectors, this equation has the appearance of a \"conservation law\", but there is no \"conservation\" involved. The primary significance of the Minkowski inner product is that for any two four-vectors, its value is invariant for all observers; a change of coordinates does not result in a change in value of the inner product. The components of the four-vectors change from one frame to another; A and A′ are connected by a Lorentz transformation, and similarly for B and B′, although the inner products are the same in all frames. Nevertheless, this type of expression is exploited in relativistic calculations on a par with conservation laws, since the magnitudes", "x}+{\\mathbf y}\\neq {\\mathbf 0} \\}\\\\ & = \\min\\{ w({\\mathbf z}) : {\\mathbf z} \\neq {\\mathbf 0} \\}\\text{.} \\end{align*} ### SubsectionLinear Codes From Example 8.16, it is now easy to check that the minimum nonzero weight is $$3\\text{;}$$ hence, the code does indeed detect and correct all single errors. We have now reduced the problem of finding “good” codes to that of generating group codes. One easy way to generate group codes is to employ a bit of matrix theory. Define the inner product of two binary $$n$$-tuples to be \\begin{equation*} {\\mathbf x} \\cdot {\\mathbf y} = x_1 y_1 + \\cdots + x_n y_n\\text{,} \\end{equation*} where $${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^\\transpose$$ and $${\\mathbf y} = (y_1, y_2, \\ldots, y_n)^\\transpose$$ are column vectors. 5 For example, if $${\\mathbf x} = (011001)^\\transpose$$ and $${\\mathbf y} = (110101)^\\transpose\\text{,}$$ then $${\\mathbf x} \\cdot {\\mathbf y} = 0\\text{.}$$ We can also look at an inner product as the product of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^\\transpose {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}\\text{.} \\end{align*} Since we will be working with matrices, we will write binary", "# Normalize Lengths to Obtain an Orthonormal Basis ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Solution. We begin by computing \\begin{align*} \\mathbf{v}_{1}\\cdot\\mathbf{v}_{2} &= \\mathbf{v}_{1}^{T}\\mathbf{v}_{2} = \\begin{bmatrix} 1 & 1 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = 1\\cdot 1+1\\cdot -1 = 1-1 = 0, \\\\ \\mathbf{v}_{1}\\cdot\\mathbf{v}_{1} &= \\begin{bmatrix} 1 & 1 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} = 1+1 = 2, \\\\ \\mathbf{v}_{2}\\cdot\\mathbf{v}_{2} &= \\begin{bmatrix} 1 & -1 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = 1+1 = 2. \\end{align*} Since $\\mathbf{v}_{1}\\cdot\\mathbf{v}_{2}=0$, the vectors $\\mathbf{v}_1$ and $\\mathbf{v}_2$ are orthogonal. Since $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ are nonzero orthogonal vectors, they are linearly independent, and it follows that $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthogonal basis for $V$. However, since $\\mathbf{v}_{i}\\cdot\\mathbf{v}_{i}=2\\neq 1$ for $i=1,2$, we know that $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ do not form an orthonormal basis for $V$. To find an orthonormal basis for $V$, note that for any scalars $a$ and $b$, $(a\\mathbf{v}_{1})\\cdot(b\\mathbf{v}_{2})=ab(\\mathbf{v}_{1}\\cdot\\mathbf{v}_{2})=ab\\cdot 0=0$. Therefore, $a\\mathbf{v}_{1}$ and $b\\mathbf{v}_{2}$ will always form an orthogonal basis for $V$. All we need to do is choose $a$ and $b$ so that $a\\mathbf{v}_{1}$ and $b\\mathbf{v}_{2}$ form an orthonormal set. For $a$, we", "coefficients in the $i$th column of the matrix. When we compute $B\\mathbf{v}_1,\\ldots,B\\mathbf{v}_k$, each of these will lie in $E_{\\lambda}$. Therefore, each of these can be expressed as a linear combination of $\\mathbf{v}_1,\\ldots,\\mathbf{v}_k$ (since they form a basis for $E_{\\lambda}$. So, to express them as linear combinations of $\\beta$, we just add $0$s; we will have: \\begin{align*} B\\mathbf{v}_1 &= b_{11}\\mathbf{v}_1 + b_{21}\\mathbf{v}_2+\\cdots+b_{k1}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n\\\\ B\\mathbf{v}_2 &= b_{12}\\mathbf{v}_1 + b_{22}\\mathbf{v}_2 + \\cdots +b_{k2}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n\\\\ &\\vdots\\\\ B\\mathbf{v}_k &= b_{1k}\\mathbf{v}_1 + b_{2k}\\mathbf{v}_2 + \\cdots + b_{kk}\\mathbf{v}_k + 0\\mathbf{v}_{k+1}+\\cdots + 0\\mathbf{v}_n \\end{align*} where $b_{ij}$ are some scalars (some possibly equal to $0$). So the matrix of $B$ relative to $\\beta$ would start off something like: $$\\left(\\begin{array}{ccccccc} b_{11} & b_{12} & \\cdots & b_{1k} & * & \\cdots & *\\\\ b_{21} & b_{22} & \\cdots & b_{2k} & * & \\cdots & *\\\\ \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ b_{k1} & b_{k2} & \\cdots & b_{kk} & * & \\cdots & *\\\\ 0 & 0 & \\cdots & 0 & * & \\cdots & *\\\\ \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & 0 & * & \\cdots & * \\end{array}\\right).$$ So, now suppose that you have a basis for", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "& 1 \\end{array}} \\right| = 1\\cdot\\left| {\\begin{array}{*{20}{c}} -1 & -1\\\\ 1 & 1 \\end{array}} \\right| - 2\\cdot\\left| {\\begin{array}{*{20}{c}} 1 & -1\\\\ 1 & 1 \\end{array}} \\right| + 3\\cdot\\left| {\\begin{array}{*{20}{c}} 1 & -1\\\\ 1 & 1 \\end{array}} \\right| = 1\\cdot0 - 2\\cdot2 + 3\\cdot2 = 2.$ ### Example 2. Show that four points $$A\\left({-1,2,1}\\right),$$ $$B\\left({0,4,2}\\right),$$ $$C\\left({-1,5,5}\\right),$$ and $$D\\left({2,5,0}\\right)$$ lie in the same plane. Solution. Compute the coordinates of vectors $$\\mathbf{a} = \\mathbf{AB},$$ $$\\mathbf{b} = \\mathbf{AC},$$ and $$\\mathbf{c} = \\mathbf{AD}:$$ $\\mathbf{a} = \\mathbf{AB} = \\left({1,2,1}\\right),\\;\\mathbf{b} = \\mathbf{AC} = \\left({0,3,4}\\right),\\;\\mathbf{c} = \\mathbf{AD} = \\left({3,3,-1}\\right).$ Check that these vectors are coplanar (this will mean that they lie in the same plane). Calculate the determinant: $\\left| {\\begin{array}{*{20}{c}} 1 & 2 & 1\\\\ 0 & 3 & 4\\\\ 3 & 3 & -1 \\end{array}} \\right| = 1\\cdot\\left| {\\begin{array}{*{20}{c}} 3 & 4\\\\ 3 & -1 \\end{array}} \\right| + 3\\cdot\\left| {\\begin{array}{*{20}{c}} 2 & 1\\\\ 3 & 4 \\end{array}} \\right| = 1\\cdot\\left({-15}\\right) + 3\\cdot5 = 0.$ The determinant is zero. Hence, the vectors are coplanar. ### Example 3. Show that vectors $$\\mathbf{a} = 2\\mathbf{i} + 3\\mathbf{k},$$ $$\\mathbf{b} = \\mathbf{i} - \\mathbf{j}$$ and $$\\mathbf{c} = 2\\mathbf{j} + 3\\mathbf{k}$$ are coplanar and express vector $$\\mathbf{c}$$ in terms of $$\\mathbf{a}$$ and $$\\mathbf{b}.$$ Solution. The coordinates of the vectors $$\\mathbf{a},$$ $$\\mathbf{b},$$ and $$\\mathbf{c}$$ are equal $\\mathbf{a} = 2\\mathbf{i} + 3\\mathbf{k} =", "of $$X$$, $$f_{X}(x)$$, is then \\begin{align*} f_{X}(0) &= \\mathbf{P}(X = 0) \\\\ &= \\mathbf{P}(\\mathrm{Hit\\:E})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 81/625 \\\\ f_{X}(1) &= \\mathbf{P}(X = 1) \\\\ &= 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 126/625 \\\\ f_{X}(2) &= \\mathbf{P}(X = 2) \\\\ &= \\begin{align*}& 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:C})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &+ \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:D}) \\end{align*} \\\\ &= 139/625 \\end{align*} and so on. From (2), the expected value of $$X$$ is $$\\mathbf{E}(X) = 0.81/625 + 1 \\cdot 126/625 + 2 \\cdot 139/625 + 3 \\cdot 124/625 + ... = 2.4$$. From (4), the variance of $$X$$ is $$\\mathrm{Var}(X) = 2.4^{2} \\cdot 81/625 + 1.4^{2} \\cdot 126/625 + 0.4^{2} \\cdot 139/625 + 0.6^{2} \\cdot 124/625 + ... = 2.72$$. The distribution function of $$X$$, $$F_{X}(x)$$, is then \\begin{align*} F_{X}(0) &= \\mathbf{P}(X \\le 0) \\\\ &= f_{X}(0) \\\\ F_{X}(1) &= \\mathbf{P}(X \\le 1) \\\\ &= f_{X}(1) + f_{X}(0) \\\\ F_{X}(2) &= \\mathbf{P}(X \\le 2) \\\\ &= f_{X}(2) + f_{X}(1) + f_{X}(0) \\end{align*} and so on. The distribution function $$F_{X}(x)$$ is shown in Figure 4. Figure 4: The distribution function $$F_{X}$$ of $$X$$ for the archery target.", "would be linearly dependent and $\\det(\\mathbf A\\cdot\\mathbf B)=0$. Therefore $\\boldsymbol{\\text{rank}\\,(\\mathbf A)=2}$. Next let us define a matrix $\\mathbf D$ such that $\\mathbf D=\\mathbf B\\cdot\\mathbf A= \\begin{pmatrix} d_{11} & d_{12} & d_{13}\\\\ d_{21} & d_{22} & d_{23}\\\\ d_{31} & d_{32} & d_{33} \\end{pmatrix}$ Similarly as in the previous cases we have $(d_{11},d_{12},d_{13})&=b_{11}(a_{11},a_{12},a_{13})+b_{12}(a_{21},a_{22},a_{23})\\\\ (d_{21},d_{22},d_{23})&=b_{21}(a_{11},a_{12},a_{13})+b_{22}(a_{21},a_{22},a_{23})\\\\ (d_{31},d_{32},d_{33})&=b_{31}(a_{11},a_{12},a_{13})+b_{32}(a_{21},a_{22},a_{23}).$ From the fact that $\\text{rank}\\,(\\mathbf A)=2$ we deduce that the vectors $(d_{11},d_{12},d_{13})$, $(d_{21},d_{22},d_{23})$, $(d_{31},d_{32},d_{33})$ are linearly dependent which implies $\\color{blue}\\det(\\mathbf B\\cdot\\mathbf A)=0$ Backslash je v TeXu tak dùležitý jako nekoneèno pøi dìlení nulou v tìlesech charakteristiky 0. Offline #5 08. 03. 2016 17:04 — Editoval vanok (08. 03. 2016 17:06) vanok Pøíspìvky: 13324 Reputace: 722 Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix Hi ↑ stuart clark:, As regards possible complements for the given solution, there is 2 ways. The first one is to take advantage your first course of linear algebra (normally made in first year of university); and specially call back it two theorems The determinant of a matrix M, of type (n, n) is no zero iff his rank $rg(M)$ is n. And also $rg (AB) \\leq min (rg (A), rg (B))$ A,B matrices of type (m,n),(n,m) resp. and simply to apply. (It is my method). Other method, (for the high of grammar schools) is the one" ]

[ "# Thread: Left inverse of matrix 1. ## Left inverse of matrix Consider the matrix $\\mathbf{C} = \\begin{pmatrix}1 & -1\\\\ 1&1 \\\\ 0&1 \\end{pmatrix}$ Find a left-inverse of C. Attempt: $\\begin{pmatrix}x1 & y1 & z1\\\\ x2&y2 &z2 \\end{pmatrix}\\begin{pmatrix}1 & -1\\\\ 1&1\\\\ 0&1 \\end{pmatrix} = \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ $\\begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\\\ x2 + y2& -x2 + y2 + z2 \\end{pmatrix}= \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ I know that you suppose to somehow use gauss reduction, and thought about gauss reducing: $x1 + y1 = 1$ $-x1 + y1 + z1 = 0$ and $x2 + y2 = 0$ $-x2 + y2 + z2 = 1$ but didnt get any luck 2. Only square matrices have inverses. To solve a matrix equation of the form $\\mathbf{Y} = \\mathbf{AX}$ for $\\mathbf{X}$ where $\\mathbf{A}$ is not square, you need to premultiply both sides by $\\mathbf{A}^T$ first. $\\mathbf{Y} = \\mathbf{AX}$ $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$. Now $\\mathbf{A}^T\\mathbf{A}$ is a square matrix, so you can premultiply both sides by $(\\mathbf{A}^T\\mathbf{A})^{-1}$. $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = (\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{IX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{X}$. 3. Originally Posted by Prove It Only square matrices have inverses. Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both. 4. You have", "x}+{\\mathbf y}\\neq {\\mathbf 0} \\}\\\\ & = \\min\\{ w({\\mathbf z}) : {\\mathbf z} \\neq {\\mathbf 0} \\}\\text{.} \\end{align*} ### SubsectionLinear Codes From Example 8.16, it is now easy to check that the minimum nonzero weight is $$3\\text{;}$$ hence, the code does indeed detect and correct all single errors. We have now reduced the problem of finding “good” codes to that of generating group codes. One easy way to generate group codes is to employ a bit of matrix theory. Define the inner product of two binary $$n$$-tuples to be \\begin{equation*} {\\mathbf x} \\cdot {\\mathbf y} = x_1 y_1 + \\cdots + x_n y_n\\text{,} \\end{equation*} where $${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^\\transpose$$ and $${\\mathbf y} = (y_1, y_2, \\ldots, y_n)^\\transpose$$ are column vectors. 5 For example, if $${\\mathbf x} = (011001)^\\transpose$$ and $${\\mathbf y} = (110101)^\\transpose\\text{,}$$ then $${\\mathbf x} \\cdot {\\mathbf y} = 0\\text{.}$$ We can also look at an inner product as the product of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^\\transpose {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}\\text{.} \\end{align*} Since we will be working with matrices, we will write binary", "pointing away from the origin over a circle of starting points in black and the new direction that results from multiplying by the matrix m in red. Program 1: Visualising The Eigenvectors Note that the coincident directions of the vectors lying above the top-left to bottom-right diagonal are close to $$\\pm\\mathbf{v}_1$$ and the nearly coincident vectors lying above and below the bottom-left to top-right diagonal are close to $$\\pm\\mathbf{v}_0$$. Now this visualisation is quite a handy way to illustrate some special cases. Firstly, try changing m to $\\begin{pmatrix} 2 & 0\\\\ 0 & 2 \\end{pmatrix}$ Notice how every vector is direction invariant? This is because both of the eigenvalues are equal and, for such matrices, given a pair of independent eigenvectors $$\\mathbf{v}_0$$ and $$\\mathbf{v}_1$$ we have \\begin{align*} \\mathbf{M} \\times \\left(a \\mathbf{v}_0 + b \\mathbf{v}_1\\right) &=\\mathbf{M} \\times \\left(a \\mathbf{v}_0\\right) + \\mathbf{M} \\times \\left(b \\mathbf{v}_1\\right)\\\\ &=a \\mathbf{M} \\mathbf{v}_0 + b \\mathbf{M} \\mathbf{v}_1\\\\ &=a \\lambda \\mathbf{v}_0 + b \\lambda \\mathbf{v}_1\\\\ &=\\lambda \\left(a \\mathbf{v}_0 + b \\mathbf{v}_1\\right) \\end{align*} and so every linear combination of them, and consequently every vector in the plain, is also an eigenvector. Next, try the matrix $\\begin{pmatrix} 1 & -4\\\\ 1 & 1 \\end{pmatrix}$ This time none of the vectors have invariant directions! This is because the characteristic equation of this matrix has complex roots \\begin{align*} \\lambda_0 &=", "{\\mathbf y} \\|^2 - \\|A {\\mathbf x} \\|^2 - \\|A {\\mathbf y} \\|^2 \\right]\\\\ & = \\frac{1}{2} \\left[ \\|A ( {\\mathbf x} + {\\mathbf y} ) \\|^2 - \\|A {\\mathbf x} \\|^2 - \\|A {\\mathbf y} \\|^2 \\right]\\\\ & = \\frac{1}{2} \\left[ \\|{\\mathbf x} + {\\mathbf y}\\|^2 - \\|{\\mathbf x}\\|^2 - \\|{\\mathbf y}\\|^2 \\right]\\\\ & = \\langle {\\mathbf x}, {\\mathbf y} \\rangle. \\end{align*} ###### Example12.10 Let us examine the orthogonal group on ${\\mathbb R}^2$ a bit more closely. An element $T \\in O(2)$ is determined by its action on ${\\mathbf e}_1 = (1, 0)^\\transpose$ and ${\\mathbf e}_2 = (0, 1)^\\transpose\\text{.}$ If $T({\\mathbf e}_1) = (a,b)^\\transpose\\text{,}$ then $a^2 + b^2 = 1$ and $T({\\mathbf e}_2) = (-b, a)^\\transpose\\text{.}$ Hence, $T$ can be represented by \\begin{equation*} A = \\begin{pmatrix} a & -b \\\\ b & a \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix}, \\end{equation*} where $0 \\leq \\theta \\lt 2 \\pi\\text{.}$ A matrix $T$ in $O(2)$ either reflects or rotates a vector in ${\\mathbb R}^2$ (Figure 12.9). A reflection about the horizontal axis is given by the matrix \\begin{equation*} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}, \\end{equation*} whereas a rotation by an angle $\\theta$ in a counterclockwise direction must come from a matrix of the form \\begin{equation*} \\begin{pmatrix} \\cos", "the rows of AT • reflect A by its main diagonal (which starts from the top left) to obtain AT Formally, the transpose of an m × n matrix A is the n × m matrix \\mathbf^\\mathrm_ = \\mathbf_ for 1 \\le i \\le n, 1 \\le j \\le m. Examples • \\begin 1 & 2 \\\\ 3 & 4 \\end^ \\!\\! \\;\\! = \\, \\begin 1 & 3 \\\\ 2 & 4 \\end. \\begin 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \\end^ \\!\\! \\;\\! = \\, \\begin 1 & 3 & 5\\\\ 2 & 4 & 6 \\end. \\; Properties For matrices A, B and scalar c we have the following properties of transpose: \\left( \\mathbf^\\mathrm \\right) ^\\mathrm = \\mathbf \\quad \\, Taking the transpose is an involution (self inverse). (\\mathbf+\\mathbf) ^\\mathrm = \\mathbf^\\mathrm + \\mathbf^\\mathrm \\, The transpose is a linear map from the space of m × n matrices to the space of all n × m matrices. \\left( \\mathbf \\right) ^\\mathrm = \\mathbf^\\mathrm \\mathbf^\\mathrm \\, Note that the order of the factors reverses. From this one can deduce that a square matrix A is invertible if and only if AT is invertible, and in this case we have (A−1)T = (AT)−1. It is relatively easy to extend this result to", "Matrix determinant lemma In mathematics, in particular linear algebra, the matrix determinant lemma computes the determinant of the sum of an invertible matrix A and the dyadic product, uvT, of a column vector u and a row vector vT.[1][2] Statement Suppose A is an invertible square matrix and u, v are column vectors. Then the matrix determinant lemma states that ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\left(1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {A} ^{-1}\\mathbf {u} \\right)\\,\\det \\left(\\mathbf {A} \\right)\\,.}$ Here, uvT is the outer product of two vectors u and v. The theorem can also be stated in terms of the adjugate matrix of A: ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\det \\left(\\mathbf {A} \\right)+\\mathbf {v} ^{\\textsf {T}}\\mathrm {adj} \\left(\\mathbf {A} \\right)\\mathbf {u} \\,,}$ in which case it applies whether or not the square matrix A is invertible. Proof First the proof of the special case A = I follows from the equality:[3] ${\\displaystyle {\\begin{pmatrix}\\mathbf {I} &0\\\\\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} +\\mathbf {uv} ^{\\textsf {T}}&\\mathbf {u} \\\\0&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} &0\\\\-\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {I} &\\mathbf {u} \\\\0&1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {u} \\end{pmatrix}}.}$ The determinant of the left hand side is the product of the determinants of the three matrices. Since the first and third matrix are triangular matrices with unit diagonal, their determinants are just 1. The determinant of", "a matrix :math:\\mathbf{A} is the matrix :math:\\mathbf{B} such that :math:\\mathbf{AB}=\\mathbf{I} where :math:\\mathbf{I} is the identity matrix consisting of ones down the main diagonal. -Usually :math:\\mathbf{B} is denoted :math:\\mathbf{B}=\\mathbf{A}^{-1} . In SciPy, the matrix inverse of the Numpy array, A, is obtained -using :obj:linalg.inv (A) , or using A.I if A is a Matrix. For example, let +The inverse of a matrix :math:\\mathbf{A} is the matrix +:math:\\mathbf{B} such that :math:\\mathbf{AB}=\\mathbf{I} where +:math:\\mathbf{I} is the identity matrix consisting of ones down the +main diagonal. Usually :math:\\mathbf{B} is denoted +:math:\\mathbf{B}=\\mathbf{A}^{-1} . In SciPy, the matrix inverse of +the Numpy array, A, is obtained using :obj:linalg.inv (A) , or +using A.I if A is a Matrix. For example, let .. math:: :nowrap: @@ -2011,7 +2080,13 @@ Finding Determinant ^^^^^^^^^^^^^^^^^^^ -The determinant of a square matrix :math:\\mathbf{A} is often denoted :math:\\left|\\mathbf{A}\\right| and is a quantity often used in linear algebra. Suppose :math:a_{ij} are the elements of the matrix :math:\\mathbf{A} and let :math:M_{ij}=\\left|\\mathbf{A}_{ij}\\right| be the determinant of the matrix left by removing the :math:i^{\\textrm{th}} row and :math:j^{\\textrm{th}} column from :math:\\mathbf{A} . Then for any row :math:i, +The determinant of a square matrix :math:\\mathbf{A} is often denoted +:math:\\left|\\mathbf{A}\\right| and is a quantity often used in linear +algebra. Suppose :math:a_{ij} are the elements of the matrix +:math:\\mathbf{A} and let :math:M_{ij}=\\left|\\mathbf{A}_{ij}\\right| +be the determinant of the matrix", "a matrix, $$\\mathbf{M}^\\mathrm{T}$$, which we get by swapping the rows and columns of a matrix $$\\mathbf{M}$$ $\\mathbf{M}^\\mathrm{T}_{ij} = \\mathbf{M}_{ji}$ For example $\\begin{pmatrix} a & b & c\\\\ d & e & f \\end{pmatrix}^\\mathrm{T} = \\begin{pmatrix} a & d\\\\ b & e\\\\ c & f \\end{pmatrix}$ Note that the transpose and the inverse have a similar property with regard to multiplication, in that \\begin{align*} \\left(\\mathbf{M} \\times \\mathbf{N}\\right)^{-1} &= \\mathbf{N}^{-1} \\times \\mathbf{M}^{-1}\\\\ \\left(\\mathbf{M} \\times \\mathbf{N}\\right)^\\mathrm{T} &= \\mathbf{N}^\\mathrm{T} \\times \\mathbf{M}^\\mathrm{T} \\end{align*} Now the product of a matrix and its transpose is trivially square and, providing it isn't singular, can therefore be inverted. We can consequently define $$\\mathbf{L}$$ and $$\\mathbf{R}$$ as \\begin{align*} \\mathbf{L} &= \\left(\\mathbf{M}^\\mathrm{T} \\times \\mathbf{M}\\right)^{-1} \\times \\mathbf{M}^\\mathrm{T}\\\\ \\mathbf{R} &= \\mathbf{M}^\\mathrm{T} \\times \\left(\\mathbf{M} \\times \\mathbf{M}^\\mathrm{T}\\right)^{-1} \\end{align*} The right inverse can be used to find the RMS smallest set of variables that solve a system of linear equations in which there are fewer equations than unknowns. For example \\begin{align*} 2x + 3y - 2z &= 2\\\\ 3x - 2y + 3z &= 4 \\end{align*} can be rewritten as $\\begin{pmatrix} 2 & 3 & -2\\\\ 3 & -2 & 3 \\end{pmatrix} \\times \\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = \\begin{pmatrix} 2\\\\ 4 \\end{pmatrix}$ and solved with \\begin{align*} \\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} &= \\begin{pmatrix} 2 & 3\\\\ 3 & -2\\\\ -2 & 3 \\end{pmatrix}", "# Matrix determinant lemma In mathematics, in particular linear algebra, the matrix determinant lemma computes the determinant of the sum of an invertible matrix A and the dyadic product, uvT, of a column vector u and a row vector vT.[1][2] ## Statement Suppose A is an invertible square matrix and u, v are column vectors. Then the matrix determinant lemma states that ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\left(1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {A} ^{-1}\\mathbf {u} \\right)\\,\\det \\left(\\mathbf {A} \\right)\\,.}$ Here, uvT is the outer product of two vectors u and v. The theorem can also be stated in terms of the adjugate matrix of A: ${\\displaystyle \\det \\left(\\mathbf {A} +\\mathbf {uv} ^{\\textsf {T}}\\right)=\\det \\left(\\mathbf {A} \\right)+\\mathbf {v} ^{\\textsf {T}}\\mathrm {adj} \\left(\\mathbf {A} \\right)\\mathbf {u} \\,,}$ in which case it applies whether or not the square matrix A is invertible. ## Proof First the proof of the special case A = I follows from the equality:[3] ${\\displaystyle {\\begin{pmatrix}\\mathbf {I} &0\\\\\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} +\\mathbf {uv} ^{\\textsf {T}}&\\mathbf {u} \\\\0&1\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {I} &0\\\\-\\mathbf {v} ^{\\textsf {T}}&1\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {I} &\\mathbf {u} \\\\0&1+\\mathbf {v} ^{\\textsf {T}}\\mathbf {u} \\end{pmatrix}}.}$ The determinant of the left hand side is the product of the determinants of the three matrices. Since the first and third matrix are triangular matrices with unit diagonal, their determinants are just 1.", "\\end{pmatrix}^k = \\begin{pmatrix} A_{11}^k & 0 & \\cdots & 0 \\\\ 0 & A_{22}^k & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & A_{nn}^k \\end{pmatrix}$$ meaning its easy to raise a diagonal matrix to a power. When raising an arbitrary matrix (not necessarily a diagonal matrix) to a power, it is often helpful to exploit this property by diagonalizing the matrix first. The inner and outer products Given two column vectors a and b, the Euclidean inner product and outer product are the simplest special cases of the matrix product, by transposing the column vectors into row vectors.[5] The inner product is a column vector multiplied on the left by a row vector: $$\\mathbf{a}\\cdot \\mathbf{b} = \\mathbf{a}^\\mathrm{T}\\mathbf{b}$$ More explicitly, $$\\mathbf{a}\\cdot \\mathbf{b} = \\begin{pmatrix}a_1 & a_2 & \\cdots & a_n\\end{pmatrix} \\begin{pmatrix}b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n\\end{pmatrix} = a_1b_1+a_2b_2+\\cdots+a_nb_n = \\sum_{i=1}^n a_ib_i$$ The outer product is a row vector multiplied on the left by a column vector: $$\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$$ where $$\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots", "\\end{pmatrix}\\nonumber$ with $$n=1$$ • A row vector:$\\begin{pmatrix} 3 &-2 &4 \\\\ \\end{pmatrix}\\nonumber$ with $$m=1$$ • The identity matrix:$\\begin{pmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0&0 &1 \\end{pmatrix}\\nonumber$ with $$a_{ij}=\\delta_{i,j}$$, where $$\\delta_{i,j}$$ is a function defined as $$\\delta_{i,j}=1$$ if $$i=j$$ and $$\\delta_{i,j}=0$$ if $$i\\neq j$$. • A diagonal matrix:$\\begin{pmatrix} a &0 &0 \\\\ 0 &b &0 \\\\ 0&0 &c \\end{pmatrix}\\nonumber$ with $$a_{ij}=c_i \\delta_{i,j}$$. • An upper triangular matrix:$\\begin{pmatrix} a &b &c \\\\ 0 &d &e \\\\ 0&0 &f \\end{pmatrix}\\nonumber$ All the entries below the main diagonal are zero. • A lower triangular matrix:$\\begin{pmatrix} a &0 &0 \\\\ b &c &0 \\\\ d&e &f \\end{pmatrix}\\nonumber$ All the entries above the main diagonal are zero. • A triangular matrix is one that is either lower triangular or upper triangular. ### The Trace of a Matrix The trace of an $$n\\times n$$ square matrix $$\\mathbf{A}$$ is the sum of the diagonal elements, and formally defined as $$Tr( \\mathbf{A})=\\sum_{i=1}^{n}a_{ii}$$. For example, $\\mathbf{A}=\\begin{pmatrix} 3 &-2 &4 \\\\ 5 &3i &3 \\\\ -i & 1/2 &9 \\end{pmatrix}\\; ; Tr(\\mathbf{A})=12+3i \\nonumber$ ### Singular and Nonsingular Matrices A square matrix with nonzero determinant is called nonsingular. A matrix whose determinant is zero is called singular. (Note that you cannot calculate the determinant of a non-square matrix). For a 2x2 matrix, $\\mathbf{B}=\\begin{pmatrix} a & b \\\\", "x} = (1,0)^\\transpose$ and ${\\mathbf y} = (0,1)^\\transpose$ is taken by $A$ to the parallelogram with sides $(1,0)^\\transpose$ and $(1, 1)^\\transpose\\text{;}$ that is, $A {\\mathbf x} = (1,0)^\\transpose$ and $A {\\mathbf y} = (1, 1)^\\transpose\\text{.}$ Notice that these two parallelograms have the same area. ### SubsectionThe Orthogonal Group $O(n)$ Another subgroup of $GL_n({\\mathbb R})$ is the orthogonal group. A matrix $A$ is orthogonal if $A^{-1} = A^\\transpose\\text{.}$ The orthogonal group consists of the set of all orthogonal matrices. We write $O(n)$ for the $n \\times n$ orthogonal group. We leave as an exercise the proof that $O(n)$ is a subgroup of $GL_n( {\\mathbb R})\\text{.}$ ###### Example12.5 The following matrices are orthogonal: \\begin{equation*} \\begin{pmatrix} 3/5 & -4/5 \\\\ 4/5 & 3/5 \\end{pmatrix}, \\quad \\begin{pmatrix} 1/2 & -\\sqrt{3}/2 \\\\ \\sqrt{3}/2 & 1/2 \\end{pmatrix}, \\quad \\begin{pmatrix} -1/\\sqrt{2} & 0 & 1/ \\sqrt{2} \\\\ 1/\\sqrt{6} & -2/\\sqrt{6} & 1/\\sqrt{6} \\\\ 1/ \\sqrt{3} & 1/ \\sqrt{3} & 1/ \\sqrt{3} \\end{pmatrix}. \\end{equation*} There is a more geometric way of viewing the group $O(n)\\text{.}$ The orthogonal matrices are exactly those matrices that preserve the length of vectors. We can define the length of a vector using the Euclidean inner product, or dot product, of two vectors. The Euclidean inner product of two vectors ${\\mathbf x}=(x_1, \\ldots, x_n)^\\transpose$ and ${\\mathbf y}=(y_1, \\ldots, y_n)^\\transpose$ is \\begin{equation*} \\langle", "$$\\mathbf{A}$$. This defines the elements of the diagonal matrix $$\\mathbf{W}$$ that is multiplied by $$\\mathbf{A}$$ matrix, i.e., $$\\mathbf{A} \\rightarrow \\mathbf{WA}$$. • lapack_driver ({'gesvd', 'gesdd'}, optional) -- Whether to use the more efficient divide-and-conquer approach ('gesdd') or the more robust general rectangular approach ('gesvd') to compute the singular-value decomposition with scipy.linalg.svd. Returns res -- The Procrustes result represented as a class:utils.ProcrustesResult object. Return type ProcrustesResult Notes Given matrix $$\\mathbf{A}_{n \\times n}$$ and a reference $$\\mathbf{B}_{n \\times n}$$, find a permutation of rows/columns of $$\\mathbf{A}_{n \\times n}$$ that makes it as close as possible to $$\\mathbf{B}_{n \\times n}$$. I.e., $\\begin{split}&\\underbrace{\\text{min}}_{\\left\\{\\mathbf{P} \\left| {p_{ij} \\in \\{0, 1\\} \\atop \\sum_{i=1}^n p_{ij} = \\sum_{j=1}^n p_{ij} = 1} \\right. \\right\\}} \\|\\mathbf{P}^\\dagger \\mathbf{A} \\mathbf{P} - \\mathbf{B}\\|_{F}^2\\\\ = &\\underbrace{\\text{min}}_{\\left\\{\\mathbf{P} \\left| {p_{ij} \\in \\{0, 1\\} \\atop \\sum_{i=1}^n p_{ij} = \\sum_{j=1}^n p_{ij} = 1} \\right. \\right\\}} \\text{Tr}\\left[\\left(\\mathbf{P}^\\dagger\\mathbf{A}\\mathbf{P} - \\mathbf{B} \\right)^\\dagger \\left(\\mathbf{P}^\\dagger\\mathbf{A}\\mathbf{P} - \\mathbf{B} \\right)\\right] \\\\ = &\\underbrace{\\text{max}}_{\\left\\{\\mathbf{P} \\left| {p_{ij} \\in \\{0, 1\\} \\atop \\sum_{i=1}^n p_{ij} = \\sum_{j=1}^n p_{ij} = 1} \\right. \\right\\}} \\text{Tr}\\left[\\mathbf{P}^\\dagger\\mathbf{A}^\\dagger\\mathbf{P}\\mathbf{B} \\right]\\\\\\end{split}$ Here, $$\\mathbf{P}_{n \\times n}$$ is the permutation matrix. Given an intial guess, the best local minimum can be obtained by the iterative procedure, $p_{ij}^{(n + 1)} = p_{ij}^{(n)} \\sqrt{ \\frac{2\\left[\\mathbf{T}^{(n)}\\right]_{ij}}{\\left[ \\mathbf{P}^{(n)} \\left( \\left(\\mathbf{P}^{(n)}\\right)^T \\mathbf{T} + \\left( \\left(\\mathbf{P}^{(n)}\\right)^T \\mathbf{T} \\right)^T \\right) \\right]_{ij}} }$ where, $\\mathbf{T}^{(n)} = \\mathbf{A} \\mathbf{P}^{(n)} \\mathbf{B}$ Using an initial guess, the iteration can", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "$\\lambda_{\\min} \\left( DD^{T} \\right)$ instead of relying on iterative method. Otherwise, having explicit formula for $\\lambda_{\\min} \\left( D_{1}D_{1}^{T} \\right)$ and $\\lambda_{\\min} \\left( D_{2}D_{2}^{T} \\right)$ would also be really nice. ## UPDATE I derived the exact formula of $D_{1}D_{1}^{T}$ and $D_{2}D_{2}^{T}$ when $m,n > 2$. Hope that it would be helpful. 1. $D_{1}D_{1}^{T}$ is a $(m-1)n \\times (m-1)n$ matrix contains $n$-blocks in the main diagonal $$D_{1}D_{1}^{T} = \\begin{pmatrix} B \\\\ & B \\\\ & & \\ddots \\\\ & & & B \\end{pmatrix}$$ where $B$ is a $(m-1) \\times (m-1)$ matrix defined as $$B = \\begin{pmatrix} 2 & -1 \\\\ -1 & 2 & -1 \\\\ & \\ddots & \\ddots & \\ddots \\\\ & & -1 & 2 & -1 \\\\ & & & -1 & 2 \\end{pmatrix} .$$ 2. $D_{2}D_{2}^{T}$ is a $m(n-1) \\times m(n-1)$ matrix with $m$ diag contains $2$ in the main diagonal and $-1$ at the $m^{th}$-upper and lower diagonal $$D_{2}D_{2}^{T} \\left[ r , c \\right] = \\begin{cases} 2 & \\textrm{ if } r = c , \\\\ -1 & \\textrm{ if } \\left\\lvert r-c \\right\\rvert = m , \\\\ 0 & \\textrm{ otherwise } . \\end{cases}$$ • Are you using the notation $A[i;j]$ to mean $A_{ij}$? Could you make the definitions of $D_1$ and D_2$clearer, please? – Ben McKay Jun 9 '18 at 14:17 •", "+ m_{44} v_w \\end{pmatrix} = \\mathbf{m_{|1}} v_x + \\mathbf{m_{|2}} v_y + \\mathbf{m_{|3}} v_z + \\mathbf{m_{|4}} v_w$ where $\\mathbf{m_{|i}}$ denotes the i-th matrix column. The multiplication of a vector with a matrix from the right corresponds thus to the sum of the matrix columns, weighted by the vector’s components. Conversely, multiplying a matrix with a vector from the left can be seen as a weighted sum of the matrix rows: $\\mathbf{v}^T \\mathbf{M} = \\begin{pmatrix}m_{11} v_x + m_{21} v_y + m_{31} v_z + m_{41} v_w \\\\ m_{12} v_x + m_{22} v_y + m_{32} v_z + m_{42} v_w\\\\ m_{13} v_x + m_{23} v_y + m_{33} v_z + m_{43} v_w\\\\ m_{14} v_x + m_{24} v_y + m_{34} v_z + m_{44} v_w \\end{pmatrix} = \\mathbf{m_{\\overline{1}}} v_x + \\mathbf{m_{\\overline{2}}} v_y + \\mathbf{m_{\\overline{3}}} v_z + \\mathbf{m_{\\overline{4}}} v_w$ where $\\mathbf{m_{\\overline{i}}}$ denotes the i-th matrix row. Since the transpose operator swaps a matrix’s rows and columns, we can conclude that multiplying a vector by a matrix from the right is equivalent to multiplying the same vector with the matrix’s transpose from the left. This works in accordance with the transposition rules $(\\mathbf{M_1} \\mathbf{M_2})^T = \\mathbf{M_2}^T \\mathbf{M_1}^T\\quad\\text{and}\\quad(\\mathbf{M}^T)^T = \\mathbf{M}$ Given these definitions, lets now look at the matrix type we encounter most in graphics programming: Combined rotation and translation transforms. Matrices of this form consist of a 3×3", "+ m_{44} v_w \\end{pmatrix} = \\mathbf{m_{|1}} v_x + \\mathbf{m_{|2}} v_y + \\mathbf{m_{|3}} v_z + \\mathbf{m_{|4}} v_w$ where $\\mathbf{m_{|i}}$ denotes the i-th matrix column. The multiplication of a vector with a matrix from the right corresponds thus to the sum of the matrix columns, weighted by the vector’s components. Conversely, multiplying a matrix with a vector from the left can be seen as a weighted sum of the matrix rows: $\\mathbf{v}^T \\mathbf{M} = \\begin{pmatrix}m_{11} v_x + m_{21} v_y + m_{31} v_z + m_{41} v_w \\\\ m_{12} v_x + m_{22} v_y + m_{32} v_z + m_{42} v_w\\\\ m_{13} v_x + m_{23} v_y + m_{33} v_z + m_{43} v_w\\\\ m_{14} v_x + m_{24} v_y + m_{34} v_z + m_{44} v_w \\end{pmatrix} = \\mathbf{m_{\\overline{1}}} v_x + \\mathbf{m_{\\overline{2}}} v_y + \\mathbf{m_{\\overline{3}}} v_z + \\mathbf{m_{\\overline{4}}} v_w$ where $\\mathbf{m_{\\overline{i}}}$ denotes the i-th matrix row. Since the transpose operator swaps a matrix’s rows and columns, we can conclude that multiplying a vector by a matrix from the right is equivalent to multiplying the same vector with the matrix’s transpose from the left. This works in accordance with the transposition rules $(\\mathbf{M_1} \\mathbf{M_2})^T = \\mathbf{M_2}^T \\mathbf{M_1}^T\\quad\\text{and}\\quad(\\mathbf{M}^T)^T = \\mathbf{M}$ Given these definitions, lets now look at the matrix type we encounter most in graphics programming: Combined rotation and translation transforms. Matrices of this form consist of a 3×3", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "the matrix product in terms of the outer product. The decomposition is done the other way around, the first matrix A is decomposed into column vectors ai and the second matrix B into row vectors bi: {\\displaystyle {\\begin{aligned}\\mathbf {AB} &={\\begin{pmatrix}\\mathbf {\\bar {a}} _{1}&\\mathbf {\\bar {a}} _{2}&\\cdots &\\mathbf {\\bar {a}} _{m}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {\\bar {b}} _{1}\\\\\\mathbf {\\bar {b}} _{2}\\\\\\vdots \\\\\\mathbf {\\bar {b}} _{m}\\end{pmatrix}}\\\\&=\\mathbf {\\bar {a}} _{1}\\otimes \\mathbf {\\bar {b}} _{1}+\\mathbf {\\bar {a}} _{2}\\otimes \\mathbf {\\bar {b}} _{2}+\\cdots +\\mathbf {\\bar {a}} _{m}\\otimes \\mathbf {\\bar {b}} _{m}\\\\&=\\sum _{i=1}^{m}\\mathbf {\\bar {a}} _{i}\\otimes \\mathbf {\\bar {b}} _{i}\\end{aligned}}} where this time ${\\displaystyle \\mathbf {\\bar {a}} _{i}={\\begin{pmatrix}A_{1i}\\\\A_{2i}\\\\\\vdots \\\\A_{ni}\\end{pmatrix}}\\,,\\quad \\mathbf {\\bar {b}} _{i}={\\begin{pmatrix}B_{i1}&B_{i2}&\\cdots &B_{ip}\\end{pmatrix}}\\,.}$ This method emphasizes the effect of individual column/row pairs on the result, which is a useful point of view with e.g. covariance matrices, where each such pair corresponds to the effect of a single sample point.[citation needed] {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}{\\color {Brown}1}&{\\color {Orange}2}&{\\color {Violet}3}\\\\{\\color {Brown}4}&{\\color {Orange}5}&{\\color {Violet}6}\\\\{\\color {Brown}7}&{\\color {Orange}8}&{\\color {Violet}9}\\\\\\end{pmatrix}}{\\begin{pmatrix}{\\color {Brown}a}&{\\color {Brown}d}\\\\{\\color {Orange}b}&{\\color {Orange}e}\\\\{\\color {Violet}c}&{\\color {Violet}f}\\\\\\end{pmatrix}}&={\\begin{pmatrix}{\\color {Brown}1}\\\\{\\color {Brown}4}\\\\{\\color {Brown}7}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Brown}{a}}&{\\color {Brown}{d}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Orange}2}\\\\{\\color {Orange}5}\\\\{\\color {Orange}8}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Orange}{b}}&{\\color {Orange}{e}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Violet}3}\\\\{\\color {Violet}6}\\\\{\\color {Violet}9}\\\\\\end{pmatrix}}\\otimes {\\begin{pmatrix}{\\color {Violet}c}&{\\color {Violet}f}\\\\\\end{pmatrix}}\\\\&={\\begin{pmatrix}{\\color {Brown}{1a}}&{\\color {Brown}{1d}}\\\\{\\color {Brown}{4a}}&{\\color {Brown}{4d}}\\\\{\\color {Brown}{7a}}&{\\color {Brown}{7d}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Orange}{2b}}&{\\color {Orange}{2e}}\\\\{\\color {Orange}{5b}}&{\\color {Orange}{5e}}\\\\{\\color {Orange}{8b}}&{\\color {Orange}{8e}}\\\\\\end{pmatrix}}+{\\begin{pmatrix}{\\color {Violet}{3c}}&{\\color {Violet}{3f}}\\\\{\\color {Violet}{6c}}&{\\color {Violet}{6f}}\\\\{\\color {Violet}{9c}}&{\\color {Violet}{9f}}\\\\\\end{pmatrix}}\\\\&={\\begin{pmatrix}{\\color {Brown}{1a}}+{\\color {Orange}{2b}}+{\\color {Violet}{3c}}&{\\color {Brown}{1d}}+{\\color {Orange}{2e}}+{\\color {Violet}{3f}}\\\\{\\color {Brown}{4a}}+{\\color {Orange}{5b}}+{\\color {Violet}{6c}}&{\\color {Brown}{4d}}+{\\color {Orange}{5e}}+{\\color {Violet}{6f}}\\\\{\\color {Brown}{7a}}+{\\color {Orange}{8b}}+{\\color {Violet}{9c}}&{\\color {Brown}{7d}}+{\\color {Orange}{8e}}+{\\color {Violet}{9f}}\\\\\\end{pmatrix}}.\\end{aligned}}} ## Algorithms for efficient matrix multiplicationEdit", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you" ]

[ "-1 & 2\\\\ -1 & 3 & -1\\\\ 2 & -1 & 3\\\\ \\end{pmatrix}, \\quad M_3 =\\frac{1}{9} \\begin{pmatrix} 3 & -1 & -1\\\\ -1 & 3 & 2\\\\ -1 & 2 & 3\\\\ \\end{pmatrix}.$$", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "+ e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... \\\\ \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{X^2} = [\\mathbf{1}, \\mathbf{A^1}] = \\begin{bmatrix} 1 & \\frac{1}{1 + e^{-z_{11}}} & \\frac{1}{1 + e^{-z_{12}}} & \\frac{1}{1 + e^{-z_{13}}} \\\\ 1 & \\frac{1}{1 + e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... & ... \\\\ 1 & \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{Z^2} = \\mathbf{X^2}\\mathbf{W^2} = \\begin{bmatrix} z_{11} & z_{12} \\\\ z_{21} & z_{22} \\\\ ... & ... \\\\ z_{N1} & z_{N2} \\end{bmatrix}$ $\\mathbf{\\hat{Y}} = \\mathbf{A^2} = softmax(\\mathbf{Z^2}) = \\begin{bmatrix} \\hat{y}_{11} & \\hat{y}_{12} \\\\ \\hat{y}_{21} & \\hat{y}_{22} \\\\ ... & ... \\\\ \\hat{y}_{N1} & \\hat{y}_{N2} \\end{bmatrix}$ Here, $sigmoid$ is the sigmoid function and similarly $softmax$ is the row-wise softmax function. (Also, note that I've used superscripts to denote the layer of the network for each element.) My goal is to use gradient descent to find the weight matrices $\\mathbf{W^1}$ and $\\mathbf{W^2}$ that minimize categorical cross entropy. Can anyone walk me through this process? I have seen it done here, but I can't understand how the solution generalizes to matrix equations.", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "\\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {11}{12} & \\frac {-5}{12}\\\\ \\frac {22}{9} & \\frac {5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {11 - 5}{12}\\\\ \\frac {22 + 5}{9}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {6}{12}\\\\ \\frac {27}{9}\\\\ \\end {pmatrix}$$ ∴X = $$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ i.e.$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{2}\\\\ 3\\\\ \\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, 4x + $$\\frac y5$$ = 7 or, $$\\frac {20x + y}{5}$$ = 7 or, 20x + y = 35.........................................(1) 3x + $$\\frac y4$$ = 5 or, $$\\frac {12x + y}{4}$$ = 5 or, 12x + y = 20..........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 20 & 1\\\\ 12 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 20 & 1\\\\ 12 &", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "\\mathbf{w}}\\right)^T\\left[I(\\mathbf{w}(\\mathbf{x},\\mathbf{p})) - T(\\mathbf{w}(\\mathbf{x},\\mathbf{0}))\\right] \\\\ \\end{align} For the affine warp we have, \\begin{align} \\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}} &= \\frac{\\partial (\\mathbf{A}\\mathbf{x} + \\mathbf{b})}{\\partial \\mathbf{p}} \\\\ &= \\frac{\\partial}{\\partial \\mathbf{p}}\\left[ \\begin{pmatrix}1+p_0 && p_1 \\\\ p_2 && 1+p_3\\end{pmatrix}\\begin{pmatrix}x \\\\ y\\end{pmatrix} \\right] + \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}p_4 \\\\ p_5\\end{pmatrix} \\\\ &= \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}x+p_0x+p_1y \\\\ p_2x+y+p_3y\\end{pmatrix} + \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}p_4 \\\\ p_5\\end{pmatrix} \\\\ &= \\begin{pmatrix}x && y && 0 && 0 && 0 && 0 \\\\0 && 0 && x && y && 0 && 0\\end{pmatrix} + \\begin{pmatrix}0 && 0 && 0 && 0 && 1 && 0 \\\\ 0 && 0 && 0 && 0 && 0 && 1\\end{pmatrix} \\\\ &= \\begin{pmatrix}x && y && 0 && 0 && 1 && 0 \\\\0 && 0 && x && y && 0 && 1\\end{pmatrix} \\end{align} If we let, \\begin{align} \\frac{\\partial T}{\\partial \\mathbf{w}} = \\begin{pmatrix}T_x && T_y\\end{pmatrix} \\\\ \\end{align} We have, \\begin{align} \\frac{\\partial T}{\\partial \\mathbf{w}} \\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}} &= \\begin{pmatrix}T_x && T_y\\end{pmatrix} \\begin{pmatrix}x && y && 0 && 0 && 1 && 0 \\\\0 && 0 && x && y && 0 && 1\\end{pmatrix} \\\\ &= \\begin{pmatrix}T_xx && T_xy && T_yx && T_yy && T_x && T_y\\end{pmatrix} \\\\ \\end{align} Detecting features Equation $$\\eqref{deltap1}$$ gives an indication of which features would be good to track. The matrix, \\begin{align} \\mathbf{H} &= \\sum_{\\mathbf{x} \\in \\mathbf{R}} \\left(\\frac{\\partial T}{\\partial \\mathbf{w}}\\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}}\\right)^T\\frac{\\partial", "\\\\ \\mathbf{\\hat y} \\end{pmatrix} = \\begin{pmatrix} \\cos \\phi & -\\sin \\phi \\\\ \\sin \\phi & \\cos \\phi \\end{pmatrix} \\begin{pmatrix} \\boldsymbol{\\hat \\rho} \\\\ \\boldsymbol{\\hat\\phi} \\end{pmatrix}$$", "can differentiate $$y = \\sqrt{3x^2 - 4x + 7}$$ as follows: Let $$u(x) = 3x^2 - 4x + 7$$, then: $y = \\sqrt{u}$ $\\frac{dy}{du} = \\frac{1}{2\\sqrt{u}}$ and $\\frac{du}{dx} = 6x - 4$ and using the chain rule we can write: \\begin{aligned} \\frac{dy}{dx} & = \\frac{dy}{du}.\\frac{dy}{dx} \\\\ & = \\frac{1}{2\\sqrt{u}}. \\begin{pmatrix} 6x - 4 \\end{pmatrix} \\\\ & = \\frac{6x - 4}{2\\sqrt{u}} \\\\ & = \\frac{2 \\begin{pmatrix} 3x - 2 \\end{pmatrix}}{2\\sqrt{u}} \\\\ & = \\frac{3x - 2}{\\sqrt{u}}\\\\ \\frac{dy}{dx} &= \\frac{3x-2}{\\sqrt{3x^2 - 4x + 7}} \\end{aligned} 3. The function $$f(x) = \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^7$$ can be written $$f(x) = g \\begin{bmatrix} h(x) \\end{bmatrix}$$ where $g(x) = x^7 \\quad \\text{and} \\quad h(x) = x^2 + 3$ and $g'(x) = 7x^6 \\quad \\text{and} \\quad h'(x) = 2x$ So, using the chain rule, we find the derivative as follows: \\begin{aligned} f'(x) & = h'(x) \\times g' \\begin{bmatrix} h(x) \\end{bmatrix} \\\\ & = 2x \\times 7 \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6 \\\\ f'(x) & = 14x \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6 \\end{aligned} 4. We differentiate $$y = e^{3x^2-1}$$ as follows: Let $u(x) = 3x^2 - 1$ then $y = e^u$ $\\frac{dy}{du} = e^u$ and $\\frac{du}{dx} = 6x$ Now, using the chain rule, we find the derivative: \\begin{aligned} \\frac{dy}{dx} & = \\frac{dy}{du}.\\frac{du}{dx} \\\\ & = e^u.6x \\\\ \\frac{dy}{dx} & = 6x.e^{3x^2-1} \\end{aligned} 5. The", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0\\},\\quad N=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, 2x+y=0\\}$ 3. $\\begin{pmatrix} 3 \\\\ 0 \\\\ 1 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x+y=0\\},\\quad N=\\{c\\cdot\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\\,\\big|\\, c\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 2 Find $M^\\perp$. 1. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x+y=0 \\}$ 2. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, -2x+3y=0 \\}$ 3. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0 \\}$ 4. $M=\\{\\vec{0}\\,\\}$ 5. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x=0 \\}$ 6. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, -x+3y+z=0 \\}$ 7. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x=0\\text{ and } y+z=0\\}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane $P$ specified by $3x+2y-z=0$ in $\\mathbb{R}^3$. 1. Find a basis for $P$. 2. Find $P^\\perp$ and a basis for $P^\\perp$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. $\\vec{v}=\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 4. Find the orthogonal projection of $\\vec{v}$ onto $P$ by keeping only the $P$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. This exercise is recommended for all", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary" ]

[ "+ \\sqrt{2} & 2 - \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} 2 + \\sqrt{2} & 0 \\\\ 0 & 2 - \\sqrt{2} \\end{pmatrix}^n \\begin{pmatrix} \\frac{1}{2}-\\frac{1}{\\sqrt{2}} & \\frac{1}{2\\sqrt{2}} \\\\ \\frac{1}{2}+\\frac{1}{\\sqrt{2}} & -\\frac{1}{2\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} 1/4 \\\\ 1 \\end{pmatrix} \\nonumber\\\\ &= \\begin{pmatrix} \\frac{1}{8}\\left((2+\\sqrt{2})^n(1+\\sqrt{2}) + (2-\\sqrt{2})^n(1-\\sqrt{2})\\right) \\\\ \\frac{1}{8}\\left((2+\\sqrt{2})^{n+1}(1+\\sqrt{2}) + (2-\\sqrt{2})^{n+1}(1-\\sqrt{2})\\right) \\end{pmatrix} \\label{eqn:3b_c_matrix} \\end{align} by diagonalization. Using Equation \\ref{eqn:3b_c_matrix}, Equation \\ref{eqn:3b_p_lim_2} becomes $$\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M) = \\frac{1}{2}(1 - 2a_{n})\\left((1+\\sqrt{2}) + \\left(\\frac{2-\\sqrt{2}}{2+\\sqrt{2}}\\right)^n(1-\\sqrt{2})\\right), \\label{eqn:3b_p_lim_3}$$ where we recall that $a = \\frac{1}{2 + \\sqrt{2}}.$ $|2-\\sqrt{2}| < 1$ and so, taking the limit as $n \\rightarrow \\infty$ of Equation \\ref{eqn:3b_p_lim_3}, the second term goes to $0$: $$\\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M) = \\frac{1-2a}{2}(1+\\sqrt{2}) = \\frac{\\sqrt{2} - 1}{2}(1+\\sqrt{2}) = \\frac{1}{2}. \\label{eqn:3b_p_lim_lim}$$ And finally, by symmetry, \\begin{align} \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 1 \\mid \\tau > M) &= \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 3 \\mid \\tau > M) \\nonumber\\\\ &= \\frac{1 - \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M)}{2} \\nonumber\\\\ &= \\frac{1}{4}. \\label{eqn:3b_p_1_3} \\end{align} Equations \\ref{eqn:3b_p_lim_lim} and \\ref{eqn:3b_p_1_3} combine to give us the limiting distribution. ## Graphs in $\\LaTeX$ Here's the code for the graph. I used a package called tikz. \\documentclass[10pt]{article} \\usepackage[utf8]{inputenc} \\usepackage[T1]{fontenc} \\usepackage{tikz} \\usetikzlibrary{arrows} \\usepackage[active, tightpage]{preview} \\PreviewEnvironment{tikzpicture} \\renewcommand\\PreviewBbAdjust {-0bp -25bp 0bp 25bp} \\begin{document} \\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.75cm, thick,main node/.style={circle,draw}] \\node[main node]", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "{\\mathbf y} \\|^2 - \\|A {\\mathbf x} \\|^2 - \\|A {\\mathbf y} \\|^2 \\right]\\\\ & = \\frac{1}{2} \\left[ \\|A ( {\\mathbf x} + {\\mathbf y} ) \\|^2 - \\|A {\\mathbf x} \\|^2 - \\|A {\\mathbf y} \\|^2 \\right]\\\\ & = \\frac{1}{2} \\left[ \\|{\\mathbf x} + {\\mathbf y}\\|^2 - \\|{\\mathbf x}\\|^2 - \\|{\\mathbf y}\\|^2 \\right]\\\\ & = \\langle {\\mathbf x}, {\\mathbf y} \\rangle. \\end{align*} ###### Example12.10 Let us examine the orthogonal group on ${\\mathbb R}^2$ a bit more closely. An element $T \\in O(2)$ is determined by its action on ${\\mathbf e}_1 = (1, 0)^\\transpose$ and ${\\mathbf e}_2 = (0, 1)^\\transpose\\text{.}$ If $T({\\mathbf e}_1) = (a,b)^\\transpose\\text{,}$ then $a^2 + b^2 = 1$ and $T({\\mathbf e}_2) = (-b, a)^\\transpose\\text{.}$ Hence, $T$ can be represented by \\begin{equation*} A = \\begin{pmatrix} a & -b \\\\ b & a \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix}, \\end{equation*} where $0 \\leq \\theta \\lt 2 \\pi\\text{.}$ A matrix $T$ in $O(2)$ either reflects or rotates a vector in ${\\mathbb R}^2$ (Figure 12.9). A reflection about the horizontal axis is given by the matrix \\begin{equation*} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}, \\end{equation*} whereas a rotation by an angle $\\theta$ in a counterclockwise direction must come from a matrix of the form \\begin{equation*} \\begin{pmatrix} \\cos", "2\\mathbf {r} _{1}\\to \\mathbf {r} _{1}}$ and ${\\displaystyle 2\\mathbf {r} _{1}+\\mathbf {r} _{2}\\to \\mathbf {r} _{2}}$ are ${\\displaystyle \\mathbf {r} _{1}\\leftrightarrow \\mathbf {r} _{2}}$, ${\\displaystyle {\\frac {1}{2}}\\mathbf {r} _{1}\\to \\mathbf {r} _{1}}$ and ${\\displaystyle -2\\mathbf {r} _{1}+\\mathbf {r} _{2}\\to \\mathbf {r} _{2}}$, the inverses of the elementary matrices ${\\displaystyle {\\begin{pmatrix}0&1\\\\1&0\\\\\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}2&0\\\\0&1\\\\\\end{pmatrix}}}$ and ${\\displaystyle {\\begin{pmatrix}1&0\\\\2&1\\\\\\end{pmatrix}}}$ are ${\\displaystyle {\\begin{pmatrix}0&1\\\\1&0\\\\\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}{\\frac {1}{2}}&0\\\\0&1\\\\\\end{pmatrix}}}$ and ${\\displaystyle {\\begin{pmatrix}1&0\\\\-2&1\\\\\\end{pmatrix}}}$ respectively. In particular, the inverse of type I elementary matrix is itself. Exercise. It is given that matrix ${\\displaystyle B={\\begin{pmatrix}1&3&8\\\\3&7&2\\\\0&2&2\\\\\\end{pmatrix}}}$ is obtained from matrix ${\\displaystyle A}$ by performing the EROs ${\\displaystyle \\mathbf {r} _{1}\\leftrightarrow \\mathbf {r} _{3},3\\mathbf {r} _{2}\\to \\mathbf {r} _{2},-\\mathbf {r} _{1}+\\mathbf {r} _{3}\\to \\mathbf {r} _{3}}$, in this order, and ${\\displaystyle B^{-1}={\\frac {1}{20}}{\\begin{pmatrix}5&5&-25\\\\-3&1&11\\\\3&-1&-1\\\\\\end{pmatrix}}}$. 1 Find ${\\displaystyle A}$. ${\\displaystyle {\\begin{pmatrix}1&5&10\\\\1&{\\frac {7}{3}}&{\\frac {2}{3}}\\\\1&3&8\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}0&2&2\\\\9&21&6\\\\1&1&6\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}-1&-1&-6\\\\9&21&6\\\\1&3&8\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}7&9&1\\\\-1&21&3\\\\2&6&0\\\\\\end{pmatrix}}}$ 2 Find ${\\displaystyle A^{-1}}$. ${\\displaystyle {\\frac {1}{20}}{\\begin{pmatrix}-25&15&30\\\\11&3&-14\\\\-1&-3&4\\end{pmatrix}}}$ ${\\displaystyle {\\frac {1}{20}}{\\begin{pmatrix}-30&15&5\\\\41&3&-3\\\\-4&-3&3\\end{pmatrix}}}$ ${\\displaystyle {\\frac {1}{20}}{\\begin{pmatrix}8&4&-26\\\\-1&{\\frac {1}{3}}&{\\frac {11}{3}}\\\\5&5&-25\\end{pmatrix}}}$ Then, we will state a simplified version of invertible matrix theorem, in which some results from the complete version of invertible matrix theorem are removed. Theorem. (Simplified invertible matrix theorem) Let ${\\displaystyle A}$ be an ${\\displaystyle n\\times n}$ matrix. Then, the following are equivalent. (i) ${\\displaystyle A}$ is invertible (ii) the homogeneous SLE ${\\displaystyle A\\mathbf {x} =\\mathbf {0} }$ only has the trivial solution ${\\displaystyle \\mathbf {x} =\\mathbf {0}", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "\\text{L}}{\\mathbf{a}^{[1]}} = \\mathbf{W}^{[2]}.T \\frac{d \\text{L}}{d \\mathbf{z}^{[2]}}$$ Now we only missing $\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}}$ and $\\frac{d \\text{L}}{db^{[1]}}$. Let's look how we calculate $\\mathbf{z}^{[1]}$: $$\\begin{matrix} (\\mathbf{W}^{[1]}) \\\\ \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\\\ 0 & 0 \\\\ \\end{bmatrix} \\end{matrix} \\begin{matrix} (\\mathbf{x}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} + \\begin{matrix} (\\mathbf{b}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} = \\begin{matrix} (\\mathbf{z}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix}$$ Now, let's look at: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1](1,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](1)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](2,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](2)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](3,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](3)}} \\mathbf{x}.\\text{T}$$ or equivalently: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}} \\mathbf{x}.\\text{T}$$ and as expected: $$\\frac{d \\text{L}}{d \\mathbf{b}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}}$$ Let's apply the backprop altogether! ALPHA = 0.4 # learning rate def get_loss(y, a): return -1 * (y * np.log(a) + (1-y) * np.log(1-a)) def get_loss_numerically_stable(y, z): return -1 * (y * -1 * np.log(1 + np.exp(-z)) + (1-y) * (-z - np.log(1 + np.exp(-z)))) def get_gradients_loops(z1, a1, z2, a2, z3, a3, x, y, W1, b1, W2, b2, W3, b3): dz3 = a3 - y db3 = dz3 dW3 = dz3 * a2.T da2 = dz3 * W3.T dz2 = da2 * (a2 * (1-a2)) db2 = dz2 dW2 = np.matmul(dz2, a1.T)", "x}+{\\mathbf y}\\neq {\\mathbf 0} \\}\\\\ & = \\min\\{ w({\\mathbf z}) : {\\mathbf z} \\neq {\\mathbf 0} \\}\\text{.} \\end{align*} ### SubsectionLinear Codes From Example 8.16, it is now easy to check that the minimum nonzero weight is $$3\\text{;}$$ hence, the code does indeed detect and correct all single errors. We have now reduced the problem of finding “good” codes to that of generating group codes. One easy way to generate group codes is to employ a bit of matrix theory. Define the inner product of two binary $$n$$-tuples to be \\begin{equation*} {\\mathbf x} \\cdot {\\mathbf y} = x_1 y_1 + \\cdots + x_n y_n\\text{,} \\end{equation*} where $${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^\\transpose$$ and $${\\mathbf y} = (y_1, y_2, \\ldots, y_n)^\\transpose$$ are column vectors. 5 For example, if $${\\mathbf x} = (011001)^\\transpose$$ and $${\\mathbf y} = (110101)^\\transpose\\text{,}$$ then $${\\mathbf x} \\cdot {\\mathbf y} = 0\\text{.}$$ We can also look at an inner product as the product of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^\\transpose {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}\\text{.} \\end{align*} Since we will be working with matrices, we will write binary", "x^2 $$\\iint_{i=1}^\\infty x^2$$ = \\iint_{i=1}^\\infty x^2 $$\\prod_{i=1}^n x^2$$ = \\prod_{i=1}^n x^2 ## Fractions $$\\frac ab$$ = \\frac ab $$\\dfrac ab$$ = \\dfrac ab = in displaymode! $$\\frac{a+1}{b+1}$$ = \\frac{a+1}{b+1} $$\\sqrt[3] \\frac{x^3}{b}$$ = \\sqrt[3] \\frac{x^3}{b} $$x^\\frac 23$$ = x^\\frac 23 $$\\lim_{x\\to 0}$$ =\\lim_{x\\to 0} $$\\displaystyle\\lim_{x\\to 0}$$ =\\displaystyle\\lim_{x\\to 0} $$\\hat x$$ = \\hat x $$\\widehat {xy}$$ = \\widehat {xy} $$\\bar x$$ = \\bar x $$\\vec x$$ = \\vec x $$\\overrightarrow {xyz}$$ = \\overrightarrow {xyz} $$\\lvert x \\rvert$$ = \\lvert x \\rvert $$\\lVert x \\rVert$$ = \\lVert x \\rVert $$_5C_3$$ = _5C_3 ## Matrices ### matrix $$\\begin{matrix} 1 & x \\\\ 1 & y \\end{matrix}$$ = \\begin{matrix} 1 & x \\\\ 1 & y \\end{matrix} ### pmatrix (parentheses) $$\\begin{pmatrix} 1 & x \\\\ 1 & y \\end{pmatrix}$$ = \\begin{pmatrix} 1 & x \\\\ 1 & y \\end{pmatrix} ### bmatrix (curved bracket) $$\\begin{bmatrix} 1 & x \\\\ 1 & y \\end{bmatrix}$$ = \\begin{bmatrix} 1 & x \\\\ 1 & y \\end{bmatrix} ### Bmatrix (square Bracket) $$\\begin{Bmatrix} 1 & x \\\\ 1 & y \\end{Bmatrix}$$ = \\begin{Bmatrix} 1 & x \\\\ 1 & y \\end{Bmatrix} ### vmatrix (vertical line) $$\\begin{vmatrix} 1 & x \\\\ 1 & y \\end{vmatrix}$$ = \\begin{vmatrix} 1 & x \\\\ 1 & y \\end{vmatrix} ### Vmatrix (Vertical lines) $$\\begin{Vmatrix} 1 & x \\\\ 1 & y \\end{Vmatrix}$$ = \\begin{Vmatrix}", "\\begin{pmatrix} 1 & \\pmb{0} \\\\ \\pmb{0} & Q_2 \\end{pmatrix}\\quad \\text{and}\\quad U= \\begin{pmatrix} \\lambda_1 & \\pmb{0} \\\\ \\pmb{0} & U_2 \\end{pmatrix}.$$ Then $U$ is upper triangular and $Q$ is orthogonal and $A=QUQ^\\mathrm{T}$. After you prove the decomposition theorem, we can then show that if $A$ corresponds to your linear operator $T$, then $A=QUQ^{\\mathrm{T}}$ and so using $AA^\\mathrm{T}=A^\\mathrm{T}A$, we get that $UU^{\\mathrm{T}}=U^\\mathrm{T}U$. Suppose that $U$ is an upper triangle matrix with $UU^\\mathrm{T}=U^\\mathrm{T}U$, then $U$ is diagonal. Proof. Denote the $ij$-entry of $U$ by $u_{ij}$. Then the for any $i$, we have $$\\sum_{j=1}^n u_{ij}u_{ij} = \\sum_{j=1} u_{ji}u_{ji}\\implies \\sum_{j=1}u_{ij}^2 =\\sum_{j=1}^n u_{ji}^2. \\tag1$$ But because $U$ is upper triangular, Let $i$ be $1$ in Eq.(1), then we get that on the first row of $U$, every entry except the one on the first column is zero. Now, let $i$ be $2$ in Eq. (1), we immediately get that on the second row of $U$, every entry except the one on the second column is zero. This carries on. So $U$ has to be diagonal. Now, since $U$ is diagonal, it follows that $A^\\mathrm{T}=QU^\\mathrm{T}Q^\\mathrm{T}= QUQ^\\mathrm{T}=A$. This shows that $A$ is indeed Hermitian.", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "\\end{pmatrix}\\nonumber$ with $$n=1$$ • A row vector:$\\begin{pmatrix} 3 &-2 &4 \\\\ \\end{pmatrix}\\nonumber$ with $$m=1$$ • The identity matrix:$\\begin{pmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0&0 &1 \\end{pmatrix}\\nonumber$ with $$a_{ij}=\\delta_{i,j}$$, where $$\\delta_{i,j}$$ is a function defined as $$\\delta_{i,j}=1$$ if $$i=j$$ and $$\\delta_{i,j}=0$$ if $$i\\neq j$$. • A diagonal matrix:$\\begin{pmatrix} a &0 &0 \\\\ 0 &b &0 \\\\ 0&0 &c \\end{pmatrix}\\nonumber$ with $$a_{ij}=c_i \\delta_{i,j}$$. • An upper triangular matrix:$\\begin{pmatrix} a &b &c \\\\ 0 &d &e \\\\ 0&0 &f \\end{pmatrix}\\nonumber$ All the entries below the main diagonal are zero. • A lower triangular matrix:$\\begin{pmatrix} a &0 &0 \\\\ b &c &0 \\\\ d&e &f \\end{pmatrix}\\nonumber$ All the entries above the main diagonal are zero. • A triangular matrix is one that is either lower triangular or upper triangular. ### The Trace of a Matrix The trace of an $$n\\times n$$ square matrix $$\\mathbf{A}$$ is the sum of the diagonal elements, and formally defined as $$Tr( \\mathbf{A})=\\sum_{i=1}^{n}a_{ii}$$. For example, $\\mathbf{A}=\\begin{pmatrix} 3 &-2 &4 \\\\ 5 &3i &3 \\\\ -i & 1/2 &9 \\end{pmatrix}\\; ; Tr(\\mathbf{A})=12+3i \\nonumber$ ### Singular and Nonsingular Matrices A square matrix with nonzero determinant is called nonsingular. A matrix whose determinant is zero is called singular. (Note that you cannot calculate the determinant of a non-square matrix). For a 2x2 matrix, $\\mathbf{B}=\\begin{pmatrix} a & b \\\\", "+ y^2 + z^2 \\\\ \\text{A2g: Rz} \\\\ \\text{B1g} \\\\ \\text{B2g} \\\\ \\text{E1g: (Rx, Ry), (xz, yz)} \\\\ \\text{E2g: }(x^2 - y^2,~xy) \\\\ \\text{A1u:} \\\\ \\text{A2u: z} \\\\ \\text{B1u:} \\\\ \\text{B2u:} \\\\ \\text{E1u: (x, y)} \\\\ \\text{E2u:} \\end{array} & \\Gamma_{uma} = \\begin{pmatrix} 12 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 4 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 12 \\\\ 0 \\\\ 4 \\end{pmatrix} \\end{matrix}$, $\\begin{matrix} \\Gamma_{trans} = (CD6h^T)^{} + (CD6h^T)^{<11>} & \\Gamma_{rot} = (CD6h^T)^{<2>} + (CD6h^T)^{<5>} & \\Gamma_{tot} = \\overrightarrow{( \\Gamma_{uma} \\Gamma_{trans})} \\end{matrix}$, $\\begin{matrix} \\Gamma_{vib} = \\Gamma_{rot} - \\Gamma_{trans} - \\Gamma_{rot} & i = 1 .. 12 & \\text{Vib}_i = \\frac{ \\sum \\overrightarrow{ \\left[ D6h (CD6h^T )^{ } \\Gamma_{vib} \\right] }}{h} \\end{matrix}$, $\\text{Vib}^T = \\begin{pmatrix} 2 & 1 & 0 & 2 & 1 & 4 & 0 & 1 & 2 & 2 & 3 & 2 \\end{pmatrix}$, $\\Gamma_{vib} = 2A_{1g} + A_{2g} + 2B_{2g} + E_{1g} + 4E_{2g} + A_{2u} + 2B_{1u} + 1B_{2u} + 3E_{1u} + 2E_{2u}$. ΔR = 0 “vertical transitions” ΔP = 0 no change iktin momentum. The total change in energy associated with a molecular transition (emission or absorption), can be described by the following: = − = In this equation, is the energy of the photon which is equal to the difference in energy associated with the molecular", "0 & -x \\\\ -y & x & 0 \\\\ \\end{bmatrix} \\\\ &= \\begin{bmatrix} -y^2 - z^2 & xy & xz \\\\ xy & -x^2 - z^2 & yz \\\\ xz & yz & -x^2 - y^2 \\\\ \\end{bmatrix} \\\\ \\end{align} Therefore, the rotation matrix could be expressed using $\\mathbf{q} = (x, y, z, w)$. \\begin{align} \\mathbf{R}(\\hat{\\mathbf{n}}, \\theta) &= \\mathbf{I} + 2w \\left( [\\hat{\\mathbf{n}}]_{\\times} \\sin \\frac{\\theta}{2} \\right) + 2 \\left( [\\hat{\\mathbf{n}}]_{\\times} \\sin \\frac{\\theta}{2} \\right)^2 \\\\ &= \\mathbf{I} + 2w [\\hat{\\mathbf{v}}]_{\\times} + 2 [\\hat{\\mathbf{v}}]_{\\times}^2 \\\\ &= \\begin{bmatrix} 1 - 2(y^2 + z^2) & 2(xy - zw) & 2(xz + yw) \\\\ 2(xy + zw) & 1 - 2(x^2 + z^2) & 2(yz- xw) \\\\ 2(xz - yw) & 2(yz + xw) & 1 - 2(x^2 + y^2) \\\\ \\end{bmatrix} \\\\ \\end{align} Formally, the rotation matrix $\\mathbf{R}(\\mathbf{q})$ could be expressed as \\begin{align} \\mathbf{R}(\\mathbf{q}) &= \\begin{bmatrix} 1 - 2(y^2 + z^2) & 2(xy - zw) & 2(xz + yw) \\\\ 2(xy + zw) & 1 - 2(x^2 + z^2) & 2(yz- xw) \\\\ 2(xz - yw) & 2(yz + xw) & 1 - 2(x^2 + y^2) \\\\ \\end{bmatrix} \\\\ \\end{align} Notice that \\begin{align} 1 - 2(y^2 + z^2) &= \\left( 1 - (y^2 + z^2) \\right) - (y^2 + z^2) \\\\ &= \\left( 1 - \\sin^2 \\frac{\\theta}{2}(\\hat{n}_2^2 + \\hat{n}_3^2) \\right) -", "2-i\\\\ -i\\\\ -1+2i\\end{array}\\right)$$\\mathbf{x}=\\begin{pmatrix} 1 \\\\ 2-i \\\\ -i \\\\ -1+2i \\end{pmatrix}$ We need to find the column vector $\\mathbf{X}$$\\mathbf{X}$. The general method is already shown on the Wikipedia page. But we will develop a matrix notation for the same. $\\mathbf{X}$$\\mathbf{X}$ can be easily obtained by pre multiplying $\\mathbf{x}$$\\mathbf{x}$ by the matrix: $M=\\frac{1}{\\sqrt{N}}\\left(\\begin{array}{cccc}1& 1& 1& 1\\\\ 1& w& {w}^{2}& {w}^{3}\\\\ 1& {w}^{2}& {w}^{4}& {w}^{6}\\\\ 1& {w}^{3}& {w}^{6}& {w}^{9}\\end{array}\\right)$ where $w$$w$ is ${e}^{\\frac{-2\\pi i}{N}}$$e^{\\frac{-2\\pi i}{N}}$. Each element of the matrix is basically ${w}^{ij}$$w^{ij}$. $\\frac{1}{\\sqrt{N}}$$\\frac{1}{\\sqrt{N}}$ is simply a normalization constant. Finally, $\\mathbf{X}$$\\mathbf{X}$ turns out to be: $\\frac{1}{2}\\left(\\begin{array}{c}2\\\\ -2-2i\\\\ -2i\\\\ 4+4i\\end{array}\\right)$$\\frac{1}{2}\\begin{pmatrix} 2 \\\\ -2-2i \\\\ -2i \\\\ 4+4i \\end{pmatrix}$. Now, sit back for a while and notice a few important properties: • All the columns of the matrix $M$$M$ are orthogonal to each other. • All the columns of $M$$M$ have magnitude $1$$1$. • If you post multiply $M$$M$ with a column vector having lots of zeroes (large spread) you'll end up with a column vector with only a few zeroes (narrow spread). The converse also holds true. (Check!) It can be very simply noticed that the classical DFT has a time complexity $\\mathcal{O}\\left({N}^{2}\\right)$$\\mathcal O(N^2)$. That is because for obtaining every row of $\\mathbf{X}$$\\mathbf{X}$, $N$$N$ operations need to be performed. And there are $N$$N$ rows in $\\mathbf{X}$$\\mathbf{X}$. ## The Fast fourier transform:", "R^{-1}= \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix}$$ • #### Result The given matrix can be factorized into Jordan normal form as $$\\mathbf A= \\begin{pmatrix} 1 & 1 & 1\\\\ 0 & 1 & 0\\\\ -1 & 0 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 2\\\\ 0 & 0 & -1\\\\ 1 & 1 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 1 & 0\\\\ 0 & 2 & 0\\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 2 & 0\\\\ -1 & -1 & 1\\\\ 0 & -1 & 0 \\end{pmatrix} = \\mathbf R\\mathbf J\\mathbf R^{-1}$$ • #### Variant $$\\mathbf A= \\begin{pmatrix} 2 & 1 & 0 & 2\\\\ 0 & 2 & 0 & -2\\\\ 0 & 0 & 2 & 2\\\\ 0 & 0 & 0 & 2 \\end{pmatrix}$$ • #### Solution The matrix has a single eigenvalue $$\\lambda = 2$$ of algebraicmultiplicity 4, eigenvectors are nontrivial solutions of a system of equation with matrix $\\mathbf A-\\lambda \\mathbf I = \\begin{pmatrix} 0&1&0&2\\\\ 0&0&0&-2\\\\ 0&0&0&2\\\\ 0&0&0&0 \\end{pmatrix} \\sim \\begin{pmatrix} 0&1&0&0\\\\ 0&0&0&1\\\\ \\end{pmatrix}$ The dimension of $$\\ker(\\mathbf A-\\lambda \\mathbf I)$$ is 2, so the matrix $$\\mathbf A$$ has two linearly independent eigenvectors, e.g. $$\\mathbf x_1=(0{,}0,1{,}0)^T$$ and $$\\mathbf x_2=(1{,}0,0{,}0)^T$$ To get a regular matrix $$\\mathbf R$$ we", "each of the singular values of $$\\mathbf{A}$$ twice, along two equidistant diagonals off its main diagonal. We are going to leverage this structure to diagonalize $$\\mathbf{C}$$. For this, define $$\\Pi$$ as a permutation matrix such that $$\\Pi \\mathbf{C} \\Pi = \\begin{pmatrix} \\mathbf{S}_1 & & & & \\\\ & \\mathbf{S}_2 & & & \\\\ & & \\ddots & & \\\\ & & & \\mathbf{S}_l & \\\\ & & & &\\mathbf{0} \\\\ \\end{pmatrix}$$ is a block diagonal matrix where $$\\mathbf{S}_j = \\begin{pmatrix} 0 & \\sigma_j \\\\ \\sigma_j & 0 \\end{pmatrix} \\in \\mathbb{R}^{2\\times2}$$, $$l = \\min{(m, n)}$$, and $$1 \\leq j \\leq l$$. Hence, diagonalizing $$\\mathbf{C}$$ is equivalent to diagonalizing each of the $$\\mathbf{S}_j$$ blocks. For this purpose, notice that $$\\mathbf{S}_j$$ has two eigenvalues (found from its characteristic equation $$(\\mathbf{S}_j – \\lambda\\mathbf{I}) = 0$$): $$\\lambda = \\pm\\sigma_j$$. Furthermore, the corresponding eigenvectors are $$\\frac{1}{\\sqrt{2}}\\begin{bmatrix}1 \\\\ 1\\end{bmatrix}$$ and $$\\frac{1}{\\sqrt{2}}\\begin{bmatrix}1 \\\\ -1\\end{bmatrix}$$ for $$\\lambda = \\sigma_j$$ and $$\\lambda = -\\sigma_j$$, respectively. The diagonalization of $$\\mathbf{S}_j$$ is then given by: $$\\mathbf{S}_j = \\begin{pmatrix} 0 & \\sigma_j \\\\ \\sigma_j & 0 \\end{pmatrix} = \\left(\\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 & 1 \\\\ 1 & -1 \\end{pmatrix} \\right) \\begin{pmatrix} \\sigma_j & 0 \\\\ 0 & -\\sigma_j \\end{pmatrix} \\left(\\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 & 1 \\\\ 1 & -1 \\end{pmatrix} \\right)^T = \\tau \\Lambda_j \\tau^T$$ where $$\\tau^T\\tau = \\tau\\tau^T = \\mathbf{I}$$. This result", "$\\lambda_{\\min} \\left( DD^{T} \\right)$ instead of relying on iterative method. Otherwise, having explicit formula for $\\lambda_{\\min} \\left( D_{1}D_{1}^{T} \\right)$ and $\\lambda_{\\min} \\left( D_{2}D_{2}^{T} \\right)$ would also be really nice. ## UPDATE I derived the exact formula of $D_{1}D_{1}^{T}$ and $D_{2}D_{2}^{T}$ when $m,n > 2$. Hope that it would be helpful. 1. $D_{1}D_{1}^{T}$ is a $(m-1)n \\times (m-1)n$ matrix contains $n$-blocks in the main diagonal $$D_{1}D_{1}^{T} = \\begin{pmatrix} B \\\\ & B \\\\ & & \\ddots \\\\ & & & B \\end{pmatrix}$$ where $B$ is a $(m-1) \\times (m-1)$ matrix defined as $$B = \\begin{pmatrix} 2 & -1 \\\\ -1 & 2 & -1 \\\\ & \\ddots & \\ddots & \\ddots \\\\ & & -1 & 2 & -1 \\\\ & & & -1 & 2 \\end{pmatrix} .$$ 2. $D_{2}D_{2}^{T}$ is a $m(n-1) \\times m(n-1)$ matrix with $m$ diag contains $2$ in the main diagonal and $-1$ at the $m^{th}$-upper and lower diagonal $$D_{2}D_{2}^{T} \\left[ r , c \\right] = \\begin{cases} 2 & \\textrm{ if } r = c , \\\\ -1 & \\textrm{ if } \\left\\lvert r-c \\right\\rvert = m , \\\\ 0 & \\textrm{ otherwise } . \\end{cases}$$ • Are you using the notation $A[i;j]$ to mean $A_{ij}$? Could you make the definitions of $D_1$ and D_2$clearer, please? – Ben McKay Jun 9 '18 at 14:17 •", "0 & 2(r_wu_z - r_zu_w) \\\\ 0 & 0 & 0 & 0\\end{bmatrix}$$ . Then the corresponding 4×4 matrix $$\\mathbf M$$ that transforms a point $$\\mathbf p$$, regarded as a column matrix, as $$\\mathbf p' = \\mathbf{Mp}$$ is given by $$\\mathbf M = \\mathbf A + \\mathbf B$$ . The inverse of $$\\mathbf M$$, which transforms a plane $$\\mathbf f$$, regarded as a row matrix, as $$\\mathbf f' = \\mathbf{fM^{-1}}$$ is given by $$\\mathbf M^{-1} = \\mathbf A - \\mathbf B$$ . ## Conversion from Matrix to Motor Let $$\\mathbf M$$ be an orthogonal 4×4 matrix with determinant +1 having the form $$\\mathbf M = \\begin{bmatrix} M_{00} & M_{01} & M_{02} & M_{03} \\\\ M_{10} & M_{11} & M_{12} & M_{13} \\\\ M_{20} & M_{21} & M_{22} & M_{23} \\\\ 0 & 0 & 0 & 1\\end{bmatrix}$$ . Then, by equating the entries of $$\\mathbf M$$ to the entries of $$\\mathbf A + \\mathbf B$$ from above, we have the following four relationships based on the diagonal entries of $$\\mathbf M$$: $$M_{00} - M_{11} - M_{22} + 1 = 4r_x^2$$ $$M_{11} - M_{22} - M_{00} + 1 = 4r_y^2$$ $$M_{22} - M_{00} - M_{11} + 1 = 4r_z^2$$ $$M_{00} + M_{11} + M_{22} + 1 = 4(1 - r_x^2 - r_y^2 - r_z^2) = 4r_w^2$$ And we have", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you" ]

[ "+0 # Pls help me +1 32 1 +57 The vectors $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}$ satisfy $\\|\\mathbf{a}\\| = \\|\\mathbf{b}\\| = 1,$ $\\|\\mathbf{c}\\| = 2,$ and $\\mathbf{a} \\times (\\mathbf{a} \\times \\mathbf{c}) + \\mathbf{b} = \\mathbf{0}.$If $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{c},$ then find all possible values of $\\theta,$ in degrees. Apr 25, 2021", "Vectors $\\mathbf{a}$ and $\\mathbf{b}$ must be going in the same direction.", "If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.", "views Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy$$ \\cos \\ ... Let $\\mathbf{r}_{0}=\\left(x_{0}, y_{0}\\right)$ be a fixed vector in $R^{2}$. In each part, describe in words the set of all vectors $\\mathbf{r}=(x, y)$ that satisfy the stated condition.Let $\\mathbf{r}_{0}=\\left(x_{0}, y_{0}\\right)$ be a fixed vector in $R^{2}$. In each part, describe in words the set of all vectors \\$\\mathbf{r}=(x, y) ...", "and $C$ have position vectors $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ respectively.", "Vectors Question #2 If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.", "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "## Angle between two vectors Determine the angle between the following pairs of real vectors (if possible, determine the exact value, otherwise the cosine). Decide whether the angle is acute or obtuse. • #### Variant 1 $$\\mathbf x^T=(1, -4)$$, $$\\mathbf y^T=(8, 2)$$. • #### Variant 2 $$\\mathbf x^T=(3, 2, -2)$$, $$\\mathbf y^T=(0, 4, 1)$$. • #### Variant 3 $$\\mathbf x^T=(0, 0, 1)$$, $$\\mathbf y^T=(1, 0, -1)$$. • #### Variant 4 $$\\mathbf x^T=(3, 4)$$, $$\\mathbf y^T=(-1, 0)$$.", "Browse Questions # Find the angle between two vectors $\\overrightarrow a \\: and \\: \\overrightarrow b$ if $|\\overrightarrow a |=3, |\\overrightarrow b|=4\\: and \\: |\\overrightarrow a * \\overrightarrow b |=6$ Can you answer this question?", "Review question # Can we find the angle between the planes $ABC$ and $ABD$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R6449 ## Question The position vectors of four points $A$, $B$, $C$, $D$ relative to an origin $O$ are given below. The vectors $\\mathbf{i}$, $\\mathbf{j}$, $\\mathbf{k}$ are mutually perpendicular unit vectors. \\begin{align*} A \\colon\\;& 2\\mathbf{i} + 3\\mathbf{j} + \\mathbf{k}, \\\\ B \\colon\\;& \\phantom{2}\\mathbf{i} + \\phantom{3}\\mathbf{j} - \\mathbf{k}, \\\\ C \\colon\\;& \\phantom{2}\\mathbf{i} \\phantom{{}+{}3\\mathbf{i}} + \\mathbf{k}, \\\\ D \\colon\\;& \\phantom{2\\mathbf{i}{}+{}} 3\\mathbf{j}.\\phantom{{}+{}\\mathbf{k}} \\end{align*} Find 1. the equation (in any form) of the line $AB$, 2. the shortest distance between $AB$ and $CD$, 3. the equation (in any form) of the plane $ABC$, 4. the angle between the planes $ABC$ and $ABD$.", "The bearing angle between two vectors? What is the best way of showing the angle between two vectors defined by start and end points? E.g., Vector 1 points from point A to point B. Vector 2 points from point C to point D. I have tried: $\\angle (\\vec{AB}, \\vec{CD})$ But it is ugly. In addition, the abovehead arrowhead is not long enough to cap the long expression. How may I fix this? $\\angle (\\overrightarrow{AB}, \\overrightarrow{CD})$ $\\mathbf{a} = \\vec{AB}$ $\\mathbf{b} = \\vec{CD}$ $\\alpha = \\angle (\\mathbf{a}, \\mathbf{b})$", "## Angle between two vectors ### Task number: 2689 Determine the angle between the following pairs of real vectors (if possible, determine the exact value, otherwise the cosine). Decide whether the angle is acute or obtuse. • #### Variant 1 $$\\mathbf x^T=(1, -4)$$, $$\\mathbf y^T=(8, 2)$$. • #### Variant 2 $$\\mathbf x^T=(3, 2, -2)$$, $$\\mathbf y^T=(0, 4, 1)$$. • #### Variant 3 $$\\mathbf x^T=(0, 0, 1)$$, $$\\mathbf y^T=(1, 0, -1)$$. • #### Variant 4 $$\\mathbf x^T=(3, 4)$$, $$\\mathbf y^T=(-1, 0)$$.", "# Problem: Find the angle between each of the following pairs of vectors A =Axi + Ayj and B =Bxi + ByjAx3= -4.00, Ay3 = 2.00; Bx3 = 7.00, By3 = 14.00. ###### FREE Expert Solution Vector dot product: $\\overline{)\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}{\\mathbf{·}}\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}{\\mathbf{=}}{\\mathbf{|}}{\\mathbf{A}}{\\mathbf{|}}{\\mathbf{|}}{\\mathbf{B}}{\\mathbf{|}}{\\mathbf{c}}{\\mathbf{o}}{\\mathbf{s}}{\\mathbf{\\theta }}}$ Vector magnitude: $\\overline{)\\mathbf{|}\\stackrel{\\mathbf{⇀}}{\\mathbit{A}}\\mathbf{|}{\\mathbf{=}}\\sqrt{{{\\mathbit{A}}_{\\mathbit{x}}}^{\\mathbf{2}}\\mathbf{+}{{\\mathbit{A}}_{\\mathbit{y}}}^{\\mathbf{2}}}}$ $\\begin{array}{rcl}{\\mathbf{\\theta }}_{{\\mathbf{A}}_{\\mathbf{3}}{\\mathbf{B}}_{\\mathbf{3}}}& \\mathbf{=}& {\\mathbf{cos}}^{\\mathbf{-}\\mathbf{1}}\\mathbf{\\left(}\\frac{{\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}}_{\\mathbf{3}}\\mathbf{·}{\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}}_{\\mathbf{3}}}{\\mathbf{|}{\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}}_{\\mathbf{3}}\\mathbf{|}\\mathbf{|}{\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}}_{\\mathbf{3}}\\mathbf{|}}\\mathbf{\\right)}\\end{array}$ ###### Problem Details Find the angle between each of the following pairs of vectors =Ax+ Ayj and B =Bx+ Byj Ax3= -4.00, Ay3 = 2.00; Bx3 = 7.00, By3 = 14.00.", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "$$\\|\\mathbf{u}+\\mathbf{v}\\|^{2}=\\|\\mathbf{u}\\|^{2}+\\|\\mathbf{v}\\|^{2}$$If$\\mathbf{u}$and$\\mathbf{v}$are orthogonal vectors in$R^{n}$with the Euclidean inner product, then prove that$\\$ \\|\\mathbf{u}+\\mathbf{v}\\|^{2}= ...", "Find $\\mathbf{a} \\cdot \\mathbf{b}$ if $|\\mathbf{a}| = 6$, $|\\mathbf{b}| = 6$, and the angle between $\\mathbf{a}$ and $\\mathbf{b}$ is $\\pi/2$.", "## Trigonometry (11th Edition) Clone $\\textbf{p}$ and $\\textbf{q}$ $\\textbf{r}$ and $\\textbf{s}$ $\\textbf{n}$ and $\\textbf{s}$ Two vectors are opposites when they have the same magnitude but different directions. According to this definition, the pair of vectors that appear to be opposite are $\\textbf{p}$ and $\\textbf{q}$,$\\textbf{r}$ and $\\textbf{s}$ and $\\textbf{n}$ and $\\textbf{s}$.", "# angles_points_xy compas.geometry.angles_points_xy(a, b, c, deg=False)[source] Compute the two angles between the two vectors defined by the XY components of three points. Parameters Returns • float – The smallest angle in radians, or in degrees if deg == True. • float – The other angle. Notes The vectors are defined in the following way $\\begin{split}\\mathbf{u} = \\mathbf{b} - \\mathbf{a} \\\\ \\mathbf{v} = \\mathbf{c} - \\mathbf{a}\\end{split}$ Z components may be provided, but are simply ignored. Examples >>>", "the magnitudes of the vectors times the sine of the angle between the vectors and it is a vector perpendicular to ${\\bf a}$ and ${\\bf b}$." ]

[ "$\\cos\\theta_1=\\mathbf{n}\\cdot(-\\mathbf{l})$ $\\cos\\theta_2=\\sqrt{1-\\left(\\frac{n_1}{n_2}\\right)^2\\left(1-\\left(\\cos\\theta_1\\right)^2\\right)}$ $\\mathbf{v}_{\\mathrm{reflect}}=\\mathbf{l}-\\left(2\\cos\\theta_1\\right)\\mathbf{n}$ $\\mathbf{v}_{\\mathrm{refract}}=\\left(\\frac{n_1}{n_2}\\right)\\mathbf{l} + \\left( \\frac{n_1}{n_2}\\cos\\theta_1 - \\cos\\theta_2\\right)\\mathbf{n}$ Note: $\\mathbf{n}\\cdot(-\\mathbf{l})$ must be positive. Otherwise, use $\\mathbf{v}_{\\mathrm{refract}}=\\left(\\frac{n_1}{n_2}\\right)\\mathbf{l} + \\left(\\frac{n_1}{n_2}\\cos\\theta_1 + \\cos\\theta_2\\right)\\mathbf{n}.$ Example: $\\mathbf{l}=\\{0.707107, -0.707107\\},~\\mathbf{n}=\\{0,1\\},~\\frac{n_1}{n_2}=1.1$ $\\mathbf{~}\\cos\\theta_1=0.707107,~\\cos\\theta_2=0.62849$ $\\mathbf{v}_{\\mathrm{reflect}}=\\{0.707107, 0.707107\\} ,~\\mathbf{v}_{\\mathrm{refract}}=\\{0.777817, -0.62849\\}$ The cosines may be recycled and used in the Fresnel equations for working out the intensity of the resulting rays. During total internal reflection an evanescent wave is produced, which rapidly decays from the surface into the second medium. Conservation of energy is maintained by the circulation of energy across the boundary, averaging to zero net energy transmission. Dispersion In many wave-propagation media, wave velocity changes with frequency or wavelength of the waves; this is true of light propagation in most transparent substances other than a vacuum. These media are called dispersive. The result is that the angles determined by Snell's law also depend on frequency or wavelength, so that a ray of mixed wavelengths, such as white light, will spread or disperse. Such dispersion of light in glass or water underlies the origin of rainbows, in which different wavelengths appear as different colors. In optical instruments, dispersion leads to chromatic aberration, a color-dependent blurring that sometimes is the resolution-limiting effect. This was especially true in refracting telescopes, before the invention of achromatic objective lenses. 5. Kwan, A., Dudley, J., and Lantz, E. (2002). \"Who really discovered Snell's law?\".", "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "number $c$, namely one satisfying, $$(A-cB) \\cdot B = 0$$ We then get $$\\textbf{A} \\cdot \\textbf{B} = c\\textbf{B} \\cdot \\textbf{B}$$ and therefore $c=\\frac{ \\textbf{A} \\cdot \\textbf{B}}{\\textbf{B} \\cdot \\textbf{B}}$. From this construction and use of what we know about planee geometry, $$\\cos(\\theta) = \\frac{c||\\textbf{B}||}{||\\textbf{A}||}$$ and substituting in our value for $c$ we get, $$A \\cdot B = ||\\textbf{A}|| ||\\textbf{B}|| \\cos(\\theta)$$ \\ $\\textbf{Picture for our Construction}$ If $v=\\|v\\|\\cdot(\\cos\\theta_1,\\sin\\theta_1)$ and $w=\\|w\\|\\cdot(\\cos\\theta_2,\\sin\\theta_2)$, then $$\\begin{eqnarray} \\frac{v\\cdot w}{\\|v\\| \\|w\\|}&=&(\\cos\\theta_1,\\sin\\theta_1)\\cdot (\\cos\\theta_2,\\sin\\theta_2)\\\\ &=& \\cos\\theta_1 \\cos\\theta_2 +\\sin\\theta_1 \\sin\\theta_2 \\\\&=&\\cos\\left(\\theta_2 - \\theta_1\\right); \\end{eqnarray}$$ and $\\theta_2-\\theta_1$ is the angle between $v$ and $w$. Suppose you have two unit vectors $\\hat a$ and $\\hat b$ and that the angle between them is $\\pi/2$. Then imagine the unit circle, with $\\theta = 0$ corresponding to the $\\hat a$ direction and $\\theta = \\pi/2$ corresponding to the $\\hat b$ direction. Now, suppose there is a unit vector $\\hat c$ in the plane spanned by $\\hat a, \\hat b$. It lies on the unit circle, so you can write $\\hat c = \\hat a \\cos \\theta + \\hat b \\sin \\theta$. Now then, how can you solve for $\\cos \\theta$? By taking a dot product: see, since $\\hat a \\perp \\hat b$, you know that $\\hat a \\cdot \\hat b = 0$. So you solve for the cosine as follows: $$\\hat c \\cdot", "}$, which option is closest to the exact value of $\\mathbf{u}\\cdot \\mathbf{v}$ ? Exactly one option must be correct) a) 4.5 b) 6.2 c) 4.2 d) 5.1 Choice (a) is incorrect Choice (b) is incorrect Choice (c) is correct! Since $\\mathbf{u}\\cdot \\mathbf{v}=||\\mathbf{u}||\\phantom{\\rule{0.3em}{0ex}}||\\mathbf{v}||cos\\theta$, where $\\theta$ is the angle between the vectors when placed tail to tail, we have $\\mathbf{u}\\cdot \\mathbf{v}=2×3×cos4{5}^{\\circ }\\approx 4.24.$ Choice (d) is incorrect What is the approximate angle between $\\mathbf{a}$ and $\\mathbf{b}$ if $\\mathbf{a}\\cdot \\mathbf{b}=3$, $||\\mathbf{a}||=2$, and $||\\mathbf{b}||=2.6$? Exactly one option must be correct) Choice (a) is incorrect Choice (b) is incorrect Choice (c) is incorrect Choice (d) is correct! If $\\theta$ is the required angle, then $cos\\theta =\\frac{\\mathbf{a}\\cdot \\mathbf{b}}{||\\mathbf{a}||\\phantom{\\rule{0.3em}{0ex}}||\\mathbf{b}||}=\\frac{3}{5.2}\\approx 0.577$ and hence $\\theta \\approx 0.955$ radians. Find non-zero scalars $\\alpha$, $\\beta$ such that for all vectors $\\mathbf{a}$ and $\\mathbf{b}$, $\\alpha \\left(\\mathbf{a}+2\\mathbf{b}\\right)-\\beta \\mathbf{a}+\\left(4\\mathbf{b}-\\mathbf{a}\\right)=\\mathbf{0}.$ Exactly one option must be correct) a) $\\alpha =2,\\phantom{\\rule{0.3em}{0ex}}\\beta =1$ b) $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =-3$ c) $\\alpha =1,\\phantom{\\rule{0.3em}{0ex}}\\beta =3$ d) $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =3$ Choice (a) is incorrect Choice (b) is correct! If the vector equation is simplified, we get $\\left(\\alpha -\\beta -1\\right)\\mathbf{a}+\\left(2\\alpha +4\\right)\\mathbf{b}=\\mathbf{0}.$ Since this holds for all $\\mathbf{a}$ and $\\mathbf{b}$, it will hold when $\\mathbf{a}$ and $\\mathbf{b}$ are set equal to $\\mathbf{0}$ in turn. This gives the two conditions $\\alpha -\\beta -1=0$ and $2\\alpha +4=0$, whose solution is $\\alpha =-2,\\phantom{\\rule{0.3em}{0ex}}\\beta =-3.$ Choice (c)", "3. You should know that $\\mathbf{a}\\cdot\\mathbf{b} = |\\mathbf{a}||\\mathbf{b}|\\cos{\\theta}$, where $\\theta$ is the angle between them. If $\\mathbf{x} = 6\\mathbf{i} + 3\\mathbf{j} -2\\mathbf{k}$ then $|\\mathbf{x}| = \\sqrt{6^2 + 3^2 + (-2)^2}$ $= \\sqrt{36 + 9+ 4}$ $= \\sqrt{49}$ $= 7$. If $\\mathbf{y} = -2\\mathbf{i} + t\\mathbf{j} - 4\\mathbf{k}$ then $|\\mathbf{y}| = \\sqrt{(-2)^2 + t^2 + (-4)^2}$ $= \\sqrt{4 + t^2 + 16}$ $= \\sqrt{t^2 + 20}$. $\\mathbf{x}\\cdot \\mathbf{y} = (6\\mathbf{i} + 3\\mathbf{j} -2\\mathbf{k})\\cdot (-2\\mathbf{i} + t\\mathbf{j} - 4\\mathbf{k})$ $= -12 + 3t + 8$ $= 3t - 4$. Since $\\mathbf{x} \\cdot \\mathbf{y} = |\\mathbf{x}||\\mathbf{y}|\\cos{\\theta}$ $3t - 4 = 7\\sqrt{t^2 + 20}\\cos{\\left(\\arccos{\\frac{4}{21}}\\right)}$ $3t - 4 = 7\\sqrt{t^2 + 20}\\cdot \\frac{4}{21}$ $3t - 4 = \\frac{4}{3}\\sqrt{t^2 + 20}$ $\\frac{3}{4}(3t - 4) = \\sqrt{t^2 + 20}$ $\\frac{9}{4}t - 3 = \\sqrt{t^2 + 20}$ $\\left(\\frac{9}{4}t - 3\\right)^2 = t^2 + 20$ $\\frac{81}{16}t^2 - \\frac{27}{2}t + 9 = t^2 + 20$ $\\frac{65}{16}t^2 - \\frac{27}{2}t - 11 = 0$. Solve for $t$. 5. Originally Posted by Paymemoney Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds? P.S Seeing as what you have posted is a unit of speed and not a unit of time, you can't convert it into minutes OR seconds. 6. Originally Posted by Prove It The force is the vector that has magnitude $84$ and is", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "we can write $$D=D^\\prime + |DD^\\prime| \\frac{( C-B )\\times ( B \\times C )}{|BC|\\,|B\\times C|} \\tag{1}$$ where $$D^\\prime := \\frac12(B+C)$$ is the midpoint of $$\\overline{BC}$$. Further, by the Inscribed Angle Theorem, $$\\angle BOC\\cong\\angle BDC$$, and we conclude that $$\\overline{DD^\\prime}$$ is the altitude of an isosceles triangle with vertex angle $$\\alpha$$ and base $$|BC|$$. Therefore, $$|DD^\\prime| = \\frac12|BC|\\,\\cot\\frac12\\alpha$$, so we have $$\\frac{|DD^\\prime|}{|BC|\\,|B\\times C|} = \\frac{\\cot\\frac12\\alpha}{2bc\\sin\\alpha} = \\frac{\\cos\\frac12\\alpha\\,/\\,\\sin\\frac12\\alpha}{4bc\\sin\\frac12\\alpha\\cos\\frac12\\alpha}=\\frac{1}{4bc\\sin^2\\frac12\\alpha} = \\frac{1}{2bc(1-\\cos\\alpha)} \\tag{2}$$ Further, via a cross product identity, \\begin{align} (C-B)\\times(B\\times C) &= \\phantom{-}B\\,((C-B)\\cdot B) - C\\,((C-B)\\cdot C) \\\\ &=\\phantom{-}B\\left(B\\cdot C - |B|^2\\right)-C\\left(|C|^2 - B\\cdot C\\right) \\\\ &=-B\\,b(b-c\\cos\\alpha) - C\\,c(c-b\\cos\\alpha) \\end{align}\\tag{3} Altogether, this gives us \\begin{align}D\\;2bc(1-\\cos\\alpha) &= (B+C)bc(1-\\cos\\alpha)-B\\,b(b-c\\cos\\alpha)-C\\,c(c-b\\cos\\alpha) \\\\ &=Bb(c-b)+Cc(b-c) \\\\ &=(c-b)(B b-C c) \\tag{4a} \\end{align} (Note that, if $$b=c$$, then $$D=0$$, as we would expect. This confirms that we got our cross-product vector directions correct in $$(1)$$.) Likewise, \\begin{align} E\\,2ca(1-\\cos\\beta) &= (a-c)(C c-A a) \\tag{4b} \\\\ F\\,2ab(1-\\cos\\gamma) &= (b-a)(A a-B b) \\tag{4c} \\end{align} Finally, observe that, for $$a$$, $$b$$, $$c$$ not all equal (the only case with which we need concern ourselves), $$(\\text{eq }4a)\\;(a-c)(b-a) + (\\text{eq }4b)\\;(c-b)(b-a)+(\\text{eq }4c)\\;(a-c)(c-b) \\tag{5}$$ gives a non-trivial linear combination of $$D$$, $$E$$, $$F$$ that vanishes. Consequently, $$D$$, $$E$$, $$F$$ are linearly dependent; that is, they lie on a common plane through $$O$$, and the result follows. $$\\square$$ • Pardon my ignorance but I'm confused as to", "and }y=3.$ Code: B * : | * : | * y | * : | * : | @ * R *-----------------* : - - - x - - - A Blunt Rock is at $\\,R.$ $\\text{Ship A is at }\\,A\\!:\\;x = AR,\\;\\dfrac{dx}{dt} = \\text{-}6\\text{ kph}$ $\\text{Ship B is at }\\,B\\!:\\;y = BR,\\;\\dfrac{dy}{dt} = 10\\text{ kph}$ $\\,\\theta \\,=\\,\\angle A$ Note: .when $x = 4,\\;y = 3\\!:\\;\\cos\\theta \\:=\\:\\frac{4}{5}$ $\\text{We have: }\\;\\tan\\theta \\:=\\:\\dfrac{y}{x}$ $\\text{Then: }\\;\\sec^2\\!\\theta\\,\\dfrac{d\\theta}{dt} \\;=\\;\\dfrac{x\\frac{dy}{dt} - y \\frac{dx}{dt}}{x^2}$ . . . . . . . . . . $\\dfrac{d\\theta}{dt} \\;=\\;\\dfrac{x\\frac{dy}{dt} - y\\frac{x}{dt}}{x^2}\\cos^2\\!\\theta$ . . . . . . . . . . $\\dfrac{d\\theta}{dt} \\;=\\;\\dfrac{4(10) - 3(\\text{-}15)}{4^2}\\cdot \\left(\\frac{4}{5}\\right)^2$ . . . . . . . . . . $\\displaystyle \\frac{d\\theta}{dt} \\;=\\;\\frac{85}{16}\\cdot\\frac{16}{25} \\;=\\;\\frac{17}{5} \\;=\\;3\\tfrac{2}{5}\\text{ radians/hour}$", "= \\mathbf{a} \\cdot \\mathbf{c} + \\mathbf{a} \\cdot \\mathbf{d} + \\mathbf{b} \\cdot \\mathbf{c} + \\mathbf{b} \\cdot \\mathbf{d}$ Consider a triangle $OAB$ with $\\mathbf{a}, \\mathbf{b}$ the position vectors of points $A$ and $B$. Let $\\theta$ be the angle $AOB$. By the cosine rule, $AB^2 = OA^2 + OB^2 - 2 \\, OA \\cdot OB \\cos \\theta$. $\\begin{gather} | \\mathbf{b}-\\mathbf{a} |^2 = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta \\tag{4} \\end{gather}$ By using formula 1, $|\\mathbf{b}-\\mathbf{a}|^2 = (\\mathbf{b}-\\mathbf{a}) \\cdot (\\mathbf{b}-\\mathbf{a})$. Expanding using formula 3 gives us $\\mathbf{b} \\cdot \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} - \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{a}$. The use of formulas 1 and 2 simplifies the expression to $|\\mathbf{b}|^2 - 2 \\mathbf{a}\\cdot\\mathbf{b} + |\\mathbf{a}|^2$. Plugging this into equation 4 gives us $|\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 \\mathbf{a} \\cdot \\mathbf{b} = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta$ $-2\\mathbf{a} \\cdot \\mathbf{b} = -2 |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$. ### Angles and perpendicular vectors The dot product formula is especially handy in two situations. The first is in the computation of angles. Given two vectors $\\mathbf{a}$ and $\\mathbf{b}$, the values of $\\mathbf{a} \\cdot \\mathbf{b}$, $|\\mathbf{a}|$ and $|\\mathbf{b}|$ are all relatively simple to compute. Substituting them into our formula allows us to calculate the angle between them. The second situation is when vectors", "{\\overline{C}}^{2}|N,u \\right\\rbrack|\\mathbf{x} \\right\\rbrack,$$ and because $$\\mathbb{E}\\left\\lbrack \\overline{C}|N,u,\\mathbf{x} \\right\\rbrack$$ $$= \\mu e^{n\\theta + u}$$ and $$Var(\\overline{C}|N,u,\\mathbf{x})$$ $$= \\phi\\mu^{2}e^{2(n\\theta + u)}/n,$$ we can get $\\mathbb{E}\\left\\lbrack {\\overline{C}}^{2}|N,u,\\mathbf{x} \\right\\rbrack = (\\phi/n + 1)\\mu^{2}e^{2(n\\theta + u)},$ \\begin{align} \\mathbb{E}\\left\\lbrack N^{2}\\mathbb{E}\\left\\lbrack {\\overline{C}}^{2}|N,u \\right\\rbrack|\\mathbf{x} \\right\\rbrack = &\\mu^{2}\\mathbb{E}\\left\\lbrack (\\phi n + n^{2})e^{2(n\\theta + u)}|\\mathbf{x} \\right\\rbrack \\\\ = &\\mu^{2}(\\phi\\mathbb{E}\\left\\lbrack M'_{N|b,\\mathbf{x}}(2\\theta) \\right\\rbrack \\\\ &+ \\frac{1}{4}\\mathbb{E}\\left\\lbrack M''_{N|b,\\mathbf{x}}(2\\theta) \\right\\rbrack)e^{2\\sigma_{u}^{2}}, \\end{align} and \\begin{align} \\mathbb{E}\\left\\lbrack N^{2}e^{{nt}}|b,\\mathbf{x} \\right\\rbrack = &M''_{N|b,\\mathbf{x}}(t) \\\\ = &\\nu^{2}e^{2b + 2t}(1 + 1/r)\\lbrack 1 - (\\nu e^{b}/r)(e^{t} - 1)\\rbrack^{- r - 2} \\\\ &+ M'_{N|b,\\mathbf{x}}(t). \\end{align} Finally, combining the expressions for $$\\mathbb{E}\\left\\lbrack S|\\mathbf{x} \\right\\rbrack$$ and $$\\mathbb{E}\\left\\lbrack S^{2}|\\mathbf{x} \\right\\rbrack,$$ we have \\begin{align} Var(S|\\mathbf{x}) = &\\mathbb{E}\\left\\lbrack S^{2}|\\mathbf{x} \\right\\rbrack - \\left( \\mathbb{E}\\left\\lbrack S|\\mathbf{x} \\right\\rbrack \\right)^{2} \\\\ = &\\mu^{2}e^{\\sigma_{u}^{2}}(\\phi\\mathbb{E}\\left\\lbrack M'_{N|b,\\mathbf{x}}(2\\theta) \\right\\rbrack e^{\\sigma_{u}^{2}} \\\\ &+ \\frac{1}{4}\\mathbb{E}\\left\\lbrack M''_{N|b,\\mathbf{x}}(2\\theta) \\right\\rbrack e^{\\sigma_{u}^{2}} \\\\ &- \\mathbb{E}\\left\\lbrack M'_{N|b,\\mathbf{x}}(\\theta) \\right\\rbrack^{2}) \\\\ = &\\mu^{2}e^{\\sigma_{u}^{2} + 2\\theta}(\\phi\\mathbb{E}\\bigl\\lbrack \\nu e^{b}\\lbrack 1 - (\\nu e^{b}/r) \\\\ &\\times (e^{2\\theta} - 1)\\rbrack^{- r - 1} \\bigr\\rbrack e^{\\sigma_{u}^{2}} \\\\ &+ \\mathbb{E}\\bigl\\lbrack \\nu^{2}e^{2b + 2\\theta}(1 + 1/r) \\\\ &\\times \\lbrack 1 - (\\nu e^{b}/r)(e^{2\\theta} - 1)\\rbrack^{- r - 2} \\bigr\\rbrack e^{\\sigma_{u}^{2}} \\\\ &+ \\mathbb{E}\\bigl\\lbrack \\nu e^{b}\\lbrack 1 - (\\nu e^{b}/r) \\\\ &\\times (e^{2\\theta} - 1)\\rbrack^{- r - 1} \\bigr\\rbrack e^{\\sigma_{u}^{2}} \\\\ &- \\mathbb{E}\\bigl\\lbrack \\nu e^{b}\\lbrack 1 - (\\nu e^{b}/r)\\\\ &\\times (e^{\\theta} - 1)\\rbrack^{- r - 1} \\bigr\\rbrack^{2}). \\end{align} \\tag{B.4} Note that if we again set b = 0 and", "i\\mathbf{q}\\cdot\\mathbf{r}}\\right) = \\frac{\\hbar}{2m}e^{+ i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)$$ and $$[\\mathcal{H},e^{- i\\mathbf{q}\\cdot\\mathbf{r}}]= -\\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{p}+\\mathbf{p}e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\right) =- \\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)e^{- i\\mathbf{q}\\cdot\\mathbf{r}}.$$ Putting either of these into the corresponding formula above, you get $$\\sum_n (\\mathcal{E_n}-\\mathcal{E_s})|\\langle n|e^{i\\mathbf{q}\\cdot\\mathbf{r}}|s \\rangle|^2 = \\frac{\\hbar}{2m}\\langle s|\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)|s \\rangle = \\frac{\\hbar^2}{2m}\\mathbf{q}^2+\\frac{\\hbar}{m}\\langle s|\\mathbf{q}\\cdot\\mathbf{p}|s \\rangle.$$ You can then always force the mean value of $\\mathbf{p}$ to vanish. - I) Let us work in the Schrödinger picture. Assume that the Hamiltonian operator $$\\tag{1} \\hat{H}~=~T(\\hat{\\bf p})+ V(\\hat{\\bf r}), \\qquad T(\\hat{\\bf p})~:=~ \\frac{\\hat{\\bf p}^2}{2m},$$ does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\\psi_n({\\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE $$\\tag{2} \\hat{H}\\psi_n({\\bf r})~=~E_n \\psi_n({\\bf r}).$$ We may assume that $\\psi_n({\\bf r})\\in{\\mathbb{R}}$ is a real function, see also this Phys.SE post. Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions $$\\tag{3} \\langle n, t |{\\bf r} \\rangle~=~ \\Psi_n({\\bf r},t)~=~ \\psi_n({\\bf r})e^{-\\frac{i}{\\hbar}E_n (t-t_0)}$$ that becomes real at the time $t=t_0$. It follows that for any (possibly complex) function $F=F({\\bf r})$ of position ${\\bf r}$, that the matrix element $$\\tag{4} \\langle n, t_0 |F(\\hat{\\bf r}) | m, t_0 \\rangle ~=~ \\int \\! d^3r ~ \\psi^{*}_n({\\bf r}) ~F({\\bf r})~\\psi_m({\\bf r})~=~ n \\leftrightarrow m$$ is symmetric in $n\\leftrightarrow m$ at $t=t_0$. The time dependence of the matrix element is just", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "f}{\\partial\\theta_2} = \\frac{\\partial f}{\\partial \\bar{\\theta_2}}\\cdot\\frac{\\partial \\bar{\\theta_2}}{\\partial \\theta_2} \\end{align} \\begin{align} \\tag{8.1.1} \\frac{\\partial \\bar{\\theta_2}}{\\partial \\theta_2} = \\log_{10} e \\cdot \\frac{1}{\\theta_2} = \\frac{1}{\\ln10\\cdot\\theta_2} \\end{align} and by equation (5) we arrive at \\begin{align} \\tag{8.2} \\frac{\\partial f}{\\partial\\theta_2} = \\frac{\\partial f}{\\partial \\bar{\\theta_2}}\\cdot\\frac{\\partial \\bar{\\theta_2}}{\\partial \\theta_2} \\end{align} \\begin{align} \\tag{8.3} = \\frac{\\theta_4 - \\theta_1}{(1 + \\bar{\\rho})^2}\\cdot \\ln(10)\\cdot\\bar{\\rho}\\cdot\\theta_3\\cdot\\frac{1}{\\ln10\\cdot\\theta_2} \\end{align} \\begin{align} =\\frac{\\theta_3}{\\theta_2}\\cdot\\frac{\\theta_4 - \\theta_1}{(1 + \\rho)^2}\\cdot \\rho \\end{align} ## Second Derivatives of $$\\bar{f}(x|\\bar{\\theta})$$ Due to the equivalence of equation (0.4), we do not need to compute all the the partial derivatives, only 10 of the 16 combinations. Again, I will start with equation (3) and then show the differences with equation (1) when applicable. This only is necessary when working with $$\\bar{\\theta_2}$$ which would otherwise be $$\\theta_2$$ if using equation (1). Again, to make computation easier lets remember the partial derivatives of $$\\bar{\\rho}$$, specifically equation (3.2) and (3.3). for brevity, I will refer to equation (3) via $$f$$ instead of $$\\bar{f}$$ ### Derivative: $$\\frac{\\partial}{\\partial \\theta_1}[\\frac{\\partial f}{\\partial \\theta_1}]$$ \\begin{align} \\tag{8} \\frac{\\partial^2 f}{\\partial \\theta_1^2} = 0 \\end{align} proof \\begin{align} \\tag{8.1} \\frac{\\partial^2 f}{\\partial \\theta_1^2} = \\frac{\\partial}{\\partial \\theta_1}[\\frac{\\partial f}{\\partial \\theta_1}] \\end{align} \\begin{align} \\tag{8.2} = \\frac{\\partial}{\\partial \\theta_1}[1 - \\frac{1}{1 + \\bar{\\rho}}] \\end{align} \\begin{align} \\tag{8.3} = 0 \\end{align} ### Derivative: $$\\frac{\\partial}{\\partial \\theta_1}[\\frac{\\partial f}{\\partial \\bar{\\theta_2}}]$$ \\begin{align} \\tag{9} \\frac{\\partial^2 f}{\\partial \\theta_1\\partial\\bar{\\theta_2}} = \\frac{\\partial \\bar{\\rho}}{\\partial\\bar{\\theta_2}}\\cdot\\frac{1}{(1+\\bar{\\rho})^2} \\end{align} proof \\begin{align} \\tag{9.1} \\frac{\\partial^2 f}{\\partial \\theta_1\\partial\\bar{\\theta_2}} = \\frac{\\partial}{\\partial \\theta_1}[\\frac{\\partial f}{\\partial \\bar{\\theta_2}}] \\end{align}", "u_B =-\\frac1{1+\\mu}v_A\\mathbf i.$$ If in the picture above $A$ is the ball on the left and $B$ is the ball on the right, after the collision we have velocities \\begin{align}\\mathbf u_A' &=-\\frac\\mu{1+\\mu}v_A\\cos(2\\theta)\\mathbf i - \\frac\\mu{1+\\mu}v_A\\sin(2\\theta)\\mathbf j,\\\\\\mathbf u_B'&=\\frac1{1+\\mu}v_A\\mathbf\\cos(2\\theta)\\mathbf i + \\frac1{1+\\mu}v_A\\mathbf\\sin(2\\theta)\\mathbf j.\\end{align} The last step is to bring everything back to the original frame by adding the velocity of the centre of mass back to all the computed velocities, thus obtaining \\begin{align}\\mathbf v_A' &=\\left(\\frac1{1+\\mu}-\\frac\\mu{1+\\mu}\\cos(2\\theta)\\right)v_A\\mathbf i-\\frac\\mu{1+\\mu}v_A\\sin(2\\theta)\\mathbf j,\\\\\\mathbf v_B'&=\\frac1{1+\\mu}\\left(1+\\cos(2\\theta)\\right)v_A\\mathbf i +\\frac1{1+\\mu}v_A\\mathbf\\sin(2\\theta)\\mathbf j.\\end{align} One easily sees that the angle of ball $B$ is the same as in the case $\\mu=1$, since $$\\tan(\\theta_B) = \\frac{\\sin(2\\theta)}{1+\\cos(2\\theta)} =\\tan\\theta,$$ while the angle of ball $A$ now depends on $\\mu$ through $$\\tan(\\theta_A) = \\frac{\\sin(2\\theta)}{\\frac1\\mu-\\cos(2\\theta)}.$$ One can check that, for $\\mu=1$, the expected result is obtained back, namely $$\\mu = 1\\quad\\Rightarrow\\quad\\tan(\\theta_A) = \\cot(\\theta).$$ Remark It goes without saying that those angles must be corrected in order to fall into the right quadrant, so one has to take the signs of the component of the velocities into account as well, and this is not done in the above formulae. The above steps show how passing to the centre of mass drastically simplifies the original problem. Here is the solution of $\\theta_A$ vs $\\theta$ for the case $\\mu=3$. At $\\theta = 60^\\circ$ the angle $\\theta_A$ is roughly $46^\\circ$, while at", "1$. Prove that: $$\\sqrt{(a^2+b^2+2abx)(c^2+d^2+2cdx)}+\\sqrt{(a^2+d^2-2adx)(b^2+c^2-2bcx)}\\le 2\\sqrt{5}$$ (Note: here, $x$ is taken to be the nonnegative cosine of the angle between the two diagonal, WLOG supposing that the angle between the segments with length $a$ and $b$ is acute.) First, we manipulate the expressions under the radicals, call them $L$ and $R$: \\begin{align*} L&=(a^2+b^2+2abx)(c^2+d^2+2cdx) \\\\ &=(a^2+b^2)(c^2+d^2)+2x(ab(c^2+d^2)+cd(a^2+b^2))+4x^2abcd \\\\ &=(ac-bd)^2+(ad+bc)^2+2x(ad+bc)(ac+bd)+x^2[(ac+bd)^2-(ac-bd)^2] \\\\ &=[x(ac+bd)+(ad+bc)]^2+(1-x^2)(ac-bd)^2 \\end{align*} In a similar manner, we see that: $$R=[x(ac+bd)-(ab+cd)]^2+(1-x^2)(ac-bd)^2$$ Now let $p=x(ac+bd)+\\frac{(ad+bc)-(ab+cd)}{2}$ and $q=|ac-bd|\\sqrt{1-x^2}$. Using the fact that $\\frac{ad+bc+ab+cd}{2}=\\frac{(a+c)(b+d)}{2}=2$, our inequality can be rewritten as: $$\\sqrt{(p+2)^2+q^2}+\\sqrt{(p-2)^2+q^2}\\le 2\\sqrt{5}$$ We will now prove the key lemma: Lemma: $p^2+5q^2\\le 5$ Proof: Using the fact that $ad+bc-(ab+cd)=(a-c)(d-b)=4(a-1)(b-1)$, this inequality can be rewritten as: $$5[a(2-a)-b(2-b)]^2(1-x^2)+[x(a(2-a)+b(2-b))+2(a-1)(b-1)]^2\\le 5$$ Upon the translation $a\\Rightarrow 1+y, b\\Rightarrow 1+z$ for $y, z\\in [-1, 1]$, this simplifies to: y, z\\ge 9$$5(y^2-z^2)^2(1-x^2)+[x(2-y^2-z^2)^2-2yz]^2\\le 5$$ $$\\Leftrightarrow 4x^2(1+3y^2z^2-y^2-y^4-z^2-z^4)+4xyz(2-y^2-z^2)+5y^4+5z^4-6y^2z^2\\le 5$$ Let this expression be $F(x, y, z)$. From this formulation, we see that since $F(x, y, z)\\le F(x, |y|, |z|)$, we can say $y, z\\ge 0$ WLOG. Now, we take two cases: Case 1: $x\\le \\frac{3}{4}$. In this case, note that: \\begin{align*}F(x, 1, yz)-F(x, y, z)&= (5-4x^2)(1-y^4)(1-z^4)-(4x^2+4xyz)(1-y^2)(1-z^2) \\\\ &=(1-y^2)(1-z^2)[(5-4x^2)(1+y^2)(1+z^2)-4x^2-4xyz] \\\\ &\\ge (1-y^2)(1-z^2)[(5-\\frac{9}{4})(1+y^2)(1+z^2)-\\frac{9}{4}-3yz] \\\\ &= (1-y^2)(1-z^2)[\\frac{1}{2}+\\frac{11}{4}(y-z)^2+\\frac{5}{2}yz+\\frac{11}{4}y^2z^2] \\\\ &\\ge 0 \\end{align*} It follows that it suffices to prove the inequality for $z=1$. But we see that: \\begin{align*}F(x, y, 1)&=4x^2(-y^4+2y^2-1)+4xy(1-y^2)+5y^4-6y^2+5 \\\\ &=5-5y^2(1-y^2)-[2x(1-y^2)-y]^2 \\\\ &\\le 5 \\end{align*} And so this case", "are perpendicular . If $\\mathbf{a}$ and $\\mathbf{b}$ are perpendicular, then $\\theta=90^\\circ$, $\\cos \\theta = 0$ and $\\mathbf{a} \\cdot \\mathbf{b} = 0$. Conversely, if $\\mathbf{a}\\cdot \\mathbf{b}=0$, then either $\\theta=90^\\circ$, $\\mathbf{a}=\\mathbf{0}$ or $\\mathbf{b}=\\mathbf{0}$. ### Solution to examples: $\\mathbf{a}\\cdot \\mathbf{b} = |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$ $\\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix} \\cdot \\begin{pmatrix} 5 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\left | \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix} \\right | \\, \\left | \\begin{pmatrix} 5 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right | \\cos \\theta$ $5+0-3 = \\sqrt{14} \\sqrt{26} \\cos \\theta$ $\\theta = \\cos^{-1} \\left ( \\frac{2}{\\sqrt{14} \\sqrt{26}} \\right ) =$ Since $\\mathbf{a}$ is perpendicular to $\\mathbf{c}$, $\\mathbf{a} \\cdot \\mathbf{c} = 0$ $-1 + 2 - 3z = 0$ $\\mathbf{a} \\cdot \\mathbf{d} = k+2 + 2(k-1) -3k = k+2+2k-2-3k = 0$. Hence $\\mathbf{a}$ is perpendicular to $\\mathbf{d}$ for all $k \\in \\mathbb{R}$.", "# Problem 4132 from Crux Mathematicorum ### Solution 1 Case 1: $\\mathbf{2a^3+b^2=a^2+b^2}$ In this case, $2a^3=a^2\\,$ and, since $a\\in\\mathbb{Z},\\,$ $a=0.\\,$ It follows that $2a^3b^2+ab^2+2b^4=3b^4\\,$ which is divisible by $b^2=a^2+b^2.$ Case 2: $\\mathbf{2a^3+b^2\\ne a^2+b^2}$ In this case, $2a^3+b^2)=k(a^2+b^2),\\,$ where $k\\in\\mathbb{Z}\\setminus\\{1\\},\\,$ implying $\\displaystyle b^2=\\frac{a^2(2a-k)}{k-1},\\,$ or $\\displaystyle a^2+b^2=\\frac{a^2(2a-1)}{k-1}.\\,$ So, we have \\displaystyle\\begin{align} 2a^3b^2+ab^2+3b^4 &= b^2(2a^3+a^3b^2)\\\\ &=\\frac{a^2(2a-k)}{k-1}\\left[2a^3+a+\\frac{3a^2(2a-k)}{k-1}\\right]\\\\ &=\\frac{a^3(2a-k)}{k-1}\\cdot\\frac{2(k+2)a^2-3ak+k-1}{k-1}\\\\ &=\\frac{a^3(2a-k)}{k-1}\\cdot\\frac{(2a-1)[(k+2)a-(k-1)]}{k-1}\\\\ &=\\frac{a^2(2a-1)}{k-1}\\cdot\\frac{(2a-k)[(k+2)a^2-(k-1)a]}{k-1}\\\\ &=(a^2+b^2)\\cdot\\frac{(2a-k)[(k+2)a^2-(k-1)a]}{k-1}. \\end{align} ### Solution 2 Since $b^2-2ab^2=(2a^3+b^2)-2a(a^2+b^2),\\,$ it follows from $(a^2+b^2)|(2a^3+b^2)\\,$ that (1) $(a^2+b^2)|b^2(1-2a).$ Further, $ab^2+2b^4=ab^2(1-2a)+2b^2(a^2+b^2),\\,$ thus, by (1), (2) $(a^2+b^2)|(ab^2+2b^4).$ Finally, $2a^3b^2+ab^2+3b^4 = (ab^2+2b^4)+b^2(2a^3+b^2)$ so that $(a^2+b^2)|(2a^3b^2+ab^2+3b^4)\\,$ by (2) and by the hypothesis. ### Acknowledgment The above problem, with a solution (Solution 1), has been kindly communicated to me by Leo Giugiuc. Solution 2 is by Lorenzo Villa.", "b^2}}, 0)$ is a perpendicular unit vector. We can find another unit vector perpendicular to $\\mathbf{A}$ and $\\mathbf{B}$ by taking their cross product. $$\\mathbf{C} = \\mathbf{A} \\times \\mathbf{B} = \\left|\\begin{array}{ccc} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ a & b & c \\\\ \\frac{-b}{\\sqrt{a^2 + b^2}} & \\frac{a}{\\sqrt{a^2 + b^2}}& 0\\\\ \\end{array}\\right| = \\left(\\frac{-ac}{\\sqrt{a^2 + b^2}}, \\frac{-bc}{\\sqrt{a^2 + b^2}}, \\sqrt{a^2 + b^2})\\right)$$ Now we can parameterize over all unit-length linear combinations of $\\mathbf{B}$ and $\\mathbf{C}$. \\begin{align}\\mathbf{D}(\\theta) & = \\mathbf{B} \\sin \\theta + \\mathbf{C} \\cos \\theta\\\\ & = \\left(-\\frac{b \\sin \\theta + ac \\cos \\theta}{\\sqrt{a^2 + b^2}}, \\frac{a \\sin \\theta - bc \\cos \\theta}{\\sqrt{a^2 + b^2}}, \\sqrt{a^2 + b^2} \\cos \\theta\\right) \\end{align} Now vary $\\theta$ from $0$ to $2\\pi$ and you will cover all unit vectors perpendicular to $\\mathbf{A}$. Checking: \\begin{align} \\mathbf{A} \\cdot \\mathbf{D}(\\theta) & = (a,b,c) \\cdot \\left(-\\frac{b \\sin \\theta + ac \\cos \\theta}{\\sqrt{a^2 + b^2}}, \\frac{a \\sin \\theta - bc \\cos \\theta}{\\sqrt{a^2 + b^2}}, \\sqrt{a^2 + b^2} \\cos \\theta\\right)\\\\ & = -\\frac{ab \\sin \\theta + a^2c \\cos \\theta}{\\sqrt{a^2 + b^2}}+ \\frac{ab \\sin \\theta - b^2c \\cos \\theta}{\\sqrt{a^2 + b^2}} + c\\sqrt{a^2 + b^2} \\cos \\theta\\\\ & = \\frac{ab \\sin \\theta - ab \\sin \\theta}{\\sqrt{a^2 + b^2}} - \\frac{a^2 c \\cos \\theta + b^2 c \\cos \\theta}{\\sqrt{a^2 + b^2}} + c \\sqrt{a^2 + b^2} \\cos \\theta\\\\ & = 0", "# If A = <2 ,4 ,-3 >, B = <3 ,4 ,1 > and C=A-B, what is the angle between A and C? Mar 24, 2018 The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${63.23}^{0}$ #### Explanation: $\\vec{C} = \\vec{A} - \\vec{B} = \\left(< 2 , 4 , - 3 >\\right) - \\left(< 3 , 4 , 1 >\\right)$ $= \\left(< \\left(2 - 3\\right) , \\left(4 - 4\\right) , \\left(- 3 - 1\\right) >\\right) = < - 1 , 0 , - 4 >$ $\\vec{A} = < 2 , 4 , - 3 > \\mathmr{and} \\vec{C} = < - 1 , 0 , - 4 >$ Let $\\theta$ be the angle between them ; then we know $\\cos \\theta = \\frac{\\vec{A} \\cdot \\vec{C}}{| | \\vec{A} | | \\cdot | | \\vec{C} | |}$ =(2* (-1)+(4*0)+(-3*(-4)))/(sqrt(2^2+4^2+(-3)^2)* (sqrt((-1)^2+0^2+(-4)^2)) $= \\frac{10}{\\sqrt{29} \\cdot \\sqrt{17}} = \\frac{10}{22.2} \\approx 0.4504$ $\\therefore \\theta = {\\cos}^{-} 1 \\left(0.4504\\right) \\approx {63.23}^{0}$ The angle between $\\vec{A} \\mathmr{and} \\vec{C}$ is ${63.23}^{0}$ [Ans]" ]

[ "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "number $c$, namely one satisfying, $$(A-cB) \\cdot B = 0$$ We then get $$\\textbf{A} \\cdot \\textbf{B} = c\\textbf{B} \\cdot \\textbf{B}$$ and therefore $c=\\frac{ \\textbf{A} \\cdot \\textbf{B}}{\\textbf{B} \\cdot \\textbf{B}}$. From this construction and use of what we know about planee geometry, $$\\cos(\\theta) = \\frac{c||\\textbf{B}||}{||\\textbf{A}||}$$ and substituting in our value for $c$ we get, $$A \\cdot B = ||\\textbf{A}|| ||\\textbf{B}|| \\cos(\\theta)$$ \\ $\\textbf{Picture for our Construction}$ If $v=\\|v\\|\\cdot(\\cos\\theta_1,\\sin\\theta_1)$ and $w=\\|w\\|\\cdot(\\cos\\theta_2,\\sin\\theta_2)$, then $$\\begin{eqnarray} \\frac{v\\cdot w}{\\|v\\| \\|w\\|}&=&(\\cos\\theta_1,\\sin\\theta_1)\\cdot (\\cos\\theta_2,\\sin\\theta_2)\\\\ &=& \\cos\\theta_1 \\cos\\theta_2 +\\sin\\theta_1 \\sin\\theta_2 \\\\&=&\\cos\\left(\\theta_2 - \\theta_1\\right); \\end{eqnarray}$$ and $\\theta_2-\\theta_1$ is the angle between $v$ and $w$. Suppose you have two unit vectors $\\hat a$ and $\\hat b$ and that the angle between them is $\\pi/2$. Then imagine the unit circle, with $\\theta = 0$ corresponding to the $\\hat a$ direction and $\\theta = \\pi/2$ corresponding to the $\\hat b$ direction. Now, suppose there is a unit vector $\\hat c$ in the plane spanned by $\\hat a, \\hat b$. It lies on the unit circle, so you can write $\\hat c = \\hat a \\cos \\theta + \\hat b \\sin \\theta$. Now then, how can you solve for $\\cos \\theta$? By taking a dot product: see, since $\\hat a \\perp \\hat b$, you know that $\\hat a \\cdot \\hat b = 0$. So you solve for the cosine as follows: $$\\hat c \\cdot", "# On the integral $\\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$ Let $a, b$ be arbitrary vectors in $\\mathbb{R}^3$ and define $\\partial B_1(0):=\\{r\\in\\mathbb{R}^3:|r|=1\\}$ as the boundary of the three-dimensional unit sphere. I am trying to evaluate the following integral: $$I = \\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$$ So far, I've said that we can, given a vector $a$ and a fixed vector $b$, write the following: $$a = c b + x$$ where $c = \\frac{a\\cdot b}{|b|}$ and $x\\in\\mathbb{R}^3$ such that $x$ orthogonal to $b$. Then, if we choose our coordinate system so that the positive $z$-axis runs parallel to the vector $b$, then, considering spherical coordinates, the angle between $r$ and $b$ is given by $\\theta$, and we have: \\begin{align} I = \\int_0^\\pi d\\theta\\sin\\theta\\int_0^{2\\pi}d\\phi\\frac{1}{\\left(1+|a|\\cos\\theta\\right)\\left(1+c\\cos\\theta + x\\cdot r\\right)} \\end{align} Now, I know that, since $x\\cdot b = 0$, $x$ must lie in the $x-y-$plane, so we can express it as $(\\rho\\cos\\zeta, \\rho\\sin\\zeta, 0)$ for some modulus $\\rho$ and some polar angle $\\zeta$. Thus: $$x\\cdot r = \\rho\\sin\\theta\\left(\\cos\\zeta\\cos\\phi+\\sin\\zeta\\sin\\phi\\right) = \\rho\\sin\\theta\\cos\\left(\\phi-\\zeta\\right)$$ Letting $\\phi-\\zeta=:\\xi$, we have $d\\xi = d\\phi$ and thus, since all periods of the cosine function are equal, we have: \\begin{align} I&=\\int_0^\\pi d\\theta\\frac{\\sin\\theta}{1+|a|\\cos\\theta}\\int_{-\\zeta}^{2\\pi-\\zeta}\\frac{d\\xi}{1+c\\cos\\theta+\\rho\\sin\\theta\\cos\\xi}\\\\ &=\\int_{-1}^1 \\frac{du}{1+|a|u}\\int_{0}^{2\\pi}\\frac{d\\xi}{1+cu+\\rho\\sqrt{1-u^2}\\cos\\xi} \\end{align} And this is where I'm stumped. I've been thinking of maybe converting the second integral into a contour integral over the unit circle, which reduces to evaluating $$\\oint_{|z|=1}\\frac{dz}{z^2+\\alpha", "\\begin{align}\\langle \\mathbf{R}^2 \\rangle&=\\left\\langle\\left(\\sum_{i=1}^n \\mathbf{r}_i^2\\right)\\cdot\\left(\\sum_{j=1}^n \\mathbf{r}_j^2\\right)\\right\\rangle\\\\ &=\\sum_i\\sum_j\\langle \\mathbf{r}_i\\cdot\\mathbf{r}_j\\rangle\\end{align} 3. Calculation of $\\langle\\mathbf{r}_i\\cdot\\mathbf{r}_j\\rangle$: Now consider a local coordinate frame that $\\mathbf{r}_i=(b,0,0)^T$, then $\\mathbf{r}_{i+1}$ could be considered as rotating $\\mathbf{r}_i$ $\\theta$ by $z$-axis for bond angle and $\\phi_{i}$ by $x$-axis for torsion. The rotation matrix is thus $$\\mathbf{T}_i:=\\left( \\begin{array}{ccc} \\cos (\\theta ) & -\\sin (\\theta ) & 0 \\\\ \\sin (\\theta ) \\cos (\\phi_i) & \\cos (\\theta ) \\cos (\\phi_i) & -\\sin (\\phi_i) \\\\ \\sin (\\theta ) \\sin (\\phi_i ) & \\cos (\\theta ) \\sin (\\phi_i ) & \\cos (\\phi_i ) \\\\ \\end{array}\\right)$$ or $\\pi-\\theta$, $\\pi - \\phi$ according to the definitions of bond angle $\\theta$ and torsion angle $\\phi$. In every reference coordinate frame $R_i$, the bond vector is $(b,0,0)^T$, and in $R_i$, $\\mathbf{r}_{i+1}=\\mathbf{T}_i\\cdot\\mathbf{r}_i=(b \\cos (\\theta ),b \\sin (\\theta ) \\cos (\\phi_i ),b \\sin (\\theta ) \\sin (\\phi_i ))^T$. Now consider a simple case of $\\mathbf{r}_i\\cdot\\mathbf{r}_{i+2}$, the $\\mathbf{r}_{i+2}$ is calculated in the $(i+1)$th reference frame $R_{i+1}$ where $\\mathbf{r}_{i+1}$ is $(b,0,0)^T$ and $\\mathbf{r}_{i+2}=\\mathbf{T}_{i+1}\\cdot\\mathbf{r}_{i+1}$, here we must 'rotate' one of the vector into the other frame to do the inner product, i.e. rotate $\\mathbf{r}_i=(b,0,0)^T$ from $R_i$ to $R_{i+1}$ or vice versa. Now note that $\\mathbf{r}_{i+1}$ in the $R_i$ frame is $(b \\cos (\\theta ),b \\sin (\\theta ) \\cos (\\phi_i ),b \\sin (\\theta ) \\sin (\\phi_i ))^T$ and $(b,0,0)^T$ in the $R_{i+1}$ frame,", "\\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t = \\int_{\\theta(t_1)}^{\\theta(t_2)} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!\\theta}\\right)^2 + r^2 } \\operatorname{d}\\!\\theta.$ Now let $\\mathbf{C}(t) = (r(t), \\theta(t), \\phi(t))$ be a curve expressed in spherical coordinates where $\\theta$ is the polar angle measured from the positive $z$-axis and $\\phi$ is the azimuthal angle. The mapping that transforms from spherical coordinates to rectangular coordinates is $\\mathbf{x}(r,\\theta,\\phi) = (r\\sin\\theta\\cos\\phi, r\\sin\\theta\\sin\\phi, r\\cos\\theta).$ Using the chain rule again shows that $D(\\mathbf{x}\\circ\\mathbf{C}) = \\mathbf{x}_r r' + \\mathbf{x}_{\\theta}\\theta' + \\mathbf{x}_{\\phi}\\phi'$. All dot products $\\mathbf{x}_i \\cdot \\mathbf{x}_j$ where $i$ and $j$ differ are zero, so the squared norm of this vector is $(\\mathbf{x}_r\\cdot \\mathbf{x}_r )(r'^2) + (\\mathbf{x}_{\\theta} \\cdot \\mathbf{x}_{\\theta})(\\theta')^2 + (\\mathbf{x}_{\\phi}\\cdot \\mathbf{x}_{\\phi})(\\phi')^2 = (r')^2 + r^2(\\theta')^2 + r^2 \\sin^2\\theta (\\phi')^2.$ So for a curve expressed in spherical coordinates, the arc length is $\\int_{t_1}^{t_2} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\sin^2\\theta \\left({\\operatorname{d}\\!\\phi\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t.$ A very similar calculation shows that the arc length of a curve expressed in cylindrical coordinates is $\\int_{t_1}^{t_2} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 + \\left({\\operatorname{d}\\!z\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t.$ ## Simple cases ### Arcs of circles Arc lengths are denoted by s, since the Latin word for length (or size) is spatium. In the following lines, $r$ represents the radius of a circle, $d$ is its diameter, $C$ is its circumference, $s$ is the length of an arc of the circle, and $\\theta$ is the angle which the", "\\cdot (b_2\\textbf{j}) + (a_2\\textbf{j}) \\cdot (b_1\\textbf{i}) +(a_2\\textbf{j}) \\cdot (b_2\\textbf{j}) \\\\ &= a_1b_1(\\textbf{i} \\cdot \\textbf{i}) + a_1b_2(\\textbf{i} \\cdot \\textbf{j}) + a_2b_1(\\textbf{j} \\cdot \\textbf{i}) + a_2b_2(\\textbf{j} \\cdot \\textbf{j}) \\end{align*} Given the properties of the scalar product and the fact both $\\textbf{i}$ and $\\textbf{j}$ have magnitude 1, we get following definition #### Scalar product of vectors in terms of components If $\\textbf{a} = a_1\\textbf{i} + a_2\\textbf{j} + a_3\\textbf{k}$ and $\\textbf{b} = b_1\\textbf{i} + b_2\\textbf{j} + b_3\\textbf{k}$, then $$\\textbf{a} \\cdot \\textbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$ In column notation, if $\\textbf{a} = \\begin{pmatrix}a_1 \\ a_2 \\ a_3\\end{pmatrix}$ and $\\textbf{b} = \\begin{pmatrix}b_1 \\ b_2 \\ b_3\\end{pmatrix}$, then $\\textbf{a} \\cdot \\textbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ ## 7.2 Finding the angle between two vectors Rearranging the definition of the scalar product gives following fact #### Angle between two vectors The angle $\\theta$ between any two non-zero vectors $\\textbf{a}$ and $\\textbf{b}$ is given by $$\\cos\\theta = \\frac{\\textbf{a} \\cdot \\textbf{b}}{|\\textbf{a}||\\textbf{b}|}$$", "## Precalculus (6th Edition) Blitzer Assume the two vectors A and B subtend at angle $\\theta$ such that; $\\mathbf{A}=\\mathbf{i}-\\mathbf{j}+\\mathbf{k}$ And, $\\mathbf{B}=-\\mathbf{i}+\\mathbf{j}-\\mathbf{k}$ Now consider the dot product of two vectors A and B. \\begin{align} & \\mathbf{A}\\cdot \\mathbf{B}=\\left( \\mathbf{i}-\\mathbf{j}+\\mathbf{k} \\right)\\cdot \\left( -\\mathbf{i}+\\mathbf{j}-\\mathbf{k} \\right) \\\\ & AB\\cos \\theta =-1-1-1 \\\\ & \\sqrt{{{\\left( 1 \\right)}^{2}}+{{\\left( -1 \\right)}^{2}}+{{\\left( 1 \\right)}^{2}}}\\times \\sqrt{{{\\left( 1 \\right)}^{2}}+{{\\left( 1 \\right)}^{2}}+{{\\left( -1 \\right)}^{2}}}\\cos \\theta =-3 \\\\ & \\sqrt{3}\\times \\sqrt{3}\\cos \\theta =-3 \\end{align} Simplify further, \\begin{align} & 3\\cos \\theta =-3 \\\\ & \\cos \\theta =-\\frac{3}{3} \\\\ & \\cos \\theta =-1 \\\\ & \\theta =\\pi \\end{align} Since, the angle between vectors A and B is $\\theta =\\pi$, hence when the dot product of two vectors is negative, the angle between them is obtuse. Therefore the given statement makes sense.", "$\\mathbf n_\\theta=(\\cos\\theta,\\sin\\theta)$ and fixed distance $d\\ge0$ from the origin. Here $\\theta$ represents the angle that $\\mathbf n_\\theta$ makes with the positive $x$-axis. If you draw a perpendicular to this line through $\\mathbf c$, it will be parallel to $\\mathbf n_\\theta$, so it, too, makes an angle of $\\theta$ with the positive $x$-axis. Call the angle between this perpendicular and the tone arm, represented by the radial segment from $\\mathbf c$ to the intersection of the line and circle, $\\alpha$. The tone-arm angle is clearly $\\phi=\\theta-\\alpha$. The signed distance from our rotating line to the tone-arm pivot $\\mathbf c$ is given by $d-\\mathbf n_\\theta\\cdot\\mathbf c$ and so we have $$\\cos\\alpha={d-\\mathbf n_\\theta\\cdot\\mathbf c\\over R}=\\frac d R+\\cos\\theta.\\tag{1}$$ Equipped with this relationship between turntable rotation and tone-arm angle, we can proceed with the specific problem. Given the points $A=(x_a,y_a)$ and $B=(x_b,y_b)$, the unit normal of the line through them is $$\\mathbf n=\\pm\\left({y_a-y_b\\over\\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}},{x_b-x_a\\over\\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}}\\right).\\tag{2}$$ The distance of this line from the origin is given by $\\mathbf n\\cdot A$ (also by $\\mathbf n\\cdot B$). Choose the sign for $\\mathbf n$ that makes this quantity positive. Note that this sign choice matches the orientation of the triangle $\\triangle{OBA}$, which is also the sign of the “cross product:” $s=\\operatorname{sgn}(x_by_a-x_ay_b)$. This sign also tells us which way we’ll be rotating the turntable (positive being counterclockwise per the usual", "1$. Prove that: $$\\sqrt{(a^2+b^2+2abx)(c^2+d^2+2cdx)}+\\sqrt{(a^2+d^2-2adx)(b^2+c^2-2bcx)}\\le 2\\sqrt{5}$$ (Note: here, $x$ is taken to be the nonnegative cosine of the angle between the two diagonal, WLOG supposing that the angle between the segments with length $a$ and $b$ is acute.) First, we manipulate the expressions under the radicals, call them $L$ and $R$: \\begin{align*} L&=(a^2+b^2+2abx)(c^2+d^2+2cdx) \\\\ &=(a^2+b^2)(c^2+d^2)+2x(ab(c^2+d^2)+cd(a^2+b^2))+4x^2abcd \\\\ &=(ac-bd)^2+(ad+bc)^2+2x(ad+bc)(ac+bd)+x^2[(ac+bd)^2-(ac-bd)^2] \\\\ &=[x(ac+bd)+(ad+bc)]^2+(1-x^2)(ac-bd)^2 \\end{align*} In a similar manner, we see that: $$R=[x(ac+bd)-(ab+cd)]^2+(1-x^2)(ac-bd)^2$$ Now let $p=x(ac+bd)+\\frac{(ad+bc)-(ab+cd)}{2}$ and $q=|ac-bd|\\sqrt{1-x^2}$. Using the fact that $\\frac{ad+bc+ab+cd}{2}=\\frac{(a+c)(b+d)}{2}=2$, our inequality can be rewritten as: $$\\sqrt{(p+2)^2+q^2}+\\sqrt{(p-2)^2+q^2}\\le 2\\sqrt{5}$$ We will now prove the key lemma: Lemma: $p^2+5q^2\\le 5$ Proof: Using the fact that $ad+bc-(ab+cd)=(a-c)(d-b)=4(a-1)(b-1)$, this inequality can be rewritten as: $$5[a(2-a)-b(2-b)]^2(1-x^2)+[x(a(2-a)+b(2-b))+2(a-1)(b-1)]^2\\le 5$$ Upon the translation $a\\Rightarrow 1+y, b\\Rightarrow 1+z$ for $y, z\\in [-1, 1]$, this simplifies to: y, z\\ge 9$$5(y^2-z^2)^2(1-x^2)+[x(2-y^2-z^2)^2-2yz]^2\\le 5$$ $$\\Leftrightarrow 4x^2(1+3y^2z^2-y^2-y^4-z^2-z^4)+4xyz(2-y^2-z^2)+5y^4+5z^4-6y^2z^2\\le 5$$ Let this expression be $F(x, y, z)$. From this formulation, we see that since $F(x, y, z)\\le F(x, |y|, |z|)$, we can say $y, z\\ge 0$ WLOG. Now, we take two cases: Case 1: $x\\le \\frac{3}{4}$. In this case, note that: \\begin{align*}F(x, 1, yz)-F(x, y, z)&= (5-4x^2)(1-y^4)(1-z^4)-(4x^2+4xyz)(1-y^2)(1-z^2) \\\\ &=(1-y^2)(1-z^2)[(5-4x^2)(1+y^2)(1+z^2)-4x^2-4xyz] \\\\ &\\ge (1-y^2)(1-z^2)[(5-\\frac{9}{4})(1+y^2)(1+z^2)-\\frac{9}{4}-3yz] \\\\ &= (1-y^2)(1-z^2)[\\frac{1}{2}+\\frac{11}{4}(y-z)^2+\\frac{5}{2}yz+\\frac{11}{4}y^2z^2] \\\\ &\\ge 0 \\end{align*} It follows that it suffices to prove the inequality for $z=1$. But we see that: \\begin{align*}F(x, y, 1)&=4x^2(-y^4+2y^2-1)+4xy(1-y^2)+5y^4-6y^2+5 \\\\ &=5-5y^2(1-y^2)-[2x(1-y^2)-y]^2 \\\\ &\\le 5 \\end{align*} And so this case", "# Using the equation $\\mathbf{A} \\cdot \\mathbf{B} = AB\\cos(\\theta)$ to show that the dot product is distributive when the three vectors are coplanar. I am asked to use the equation $$\\mathbf{A} \\cdot \\mathbf{B} = AB\\cos(\\theta)$$ to show that the dot product is distributive when the three vectors are coplanar. It seems that the typical way in which this is proven is with reference to a diagram, as was done in this answer; however, I want to try and prove this without relying on a diagram. This equation is still valid for more than 2 vectors, since we know that the sum of two vectors (say, $$\\mathbf{B}$$ and $$\\mathbf{C}$$) is itself a vector: $$\\mathbf{A} \\cdot (\\mathbf{B} + \\mathbf{C}) = A(B + C)\\cos(\\theta)$$ So let $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}$$ be coplanar vectors. Since we have that 2 vectors span a plane, this implies that one of the vectors must be a linear combination of the other two. Let that vector be $$\\mathbf{C} = \\alpha \\mathbf{A} + \\beta \\mathbf{B} = \\alpha(a_1, a_2, a_3) + \\beta(b_1, b_2, b_3)$$, where $$\\alpha, \\beta \\in \\mathbb{R}$$. So we have that \\begin{align} \\mathbf{A} \\cdot ( \\mathbf{B} + \\mathbf{C}) &= A(B + C) \\cos(\\theta) \\\\ &= \\sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2}\\sqrt{(b_1 + c_1)^2 + (b_2 + c_2)^2 + (b_3 + c_3)^2} \\cos(\\theta) \\end{align} We can continue by factoring", "+0 # Let $\\theta$ be the angle between $\\bold{a}$ and $\\bold{b}$, where $\\bold{a}$ and $\\bold{b}$ are the vectors defined in part (a). Find \\$\\cos 0 171 2 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$. Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\cos(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{\\vec{a}\\cdot \\vec{b}}{|\\vec{a}|\\cdot |\\vec{b}|} } \\quad | \\quad |\\vec{a}| = \\sqrt{2^2+(-6)^2+5^2}=\\sqrt{65},\\ |\\vec{b}| = \\sqrt{(-1)^2+(-2)^2+0^2}=\\sqrt{5} \\\\\\\\ \\cos(\\theta) &=& \\dfrac{\\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix}\\cdot \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix} } {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{2\\cdot (-1)+(-6)\\cdot(-2)+5\\cdot 0} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{10} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {65\\cdot 5 } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {325 } \\\\\\\\ \\mathbf{\\cos^2(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{4}{13}} \\\\ \\hline \\end{array}$$ Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix}", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "# Calculating the following integral . I have found the following integral to be zero, but i don't think its correct. $$C = \\oint_K d\\mathbf r\\cdot \\mathbf A$$ Where $\\mathbf A = \\frac 1 2 \\mathbf n \\times \\mathbf r$ and $\\mathbf n \\cdot \\mathbf n=1$. Taking $K$ as a circle with radius $R$ and $\\mathbf n$ is the normal to the plane where the circle lives. Any kind of help is appreciated . The problem is that when i found $A$ it came out to be zero. i found normal as $\\nabla f$ where $f=x^2+y^2-1$ . Progress : I applied stokes theorem, and also $\\nabla \\times \\mathbf A =2\\mathbf n$ , then i get $C=\\pi R^2$ . am i right ?? how do i do it using line integral ? - In the following I assume the path $K$ is counter-clockwise with respect to the orientation $\\mathbf n$ of the plane. Let $$K\\equiv \\gamma(t) := R(\\cos(t), \\sin(t), 0)\\quad t\\in[0, 2\\pi]$$ we have $$\\mathbf A(\\gamma(t)) = \\frac 1 2 \\mathbf n\\times \\gamma(t) = \\frac R 2 (0, 0, 1)\\times (\\cos(t), \\sin(t), 0) = \\frac R 2 (-\\sin(t), \\cos(t), 0)$$ and so \\begin{align} \\oint_K d\\mathbf r \\cdot \\mathbf A &= \\int_0^{2\\pi} \\mathbf A(\\gamma(t))\\cdot \\dot\\gamma(t)dt \\\\ &= \\frac {R^2} 2 \\int_0^{2\\pi} (\\sin(t)^2 + \\cos(t)^2)dt\\\\ &= \\pi R^2 \\end{align} First note", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "A) $\\frac{1}{2|b|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ B) $\\frac{1}{2|a|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ C) $\\frac{1}{2|d|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ D) None of these • question_answer139) The value of $'a'$ for which the function $f(x)=(4a-3)(x+\\log 5)+2(a-7)$ $\\cot \\frac{x}{2}\\cdot {{\\sin }^{2}}\\frac{x}{2}$does not possess critical points is A) $(-\\infty ,\\,\\,2)$ B) $(-\\infty ,\\,\\,-1)$ C) $[1,\\,\\,\\infty )$ D) $\\left( -\\infty ,\\,\\,\\frac{-4}{3} \\right)\\cup (2,\\,\\,\\infty )$ • question_answer140) $\\mathbf{a}$and $\\mathbf{c}$ are unit vectors and$|\\mathbf{b}|\\,\\,=4$. If angle between $\\mathbf{b}$ and $\\mathbf{c}$ is and ${{\\cos }^{-1}}\\left( \\frac{1}{4} \\right)$$\\text{and}$$\\mathbf{a}\\times \\mathbf{b}=2\\mathbf{a}\\times \\mathbf{c}$, then$\\mathbf{b}=\\lambda \\mathbf{a}+2\\mathbf{c}$, where$\\lambda$is equal to A) $\\pm \\frac{1}{4}$ B) $\\pm \\frac{1}{2}$ C) $\\pm \\,\\,1$ D) None of the above • question_answer141) If the planes$x-cy-bz=0$, $cx-y+az=0$and$bx+ay-z=0$ pass through a straight line, then the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$is A) $-1$ B) $2$ C) $1$ D) $0$ • question_answer142) The plane $ax+by=0$ is rotated through an angle $\\alpha$ about its line of intersection with the plane$z=0$, then the equation to the plane in new position is A) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ B) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ C) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ D) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ • question_answer143) If$y=a\\cos (\\log x)-b\\sin (\\log x)$, then the value of${{x}^{2}}\\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\\frac{dy}{dx}+y$is A) $0$ B) $1$ C) $2$ D) $3$ • question_answer144) The inverse of the function$f(x)=\\log ({{x}^{2}}+3x+1),\\,\\,x\\in [1,\\,\\,3]$, assuming it to be an onto function, is A) $\\frac{-3+\\sqrt{5+4{{e}^{x}}}}{2}$ B) $\\frac{-3\\pm \\sqrt{5+4{{e}^{x}}}}{2}$ C) $\\frac{-3-\\sqrt{5+4{{e}^{x}}}}{2}$ D) None of the above • question_answer145) $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{\\sqrt{1-\\cos", "\\right\\|=\\sqrt{{{a}^{2}}+{{b}^{2}}}$ The vector $\\mathbf{w}-\\mathbf{v}$ is $29.9\\mathbf{i}+77.2\\mathbf{j}$. Comparing it with the above equation gives \\begin{align} & a=29.9\\mathbf{i} \\\\ & b=77.2\\mathbf{j} \\end{align} Then magnitude of $\\mathbf{w}-\\mathbf{v}$ is given by $\\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|$. \\begin{align} & \\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|=\\sqrt{{{\\left( 29.9 \\right)}^{2}}+{{\\left( 77.2 \\right)}^{2}}} \\\\ & \\approx 82.8 \\\\ & \\approx 83 \\end{align} Then magnitude of $\\mathbf{w}-\\mathbf{v}$ is $83$. The speed of the plane is $83\\text{ mph}$. If the vector is $\\mathbf{F}=a\\mathbf{i}+b\\mathbf{j}$, then magnitude of $\\mathbf{F}$ is given by $\\left\\| \\mathbf{F} \\right\\|$. $\\left\\| \\mathbf{F} \\right\\|=\\sqrt{{{a}^{2}}+{{b}^{2}}}$ And the direction angle is given as $\\cos \\theta =\\frac{a}{\\left\\| \\mathbf{F} \\right\\|}$ The vector $\\mathbf{w}-\\mathbf{v}$ is $29.9\\mathbf{i}+77.2\\mathbf{j}$. Then direction angle of $\\mathbf{w}-\\mathbf{v}$ is given by \\begin{align} & \\cos \\theta =\\frac{a}{\\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|} \\\\ & \\cos \\theta =\\frac{29.9}{82.8} \\\\ & \\theta ={{\\cos }^{-1}}\\left( \\frac{29.9}{82.8} \\right) \\\\ & \\theta =68.8{}^\\circ \\end{align}", "= \\mathbf{a} \\cdot \\mathbf{c} + \\mathbf{a} \\cdot \\mathbf{d} + \\mathbf{b} \\cdot \\mathbf{c} + \\mathbf{b} \\cdot \\mathbf{d}$ Consider a triangle $OAB$ with $\\mathbf{a}, \\mathbf{b}$ the position vectors of points $A$ and $B$. Let $\\theta$ be the angle $AOB$. By the cosine rule, $AB^2 = OA^2 + OB^2 - 2 \\, OA \\cdot OB \\cos \\theta$. $\\begin{gather} | \\mathbf{b}-\\mathbf{a} |^2 = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta \\tag{4} \\end{gather}$ By using formula 1, $|\\mathbf{b}-\\mathbf{a}|^2 = (\\mathbf{b}-\\mathbf{a}) \\cdot (\\mathbf{b}-\\mathbf{a})$. Expanding using formula 3 gives us $\\mathbf{b} \\cdot \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} - \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{a}$. The use of formulas 1 and 2 simplifies the expression to $|\\mathbf{b}|^2 - 2 \\mathbf{a}\\cdot\\mathbf{b} + |\\mathbf{a}|^2$. Plugging this into equation 4 gives us $|\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 \\mathbf{a} \\cdot \\mathbf{b} = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta$ $-2\\mathbf{a} \\cdot \\mathbf{b} = -2 |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$. ### Angles and perpendicular vectors The dot product formula is especially handy in two situations. The first is in the computation of angles. Given two vectors $\\mathbf{a}$ and $\\mathbf{b}$, the values of $\\mathbf{a} \\cdot \\mathbf{b}$, $|\\mathbf{a}|$ and $|\\mathbf{b}|$ are all relatively simple to compute. Substituting them into our formula allows us to calculate the angle between them. The second situation is when vectors", "because the path is not closed. For the work you need to calculate the path integral: \\begin{aligned}W&=\\int_C \\mathbf F \\cdot d\\mathbf s \\\\ &= \\int_0^{\\pi/2} (a\\cos\\theta, a\\sin\\theta) \\cdot (-a\\sin\\theta, a\\cos\\theta) d\\theta \\\\ &= \\int_0^{\\pi/2} (-a\\cos\\theta\\cdot a\\sin\\theta + a\\sin\\theta \\cdot a\\cos\\theta) d\\theta \\\\ &= \\int_0^{\\pi/2} 0 d\\theta \\\\ &= 0\\end{aligned} You can make it a bit easier by observing that the force $\\mathbf F$ is perpendicular to the path C everywhere. Therefore $\\mathbf F$ does not do any work. So: $$W = 0$$ Since the curve is not closed, we cannot apply that theorem. Assuming that with flux you mean the integration of the perpendicular vector over the path, we would have: \\begin{aligned}\\text{Flux}&=\\int_C \\mathbf F \\cdot \\mathbf {\\hat n} ds \\\\ &= \\int_0^{\\pi/2} (a\\cos\\theta, a\\sin\\theta) \\cdot (\\cos\\theta, \\sin\\theta) a d\\theta \\\\ &= \\int_0^{\\pi/2} (a\\cos^2\\theta + a\\sin^2\\theta) a d\\theta \\\\ &= \\int_0^{\\pi/2} a^2 d\\theta \\\\ &= \\frac{\\pi a^2}{2}\\end{aligned} Alternatively, we can see that the curve has radius $a$ and length $\\frac {\\pi a} 2$. The force is perpendicular and has size $a$. So this flux is $\\frac {\\pi a} 2 \\cdot a = \\frac {\\pi a^2} 2$. I calculated it as followed: \\begin{aligned}\\text{Flux}&=\\int_C \\mathbf F \\cdot \\mathbf {\\hat n} ds \\\\ &= \\int_0^{\\pi/2} (a\\cos\\theta, a\\sin\\theta) \\cdot (\\cos\\theta, \\sin\\theta) d\\theta \\\\ &= \\int_0^{\\pi/2} (a\\cos^2\\theta + a\\sin^2\\theta) d\\theta \\\\ &= \\int_0^{\\pi/2} a", "b are equal to the cos and sin of the angles between the two vectors. In which case we have: $a = cos(\\theta)\\\\ b = sin(\\theta)$ and after substitution: \\begin{align*} &(cos(\\theta) x - sin(\\theta) y) \\mathbf{e_1} +\\\\ &(sin(\\theta) x + cos(\\theta) y) \\mathbf{e_2} \\end{align*} ## Complex Numbers Revisited You can probably see where this is going. Let's try multiplying two of these 'scalar plus scalar times base pairs' together. $(a_1 + b_1 \\mathbf{e_1 e_2}) (a_2 + b_2 \\mathbf{e_1 e_2})$ Again, using FOIL we get: $(a_1 a_2) + (a_1 b_2) \\mathbf{e_1 e_2} + b_1 \\mathbf{e_1 e_2} a_2 + b_1 \\mathbf{e_1 e_2} b_2 \\mathbf{e_1 e_2}$ It is fine to move the scalars around, so let's clean this up: $(a_1 a_2) + (a_1 b_2) \\mathbf{e_1 e_2} + (a_2 b_1) \\mathbf{e_1 e_2} + (b_1 b_2) \\mathbf{e_1 e_2 e_1 e_2}$ We can combine the two $$\\mathbf{e_1 e_2}$$ terms: $(a_1 a_2) + (a_1 b_2 + a_2 b_1) \\mathbf{e_1 e_2} + (b_1 b_2) \\mathbf{e_1 e_2 e_1 e_2}$ Next we deal with the $$\\mathbf{e_1 e_2 e_1 e_2}$$ term. If we swap the middle $$\\mathbf{e_2 e_1}$$ into $$\\mathbf{e_1 e_2}$$ and negate the whole thing we get $$-\\mathbf{e_1 e_1 e_2 e_2}$$. And since a base times itself is $$1$$ we have $$-1 * 1$$ or $$-1$$. So, the last term can be replaced with $$-(b_1 b_2)$$, causing the" ]

[ "$t = \\frac{3 \\pi}{16} + \\frac{k \\pi}{4}$, or in degrees --> $t = {33}^{\\circ} 75 + k {90}^{\\circ}$ Check by calculator. t = 33.75 + 90 = 123.75 --> 4t = 495 --> tan 4t = - 1. Proved.", "$$\\theta=\\frac{7\\pi}{8}.$$", "\\frac{16 \\sqrt{3}}{9} {\\tan}^{-} 1 \\left(\\frac{\\sqrt{3}}{9} \\left(2 x + 2\\right)\\right) + C$", "- 4}{\\pi + 4}$", "side as 1. The $\\tan(\\theta) = \\frac x 1$. Now find $sin(\\theta)$.", "(h_4 * tan(\\frac{\\pi}{6})) = (h_2 * 2) + (h_4 * \\frac{1}{\\sqrt{3}})$", "+ 4\\pi n}$", "# How do you find tan theta if sin theta=sqrt7/4, cos theta =3/4? In this way, remembering that $\\tan \\theta = \\sin \\frac{\\theta}{\\cos} \\theta$. $\\tan \\theta = \\frac{\\frac{\\sqrt{7}}{4}}{\\frac{3}{4}} = \\frac{\\sqrt{7}}{4} \\cdot \\frac{4}{3} = \\frac{\\sqrt{7}}{3}$.", "\\frac{\\tan^7 x}{7} + \\frac{\\tan^9 x}{9} +C.$$", "\\right)\\tan \\frac{C}{2}=\\frac{\\Delta }{s}$$.", "this form: $\\int \\frac{1}{a^2 + u^2} \\ du = \\frac{1}{a} tan^{-1} \\left( \\frac{u}{a} \\right) + C$ So in your case we have: $\\int \\frac{1}{4 + n^2} \\ dn = \\frac{1}{2} tan^{-1} \\left( \\frac{n}{2} \\right) + C$", "correct? but i tried tangent for locating the adjacent side and my answer was 6.40 which is incorrect what happened? • Aug 2nd 2010, 02:50 PM Math Major $tan(58) = \\frac{4}{y} \\implies y = \\frac{4}{tan(58)} =~ 2.5$", "P \\over 4 \\pi r^2}$ ?", "# Question #be72f $\\tan \\theta = - \\frac{3}{4}$ This is a 3,4,5 triangle so the adjacent is 4 hence $\\tan \\theta = \\frac{o p p o s i t e}{a \\mathrm{dj} a c e n t} = - \\frac{3}{4}$", "\\pi) = \\frac {\\tan \\arccos x + \\tan \\pi}{1-(\\tan\\arccos x)(\\tan \\pi)}$ and of course $\\tan \\pi = 0$", "$x=\\frac{{\\pi}}{12},\\ \\frac{5{\\pi}}{12},\\ \\frac{9{\\pi}}{12}$ Then t=tanx", "$$\\frac{\\tan x}{\\tan y} = 3$$", "\\frac{4}{9} $", "$$tan(\\frac{3\\pi}{11}) +4tan(\\frac{2\\pi}{11})$$ is not equal to the square root of $$11$$. Am I somehow misunderstanding your problem? · 4 years, 1 month ago", "1. ## Easy Question sin θ = 8/17 and cos θ = 15/17, find tan θ. 2. Originally Posted by Joker37 sin θ = 8/17 and cos θ = 15/17, find tan θ. Hi Joker37, Remember: $\\sin \\theta=\\frac{y}{r}$ $\\cos \\theta=\\frac{x}{r}$ $\\tan \\theta=\\frac{y}{x}$ $r^2=x^2+y^2$ $15^2+8^2=17^2$ $x = 15$, $y= 8$, $r = 17$ Therefore, $\\tan \\theta = \\frac{8}{15}$" ]

[ "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The cosine function cos : R -> R The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The tangent function tan : R -> R The range of the function is R . Now, the period is $pi$ and the function is undefined at x = ($pi$/2) + k$pi$, k=0,1,2. The graph of the tangent function on the interval 0 - $pi$ #### The cotangent function cot : R -> R The range of the function is R . The period is $pi$ and that the function is undefined at x = k$pi$, k=0,1,2. #### The values of sin, cos, tan, cot at the angles of 0°, 30°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° $A^o$ ^o30^o45^o60^o90^o120^o135^o150^o180^o210^o225^o240^o270^o300^o315^o330^o360^oA rad $frac<6>$ $frac<4>$ $frac<3>$ $frac<2>$ $frac<2pi><3>$ $frac<3pi><4>$ $frac<5pi><6>$ $pi$ $frac<7pi><6>$ $frac<5pi><4>$ $frac<4pi><3>$ $frac<3pi><2>$ $frac<5pi><3>$ $frac<7pi><4>$ $frac<11pi><6>$ $2pi$ $sin A$ frac<1><2>frac><2>frac><2>1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ cos A1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ frac<1><2>frac><2>frac><2>1 an A $frac><3>$ $1$ $sqrt<3>$ $-$ $-sqrt<3>$ $-1$ $-frac><3>$ frac><3>1sqrt<3>--sqrt<3>-1-frac><3> $cot A$ $-$ $sqrt<3>$ $1$ $frac><3>$ -frac><3>-1-sqrt<3>-sqrt<3>1frac><3> $-frac><3>$ $-1$ $-sqrt<3>$ $-$ The easiest way to remember the basic values of sin and cos at the angles", "period of $6\\pi$ . 1. The period is $\\frac{\\pi}{\\frac{3}{2}}=\\pi \\cdot \\frac{2}{3}=\\frac{2\\pi}{3}$ . 2. The zeros are where $y$ is zero. $0 &=-4\\tan \\frac{3}{2}x \\\\0 &=\\tan \\frac{3}{2}x \\\\\\frac{3}{2}x &=\\tan ^{-1}0=0, \\pi, 2\\pi, 3\\pi \\\\x &=\\frac{2}{3}(0, \\pi, 2\\pi, 3\\pi ) \\\\x &=0, \\frac{2\\pi}{3}, \\frac{4\\pi}{3}, 2\\pi$ 3. The general equation is $y=a\\tan bx$ . We know that $a = 8$ . Let’s use the period to solve for the frequency, or $b$ . $\\frac{\\pi}{b} &=6\\pi \\\\b &=\\frac{\\pi}{6\\pi}=\\frac{1}{6}$ The equation is $y=8\\tan \\frac{1}{6}x$ . ### Vocabulary Tangent Function Defined by the coordinates $(\\theta, \\tan \\theta)$ , where $\\theta$ is the central angle from the unit circle and tangent is the ratio of the sine and cosine functions. ### Practice Graph the following tangent functions over $[0, 4\\pi]$ . Determine the period, domain, and range. 1. $y=2\\tan x$ 2. $y=-\\frac{1}{3}\\tan x$ 3. $y=-\\tan 3x$ 4. $y=4\\tan 2x$ 5. $y=\\frac{1}{2}\\tan 4x$ 6. $y=-\\tan \\frac{1}{2}x$ 7. $y=4+\\tan x$ 8. $y=-3+\\tan 3x$ 9. $y=1+\\frac{2}{3}\\tan \\frac{1}{2}x$ 10. Find the zeros of the function from #1. 11. Find the zeros of the function from #3. 12. Find the zeros of the function from #5. Write the equation of the tangent function, in the form $y=a\\tan bx$ , with the given amplitude and period. 1. Amplitude: 3 Period: $\\frac{3\\pi}{2}$ 2. Amplitude: $\\frac{1}{4}$ Period: $2\\pi$ 3. Amplitude: -2.5 Period: 8", "brb 20. anonymous thats not any of the answer choices 21. Mertsj |dw:1443103740128:dw| 22. anonymous yes then what @Mertsj 23. Mertsj $\\tan 2511/60=\\frac{x}{19427.5}$ 24. anonymous i have it set up like that i just dont remeber how to solve it thats what ive been trying to ask 25. Mertsj $.89567=\\frac{x}{19427.5}$ 26. Mertsj $.89567(19427.5)=x$ 27. anonymous i got 17,400 28. Mertsj x= 17400.72 ft 29. anonymous but that cant be the answer because its not any of the answer choices 30. Mertsj 31. anonymous 9135, 9235, 91351, 913515 32. anonymous and 17400 isnt close to any 33. Mertsj whoops. I must have misused the tangent function on my calculator. I now get .47020 for the tangent of 25 11/50 degrees. 34. anonymous yeah i just figured it out thank you 35. Mertsj $.47021=\\frac{x}{19427.5}$ 36. anonymous i got it its 9141 37. Mertsj $x= .47021(19427.5)=9134.9859$ 38. Mertsj yw", "\\leq \\pi$$, or $$0^{\\circ} \\leq x \\leq 180^{\\circ}$$ in degree mode. Note that we could also have taken $$-\\pi \\leq x \\leq 0$$, but for convenience we take $$x$$ to be a positive angle. Lastly, for $$\\tan(x)$$ we can take a full period around the origin, so $$-\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2}$$, or $$-90^{\\circ} \\leq x \\leq 90^{\\circ}$$ in degree mode.", "NZ Level 7 (NZC) Level 2 (NCEA) Period changes for tangents Lesson We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\\sin kx$sinkx and $\\cos kx$coskx. The fact that the sine and cosine functions both have period $360^\\circ$360° is expressed by the statements $\\sin(x+360^\\circ n)=\\sin x$sin(x+360°n)=sinx and $\\cos(x+360^\\circ n)=\\cos x$cos(x+360°n)=cosx for all integers $n$n. So it must be true that since $\\tan x=\\frac{\\sin x}{\\cos x}$tanx=sinxcosx, then $\\tan(x+360^\\circ n)=\\tan x$tan(x+360°n)=tanx. However, $360^\\circ$360° is not the smallest interval at which the tangent function repeats. A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\\circ$180°. That is, $\\tan(x+180^\\circ n)=\\tan x$tan(x+180°n)=tanx for all integers $n$n. The fact that the tangent function repeats at intervals of $180^\\circ$180° can be verified by considering the unit circle diagram. If $180^\\circ$180° is added to an angle $\\alpha$α, then the diagram shows that $\\sin(\\alpha+180^\\circ)$sin(α+180°) has the same magnitude as $\\sin\\alpha$sinα but opposite sign. The same relation holds between $\\cos(\\alpha+180^\\circ)$cos(α+180°) and $\\cos\\alpha$cosα. We make use of the definition: $\\tan\\alpha=\\frac{\\sin\\alpha}{\\cos\\alpha}$tanα=sinαcosα $\\tan(\\alpha+180^\\circ)$tan(α+180°) $=$= $\\frac{\\sin(\\alpha+180^\\circ)}{\\cos(\\alpha+180^\\circ)}$sin(α+180°)cos(α+180°)​ $=$= $\\frac{-\\sin\\alpha}{-\\cos\\alpha}$−sinα−cosα​ $=$= $\\tan\\alpha$tanα We are now in a position to determine the period of the function $\\tan kx$tankx where $k$k is any number. We can define a new", "your calculator. Round to 4 decimal places. a) $\\sin 78^\\circ$ b) $\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Solution: Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.9781$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.2679$ ## Finding the Sides of a Triangle using Trig Ratios One application of the trigonometric ratios is to use them to find the missing sides of a right triangle. Example 5: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the hypotenuse. Use sine to find $b$, and cosine to find $a$. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example 6: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the adjacent leg to $42^\\circ$. To find $c$, use cosine and tangent to find $d$. $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &=", "## Trigonometry (11th Edition) Clone $-\\sqrt3$ Convert the angle measure to degrees to obtain: $=\\frac{5\\pi}{3} \\cdot \\frac{180^o}{\\pi} = 5(60^o)=300^0$ Thus, $\\tan{(\\frac{5\\pi}{3})} \\\\= \\tan{300^o}$ 300 degrees is in Quadrant IV so : (i) its reference angle is $360^o-300^o=60^o$ (ii) its tangent is negative From Section 2.1 (page 50) , we learned that: $\\tan{60^o} = \\sqrt3$ This means that: $\\tan{300^o} = -\\tan{60^o} = -\\sqrt3$.", "in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round your answers to the nearest tenth:", "tangent function when the angle is 90 degrees (you can change the scale to zoom in and out). What does this really mean? What are the opposite and adjacent lengths? Similarly look at the cosine of 0 or the sine of 90. If you want to see how this interactive works you can watch this short video. #### Worked Examples ##### QUESTION 1 Round $62^\\circ$62°$19$19' to the nearest degree. 1. $\\editable{}$° ##### QUESTION 2 Convert $13$13°$36$36'$35$35'' to degrees correct to 2 decimal places. ##### QUESTION 3 Evaluate $12\\cos$12cos$25$25°$10$10' correct to 2 decimal places. ##### QUESTION 4 Evaluate the following expression, writing your answer correct to two decimal places. 1. $\\frac{\\cos16^\\circ}{\\tan30^\\circ+\\cos18^\\circ}$cos16°tan30°+cos18° ## Degrees minutes and seconds in angles Angles as we know are measured in degrees. But did you know that if we break an angle up into its smaller parts we get minutes and seconds. 1 degree is made up of 60 minutes. And every 1 minute is made up of 60 seconds. This means that this is a base $60$60 system. Instead of a base $10$10 system for our decimal numbers, to move between the 'columns' multiply or divide by $60$60. And to move 2 columns we move $3600$3600 ($60\\times60$60×60). ### Conversions between notations A base 60 system means that we need to be extra careful when", "degrees, get what they deserve. Let $\\theta=255\\cdot\\frac{\\pi}{180}$ then $r=\\frac{9}{\\theta}$. You should get $2.02220398281467$.", ") = \\frac{y}{{7.5}}$$ $7.5 \\times \\tan (64^\\circ ) = y$ $y = 15.37727...$ $y = 15.377cm(to\\,3\\,d.p.)$", "1.39 Exercise 1.39: A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert: $tan x = \\frac{x}{1 - \\frac {x^2}{3 - \\frac{x^2}{5-\\cdots}}}$ where x is in radians. Define a procedure (tan-cf x k) that computes an approximation to the tangent function based on Lambert’s formula. k specifies the number of terms to compute, as in Exercise 1.37. Let's first review the general continued fraction: $f = \\frac{N_1}{D_1 + \\frac{N_2}{D_2 + \\frac{N_3}{D_3 + \\cdots}}}$ Here, $f = tan, f(x) = tan(x)$. And $D_1 = 1, D_2 = 3, D_3 = 5, \\cdots$, while $N_1(x) = x^1, N_2(x)=-x^2, N_3(x)=-x^2, \\cdots$. Firstly we can define d as: (define (d i) (- (* 2 i) 1)) (d 3) Then we can define n as: (define (n i x) (if (= i 1) x (- (* x x)))) (n 1 3) (n 2 3) (n 3 3) (n 4 3) Now, let's define tan-cf based on cont-frac: (define (cont-frac n d k x) (define (g i k) (if (= i k) (/ (n k x) (d k)) (/ (n i x) (+ (d i) (g (+ i 1) k))) ) ) (g 1 k) ) (define (tan-cf x k) (cont-frac n d k x) ) (tan-cf 3 30) Compare the result with (tan 3)", "$\\frac{180^\\circ}{\\pi}$ (1) $-\\frac{7\\pi}{6}\\cdot\\frac{180^\\circ}{\\pi}=-210^\\circ$ 3. Great! Following your formula I got 78 degrees for the other problem which is correct, so I think I've got the hang of it now. Thanks so much! -Cillian3", "the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round", "$0 < x < 180$ as was stated in your original post. 7. Jun 15, 2015 ### Tangeton Ahh okay I forgot about that bit.. I guess I would have to remember to use the CAST diagram for my answer. Thanks for all help, everyone. 8. Jun 15, 2015 ### ehild You know that the tangent function is periodic with period of 180°. And you need to find solution in the range 0<θ<180°. Adding 180° to your result is also solution, just in the desired range.", "range of inverse tangent is $\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$ ... $(-90^\\circ,90^\\circ)$ , quads I and IV 6. ## Thank you! Thanks so much!", "inverse tangent of that you get \\begin{align} x_1 &= \\tan^{-1} y_1 =\\tan^{-1} 1.0000 = 45^\\circ\\\\ x_2 &= \\tan^{-1} y_2 =\\tan^{-1} 0.2679= 15^\\circ \\end{align} • I think you missed multiplying 1+tanx on the R.H.S – NoLand'sMan May 27 at 8:24 • right .. ill edit – Ahmad Bazzi May 27 at 8:25 $$\\frac{\\sqrt{3} +1}{2\\sqrt{2}}=\\frac{\\sqrt{3}\\sqrt{2} +\\sqrt{2}}{4}=\\frac{\\sqrt{2}}{2}\\cdot\\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{2}}{2}\\cdot\\frac{1}{2}=\\sin{45^\\circ}\\cos{30^\\circ}+\\cos{45^\\circ}\\sin{30^\\circ}=\\sin{(45^\\circ+30^\\circ)}=\\sin{75^\\circ}$$", "$t = \\frac{3 \\pi}{16} + \\frac{k \\pi}{4}$, or in degrees --> $t = {33}^{\\circ} 75 + k {90}^{\\circ}$ Check by calculator. t = 33.75 + 90 = 123.75 --> 4t = 495 --> tan 4t = - 1. Proved.", "# How do you find the period of y = 2 tan 3x? $\\frac{\\pi}{3}$ The period of the tangent function is $\\pi$. Therefore, the period of $\\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ ($3 x$ will increase by $\\pi$ units anytime $x$ increases by $\\frac{\\pi}{3}$ units). This means the period of $y = 2 \\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ as well. The \"2\" is irrelevant." ]

[ "# Tangent of a complex number Calculator for the tangent of a complex number ## Tangent calculator This function returns the tangent to an angle specified as a complex number. To perform the calculation, enter the complex number. Then click on the 'Calculate' button. Complex number tangent calculator Input Complex number + i Decimal places 0 1 2 3 4 6 8 10 Result Cosine ## Formulas for tangent of complex number In the following description, $$z$$ stands for the complex number. $$x$$ stands for the real value $$Re$$ and $$y$$ for the imaginary value $$Im$$. $$\\displaystyle tan(z) = \\frac{sin(2x)}{cos(2x)+cosh(2y)} + \\frac{sinh(2y)}{cos(2x)+cosh(2y)}i$$ Example $$tan(z) = sinh(0.3+0.5i)$$ $$\\displaystyle tan(z) = \\frac{sin(2\\cdot0.3)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)} + \\frac{sinh(2\\cdot0.5)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)}i$$ $$\\displaystyle tan(z) = \\frac{sin(0.6)}{cos(0.6)+cosh(1)} + \\frac{sinh(1)}{cos(0.6)+cosh(1)}i$$ $$\\displaystyle tan(z) = \\frac{0.565}{2.368} + \\frac{1.175}{2.368}i$$ $$\\displaystyle tan(z) = tan(0.3+0.5i) = 0.238+0.496i$$", "your calculator. Round to 4 decimal places. a) $\\sin 78^\\circ$ b) $\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Solution: Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.9781$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.2679$ ## Finding the Sides of a Triangle using Trig Ratios One application of the trigonometric ratios is to use them to find the missing sides of a right triangle. Example 5: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the hypotenuse. Use sine to find $b$, and cosine to find $a$. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example 6: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the adjacent leg to $42^\\circ$. To find $c$, use cosine and tangent to find $d$. $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &=", "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "NZ Level 7 (NZC) Level 2 (NCEA) Period changes for tangents Lesson We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\\sin kx$sinkx and $\\cos kx$coskx. The fact that the sine and cosine functions both have period $360^\\circ$360° is expressed by the statements $\\sin(x+360^\\circ n)=\\sin x$sin(x+360°n)=sinx and $\\cos(x+360^\\circ n)=\\cos x$cos(x+360°n)=cosx for all integers $n$n. So it must be true that since $\\tan x=\\frac{\\sin x}{\\cos x}$tanx=sinxcosx, then $\\tan(x+360^\\circ n)=\\tan x$tan(x+360°n)=tanx. However, $360^\\circ$360° is not the smallest interval at which the tangent function repeats. A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\\circ$180°. That is, $\\tan(x+180^\\circ n)=\\tan x$tan(x+180°n)=tanx for all integers $n$n. The fact that the tangent function repeats at intervals of $180^\\circ$180° can be verified by considering the unit circle diagram. If $180^\\circ$180° is added to an angle $\\alpha$α, then the diagram shows that $\\sin(\\alpha+180^\\circ)$sin(α+180°) has the same magnitude as $\\sin\\alpha$sinα but opposite sign. The same relation holds between $\\cos(\\alpha+180^\\circ)$cos(α+180°) and $\\cos\\alpha$cosα. We make use of the definition: $\\tan\\alpha=\\frac{\\sin\\alpha}{\\cos\\alpha}$tanα=sinαcosα $\\tan(\\alpha+180^\\circ)$tan(α+180°) $=$= $\\frac{\\sin(\\alpha+180^\\circ)}{\\cos(\\alpha+180^\\circ)}$sin(α+180°)cos(α+180°)​ $=$= $\\frac{-\\sin\\alpha}{-\\cos\\alpha}$−sinα−cosα​ $=$= $\\tan\\alpha$tanα We are now in a position to determine the period of the function $\\tan kx$tankx where $k$k is any number. We can define a new", "period of $6\\pi$ . 1. The period is $\\frac{\\pi}{\\frac{3}{2}}=\\pi \\cdot \\frac{2}{3}=\\frac{2\\pi}{3}$ . 2. The zeros are where $y$ is zero. $0 &=-4\\tan \\frac{3}{2}x \\\\0 &=\\tan \\frac{3}{2}x \\\\\\frac{3}{2}x &=\\tan ^{-1}0=0, \\pi, 2\\pi, 3\\pi \\\\x &=\\frac{2}{3}(0, \\pi, 2\\pi, 3\\pi ) \\\\x &=0, \\frac{2\\pi}{3}, \\frac{4\\pi}{3}, 2\\pi$ 3. The general equation is $y=a\\tan bx$ . We know that $a = 8$ . Let’s use the period to solve for the frequency, or $b$ . $\\frac{\\pi}{b} &=6\\pi \\\\b &=\\frac{\\pi}{6\\pi}=\\frac{1}{6}$ The equation is $y=8\\tan \\frac{1}{6}x$ . ### Vocabulary Tangent Function Defined by the coordinates $(\\theta, \\tan \\theta)$ , where $\\theta$ is the central angle from the unit circle and tangent is the ratio of the sine and cosine functions. ### Practice Graph the following tangent functions over $[0, 4\\pi]$ . Determine the period, domain, and range. 1. $y=2\\tan x$ 2. $y=-\\frac{1}{3}\\tan x$ 3. $y=-\\tan 3x$ 4. $y=4\\tan 2x$ 5. $y=\\frac{1}{2}\\tan 4x$ 6. $y=-\\tan \\frac{1}{2}x$ 7. $y=4+\\tan x$ 8. $y=-3+\\tan 3x$ 9. $y=1+\\frac{2}{3}\\tan \\frac{1}{2}x$ 10. Find the zeros of the function from #1. 11. Find the zeros of the function from #3. 12. Find the zeros of the function from #5. Write the equation of the tangent function, in the form $y=a\\tan bx$ , with the given amplitude and period. 1. Amplitude: 3 Period: $\\frac{3\\pi}{2}$ 2. Amplitude: $\\frac{1}{4}$ Period: $2\\pi$ 3. Amplitude: -2.5 Period: 8", "The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The cosine function cos : R -> R The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The tangent function tan : R -> R The range of the function is R . Now, the period is $pi$ and the function is undefined at x = ($pi$/2) + k$pi$, k=0,1,2. The graph of the tangent function on the interval 0 - $pi$ #### The cotangent function cot : R -> R The range of the function is R . The period is $pi$ and that the function is undefined at x = k$pi$, k=0,1,2. #### The values of sin, cos, tan, cot at the angles of 0°, 30°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° $A^o$ ^o30^o45^o60^o90^o120^o135^o150^o180^o210^o225^o240^o270^o300^o315^o330^o360^oA rad $frac<6>$ $frac<4>$ $frac<3>$ $frac<2>$ $frac<2pi><3>$ $frac<3pi><4>$ $frac<5pi><6>$ $pi$ $frac<7pi><6>$ $frac<5pi><4>$ $frac<4pi><3>$ $frac<3pi><2>$ $frac<5pi><3>$ $frac<7pi><4>$ $frac<11pi><6>$ $2pi$ $sin A$ frac<1><2>frac><2>frac><2>1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ cos A1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ frac<1><2>frac><2>frac><2>1 an A $frac><3>$ $1$ $sqrt<3>$ $-$ $-sqrt<3>$ $-1$ $-frac><3>$ frac><3>1sqrt<3>--sqrt<3>-1-frac><3> $cot A$ $-$ $sqrt<3>$ $1$ $frac><3>$ -frac><3>-1-sqrt<3>-sqrt<3>1frac><3> $-frac><3>$ $-1$ $-sqrt<3>$ $-$ The easiest way to remember the basic values of sin and cos at the angles", "# How do you find the period of y = 2 tan 3x? $\\frac{\\pi}{3}$ The period of the tangent function is $\\pi$. Therefore, the period of $\\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ ($3 x$ will increase by $\\pi$ units anytime $x$ increases by $\\frac{\\pi}{3}$ units). This means the period of $y = 2 \\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ as well. The \"2\" is irrelevant.", "obtain 15° by subtracting 45° - 30°. We will now convert back to radians. $$45°=\\frac{π}{180 °}=\\frac{π}{4}$$ $$30°=\\frac{π}{180 °}=\\frac{π}{6}$$ For reference, let's obtain the tangent of each: $$\\text{tan}\\frac{π}{4}=1$$ $$\\text{tan}\\frac{π}{6}=\\frac{\\sqrt{3}}{3}$$ Now, we return to our problem. Let's use our difference identity for tangent: $$\\text{tan}\\frac{π}{12}$$ $$=\\text{tan}\\left(\\frac{π}{4}- \\frac{π}{6}\\right)$$ $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}$$ Simplify: $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}\\cdot \\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 - \\large{\\frac{\\sqrt{3}}{3}}}$$ $$=\\frac{1 - \\large{\\frac{2\\sqrt{3}}{3}}+ \\large{\\frac{1}{3}}}{1 - \\large{\\frac{1}{3}}}$$ $$=\\frac{\\Large{\\frac{4 - 2 \\sqrt{3}}{3}}}{\\Large{\\frac{2}{3}}}$$ $$=\\require{cancel}\\frac{4 - 2\\sqrt{3}}{\\cancel{3}}\\cdot \\frac{\\cancel{3}}{2}$$ $$=\\frac{\\cancel{2}(2 - \\sqrt{3})}{\\cancel{2}}$$ $$=2 - \\sqrt{3}$$ We will also see problems that ask us to simplify. Let's look at an example. Example #3: Simplify. $$\\text{sin}\\hspace{.1em}3 θ \\hspace{.1em}\\text{cos}({-}2θ) - \\text{cos}\\hspace{.1em}3θ \\hspace{.1em}\\text{sin}(-2 θ)$$ To simplify, we use our difference identity for sine: $$\\text{sin}(3θ - (-2θ))=\\text{sin}\\hspace{.1em}5 θ$$ ### Finding Function Values and the Quadrant of A + B Another type of problem involves finding trigonometric function values using the sum and difference identities. Let's look at some examples. Example #4: Find sin(θ + β), tan(θ + β), and the quadrant of A + B $$\\text{sin}\\hspace{.1em}θ=\\frac{4}{5}$$ θ is in quadrant II. $$\\text{cos}\\hspace{.1em}β=-\\frac{5}{13}$$ β is in quadrant III. To find the sine of (θ + β), we need to obtain the sine of β and the cosine of θ: $$\\text{cos}^2 β + \\text{sin}^2 β=1$$ $$\\left(-\\frac{5}{13}\\right)^2 + \\text{sin}^2 β=1$$ $$\\frac{25}{169}+ \\text{sin}^2 β=1$$ $$\\text{sin}^2 β=\\frac{169}{169}- \\frac{25}{169}$$ $$\\text{sin}^2 β=\\frac{144}{169}$$ $$\\text{sin}\\hspace{.1em}β=\\pm \\sqrt{\\frac{144}{169}}$$ Since", "$t = \\frac{3 \\pi}{16} + \\frac{k \\pi}{4}$, or in degrees --> $t = {33}^{\\circ} 75 + k {90}^{\\circ}$ Check by calculator. t = 33.75 + 90 = 123.75 --> 4t = 495 --> tan 4t = - 1. Proved.", "the function $y=0.5\\tan(\\frac{\\pi}{2}x)\\\\$. Solution First, we identify A and B. Figure 2 Because $A=0.5$ and $B=\\frac{\\pi}{2}\\\\$, we can find the stretching/compressing factor and period. The period is $\\frac{\\pi}{\\frac{\\pi}{2}}=2\\\\$, so the asymptotes are at $x=\\pm 1$. At a quarter period from the origin, we have $\\begin{array}f(0.5)=0.5\\tan(\\frac{0.5\\pi}{2}) \\hfill& \\\\ =0.5\\tan(\\frac{\\pi}{4}) \\hfill& \\\\ =0.5 \\end{array}\\\\$ This means the curve must pass through the points(0.5,0.5),(0,0),and(−0.5,−0.5).The only inflection point is at the origin. Figure shows the graph of one period of the function. Figure 3 Try It 1 Sketch a graph of $f(x)=3\\tan\\left(\\frac{\\pi}{6}x\\right)\\\\$. Solution Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. $f(x)=A\\tan(Bx−C)+D\\\\$ The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: A General Note: Features of the Graph of y = Atan(Bx−C)+D • The stretching factor is |A|. • The period is $\\frac{\\pi}{|B|}\\\\$. • The domain is $x\\ne\\frac{C}{B}+\\frac{\\pi}{|B|}k\\\\$, where k is an integer. • The range is (−∞,−|A|] ∪ [|A|, ∞). • The vertical asymptotes occur at $x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k\\\\$, where k is an odd integer. • There is no amplitude. • $y=A\\tan(Bx)$ is an", "= \\frac {2}{3}\\\\ \\tan\\theta = \\frac {2}{3}\\\\ \\theta = \\arctan (\\frac {2}{3})$ This is not one of your special angles, so you will need a calculator. $\\theta = 33.69^\\circ$ $\\tan \\theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\\circ)$. There is second answer in the interval $[0,360^\\circ)$. $\\tan \\theta = \\tan (\\theta + 180)$ $213.69^\\circ$ is also in the solution set. The next one might even be a little bit trickier. $\\cos 3\\theta = -0.5\\\\ 3\\theta = \\arccos (-0.5)\\\\ 3\\theta = 120^\\circ$ but again, this isn't the only answer. $\\theta = \\arccos (-0.5)$ has two answers in the 360 degree interval, and $3\\theta = \\arccos (-0.5)$ will have $6.$ $\\cos \\theta = \\cos (-\\theta)$ and $\\cos \\theta = \\cos (360+\\theta)$ using these identities we find other values of theta. $3\\theta = 240^\\circ, 480^\\circ, 600^\\circ, 840^\\circ, 960^\\circ$ are all viable, as well. $\\theta \\in \\{40, 80, 160, 200, 280, 320\\}$ a) Reduce to $$\\tan\\theta=\\frac23.$$ Then as the tangent function has period $180°$ and is monotonic in a period, $$\\tan\\theta=\\tan(\\theta+k180°)$$ and all solutions are of the form $$\\theta+k180°=\\arctan\\frac23.$$ Find all $k$ such that $0\\le\\theta\\le360°$. b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that $$\\cos3\\theta=\\cos(k360°\\pm3\\theta).$$ Then all solutions are of the form $$k120°\\pm\\theta=\\frac13\\arccos(-0.5).$$ Again, find all $k$", "\\leq \\pi$$, or $$0^{\\circ} \\leq x \\leq 180^{\\circ}$$ in degree mode. Note that we could also have taken $$-\\pi \\leq x \\leq 0$$, but for convenience we take $$x$$ to be a positive angle. Lastly, for $$\\tan(x)$$ we can take a full period around the origin, so $$-\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2}$$, or $$-90^{\\circ} \\leq x \\leq 90^{\\circ}$$ in degree mode.", "# How do you find the period of y = tan (x + pi/3)? Jan 19, 2016 Use the coefficient on the variable x to find the period ... #### Explanation: For this problem, the coefficient on x is $1$. Next, use this formula for finding the period for tangent ... Period $= \\frac{\\pi}{\\text{coefficient}} = \\frac{\\pi}{1} = \\pi$ So, this function has the same period as a standard tangent function except it has a phase shift of $\\frac{\\pi}{3}$ . See graph below. Hope that helped", "in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round your answers to the nearest tenth:", "brb 20. anonymous thats not any of the answer choices 21. Mertsj |dw:1443103740128:dw| 22. anonymous yes then what @Mertsj 23. Mertsj $\\tan 2511/60=\\frac{x}{19427.5}$ 24. anonymous i have it set up like that i just dont remeber how to solve it thats what ive been trying to ask 25. Mertsj $.89567=\\frac{x}{19427.5}$ 26. Mertsj $.89567(19427.5)=x$ 27. anonymous i got 17,400 28. Mertsj x= 17400.72 ft 29. anonymous but that cant be the answer because its not any of the answer choices 30. Mertsj 31. anonymous 9135, 9235, 91351, 913515 32. anonymous and 17400 isnt close to any 33. Mertsj whoops. I must have misused the tangent function on my calculator. I now get .47020 for the tangent of 25 11/50 degrees. 34. anonymous yeah i just figured it out thank you 35. Mertsj $.47021=\\frac{x}{19427.5}$ 36. anonymous i got it its 9141 37. Mertsj $x= .47021(19427.5)=9134.9859$ 38. Mertsj yw", "the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round", "$\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ #### Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ #### Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that", "\\pi, 2\\pi, 3\\pi, 4\\pi \\\\ x &=0, \\frac{\\pi}{2}, \\pi, \\frac{3\\pi}{2}, 2\\pi$$ Example C Graph $y=\\frac{1}{4}\\tan\\frac{1}{4}x$ from $[0, 4\\pi]$ and state the domain and range. Solution: This function has a period of $\\frac{\\pi}{\\frac{1}{4}}=4\\pi$. The domain is all real numbers, except $2\\pi, 6\\pi, 10\\pi, 2\\pi \\pm 4\\pi n$, where $n$ is any integer. The range is all real numbers. Guided Practice 1. Find the period of the function $y=-4\\tan \\frac{3}{2}x$. 2. Find the zeros of the function from #1, from $[0, 2\\pi]$. 3. Find the equation of the tangent function with an amplitude of 8 and a period of $6\\pi$. 1. The period is $\\frac{\\pi}{\\frac{3}{2}}=\\pi \\cdot \\frac{2}{3}=\\frac{2\\pi}{3}$. 2. The zeros are where $y$ is zero. $$0 &=-4\\tan \\frac{3}{2}x \\\\ 0 &=\\tan \\frac{3}{2}x \\\\ \\frac{3}{2}x &=\\tan ^{-1}0=0, \\pi, 2\\pi, 3\\pi \\\\ x &=\\frac{2}{3}(0, \\pi, 2\\pi, 3\\pi ) \\\\ x &=0, \\frac{2\\pi}{3}, \\frac{4\\pi}{3}, 2\\pi$$ 3. The general equation is $y=a\\tan bx$. We know that $a = 8$. Let’s use the period to solve for the frequency, or $b$. $$\\frac{\\pi}{b} &=6\\pi \\\\ b &=\\frac{\\pi}{6\\pi}=\\frac{1}{6}$$ The equation is $y=8\\tan \\frac{1}{6}x$. Vocabulary Tangent Function Defined by the coordinates $(\\theta, \\tan \\theta)$, where $\\theta$ is the central angle from the unit circle and tangent is the ratio of the sine and cosine functions. Problem Set Graph the following tangent functions over $[0, 4\\pi]$. Determine the period,", "your answers to the nearest tenth: . 3. Find the length of the missing sides and round your answers to the nearest tenth: . 1. Using your calculator, you should find that $\\tan 45^\\circ=1$ ? 2. Use tangent for $x$ and cosine for $y$ . $\\tan 28^\\circ &= \\frac{x}{11} && \\quad \\ \\ \\cos 28^\\circ = \\frac{11}{y}\\\\11 \\cdot \\tan 28^\\circ &= x && \\frac{11}{\\cos 28^\\circ} = y\\\\x & \\approx 5.8 && \\qquad \\quad \\ \\ \\ y \\approx 12.5$ 3. Use tangent for $y$ and cosine for $x$ . $\\tan 40^\\circ &= \\frac{y}{16} && \\quad \\ \\ \\cos 40^\\circ = \\frac{16}{x}\\\\16 \\cdot \\tan 40^\\circ &= y && \\frac{16}{\\cos 40^\\circ} = x\\\\y & \\approx 13.4 && \\qquad \\quad \\ \\ \\ x \\approx 20.9$ Practice Use your calculator to find the value of each trig function below. Round to four decimal places. 1. $\\sin 24^\\circ$ 2. $\\cos 45^\\circ$ 3. $\\tan 88^\\circ$ 4. $\\sin 43^\\circ$ 5. $\\tan 12^\\circ$ 6. $\\cos 79^\\circ$ 7. $\\sin 82^\\circ$ Find the length of the missing sides. Round your answers to the nearest tenth.", "the equation for the hill is $$Y = X \\cdot tan(20^\\circ)$$ Substituting for Y in the parabola equation (equation A) gives us $$X \\cdot tan(20^\\circ) = X - 4.9 \\cdot \\left( \\frac{X^2}{V_{ix}^2} \\right)$$ $$tan(20^\\circ) = 1 - 4.9 \\cdot \\left( \\frac{X}{V_{ix}^2} \\right)$$ $$\\left(\\frac{4.9}{V_{ix}^2}\\right) \\cdot X = 1 - tan(20^\\circ)$$ $$X = \\frac{V_{ix}^2(1 - tan(20^\\circ))}{4.9} = \\frac{(50 \\cdot cos(45^\\circ))^2 \\cdot (1 - tan(20^\\circ))}{4.9}$$ $$X = \\frac{35.355^2 \\cdot 0.636}{4.9} = \\frac{1250 \\cdot 0.636}{4.9} = \\frac{795.037}{4.9} = 162.252$$ Plug this value back into equation A, above, and you get Y = 59.055 The distance up the his is found using the Pythagorean Theorem: $$D^2 = X^2 + Y^2 = 162.252^2 + 59.055^2 = 29813.373$$ and D = 172.665 m (note: I only rounded off in the text, not in my actual calculations) Last edited: Jun 12, 2010" ]

[ "of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(1) = \\frac{1}{2}$ and maximum value of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(0)=1$ so, range of f(x)=[1/2,1] I think there is any mistake in the book. As you can see the detailed solution of the question and if there is any other doubt related to this question you can ask", "# find the Range Geometry Level pending What is the maximum value of $$f(x) = 4\\sin^4 x - \\sin^2 2x$$ for all real $$x$$? ×", "$0$ to $-1$, and for $\\pi/2 \\le x \\le \\pi,\\, \\sin x$ decreases from $1$ to $0$. Hence in these ranges $f(x)$ is increasing, and therefore $f'(x) = +1$. The graph is as shown in the attached diagram.", "are in the range of the function $f(x) = \\frac{4x^2+75}{2x^2+3}$?", "# How do you calculate f '(x) and use calculus to find the maximum value of sin (ln x) on the interval [1, 10]? The answer is: $1$. This function is in the range $\\left[- 1 , 1\\right]$ and oscillates continually between these two values.", "$f(x) = x^2$ defined on $[-2,2]$. Then you are saying that the range is $[4,4] = \\{4\\}$ ?! Apr 4, 2014 at 23:41", "x^2 . The range of $f(x)= x^2$ is the set of all non-negative numbers. 8. Originally Posted by thereddevils How about only one of its side is negative ? $-5 TO find $x^2$, do i do that this way : math]25>x^2>4 [/tex] Then rearrange $4 Originally Posted by mr fantastic x = 0 satisfies $-5. Does x = 0 satisfy $4 ....? Originally Posted by thereddevils THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 . Draw a graph of y = x^2 for -5 < x < 2. Now look at the range of values of x^2.", "# What is the range of f(x) = -3^x + 4? May 6, 2015 Write $y = - {3}^{x} + 4$ $\\implies {3}^{x} = 4 - y$ Take $\\ln$ of both sides $\\implies \\ln {3}^{x} = \\ln \\left(4 - y\\right)$ $\\implies x = \\ln \\frac{4 - y}{\\ln} 3$ Now notice that $\\left(4 - y\\right)$ cannot be negative nor zero! $\\implies 4 - y > 0 \\implies y < 4$ Hence the range of $f \\left(x\\right)$ is $f \\left(x\\right) < 4$", "Math Help - range 1. range Find Range of $f(x) = \\sin^2 x-5\\sin x+6$ where $x\\in \\left[0,\\pi\\right]$ 2. Re: range Originally Posted by jacks Find Range of $f(x) = \\sin^2 x-5\\sin x+6$ where $x\\in \\left[0,\\pi\\right]$ $f'(x)=2\\sin x \\cdot \\cos x-5 \\cos x=\\cos x \\cdot (2\\sin x-5)$ In order to find minimum of function we have to solve following equation : $f'(x)=0 \\Rightarrow \\cos x=0 \\Rightarrow x= \\frac{\\pi}{2} \\Rightarrow \\sin x = 1$ So ,we can conclude that : $f_{min}=2$ On the other hand function has maximum for $x=0$ , and $x=\\pi$ , therefore : $f_{max}=6$ So for $x \\in [0,\\pi]$ we have : $f(x) \\in [2,6]$", "+0 0 145 1 +102 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "# How do you find the range of f(x)=5cos(x + pi) + 3? $\\cos \\left(x + \\pi\\right)$ has a range of $\\left[- 1 , + 1\\right]$ (because cos(anything) has a range of $\\left[- 1 , + 1\\right]$) $5 \\cos \\left(x + \\pi\\right)$ has a range of $\\left[- 5 , + 5\\right]$ $f \\left(x\\right) = 5 \\cos \\left(x + \\pi\\right) + 3 h a s a r a n \\ge o f$[-2,+8]#", "# Think Before you Act Geometry Level 4 Find the range of the function $$f(x)=\\sin^{-1}x+\\cos^{-1}x+\\tan^{-1}x$$. ×", "# if f(x)=sin4x + cos4x, then what is the range of f(x) Vijay Mukati 6 years ago Dear Student, Differnetiate the functions and equate the result with zero to get the point of minima and maximua. You will get x = 0, 45 and 90 degree. From here we can easily find the range and it will come out to be [1/2, 1] Thanks. Chirag Jain 36 Points 6 years ago Sir, the answer is [3/4,1] as stated in the book and the book is Objective Maths by Purshottam Kumar Sharma Chirag Jain 36 Points 6 years ago The service of askiitians is very slow, posted a question 3 days ago, but till now haven’t got any satisfactory rreply. How can someone invest in such a slow service? I am sorry if you think I have been rude in my comment. arun 123 Points 6 years ago in this question we have to find the range for $\\sin ^4x + \\cos ^4x$ now,$\\sin ^4x + \\cos ^4x = (\\sin ^2x)^2+(\\cos ^2x)^2$ $\\Rightarrow \\sin ^4x + \\cos ^4x = (\\sin ^2x + \\cos ^2x)^2-2\\sin ^2x\\cos ^2x$ $\\Rightarrow \\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(4\\sin^2x\\cos^2x)$ $\\Rightarrow \\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}\\sin^22x$ as $\\sin^2\\theta$ have a maximum value of 1 and a minimum value of 0 so, minimum value", "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "\\sin \\theta } \\right) = 4$.", "# Let's get a little trigy Geometry Level 3 What's the range of the function $f(x)=\\dfrac{\\sin(x)-1}{\\sqrt{3-2\\cos(x)-2\\sin(x)}}$ for $0≤x≤2π$? ×", "x+4 \\sin x+1$", "# Math Help - Range of values 1. ## Range of values Find the range of values of $x$ bewtween $0$ and $\\pi$ inclusive which satisfies i) $sin\\frac{x}{2}>0.5$ ii) $cos2x<0$ 2. Originally Posted by Punch Find the range of values of $x$ bewtween $0$ and $\\pi$ inclusive which satisfies i) $sin\\frac{x}{2}>0.5$ ii) $cos2x<0$ i) $SinA=0.5\\Rightarrow\\ A=30^o,\\;\\; A=150^o$ $0.5<\\ SinA\\le\\ 1\\Rightarrow\\ 30^o $x=2A\\Rightarrow\\ 60^o<2A<300^o$ $\\displaystyle\\ Sin\\left(\\frac{x}{2}\\right)>0.5\\Rightarrow\\ 60^o For $\\displaystyle\\ 0^o0.5\\Rightarrow\\ x>60^o$ Try the second one.", "$f(x)={{\\sin }^{-1}}\\left( {{\\cos }^{-1}}\\left[ x \\right] \\right)$ $f(x)$ is defined when $\\frac{x\\in [1,2)}{{{\\cos }^{-1}}[x]}=\\frac{\\pi }{2}>1$ as, when $x\\in [1,2)$; Range of $f(x)$ is $0$", "Mathematics # What is the greatest value in the range of f(x)=x^2-5 for the domain {-5, -3, 0, 1, 2}? $f(x)=x^2-5;\\ x\\in\\{-5;-3;\\ 0;\\ 1;\\ 2\\}\\\\\\\\f(-5)=(-5)^2-5=25-5=20\\\\f(-3)=(-3)^2-5=9-5=4\\\\f(0)=0^2-5=-5\\\\f(1)=1^2-5=1-5=-4\\\\f(2)=2^2-5=4-5=-1\\\\\\\\Answer:f(-5)=20$" ]

[ "# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and", "49 views If $f(x) = \\dfrac{\\sqrt{3} \\sin x}{2+\\cos x}$, then the range of $f(x)$ is 1. the interval $[-1 , \\sqrt{3}{/2}]$ 2. the interval $[-\\sqrt{3}{/2}, 1]$ 3. the interval $[-1, 1]$ 4. none of these in Calculus recategorized | 49 views Answer: $\\mathbf C$ $f^2(x) =\\dfrac{3-3\\cos^2x}{\\cos^2x+4\\cos x+4}$ Now, Let t = $\\cos x$ then, $f(t) = \\dfrac{3-3t^2}{t^2+4t+4}$ $f_{max}.f'(t) = \\frac{-6(2t+1)}{t+2} = 0$, when$t = \\frac{-1}{2}$ $\\therefore f(-1) =0, f(\\frac{-1}{2}) = 1, \\;and\\; f(1) = 0$ So, $f^2\\text{max} = 1 \\implies f^2(x) \\leq 1 \\implies -1\\leq f(x) \\leq 1$ So, the range of $f$ is $[-1,1]$ $\\therefore \\mathbf C$ is the right option. by Boss (18.9k points) edited by To find the range of $f(x)$, actually we need to find the maximum and minimum value. Now \\begin{align}f(x)&=\\frac{\\sqrt{3}\\sin x}{2+\\cos x}\\\\ \\Rightarrow f'(x)&=\\sqrt{3}\\cos x \\cdot \\frac{1}{2+\\cos x}+\\sqrt{3}\\sin x \\cdot \\frac{-\\sin x}{(2+\\cos x)^2}\\\\ &=\\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} \\end{align} At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \\begin{align}\\therefore f'(x)&=0\\\\ \\Rightarrow \\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} &=0\\\\ \\Rightarrow 2\\cos x +1&=0;~[\\because \\sin^2 x+\\cos^2 x=1] \\\\ \\Rightarrow \\cos x &= -\\frac{1}{2}\\\\ \\Rightarrow x &= \\frac{2\\pi}{3}, \\frac{4\\pi}{3}\\end{align} Now putting those values of $x$ to $f(x)$, we obtain $f(\\frac{2\\pi}{3})=\\frac{\\sqrt{3}(\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=1$ and $f(\\frac{4\\pi}{3})=\\frac{\\sqrt{3}(-\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$. $$\\therefore f(x) \\in [-1,1]$$ So the correct", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "&= \\int_0^1 ((x-c)^2+c^2)dx\\\\ &=\\left[ \\frac{(x-c)^3}{3} +c^2x \\right]_0^1\\\\ &=\\frac{(1-c)^3}{3} + c^2 + \\frac{c^3}{3}\\\\ &=2c^2-c+\\frac{1}{3}. \\end{align*} 1. What is the minimum value of $I(c)$ (as $c$ varies)? Using the expression for $I(c)$ we calculated above, and completing the square, we have \\begin{align*} I(c) &= 2c^2-c+\\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 - \\frac{1}{8} + \\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 + \\frac{5}{24}. \\end{align*} So the minimum value that $I(c)$ can take is $5/24$. (This occurs when $c=1/4$.) 1. What is the maximum value of $I(\\sin \\theta)$ as $\\theta$ varies? Note that from our work above, the MINIMUM value of $I(\\sin\\theta)$ happens when $\\theta = \\arcsin (0.25)$. Now $c=\\sin \\theta$ varies between between $-1$ and $1$, so we need to find the maximum value of $I$ over this range. We worked out earlier that the overall minimum of $I$ occurs at $c=1/4$, and since $I$ is quadratic, it is symmetric about this point, and increasing for larger and smaller values of $c$. So the maximum value of $I(c)$ over the range $-1 \\le c \\le 1$ occurs at $c=-1$, the furthest point from $x=1/4$ in the range. So the maximum value of $I(\\sin \\theta)$ is $I(-1)=10/3$.", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "- \\frac {x}{2},\\\\f_n'(x) &= - \\mbox{Re}\\left(\\sum_{n = 1}^{n}z^n\\right) - \\frac12.\\end{align*} After some simplifications we get $$f_n'(x) = \\frac {( - 1)^{n + 1}}{2}((1 - \\cos(x))\\frac {\\sin((n + 1)x)}{\\sin(x)} + \\cos((n + 1)x))$$ and $$f_n''(x) = \\frac {( - 1)^{n}}{2}\\frac {(n + 1)\\sin(nx) + n\\sin((n + 1)x)}{1 + \\cos(x)}.$$ The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \\frac {\\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \\frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\\frac {\\pi}{n + 1}) > 0$. We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even. Let's restrict to the case $n$ is even. We prove $f_{2n}(\\frac {\\pi}{2n + 1}) < 0$. \\begin{align*}f_{2n}\\left( \\frac{\\pi}{2n+1} \\right) &=\\sum_{k=1}^{2n}{\\left( -1 \\right)}^{k+1}\\frac{\\sin \\left( \\frac{k\\pi}{2n+1} \\right)}{k}-\\frac{\\pi}{2\\left( 2n+1 \\right)}\\\\&=\\frac{\\pi}{2n+1}\\left( \\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{\\left( 2k-1 \\right) \\pi}{2n+1} \\right)}{\\frac{\\left( 2k-1 \\right) \\pi}{2n+1}}}-\\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{2k\\pi}{2n+1} \\right)}{\\frac{2k\\pi}{2n+1}}} \\right) -\\frac{\\pi}{2\\left( 2n+1 \\right)}.\\end{align*} The function $x \\mapsto \\frac {\\sin(x)}{x}$ is descending on $[0,\\pi]$, thus both sums lay between $a$ and $a + \\frac {2\\pi}{2n + 1}$, where $a = \\int_0^{\\pi}\\frac {\\sin(x)}{x}\\,dx$. Thus $$f_{2n}\\left(\\frac {\\pi}{2n + 1}\\right) < \\frac {\\pi}{2n + 1}\\cdot\\frac {2\\pi}{2n + 1} - \\frac {\\pi}{2(2n + 1)} < 0.$$ ## Euler-Maclaurin求和公式估计梯形积分公式的误差 $\\mathop {\\lim }\\limits_{n \\to \\infty }", "- {\\cot ^{ - 1}}x$ We have seen that $${f^{ - 1}}(f(x)) = f({f^{ - 1}}(x)) = x$$. For example, \\begin{align}&\\sin \\left( {{{\\sin }^{ - 1}}\\left( {\\frac{1}{2}} \\right)} \\right) = \\sin \\left( {\\frac{\\pi }{6}} \\right) = \\frac{1}{2} \\\\ &{\\sin ^{ - 1}}\\left( {\\sin \\left( {\\frac{\\pi }{6}} \\right)} \\right) = {\\sin ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{6}\\quad etc. \\\\ \\end{align} But this relation must be used with caution, since it is valid only if x lies in the appropriate domain. For example, ${\\sin ^{ - 1}}\\left( {\\sin \\left( {\\frac{{5\\pi }}{6}} \\right)} \\right) = {\\sin ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{6}\\;\\;{\\text{and}}\\;{\\mathbf{not}}\\;\\frac{{5\\pi }}{6}$ Therefore $${\\sin ^{ - 1}}(\\sin \\theta )$$, will not always be $$\\theta$$, since the output of $${\\sin ^{ - 1}}$$ will necessarily lie in the range \\begin{align}\\left[ {\\frac{{ - \\pi }}{2},\\;\\frac{\\pi }{2}} \\right]\\end{align}. To adjust for this fact, note the following: \\begin{align}&{\\sin ^{ - 1}}(\\sin \\theta ) = \\left\\{ \\begin{gathered}- \\pi - \\theta ,\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{{ - 3\\pi }}{2},\\;\\frac{{ - \\pi }}{2}} \\right] \\\\ \\;\\;\\;\\;\\;\\theta ,\\;\\;\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{{ - \\pi }}{2},\\;\\frac{\\pi }{2}} \\right] \\\\ \\pi - \\theta ,\\;\\;\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{\\pi }{2},\\;\\frac{{3\\pi }}{2}} \\right] \\\\ - 2\\pi + \\theta ,\\mathop {{\\text{if}}}\\limits_ \\vdots \\;\\theta \\in \\;3\\pi /2,\\;5\\pi /2 \\\\ \\end{gathered} \\right\\} \\qquad\\left({{\\text{Verify!}}} \\right) \\\\ &\\qquad\\qquad\\qquad\\qquad\\qquad{\\text{and}}\\;{\\text{so}}\\;{\\text{on}} \\\\ \\end{align} That is, we have adjusted the", "say$\\displaystyle {} > \\frac 1 {\\sqrt{1+x^2}}$ unless we have $\\tan c < x. \\qquad$ Oct 18, 2020 at 17:02 • @MichaelHardy I don't think we have an assumption of $\\tan c < x$. But I have attached the snapshot of the question from the book here: imgur.com/a/UHtGZGb The question number is 12(b) and I hope I haven't missed anything. – Sibi Oct 18, 2020 at 17:10 • \\begin{align} & \\frac{\\sin x} x = \\cos c \\\\ {} \\\\ > {} & \\cos x \\text{ since the cosine function decreases on this interval} \\\\ {} \\\\ = {} & \\sqrt{1-\\sin^2 x} \\\\ {} \\\\ > {} & \\sqrt{1-x^2} \\text{ as shown in part (a) of the same exercise} \\\\{} \\\\ \\not> {} & \\frac 1 {\\sqrt{1+x^2}}\\quad \\text{So this is not there yet.} \\end{align} Oct 18, 2020 at 17:30 You are on the right path: square both sides of $$\\sin(x)>x (1-\\sin^2(x))^{1/2}$$ and get $$\\sin^2(x)>x^2(1-\\sin^2(x))=x^2-x^2\\sin^2(x)$$, then add $$x^2\\sin^2(x)$$ to both sides to obtain $$\\sin^2(x)(1+x^2)>x^2$$. Now divide both sides by $$(1+x^2)$$ and take the square root. • Can you expand on how squaring has given you $\\sin^2(x)(1+x^2)>x$ ? • @Sibi, I think it's supposed to be an $x^2$ on the right hand side of the $\\gt$. Oct 18, 2020 at 23:14 • @BarryCipra Thanks! If that's the case, then do you where", "the range $R = \\left[ {\\frac{\\pi }{6},\\frac{\\pi }{2}} \\right)$ (b) y =\\begin{align} \\frac{1}{{{{\\sin }^4}x + {{\\cos }^4}x}} = \\frac{1}{{{{({{\\sin }^2}x + {{\\cos }^2}x)}^2} - 2{{\\sin }^2}x{{\\cos }^2}x}}\\end{align} =\\begin{align} \\frac{1}{{1 - 2{{\\sin }^2}{{\\cos }^2}x}} = \\frac{1}{{1 - \\frac{1}{2}{{(\\sin 2x)}^2}}}\\end{align} Now $$- 1 \\le \\sin 2x \\le 1$$ $$\\Rightarrow$$ $$0 \\le {(\\sin 2x)^2} \\le 1$$ $$\\Rightarrow$$ $$0 \\le \\frac{1}{2}{(\\sin 2x)^2} \\le \\frac{1}{2}$$ $$\\Rightarrow$$ $$\\frac{1}{2} \\le 1 - \\frac{1}{2}{(\\sin 2x)^2} \\le 1$$ $$\\Rightarrow$$ \\begin{align}1 \\le \\frac{1}{{1 - \\frac{1}{2}{{(\\sin 2x)}^2}}} \\le 2\\end{align} Therefore, the range is [1, 2]. (c) y = $$\\sqrt {2 - x} + \\sqrt {1 + x}$$ = $$\\sqrt {2 - x} + \\sqrt {x - ( - 1)}$$ This is of the form $$\\sqrt {a - x} + \\sqrt {x - b}$$ , which we discussed earlier. The minimum for y comes at $$x = {\\rm{ }}-1,{\\rm{ }}2$$ : y = $$\\sqrt 3$$ The maximum comes at x = \\begin{align}\\frac{{ - 1 + 2}}{2}\\,\\,\\,\\,:\\,\\,\\,\\,y = \\sqrt 6\\end{align} The range is R = $$\\left[ {\\sqrt 3 ,\\sqrt 6 } \\right]$$ (d) $$y = \\sqrt {x - 1} + 2\\sqrt {3 - x}$$ This function is no longer symmetric, so we carry out a standard analysis. The domain is [1, 3]. Also, y > 0. Now $${y^2}$$ = $$(x - 1) + 4(3 - x) + 4\\sqrt {(x - 1)(3", "- 2,2} \\right]$$ is $$- 8$$ occurring at $$x = - 2$$. (ii) The given function is $$f\\left( x \\right) = \\sin x + \\cos x$$ Therefore, $$f'\\left( x \\right) = \\cos x - \\sin x$$ Now, \\begin{align}&f'\\left( x \\right) = 0\\\\& \\Rightarrow \\; \\cos x - \\sin x = 0\\\\ &\\Rightarrow \\; \\sin x = \\cos x\\\\ &\\Rightarrow \\; \\tan x = 1\\\\& \\Rightarrow \\; x = \\frac{\\pi }{4}\\end{align} Then, we evaluate the value of $$f$$ at critical point $$x = \\frac{\\pi }{4}$$ and at the end points of the interval $$\\left[ {0,\\pi } \\right]$$. Therefore, \\begin{align}f\\left( {\\frac{\\pi }{4}} \\right) &= \\sin \\left( {\\frac{\\pi }{4}} \\right) + \\cos \\left( {\\frac{\\pi }{4}} \\right)\\\\ &= \\frac{1}{{\\sqrt 2 }} + \\frac{1}{{\\sqrt 2 }} = \\frac{2}{{\\sqrt 2 }}\\\\ &= \\sqrt 2 \\end{align} \\begin{align}f\\left( 0 \\right) &= \\sin 0 + \\cos 0\\\\ &= 0 + 1\\\\& = 1\\\\f\\left( \\pi \\right) &= \\sin \\pi + \\cos \\pi \\\\& = 0 - 1\\\\ &= - 1\\end{align} Hence, we can conclude that the absolute maximum value of $$f$$ on $$\\left[ {0,\\pi } \\right]$$ is $$\\sqrt 2$$ occurring at $$x = \\frac{\\pi }{4}$$. Also, the absolute minimum value of $$f$$ on $$\\left[ {0,\\pi } \\right]$$ is $$- 1$$ occurring at $$x = \\pi$$. (iii) The given function is $$f\\left( x \\right) = 4x - \\frac{1}{2}{x^2}$$ Therefore,", "## Precalculus (6th Edition) Blitzer The values of x are $\\frac{\\pi }{6},\\frac{5\\pi }{6},\\,\\text{ and }\\,\\frac{3\\pi }{2}$. $\\cos 2x-\\sin x=0,\\,0\\le x\\,<2\\pi$ …… (1) Substituting the value of $\\cos 2x$ as $\\cos 2x=1-2{{\\sin }^{2}}x$ in equation (1) we get, \\begin{align} & 1-2{{\\sin }^{2}}x-\\sin x=0 \\\\ & -2{{\\sin }^{2}}x-\\sin x+1=0 \\end{align} Multiplying the above equation by $-1$ we get \\begin{align} & -2{{\\sin }^{2}}x-\\sin x+1=0 \\\\ & 2{{\\sin }^{2}}x+\\sin x-1=0 \\end{align} We can also express the equation as below: \\begin{align} & 2{{\\sin }^{2}}x+\\sin x-1=0 \\\\ & 2{{\\sin }^{2}}x+2\\sin x-\\sin x-1=0 \\end{align} Now, factorize the equation. \\begin{align} & 2\\sin x\\left( \\sin x+1 \\right)-1\\left( \\sin x+1 \\right)=0 \\\\ & \\left( 2\\sin x-1 \\right)\\left( \\sin x+1 \\right)=0 \\end{align} Solve for x. $2\\sin x-1=0$ or $\\sin x+1=0$ $2\\sin x=1$ $\\sin x=-1$ $\\sin x=\\frac{1}{2}$ $x=\\frac{3\\pi }{2}$ $x=\\frac{\\pi }{6}$ In the range $0\\le x\\,<2\\pi$, the values of x for $2\\sin x-1=0$ are $x=\\pi -\\frac{\\pi }{6}=\\frac{5\\pi }{6}$.", "}{4} - \\cos \\frac{\\pi }{4}$ $f\"\\left( {\\frac{\\pi }{4}} \\right) = - \\frac{1}{{\\sqrt 2 }} - \\frac{1}{{\\sqrt 2 }} < 0$ ⇒ f is maximum at x = π/4 $f\\left( {\\frac{\\pi }{4}} \\right) = \\sin \\frac{\\pi }{4} + \\cos \\frac{\\pi }{4}$ $f\\left( {\\frac{\\pi }{4}} \\right) = \\frac{1}{{\\sqrt 2 }} + \\frac{1}{{\\sqrt 2 }} = \\sqrt 2$ f(0) = sin0+cos0 = 0+1 = 1 f(π) = sinπ + cosπ = 0 +(-1) = -1 Absolute min = min{-1, 1, √2} = -1 at x=π Absolute max = max{-1, 1, √2} = √2 at x=π/4 $(iii)f\\left( x \\right) = 4x - \\frac{1}{2}{x^2},x \\in \\left[ { - 2,\\frac{9}{2}} \\right]$ $f'\\left( x \\right) = 4 - \\frac{1}{\\cancel{2}} \\cdot \\cancel{2}x$ f'(x) = 4 -x f\"(x) = -1 f'(x) = 0 4-x = 0 x = 4 f is maximum at x=4 f(4) = 4(4) - ½(4)2 = 8 local maximum value = 8 at x = 4 f(-2) = 4(-2) - ½ (-2)2 f(-2) = -8-2 = -10 $f\\left( {\\frac{9}{2}} \\right) = 4\\left( {\\frac{9}{2}} \\right) - \\frac{1}{2}{\\left( {\\frac{9}{2}} \\right)^2}$ $f\\left( {\\frac{9}{2}} \\right) = 18 - \\frac{{81}}{8}$ $f\\left( {\\frac{9}{2}} \\right) = 18 - 10.125 = 7.875$ Absolute max. = max{-10, 8, 7.875} = 8 Absolute min. = min { -10, 8. 7.875} = -10 (iv) f(x) = (x-1)2 + 3, x ∈ [-3, 1] f'(x)", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. • Angle Trisectors on Circumcircle • Equilateral Triangles On Sides of a Parallelogram • Pompeiu's Theorem • Pairs of Areas in Equilateral Triangle • The Eutrigon Theorem • Equilateral Triangle in Equilateral Triangle • Seven Problems in Equilateral Triangle • Spiral Similarity Leads to Equilateral Triangle • Parallelogram and Four Equilateral Triangles • Miguel Ochoa's van Schooten Like Theorem • Two Conditions for a Triangle to Be Equilateral • Incircle in Equilateral Triangle • When Is Triangle Equilateral: Marian Dinca's Criterion • Barycenter of Cevian Triangle • Excircle in Equilateral Triangle • Converse Construction in Pompeiu's Theorem • Wonderful Trigonometry In Equilateral Triangle • 60o Angle And Importance of Being The Other End of a Diameter • One More Property of Equilateral", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. [an error occurred while processing this directive]", "## Calculus (3rd Edition) The maximum is $f(2)=4\\sqrt{2}$ and the minimum is $f\\left( 0\\right) =2\\sqrt{6}$ Given $$y=(2+x) \\sqrt{2+(2-x)^{2}}, \\quad[0,2]$$ Since \\begin{align*} \\frac{dy}{dx}&=\\frac{d}{dx}\\left(2+x\\right)\\sqrt{2+\\left(2-x\\right)^2}+\\frac{d}{dx}\\left(\\sqrt{2+\\left(2-x\\right)^2}\\right)\\left(2+x\\right)\\\\ &=\\frac{2x^2-4x+2}{\\sqrt{x^2-4x+6}} \\end{align*} Find critical points \\begin{align*} f'(x)&=0\\\\ \\frac{2x^2-4x+2}{\\sqrt{x^2-4x+6}}&=0\\\\ 2x^2-4x+2&=0\\\\ 2(x-1)^2&=0 \\end{align*} Then $x=1$. Because $1\\in [0,2]$, then $f(x)$ has a critical point at $x= 1$, since \\begin{align*} f(0)&=2\\sqrt{6}\\\\ f(1)&= 3\\sqrt{3}\\\\ f(2)&=4\\sqrt{2} \\end{align*} Then the maximum is $f(2)=4\\sqrt{2}$ and the minimum is $f\\left( 0\\right) =2\\sqrt{6}$", "# Min and Max of $f(x) = 2 \\sin(x) + \\cos^2 (x)$ on $[0, 2\\pi]$ Find the absolute minimum and maximum values of, $$f(x) = 2 \\sin(x) + \\cos^2 (x) \\text{ on } [0, 2\\pi]$$ What I did so far is $$f'(x) = 2\\cos(x) -2 \\cos(x) \\sin(x)$$ • So you're looking for where $f'$ vanishes. Note $f'(x)=2\\cos x(1-\\sin x)$. Also, don't forget about the endpoints of the interval. – MPW Mar 22 '18 at 1:50 • This may be helpful: mathsisfun.com/calculus/maxima-minima.html – Nash J. Mar 22 '18 at 1:51 • $f'(x) = 2(\\cos x)(1-\\sin x).$ This is $0$ if $\\cos x=0$ or if $1-\\sin x=0. \\qquad$ – Michael Hardy Mar 22 '18 at 2:00 Method $1$: Continue from what you have so far, $$\\cos(x)((1-\\sin(x))=0$$ Find the stationary point, evaluate the function values at the stationary point as well as the boundaries and conclude the minimal and maximal point. Method $2$: \\begin{align} f(x)&=2\\sin(x)+\\cos^2(x)\\\\ &=2 \\sin(x)+1-\\sin^2(x)\\\\ &=-\\sin^2(x)+2\\sin(x)+1 \\\\ &=-(\\sin(x)-1)^2+2 \\end{align} $$-(-1-1)^2+2\\le-(\\sin(x)-1)^2+2 \\le -(1-1)^2+2$$ Without differentiation: $$f(x)=-(1-\\sin x)^2+2; \\\\ f\\left(\\frac{3\\pi}{2}\\right)=-2 (\\text{min}); \\\\ f\\left(\\frac{\\pi}{2}\\right)=2 (\\text{max}).$$ hint: $\\cos^2x = 1 -\\sin^2x \\implies f(x) = 2-\\left(1-\\sin x\\right)^2$. Then use the fact that $-1 \\le \\sin x \\le 1$ to obtain the min and max.", "& =0.9980 \\end{align} At $x=-0.0001$ Substitute, $x=-0.0001$ in the provided limit notation, $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}$. $\\underset{x\\to -0.0001}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=\\frac{2{{\\left( 0.0001 \\right)}^{2}}+\\left( -0.0001 \\right)}{\\sin \\left( -0.0001 \\right)}$ Apply the trigonometric property, $\\sin \\left( -\\theta \\right)=-\\sin \\left( \\theta \\right)$ \\begin{align} & \\underset{x\\to -0.0001}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=\\frac{2\\left( 1\\times {{10}^{-8}} \\right)-0.0001}{-\\sin \\left( 0.0001 \\right)} \\\\ & =\\frac{\\left( 2\\times {{10}^{-8}} \\right)-0.0001}{-\\left( 9.99\\times {{10}^{-5}} \\right)} \\\\ & =\\frac{-9.998\\times {{10}^{-5}}}{-\\left( 9.99\\times {{10}^{-5}} \\right)} \\\\ & =0.9998 \\end{align} Now, as x approaches $0$ from the right, arbitrarily start with $x=0.01$. Then select two additional values of x that are closer to 0, but still greater than 0; we choose $0.001$ and $0.0001$. At $x=0.01$ Substitute, $x=0.01$ in the provided limit notation, $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}$. Therefore, $\\underset{x\\to 0.01}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=\\frac{2{{\\left( 0.01 \\right)}^{2}}+\\left( 0.01 \\right)}{\\sin \\left( 0.01 \\right)}$ \\begin{align} & \\underset{x\\to 0.01}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=\\frac{2\\left( 0.0001 \\right)+0.01}{\\sin \\left( 0.01 \\right)} \\\\ & =\\frac{0.0002+0.01}{0.0099} \\\\ & =\\frac{0.0102}{0.0099} \\\\ & =1.0200 \\end{align} Now, at $x=0.001$ Substitute, $x=0.001$ in the provided limit notation, $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}$ \\begin{align} & \\underset{x\\to 0.001}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=\\frac{2{{\\left( 0.001 \\right)}^{2}}+\\left( 0.001 \\right)}{\\sin \\left( 0.001 \\right)} \\\\ & =\\frac{2\\left( 1\\times {{10}^{-6}} \\right)+0.001}{9.99\\times {{10}^{-4}}} \\\\ & =\\frac{\\left( 2\\times {{10}^{-6}} \\right)+0.001}{9.99\\times {{10}^{-4}}} \\\\ & =\\frac{1.002\\times {{10}^{-3}}}{9.99\\times {{10}^{-4}}} \\end{align} Further solve, $\\underset{x\\to 0.001}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}=1.0020$ At $x=0.0001$ Substitute, $x=0.0001$ in the provided limit notation, $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{2{{x}^{2}}+x}{\\sin x}$. Therefore, \\begin{align}", "\\begin{align} \\text{Range}(g) &= \\left[ \\frac{-1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)},\\frac{1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)} \\right] \\\\ &= [-1.54291482 \\ldots,0.54291482 \\ldots] \\\\ &\\subseteq [-2,2]. \\end{align} Conclusion: $\\displaystyle \\left| \\sum_{k=1}^{n} \\sin(k) \\right| < 2$ and $\\displaystyle \\left| \\sum_{k=1}^{n} \\cos(k) \\right| < 2$ for all $n \\in \\mathbb{N}$. - You have the formula $\\frac{1-x^{n+1}}{1-x}=\\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\\frac{1-e^{i(n+1)\\theta}}{1-e^{i\\theta}}$. – minimalrho Feb 1 '13 at 22:03 +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\\,2\\,$ – DonAntonio Feb 1 '13 at 22:15 @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... – DonAntonio Feb 1 '13 at 23:05 @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had", "the Type $\\int\\sin^mx\\cos^nxdx$ Where $m,n$ Are Positive Integers Case 1. One of the integer powers, say $m$, is odd. $m=2k+1$ for some integer $k$. So, \\begin{align*} \\sin^mx&=\\sin^{2k+1}x\\\\ &=(\\sin^2x)^k\\sin x\\\\ &=(1-\\cos^2x)^k\\sin x. \\end{align*} Use the substitution $u=\\cos x$ in this case. Example. Evaluate $\\int\\sin^3x\\cos^2xdx$. Solution. \\begin{align*} \\int\\sin^3x\\cos^2xdx&=\\int \\sin^2x\\sin x\\cos^2xdx\\\\ &=\\int(1-\\cos^2x)\\cos^2x\\sin xdx\\\\ &=-\\int(1-u^2)u^2du\\ (\\mbox{substition}\\ u=\\cos x)\\\\ &=\\frac{u^5}{4}-\\frac{u^3}{3}+C\\\\ &=\\frac{\\cos^5x}{5}-\\frac{\\cos^3x}{3}+C, \\end{align*} where $C$ is a constant. Example. Evaluate $\\int\\cos^3xdx$. Solution. \\begin{align*} \\int\\cos^3xdx&=\\int\\cos^2x\\cos xdx\\\\ &=\\int(1-\\sin^2x)\\cos xdx\\\\ &=\\int(1-u^2)du\\ (\\mbox{substitution}\\ u=\\sin x)\\\\ &=u-\\frac{u^3}{3}+C\\\\ &=\\sin x-\\frac{\\sin^3x}{3}+C, \\end{align*} where $C$ is a constant. Case 2. If both $m$ and $n$ are even. In this case, use the trigonometric identities $$\\sin^2x=\\frac{1-\\cos 2x}{2},\\ \\cos^2x=\\frac{1+\\cos 2x}{2}.$$ Example. Evaluate $\\int\\sin^2x\\cos^4xdx$. $$\\frac{1}{16}\\left(x-\\frac{1}{4}\\sin 4x+\\frac{1}{3}\\sin^32x\\right)+C,$$ where $C$ is a constant. Integrals of Powers of $\\tan x$ and $\\sec x$ This type of integrals can be mostly done by using the trigonometric identity $$1+\\tan^2x=\\sec^2x.$$ Example. Evaluate $\\int\\tan^4xdx$. Solution. \\begin{align*} \\int\\tan^4xdx&=\\int\\tan^2x\\tan^2xdx\\\\ &=\\int\\tan^2x(\\sec^2x-1)dx\\\\ &=\\int\\tan^2x\\sec^2xdx-\\int\\tan^2xdx\\\\ &=\\int u^2du-\\int(\\sec^2x-1)dx\\ (\\mbox{substition}\\ u=\\tan x)\\\\ &=\\frac{\\tan^3x}{3}-\\tan x+x+C, \\end{align*} where $C$ is a constant. Example. Evaluate $\\int\\sec^3xdx$. Solution. \\begin{align*} \\int\\sec^3xdx&=\\int\\sec x\\sec^2xdx\\\\ &=\\sec x\\tan x-\\int\\tan^2x\\sec xdx\\ (\\mbox{integration by parts})\\\\ &=\\sec x\\tan x-\\int(\\sec^2x-1)\\sec xdx\\\\ &=\\sec x\\tan x-\\int\\sec^3xdx+\\int\\sec xdx+C’\\\\ &=\\sec x\\tan x-\\int\\sec^3xdx+\\ln|\\sec x+\\tan x|+C’, \\end{align*} where $C’$ is a constant. Hence, $$\\int\\sec^3xdx=\\frac{1}{2}\\sec x\\tan x+\\frac{1}{2}\\ln|\\sec x+\\tan x|+C,$$ where $C=\\frac{C’}{2}$. Products of Sines and Cosines This type of integrals include $\\int\\sin mx\\cos nxdx$, $\\int\\sin mx\\cos nxdx$, and $\\int\\cos mx\\cos", "\\sin x-\\cos x \\right)}^{2}} \\\\ \\end{align} One can go wrong while relating the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. (45 + x) has no effect on maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. As $\\cos \\theta$ always lies in [-1,1]. So, one can confuse here to get the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$." ]

[ "# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and", "49 views If $f(x) = \\dfrac{\\sqrt{3} \\sin x}{2+\\cos x}$, then the range of $f(x)$ is 1. the interval $[-1 , \\sqrt{3}{/2}]$ 2. the interval $[-\\sqrt{3}{/2}, 1]$ 3. the interval $[-1, 1]$ 4. none of these in Calculus recategorized | 49 views Answer: $\\mathbf C$ $f^2(x) =\\dfrac{3-3\\cos^2x}{\\cos^2x+4\\cos x+4}$ Now, Let t = $\\cos x$ then, $f(t) = \\dfrac{3-3t^2}{t^2+4t+4}$ $f_{max}.f'(t) = \\frac{-6(2t+1)}{t+2} = 0$, when$t = \\frac{-1}{2}$ $\\therefore f(-1) =0, f(\\frac{-1}{2}) = 1, \\;and\\; f(1) = 0$ So, $f^2\\text{max} = 1 \\implies f^2(x) \\leq 1 \\implies -1\\leq f(x) \\leq 1$ So, the range of $f$ is $[-1,1]$ $\\therefore \\mathbf C$ is the right option. by Boss (18.9k points) edited by To find the range of $f(x)$, actually we need to find the maximum and minimum value. Now \\begin{align}f(x)&=\\frac{\\sqrt{3}\\sin x}{2+\\cos x}\\\\ \\Rightarrow f'(x)&=\\sqrt{3}\\cos x \\cdot \\frac{1}{2+\\cos x}+\\sqrt{3}\\sin x \\cdot \\frac{-\\sin x}{(2+\\cos x)^2}\\\\ &=\\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} \\end{align} At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \\begin{align}\\therefore f'(x)&=0\\\\ \\Rightarrow \\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} &=0\\\\ \\Rightarrow 2\\cos x +1&=0;~[\\because \\sin^2 x+\\cos^2 x=1] \\\\ \\Rightarrow \\cos x &= -\\frac{1}{2}\\\\ \\Rightarrow x &= \\frac{2\\pi}{3}, \\frac{4\\pi}{3}\\end{align} Now putting those values of $x$ to $f(x)$, we obtain $f(\\frac{2\\pi}{3})=\\frac{\\sqrt{3}(\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=1$ and $f(\\frac{4\\pi}{3})=\\frac{\\sqrt{3}(-\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$. $$\\therefore f(x) \\in [-1,1]$$ So the correct", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "# Range of $(\\sin x)^6+ (\\cos x)^6$ I need to find the range of the function $y = (\\sin x)^6+ (\\cos x)^6$ I did find the answer but working in a crude way rather than a methodical step by step approach. I give below the steps I used , please help with a methodical approach to such problems. 1) To find the max value of the function I noticed that in the range where x is $[0,2\\pi]$ , when $\\cos x$ hits $+1$ then $\\sin x$ is $0$ , when $\\cos x$ is $0$ then $\\sin x$ is $+1$ ..etc so the max value at any of these points could be either $1^6+0^6$ or $(-1)^6+0^6$ so the max value is 1. to find the minimum I differentiated the function $f' (x) = 6(\\sin x)^5\\cos x- 6(\\cos x^5)\\sin x =0$ , equating this to zero we have $(\\sin x)^4 = (\\cos x)^4 => x = \\pi/4 =>$ min value of function is $(\\sin(\\pi/4))^6 + (\\cos(\\pi/4))^6 = 1/4$ . so the range is $(\\frac{1}{4},1)$. Please can someone help with how can this type of problems be methodically approached ? - Thanks. • Well, you can do some basic trig manipulation to view it as $\\frac18(3\\cos(4x)+5)$. This gives you a maximum of $\\frac18(3+5)=1$ and a minimum of $\\frac18(-3+5)=\\frac14$. So the range", "of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(1) = \\frac{1}{2}$ and maximum value of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(0)=1$ so, range of f(x)=[1/2,1] I think there is any mistake in the book. As you can see the detailed solution of the question and if there is any other doubt related to this question you can ask", "# Find the range of function $f(x) =\\cos\\sin\\ln\\left(\\frac{x^2+e}{x^2+1} \\right)+\\sin\\cos\\ln\\left(\\frac{x^2+e}{x^2+1} \\right)$ Problem : Find the range of function $$f\\left( x \\right) =\\cos \\left( \\sin \\left( \\ln \\left( \\frac{x^2+e}{x^2+1} \\right) \\right) \\right) +\\sin \\left( \\cos \\left( \\ln \\left( \\frac{x^2+e}{x^2+1} \\right) \\right) \\right)$$ My approach : maximum value of the function is when denominator term is minimum i.e. $$x^2+1$$ is minimum. It is minimum when $$x^2 =0$$ therefore, maximum value of function $$\\cos(\\sin(\\ln e))+\\sin(\\cos(\\ln e))$$ $$\\cos(\\sin(1))+\\sin(\\cos(1))$$ now how to find the minimum value of the function please suggest . Thanks... • \"maximum value of the function is when denominator term is minimum \" of which function? Oct 24, 2014 at 18:01 • Maybe you are just supposed to graph it? Oct 24, 2014 at 18:12 Well, first you need to see the range of $\\frac{x^2+e}{x^2+1}$, which you can easily verify to be $(1,e]$. Then, the range of $\\log$ in this domain is $(0,1]$. As pointed out in a comment: Since $\\sin$ is increasing and $\\cos$ decreasing in $(0,1]$, then $\\sin(0,1] =(0,\\sin(1)]$ and $\\cos(0,1] =[\\cos(1),1)$. Finally, since in $(0,\\sin(1)]$ the cosine is decreasing and in $[\\cos(1),1)$ the sine is increasing, we get that the range of each part of the sum is $[\\cos(\\sin(1)),1)$ and $[\\sin(\\cos(1)),1)$. So the range of the complete function is $$[\\cos(\\sin(1))+\\sin(\\cos(1)), 2)$$ Another way could be using the fact", "decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12. lurflurf Homework Helper This is a common limit. in fact it arises in finding the derivative of sine $$\\frac{d}{dx}\\sin(x)=\\lim_{h\\rightarrow 0} \\frac{\\sin(x)}{x}\\cos(x+\\frac{h}{2})$$ the limit can be shown several ways including showing that cos(x)<sin(x)/x<1 for |x|<pi/2 or seeing that for the function f $$f(x):=\\int_0^1 \\cos(x t) dt$$ f is every where continuos and f(x)=sin(x)/x for all x except x=0 (also xf(x)=sin(x))", "# Min and Max of $f(x) = 2 \\sin(x) + \\cos^2 (x)$ on $[0, 2\\pi]$ Find the absolute minimum and maximum values of, $$f(x) = 2 \\sin(x) + \\cos^2 (x) \\text{ on } [0, 2\\pi]$$ What I did so far is $$f'(x) = 2\\cos(x) -2 \\cos(x) \\sin(x)$$ • So you're looking for where $f'$ vanishes. Note $f'(x)=2\\cos x(1-\\sin x)$. Also, don't forget about the endpoints of the interval. – MPW Mar 22 '18 at 1:50 • This may be helpful: mathsisfun.com/calculus/maxima-minima.html – Nash J. Mar 22 '18 at 1:51 • $f'(x) = 2(\\cos x)(1-\\sin x).$ This is $0$ if $\\cos x=0$ or if $1-\\sin x=0. \\qquad$ – Michael Hardy Mar 22 '18 at 2:00 Method $1$: Continue from what you have so far, $$\\cos(x)((1-\\sin(x))=0$$ Find the stationary point, evaluate the function values at the stationary point as well as the boundaries and conclude the minimal and maximal point. Method $2$: \\begin{align} f(x)&=2\\sin(x)+\\cos^2(x)\\\\ &=2 \\sin(x)+1-\\sin^2(x)\\\\ &=-\\sin^2(x)+2\\sin(x)+1 \\\\ &=-(\\sin(x)-1)^2+2 \\end{align} $$-(-1-1)^2+2\\le-(\\sin(x)-1)^2+2 \\le -(1-1)^2+2$$ Without differentiation: $$f(x)=-(1-\\sin x)^2+2; \\\\ f\\left(\\frac{3\\pi}{2}\\right)=-2 (\\text{min}); \\\\ f\\left(\\frac{\\pi}{2}\\right)=2 (\\text{max}).$$ hint: $\\cos^2x = 1 -\\sin^2x \\implies f(x) = 2-\\left(1-\\sin x\\right)^2$. Then use the fact that $-1 \\le \\sin x \\le 1$ to obtain the min and max.", "Math Help - range 1. range Find Range of $f(x) = \\sin^2 x-5\\sin x+6$ where $x\\in \\left[0,\\pi\\right]$ 2. Re: range Originally Posted by jacks Find Range of $f(x) = \\sin^2 x-5\\sin x+6$ where $x\\in \\left[0,\\pi\\right]$ $f'(x)=2\\sin x \\cdot \\cos x-5 \\cos x=\\cos x \\cdot (2\\sin x-5)$ In order to find minimum of function we have to solve following equation : $f'(x)=0 \\Rightarrow \\cos x=0 \\Rightarrow x= \\frac{\\pi}{2} \\Rightarrow \\sin x = 1$ So ,we can conclude that : $f_{min}=2$ On the other hand function has maximum for $x=0$ , and $x=\\pi$ , therefore : $f_{max}=6$ So for $x \\in [0,\\pi]$ we have : $f(x) \\in [2,6]$", "- \\frac {x}{2},\\\\f_n'(x) &= - \\mbox{Re}\\left(\\sum_{n = 1}^{n}z^n\\right) - \\frac12.\\end{align*} After some simplifications we get $$f_n'(x) = \\frac {( - 1)^{n + 1}}{2}((1 - \\cos(x))\\frac {\\sin((n + 1)x)}{\\sin(x)} + \\cos((n + 1)x))$$ and $$f_n''(x) = \\frac {( - 1)^{n}}{2}\\frac {(n + 1)\\sin(nx) + n\\sin((n + 1)x)}{1 + \\cos(x)}.$$ The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \\frac {\\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \\frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\\frac {\\pi}{n + 1}) > 0$. We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even. Let's restrict to the case $n$ is even. We prove $f_{2n}(\\frac {\\pi}{2n + 1}) < 0$. \\begin{align*}f_{2n}\\left( \\frac{\\pi}{2n+1} \\right) &=\\sum_{k=1}^{2n}{\\left( -1 \\right)}^{k+1}\\frac{\\sin \\left( \\frac{k\\pi}{2n+1} \\right)}{k}-\\frac{\\pi}{2\\left( 2n+1 \\right)}\\\\&=\\frac{\\pi}{2n+1}\\left( \\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{\\left( 2k-1 \\right) \\pi}{2n+1} \\right)}{\\frac{\\left( 2k-1 \\right) \\pi}{2n+1}}}-\\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{2k\\pi}{2n+1} \\right)}{\\frac{2k\\pi}{2n+1}}} \\right) -\\frac{\\pi}{2\\left( 2n+1 \\right)}.\\end{align*} The function $x \\mapsto \\frac {\\sin(x)}{x}$ is descending on $[0,\\pi]$, thus both sums lay between $a$ and $a + \\frac {2\\pi}{2n + 1}$, where $a = \\int_0^{\\pi}\\frac {\\sin(x)}{x}\\,dx$. Thus $$f_{2n}\\left(\\frac {\\pi}{2n + 1}\\right) < \\frac {\\pi}{2n + 1}\\cdot\\frac {2\\pi}{2n + 1} - \\frac {\\pi}{2(2n + 1)} < 0.$$ ## Euler-Maclaurin求和公式估计梯形积分公式的误差 $\\mathop {\\lim }\\limits_{n \\to \\infty }", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "minimum of sin^2(x)=(sin(x))^2=0. 22. thomas5267 $\\lim_{x\\to 0+}\\sqrt{x}\\left(1+\\sin^2\\left(\\frac{2\\pi}{x}\\right)\\right)={\\lim_{x\\to 0+}\\sqrt{x}}+\\lim_{x\\to 0+}\\sqrt{x}\\sin^2\\left(\\frac{2\\pi}{x}\\right)$ Both limit exist so you can expand the brackets 23. thomas5267 I had quite a lot of brain fart in this question. 24. zepdrix $\\large\\rm 0\\lt \\sin^2\\left(\\frac{2\\pi}{x}\\right)\\le1$$\\large\\rm 1\\lt 1+\\sin^2\\left(\\frac{2\\pi}{x}\\right)\\le2$$\\large\\rm \\sqrt{x}\\lt \\sqrt{x}\\left[1+\\sin^2\\left(\\frac{2\\pi}{x}\\right)\\right]\\le2\\sqrt{x}$Why is the +1 messing you up nini? :o You still get zeroes on both ends of this inequality, ya? 25. thomas5267 *Both limits Plural singular brain fart. 26. Babynini ooh it all just moves up by 1? 27. Babynini hahah @thomas5267 you and I both! I'm not sure why this problem is being so troublesome xD 28. Empty Maybe it'll make sense to look at the range of $$\\sin x$$: $-1 \\le \\sin x \\le 1$ The range is just $$[-1,1]$$ But when we square it, the negative part of the range goes away: $0 \\le \\sin^2 x \\le 1$ since $$[0,1]$$ is the new range. 29. Babynini And then that range [0,1] becomes [1,2] because of the 1+ in the equation in front of sine? 30. Babynini (Yes, thank you, that made it make so much more sense xD) 31. Babynini 32. Babynini is that..correct? 33. Empty Yeah looks perfect to me! 34. Empty Actually one tiny thing, $0 \\le \\sin^2x$ since they can be equal! 35. zepdrix I think my inequality was wrong. For some reason", "# if f(x)=sin4x + cos4x, then what is the range of f(x) Vijay Mukati 6 years ago Dear Student, Differnetiate the functions and equate the result with zero to get the point of minima and maximua. You will get x = 0, 45 and 90 degree. From here we can easily find the range and it will come out to be [1/2, 1] Thanks. Chirag Jain 36 Points 6 years ago Sir, the answer is [3/4,1] as stated in the book and the book is Objective Maths by Purshottam Kumar Sharma Chirag Jain 36 Points 6 years ago The service of askiitians is very slow, posted a question 3 days ago, but till now haven’t got any satisfactory rreply. How can someone invest in such a slow service? I am sorry if you think I have been rude in my comment. arun 123 Points 6 years ago in this question we have to find the range for $\\sin ^4x + \\cos ^4x$ now,$\\sin ^4x + \\cos ^4x = (\\sin ^2x)^2+(\\cos ^2x)^2$ $\\Rightarrow \\sin ^4x + \\cos ^4x = (\\sin ^2x + \\cos ^2x)^2-2\\sin ^2x\\cos ^2x$ $\\Rightarrow \\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(4\\sin^2x\\cos^2x)$ $\\Rightarrow \\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}\\sin^22x$ as $\\sin^2\\theta$ have a maximum value of 1 and a minimum value of 0 so, minimum value", "the range $R = \\left[ {\\frac{\\pi }{6},\\frac{\\pi }{2}} \\right)$ (b) y =\\begin{align} \\frac{1}{{{{\\sin }^4}x + {{\\cos }^4}x}} = \\frac{1}{{{{({{\\sin }^2}x + {{\\cos }^2}x)}^2} - 2{{\\sin }^2}x{{\\cos }^2}x}}\\end{align} =\\begin{align} \\frac{1}{{1 - 2{{\\sin }^2}{{\\cos }^2}x}} = \\frac{1}{{1 - \\frac{1}{2}{{(\\sin 2x)}^2}}}\\end{align} Now $$- 1 \\le \\sin 2x \\le 1$$ $$\\Rightarrow$$ $$0 \\le {(\\sin 2x)^2} \\le 1$$ $$\\Rightarrow$$ $$0 \\le \\frac{1}{2}{(\\sin 2x)^2} \\le \\frac{1}{2}$$ $$\\Rightarrow$$ $$\\frac{1}{2} \\le 1 - \\frac{1}{2}{(\\sin 2x)^2} \\le 1$$ $$\\Rightarrow$$ \\begin{align}1 \\le \\frac{1}{{1 - \\frac{1}{2}{{(\\sin 2x)}^2}}} \\le 2\\end{align} Therefore, the range is [1, 2]. (c) y = $$\\sqrt {2 - x} + \\sqrt {1 + x}$$ = $$\\sqrt {2 - x} + \\sqrt {x - ( - 1)}$$ This is of the form $$\\sqrt {a - x} + \\sqrt {x - b}$$ , which we discussed earlier. The minimum for y comes at $$x = {\\rm{ }}-1,{\\rm{ }}2$$ : y = $$\\sqrt 3$$ The maximum comes at x = \\begin{align}\\frac{{ - 1 + 2}}{2}\\,\\,\\,\\,:\\,\\,\\,\\,y = \\sqrt 6\\end{align} The range is R = $$\\left[ {\\sqrt 3 ,\\sqrt 6 } \\right]$$ (d) $$y = \\sqrt {x - 1} + 2\\sqrt {3 - x}$$ This function is no longer symmetric, so we carry out a standard analysis. The domain is [1, 3]. Also, y > 0. Now $${y^2}$$ = $$(x - 1) + 4(3 - x) + 4\\sqrt {(x - 1)(3", "Que.7. The maximum value for the function sin 2y can be attained at what point in the interval (0,2n). Ans. f(y) = sin 2y ${f}'$(y) = 2cos 2y ${f}'$(y) = 0 cos 2y = 0 2y = $\\frac{\\pi }{2}$, $\\frac{3\\pi }{2}$, $\\frac{5\\pi }{2}$, $\\frac{7\\pi }{2}$ y = $\\frac{\\pi }{4}$, $\\frac{3\\pi }{4}$, $\\frac{5\\pi }{4}$, $\\frac{7\\pi }{4}$ f($\\frac{\\pi }{4}$) = sin $\\frac{\\pi }{2}$ = 1 f($\\frac{3\\pi }{4}$) = sin $\\frac{3\\pi }{2}$ = -1 f($\\frac{5\\pi }{4}$) = sin $\\frac{5\\pi }{2}$ = 1 f($\\frac{7\\pi }{2}$) = sin $\\frac{7\\pi }{2}$ = -1 f(0) = sin 0 = 0, f(2$\\pi$) = sin 2$\\pi$ = 0 Thus, the absolute maximum is occurring at y = $\\frac{\\pi }{4}$ and y = ($\\frac{5\\pi }{4}$). Que.8. In a function sin y + cos y, find the maximum value. f(y) = sin y + cos y Ans. ${f}'$(y) = cos y – sin y ${f}'$(y) = 0 sin y = cos y = – (sin y + cos y) $\\Rightarrow$ tan y = 1 y = $\\frac{\\pi }{4}$, $\\frac{5\\pi }{4}$ ${f}”$(y) = – sin y – cos y = – (sin y + cos y) When (sin y + cos y) is positive, ${f}”$(y) will be negative. ${f}”$(y) is negative when y $\\epsilon$ (0, $\\frac{\\pi }{2}$) Thus, y = $\\frac{\\pi }{4}$ ${f}”$( $\\frac{\\pi }{4}$) = – (sin $\\frac{\\pi }{4}$", "&= \\int_0^1 ((x-c)^2+c^2)dx\\\\ &=\\left[ \\frac{(x-c)^3}{3} +c^2x \\right]_0^1\\\\ &=\\frac{(1-c)^3}{3} + c^2 + \\frac{c^3}{3}\\\\ &=2c^2-c+\\frac{1}{3}. \\end{align*} 1. What is the minimum value of $I(c)$ (as $c$ varies)? Using the expression for $I(c)$ we calculated above, and completing the square, we have \\begin{align*} I(c) &= 2c^2-c+\\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 - \\frac{1}{8} + \\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 + \\frac{5}{24}. \\end{align*} So the minimum value that $I(c)$ can take is $5/24$. (This occurs when $c=1/4$.) 1. What is the maximum value of $I(\\sin \\theta)$ as $\\theta$ varies? Note that from our work above, the MINIMUM value of $I(\\sin\\theta)$ happens when $\\theta = \\arcsin (0.25)$. Now $c=\\sin \\theta$ varies between between $-1$ and $1$, so we need to find the maximum value of $I$ over this range. We worked out earlier that the overall minimum of $I$ occurs at $c=1/4$, and since $I$ is quadratic, it is symmetric about this point, and increasing for larger and smaller values of $c$. So the maximum value of $I(c)$ over the range $-1 \\le c \\le 1$ occurs at $c=-1$, the furthest point from $x=1/4$ in the range. So the maximum value of $I(\\sin \\theta)$ is $I(-1)=10/3$.", "}{4} - \\cos \\frac{\\pi }{4}$ $f\"\\left( {\\frac{\\pi }{4}} \\right) = - \\frac{1}{{\\sqrt 2 }} - \\frac{1}{{\\sqrt 2 }} < 0$ ⇒ f is maximum at x = π/4 $f\\left( {\\frac{\\pi }{4}} \\right) = \\sin \\frac{\\pi }{4} + \\cos \\frac{\\pi }{4}$ $f\\left( {\\frac{\\pi }{4}} \\right) = \\frac{1}{{\\sqrt 2 }} + \\frac{1}{{\\sqrt 2 }} = \\sqrt 2$ f(0) = sin0+cos0 = 0+1 = 1 f(π) = sinπ + cosπ = 0 +(-1) = -1 Absolute min = min{-1, 1, √2} = -1 at x=π Absolute max = max{-1, 1, √2} = √2 at x=π/4 $(iii)f\\left( x \\right) = 4x - \\frac{1}{2}{x^2},x \\in \\left[ { - 2,\\frac{9}{2}} \\right]$ $f'\\left( x \\right) = 4 - \\frac{1}{\\cancel{2}} \\cdot \\cancel{2}x$ f'(x) = 4 -x f\"(x) = -1 f'(x) = 0 4-x = 0 x = 4 f is maximum at x=4 f(4) = 4(4) - ½(4)2 = 8 local maximum value = 8 at x = 4 f(-2) = 4(-2) - ½ (-2)2 f(-2) = -8-2 = -10 $f\\left( {\\frac{9}{2}} \\right) = 4\\left( {\\frac{9}{2}} \\right) - \\frac{1}{2}{\\left( {\\frac{9}{2}} \\right)^2}$ $f\\left( {\\frac{9}{2}} \\right) = 18 - \\frac{{81}}{8}$ $f\\left( {\\frac{9}{2}} \\right) = 18 - 10.125 = 7.875$ Absolute max. = max{-10, 8, 7.875} = 8 Absolute min. = min { -10, 8. 7.875} = -10 (iv) f(x) = (x-1)2 + 3, x ∈ [-3, 1] f'(x)", "# Question #143f6 Dec 14, 2017 The range is the closed interval $= \\left[2 \\sinh \\left(- 1\\right) , 2 \\sinh \\left(2\\right)\\right] = \\left[\\frac{1 - {e}^{2}}{e} , \\frac{{e}^{4} - 1}{{e}^{2}}\\right] \\approx \\left[- 2.3504 , 7.2537\\right]$. #### Explanation: Let $f \\left(x\\right) = {e}^{x} - {e}^{- x} = 2 \\sinh \\left(x\\right)$, where $\\sinh \\left(x\\right)$ is the hyperbolic sine function . Then $f ' \\left(x\\right) = {e}^{x} + {e}^{- x} = 2 \\cosh \\left(x\\right)$, where $\\cosh \\left(x\\right)$ is the hyperbolic cosine function. Clearly $f ' \\left(x\\right) > 0$ for all $x \\setminus \\in \\mathbb{R}$, implying that $f \\left(x\\right)$ is strictly increasing over $\\mathbb{R}$, and also over $\\left[- 1 , 2\\right]$. Therefore, the minimum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(- 1\\right) = 2 \\sinh \\left(- 1\\right) = {e}^{- 1} - {e}^{1} = \\frac{1 - {e}^{2}}{e} \\approx - 2.3504$ and the maximum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(2\\right) = 2 \\sinh \\left(2\\right) = {e}^{2} - {e}^{- 2} = \\frac{{e}^{4} - 1}{{e}^{2}} \\approx 7.2537$. This implies that the range is the closed interval $= \\left[2 \\sinh \\left(- 1\\right) , 2 \\sinh \\left(2\\right)\\right] = \\left[\\frac{1 - {e}^{2}}{e} , \\frac{{e}^{4} - 1}{{e}^{2}}\\right] \\approx \\left[- 2.3504 , 7.2537\\right]$.", "\\sin x-\\cos x \\right)}^{2}} \\\\ \\end{align} One can go wrong while relating the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. (45 + x) has no effect on maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. As $\\cos \\theta$ always lies in [-1,1]. So, one can confuse here to get the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$." ]

[ "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "with the parameter $t$ has its whole body in the line— it is a direction vector for the line. Note that points on the line to the left of $x=1$ are described using negative values of $t$. In $\\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set of (endpoints of) vectors of this form and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$ consists of (endpoints of) the vectors in $\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} +t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} +s\\cdot\\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R} \\}$ (the column vectors associated with the parameters $\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ -3 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} \\qquad \\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ 4 \\\\ 0.5 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix}$ are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative $t$'s or negative $s$'s or", "be a vector parallel to the line. The line can then be writen as $\\vec{p_0} + t\\vec{v}$ where $\\vec{p_0}$ is a point on the line, $\\vec{v}$ is a vector parallel to the line, and $t$ is the parameter. Therefore, one possible parametrization is: $$(x,y) = (c/d_1,0) + t\\cdot (c/d_1,-c/d_2).$$", "# Math Help - explicit description of a line? 1. ## explicit description of a line? Find an explicit description of the line 3x+2y = 0 in R2 through the origin. span {[2, -3]} , but how? 2. ## Re: Explicit description of a line? The line $3x+2y=0$ in $\\mathbb R^2$ is spanned by any nonzero vector $\\begin{pmatrix}x \\\\ y\\end{pmatrix}$ such that $(x,y)$ lies on the line. $\\begin{pmatrix}2 \\\\ -3\\end{pmatrix}$ is one such vector; another would be $\\begin{pmatrix}-1 \\\\ \\frac32\\end{pmatrix}$. 3. ## Re: explicit description of a line? Originally Posted by yangx Find an explicit description of the line 3x+2y = 0 in R2 through the origin. the answer is: span {[2, -3]} , but how? I do not necessarily disagree with reply #2. However, the point $(-2,3)$ is on the line $3x+2y=0~.$ That line contains $(0,0)$, so its direction vector is $<-2,3>$. Thus in traditional vector geometry its equation is $<0,0>+t<-2,3>~.$ So notation is everything here. Unless you tell us about the notation in use in your notes, we cannot really answer.", "is the same you got setting t values <0,1,1> then set up the dot product and solve for t and input t into the direction vector I got from taking the point minus the line, then write the new parametric equations based off my original point and the direction vector I'll do that way but for sake of completeness (1) the book suggests find the intersection (2) can I use projection to find it as well?? 5. ## Re: equation of line through point orthogonal to line I got two points along the given line by plugging in $t=0$ and $t=1$. When $t=0$, I get the point $(-1,-2+0,-1+0) = (-1,-2,-1)$. When $t=1$, I get the point $(-1,-2+1,-1+1) = (-1,-1,0)$. So, the vector from the first point to the second point is $\\begin{pmatrix}-1-(-1) \\\\ -1-(-2) \\\\ 0-(-1)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. That is a vector in the direction of the given line (Essentially, this a vector containing some multiple of the coefficients of $t$ in the parametric equation of the given line). You can rewrite the line as $\\begin{pmatrix}-1 \\\\ -2 \\\\ -1\\end{pmatrix} + t\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. This immediately gives a vector in the direction of the given line to be $\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Essentially, what I did was I found all possible", "# explicit description of a line? • Mar 13th 2013, 12:24 AM yangx explicit description of a line? Find an explicit description of the line 3x+2y = 0 in R2 through the origin. span {[2, -3]} , but how? • Mar 13th 2013, 06:03 PM Nehushtan Re: Explicit description of a line? The line $3x+2y=0$ in $\\mathbb R^2$ is spanned by any nonzero vector $\\begin{pmatrix}x \\\\ y\\end{pmatrix}$ such that $(x,y)$ lies on the line. $\\begin{pmatrix}2 \\\\ -3\\end{pmatrix}$ is one such vector; another would be $\\begin{pmatrix}-1 \\\\ \\frac32\\end{pmatrix}$. • Mar 13th 2013, 06:23 PM Plato Re: explicit description of a line? Quote: Originally Posted by yangx Find an explicit description of the line 3x+2y = 0 in R2 through the origin. the answer is: span {[2, -3]} , but how? I do not necessarily disagree with reply #2. However, the point $(-2,3)$ is on the line $3x+2y=0~.$ That line contains $(0,0)$, so its direction vector is $<-2,3>$. Thus in traditional vector geometry its equation is $<0,0>+t<-2,3>~.$ So notation is everything here. Unless you tell us about the notation in use in your notes, we cannot really answer.", "which also lies on the line, there is another scalar, $t_2$, which satisfies: $\\overrightarrow{SQ} = t_2 \\vc{d}$. Since $Q$ is located in the opposite direction of $\\vc{d}$, it is clear that $t_2<0$. For $R$, on the other hand, there is no scalar, $t_3$, that satisfies a similar relation. That is, $\\overrightarrow{SR} \\neq t_3 \\vc{d}$ for all values of $t_3$. The only difference between $P$ and $Q$ is that they use different scalars, $t_1$ and $t_2$. Hence, it makes sense that any point on the line can be described by using a particular scalar, $t$. Therefore, we use $P(t)$ to denote a function of a scalar, $t$, which returns a point, $P$. That is, for different values of $t$, different points, $P$, will be generated. This expression can be rewritten as \\begin{gather} \\overrightarrow{SP(t)} = t\\vc{d} \\\\ \\Longleftrightarrow \\\\ P(t) - S = t\\vc{d} \\\\ \\Longleftrightarrow \\\\ P(t) = S + t\\vc{d}. \\end{gather} (3.56) Note that all points, $P(t)$, on the line are described by starting at $S$ and then adding a scaled direction vector, $t\\vc{d}$, to reach $P(t)$. Some examples: $P(0) = S$, $P(1) = S + \\vc{d}$, and $P(-2.5) = S -2.5\\vc{d}$. This type of parameterized line is summarized in the following definition. Definition 3.6: A Parameterized Line A line parameterized by $t\\in \\R$ can be described by", "parameterization for the $y=-1$ trace of $f\\text{.}$ What type of curve does this trace describe? 3. Find a parameterization for the level curve $f(x,y) = 25\\text{.}$ What type of curve does this trace describe? 4. How do your responses change to all three of the preceding questions if you instead consider the function $g$ defined by $g(x,y) = x^2 - y^2\\text{?}$ (Hint for generating one of the parameterizations: $\\sec^2(t)-\\tan^2(t) = 1\\text{.}$) ### Subsection9.6.2Summary • A vector-valued function is a function whose input is a real parameter $t$ and whose output is a vector that depends on $t\\text{.}$ The graph of a vector-valued function is the set of all terminal points of the output vectors with their initial points at the origin. • Every vector-valued function provides a parameterization of a curve. In $\\R^2\\text{,}$ a parameterization of a curve is a pair of equations $x = x(t)$ and $y = y(t)$ that describes the coordinates of a point $(x,y)$ on the curve in terms of a parameter $t\\text{.}$ In $\\R^3\\text{,}$ a parameterization of a curve is a set of three equations $x = x(t)\\text{,}$ $y=y(t)\\text{,}$ and $z = z(t)$ that describes the coordinates of a point $(x,y,z)$ on the curve in terms of a parameter $t\\text{.}$ ### Exercises9.6.3Exercises ###### 11. A standard parameterization for the unit circle is $\\langle", "Math Lines in three dimensions Point on a line segment # Point on a line segment? Checking whether a point is on a line segment is like finding out if a point is on a line but with one additional condition. i ### Method Point $P$ on line segment $\\overline{AB}$? 1. Set up linear equation from $A$ and $B$ 2. Finding out if a point is on a line with $P$ on $g_{AB}$ 3. $r$ must be between 0 and 1* *The requirement is that the linear equation is $\\vec{x} = \\vec{OA} + r \\cdot \\vec{AB}$ (no other support vector or direction vector). ### Example Is the point $P(-3|14|10)$ on the line segment $\\overline{AB}$ ? $A(3|4|6)$ and $B(0|9|8)$ 1. #### Set up $g_{AB}$ Set up the linear equation from two points. $\\text{g: } \\vec{x} = \\vec{OA} + r \\cdot \\vec{AB}$ $\\text{g: } \\vec{x} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 2. #### Point on the line The position vector (vector with the coordinates of the point) of $A$ must be inserted for $\\vec{x}$ in $g$. $\\begin{pmatrix} -3 \\\\ 14 \\\\ 10 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Now we set up an equation system", "}t$ to the 3D points (0,0,0), (1,1,1) and (2,2,2) at time points 0, 1 and 2 with the weights 1, 1 and 1. The fitted parameters $\\vec{ a }$ and $\\vec{ b }$ of the line are $\\vec{ a } = (0,0,0)^T$ $\\vec{ b } = (1,1,1)^T$ Fits!", "a general equation for a shape and substituting it’s coordinates with a parameterised line. Starting with what a parameterised line is. It is defined solely by an origin and a direction that scales with a parameter (a number that can be anywhere between negative and positive infinity). The points on this line are given by the equation $\\overrightarrow{P} = \\overrightarrow{O_L} + \\overrightarrow{D_L} t$. Here $\\overrightarrow{O_L}$ is the origin, $\\overrightarrow{D_L}$ the direction and $t$ the parameter. Our parameter $t$ thus denotes the distance from our line’s origin along the direction and can be anywhere between negative and positive infinity. For those unfamiliar with the notation, the arrows denote vectors with an x, y and z component, like $\\overrightarrow{P} = \\begin{pmatrix} P_x \\\\ P_y \\\\ P_z \\end{pmatrix} = \\begin{pmatrix} O_x + D_x t \\\\ O_y + D_y t \\\\ O_z + D_z t \\end{pmatrix}$. Now to find the intersections between this line and a shape we need to know the shape’s equation, substitute it’s x, y and z component with $\\overrightarrow{P}$, and solve for $t$. For example, the equation of a sphere would be $x^2 + y^2 + z^2 = r^2$, with r the radius. To find the intersections we would substitute $x \\rightarrow P_x$, $y \\rightarrow P_y$ and $z \\rightarrow P_z$. This would result in an equation that has", "## Dividing a Line Segment in a Given Ratio Suppose the line between $A(2,3)$ and $B((9,20)$ is to be divided in the ratio $2:3$ . Ket the point that divides the line segment in this ratio be $P$ . The vector from $A$ to $B$ is $\\vec{AB}=\\vec{OB}=\\vec{OA}= \\begin{pmatrix}9\\\\20\\end{pmatrix}-\\begin{pmatrix}2\\end{pmatrix}=\\begin{pmatrix}7\\\\17\\end{pmatrix}$ . The vector $\\vec{AP}=\\frac{2}{5} \\vec{AB}=\\frac{2}{5}\\begin{pmatrix}7\\\\17\\end{pmatrix}=\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}$ . Then $\\vec{OP}=\\vec{OA}+ \\vec{AP}=\\begin{pmatrix}2\\\\3\\end{pmatrix}+\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}=\\begin{pmatrix}24/5\\\\49/5\\end{pmatrix}$ . The point $P$ is $(24/5,49/5)$ .", "Free Version Easy # Diagonalize Parameterized Matrix LINALG-E3IXE5 Let: $$A_t=\\begin{pmatrix} 1 & 0 & 0\\\\\\ 0 & 1 &-1\\\\\\ t^2 & t & 1\\end{pmatrix}$$ ...where $t\\in\\mathbb{R}$. For which values of $t$ is $A_t$ diagonalizable over $\\mathbb{R}$? A $t\\in [-1,1]$ only B $t<0$ C $t\\geq 0$ D For all $t\\in\\mathbb{R}$", "dt ] + j [ d(ct^2 + d) / dt ] = ia + 2jct = 1.7 m/s i + 0.5002 m/s j (c) The \"speed\" in (b) is sqrt(1.7^2 + 0.5002^2) = use calculator, should be about 1.8 m/s • discover the area vector from the given equation, utilising the following substitutions x=t y = t²/4 r(t) = <t, t²/4> what's the fee of the parameter at (2,a million)? x = t t = 2 make sure with y(t) y = t²/4 = 2²/4 y = a million So, at (2,a million) the parameter \"t\" = 2 r'(t) = <a million, a million/2t> discover the unit tangent vector \"T\"; the cost vector is then 3T cm/s. T=r'(t)/|r'(t)| T= <a million, a million/2t>/?(a million+t²/4) Then for t=2 T(2) = <a million/?2, a million/?2> finally v = 3T v = <3/?2, 3/?2>cm/s Hottest videos Related Posts", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "$t$ and the variable given in the answer $t'$. Then let's see when the two vectors are equal. $\\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-\\tfrac{1}{2} \\\\ \\tfrac{1}{2} \\\\ 1\\end{pmatrix}t = \\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}t'$ when $t=2t'$. Since the map $t \\mapsto 2t$ is a bijection, you wind up with the same line either way. Same thing for line 2. so if i i do my ans in this way. wouldnt my ans be wrong? 4. ## Re: vector (simplified b1 and b2 ) Originally Posted by delso so if i i do my ans in this way. wouldnt my ans be wrong? I just showed that you get the same line either way. Why would you get the wrong answer that way? If two line intersect and you have two different formulas for each line, it is still the same two lines. It doesn't matter that the formulas are not identical. They still represent the same lines, so they will still intersect in the same place. 5. ## Re: vector (simplified b1 and b2 ) do u mean t=2t\" just different in terms of length ? but it's still the same line with same gradient?", "1. ## Vector problem: \"The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k are given. Find the value of t such that the three vectors u, v and w are coplanar. \" I have no idea how to solve this one!Where do I start? 2. let's take $v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$ Using Gauss method on vectors,we should have $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 2\\\\ 2\\\\ 3 \\end{pmatrix}$ $,\\begin{pmatrix} 1\\\\ 2-t\\\\ t+1 \\end{pmatrix}$ $\\Rightarrow$ $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix}$ for those three victors to be coplanar we must have $rang(\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ , $\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix} )$ $=2$ that is $\\frac{-7t+8}{2}=0$ which means $t=\\frac{7}{8}$ hope that's right 3. Originally Posted by Raoh let's take $v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$ Using Gauss method on vectors,we should have $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 2\\\\ 2\\\\ 3 \\end{pmatrix}$ $,\\begin{pmatrix} 1\\\\ 2-t\\\\ t+1 \\end{pmatrix}$ $\\Rightarrow$ $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix}$ for those three victors to be coplanar we must have $rang(\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ , $\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix} )$ $=2$ that is", "the following vector-valued functions are parameterized by arc length? $t\\vector {11/61,60/61}$ $\\vector {3\\sin (t/3),3\\cos (t/3)}$ $t\\vector {16/113,112/113}$ $\\vector {7,t17/145,t144/145}$ $\\vector {3\\cos (t),3\\sin (t)}$ Consider $\\vec {f}(t) = \\vector {3\\sin (a t),t/2,3\\cos (a t)}$ for $0\\le t$. Find $a$ that makes this parameterized by arc length. Set $|\\vec {f}'(t)| = 1$ and solve for $a$. Often given a curve one wishes to have an arc length parameterization of the curve. We proceed by discussing several special cases, and then by giving a general method. ### Disguised lines Sometimes you have a vector-valued function that is merely a line in disguise. How could this be? Well consider the vector-valued function: This doesn’t look very much like a line, for one thing it has the function $\\sin (t)$ in each component. On the other hand, if we look at $\\vec {f}'$, we see Ah, we can now factor a $\\cos (t)$ out of each component to get: this is a scalar-function times a constant vector. The fact that we can “pull-out” the scalar function, and are left with a constant vector tells us that the line segment plotted by $\\vec {f}$ for $-\\pi /2\\le t\\le \\pi /2$ is identical to the line segment plotted by: Which of the following are line segments in disguise? $\\vector {2+e^t,4,-2-e^t}$ for $0\\le t\\le 2$ $\\vector", "the vector $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ 3. the vector $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ 4. the vector $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ Note: you can download a sheet of gridded paper here. Note: this exercise can be downloaded as a worksheet to practice with: ### Solutions 1. For $$\\vec{a} = \\begin{pmatrix} 5 \\\\ 2 \\end{pmatrix}$$ we find: 2. For $$\\vec{b} = \\begin{pmatrix} -3 \\\\ 4 \\end{pmatrix}$$ we find: 3. For $$\\vec{c} = \\begin{pmatrix} 6 \\\\ -3 \\end{pmatrix}$$ we find: 4. For $$\\vec{d} = \\begin{pmatrix} 0 \\\\ 5 \\end{pmatrix}$$ we find: 5. For $$\\vec{e} = \\begin{pmatrix} -2 \\\\ -5 \\end{pmatrix}$$ we find: 6. For $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ we find: 7. For $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ we find: 8. For $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ we find: ### Tutorial: How to find a vector's components In this tutorial we learn what a vector is. We learn how to write vectors as both column and row vectors as well as how to represent vectors graphically; that is how to draw vectors. ## Exercise 2 Writing your answers as column vectors, find the coordinates of each of the vectors drawn here: 1. vector $$\\vec{a}$$: 2. vector $$\\vec{b}$$: 3. vector $$\\vec{c}$$: 4. vector $$\\vec{d}$$: 5. vector $$\\vec{e}$$:", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\" ]

[ "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "the original system using another change of coordinates. Suppose that we consider a linear system $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\tag{3.6.2}$$ where $$T$$ is an invertible matrix. If $${\\mathbf y}(t)$$ is a solution of (3.6.2), we claim that $${\\mathbf x}(t) = T {\\mathbf y}(t)$$ solves the equation $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ Indeed, \\begin{align*} {\\mathbf x}'(t) & = (T {\\mathbf y})'(t)\\\\ & = T {\\mathbf y}'(t)\\\\ & = T( (T^{-1} A T) {\\mathbf y}(t))\\\\ & = A (T {\\mathbf y}(t))\\\\ & = A {\\mathbf x}(t). \\end{align*} We can think of this in two ways. 1. A linear map $$T$$ converts solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}$$ to solutions of $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ 2. The inverse of a linear map $$T$$ takes solutions of $${\\mathbf x}' = A {\\mathbf x}$$ to solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\text{.}$$ We are now in a position to solve our problem of finding solutions of an arbitrary linear system \\begin{equation*} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\end{pmatrix}. \\end{equation*} ### Subsection3.6.4Distinct Real Eigenvalues Consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where $$A$$ has two real, distinct eigenvalues $$\\lambda_1$$ and $$\\lambda_2$$ with eigenvectors $${\\mathbf v}_1$$", "& 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 3\\mod 13$ and $y\\equiv 8\\mod 13$, $x$ and $y$ can be written as $x=3+13k_1$ and $y=8+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Suppose that $k_1+k_2$ is even, so that $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 15+13(k_1+k_2)\\mod 26\\\\ &\\equiv 15\\mod 26\\\\ &\\not\\equiv 2\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are no solutions for part 2. 3. Similarly to part 2, the solution in $\\mathbb{Z}/13\\mathbb{Z}$ is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$", "1 \\\\ -4 & -2 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -4 \\end{pmatrix} = 2 {\\mathbf v}_1. \\end{equation*} Thus, we can take ${\\mathbf v}_2 = (1/2)\\mathbf w = (1/2, 0)\\text{,}$ and our second solution is \\begin{equation*} {\\mathbf x}_2 = e^{\\lambda t} ({\\mathbf v}_2 + t {\\mathbf v}_1) = e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix} \\end{equation*} Thus, our general solution is \\begin{equation*} {\\mathbf x} = c_1 {\\mathbf x}_1 + c_2 {\\mathbf x}_2 = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix}. \\end{equation*} If the eigenvalue is positive, we will have a nodal source. If it is negative, we will have a nodal sink. Notice that we have only given a recipe for finding a solution to $\\mathbf x' = A \\mathbf x\\text{,}$ where $A$ has a repeated eigenvalue and any two eigenvectors are linearly dependent. We will justify our procedure in the next section (Section 3.6). ### Subsection3.5.3Important Lessons • If \\begin{equation*} A = \\begin{pmatrix} \\lambda & 1 \\\\ 0 & \\lambda \\end{pmatrix}, \\end{equation*} then $A$ has one repeated real eigenvalue. The general solution to the system ${\\mathbf x}' = A {\\mathbf x}$ is \\begin{equation*} {\\mathbf x}(t) = \\alpha e^{\\lambda t} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + \\beta e^{\\lambda t} \\begin{pmatrix} t \\\\ 1", "as the splitting (6), where \\begin{align} & \\mathbf{D} = \\begin{pmatrix} 6 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 5 \\end{pmatrix}, \\quad \\mathbf{U} = \\begin{pmatrix} 0 & 2 & 3 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 0 \\end{pmatrix}, \\quad \\mathbf{L} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 1 & 0 & 0 \\\\ 3 & 1 & 0 \\end{pmatrix}. \\quad (14) \\end{align} We then have \\begin{align} & \\mathbf{(D-L)^{-1}} = \\frac{1}{120} \\begin{pmatrix} 20 & 0 & 0 \\\\ 5 & 30 & 0 \\\\ 13 & 6 & 24 \\end{pmatrix}, \\end{align} \\begin{align} & \\mathbf{(D-L)^{-1}U} = \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix}, \\quad \\mathbf{(D-L)^{-1}k} = \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}. \\end{align} The Gauss-Seidel method (8) applied to the problem (10) takes the form \\bold x^{(m+1)} = \\begin{align} & \\frac{1}{120} \\begin{pmatrix} 0 & 40 & 60 \\\\ 0 & 10 & 75 \\\\ 0 & 26 & 51 \\end{pmatrix} \\bold x^{(m)} + \\frac{1}{120} \\begin{pmatrix} 100 \\\\ -335 \\\\ 233 \\end{pmatrix}, \\end{align} \\quad m = 0, 1, 2, \\ldots \\quad (15) The first few iterates for equation (15) are listed in the table below, beginning with x(0) = (0.0, 0.0, 0.0)T. From", "\\lambda_1 = \\frac{-b + \\sqrt{b^2 - 4ac}}{2a} \\quad \\text{and} \\quad \\lambda_2 = \\frac{-b - \\sqrt{b^2 - 4ac}}{2a}. \\end{equation*} We can find eigenvectors \\begin{align*} \\mathbf v_1 \\amp = \\begin{pmatrix} 1 \\\\ (-b + \\sqrt{b^2 - 4ac}\\,)/2a \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\lambda_1 \\end{pmatrix}\\\\ \\mathbf v_1 \\amp = \\begin{pmatrix} 1 \\\\ (-b - \\sqrt{b^2 - 4ac}\\,)/2a \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\lambda_2 \\end{pmatrix} \\end{align*} for $\\lambda_1$ and $\\lambda_2\\text{,}$ respectively. Thus, the general solution to the system of differential equations $\\mathbf x' = A \\mathbf x$ is \\begin{equation*} \\mathbf x(t) = \\begin{pmatrix} x(t) \\\\ y(t) \\end{pmatrix} = \\begin{pmatrix} x(t) \\\\ x'(t) \\end{pmatrix} = c_1 e^{\\lambda_1 t} \\begin{pmatrix} 1 \\\\ \\lambda_1 \\end{pmatrix} + c_2 e^{\\lambda_2 t} \\begin{pmatrix} 1 \\\\ \\lambda_2 \\end{pmatrix}, \\end{equation*} which agrees with (4.5). ###### Example4.3 Now let us solve the initial value problem \\begin{align*} x'' + 4 x' + 5 x & = 0\\\\ x(0) & = 1\\\\ x'(0) & = 1. \\end{align*} Again, we will assume that our solution has the form $x(t) = e^{rt}\\text{.}$ Substituting this function into our differential equation, we find that \\begin{equation*} 0 = x'' + 4 x' + 5 x = r^2 e^{rt} + 4r e^{rt} + 5 e^{rt} = (r^2 + 4r + 5) e^{rt}. \\end{equation*} As with the real case, $r^2 + 4r + 5 = 0$ or \\begin{equation*} r =", "we have $\\mathbf{i} + \\mathbf{j} + \\mathbf{k} + a_i \\mathbf{i} + a_j \\mathbf{j} + a_k \\mathbf{k} = 2\\mathbf{i} - 3\\mathbf{j} - \\mathbf{k}$ Comparing co-efficients, we have $\\begin{cases} 1 + a_i = 2 \\\\ 1 + a_j = -3 \\\\ 1 + a_k = -1 \\end{cases}$ Thus $$a_i = 1$$, $$a_j = -4$$, $$a_k = -2$$, and therefore $$\\mathbf{x} = \\mathbf{i} - 4 \\mathbf{j} - 2 \\mathbf{k}$$ Exercise 10 $$\\mathbf{0} = \\begin{pmatrix}0 \\\\ 0 \\\\ 0\\end{pmatrix}\\quad \\mathbf{i} = \\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix} \\quad \\mathbf{j} = \\begin{pmatrix}0 \\\\ 1 \\\\ 0\\end{pmatrix} \\quad \\mathbf{k} = \\begin{pmatrix}0 \\\\ 0 \\\\ 1\\end{pmatrix}$$ ### Vector Equation of a Straight Line If $$A$$ and $$B$$ are any two distinct points on a straight line in space with position vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ respectively, then the vector equation of a straight line is $\\mathbf{r} = (1 - s)\\mathbf{a} + s \\mathbf{b}$ where $$\\mathbf{r}$$ represents the position vector of any point on the line. In Cartesian coordinate form, with $\\mathbf{a} = a_1 \\mathbf{i} + a_2 \\mathbf{j} + a_3 \\mathbf{k}$ $\\mathbf{b} = b_1 \\mathbf{i} + b_2 \\mathbf{j} + b_3 \\mathbf{k}$ $\\mathbf{r} = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ the equation of the line is $x = a_1 + s (b_1 - a_1)\\quad,\\quad y = a_2 + s (b_2 - a_2)\\quad,\\quad z = a_3 + s", "form: \\begin{align*} \\begin{pmatrix}t\\\\t\\end{pmatrix} = t\\begin{pmatrix}1\\\\1\\end{pmatrix} \\end{align*} So we may take $$\\boldsymbol{v}_2 = \\begin{pmatrix}1\\\\1\\end{pmatrix}.$$ ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L5E1 2. Find the eigenvalues and eigenvectors of the following matrix: \\begin{align*} \\begin{pmatrix} -2 & 1\\\\ 1 & -2 \\end{pmatrix} \\end{align*} First we find the eigenvalues: \\begin{align*} =\\det\\begin{pmatrix}-2-\\lambda & 1\\\\ 1 & -2-\\lambda\\end{pmatrix} =(\\lambda +2)^2 - 1. \\end{align*} So the eigenvalues are $$\\lambda_1 = -3$$ and $$\\lambda_2 = -1$$. Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue. First, for $$\\lambda_1$$ we find (non–zero) solutions to: \\begin{align*} \\begin{pmatrix} 1 & 1\\\\1 & 1 \\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\end{pmatrix} =\\begin{pmatrix}0\\\\0\\end{pmatrix}. \\end{align*} The solutions are $$x_1 + x_2 = 0$$ or $$x_1 = -x_2$$. That is, vectors of the form: \\begin{align*} \\begin{pmatrix}t\\\\-t\\end{pmatrix}=t\\begin{pmatrix}1\\\\-1\\end{pmatrix}. \\end{align*} So we may take $$\\boldsymbol{v}_1 = \\begin{pmatrix}1\\\\-1\\end{pmatrix}$$. Next, for $$\\lambda_2$$ we find (non–zero) solutions to: \\begin{align*} \\begin{pmatrix} -1 & 1\\\\1 & -1 \\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\end{pmatrix} =\\begin{pmatrix}0\\\\0\\end{pmatrix}. \\end{align*} The solutions are $$x_1 - x_2 = 0$$ or $$x_1 = x_2$$. That is, vectors of the form: \\begin{align*} \\begin{pmatrix}t\\\\t\\end{pmatrix} = t\\begin{pmatrix}1\\\\1\\end{pmatrix} \\end{align*} So we may take $$\\boldsymbol{v}_2 = \\begin{pmatrix}1\\\\1\\end{pmatrix}.$$ ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L5E2 3. Find the eigenvalues and eigenvectors of the following matrix:", "+ b \\begin{pmatrix} \\sin \\beta t \\\\ \\cos \\beta t \\end{pmatrix}\\\\ & = a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $$2 \\pi / \\beta\\text{.}$$ This type of system is called a center. Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $$\\lambda = \\pm 2i\\text{.}$$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $$c_1$$ and $$c_2\\text{,}$$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.4.3). ### Subsection3.4.2Spiral Sinks and Sources Now let us consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $$\\alpha$$ and", "1\\\\ 0 & 1 & 2\\\\ \\end{pmatrix}$$ I proceed normally after here? – Kode Ch Apr 15 '18 at 14:51 • @KodeCh: Yes, if that is the way around you would normally compute the span of three length-4 vectors. My instinct would be to look at the tanspose of your matrix, and then do ordinary Gaussuan elimination where the row operations do not change the row space (aka the span), but the other way around would also work, and I believe there are textbooks that teach it that way. – Henning Makholm Apr 15 '18 at 14:54 Take an arbitrary symmetric matrix $$\\begin{pmatrix} x & y \\\\ y & z \\end{pmatrix}$$ and compute $a, b, c$ such that \\begin{align*}\\begin{pmatrix} x & y \\\\ y & z \\end{pmatrix} &= a\\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix} + b\\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} + c\\begin{pmatrix} 0 & 1 \\\\ 1 & 2 \\end{pmatrix} \\\\ &= \\begin{pmatrix} a - b & a + c \\\\ a + c & b + 2c \\end{pmatrix}. \\end{align*} Equating entries, this produces the system of linear equations, \\begin{align*} a - b &= x \\\\ a + c &= y \\\\ b + 2c &= z, \\end{align*} which can be solved using whichever methods you like. Either way, we get", "0 & 3 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\end{equation*} is \\begin{equation*} \\mathbf y(t) = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} t \\\\ 1 \\end{pmatrix}. \\end{equation*} Therefore, the general solution to \\begin{equation*} \\begin{pmatrix} dx/dt \\\\ dy/dt \\end{pmatrix} = \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\end{equation*} is \\begin{align*} \\mathbf x(t) \\amp = T \\mathbf y(t)\\\\ \\amp = c_1 e^{3t} T \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + c_2 e^{3t} T \\begin{pmatrix} t \\\\ 1 \\end{pmatrix}\\\\ \\amp = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix}. \\end{align*} This solution agrees with the solution that we found in Example 3.5.5. In practice, we find solutions to linear systems using the methods that we outlined in Sections 3.2–3.4. What we have demonstrated in this section is that those solutions are exactly the ones that we want. ### Subsection3.6.7Important Lessons • A linear map $$T$$ is invertible if and only if $$\\det T \\neq 0\\text{.}$$ • A linear map $$T$$ converts solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}$$ to solutions of $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ • The inverse of a linear map $$T$$ takes solutions of $${\\mathbf x}' = A {\\mathbf x}$$ to solutions of $${\\mathbf y}'", "\\end{equation*} ##### 2. \\begin{equation*} A = \\begin{pmatrix} -12 \\amp 30 \\\\ -5 \\amp 13 \\end{pmatrix} \\end{equation*} ##### 3. \\begin{equation*} A = \\begin{pmatrix} -9 \\amp -2 \\\\ 10 \\amp 0 \\end{pmatrix} \\end{equation*} ##### 4. \\begin{equation*} A = \\begin{pmatrix} 11 \\amp 8 \\\\ -12 \\amp -9 \\end{pmatrix} \\end{equation*} #### 5. Solve each linear systems $$d\\mathbf x/dt = A \\mathbf x$$ in Exercise Group 3.3.6.1–4 for the initial condition $$\\mathbf x(0) = (2,2)\\text{.}$$ #### 6. Consider the linear system $$d \\mathbf x/dt = A \\mathbf x\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 3 \\amp 2 \\\\ 3 \\amp -2 \\end{pmatrix}. \\end{equation*} Suppose the initial conditions for the solution curve are $$x(0) = 1$$ and $$y(0) = 1\\text{.}$$ We can use the following Sage code to plot the phase portrait of this system, including the straight-line solutions and a solution curve. x, y, t = var('x y t') #declare the variables F = [3*x + 2*y, 3*x - 2*y] #declare the system # normalize the vector fields so that all of the arrows are the same length n = sqrt(F[0]^2 + F[1]^2) # plot the vector field p = plot_vector_field((F[0]/n, F[1]/n), (x, -4, 4), (y, -4, 4), aspect_ratio = 1) # solve the system for the initial condition t = 0, x = 1, y = 1 P1 = desolve_system_rk4(F, [x, y], ics=[0,", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "+ 3 x_{22}\\\\ \\end{pmatrix} \\end{align*} So our equation is $$\\begin{pmatrix} 9 & -3 \\\\ 5 & -5 \\end{pmatrix} - \\begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\\\ \\end{pmatrix}= \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ Now we write it as a system of equations: \\begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\\\ 0&= -3-(-2x_{11} +3x_{12}) \\\\ 0&=5-(-9x_{21} +8 x_{21}) \\\\ 1&=-5 - (-2x_{21}+3x_{22}) \\\\ \\end{align*} If you need help solving this system tell me. - Thank you very much for your excellent answer! – Lukas Arvidsson Mar 15 at 9:27 You are welcome :) – Dominic Michaelis Mar 15 at 9:28 First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution. - Denote $B=\\begin{pmatrix}-9&-2\\\\8&5\\end{pmatrix}$. Then we have: $$\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E=XB$$ Observe that $\\det B=-45+16\\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\\left(\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E\\right)B^{-1}$$ Do you know how to find $B^{-1}$? - Thanks for your answer! I get the final result to be: $\\left[ \\begin{array}{cc} \\frac{-64}{29} & \\frac{-43}{29}\\\\ \\frac{-48}{29} & \\frac{-54}{29}\\end{array} \\right]$ Is that correct?", "given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}\\mod 13\\ &\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 13 \\end{align*} We test the possibilities of solutions in $\\mathbb{Z}/26\\mathbb{Z}$. Since $x\\equiv 6\\mod 13$ and $y\\equiv 7\\mod 13$, $x$ and $y$ can be written as $x=6+13k_1$ and $y=7+13k_2$ for some $k_1,k_2\\in\\mathbb{Z}$. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\ &\\equiv 1\\mod 26 \\end{align*} if and only if $k_1+k_2$ is even. Now we assume that $k_1+k_2$ is even, so that we have $x+3y\\equiv 1\\mod 26$. \\begin{align*} 7x+9y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} since $k_1+k_2$ is even. Hence, there are solutions for part 3. Let $0\\leq x,y\\leq 25$. Then $k_1, k_2=0,1$. Since $k_1+k_2$ is even, we have either $k_1=k_2=0$ or $k_1=k_2=1$. Therefore, the solutions are \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 6\\\\ 7 \\end{pmatrix}\\mod 26\\ \\mbox{or}\\ \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 19\\\\ 20 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} \\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ with $(D,N)=1$ where $D=ad-bc$ be an enciphering matrix. Then it defines the enciphering transformation $C=AP$ where $P=\\begin{pmatrix} x\\\\ y \\end{pmatrix}$ and $C=\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}$ are plaintext message unit and ciphertext, respectively. That is, $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}$$ The deciphering transformation is then given by $P=A^{-1}C$, i.e. $$\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} D^{-1}d & -D^{-1}b\\\\ -D^{-1}c &", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $2 \\pi / \\beta\\text{.}$ This type of system is called a center. ##### Example3.26 Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $\\lambda = \\pm 2i\\text{.}$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $c_1$ and $c_2\\text{,}$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.27). # Subsection3.4.2Spiral Sinks and Sources¶ permalink Now let us consider the system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $\\alpha$ and $\\beta$ are nonzero real numbers. The characteristic equation of $A$", "1 \\amp 1/2 \\\\ -2 \\amp 0 \\end{pmatrix}, \\end{equation*} and \\begin{equation*} T^{-1} A T = \\begin{pmatrix} -1/2 & 2 \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix} \\begin{pmatrix} 1 \\amp 1/2 \\\\ -2 \\amp 0\\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\ 0 & 3 \\end{pmatrix}. \\end{equation*} From Section 3.3, we know that the general solution to the system \\begin{equation*} \\begin{pmatrix} dx/dt \\\\ dy/dt \\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\ 0 & 3 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\end{equation*} is \\begin{equation*} \\mathbf y(t) = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} t \\\\ 1 \\end{pmatrix}. \\end{equation*} Therefore, the general solution to \\begin{equation*} \\begin{pmatrix} dx/dt \\\\ dy/dt \\end{pmatrix} = \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\end{equation*} is \\begin{align*} \\mathbf x(t) \\amp = T \\mathbf y(t)\\\\ \\amp = c_1 e^{3t} T \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + c_2 e^{3t} T \\begin{pmatrix} t \\\\ 1 \\end{pmatrix}\\\\ \\amp = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix}. \\end{align*} This solution agrees with the solution that we found in Example 3.37. In practice, we find solutions to linear systems using the methods that we outlined in Sections 3.23.4. What we have demonstrated in this section", "\\\\[4mm]&= \\begin{pmatrix}c_1 a^1_1 + \\cdots + c_m a^m_1 \\\\ \\vdots \\\\ c_1 a^1_n + \\cdots + c_m a^m_n \\end{pmatrix}. \\end{align*} Example. Compute $$2\\begin{pmatrix}1\\\\2\\end{pmatrix} - 3\\begin{pmatrix}10\\\\2\\end{pmatrix} + \\begin{pmatrix}2\\\\1\\end{pmatrix}$$. Solution. \\begin{align*} 2\\begin{pmatrix}1\\\\2\\end{pmatrix} - 3\\begin{pmatrix}10\\\\2\\end{pmatrix} + \\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}2\\\\4\\end{pmatrix} + \\begin{pmatrix}-30\\\\-6\\end{pmatrix} + \\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}-26\\\\-1\\end{pmatrix}. \\end{align*} Motivation from Systems of Equations To see the full video page and find related videos, click the following link. #### Video Notes Oftentimes (and we will see specific applications to ODEs later in the course), we want to solve systems of equations like: \\begin{align*} c_1\\begin{pmatrix}1\\\\1\\\\1\\end{pmatrix} + c_2\\begin{pmatrix}1\\\\0\\\\-1\\end{pmatrix} + c_3\\begin{pmatrix}0\\\\1\\\\-1\\end{pmatrix} \\begin{pmatrix} 2\\\\5\\\\2 \\end{pmatrix} \\end{align*} This means that we want to find all $$c_1,c_2,c_3$$ that satisfy this equation. So there are two questions to ask: (1) Is there a solution? and (2) If there is, is the solution unique? For this particular problem: \\begin{align*} \\begin{pmatrix} 2\\\\5\\\\2 \\end{pmatrix} = 3\\begin{pmatrix}1\\\\1\\\\1\\end{pmatrix} -1\\begin{pmatrix}1\\\\0\\\\-1\\end{pmatrix} + 2\\begin{pmatrix}0\\\\1\\\\-1\\end{pmatrix} \\end{align*} And so one solution is: \\begin{align*} c_1 = 3 \\hspace{.25in} c_2 = -1\\hspace{.25in} c_3 = 2. \\end{align*} Next, we need to answer: is there more than one solution? If $$c_1,c_2,c_3$$ satisfy the equation then the top equation implies $$c_2 = 2-c_1$$ and the middle equation implies $$c_3 = 5-c_1$$. We can now plug this in the bottom equation to get: \\begin{align*} &= c_1 - c_2 - c_3 \\\\&= c_1 - (2-c_1) - (5-c_1) \\\\&=" ]

[ "0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •", "plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 • How did you parameterize? How did you find the normal vectors? What is the upward caret operator? – rocksNwaves May 1 '18 at 14:02", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "that a direction vector is $$(3/2,1)$$. Of course, there's nothing wrong with using our method of chapter 1: if $$(a,b)$$ is a direction vector, then $$\\pm(-b,a)$$ are normal vectors, and vice versa. Example. Consider the line with point-parallel form equation $\\mathbf r(t)=\\begin{pmatrix}2\\\\-1\\end{pmatrix}+t\\begin{pmatrix}-3\\\\1\\end{pmatrix}.$ We can see that the line contains the point $$P(2,-1)$$, and that a direction vector is $$\\mathbf u=(-3,1)$$. Therefore, a normal vector is $$\\mathbf n=(1,3)$$, and from the point-normal form equation $\\left(\\begin{pmatrix}x & y\\end{pmatrix}-\\begin{pmatrix}2 & -1\\end{pmatrix}\\right)\\cdot\\begin{pmatrix}1\\\\3\\end{pmatrix}=0$ we get the standard form equation $x+3y=1.$ Note that a standard equation of a hyperplane in $$\\mathbf R^n$$ gives a $$1\\times n$$ matrix with one nonzero row. Therefore, the rank is 1, and you get an ($$n-1$$)-parameter family of solutions. This is why in $$\\mathbf R^2$$, from one normal vector you get one directional vector (unique up multiplication by a nonzero scalar). Example. Consider the plane in $$\\mathbf R^3$$ with standard equation $$2x-y+z=2$$. Solving the SLE, you get the 2-parameter family of solutions \\begin{align*} x&=s/2-t/2+2\\\\ y&=s\\\\ z&=t \\end{align*} That is, we have the parametrization $\\mathbf r(s,t)=\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}=\\begin{pmatrix} s/2-t/2+2 \\\\ s \\\\ t \\end{pmatrix}=\\begin{pmatrix}2\\\\0\\\\0\\end{pmatrix}+\\begin{pmatrix}1/2\\\\1\\\\0\\end{pmatrix}+\\begin{pmatrix}-1/2\\\\0\\\\1\\end{pmatrix}.$ You can see that a plane is determined by one point and two nonparallel direction vectors. To get the normal vector back from the two direction vectors, you need to find a nonzero vector that is", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "the original system using another change of coordinates. Suppose that we consider a linear system $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\tag{3.6.2}$$ where $$T$$ is an invertible matrix. If $${\\mathbf y}(t)$$ is a solution of (3.6.2), we claim that $${\\mathbf x}(t) = T {\\mathbf y}(t)$$ solves the equation $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ Indeed, \\begin{align*} {\\mathbf x}'(t) & = (T {\\mathbf y})'(t)\\\\ & = T {\\mathbf y}'(t)\\\\ & = T( (T^{-1} A T) {\\mathbf y}(t))\\\\ & = A (T {\\mathbf y}(t))\\\\ & = A {\\mathbf x}(t). \\end{align*} We can think of this in two ways. 1. A linear map $$T$$ converts solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}$$ to solutions of $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ 2. The inverse of a linear map $$T$$ takes solutions of $${\\mathbf x}' = A {\\mathbf x}$$ to solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\text{.}$$ We are now in a position to solve our problem of finding solutions of an arbitrary linear system \\begin{equation*} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\end{pmatrix}. \\end{equation*} ### Subsection3.6.4Distinct Real Eigenvalues Consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where $$A$$ has two real, distinct eigenvalues $$\\lambda_1$$ and $$\\lambda_2$$ with eigenvectors $${\\mathbf v}_1$$", "2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ • Can you explain how you got the parameterized result? Sep 23, 2021 at 4:46 From the coefficients of x, y and z of the general form equations, the first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try $$y=0$$(since it passes through the $$x-z$$ plane) and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "= {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}(^1_0)$$ In other words, if $A=A^T$: $$Df(\\mathbf x) = {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}$$ Last edited: Jan 24, 2013 3. Jan 24, 2013 ### libragirl79 Since I forgot a lot of my matrix algebra, when you say A (x y) it looks like x choose y, but i am assuming that's not what you meant... 4. Jan 24, 2013 ### I like Serena Since you are talking about x transpose, that means that x is a vector. A vector is commonly written as: $$\\mathbf x = \\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix}$$ I have simplified it to: $$\\mathbf x = \\begin{pmatrix}x \\\\ y\\end{pmatrix}$$ So indeed, this does not mean x choose y. ;) Furthermore: $$\\mathbf x^T = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}$$ and $$\\mathbf x^T \\mathbf x = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}\\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix} = x_1^2 + x_2^2 + \\dots + x_n^2$$ 5. Jan 24, 2013 ### libragirl79 Ok, I see! Thanks!! :)", "y(0) & = -3 \\end{align*} ##### 6. \\begin{align*} x' & = 5x + 4y\\\\ y' & = -9x - 7y\\\\ x(0) & = 2\\\\ y(0) & = 1 \\end{align*} ##### 7. \\begin{align*} x' & = -x + y\\\\ y' & = -x - 3y\\\\ x(0) & = 2\\\\ y(0) & = 2 \\end{align*} ##### 8. \\begin{align*} x' & = 2x + y\\\\ y' & = -x\\\\ x(0) & = 0\\\\ y(0) & = -5 \\end{align*} #### 9. Consider the linear system $$d \\mathbf x/dt = A \\mathbf x\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 3 \\amp 1 \\\\ -4 \\amp -1 \\end{pmatrix}. \\end{equation*} Suppose the initial conditions for the solution curve are $$x(0) = -2$$ and $$y(0) = 5\\text{.}$$ We can use the following Sage code to plot the phase portrait of this system, including a solution curve and the straight-line solution. x, y, t = var('x y t') #declare the variables F = [3*x + y, -4*x - y] #declare the system # normalize the vector fields so that all of the arrows are the same length n = sqrt(F[0]^2 + F[1]^2) # plot the vector field p = plot_vector_field((F[0]/n, F[1]/n), (x, -20, 20), (y, -20, 20), aspect_ratio = 1) # solve the system for the initial condition t = 0, x = -2, y = 5 P1", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "## Derivation of Formula For Distance Between Parallel Lines in Three Dimensions If two lines in three dimensions are parallel, the coefficients in the Cartesian equations are the same, since the coefficients of the Cartesian equation are the components of the tangent, which is the same for parallel lines. Suppose we want to find the distance between the parallel lines $\\frac{x-x_1}{a}=\\frac{y-y_1}{b}=\\frac{z-z_1}{c}=t_1$ (1) $\\frac{x-x_2}{a}=\\frac{y-y_2}{b}=\\frac{z-z_2}{c}=t_2$ (2) /> Rearrange (1) into vector form to give $\\mathbf{r}_1(t_1)= \\begin{pmatrix}x_1\\\\y_1\\\\z_1\\end{pmatrix}+ \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}t_1$ Rearrange (2) into vector form to give $\\mathbf{r}_2(t_2)= \\begin{pmatrix}x_2\\\\y_2\\\\z_2\\end{pmatrix}+ \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}t_2$ (2)-(1) gives $\\mathbf{r}_{12}= \\begin{pmatrix}x_2-x_1\\\\y_2-y_1\\\\z_2-z_1\\end{pmatrix}+ \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}(t_2-t_1)$ We can take this vector as perpendicular to the tangent vector $\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}$ for suitable values $t'_1, \\; t'_2$ of $t_1, \\; t_2$ . Hence \\begin{aligned} &( \\begin{pmatrix}x_2-x_1\\\\y_2-y_1\\\\z_2-z_1\\end{pmatrix}+ \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}(t'_2-t'_1)) \\cdot \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} \\\\ &= \\begin{pmatrix}x_2-x_1+a(t'_2-t'_1)\\\\y_2-y_1++b(t'_2-t'_1)\\\\z_2-z_1+c(t'_2-t'_1)\\end{pmatrix} \\cdot \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}=0 \\end{aligned} This gives the equation \\begin{aligned} & a(x_2-x_1)+a^2(t'_2-t'_1)+b(y_2-y_1)+b^2(t'_2-t'_1)+c(z_2-z_1)+c^2(t'_2-t'_1)=0 \\\\ & \\rightarrow t'_2-t'_1=- \\frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2} \\end{aligned} \\begin{aligned} \\left| \\mathbf{r}_{12} \\right| &= \\sqrt{(x_2-x_1+a(t'_2-t'_1))^2+(y_2-y_1+b(t'_2-t'_1))^2+(z_2-z_1+c(t'_2-t'_1))^2} \\\\ &= \\sqrt{(x_2-x_1-a( \\frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2+(y_2-y_1-b( \\frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2+(z_2-z_1-c( \\frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2} \\\\ &= \\sqrt{\\frac{(a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1))^2}{a^2+b^2+c^2}} \\end{aligned}", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "&1 &\\sqrt{2}\\\\ 0 &0 &\\sqrt{2}+1\\end{bmatrix}. $ The equations these computations represent are as follows: $\\begin{bmatrix}1 &1 &\\sqrt{2}\\\\ 0 &0 &\\sqrt{2}+1\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end {bmatrix}=\\begin{bmatrix}0\\\\0\\end{bmatrix},$ and hence we see immediately that $z=0.$ Plugging this back in, we see that $x+y=0,$ or $x=-y.$ Setting $t=y$ as our parameter, we can represent the equation of the line as $\\mathbf{r}(t)=t\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ Hence, a vector parallel to the line generated by the intersection of these two planes is given by $\\mathbf{r}(1)-\\mathbf{r}(0)=\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ Perform this same procedure with the intersection of the tilted wall and the ground, and you'll get another vector. This constitutes step 1. Does all this make sense? 3. Most of it, but where does $\\mathbf{r}(t)=t\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ come from? 4. It comes from the general equation of a line in 3D. You have $\\mathbf{r}(t)=t\\mathbf{v}+\\mathbf{b},$ where $\\mathbf{v}$ is a vector in the direction of the line, and $\\mathbf{b}$ is a point on the line. It's essentially the same thing as your $y=mx+b$. In this case, the $\\mathbf{b}=0,$ since the line goes through the origin (plug in $t=0$, and you'll see that). Make sense? 5. the procedure yes i think so...but will take a bit of going over. 6. Ok. Let me know if you have more questions. 7. So once I have the two vectors, they go into $\\displaystyle\\alpha=\\cos^{-1}\\left(\\frac{\\mathbf{x}\\cdot\\mathbf{y}}{|\\mathbf{ x}|\\,|\\mathbf{y}|}\\right)$ (step 3 is just a reordered step 2?)", "+0 # Consider the line parametrized by \\begin{align*} x&= 4t + 2,\\\\ y& = t+2. \\end 0 207 1 Consider the line parametrized by$$x= 4t + 2, y = t+2$$ Find vector $$\\begin{pmatrix}a\\\\b\\end{pmatrix}$$ that's parallel to this line and satisfies $$a+b=10$$. Feb 2, 2019 #1 +103947 +1 Here's my best effort....although there might be an easier method !!! Wnen = 0 we have the point (2, 2) When t = 1 we have the point (6, 3) So...the slope of the given line is (3 - 2) / (6 - 2) = 1/4 If we let the parallel line go through the origin....we have the line y = (1/4)x This means that every point on this line has the coordinates ( x, 1/4x) This implies that x + (1/4)x = 10 (5/4)x = 10 x = 8 and y = (1/4)x = 2 So....the required vector is < 8, 2 > Feb 2, 2019", "those vectors that form a linear combination for all vectors in $\\mathbb{C}^4$. My proof would go something like this: Let $x_1, x_2, x_3, x_4 \\in{\\mathbb{C}}$ be arbitrary. Thus, $x_1+x_2+x_3$, $x_1+x_4$, and $x_2+x_4$ are also arbitrary elements of $\\mathbb{C}$. This means that $x=\\begin{pmatrix}x_1+x_2+x_3\\\\x_1+x_4\\\\x_2+x_4\\\\x_3\\end{pmatrix} \\in{\\mathbb{C}^4}$ is also arbitrary. So a linear combination of $x$ with the given vectors is simply $x_1\\begin{pmatrix}1\\\\1\\\\0\\\\0\\end{pmatrix}+ x_2\\begin{pmatrix}1\\\\0\\\\1\\\\0\\end{pmatrix}+ x_3\\begin{pmatrix}1\\\\0\\\\0\\\\1\\end{pmatrix}+ x_4\\begin{pmatrix}0\\\\1\\\\1\\\\0\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\1\\\\0\\\\1\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\0\\\\1\\\\1\\end{pmatrix}$ Therefore, the given vectors span $\\mathbb{C}^4$ for all $x_i$. $\\blacksquare$ For the second question, I'd argue that any 4 of those given vectors can be shown to span $\\mathbb{C}^4$ in a similar manner. So there are ${6}\\choose{4}$ ways 4 of those vectors form a basis for $\\mathbb{C}^4$. • Could you define what you mean by arbitrary? – YoTengoUnLCD Jul 15 '15 at 2:43 • I might be abusing that word in my example, but I'm taking it to mean that if $z\\in{Z}$ is arbitrary , then whatever result I get for $z$ extends to all elements of the set $Z$. Would you say \"$z\\in{Z}$ is arbitrary\" is the functional equivalent of \"$\\forall z \\in{Z}$\"? – gbrlrz017 Jul 15 '15 at 2:48 • I see what you mean, I would avoid using that word altogether, it causes confusion. Also, there aren't 6 choose 4 ways to pick those vectors to form a basis, pick vectors" ]

[ "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", ")}$ or $r=\\frac{ed}{(r± e \\sin \\theta)}$ where r traces out the curve as $$r= f( \\theta )$$ and e follows an ellipse if e < 1, a parabola if e = 1, a hyperbola if e > 1. Public Topic 2 years 17 weeks ago Polar Equation for Conics", "## Curvature of functions Find the curvature of the following curve at the specified point. $r=a \\sin 3\\theta$; $\\theta =\\frac{\\pi}{12}$ I'm pretty much lost on this one. Thanks for any help!", ". . . $3\\theta \\:=\\:\\pi n$ . . . . . . . . . . . . $\\theta \\:=\\:\\frac{\\pi}{3}n$ • November 26th 2011, 11:04 AM joatmon Re: Finding Vertical Tangent of Parametric Curve Thanks. I had a feeling that I was missing something obvious.", "# Asymptotes of Hyperbolas Before discussing rectangular hyperbolas, we must first understand what asymptotes are. Asymptotes : An asymptote to a curve is a straight line, to which the tangent to the curve tends as the point of contact goes to infinity. If this sounds confusing, you can think of an asymptote as follows: an asymptote to a curve is a straight line such that the perpendicular distance of a point $$P(x,\\,y)$$ on the curve from this line tends to zero as the point P goes to infinity (along some branch of the curve). For example, the line $$y=0$$ is an asymptote to the curve \\begin{align}y=\\frac{1}{1+{{x}^{2}}}\\end{align} as shown below : Consider a hyperbola \\begin{align}\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1.\\end{align} Can this have asymptotes ? If so, what will be their equations ? Let $$y=mx+c$$ be an asymptote to the given hyperbola. We need to determine m and c. Consider any point $$P(a\\sec \\theta ,\\,b\\tan \\theta )$$ on the hyperbola. The perpendicular distance of P from our supposed asymptote $$y=mx+c$$ will be given by ${{d}^{2}}=\\frac{{{(am\\sec \\theta -b\\tan \\theta +c)}^{2}}}{1+{{m}^{2}}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\ldots \\left( 1 \\right)$ Now first observe that as $$\\theta \\to \\pm \\frac{\\pi }{2},$$ the point P tends to go away to infinity along some arm of the hyperbola. Thus, we want a value of m and c, such that in (1), As\\;\\quad\\begin{align}\\theta \\to \\pm \\frac{\\pi", "0 \\\\ 1 & 0 & 0 \\end{bmatrix}$ and it is easy to verify that MT is its own inverse. The locus of points described parametrically as $( \\cos \\theta, \\, \\sin \\theta )$ describe a circle, due to the trigonometric identity $\\cos^2 \\theta + \\sin^2 \\theta = 1 \\,$ which has the same form as the canonical equation of a circle. Applying the projectivity T yields the locus of points described parametrically by $(\\sec \\theta,\\, \\tan \\theta)$ which describe a hyperbola, due to the trigonometric identity $\\sec^2 \\theta - \\tan^2 \\theta = 1 \\,$ which has the same form as the canonical equation of a hyperbola. Notice that points $(~-1,0)$ and $(1,0)$ are fixed points. Indeed, this projectivity transforms any circle, of any radius, into a hyperbola centered at the origin with both of its foci lying on the x-axis, and vice versa. This projectivity also transforms the y-axis into the line at infinity, and vice versa: $T : (0, y) \\rightarrow \\left( {1 \\over 0}, {y \\over 0} \\right) = (\\pm \\infty, \\pm \\infty),$ $T: (\\pm \\infty, \\pm \\infty) \\rightarrow \\left( {1 \\over \\pm \\infty}, {\\pm \\infty \\over \\pm \\infty} \\right) = (0, y).$ The ratio of infinity over infinity is indeterminate which means that it can be set to any value y desired. This example", "Tangents to Polar Curve)}\\qquad \\frac{dy}{dx}=\\frac{dy/d\\theta}{dx/d\\theta}=\\frac{f'(\\theta)\\sin\\theta+f(\\theta)\\cos\\theta}{f'(\\theta)\\cos\\theta-f(\\theta)\\sin\\theta}. }", "# Find the arclength of parametric equation x(\\Theta )=\\Theta \\cos Oy(\\Theta )=\\Theta \\sin \\Theta... ## Question: Find the arclength of parametric equation {eq}x(\\Theta )=\\Theta \\cos Oy(\\Theta )=\\Theta \\sin \\Theta \\,\\, on \\left [ 0,\\pi /2 \\right ] {/eq} ## IArclength of parameterized curves. The arclength of a section of a curve is a measure of how long the curve section is. The arclength of a straight line segment is simply the distance between its two endpoints. For a more general curve, the arclength calculation requires calculus. Become a Study.com member to unlock this answer! Create your account", "Find an equation of the tangent to the curve given by at the point corresponding to $\\theta=0.$ Your answer should be in the form of $y=f(x)$ without $\\theta$'s. $y(x)$ =", "# General Parametric Curves In this post, we will take a look at some of the general parametric curves in $\\mathbb{R}^{2}$ as well as $\\mathbb{R}^{3}$, viz., circles, cycloids, and helixes; explore their curves to see how and why their respective equations describe such paths. # Circles A circle can be defined as the locus of all points that satisfy the equations $$x = a\\cos{\\theta}\\quad y = a\\sin{\\theta}$$ where $x$, $y$ are the coordinates of any point on the circle, $a$ is the radius of the circle and $\\theta$ is the parameter - the angle subtended by the point at the centre of the circle. 0 0 Figure 1: Circle as a parametric curve Another way to parameterize a circle by rational functions is: $$x = \\frac{1-\\theta^{2}}{1+\\theta^{2}}\\quad y = \\frac{2\\theta}{1+\\theta^{2}}$$ It is implied that any point not on the circle does not satisfy this pair of equations. If the centre of the circle does not lie on the origin, we translate the circle from the origin to its proper location, i.e.:$$x = h + a\\cos{\\theta}\\quad y = k + a\\sin{\\theta}$$ where $(h,k)$ is the centre of the circle. # Cycloids Let $P$ be a fixed point on a circle. If the circle rolls along a straight line with uniform velocity and without slipping, the path traced out by $P$", "### Solution The curve is described by the following parametric equations for x(t) and y(t), respectively: $x(\\theta) = 2 \\cos^2 (\\theta)$ $y(\\theta) = 2 \\cos (\\theta) \\sin (\\theta)$ To find the Arc Length, we must first find the integral of the derivative sum given below: $L_{arc} = \\int_{a}^{b} \\sqrt {(\\frac {dx}{d\\theta})^2 + (\\frac {dy}{d\\theta})^2} \\,d\\theta$ Input the values inside this equation. The arc length $L_{arc}$ is given as: $L_{arc} = \\int_{0}^{\\pi} \\sqrt {\\bigg(\\frac {d(2 \\cos^2 (\\theta))}{d\\theta}\\bigg)^2 + \\bigg(\\frac {d(2 \\cos (\\theta) \\sin (\\theta))}{d\\theta}\\bigg)^2} \\,d\\theta = \\int_{0}^{\\pi}2 \\,d\\theta \\approx 6.28$", "Free Version Easy # Given Slope, Find the Type of Curve APCALC-SBPLKC If the slope of a certain curve is $-\\cfrac {x}{y}$ for every point $(x, \\, y)$, then the curve is a(n) A Hyperbola. B Parabola. C Ellipse. D Circle.", "15 views The equations $x=a\\cos\\theta+b\\sin\\theta$ and $y=a\\sin\\theta+b\\cos\\theta,(0\\leq\\theta\\leq2\\pi$ and $a,b$ are arbitrary constants$)$ represent 1. a circle 2. a parabola 3. an ellipse 4. a hyperbola in Geometry recategorized | 15 views 0 Here, $x$ should be $x = a\\cos\\theta - b \\sin \\theta$ I think this is the mistake in question??", "of given hyperbola can be written as $\\frac{x^{2}}{1/4} + \\frac{y^{2}}{1/9} = 1$ So $a^{2} = 1/4$ $a= 1/2$ Similarly $b= 1/3$ General point on ellipse $(1/2 \\sec \\theta,1/3 \\tan \\theta)$ Using T=0 on ellipse We get $4px-9qy = 1$ Where (p,q) is point of contact Since it's equation of tangent, so by comparing this with given equation of tangent $4x-3y = \\sqrt{3}$ We get p= $1/\\sqrt{3}$ And q = $1/3\\sqrt{3}$ Since p = $1/2\\sec \\theta$ $1/2\\sec \\theta = 1/\\sqrt{3}$ $sec\\theta = 2/\\sqrt{3}$ So $\\Theta = %u03C0/6$ π/6 Hope it helps.", "# Math Help - Curve Tracing 1. ## Curve Tracing The equation of the curve is- r^2cos2(theta)=q^2 Sorry I don't know to write this in Natural form but I just hope that I am clear enough 2. What are you supposed to do with this equation? 3. Originally Posted by stapel What are you supposed to do with this equation? Make the cure mate.. 4. That's even more confusing! And what is \"q\"? Some constant? Do you know the identity \" $cos(2\\theta)= cos^2(\\theta)- sin^2(\\theta)$\"? You can use that to write your equation as $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q$ and use the fact that $x= r cos(\\theta)$, $y= r sin(\\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me. 5. Originally Posted by HallsofIvy That's even more confusing! And what is \"q\"? Some constant? Do you know the identity \" $cos(2\\theta)= cos^2(\\theta)- sin^2(\\theta)$\"? You can use that to write your equation as $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q$ and use the fact that $x= r cos(\\theta)$, $y= r sin(\\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me. Yes q is constant..!! and the question is $r^2cos^2(\\theta)- r^2sin^2(\\theta)= q^2$ So I think it will be a rectangular hyperbola..?? 6. Originally Posted by stapel What are you supposed to do with this equation? You have to make the curve with this.", "CJC/2021/JC1/Promo/P1/Q06 The curve $C$ is defined by the parametric equations $x=\\theta+\\frac{1}{2} \\sin 2 \\theta, y=2 \\tan \\theta \\quad \\text { where }-\\frac{\\pi}{2}<\\theta<\\frac{\\pi}{2} \\text {. }$ (i) Show that $\\frac{\\mathrm{d} y}{\\mathrm{~d} x}=\\frac{1}{\\cos ^k \\theta}$, where $k$ is an integer to be determined. [3] The lines $T$ and $N$ are the tangent and normal to $C$ at the point where $\\theta=\\frac{\\pi}{4}$ respectively. (ii) Find the equations of $T$ and $N$, leaving your answers in exact form. [3] (iii) Find the area enclosed by $T, N$ and the $y$-axis. [2]", "closed curve with larger area by hinging $$AC$$ and $$BC$$ about $$C$$ until $$\\theta=90^\\circ$$. The perimeter of the curve remains the same but the subtended area is larger. The only curve for which all points subtend $$90^\\circ$$ to $$AB$$ is a circle.", "# Tangents of polar curves ### Tangents of polar curves In this lesson, we will learn how to find the tangent line of polar curves. Just like how we can find the tangent of Cartesian and parametric equations, we can do the same for polar equations. First, we will examine a generalized formula to taking the derivative, and apply it to finding tangents. Then we will look at a few examples to finding the first derivative. Lastly, we will do some applications which involve finding tangent lines of polar curves at a specified point. #### Lessons In order to find the tangent line to polar curves, we have to take the derivative in polar coordinates. Here is the formula to take the derivative in polar coordinates: $\\frac{dy}{dx}=\\frac{\\frac{dr}{d \\theta}\\sin \\theta+r\\;\\cos \\theta}{\\frac{dr}{d \\theta}\\cos \\theta-r\\;\\sin \\theta}$ • Introduction Tangents of Polar Curves Overview • 1. Finding the Derivative Find $\\frac{dy}{dx}$ for each of the following polar equations: a) $r=\\sin \\theta + \\theta$ b) $r= \\frac{\\sin \\theta}{\\cos \\theta}$ • 2. Finding the Tangent Line Find the tangent line with the following polar curves at the specified point: $r=\\sin (3\\theta)$ at $\\theta = \\frac{\\pi}{4}$ • 3. Finding the Tangent Line Find the tangent line with the following polar curves at the specified point: $r=\\theta \\cos \\theta$ at $\\theta =0$", "# Thread: find length of curve 1. ## find length of curve find the length of the curve r = (theta)^2 ... 0<= theta <= 2pi just wanted the equations or formulas involved in this equation please. also, i believe i use will take an integral for this equation. thankyou. 2. Originally Posted by rcmango find the length of the curve r = (theta)^2 ... 0<= theta <= 2pi just wanted the equations or formulas involved in this equation please. also, i believe i use will take an integral for this equation. thankyou. $S = \\int_{\\theta_1}^{\\theta_2} \\sqrt{r^2 + \\left(\\frac{dr}{d\\theta}\\right)^2} \\, d\\theta$", "# How do you name the curve given by the conic theta=(2pi)/3? $\\theta = \\frac{2 \\pi}{3}$ represents a ray travelling from the polar origin to $\\infty$" ]

[ "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "large negative values, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the positive numbers as the angle enters the $3^{rd}$ quadrant. The segment will again get infinitely large as it approaches $270^\\circ$ . After being undefined at $270^\\circ$ , the angle crosses into the $4^{th}$ quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full rotation. The graph $y=\\tan x$ over several rotations would look like this: Notice the $x-$ axis is measured in radians. Our asymptotes occur every $\\pi$ radians, starting at $\\frac{\\pi}{2}$ . The period of the graph is therefore $\\pi$ radians. The domain is all reals except for the asymptotes at $\\frac{\\pi}{2}, \\frac{3\\pi}{2}, -\\frac{\\pi}{2}, etc.$ and the range is all real numbers. Cotangent is the reciprocal of tangent, $\\frac{x}{y}$ , so it would make sense that where ever the tangent had an asymptote, now the cotangent will be zero. The opposite of this is also true. When the tangent is zero, now the cotangent will have an asymptote. The shape of the curve is generally the same, so the graph looks like this: When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles that have $45^\\circ$ as reference angles, which always", "\\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*} If the length of $AD$ is $200$, then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be $$\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}$$ Which is a contradiction. Therefore, our answer is $\\boxed{500}.$ ## Solution 3 (Trigonometry Bash) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,E); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$ Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on $\\Delta BOC$; let $\\theta = \\angle BOC$. We get the following equation: $$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta$$ Substituting the values in, we get $$(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta$$ Canceling out, we get $$\\cos\\theta=\\frac{3}{4}$$ Because $\\angle AOB$, $\\angle BOC$, and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$. To find", "unitsize(1cm); xaxis(\"Re\",Arrows); yaxis(\"Im\",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label(\"Im(z)=164\",F(-330),N); pair z = (-656,164); dot(Label(\"z\",align=N),z); dot(Label(\"z+n\",align=N),z+(697,0)); draw(Label(\"4x\"),z--(0,0)); draw(Label(\"x\"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$. Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\\angle\\theta_1 \\cdot r_2\\angle\\theta_2 = r_1r_2\\angle(\\theta_1+\\theta_2)$ and $\\frac{z_1}{z_2} = \\frac{r_1\\angle\\theta_1}{r_2\\angle\\theta_2} = \\frac{r_1}{r_2}\\angle(\\theta_1-\\theta_2)$, where $r$ is the magnitude and $\\theta$ is the phase, and $z_n=r_n\\angle\\theta_n$. Since $4i$ has magnitude $4$ and phase $90^\\circ$ (since the positive imaginary axis points in a direction $90^\\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$. This means that", "positive angles that lie between 0 and (2pi ). This angle will not be the only possibility of course, but we typically look for angles that meet these conditions. To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of (frac<6>) with the positive (x)-axis, then so must the angle in the fourth quadrant. So, we have two options. We could use ( - frac<6>), but again, it’s more common to use positive angles. To get a positive angle all we need to do is use the fact that the angle is (frac<6>) with the positive (x)-axis (as noted above) and a positive angle will be (t = 2pi - frac <6>= frac<<11pi >><6>). One way to remember how to get the positive form of the second angle is to think of making one full revolution from the positive (x)-axis (i.e. (2pi )) and then backing off (i.e. subtracting) (frac<6>). We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be ( - frac<6>) that we want for the solution and sometimes we will want both (or neither) of the", "a circle around $$C_{k_0} = C_k$$ with radius $$r$$. 3. On the circle do lie two other corners, pick the next one in counter-clockwise direction, $$C_{k_1}$$. 4. If $$C_1$$ does not lie between $$C_{k_0}$$ and $$C_{k_1}$$ (in counter-clockwise direction) or equals $$C_{k_1}$$: 5. Draw a line from $$C_k$$ to $$C_{k_1}$$. 6. Let $$C_{k_0} = C_{k_1}$$ and proceed with 5. 7. Else: Stop. There are three equivalent ways to create the division graph for $$N$$ edge by edge: 1. For each $$n = 1,2,...,N$$: For each $$m\\leq N$$ draw an edge between $$n$$ and $$m$$ when $$n$$ divides $$m$$. 2. For each $$n = 1,2,...,N$$: For each $$k = 1,2,...,N$$ draw an edge between $$n$$ and $$m = k\\cdot n$$ when $$m \\leq N$$. 3. For each $$k = 1,2,...,N$$: For each $$n = 1,2,...,N$$ draw an edge between $$n$$ and $$m = k\\cdot n$$ when $$m \\leq N$$. • Very pretty. The envelopes stay in place because each corresponds to connecting angle point $\\theta$ (measured anticlockwise from the bottom) to angle point $k\\theta$ for given integer $k \\ge 2$ and all possible up to $0 \\lt \\theta \\le \\frac{2\\pi}{k}$. For given $N$ you only get lines when $\\theta = 2\\pi \\frac{n}{N}$ and $0 \\lt n \\le \\frac{N}{k}$, but that has little effect on the envelopes. – Henry Jan", "N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); [/asy]$ $C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius of $C_4$. We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\\Omega$, and all points on $\\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$, and since $\\Omega$ has radius of $6$, the resulting line must be $\\frac{36}{4} = 9$ units away from $O$. $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2,", "because when $$\\cos 3\\theta < 0$$, then the radius becomes negative. On the interval $$\\theta \\in [0,2\\pi)$$, this occurs when $$\\theta \\in (\\pi/6, \\pi/2) \\cup (5\\pi/6, 7\\pi/6) \\cup (3\\pi/2, 11\\pi/6).$$ So in fact, this curve is traversed twice as $$\\theta \\in [0, 2\\pi)$$. Refer to the following diagram. As $$\\theta$$ traverses the interval, when $$r > 0$$, the radial line with angle $$\\theta$$ is blue and the point it is drawing is also blue. When $$r < 0$$, the line and point becomes red. The black circle represents the angle of the line, so as you can see, the line only makes a single complete rotation, but the curve is drawn twice.", "Solve Trigonometric Problems Remember that we’re typically looking for positive angles between 0 and $$2\\pi$$ so we’ll use the positive angle.An easy way to remember how to the positive angle here is to rotate one full revolution from the positive $$x$$-axis ( subtracting) $$\\frac$$. To do this all we need to do is divide both sides by 2. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions. Recall from the previous section and you’ll see there that we used $\\frac 2\\pi \\,n\\,,\\;\\;n = 0,\\, \\pm 1,\\, \\pm 2,\\, \\pm 3,\\, \\ldots$ to represent all the possible angles that can end at the same location on the unit circle, angles that end at $$\\frac$$. Remember that all this says is that we start at $$\\frac$$ then rotate around in the counter-clockwise direction ($$n$$ is positive) or clockwise direction ($$n$$ is negative) for $$n$$ complete rotations. The angle in the first quadrant makes an angle of $$\\frac$$ with the positive $$x$$-axis, then so must the angle in the fourth quadrant. We could use $$- \\frac$$, but again, it’s more common to use positive angles. To get a positive angle all we need to do is use the fact that", "the positive real axis. $e^{i\\frac{-3\\pi}{2}}$ lies on the unit circle, an angle $3\\pi/2$ measured clockwise from the positive real axis. Another way to see this is to recall that $x^{-1}= 1/x$ and that $1/i= -i$.", "the negative X axis corresponds to $\\varphi=\\frac{3\\pi}{2}$ and, hence, $(-1,0,0)$ does belong to the circle. • But $\\theta$ is not the angle between with the $x$-axis (in positive orientation), since $(-1, 0, 0)$ would have angle $\\theta = \\pi$ according to the picture. – user40276 Jun 1 '15 at 8:00 • @user40276 You have both $(\\pi,\\dfrac{\\pi}{2})$ and $(0,\\dfrac{3\\pi}{2})$ as coordinates that give $(-1,0,0)$ – Martigan Jun 1 '15 at 8:04 • @Martigan So how are the sets such that $\\phi$ is injective? They should induce a kind of tessellation of the plane and this is my main problem – user40276 Jun 1 '15 at 8:10 Honestly this is difficult to comment with no picture in the text, even if you linked one. The problem is that you linked two of them... And perhaps this is one of the answer you might have. In physics and mathematics the conventions are not the same. Don't try to set a value to $\\theta$ and see wether it fits your understanding of the cartesian coordinates corresponding to it. Try to take any point on the sphere and see if you can have $(\\theta,\\varphi)$ that works for it. You will see that it does, whatever your convention, and that, in turn, $\\phi$ is a covering space as you call it. • I think", "angle $$\\frac{{2\\pi }}{3}$$. That’s not on our unit circle above, however notice that $$\\frac{{2\\pi }}{3} = \\pi - \\frac{\\pi }{3}$$. So $$\\frac{{2\\pi }}{3}$$ is found by rotating up $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the second quadrant. The coordinates for $$\\frac{{2\\pi }}{3}$$ will be the coordinates for $$\\frac{\\pi }{3}$$ except the $$x$$ coordinate will be negative. Likewise, for $$- \\frac{{2\\pi }}{3}$$we can notice that $$- \\frac{{2\\pi }}{3} = - \\pi + \\frac{\\pi }{3}$$, so this angle can be found by rotating down $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\\frac{\\pi }{3}$$ except both will be negative. Both of these angles, along with the coordinates of the intersection points, are shown on the following unit circle. From this unit circle we can see that $$\\sin \\left( {\\frac{{2\\pi }}{3}} \\right) = \\frac{{\\sqrt 3 }}{2}$$and $$\\sin \\left( { - \\frac{{2\\pi }}{3}} \\right) = - \\frac{{\\sqrt 3 }}{2}$$. This leads to a nice fact about the sine function. The sine function is called an odd function", "the angle closest to the x-axis is $\\frac{7\\pi}{6}$, the next is $\\frac{5\\pi}{4}$, the next is $\\frac{4\\pi}{3}$, and finally $\\frac{3\\pi}{2}$. The last mnemonic, P for prime, actually also begins $\\frac{3\\pi}{2} radians. Each denominator in this quadrant follows the same progression, but the numerator coefficients for each are the primes, beginning with three, in increasing order. Thus, the angles are then$\\frac{5\\pi}{6}$,$\\frac{7\\pi}{4}$, and$\\frac{11\\pi}{3}$. When filling in the unit circle from memory, begin by writing$\\frac{\\pi}{6}, \\frac{\\pi}{4}, \\frac{\\pi}{3}$in each quadrant moving from the x-axis to the y-axis. Then fill in$2\\pi$and$\\pi$on the angles coterminal with the x-axis. Next, put$\\frac{\\pi}{2}$on the two angles coterminal with the y-axis. Finally, fill in the numerator coefficients of pi in the second, third, and fourth quadrants using the mnemonic. ### Left Hand Trick for Memorizing Sine and Cosine Recall that the sine and cosine of an angle are equivalent to the intersection of the terminal ray and the unit circle’s circumference. For the common angles$\\frac{\\pi}{6}, \\frac{\\pi}{4},$and$\\frac{\\pi}{3}$, these all have a denominator of two and a square root on top. Remembering these values for the first quadrant is simple with the left-hand trick. Then, remembering where sine and cosine are positive and negative helps with the rest. To use the left hand trick, hold your left hand so that the palm is towards you, your thumb is straight up, and", "# 1992 AIME Problems/Problem 11 ## Problem Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\\frac{\\pi}{70}$ and $\\frac{\\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$, the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$, and the resulting line is reflected in $l_2^{}$. Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\\left(R^{(n-1)}(l)\\right)$. Given that $l^{}_{}$ is the line $y=\\frac{19}{92}x^{}_{}$, find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$. ## Solution Let $l$ be a line that makes an angle of $\\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$. The angle between $l$ and $l_1$ is $\\theta - \\frac{\\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\\theta - \\frac{\\pi}{70}$. Thus, $l'$ makes an angle of $\\frac{\\pi}{70}-\\left(\\theta-\\frac{\\pi}{70}\\right) = \\frac{\\pi}{35}-\\theta$ with the positive $x$-axis. Similarly, since the angle between $l'$ and $l_2$ is $\\left(\\frac{\\pi}{35}-\\theta\\right)-\\frac{\\pi}{54}$, the angle between $l''$ and the positive $x$-axis is $\\frac{\\pi}{54}-\\left(\\left(\\frac{\\pi}{35}-\\theta\\right)-\\frac{\\pi}{54}\\right) = \\frac{\\pi}{27}-\\frac{\\pi}{35}+\\theta = \\frac{8\\pi}{945} + \\theta$. Thus, $R(l)$ makes an $\\frac{8\\pi}{945} + \\theta$ angle with the positive $x$-axis. So $R^{(n)}(l)$ makes an $\\frac{8n\\pi}{945} + \\theta$ angle with the positive $x$-axis. Therefore, $R^{(m)}(l)=l$ iff $\\frac{8m\\pi}{945}$ is an integral multiple of $\\pi$. Thus, $8m \\equiv 0\\pmod{945}$. Since $\\gcd(8,945)=1$, $m \\equiv 0 \\pmod{945}$, so", "### Home > PC > Chapter 3 > Lesson 3.1.3 > Problem3-47 3-47. Sketch a graph of the unit circle then draw and label the following angles. 1. $\\frac{\\pi}{4}$ 1. $- \\frac { \\pi } { 3 }$ Wrap clockwise from $\\left(1, 0\\right)$ because of the $−$ sign. 1. $3 \\pi$ Wrap counter-clockwise more than once. How much more?", "look at the following unit circle. From quick inspection we can see that $$t = \\frac{\\pi }{6}$$ is a solution. However, as I have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and $$2\\pi$$. This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions. To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of $$\\frac{\\pi }{6}$$ with the positive $$x$$-axis, then so must the angle in the fourth quadrant. So, we could use $$- \\frac{\\pi }{6}$$, but again, it’s more common to use positive angles so, we’ll use $$t = 2\\pi - \\frac{\\pi }{6} = \\frac{{11\\pi }}{6}$$. We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be $$- \\frac{\\pi }{6}$$ that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem", "= \\frac{{11\\pi }}{6}$$. One way to remember how to get the positive form of the second angle is to think of making one full revolution from the positive $$x$$-axis (i.e. $$2\\pi$$) and then backing off (i.e. subtracting) $$\\frac{\\pi }{6}$$. We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be $$- \\frac{\\pi }{6}$$ that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions. This is very easy to do. Recall from the previous section and you’ll see there that we used $\\frac{\\pi }{6} + 2\\pi \\,n\\,,\\;\\;n = 0,\\, \\pm 1,\\, \\pm 2,\\, \\pm 3,\\, \\ldots$ to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at $$\\frac{\\pi }{6}$$. Remember that all this says is that we start at $$\\frac{\\pi }{6}$$ then rotate around in the counter-clockwise direction ($$n$$ is positive) or clockwise direction ($$n$$ is negative) for $$n$$ complete rotations. The same thing can be done for", "SubsectionB.2.3Unit Circle and Standard Position In order to have a universal reference for an angle, we introduce the standard position of an angle in terms of a unit circle. The equation for a circle with radius $r=1$ and a center at the origin $(0,0)$ is $x^2+y^2=1\\text{.}$ Standard position for an angle always forms an arc on the unit circle that starts on the $x$-axis at $(1,0)$ and moves along the circle in the counter clockwise direction for positive angles (clockwise for negative angles). An angle greater than $2 \\pi$ will wrap completely around the circle and continue. Multiples of $2 \\pi$ ($0, \\pm 2 \\pi, \\pm 4 \\pi, \\ldots$) all end at the same point $(1,0)$ on the unit circle but correspond to a different number of rotations. In fact, every point on the unit circle has infinitely many different angles (that differ by multiples of $2 \\pi$) that end at that point when in standard position. ###### ExampleB.2.2 Find all angles that terminate at the point $(0,1)\\text{.}$ Solution The point $(0,1)$ is a quarter turn in the positive direction or three quarters turn in the negative direction. So the angles $\\theta = \\frac{1}{4} \\tau = \\frac{1}{4}(2 \\pi) = \\frac{\\pi}{2}$ and $\\theta = -\\frac{3}{4} \\tau = -\\frac{3}{4}(2 \\pi) = -\\frac{3\\pi}{2}$ are two of the angles desired. Notice that", "### Home > PC > Chapter 4 > Lesson 4.2.1 > Problem4-71 4-71. Find all values of $θ$ from $[0, 2\\pi]$ where the following equations are true. Sketching a picture of the unit circle may help. 1. $\\text{sin }θ =\\frac { 1 } { 2 }$ $30\\degree\\text{ or }\\frac{\\pi}{6}$ $150\\degree\\text{ or }\\frac{5\\pi}{6}$ 1. $\\text{cos }θ = −\\frac { 1 } { 2 }$ 1. $\\text{sin }θ = - \\frac { \\sqrt { 3 } } { 2 }$ What special angle(s) are involved? What does the negative sign mean? 1. $\\text{cos }θ = 0$ These angles are coterminal with the $y$-axis.", "of $k .$ We obtain $\\frac{2 \\pi}{3},-\\frac{10 \\pi}{3}$ and $\\frac{8 \\pi}{3}$ as coterminal angles. 3. Since $\\gamma=\\frac{9 \\pi}{4}$ is positive, we rotate counter-clockwise from the positive $x$-axis. One full revolution accounts for $2 \\pi=\\frac{8 \\pi}{4}$ of the radian measure with $\\frac{\\pi}{4}$ or $\\frac{1}{8}$ of a revolution remaining. We have $\\gamma$ as a Quadrant I angle. All angles coterminal with $\\gamma$ are of the form $\\theta=\\frac{9 \\pi}{4}+\\frac{8 \\pi}{4} k$, where $k$ is an integer. Working through the arithmetic, we find coterminal angles of $\\frac{\\pi}{4},-\\frac{7 \\pi}{4}$ and $\\frac{17 \\pi}{4}$.4. To graph $\\varphi=-\\frac{5 \\pi}{2}$, we begin our rotation clockwise from the positive $\\mathrm{x}$-axis. As $2 \\pi=\\frac{4 \\pi}{2}$, after one full revolution clockwise we have $\\frac{\\pi}{2}$ or $\\frac{1}{4}$ of a revolution remaining. Since the terminal side of $\\varphi$ lies on the negative $y$-axis, $\\varphi$ is a quadrantal angle. To find coterminal angles, we compute $\\theta=-\\frac{5 \\pi}{2}+\\frac{4 \\pi}{2} k$ for a few integers $k$ and obtain $-\\frac{\\pi}{2}$, $\\frac{3 \\pi}{2}$ and $\\frac{7 \\pi}{2}$ by Platinum (132,156 points) 0 like 0 dislike 1 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike" ]

[ "+0 # Polar Cordinates 0 39 1 +157 Find the curve defined by the equation $$r^2 \\cos 2 \\theta = 4.$$ (A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. Apr 16, 2020", "on from there. 3) set dr/d𝛉 equal to zero. This will give you candidates for where the curve is at a maximum. In the case of cos(2𝛉), it will give you the endpoints of each petal. In some cases you may also need to find where the tangent line is undefined. 4) sketch the pitch lines found in step two. Then plot the points found in step 3. Remember to keep in mind what the generic form of your graph looks like, and then draw your graph! I hope this helps some! :) • why is it that for rcos(theta) the radius is 0 at pie, when clearly cos(pie) is -1 and so the radius would have to be -3? where is the graph from when theta equals to pie/2 to 3pie/2? • In the polar coordinate system, r denotes the distance of the point from the origin. Having -a for r means going a distance of a in the opposite direction. Suppose that at an angle of pi, r = -3. This means that you have to go 3 in the direction. This is just (in fact exactly) the same as going -3 along the number line in the negative direction ( -(-3) = 3 ). In general, mathematically: ( -r , theta ) = ( r", "interest should find the desired result. Create a graphing window and enter your polar equation (menu –> 3:Graph Entry –> 4:Polar). Then drop a tangent line on the polar curve (menu –> 8:Geometry –> 1:Points&Lines –> 7:Tangent). You would then click on the polar curve once to select the curve and a second time to place the tangent line. Then press ESC to exit the Tangent Line command. To get the current coordinates of the point and the equation of the tangent line, use the Coordinates & Equation tool (menu –> 1:Actions –> 8:Coordinates and Equations). Click on the point and the line to get the current location’s information. After each click, you’ll need to click again to tell the nSpire where you want the information displayed. To get the tangent line at $\\theta =\\frac{\\pi }{6}$, you could drag the point, but the graph settings seem to produce only Cartesian coordinates. Converting $\\theta =\\frac{\\pi }{6}$ on $r=2+3sin(\\theta )$ to Cartesian gives $\\left( x,y \\right) = \\left( r \\cdot cos(\\theta ), r \\cdot sin(\\theta ) \\right)=\\left( \\frac{7\\sqrt{3}}{4},\\frac{7}{4} \\right)$ . So the x-coordinate is $\\frac{7\\sqrt{3}}{4} \\approx 3.031$. Drag the point to find the approximate slope, $\\frac{dy}{dx} \\approx 8.37$. Because the slope of the tangent line changes rapidly at this location on this polar curve, this value of 8.37 will be", "graph in Quadrant IV, with all of the angle rotations starting from the negative $$x$$-axis. As $$\\theta$$ ranges from $$\\pi$$ to $$\\frac{3\\pi}{2}$$, the $$r$$ values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the $$|r|$$ for these values of $$\\theta$$ match the $$r$$ values for $$\\theta$$ in $$\\left[0, \\frac{\\pi}{2} \\right]$$, we have that the curve begins to retrace itself at this point. Proceeding further, we find that when $$\\frac{3\\pi}{2} \\leq \\theta \\leq 2\\pi$$, we retrace the portion of the curve in Quadrant IV that we first traced out as $$\\frac{\\pi}{2} \\leq \\theta \\leq \\pi$$. The reader is invited to verify that plotting any range of $$\\theta$$ outside the interval $$\\left[ 0, \\pi \\right]$$ results in retracting some portion of the curve.\\footnote{The graph of $$r=6\\cos(\\theta)$$ looks suspiciously like a circle, for good reason. See number \\ref{ris6costheta} in Example \\ref{eqnconversionex}.} We present the final graph below. Example $$\\PageIndex{1}$$:\\label{polargraphex} Graph the following polar equations. 1. \\label{limacon01} $$r = 4 - 2\\sin(\\theta)$$ 2. \\label{limacon02} $$r = 2 + 4 \\cos(\\theta)$$ 3. \\label{rose} $$r = 5\\sin(2\\theta)$$ 4. \\label{lemniscate} $$r^2 = 16 \\cos(2\\theta)$$ Solution We first plot the fundamental cycle of $$r = 4 - 2\\sin(\\theta)$$ on the $$\\theta r$$-axes. To help us visualize what is going on graphically, we divide up $$[0,2\\pi]$$ into", "large negative values, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the positive numbers as the angle enters the $3^{rd}$ quadrant. The segment will again get infinitely large as it approaches $270^\\circ$ . After being undefined at $270^\\circ$ , the angle crosses into the $4^{th}$ quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full rotation. The graph $y=\\tan x$ over several rotations would look like this: Notice the $x-$ axis is measured in radians. Our asymptotes occur every $\\pi$ radians, starting at $\\frac{\\pi}{2}$ . The period of the graph is therefore $\\pi$ radians. The domain is all reals except for the asymptotes at $\\frac{\\pi}{2}, \\frac{3\\pi}{2}, -\\frac{\\pi}{2}, etc.$ and the range is all real numbers. Cotangent is the reciprocal of tangent, $\\frac{x}{y}$ , so it would make sense that where ever the tangent had an asymptote, now the cotangent will be zero. The opposite of this is also true. When the tangent is zero, now the cotangent will have an asymptote. The shape of the curve is generally the same, so the graph looks like this: When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles that have $45^\\circ$ as reference angles, which always", "##n## for the points you have drawn. By allowing ##n## to have non-integer values larger than 2 you can plot the curve between each pair of points on consecutive circles. Once you have the formulas for ##r## and ##\\theta## in terms of ##n## you can eliminate ##n## to get a formula for ##r## in terms of ##\\theta##. Even more fun: The above gives you a curve for values of ##\\theta## in the range ##[-\\frac{\\pi}{2},\\frac{\\pi}{2})##. Now you can use the formula obtained above to extend it to ##\\theta## in the range ##[-\\infty,-\\frac{\\pi}{2})##, which corresponds to ##n\\in(0,2)##. That curve will spiral in towards the origin, without ever reaching it, as it rotates clockwise around it. Additional exercise: what will the curve do as ##n\\to\\infty##? Last edited: The function is easily written using polar coordinates. Let ##n## be the number of segments in the circle. Then write the values of ##r## and ##\\theta## in terms of ##n## for the points you have drawn. By allowing ##n## to have positive non-integer values you can plot the curve between each pair of points on consecutive circles. Once you have the formulas for ##r## and ##\\theta## in terms of ##n## you can eliminate ##n## to get a formula for ##r## in terms of ##\\theta##. Even more fun: The above gives you a curve for", "when $$e = 1$$, the graph is a parabola when $$e > 1$$, the graph is a hyperbola Taking a more intuitive approach, notice that if $$e < 1$$, the denominator $$1 + e\\cos (\\theta )$$ will always be positive and so $$r$$ will always be positive. This means that the radial distance $$r$$ is defined and finite for every value of $$\\theta$$, including $$\\dfrac{\\pi }{2}$$, with no breaks. The only conic with this characteristic is an ellipse. If $$e = 1$$, the denominator will be positive for all values of $$\\theta$$, except $$- \\pi$$ where the denominator is 0 and $$r$$ is undefined. This fits with a parabola, which has a point at every angle except at the angle pointing along the axis of symmetry away from the vertex. If $$e > 1$$, then the denominator will be zero at two angles other than $$\\pm \\dfrac{\\pi }{2}$$, and $$r$$ will be negative for a set of $$\\theta$$ values. This division of positive and negative radius values would result in two distinct branches of the graph, fitting with a hyperbola. EXAMPLE $$\\PageIndex{2}$$ For each of the following conics with focus at the origin, identify the shape, the directrix, and the eccentricity. a. $$r = \\dfrac{8}{1 - 2\\sin (\\theta )}$$ b. $$r = \\dfrac{6}{3 - 2\\cos (\\theta )}$$ c.", "denominator $$1 + e\\cos (\\theta )$$ will always be positive and so r will always be positive. This means that the radial distance r is defined and finite for every value of ?, including $$\\frac{\\pi }{2}$$, with no breaks. The only conic with this characteristic is an ellipse. If e = 1, the denominator will be positive for all values of ?, except $$- \\pi$$ where the denominator is 0 and r is undefined. This fits with a parabola, which has a point at every angle except at the angle pointing along the axis of symmetry away from the vertex. If e > 1, then the denominator will be zero at two angles other than $$\\pm \\frac{\\pi }{2}$$, and r will be negative for a set of ? values. This division of positive and negative radius values would result in two distinct branches of the graph, fitting with a hyperbola. Example 2 For each of the following conics with focus at the origin, identify the shape, the directrix, and the eccentricity. a. $$r = \\frac{8} ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/9:_Conics/9.1:_Ellipses), /content/body/div[5]/p[306]/span[1], line 1, column 6 $$ b. $$r = \\frac{6} ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/9:_Conics/9.1:_Ellipses), /content/body/div[5]/p[306]/span[2], line 1, column 6 $$ c. $$r = \\frac{8} ParseError: EOF expected (click for", "Example 1 Sketch the polar curve $r = \\sin \\theta$ on the interval $0 ≤ \\theta ≤ 2\\pi$ To begin sketching the curve, let's first look at the curve $r = \\sin \\theta$ on a $r-\\theta$-coordinate system (which is analogous to the $xy$-coordinate system): Let's find some points to plot by setting $\\theta$ equal to various angles: Θ r (r, Θ ) 0 0 (0, 0) π/4 √2/2 (√2/2, π/4) π/2 1 (1, π/2) 3π/4 √2/2 (√2/2, 3π/4) π 0 (0, π) 5π/4 -√2/2 (-√2/2, 5π/4) 3π/2 -1 (-1, 3π/2) 7π/4 -√2/2 (-√2/2, 7π/4) 0 (0, 2π) So we are going to start at the origin $(0,0)$ and have the distance the the point it from the origin increase from the interval $(0, \\frac{\\pi}{2})$ as the ray length is increasing as theta increases. Then on the interval $(\\frac{\\pi}{2}, \\pi)$, the point tracing the curve is going to return to the origin at $(0, \\pi)$. On the interval $(\\pi, \\frac{3\\pi}{2})$, $r$ becomes negative, so the point the curve subtracts $\\pi$ from each angle to make $r$ positive, and hence, the first part of the curve from $(0, \\frac{\\pi}{2})$ is retraced. The same happens on the interval $(\\frac{3\\pi}{2}, 2\\pi)$. Hence the graph of the curve in polar coordinates is as follows: Hence a circle is drawn. If were were to", "the first circle at R above the x-axis and line segment PQ at x. Find the maximum area of the triangle PQR. 31. A straight line L with negative slope passes through the point $(8,2)$ and cuts the positive co-ordinate axes at points P and Q. Find the absolute minimum value of $OP+PQ$, as L varies, where O is the origin. 32. Determine the points of maxima and minima of the function $f(x)=(1/8)\\log{x}-bx+x^{2}$, with $x>0$, where $b \\geq 0$ is a constant. 33. Let $(h,k)$ be a fixed point, where $h>0, k>0$. A straight line passing through this point cuts the positive direction of the co-ordinate axes at points P and Q. Find the minimum area of the triangle $OPQ$, O being the origin. 34. Let $-1 \\leq p \\leq 1$. Show that the equations $4x^{3}-3x-p=0$ has a unique root in the interval $[1/2,1]$ and identify it. 35. Show that the following functions have at least one zero in the given interval: (i) $f(x)=x^{4}+3x+1$, with $[-2,-1]$ (ii) $f(x)=x^{3}+\\frac{4}{x^{2}}+7$ with $(-\\infty,0)$ (iii) $r(\\theta)=\\theta + \\sin^{2}({\\theta}/3)-8$, with $(-\\infty, \\infty)$ 36. Show that all points of the curve $y^{2}=4a(x+\\sin{(x+a)})$ at which the tangent is parallel to axis of x lie on a parabola. 37. Show that the function f defined by $f(x)=|x|^{m}|x-1|^{n}$, with $x \\in \\Re$ has a maximum value $\\frac{m^{m}n^{n}}{(m+n)^{m+n}}$", "d. $$\\dfrac{7\\pi }{6}$$ e. 340$$\\mathrm{{}^\\circ}$$ 17. Find an angle $$\\theta$$ with $$0<\\theta <360{}^\\circ$$ or $$0<\\theta <2\\pi$$ that has the same cosine value as: a. $$\\dfrac{\\pi }{3}$$ b. 80$$\\mathrm{{}^\\circ}$$ c. 140$$\\mathrm{{}^\\circ}$$ d. $$\\dfrac{4\\pi }{3}$$ e. 305$$\\mathrm{{}^\\circ}$$ 18. Find an angle $$\\theta$$ with $$0<\\theta <360{}^\\circ$$ or $$0<\\theta <2\\pi$$ that has the same cosine value as: a. $$\\dfrac{\\pi }{4}$$ b. 15$$\\mathrm{{}^\\circ}$$ c. 160$$\\mathrm{{}^\\circ}$$ d. $$\\dfrac{7\\pi }{6}$$ e. 340$$\\mathrm{{}^\\circ}$$ 19. Find the coordinates of the point on a circle with radius 15 corresponding to an angle of 220$$\\mathrm{{}^\\circ}$$. 20. Find the coordinates of the point on a circle with radius 20 corresponding to an angle of 280$$\\mathrm{{}^\\circ}$$. 21. Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12 seconds to reach the northernmost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive $$y$$-axis. [UW] a. Give Marla’s coordinates at her starting point. b. Give Marla’s coordinates when she has been running for 10 seconds. c. Give Marla’s coordinates when she has been running for 901.3 seconds. 1. a. III b. II 3. $$-\\dfrac{4}{5}$$ 5. $$-\\dfrac{4\\sqrt{3}}{7}$$ 7. $$-\\dfrac{\\sqrt{55}}{8}$$ 9. a.", "# Epicycloid The red curve is an epicycloid traced as the small circle (radius r = 1) rolls around the outside of the large circle (radius R = 3). In geometry, an epicycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circle—called an epicycle—which rolls without slipping around a fixed circle. It is a particular kind of roulette. ## Equations If the smaller circle has radius r, and the larger circle has radius R = kr, then the parametric equations for the curve can be given by either: {\\displaystyle {\\begin{aligned}&x(\\theta )=(R+r)\\cos \\theta \\ -r\\cos \\left({\\frac {R+r}{r}}\\theta \\right)\\\\&y(\\theta )=(R+r)\\sin \\theta \\ -r\\sin \\left({\\frac {R+r}{r}}\\theta \\right)\\end{aligned}}} or: {\\displaystyle {\\begin{aligned}&x(\\theta )=r(k+1)\\cos \\theta -r\\cos \\left((k+1)\\theta \\right)\\\\&y(\\theta )=r(k+1)\\sin \\theta -r\\sin \\left((k+1)\\theta \\right)\\end{aligned}}} in a more concise and complex form[1] ${\\displaystyle z(\\theta )=r\\left(e^{i(k+1)\\theta }-(k+1)e^{i\\theta }\\right)}$ where • angle θ is in turns: ${\\displaystyle \\theta \\in [0,2\\pi ].}$ • smaller circle has radius r • the larger circle has radius kr ## Area (Assuming the initial point lies on the larger circle.) When k is a positive integer, the area of this epicycloid is ${\\displaystyle A=(k+1)(k+2)\\pi r^{2}.}$ It means that the epicycloid is ${\\displaystyle {\\frac {(k+1)(k+2)}{k^{2}}}}$ larger than the original stationary circle. If k is a positive integer, then the curve is closed, and has", "from that of the ellipse in having −b2 instead of b2 , it will be found that many propositions for the hyperbola are derived from those for the ellipse by changing the sign of b2 . Some results for the hyperbola $\\displaystyle \\frac{x^2}{a^2} – \\frac{y^2}{b^2} = 1$ are : (i) The tangent at any point (x1, y1) on the curve is $\\displaystyle \\frac{x x_1}{a^2} – \\frac{y y_1}{b^2} = 1$ (ii) The tangent at the point ‘ θ ‘ is $\\displaystyle \\frac{x sec\\theta}{a} – \\frac{y tan\\theta}{b } = 1$ (iii) The straight line y = mx + c is a tangent to the curve , if c2 = a2 m2 − b2 . In other words , $\\displaystyle y = m x \\pm \\sqrt{a^2 m^2 – b^2}$ touches the curve, for all those values of m when m > b/a or m < − b/a. If m = ±b/a the tangents touch at infinity and are called as asymptotes. (iv) Equation of the normal at any point (x1, y1) to the curve is $\\displaystyle \\frac{x-x_1}{x_1^2 /a^2} = \\frac{y-y_1}{y_1/(-b^2)}$ (v) The equation of the chord through the points θ1 and θ2 is $\\large \\left| \\begin{array}{ccc} x & y & 1 \\\\ a sec\\theta_1 & b tan\\theta_1 & 1 \\\\ a sec\\theta_2 & b tan\\theta_2 & 1\\end{array} \\right| =0$ (vi) The", "the circle before moving on to the classic curves and their variations. ### A General Note: Formulas for the Equation of a Circle Some of the formulas that produce the graph of a circle in polar coordinates are given by $r=a\\cos \\theta$ and $r=a\\sin \\theta$, where $a$ is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is $\\frac{|a|}{2}$, or one-half the diameter. For $r=a\\cos \\theta ,$ the center is $\\left(\\frac{a}{2},0\\right)$. For $r=a\\sin \\theta$, the center is $\\left(\\frac{a}{2},\\pi \\right)$. Figure 5 shows the graphs of these four circles. ### Example 3: Sketching the Graph of a Polar Equation for a Circle Sketch the graph of $r=4\\cos \\theta$. ### Solution First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum $|r|$ for $r=4\\cos \\theta$. First, set $r=0$, and solve for $\\theta$ . Thus, a zero occurs at $\\theta =\\frac{\\pi }{2}\\pm k\\pi$. A key point to plot is $\\left(0,\\text{ }\\text{ }\\frac{\\pi }{2}\\right)$. To find the maximum value of $r$, note that the maximum value of the cosine function is 1 when $\\theta =0\\pm 2k\\pi$. Substitute $\\theta =0$ into the equation: $\\begin{array}{c}r=4\\cos \\theta \\\\ r=4\\cos \\left(0\\right)\\\\ r=4\\left(1\\right)=4\\end{array}$ The maximum value of the equation is 4. A", "1. ## Finding polar equations Let C be a circle with center (a/2,0) and radius a/2. Let l be the line with equation x=a. Find the polar equation of the curve produced in the following manner: for every angle theta, -pi/2 < theta < pi/2, consider the ray from the origin that makes the angle theta with the positive x-axis. This ray intersects C at a point A and l at point B and let O be the origin. Then the point P on the ray is on the curve if the segment OP = the segment AB. I can see what the curve looks like (sort of an exponential line) but I cannot figure out how to find the equation for it. thanks for the help!! 2. Hello, poorm87! Let $C$ be a circle with center $\\left(\\frac{a}{2},\\,0\\right)\\text{ and radius }\\frac{a}{2}$ Let $L$ be the line with equation $x=a$ Find the polar equation of the curve produced in the following manner: For every angle $\\theta,\\; -\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}$, consider the ray from the origin . . that makes the angle $\\theta$ with the positive x-axis. The ray intersects $C$ at point $A$, and $L$ at point $B$. Let $O$ be the origin. Then the point $P$ on the ray is on the curve if: . $OP", "with $$n \\gt 1$$ and odd. When $$n = 1$$ the area is that of a circle $$\\pi a^2$$ and can be obtained from the integration result by using l’Hopital’s rule to obtain the limit since the denominator is zero when $$n = 1$$. The calculation is a straightforward but long one. In the plot, curve(a) is on the left and the right-hand figure curve (b) has $$n = 5$$ and both are plotted with $$a = 1$$. Figure 45. Polar plots, left curve (a). Note that some graphics applications will plot the small loop in (a) inside the larger one. (a) The limits to the integration are multiples of $$\\pm 2\\pi n/3$$. By plotting we find that the outer curve, starting and ending at the pole has limits $$-2\\pi/3 \\to +2\\pi/3$$ and the inner loop from \\$+2\\pi/3 \\to +4\\pi/3. The outer area is therefore $\\displaystyle A=\\frac{a^2}{2}\\int_0^{\\pi/2} r^2d\\theta = \\frac{a^2}{2}\\int_{-2\\pi/3}^{2\\pi/3} (2\\cos(\\theta)+1)^2 d\\theta= \\frac{a^2( 3\\sqrt{3}+4\\pi)}{2}$ The inner loop has area $\\displaystyle \\frac{a^2}{2}\\int_{-2\\pi/3}^{2\\pi/3} (2\\cos(\\theta)+1)^2 d\\theta= \\frac{a^2( -3\\sqrt{3}+2\\pi)}{2}$ making the area excluding the loop $$a^2(3\\sqrt{3} + \\pi)$$. The inner loop calculation using Sympy is theta, a, n = symbols('theta, a, n', positive = True) eq = a**2*(2*cos(theta) + 1)**2/2 simplify(integrate(eq,(theta,2*pi/3,4*pi/3))) (b) If $$n=1$$,the equation is that of a circle of radius $$a/2$$ with its centre at $$r = 1/2$$ and $$\\theta=", "## Graphing the Polar Equations of Conics When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine $e$ and, therefore, the shape of the curve. The next step is to substitute values for $\\theta$ and solve for $r$ to plot a few key points. Setting $\\theta$ equal to $0,\\frac{\\pi }{2},\\pi$, and $\\frac{3\\pi }{2}$ provides the vertices so we can create a rough sketch of the graph. ### Example 2: Graphing a Parabola in Polar Form Graph $r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }$. ### Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is $\\frac{1}{3}$. $\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }=\\frac{5\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+3\\left(\\frac{1}{3}\\right)\\cos \\text{ }\\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{\\frac{5}{3}}{1+\\cos \\text{ }\\theta }\\hfill \\end{array}$ Because $e=1$, we will graph a parabola with a focus at the origin. The function has a $\\cos \\text{", "0^+ }\\frac{S(\\theta)}{\\theta^3}.$ $~$ [4 points] \\begin{aligned}&\\boxed{\\text{A}}. && \\frac{1}{4} \\\\\\\\ &\\boxed{\\text{B}}. && \\frac{1}{2} \\\\\\\\ &\\boxed{\\text{C}}. && 1 \\\\\\\\ &\\boxed{\\text{D}}. && 2 \\\\\\\\ &\\boxed{\\text{E}}. && 4 \\end{aligned} . . . $\\huge 15.~$ There's a point $\\mathrm{P}(t,~-\\ln t)~(0 on the curve $y=-\\ln x.$ $\\mathrm{Q}$ is the foot of perpendicular from $\\mathrm{P}$ to the $x$-axis, and $\\mathrm{R}$ is the point of intersection of the tangent line from $\\mathrm{P}$ and the $x$-axis. What's the maximum area of triangle $\\mathrm{PQR}?$ $~$ [4 points] \\begin{aligned}&\\boxed{\\text{A}}. && \\frac{1}{e^2} \\\\\\\\ &\\boxed{\\text{B}}. && \\frac{2}{e^2} \\\\\\\\ &\\boxed{\\text{C}}. && \\frac{3}{e^2} \\\\\\\\ &\\boxed{\\text{D}}. && \\frac{4}{e^2} \\\\\\\\ &\\boxed{\\text{E}}. && \\frac{5}{e^2} \\end{aligned} . . . $\\huge 16.~$ Two fixed points, $\\mathrm{A},~\\mathrm{B},$ satisfy $\\overline{\\mathrm{AB}}=2.$ A point $\\mathrm{P}$ satisfies the below conditions. (1) $\\overrightarrow{\\mathrm{AP}}\\cdot\\overrightarrow{\\mathrm{PB}}=0.$ (2) $\\overrightarrow{\\mathrm{AB}}\\cdot\\overrightarrow{\\mathrm{AP}}\\ge2+\\sqrt{3}$ What is the length of the locus of $\\mathrm{P}?$ $~$ [4 points] \\begin{aligned}&\\boxed{\\text{A}}. && \\frac{\\pi}{3} \\\\\\\\ &\\boxed{\\text{B}}. && \\frac{\\pi}{2} \\\\\\\\ &\\boxed{\\text{C}}. && \\frac{2\\pi}{3} \\\\\\\\ &\\boxed{\\text{D}}. && \\frac{5\\pi}{6} \\\\\\\\ &\\boxed{\\text{E}}. && \\pi \\end{aligned} . . . $\\huge 17.~$ For a natural number $n\\ge2,$ there's a pocket that contains $n$ cards where each card is written a different natural number from $1$ to $n.$ Choose 2 cards randomly, at the same time, from the pocket. Let the random variable $X$ be the product of the two numbers written on them. Below is the process of finding $E(X).$ When choosing 2 cards", "$\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ whose foci are the points $S_{1}, S_{2}$, and the eccentricity is e. Prove that the locus of incentre of $\\triangle PS_{1}S_{2}$ is an ellipse whose eccentricity is $\\sqrt{\\frac{2e}{1+e}}$. Question 20: Consider the family of circles $x^{2}+y^{2}=r^{2}$, $2. If in the first quadrant, the common tangent to a circle of this family and the ellipse $4x^{2}+25y^{2}=100$ meets the coordinate axes at A and B, then find the equation of the locus of the mid-point of AB. Question 21: Let P be a point on the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$, $0. Let the line parallel to y-axis passing through P meet the circle $x^{2}+y^{2}=a^{2}$ at the point Q such that P and Q are on the same side of the x-axis. For two positive real numbers, r and s, find the locus of the point R on PQ such that $PR:RQ=r:s$ as P varies over the ellipse. Question 22: If the eccentric angles of points P and Q on the ellipse be $\\theta$ and $\\frac{\\pi}{2} + \\theta$ and $\\alpha$ be the angle between the normals at P and Q, then prove that the eccentricity e is given by $2\\sqrt{1-e^{2}}=e^{2}=e^{2}\\sin^{2}{2\\theta}\\tan{\\alpha}$. Question 23: A series of hyperbolas are such that the length of their transverse axis is 2a. Prove that the locus of a point P on each, such", "diameter, and divide the vertical diameter by $\\sqrt{2}$ to get the maximal ellipse. Its area is $\\pi$, $1/2$ the area between the curves. Note that the common tangent to the ellipse and the curve passes through the vertex of the curve. Consider a point on the curve $\\frac1{x^2+c}$, $(p, \\frac1{p^2+c}$). A ellipse is drawn which is contained in the curves and it is tangent at this point. This means that $$a\\cos\\theta=p\\\\ b\\sin\\theta=\\frac1{p^2+c} \\\\ \\implies A=\\pi ab=\\frac{2p\\pi}{(p^2+c)\\sin2\\theta}$$ For maximum and minimum area we should take the derivative wrt to $p$ and $\\theta$, and set them to $0$. $$\\frac{\\partial A}{\\partial\\theta}=\\frac{4p\\pi \\cos2\\theta}{(p^2+c)\\sin^22\\theta}=0 \\\\ \\frac{\\partial A}{\\partial p}=\\frac{2(c-p^2)\\pi}{(p^2+c)^2\\sin2\\theta}=0$$ This gives $\\theta=\\pi/4$ from the first equation and the $p= \\sqrt c$. So the area is $$A=\\pi/\\sqrt c$$ To show that this ellipse is completely contained in the curves we must show that tangent of this ellipse at $\\theta=\\pi/4$ is same as tangent at for the curve at $p=\\sqrt c$ which is true in this case. So this is the completely contained ellipse. Incomplete: Although what this is missing is, how to tell if this is maxima or minima. Plus it feels like the cases of $x=0, \\infty$ aren't well accommodated here. Expanding in Taylor series around $x=0$ $$y = b\\sqrt{1-\\left(\\frac{x}{a}\\right)^2} = b-\\frac{b x^2}{2 a^2}-\\frac{b x^4}{8 a^4}+O\\left(x^5\\right)$$ and $$y = \\frac{1}{(x^2+c)} = \\frac{1}{c}-\\frac{x^2}{c^2}+\\frac{x^4}{c^3}+O\\left(x^5\\right)$$ Now solving" ]

[ "# Difference between revisions of \"2011 AIME I Problems/Problem 13\" ## Problem A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$. ## Solution 1 Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. Then the (directed) distance from any point (x,y,z) to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, $(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100$. Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\\sqrt {98/3}$. The distance from the origin to the plane is simply $d$, which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$, so $33+294+3=330$ ## Solution 2 Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the", "and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information ### Key Concepts Plane Distance Algebra AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d $\\frac{a10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=10-d and $\\frac{b10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=11-d and $\\frac{c10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=12-d squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\\frac{98}{3}^\\frac{1}{2}$ then d=11-$\\frac{98}{3}^\\frac{1}{2}$=$\\frac{33-294^\\frac{1}{2}}{3}$ then 33+294+3=330. Categories ## Rolling ball Problem | Semicircle |AMC 8- 2013 -|Problem 25 Try this beautiful problem from Geometry based on a rolling ball on a semicircular track. ## A Ball rolling Problem from AMC-8, 2013 A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose", "10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. (2011 II #9) https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\\overline{AH}$, $\\overline{BI}$, $\\overline{CJ}$, $\\overline{DK}$, $\\overline{EL}$, and $\\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.", "# Difference between revisions of \"2011 AIME I Problems/Problem 13\" ## Problem A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$. ## Solution Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. Then the (directed) distance from any point (x,y,z) to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, $(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100$. Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\\sqrt {98/3}$. The distance from the origin to the plane is simply d, which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$, so $33+294+3=330$ ## Solution 2 Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other", "Distance. ## Planes and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information ### Key Concepts Plane Distance Algebra AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d $\\frac{a10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=10-d and $\\frac{b10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=11-d and $\\frac{c10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=12-d squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\\frac{98}{3}^\\frac{1}{2}$ then d=11-$\\frac{98}{3}^\\frac{1}{2}$=$\\frac{33-294^\\frac{1}{2}}{3}$ then 33+294+3=330.", "ranges through the positive integers. • is 107 • is 401 • is 840 • cannot be determined from the given information GCD Sequence Integers ## Check the Answer AIME I, 1985, Question 13 Elementary Number Theory by David Burton ## Try with Hints $a_n=100+n^{2}$ $a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1$ and $a_{n+1}-a_{n}=2n +1$ $d_{n}|(2n+1)$ and $d_{n}|(100 +n^{2})$ then $d_{n}|[(100+n^{2})-100(2n+1)]$ then $d_{n}|(n^{2}-200n)$ here $n^{2} -200n=0$ then n=200 then $d_{n}$=2n+1=2(200)+1=401. Categories ## Planes and distance | AIME I, 2011 | Question 13 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance. ## Planes and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information Plane Distance Algebra ## Check the Answer AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here", "https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. (2011 II #9) https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\\overline{AH}$, $\\overline{BI}$, $\\overline{CJ}$, $\\overline{DK}$, $\\overline{EL}$, and $\\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.", "with an area of $30$ square inches. Find $m+n$. Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\\overline{AB}$ and $\\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ In the accompanying figure, the outer square $S$ has side length $40$. A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S$. From each midpoint of a side of $S$, segments are drawn to the two closest vertices of $S'$. The result is a four-pointed starlike figure inscribed in $S$. The star figure is cut out and then folded to form a pyramid with base $S'$. Find the volume of this pyramid. A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$", "Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. ## Problem 13 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r - \\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r + s + t$. ## Problem 14 Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$, $M_3$, $M_5$, and $M_7$ be the midpoints of sides $\\overline{A_1 A_2}$, $\\overline{A_3 A_4}$, $\\overline{A_5 A_6}$, and $\\overline{A_7 A_8}$, respectively. For $i = 1$, 3, 5, 7, ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \\perp R_3$, $R_3 \\perp R_5$, $R_5 \\perp R_7$, and $R_7 \\perp R_1$. Pairs of rays $R_1$ and $R_3$, $R_3$ and $R_5$, $R_5$ and $R_7$, and $R_7$ and $R_1$ meet at $B_1$, $B_3$, $B_5$, $B_7$ respectively. If $B_1 B_3 = A_1 A_2$, then $\\cos", "a third of the way between vertex $A$ and $H$, the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\\sqrt{3}$. So the distance from the $\\text{centroid}$ to $A$ is $10/\\sqrt{3}$. So, the $\\Delta z$ from $A$ to the centroid is $\\frac{10}{\\sqrt{3}}\\sin\\theta=\\frac{10}{\\sqrt{3}}\\left(\\frac{7\\sqrt{2}}{10}\\right)=\\frac{7\\sqrt{6}}{3}$. Thus the distance from $A$ to the plane is $11-\\frac{7\\sqrt{6}}{3}=\\frac{33-7\\sqrt{6}}{3}=\\frac{33-\\sqrt{294}}{3}$, and $33+294+3=\\boxed{330}$.", "Categories # Planes and distance | AIME I, 2011 | Question 13 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and distance. Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance. ## Planes and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information ### Key Concepts Plane Distance Algebra AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d $\\frac{a10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=10-d and $\\frac{b10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=11-d and $\\frac{c10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=12-d squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\\frac{98}{3}^\\frac{1}{2}$ then d=11-$\\frac{98}{3}^\\frac{1}{2}$=$\\frac{33-294^\\frac{1}{2}}{3}$ then 33+294+3=330.", "2} r$. We have two sides of a triangle which has one vertex at the origin, one as a vertex of the cube on the xy-plane, and one as a top vertex of the cube. Using vertices of the cube which are on the same edge of the cube, we have that the last side of such a triangle has length $r$. Thus we just use the pythagorean theorem again: $$r^2+\\frac {r^2} 2 = a^2$$ Solving for $r$ in terms of $a$ gives the desired result.", "vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives $$\\frac{a(10)}{\\sqrt{a^2 + b^2 + c^2}} = 10-d,$$ $$\\frac{b(10)}{\\sqrt{a^2 + b^2 + c^2}} = 11-d,$$ and $$\\frac{c(10)}{\\sqrt{a^2 + b^2 + c^2}} = 12-d.$$ An easy algebraic manipulation gives the equation in the first solution.", "otherwise $$|d| < r$$, which means the plane intersects the sphere. ## Clipping triangles If the sphere-plane test isn’t enough to determine whether an object is fully in front or fully behind the clipping plane, it’s necessary to clip each triangle against it. We can classify each vertex of the triangle against the clipping plane by taking the sign of its signed distance to the plane. If the distance is zero or positive, the vertex is in front of the clipping plane, and otherwise it’s behind: There are four possible cases: • Three vertexes in front. In this case the whole triangle is in front of the clipping plane, so it’s accepted with no further clipping against this plane needed. • Three vertexes behind. In this case the whole triangle is behind the clipping plane, so it’s discarded with no further clipping necessary at all. • One vertex in front. Let $$A$$ be the vertex of the triangle $$ABC$$ that is in front of the plane. In this case, $$ABC$$ is discarded, and a new triangle $$AB'C'$$ is added, where $$B'$$ and $$C'$$ are the intersections of $$AB$$ and $$AC$$ with the clipping plane. • Two vertexes in front. Let $$A$$ and $$B$$ the vertexes of the triangle $$ABC$$ that are in front of the plane. In this", "middle rectangle is actually a square, so the total area is $(15\\sqrt{2})^2 + 4\\left(\\frac 12\\right)\\left(\\frac 52\\sqrt{2}\\right)\\left(\\frac{15}{\\sqrt{2}}\\right) = 525$. ## Solution 2 First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane $PQR$, so they cannot be skew. Therefore, they are parallel. Let the cube's vertices be $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$, with $A$, $B$, and $C$ on the bottom face as before, $H$ being the other bottom vertex, $D$ directly above $C$, $E$ above $B$, $F$ above $A$, and $G$ above $H$. Clearly, the next vertex of the intersection (starting with $P$, $Q$, $R$) will be somewhere on $DG$. Let it be $X$, and have a distance of $x$ from D, and a distance of $20 - x$ from $G$. Then, the next vertex will be somewhere on $FG$. It must be parallel to $PQ$, so this implies that it has a distance of $20 - x$ from $G$, and thus a distance of $x$ from $F$. Now, the next vertex (call it $Y$) will be somewhere on $AF$. The segment must be parallel to $QR$, so $FY$ must have length $2x$,", "# The shortest distance between a vertex in a simplex and a non-adjacent face Provided a simplex defined by four vertices $(v_1,v_2,v_3,v_4)$, and known edge lengths, for a given vertex $v_i$, how can I calculate the shortest distance between $v_i$ and the non-adjacent face of the simplex? To provide an example, if $v_i=v_1$, I would want to calculate to shortest distance to the triangle defined by the vertices $(v_2,v_3,v_4)$. If you drop a perpendicular from $v_1$ to the plane spanned by $v_2, v_3, v_4$, it will land either within the triangle $T=v_2v_3v_4$ or outside it. If it lands within, all three of the other faces of the simplex have acute angles adjacent to $T$, while if it lands outside, two of them will have an obtuse angle adjacent to $T$. You can therefore use the law of cosines to distinguish between these two possibilities while knowing only the side lengths of your simplex. The shortest distance from $v_1$ to the opposite face is the length of the perpendicular: i.e., an altitude of the simplex. So it is given by $3V/A$, where $A$ is the area of the opposite face and $V$ the volume of the simplex. If you know all the side lengths of the simplex, you can compute $A$ via Heron's formula, and $V$ via its 3-dimensional", "(v_i,H')$ is just a length of altitude from $v_2$, $\\sum_{i=0}^2 {\\rm dist}\\, (v_i,H)$ is almost the length of altitude from $v_1$, which may be less than that from $v_2$. As for your new question, the answer is still negative for triangles. Assume that a median $v_0 p$ is perpendicular to $H'=v_0v_1$. For any line $H$ passing through $v_0$ and not cutting the triangle we have $\\sum_{i=0}^2 {\\rm dist}\\,(v_i,H)=2\\,{\\rm dist}\\, (p,H)\\leqslant 2|v_0p|$ and maximum is achieved for $H=H'$. • did you mean \"both sums\" instead of \"both sides\"? – Manfred Weis Jan 16 '16 at 12:43 • Both sums (or both sides of inequality). – Fedor Petrov Jan 16 '16 at 13:01 It just occured to me that the question can be answered in the affirmative way for the following reasons: • the order relation doesn't change if the distance sum is divided by some positive constant value $c$. • if we divide the distance sums by $n+1$, then that value equals the distance of the vertices' center of mass $R$ to the same plane. • all local maxima of $R$'s distances to points on the convex hull of the vertices are attained in a vertex which also is a corner of the convex hull and consequently also the global maximum of the distances is attained in such a", "A square pyramid with base $$ABCD$$ and vertex $$E$$ has eight edges of length 4. A plane passes through the midpoints of $$AE$$, $$BC$$, and $$CD$$. The plane's intersection with the pyramid has an area that can be expressed as $$\\sqrt{p}$$. Find $$p$$. (第二十五届AIME1 2007 第13题)", "Categories ## Problem on Cube | AMC 10A, 2008 | Problem 21 Try this beautiful problem from Geometry: Problem on Cube. ## Problem on Cube – AMC-10A, 2008- Problem 21 A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$ , i • $\\frac{\\sqrt{6}}{2}$ • $\\frac{5}{4}$ • $\\sqrt{2}$ • $\\frac{5}{8}$ • $\\frac{3}{4}$ ### Key Concepts Geometry Square Pythagoras Answer: $\\frac{\\sqrt{6}}{2}$ AMC-10A (2008) Problem 21 Pre College Mathematics ## Try with Hints The above diagram is a cube and given that side length $1$ and $B$ and $D$ are the mid points .we have to find out area of the $ABCD$.Now since $A B=A D=C B=C D=\\sqrt{\\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus? Can you now finish the problem ………. The area of the rhombus can be computed by the formula $A = \\frac 12 d_1d_2$, where $d_1,\\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\\sqrt{1^{2}+1^{2}}=\\sqrt{2} \\cdot A C$ is a space diagonal,", "Let $$EFGH$$, $$EFDC$$, and $$EHBC$$ be three adjacent square faces of a cube, for which $$EC = 8$$, and let $$A$$ be the eighth vertex of the cube. Let $$I$$, $$J$$, and $$K$$, be the points on $$\\overline{EF}$$, $$\\overline{EH}$$, and $$\\overline{EC}$$, respectively, so that $$EI = EJ = EK = 2$$. A solid $$S$$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $$\\overline{AE}$$, and containing the edges, $$\\overline{IJ}$$, $$\\overline{JK}$$, and $$\\overline{KI}$$. The surface area of $$S$$, including the walls of the tunnel, is $$m + n\\sqrt {p}$$, where $$m$$, $$n$$, and $$p$$ are positive integers and $$p$$ is not divisible by the square of any prime. Find $$m + n + p$$. (第十九届AIME2 2001 第15题)" ]

[ "se the problem with the projection : \\pstTranslation[PosAngle=30]{A}{S}{O}[P] \\pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction – Alain Matthes Apr 15 '13 at 6:50 • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD. – Alain Matthes Apr 15 '13 at 17:35 Asymptote version pyramid.asy: import three; // 3D module import math; currentprojection=orthographic(camera=(55,144,80), up=(0,0,1),target=(0,0,0),zoom=1,center=true); size(300); size3(300,300,300); triple intersectionpoint(triple a,triple b,triple c,triple d){ real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/ ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x); return a*(1.0-u)+b*u; } triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy: // as South, East and North or (0,-1), (1,0) and (0,1) // and I is a sqrt(-1)=(0,1) real a=40; // side of the square; real h=50; // height, h=AS real d=sqrt(2)/2*a; // half of the diagonal O=(0,0,0); C=(0,d,0); B=(d,0,0); A=(0,-d,0); D=(-d,0,0); SS=A+(0,0,h); M=0.5(SS+A); II=0.5(SS+D); NN=intersectionpoint(SS,(SS+A-B),C,II); real phi=atan(h/a);", "# 1983 AIME Problems/Problem 11 ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label(\"A\",A,W); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); [/asy]$ ## Solutions ### Solution 1 First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); label(\"12\\sqrt{2}\",(E+F)/2,N); label(\"6\\sqrt{2}\",(A+B)/2,S); label(\"6\",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\\cdot \\left(", "# 2001 AIME II Problems/Problem 15 ## Problem Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$, be the points on $\\overline{EF}$, $\\overline{EH}$, and $\\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\\overline{AE}$, and containing the edges, $\\overline{IJ}$, $\\overline{JK}$, and $\\overline{KI}$. The surface area of $S$, including the walls of the tunnel, is $m + n\\sqrt {p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy] import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]$ $[asy] import three; currentprojection = orthographic(camera=(1/2,1/3,-1/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]$ Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little", "# Difference between revisions of \"1983 AIME Problems/Problem 11\" ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label(\"A\",A); label(\"B\",B); label(\"C\",C); label(\"D\",D); label(\"E\",E,N); label(\"F\",F,N); [/asy]$ ## Solution 1 First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to find that the height is equal to $6$. $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A); label(\"B\",B); label(\"C\",C); label(\"D\",D); label(\"E\",E,N); label(\"F\",F,N); label(\"12\\sqrt{2}\",(E+F)/2,N); label(\"6\\sqrt{2}\",(A+B)/2); label(\"6\",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find the area, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined area is", "# Difference between revisions of \"2007 AIME I Problems/Problem 13\" ## Problem A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$. A plane passes through the midpoints of $AE$, $BC$, and $CD$. The plane's intersection with the pyramid has an area that can be expressed as $\\sqrt{p}$. Find $p$. ## Contents import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2; D(A--B--C--D--A--E--B--E--C--E--D); MP(\"A\",A); MP(\"B\",B,(1,0,0)); MP(\"C\",C); MP(\"D\",D); MP(\"E\",E,N); D(MP(\"P\",P)); D(MP(\"Q\",Q,(1,0,0))); D(MP(\"R\",R)); (Error compiling LaTeX. 79ebb2a541577ae09ecbe3d167a6828243c28f28.asy: 7.2: no matching function 'D(void(flatguide3))') ## Solution ### Solution 1 Note first that the intersection is a pentagon. Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,0,2\\sqrt{2})$. Using the coordinates of the three points of intersection ($(-1,1,\\sqrt{2}),\\ (2,0,0),\\ (0,-2,0)$), it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$, and using the points we find that $2a = d \\Longrightarrow d = \\frac{a}{2}$, $-2b = d \\Longrightarrow d = \\frac{-b}{2}$, and $-a + b + \\sqrt{2}c = d \\Longrightarrow -\\frac{d}{2} - \\frac{d}{2} + \\sqrt{2}c = d \\Longrightarrow c = d\\sqrt{2}$. It is then $x - y +", "Vector3(x * 150, stage.surfaceHeight(x * 150, z * 150), z * 150); Vector3 B = new Vector3((x + 1) * 150, stage.surfaceHeight((x + 1) * 150, z * 150), z * 150); Vector3 C = new Vector3(x * 150, stage.surfaceHeight(x * 150, (z + 1) * 150), (z + 1) * 150); Vector3 D = new Vector3((x + 1) * 150, stage.surfaceHeight((x + 1) * 150, (z + 1) * 150), (z + 1) * 150); I use a simple algorithm to determine my destination plane: public bool isInUpperPlane(Vector3 A, Vector3 B, Vector3 C, Vector3 D, Vector3 point) { /* ^ y | | | | A B ----------------------------> x / / / x / C /_________ / D / / / v z */ // consider A as the origin if (Math.Abs(point.X - A.X) + Math.Abs(point.Z - A.Z) >= 150) return true; return false; } Next I construct a plane the 3 vertices (either ABC or BCD): Plane destPlane; if (isInUpperPlane(A, B, C, D, stopLocation)) { destPlane = new Plane(A, B, C); } else { destPlane = new Plane(B, C, D); } Then I project this my start location to this plane normal, and translate my avatar to this location: Vector3 normal = destPlane.Normal; Vector3 r = projectVectorToPlane(startLocation, normal); Orientation *= Matrix.CreateTranslation(startLocation); But doing this make my", "at 10:40 @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points. – Gonzalo Medina Apr 14 '13 at 14:33 The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5. – Alain Matthes Apr 15 '13 at 17:15 Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you. – Gonzalo Medina Apr 15 '13 at 17:25 Asymptote version pyramid.asy: import three; // 3D module import math; currentprojection=orthographic(camera=(55,144,80), up=(0,0,1),target=(0,0,0),zoom=1,center=true); size(300); size3(300,300,300); triple intersectionpoint(triple a,triple b,triple c,triple d){ real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/ ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x); return a*(1.0-u)+b*u; } triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy: // as South, East and North or (0,-1), (1,0) and (0,1) // and I is a sqrt(-1)=(0,1) real a=40; // side of the square; real h=50; // height, h=AS real d=sqrt(2)/2*a; // half of the diagonal O=(0,0,0); C=(0,d,0); B=(d,0,0); A=(0,-d,0); D=(-d,0,0); SS=A+(0,0,h); M=0.5(SS+A); II=0.5(SS+D); NN=intersectionpoint(SS,(SS+A-B),C,II); real phi=atan(h/a); real u=a*cos(phi)/sqrt(a^2+h^2); EE=D*(1-u)+SS*u; K=EE+B-A; real psi=atan(h/d); real u=d*cos(psi)/sqrt(d^2+h^2); H=O*(1-u)+SS*u; pair Q=intersectionpoint(project(NN--B),project(SS--D)); pen", "# Difference between revisions of \"1983 AIME Problems/Problem 11\" ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label(\"A\",A,W); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); [/asy]$ ## Solutions ### Solution 1 First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); label(\"12\\sqrt{2}\",(E+F)/2,N); label(\"6\\sqrt{2}\",(A+B)/2,S); label(\"6\",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined", "z * 150); Vector3 C = new Vector3(x * 150, stage.surfaceHeight(x * 150, (z + 1) * 150), (z + 1) * 150); Vector3 D = new Vector3((x + 1) * 150, stage.surfaceHeight((x + 1) * 150, (z + 1) * 150), (z + 1) * 150); I use a simple algorithm to determine my destination plane: public bool isInUpperPlane(Vector3 A, Vector3 B, Vector3 C, Vector3 D, Vector3 point) { /* ^ y | | | | A B ----------------------------> x / / / x / C /_________ / D / / / v z */ // consider A as the origin if (Math.Abs(point.X - A.X) + Math.Abs(point.Z - A.Z) >= 150) return true; return false; } Next I construct a plane the 3 vertices (either ABC or BCD): Plane destPlane; if (isInUpperPlane(A, B, C, D, stopLocation)) { destPlane = new Plane(A, B, C); } else { destPlane = new Plane(B, C, D); } Then I project this my start location to this plane normal, and translate my avatar to this location: Vector3 normal = destPlane.Normal; Vector3 r = projectVectorToPlane(startLocation, normal); Orientation *= Matrix.CreateTranslation(startLocation); But doing this make my avatar stays still. It's very very wrong, but I really don't know how it actually works. The 3D world is just too confusing at this moment. Could anyone help", "substituting * = —5.0 into the x and y component equations yields the intersection with the z = 0 plane, i.e., x(-5.0) = 1 - (0.4)(-5.0) = 3.0 2/(-5.0) = 0.667 - (0.267)(-5.0) = 2.0 which is the same as the intersection of the line AB with the z = 0 plane. Projection of line A'B' into the line A*B* in the z = 0 plane is given by A' B' 1 0.667 1.333 1 0.6 0.4 1.6 1 1 0 0 0 0 1 0 0 0 0 0 0 01 0 0 1. 1 0.667 0 1 0.6 0.4 0 1 A* B* An example using a simple cube is given below. PERSPECTIVE TRANSFORMATIONS 163 Example 3-18 Single-Point Perspective Transformation of a Cube Perform a perspective projection onto the z = 0 plane of the unit cube shown in Fig. 3-28a from a center of projection at zc = 10 on the z-axis. The single-point perspective factor r is r = -l/zc = -1/10 = -0.1 y H A A i VL_ ~~A f A G F B (a) D' y> H' A z_ Vl ~A / A G' F' D* H* Si* A^ G*| A*E* F* B* (b) To C.P. at z = 10 -* ^^^___^ Projector _ * ^ z* -* C, A, B", "can see here for the sphere. Nov 25, 2021 at 1:32 NOTE that: I don't know either what I am drawing. :) Drawing 3D image in Asymptote is a difficult problem for both my knowledge and my computer (it is weak). You can add mesh for surface. You can also draw two surface separately (can be done on http://asymptote.ualberta.ca/). You must have a parametrization of your surface to use the command vectorfield. I am happy if one of top asymptote answerers has a better solution, certainly. This is my try, this code is compiled with asy -f png -render=8 <name file>.asy. import graph3; import palette; currentprojection=orthographic(1,1,0.3); currentlight.background = gray(.7); //currentlight=nolight; typedef triple newtriple(pair); // https://en.wikipedia.org/wiki/Hyperboloid newtriple f(real a, real b, real c) { return new triple(pair k){ real u=k.x,v=k.y; real x,y,z; x=a*cosh(v)*cos(u); y=b*cosh(v)*sin(u); z=c*sinh(v); return (x,y,z); }; } triple F(pair z){ return f(1,1,1)(z); } // Gauss map triple g(pair z) { real u=z.x, v=z.y; triple U=(-cosh(v)*sin(u),cosh(v)*cos(u),0); triple V=(sinh(v)*cos(u),sinh(v)*sin(u),cosh(v)); return cross(U,V)/abs(cross(U,V)); } // https://trecs.se/hyperboloidOfOneSheet.php path3 vector(pair z) { real u=z.x, v=z.y; real a=1, b=1, c=1; real x,y,z; x=-b*c*cosh(v)^2*cos(u); y=-a*c*cosh(v)^2*sin(u); z=a*b*sinh(v)*cosh(v); return O--(-(x,y,z)/sqrt(x^2+y^2+z^2)); } picture pic, pic1, pic2, pic3; size(pic1,300); size(pic2,300); surface sf=surface(F,(0,-1.5),(2pi,1.5),40,Spline); sf.colors(palette(sf.map(zpart),Rainbow())); draw(pic1,sf,render(merge=true)); label(pic1,\"$(\\mathcal{H})$\",(-1,2.2,2.7),dir(45)); xaxis3(pic1,Label(\"$x$\",EndPoint),0,3.5,Arrow3); yaxis3(pic1,Label(\"$y$\",EndPoint),0,3.5,Arrow3); zaxis3(pic1,Label(\"$z$\",EndPoint),0,3,Arrow3); pair z=(pi/7,-0.5); dot(pic1,\"$P$\",F(z),dir(90),black+5bp); draw(pic1,Label(\"$N(P)$\",EndPoint),shift(F(z))*vector(z),blue,Arrow3); surface sg=surface(g,(0,-1.5),(2pi,1.5),20,Spline); sg.colors(palette(sg.map(zpart),Rainbow())); draw(pic2,sg,render(merge=true)); label(pic2,\"$N\\left(\\mathcal{H} \\right)$\",(-.5,0.5,0.8),dir(45)); dot(pic2,Label(\"$\\frac{1}{\\sqrt{2}}$\",blue),(0,0,1/sqrt(2)),dir(45),green+5bp); xaxis3(pic2,Label(\"$x$\",EndPoint),0,1.5,Arrow3); yaxis3(pic2,Label(\"$y$\",EndPoint),0,1.5,Arrow3); zaxis3(pic2,Label(\"$z$\",EndPoint),0,1,Arrow3); draw(pic3,Label(\"$N$\",MidPoint,LeftSide),(-20,0)--(20,0),Arrow); or settings.render=8; import graph3; import palette;", "while preserving $$P_1$$, and the fourth slice by the points $$P_3$$, $$P_0$$, and $$P_1$$, while preserving $$P_2$$. Each of these planes will be specified in the slicing operation by a point and the corresponding normal vector to the plane. To calculate this vector, we need to use the cross product of two vectors (implemented by the predefined cross function) and we have to verify if the normal is pointing towards the opposite direction of the point to preserve, which we can do by verifying the sign of the dot product (implemented by the predefined dot function) between the normal vector and the vector that ends at the point to preserve. normal_points(p0, p1, p2, p3) = let v0 = p1-p0, v1 = p2-p0, n = cross(v0, v1) dot(n, p3-p0) < 0 ? n : n*-1 end tetrahedron(p0, p1, p2, p3) = let pmin = xyz(min(p0.x, p1.x, p2.x, p3.x), min(p0.y, p1.y, p2.y, p3.y), min(p0.z, p1.z, p2.z, p3.z)), pmax = xyz(max(p0.x, p1.x, p2.x, p3.x), max(p0.y, p1.y, p2.y, p3.y), max(p0.z, p1.z, p2.z, p3.z)), solid = box(pmin, pmax) solid = slice(solid, p0, normal_points(p0, p1, p2, p3)) solid = slice(solid, p1, normal_points(p1, p2, p3, p0)) solid = slice(solid, p2, normal_points(p2, p3, p0, p1)) solid = slice(solid, p3, normal_points(p3, p0, p1, p2)) solid end This figure presents the several phases of the slicing process of", "regular 3/4/6-gons made of 3D grid points, however large the grid is, right? – Bubbler Oct 23 '20 at 0:34 • @Bubbler—that's right! – Peter Kagey Oct 23 '20 at 0:35 • I conjecture that the squares only appear on the planes parallel to the cube's faces, and the triangles and hexagons only appear on the (1,1,1)-lattice equivalents. Is this true? – Bubbler Oct 23 '20 at 0:57 • @Bubbler, I've edited the example of the square to provide a square that is not parallel to a face. – Peter Kagey Oct 23 '20 at 1:06 • @Bubbler, for a triangle, see $(0,4,0),(1,0,1),(4,3,1)$. – Peter Kagey Oct 23 '20 at 1:11 # Python 2, 330 313 bytes import numpy as N,itertools as I L=list D=N.dot n=input() def G((a,b,c)): A=a while 1: u=b-a;v=c-b;d=D(v,v) if D(u,u)-d:return 0 p=2*D(u,v)*v a,b,c=b,c,c+a-b+p/d if(p%d|(c<0)|(c>n)).any()or L(a)<L(A):return 0 if L(a)==L(A):return 1 print sum(map(G,I.permutations(N.indices((n+1,)*3).reshape(3,-1).T,3)))/2 Try it online! TIO spent 36.893 s calculating a(5) = 6042, so this solution is just fast enough. Given 3 consecutive points of a regular polygon a,b,c, the next point is d = c + a - b + 2*proj(b-a, c-b) (where proj is vector projection). My solution iterates through all triples of points and determines if they form a polygon using this formula. Verified up to a(7) = 28400 on my machine.", "so that in each cuboid the z axis represents the size, the x axis the color, and the y axis the shape. Suppose the reference block is in position $(0,0,0)$ in the plastic cuboid. The wooden blocks already differ from the reference block in one way, so if $(0,0,0)$ in the wooden cuboid represents 'wood medium red circle' then any wooden block with two zero coordinates satisfies the requirements. These form the edges of the cuboid which adjoin $(0,0,0)$, so there are $2+3+3=8$. The required plastic blocks must have two non-zero coordinates, so we count the blocks on the three faces around $(0,0,0)$ but not on the adjoining edges. There are $2\\cdot3+2\\cdot3+3\\cdot3=21$. In total there are $29$ blocks meeting the requirements. $[asy] import three; currentprojection=perspective((7.5,5,5),up=Z); currentlight=nolight; viewportmargin=(1mm,1mm); real c=.5; size(8cm,0); draw((0,0,0)--(0,0,4),Arrow3); draw((0,0,0)--(0,5,0),Arrow3); draw((0,0,0)--(5,0,0),Arrow3); for(int z=0 ; z<3 ; ++z) { for(int x=0 ; x<4 ; ++x) { for(int y=0 ; y<4 ; ++y) { if (((x==0)&&(y*z!=0))||((y==0)&&(x*z!=0))||((z==0)&&(x*y!=0))) {c=.5;} else {c=.75;} if (x*y*z>0) {c=1;} draw(shift(x,y,z)*unitcube,gray(c)+opacity(.6),.5bp+black); }}} [/asy]$ Solution 3 (Fastest) The amount of ways we can differ in the material, sizes, colors, and shapes category is 1, 2, 3, and 3 respectively (we take away one because we cannot choose the characteristic already chosen). We can choose to differ two of these characteristics, so our answer is $1\\cdot2 + 1\\cdot3", "# Difference between revisions of \"2001 AIME I Problems/Problem 12\" ## Problem A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ ## Solution $[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label(\"I\",I,W); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,W*-1); label(\"D\",D,W*-1); [/asy]$ The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$ The signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC. A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\\times(B-C)=\\langle 8, 12, 24 \\rangle$ Thus $\\frac{(I-C) \\cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$ \\begin{align*}\\frac{(I-C) \\cdot P}{|P|}&=-r\\\\ \\frac{(\\langle r, r, r", "# Difference between revisions of \"2012 AMC 12B Problems/Problem 19\" ## Problem 19 A unit cube has vertices $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ and $P_4'$. Vertices $P_2$, $P_3$, and $P_4$ are adjacent to $P_1$, and for $1\\le i\\le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_1'P_2'$, $P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length? $[asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype(\"4 4\")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype(\"4 4\")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label(\"P_1\", P[1], dir(P[1])); label(\"P_2\", P[2], dir(P[2])); label(\"P_3\", P[3], dir(-45)); label(\"P_4\", P[4], dir(P[4])); label(\"P'_1\", Pp[1], dir(Pp[1])); label(\"P'_2\", Pp[2], dir(Pp[2])); label(\"P'_3\", Pp[3], dir(-100)); label(\"P'_4\", Pp[4], dir(Pp[4])); [/asy]$ $\\textbf{(A)}\\ \\frac{3\\sqrt{2}}{4}\\qquad\\textbf{(B)}\\", "this to my repository as I don't know many pstricks examples and I know many others would like to have some examples, too) – Martin Thoma Jul 2 '13 at 13:51 This solution uses http://mathworld.wolfram.com/Sphere-SphereIntersection.html to get the coordinates of the intersection point P. spheres.tex: \\documentclass{article} \\usepackage{lmodern} \\usepackage[inline]{asymptote} \\begin{document} \\begin{figure} \\begin{asy} settings.prc=false; settings.render=0; import graph3; currentprojection=orthographic(camera=(150,900,412),up=(-0.4,-0.7,1.4),target=(100,111,1),zoom=0.5); currentlight=nolight; size(300);size3(300); real r1=162; real r2=100; triple v1=(0,0,0); triple v2=(250,0,0); // from http://mathworld.wolfram.com/Sphere-SphereIntersection.html : real d=arclength(v1--v2); // distance between v1,v2 real a=1/2/d*sqrt((2*d*r1)^2-(d^2-r2^2+r1^2)^2); triple P=(sqrt(r1^2-a^2),0,-a); triple fs1(pair t){ return v1+r1*(cos(t.x)*sin(t.y),sin(t.x)*sin(t.y),cos(t.y)); } triple fs2(pair t){ return v2+r2*(cos(t.x)*sin(t.y),sin(t.x)*sin(t.y),cos(t.y)); } surface s1=surface(fs1,(0,0),(2pi,pi),16,Spline); surface s2=surface(fs2,(0,0),(2pi,pi),16,Spline); draw(s1 ,darkgreen+opacity(0.2) ,render(compression=Low,merge=true) ); draw(s2 ,darkblue+opacity(0.2) ,render(compression=Low,merge=true) ); dot(v1); label(\"$V_1$\",v1,-X+Z); dot(v2); label(\"$V_2$\",v2,-X+Z); dot(P); label(\"$P$\",P,-X-Z); \\end{asy} \\end{figure} \\end{document} To process it with latexmk, create file latexmkrc: sub asy {return system(\"asy '\\$_[0]'\");} and run latexmk -pdf spheres.tex.", "1) cutCube p = clip $mapMaybe (fromHGA . meet p . subspace) cubeEs Let’s try this for one particular plane containing the axis through (0, 1/2, 0) and (1, 1/2, 1): hexample = cutCube$ subspace [[0, 0.5, 0], [1, 0.5, 1], [0, -0.5, 1]] prop_hexample = let h = 0.5 in hexample == [[0,h,0], [0,0,h], [h,0,1], [h,1,0], [1,h,1], [1,1,h]] These are the vertices of a regular hexagon. Next, let’s find the cutting plane for any given angle. We’ll use matrices so we define functions familiar to anyone with experience in computer graphics: embedding a 3x3 matrix into a 4x4 matrix for the homoegeneous model, matrix multiplication by another matrix or by a vector, and building a translation matrix: mat3To4 :: [[Double]] -> [[Double]] mat3To4 m = map (++ [0]) m ++ [[0, 0, 0, 1]] matMul :: [[Double]] -> [[Double]] -> [[Double]] matMul m n = map (zipWith f (transpose n) . repeat) m where f bs as = sum $zipWith (*) bs as matVec :: [[Double]] -> [Double] -> [Double] matVec m v = zipWith ((sum .) . zipWith (*)) m$ repeat v traMat4 :: [Double] -> [[Double]] traMat4 [x, y, z] = [ [1, 0, 0, x] , [0, 1, 0, y] , [0, 0, 1, z] , [0, 0, 0, 1] ] We still cling", "through like $$w=1$$ will not find the whole surface. So multiple slices are taken. In the case of $$\\mathbb{P}^2$$ we can find the intersection of a surface $$f(x,y,z)=0$$ with the three faces of the surrounding cube, $$\\{x\\in[-1,1], y\\in[-1,1], z=1\\}$$, $$\\{x\\in[-1,1], y=1, z\\in[-1,1]\\}$$ and $$\\{x=1, y\\in[-1,1], z\\in[-1,1]\\}$$. This involves calculating the three curves $$f(x,y,1)=0$$, $$f(x,1,y)=0$$ and $$f(1,y,z)=0$$. The three intersections can then be projected onto a plane using either perspective of stereographic projection. As antipodal points are identified there is no need to calculate intersections with the other three faces. For surfaces $$f(x,y,z,w)=0$$ in $$\\mathbb{P}^3$$ the intersection with four cubes are calculated $$\\{x\\in[-1,1], y\\in[-1,1], z\\in[-1,1], w=1\\}$$, $$\\{x\\in[-1,1], y\\in[-1,1], z=1, w\\in[-1,1]\\}$$, $$\\{x\\in[-1,1], y=1, z\\in[-1,1], w\\in[-1,1]\\}$$ and $$\\{x=1, y\\in[-1,1], z\\in[-1,1], w\\in[-1,1]\\}$$. The full algorithm is: 1. Calculate the sets $$\\{f(x,y,z,1) : x\\in[-1,1], y\\in[-1,1], z\\in[-1,1] \\}$$, $$\\{f(x,y,1,w) : x\\in[-1,1], y\\in[-1,1], w\\in[-1,1] \\}$$, $$\\{f(x,1,z,w) : x\\in[-1,1], z\\in[-1,1], w\\in[-1,1] \\}$$ and $$\\{f(1,y,z,w) : y\\in[-1,1], z\\in[-1,1], w\\in[-1,1] \\}$$; 2. Rotate in $$\\mathbb{R}^4$$; 3. Cut each surface into sets $$\\{x,y,z,w:w>0\\}$$, $$\\{x,y,z,w:w<0\\}$$; 4. Project into $$\\mathbb{R}^3$$, using either perspective of stereographic projections; 5. For perspective projection clip the result. The four dimensional rotation is applied by using three rotation matrices $$R_{xw}=\\begin{pmatrix}\\cos \\theta& 0 & 0 &\\sin \\theta\\\\0 & 1 & 0 & 0\\\\0 & 0 & 1 & 0\\\\ -\\sin \\theta & 0 & 0 & \\cos", "but with the plane at z=-2. • Alternatively, you can scale everything by -1/2 to put the projection plane at z=1, apply the formula in the notes, then scale everything by -2 to restore the size. • Note that points on the projection plane don't move. 2. Consider the cube with opposite corners at (-4,-4,-2) and (4,4,-10). What does it transform to? Show the projection vertex by vertex. E.g., what does (4,4,-10) transform to? • {$\\begin{array}{c} (-4,-4,-2)\\\\(-4,-4,-10)\\\\(-4,4,-2)\\\\(-4,4,-10)\\\\(4,-4,-2)\\\\(4,-4,-10)\\\\(4,4,-2)\\\\(4,4,-10) \\end{array} \\rightarrow \\begin{array}{c} (-4,-4,-2)\\\\(-4/5,-4/5,-2)\\\\(-4,4,-2)\\\\(-4/5,4/5,-2)\\\\(4,-4,-2)\\\\(4/5,-4/5,-2)\\\\(4,4,-2)\\\\(4/5,4/5,-2) \\end{array}$} Total: 20" ]

[ "# Difference between revisions of \"2011 AIME I Problems/Problem 13\" ## Problem A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$. ## Solution 1 Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. Then the (directed) distance from any point (x,y,z) to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, $(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100$. Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\\sqrt {98/3}$. The distance from the origin to the plane is simply $d$, which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$, so $33+294+3=330$ ## Solution 2 Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the", "at the second picture, which is what I'm drawing but with triangles instead of pentagons. All the triangles are easy to draw as we know the lengths. Oct 4, 2021 at 10:04 Update: The calculations are from the 3-D Cartesian true range multilateration (in fact, those are formulae for calculating intersections of three spheres). I also add the 3D grid on XY-plane. unitsize(1cm); import three; import grid3; triple barycentric(triple A, triple B, triple C, real a, real b, real c){return (a*A+b*B+c*C)/(a+b+c);} currentprojection=orthographic((1,1,.8),center=true,zoom=.9); // Step 1: construct the base A,B,C on the plane z=0 real a=5, b=4, c=3; triple B=(0,0,0), C=(a,0,0); // Abc is the projection of A on the segment BC triple Abc=barycentric(B,C,O,1/(a^2+c^2-b^2),1/(a^2+b^2-c^2),0); real bt=abs(Abc-B); real hbc=sqrt(c*c-bt*bt); triple A=Abc+hbc*dir(90,90); draw(A--B--C--cycle,blue+1pt); label(\"$A$\",A,plain.S); label(\"$B$\",B,dir(150)); label(\"$C$\",C,plain.NW); //label(\"$A_{bc}$\",Abc,plain.N,red); //draw(A--Abc,red); // Step 2: get the top point D real at=6, bt=7, ct=8; // H is the projection of D on the base ABC // https://en.wikipedia.org/wiki/True-range_multilateration real Hx=(bt^2-ct^2+a^2)/(2a); real Hy=(bt^2-at^2+A.x^2+A.y^2-2A.x*Hx)/(2A.y); real Dz=sqrt(bt^2-Hx^2-Hy^2); triple H=(Hx,Hy,0); triple D=(Hx,Hy,Dz); draw(D--A^^D--B^^D--C,blue+1pt); draw(D--H,red+dashed); label(\"$D$\",D,plain.N); label(\"$H$\",H,plain.E,red); dot(H,red); // grid on XY-plane limits((-2,-2,0),(7,5,0)); grid3(XYXgrid,step=1,.5gray+.5white); draw(Label(\"$x$\",EndPoint),O--8X,Arrow3()); draw(Label(\"$y$\",EndPoint),O--6Y,Arrow3()); draw(Label(\"$z$\",EndPoint),O--6Z,Arrow3()); write(\"H = (\"+string(H.x)+\",\"+string(H.y)+\",0)\"); write(\"DH = \"+string(abs(H-D))); Old answer This is a well-known construction problem of Euclidean geometry. In drawing, we can use geometric constructions, but it is better to use computation approaches. Among some computation approaches, I found one using barycentric coordinate is more", "# Difference between revisions of \"2011 AIME I Problems/Problem 13\" ## Problem A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$. ## Solution Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. Then the (directed) distance from any point (x,y,z) to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, $(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100$. Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\\sqrt {98/3}$. The distance from the origin to the plane is simply d, which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$, so $33+294+3=330$ ## Solution 2 Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other", "# Difference between revisions of \"2001 AIME I Problems/Problem 12\" ## Problem A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ ## Solution $[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label(\"I\",I,W); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,W*-1); label(\"D\",D,W*-1); [/asy]$ The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$ The signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC. A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\\times(B-C)=\\langle 8, 12, 24 \\rangle$ Thus $\\frac{(I-C) \\cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$ \\begin{align*}\\frac{(I-C) \\cdot P}{|P|}&=-r\\\\ \\frac{(\\langle r, r, r", "https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. (2011 II #9) https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\\overline{AH}$, $\\overline{BI}$, $\\overline{CJ}$, $\\overline{DK}$, $\\overline{EL}$, and $\\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.", "10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. (2011 II #9) https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\\overline{AH}$, $\\overline{BI}$, $\\overline{CJ}$, $\\overline{DK}$, $\\overline{EL}$, and $\\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.", "# A Triangle out of Three Broken Sticks ### Solution, Part 1 Assume the sticks are of length $1$ and consider the cube with vertices $A(0,0,0),$ $B(1,0,0),$ $D(0,1,0),$ $E(0,0,1),$ $C(1,1,0),$ $F(1,0,1),$ $H(0,1,1),$ and $G(1,1,1).$ The left pieces of the sticks are defined by their right points and I shall describe their lengths with the real $x,y,z\\in(0,1).$ Thus the cube constitutes the sample space for all possible combinations of the breaking points. For $x,y,z$ to form a triangle, we need three inequalities: \\begin{align} z&\\lt x+y\\\\ y&\\lt x+z\\\\ x&\\lt y+z. \\end{align} Note that, say, line $AF$ has the equation $z=x$ and line $AH$ the equation $z=y,$ so that the plane through $A,F,H$ is described by $z=x+y.$ The plane divides the space into two half-spaces. In the one that contains $G,$ $x+y\\gt z.$ Similarly, the plane $AFC$ is described by $x=y+z$ and the plane $ACH$ by $y=x+z.$ The set of points that satisfies the three inequalities belongs to the intersection of the three half-spaces and the cube: that's the figure $V=ACFHG.$ It consists of two pyramids with the base $CFH$ and apices at $A$ and $G.$ $V$ is obtained from the cube by cutting corner pyramids $ACFB,$ $ACHD,$ and $AFHE.$ Each of these has volume $\\displaystyle \\frac{1}{3}\\cdot\\frac{1}{2}\\cdot 1=\\frac{1}{6}.$ It follows that the volume of $V$ is $\\displaystyle 1-3\\cdot\\frac{1}{6}=\\frac{1}{2}.$ ### Solution, Part", "ranges through the positive integers. • is 107 • is 401 • is 840 • cannot be determined from the given information GCD Sequence Integers ## Check the Answer AIME I, 1985, Question 13 Elementary Number Theory by David Burton ## Try with Hints $a_n=100+n^{2}$ $a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1$ and $a_{n+1}-a_{n}=2n +1$ $d_{n}|(2n+1)$ and $d_{n}|(100 +n^{2})$ then $d_{n}|[(100+n^{2})-100(2n+1)]$ then $d_{n}|(n^{2}-200n)$ here $n^{2} -200n=0$ then n=200 then $d_{n}$=2n+1=2(200)+1=401. Categories ## Planes and distance | AIME I, 2011 | Question 13 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance. ## Planes and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information Plane Distance Algebra ## Check the Answer AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here", "THREE.LineBasicMaterial( { color: whatever}); var geometry = new THREE.Geometry(); var thePoints = getPointsFromBezierD(d); var curve = new THREE.CubicBezierCurve3( new THREE.Vector3(thePoints.P0.x, height-thePoints.P0.y,0), new THREE.Vector3(thePoints.P1.x, height-thePoints.P1.y,0), new THREE.Vector3(thePoints.P2.x, height-thePoints.P2.y,0), new THREE.Vector3(thePoints.P3.x, height-thePoints.P3.y,0)); geometry.vertices = curve.getPoints(20); var curveObject = new THREE.Line(geometry, material); Of course, you don't have to put the 3D curve in the $z=0$ plane - that's just what I'm doing here initially.", "Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. ## Problem 13 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\\frac{r - \\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r + s + t$. ## Problem 14 Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$, $M_3$, $M_5$, and $M_7$ be the midpoints of sides $\\overline{A_1 A_2}$, $\\overline{A_3 A_4}$, $\\overline{A_5 A_6}$, and $\\overline{A_7 A_8}$, respectively. For $i = 1$, 3, 5, 7, ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \\perp R_3$, $R_3 \\perp R_5$, $R_5 \\perp R_7$, and $R_7 \\perp R_1$. Pairs of rays $R_1$ and $R_3$, $R_3$ and $R_5$, $R_5$ and $R_7$, and $R_7$ and $R_1$ meet at $B_1$, $B_3$, $B_5$, $B_7$ respectively. If $B_1 B_3 = A_1 A_2$, then $\\cos", "of edge $$AE$$, $$BC$$, and $$GH$$, as shown. Find the shape of the section in the cube and calculate its area. 3. Given the three lines with parametric equations $$L_1: \\begin{cases} x = 1 \\\\ y = 2 - 2a \\\\ z = a + 3 \\end{cases}, L_2: \\begin{cases} x = 17 - 8b \\\\ y = 6 - 2b \\\\ z = 9 - 3b \\end{cases}, L_3: \\begin{cases} x = 1 \\\\ y = 6c - 4 \\\\ z = 3c \\end{cases}$$, where $$a, b, c$$ are real numbers, and the plane with equation $$x + 2y - 2z + 7 = 0$$: 1. Prove that the three lines are concurrent, and find the coordinates of the intersection point, $$V$$. 2. Find the intersection points of the three lines with the plane, denoted $$A, B, C$$ respectively for $$L_1, L_2, L_3$$. 3. Find the volume of tetrahedron $$VABC$$. 1. Show that the set of equations $$\\begin{cases} x = 6 - p - t \\\\ y = p \\\\ z = t \\end{cases}$$, $$0 \\leq p \\leq 2$$, $$0 \\leq t \\leq 3$$ bounds a region whose interior is in the shape of a parallelogram. 2. A parallelepiped is bordered by the pairs of parallel planes: \\begin{align*} P_1: x + 2y + 2z &= 3 & P_2: x", "# Difference between revisions of \"2007 AIME I Problems/Problem 13\" ## Problem A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$. A plane passes through the midpoints of $AE$, $BC$, and $CD$. The plane's intersection with the pyramid has an area that can be expressed as $\\sqrt{p}$. Find $p$. ## Solution ### Solution 1 Note first that the intersection is a pentagon. Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,2\\sqrt{2})$. Using the coordinates of the three points of intersection $(-1,1,\\sqrt{2}),\\ (2,0,0),\\ (0,-2,0)$, it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$, and using the points we find that $2a = d \\Longrightarrow d = \\frac{a}{2}$, $-2b = d \\Longrightarrow d = \\frac{-b}{2}$, and $-a + b + \\sqrt{2}c = d \\Longrightarrow -\\frac{d}{2} - \\frac{d}{2} + \\sqrt{2}c = d \\Longrightarrow c = d\\sqrt{2}$. It is then $x - y + 2\\sqrt{2}z = 2$. $[asy]import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2); draw(A--B--C--D--A--E--B--E--C--E--D); label(\"A\",A, SE); label(\"B\",B,(1,0,0)); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE);", "middle rectangle is actually a square, so the total area is $(15\\sqrt{2})^2 + 4\\left(\\frac 12\\right)\\left(\\frac 52\\sqrt{2}\\right)\\left(\\frac{15}{\\sqrt{2}}\\right) = 525$. ## Solution 2 First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane $PQR$, so they cannot be skew. Therefore, they are parallel. Let the cube's vertices be $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$, with $A$, $B$, and $C$ on the bottom face as before, $H$ being the other bottom vertex, $D$ directly above $C$, $E$ above $B$, $F$ above $A$, and $G$ above $H$. Clearly, the next vertex of the intersection (starting with $P$, $Q$, $R$) will be somewhere on $DG$. Let it be $X$, and have a distance of $x$ from D, and a distance of $20 - x$ from $G$. Then, the next vertex will be somewhere on $FG$. It must be parallel to $PQ$, so this implies that it has a distance of $20 - x$ from $G$, and thus a distance of $x$ from $F$. Now, the next vertex (call it $Y$) will be somewhere on $AF$. The segment must be parallel to $QR$, so $FY$ must have length $2x$,", "cities for which a flight between the cities exists. Consider any vertex$v \\in V(G)$. Let$d_4(v)$be the number of vertices of distance$4$from$v$in$G$. Let$\\{v_1, v_2, \\dots , v_p\\}$be the set of vertices of distance$1$from$v$in$G$. Now for each ... 1 I propose that the question has a typo and that the inequalities in the index of the summation cannot be strictly less than and need to be less than or equal to. Here is why. Consider the case where$n=2$and$r=3$. Clearly, the left hand side is an application of binomial theorem and we have$a_1^3+3a_1^2a_2+3a_1a_2^2+a_2^3$The right hand side is (from ... 1 The first two problems ask for Steiner Triple Systems. It is known that a Steiner Triple System exists if and only if$n \\equiv 1 \\text{ or } 3 \\pmod 6$. For$K_7$we have$7 \\equiv 1 \\pmod 6$, so one exists. An example is drawn on the Wikipedia page (with triangles are drawn as lines): (see also Fano plane). For$K_{12}$we have$12 \\equiv 0 \\pmod ... 1 first of all you should find $A$, $B$ and $C$ with initial $y(-2)=41$, $y(5)=20$ and quadratic function is minimum at $x_{min}=-\\frac{-B}{2A}=\\frac{B}{2A}$. so you have three equations and three : $41=4A+2B+C$ $20=25A-5B+C$ $2=\\frac{B}{2A}$ Solve this for $A$, $B$, $C$ and then you can find $D=y_{min}=y(x_{min})$ or ... 2 Hint: A quadratic function that has its minumum (or extremum)", "the plane that perpendicularly bisects the edge determined by $A$ and $B$, which contains the other two vertices, has equation $$X\\cdot(A-B)=\\left(\\frac{A+B}{2}\\right)\\cdot(A-B)$$ (where $\\cdot$ is the vector dot product and $X$ is a general point in 3-space). The other two vertices must also lie on the unit sphere, $$|X|=1.$$ And, the other two vertices must be the same distance from both $A$ and $B$ as $A$ and $B$ are from one another: $$|A-X|=|A-B|$$ $$|B-X|=|A-B|.$$ Solving all four (or perhaps any three) of these equations simultaneously should give the coordinates of the other two vertices. I strongly doubt that this yields anything nice in the fully general case. Even putting one vertex at $(0,0,1)$ and letting a second vertex have $y$-coordinate $0$, the expressions for the coordinates of the vertices weren't pretty. (Numerical approximations: $(0.970984, 0, -0.239146)$, $(-0.485492, 0.840896, -0.239146)$, and $(-0.485492, -0.840896, -0.239146)$.) - Given tetrahedral vertices $P$ and $Q$ on a unit sphere centered at the origin $O$, let $M$ be the midpoint of edge $PQ$, and let $N$ be the reflection of $M$ in the origin. (That is, $N := -\\frac{1}{2}(P+Q)$.) Then, the tetrahedron's other vertices, $R$ and $S$, determine vectors $NR$ and $NS$ that are (1) perpendicular to plane $OPQ$ (and, hence, are parallel to the vector $P\\times Q$), and (2) of length $\\frac{1}{2}|PQ|$. Noting that", "# 1983 AIME Problems/Problem 11 ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); MP(\"A\",A);MP(\"B\",B);MP(\"C\",C);MP(\"D\",D);MP(\"E\",E,N);MP(\"F\",F,N); (Error compiling LaTeX. D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); ^ 0707a68f45e6aa7b35cf97f6d26e107a5d73de40.asy: 8.2: no matching function 'D(void(flatguide3))') ## Solution First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to find that the height is equal to $6$. size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); D(B--Ba--Ca--C,dashed+d);D(A--Aa--Da--D,dashed+d);D(E--(E.x,E.y,0),dashed+l);D(F--(F.x,F.y,0),dashed+l); D(Aa--E--Da,dashed+d); D(Ba--F--Ca,dashed+d); MP(\"A\",A);MP(\"B\",B);MP(\"C\",C);MP(\"D\",D);MP(\"E\",E,N);MP(\"F\",F,N);MP(\"12\\sqrt{2}\",(E+F)/2,N);MP(\"6\\sqrt{2}\",(A+B)/2);MP(\"6\",(3*s/2,s/2,3),ENE); (Error compiling LaTeX. D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); ^ 4236047701a23aee0aec75276861479a62d4d1f1.asy: 9.2: no matching function 'D(void(flatguide3))') Next, we complete the figure into a triangular prism, and find the area, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined area", "Distance. ## Planes and distance- AIME I, 2011 A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\\frac{r-s^\\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \\lt 1000$, find r+s+t. • is 107 • is 330 • is 840 • cannot be determined from the given information ### Key Concepts Plane Distance Algebra AIME I, 2011, Question 13 Geometry Revisited by Coxeter ## Try with Hints Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d $\\frac{a10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=10-d and $\\frac{b10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=11-d and $\\frac{c10}{({a^{2}+b^{2}+c^{2}})^\\frac{1}{2}}$=12-d squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\\frac{98}{3}^\\frac{1}{2}$ then d=11-$\\frac{98}{3}^\\frac{1}{2}$=$\\frac{33-294^\\frac{1}{2}}{3}$ then 33+294+3=330.", "other vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives $$\\frac{a(10)}{\\sqrt{a^2 + b^2 + c^2}} = 10-d,$$ $$\\frac{b(10)}{\\sqrt{a^2 + b^2 + c^2}} = 11-d,$$ and $$\\frac{c(10)}{\\sqrt{a^2 + b^2 + c^2}} = 12-d.$$ Squaring each equation and then adding yields $100=(10-d)^2+(11-d)^2+(12-d)^2$, and we can proceed as in the first solution. ## Solution 3 Let the vertices with distance $10,11,12$ be $B,C,D$, respectively. An equilateral triangle $\\triangle BCD$ is formed with side length $10\\sqrt{2}$. We care only about the $z$ coordinate: $B=10,C=11,D=12$. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\\text{centroid}=(10+11+12)/3=11$. Designate the midpoint of $BD$ as $M$. Notice that median $CM$ is parallel to the plane because the $\\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$, and the median contains $C$ and the $\\text{centroid}$. We seek the angle $\\theta$ of the line:$(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\\triangle BCD$, $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $in\\text{ }slope$ to $BD$. Since $BD$ makes a $2-14-10\\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\\circ$. Therefore, $\\sin\\theta=\\frac{14}{10\\sqrt{2}}=\\frac{7\\sqrt{2}}{10}$. It is also known that the centroid of $\\triangle BCD$ is", "'12 at 11:45 • I just want to create one, no animation. – Djentleman Nov 28 '12 at 18:45 First, create three sets of 4 points, each set being a 3D golden rectangle in the X, Y or Z direction: Vector3[] points = new Vector3[12]; float phi = 0.5 + 0.5 * sqrt(5.0); for (int i = 0; i < 4; i++) { float x = ((i & 1) != 0) ? 0.5 : -0.5; float y = ((i & 2) != 0) ? phi * 0.5 : phi * -0.5; points[3 * i] = Vector3(x, y, 0.0); points[3 * i + 1] = Vector3(0.0, x, y); points[3 * i + 2] = Vector3(y, 0.0, x); } Then create triangles to build the icosahedron: static int[] trilist = { 0, 1, 2, 2, 4, 6, 3, 8, 1, 9, 4, 8, 7, 0, 5, 7, 11, 3, 10, 5, 6, 10, 9, 11, 0, 3, 1, 7, 3, 0, 1, 4, 2, 8, 4, 1, 2, 5, 0, 6, 5, 2, 6, 9, 10, 4, 9, 6, 7, 10, 11, 5, 10, 7, 8, 11, 9, 3, 11, 8 }; VertexPositionNormalTexture[] vertices = new VertexPositionNormalTexture[60]; for (int i = 0; i < 60; i += 3) { vertices[i].Position = points[trilist[i]]; vertices[i + 1].Position = points[trilist[i + 1]]; vertices[i", "the number of ways to reach it is $a + b + c$, where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left). Assign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$. Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$. Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$. This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$. (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus: $a_{i,j} = s(i-1,j) - c_{i-1,j}$ $b_{i,j} = s(i-1,j-1)$ $c_{i,j} = s(i,j-1) - a_{i,j-1}$ It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$. Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \\leq i, j \\leq 5$. Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) =" ]

[ "$$5-1=4$$. These points are 4 units apart.", "the point $(1,1)$. 3. Draw a line connecting these two points.", "subtract the $$x$$-coordinates, we get $$5-1=4$$. These points are 4 units apart.", "send the point at infinity to $(1:1:1)$.", "passes through points (4, -5) and (4, 4). $$\\textrm{line passing through the points }( 4, -(5) )\\textrm{ and }( 4, 4 )$$", "$$\\mathbb{Q}$$, you can find torsion points of order $$m-1=1,2,3,4,5,6,7,8,9,10$$, and $$12$$.", "at two points in $[0, 1]$.", "You are given four points $$(1, 3), (-3, 5), (-5, -3),$$ and $$(7, -1).$$ Suppose two line segments are to constructed so that each point is used exactly once. What is the greatest possible distance between the midpoints of two such line segments?", "# From A to B Geometry Level 1 Given that $$A = (-3, 6)$$ and $$B = (2, 18 )$$, what is the distance between points $$A$$ and $$B$$? ×", "# Solve this following Question: Find the distance of the point $(3,4,5)$ from the plane $\\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}-5 \\hat{\\mathrm{j}}+3 \\hat{\\mathrm{k}})=13$. Solution:", "# 3-D is beautiful! Geometry Level 1 Find the distance between the points (7; 1; 1) and (5; -1; 2) ×", "point$(3,1,2)\\$. ...", "Four points are arranged on a plane such that the distance between every pair is either $a$ or $b$. If $b>a$, find all possible values of $\\frac{b}{a}$.", "$$E(F_{5^{2n}})$$ has 4 points of order 5 (plus the point at infinity whose order == 1 divides 5)", "and Solution: Write a Python program to compute the distance between the points (x1, y1) and (x2, y2).", "get the coordinates $(-5, 0, -7)$.", "$3 - t + p = 3 - 6s$ Solve this system for t, s, and p. That will give you the two points of intersection. Then find the distance between these points. -Dan", "to fit the same distance between the points $f(x_1)$ and $f(x_2)$ in a way that fits the topology of $Y$.", "1 - 4 = 5$ $x^2 + 2x - 8 = 0$ $(x + 4)(x - 2) = 0$ $x = -4, 2$ So the two points are (-4, -1), and (2, -1).", "Browse Questions +1 vote # Find the distance between the point P (6, 5, 9) and the plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). Can you answer this question?" ]

[ "distance at least $2$, and suppose that the original list is $1,2,3,4,5,6,7,8$. After trimming we are left with $1,2,4,8$, and any two adjacent elements are at multiplicative distance at least $2$ (in this case, exactly $2$). Had we started with, say, $1,4,9,16$, no trimming would be necessary, but still adjacent elements are at multiplicative distance at least $2$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "$|b|=1$. $b^2+2a+1=0\\rightarrow b^2=-1-2a$. This is a circle in the complex plane centered at $(-1,0)$ with radius $2$ since $|a|=1$. The maximum distance from the origin is $3$ at $(-3,0)$ and similarly the minimum distance is $1$ at $(1,0)$. So $1\\le |b^2|\\le 3\\rightarrow 1\\le |b|\\le \\sqrt{3}$. Both solutions give the same lower bound, $1$. So the range is $\\sqrt{3}-1=\\boxed{\\textbf{(C) }\\sqrt{3}-1}$.", "and $n$ is the distance between $x$ and (duh) ($x-n$). $x^2-(x-n)^2=2nx-n^2$ Example: $26^2-22^2$ (without a calculator and you don't want to square $26$ and $22$)$= 2(4)(26)-4^2=192$ P.S. I'm in tenth grade, and I spent an hour figuring this out myself. This works especially well for very large numbers whose square roots are gross, and you don't have a calculator. • Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. For further information about writing math on this site see e.g. here, here, here and here. – user409521 Mar 31 '17 at 14:12 • Congratulations for having investigated this two-year old Question and drawing a valid conclusion. However the Question already had an Accepted Answer, and you have not highlighted what new information you are adding. Indeed you did not directly address the crux of the Question (when we compare $|\\sqrt{x} - \\sqrt{y}|$ with $|x-y|$). – hardmath Mar 31 '17 at 15:40", "## Algebra and Trigonometry 10th Edition We know that the distance between the numbers is: $$74+48 = 122$$", "notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$. after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$. Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta. Solving, we get $a=5$ and $b=-1$, so the distance is $a^2 + b^2 = \\boxed{026}$.", "2 & 3 & 2 & 3 & 4 & 3 (4) &4\\\\ y & 1 & 0 & 1 & 2 & 1 & 2 & 2 & 0 & 3 (1) &0\\\\ z & 0 & 0 & 1 & 0 & 0 & 2 & 1 & 0 & 0 (1) &2\\\\ x^2+y^3+z^2 & 2 & 4 & 6 & 8 & 10 & 12 & 14 &16 &18 & 20\\\\ d(a) & 1 & \\sqrt{2} & \\sqrt{3} & 2 & \\sqrt{5} & \\sqrt{6} & \\sqrt{7}& \\sqrt{8} &\\sqrt{9} &\\sqrt{10} \\end{matrix}$$ According to the link in Michael Seifert's comment, the even value of $$x^2+y^2+z^2$$ only misses $$28$$ and $$60$$ for $$x^2+y^2+z^2 < 83$$. • Note that the integers which can be expressed as the sum of three squares can be found as sequence 378 in the OEIS. You would need to drop the odd numbers in that sequence ($x + y + z \\text{ even} \\iff x^2 + y^2 + z^2 \\text{ even}$), but all the possible distances would be there. Alternately, the distances squared appear to be sequence 401, though I'm not sure from the description how it connects to lattice distances. Feb 22, 2021 at 13:24 • There's also a comment on sequence 401 that all numbers in the list there are not of", "distance is $5 - \\frac{400}{108} = \\frac{140}{108} = \\frac{70}{54} = \\frac{35}{27}.$", "## Intermediate Algebra (12th Edition) We find the differences by looking at the distance between the numbers on a number line (see plot). (a) The difference, or distance, between $0$ and $6$ on the number line is 6. (b) The difference, or distance, between $-15$ and $-24$ on the number line is $9$. (c) The difference, or distance, between $-4$ and $-20$ on the number line is $16$.", "while the distance to $(1/2,1/2,1/2,1/2)$ is $1$.", "-4\\;\\;\\;(2)$ You are required to find the distance between the two points or $d = \\sqrt{(a-4)^2+(b-3)^2}$ $= \\sqrt{a^2+b^2 - 8a -6b +25}$ Now $2\\times(2)-(1)$ gives $a^2+b^2 - 8a - 6b + 12 = 0$ so $d = \\sqrt{25-12} = \\sqrt{13}$ Edit: Earthboth's solution is so much more elegant (and of course easier!) 6. Originally Posted by Danny Now $2\\times(2)-(1)$ gives Why? How did you come up with this? 7. This is what I have $a^2+b^2-2a = 4\\;\\;\\;(1)$ $a^2+b^2 - 5a - 3b = -4\\;\\;\\;(2)$ this is what I want $a^2+b^2 -8a - 6b\\;\\;\\;(3)$ Multiply $p \\times (1) + q \\times (2) = (3)$ (LHS only) and equate like terms. As it turns out, there is a unique $p$ and $q$ solution.", "$$20\\sqrt{5}$$ (C) $$41\\sqrt{5}$$ (D) $$\\sqrt{205}$$ (E) 100 Discussed HERE Arithmetic: Approximation Many GMAT arithmetic problems require approximations, for example: The average distance between the Sun and a certain planet is approximately 2.3 x 10^14 inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately 3.9 x 10^4 inches.) (A) 7.1 x 10^8 (B) 5.9 x 10^9 (C) 1.6 x 10^10 (D) 1.6 x 10^11 (E) 5.9 x 10^11 Discussed HERE M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true? (A) 1/3 < M < 1/2 (B) 1/5 < M < 1/3 (C) 1/7 < M < 1/5 (D) 1/9 < M < 1/7 (E) 1/12 < M < 1/9 Discussed HERE If $$M=\\sqrt{4}+\\sqrt[3]{4}+\\sqrt[4]{4}$$, then the value of M is: A. Less than 3 B. Equal to 3 C. Between 3 and 4 D. Equal to 4 E. Greater than 4 Discussed HERE The product of all the prime numbers less than 20 is closest to which of the following powers of 10 ? (A) 10^9 (B) 10^8 (C) 10^7 (D) 10^6 (E) 10^5 Discussed HERE If S is the sum of reciprocals of a list of consecutive integers from 45 to 54,", "distance between $(4,5,3)$ with $(x,y,z)$. $\\sqrt{(x – 4)^2 + (y – 5)^2 + (z-3)^2 }$ $\\sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)}$ As both distances are equidistant, putting them equal to each other and simplifying: $\\sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)} = \\sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)}$ Re-writing: $10x + 8x -10y + 10y +6z +6z +59 -50 = 0$ $6x + 4z = -3$", "as it did in the other directions. What about something in between? Say, $s=(0.921,0.395)$, so then $T(s)=1.316$. So, taking a step of size $1$ led to an increase of ${\\sim}1.3$. This is between the other values, as we would kind of expect.", "## College Algebra 7th Edition $24$ The formula for the distance between two numbers is $|a-b|$. Thus, we have $|-3-21|=|-24|=24$", "## Calculus (3rd Edition) 1) d = $2\\sqrt 2$ 2) d = $3$ 3) d = $\\sqrt {13}$ 4) d = $\\sqrt {37}$ The distance between point (x_{1}, y_{1}) and the point (x_{2}, y_{2}) is: $d = \\sqrt {(x_2-x_1)^{2}+(y_2-y_1)^{2}}$ 1) Points (1,4) and (3,2): $d = \\sqrt {(3-1)^{2}+(2-4)^{2}}$ $d= 2\\sqrt2$ 2) Points (2,4) and (2,1): $d = \\sqrt {(2-2)^{2}+(1-4)^{2}}$ $d= 3$ 3) Points (-2,3) and (0,0): $d = \\sqrt {(0-(-2))^{2}+(0-3)^{2}}$ $d= \\sqrt{13}$ 4) Points (-2,3) and (-3,-3): $d = \\sqrt {(-3-(-2))^{2}+(-3-3)^{2}}$ $d= \\sqrt{37}$", "$n+8$, $n+9$, $n+10$, $n+11$, and $n+12$. This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\\sqrt{3})$, a move that repeats every 12 moves. Since $\\frac{2016}{12} = 168$, we have that $P_(2016) = (-1008, -2016-1008\\sqrt{3})$. It follows that $P_(2015) = (-1008-1008\\sqrt{3}, -1008-1008\\sqrt{3})$ so the distance to $P_0$ is $1008(\\sqrt{2} + \\sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \\boxed{\\text{(B) } 2024}$. - bobjoebilly", "$0$ and $3$ is $0-1-3$ , and the simple path between $2$ and $3$ is $2-1-3$ . All of them have a distance of $2$ .", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "much so i assumed you meant distance is halved] 32. rock_mit182 yeah that is what i meant 33. rock_mit182 distance is halved 34. IrishBoy123 so $4 \\times 10^2 = ?? \\times 5^2$" ]

[ "between points A and B is: $2\\sqrt{13}$ $\\sqrt{13}$ $4\\sqrt{3}$ $4\\sqrt{13}$", "of A and B. a. If A=-5+2i and C=3+4i, find B. C=$$\\frac{A+B}{2}$$ B=2C-A =2(3+4i)-(-5+2i) =6+8i+5-2i =11+6i b. If B=1+11i and C=-5+3i, find A. C=$$\\frac{\\boldsymbol{A}+\\boldsymbol{B}}{2}$$ A=2C-B =2(-5+3i)-(1+11i) =-10+6i-1-11i =-11-5i Question 3. Point C is the midpoint between A=4+3i and B=-6-5i. Find the distance between points C and D for each point D provided below. a. 2D=-6+8i C=$$\\frac{4-6}{2}$$+$$\\frac{3-5}{2}$$i=-1-i D=-3+4i |C-D|=|-1-i+3-4i| =|-2-5i| =$$\\sqrt{(-2)^{2}+(5)^{2}}$$ =$$\\sqrt{29}$$ b. D=-$$\\overline{\\boldsymbol{B}}$$ D=6-5i |C-D|=|-1-i-6+5i| =|-7+4i| =$$\\sqrt{(-7)^{2}+(4)^{2}}$$ =$$\\sqrt{65}$$ Question 4. The distance between points A=1+1i and B=a+bi is 5. Find the point B for each value provided below. a. a=4 5=$$\\sqrt{(1-4)^{2}+(1-b)^{2}}$$ 25=9+(1-b)2 1-b=±4 b=-3,5 B=4-3i or B=4+5i b. b=6 5=$$\\sqrt{(1-a)^{2}+(1-6)^{2}}$$ 25=(1-a)2+25 (1-a)2=0 1-a=0 a=1 B=1+6i Question 5. Draw five points in the plane A, B, C, D, E. Start at any position, P0, and leapfrog over A to a new position, P1(so, A is the midpoint of ($$\\overline{P_{0} P_{1}}$$). Then, leapfrog over B, then C, then D, then E, then A, then B, then C, then D, then E, then A again, and so on. How many jumps will it take to get back to the starting position, P0? P1=2A-P0 P2=2B-P1=2B-2A+P0 P3=2C-P2=2C-2B+2A-P0 P4=2D-P3=2D-2C+2B-2A+P0 P5=2E-P4=2E-2D+2C-2B+2A-P0 P6=2A-P5=2A-2E+2D-2C+2B-2A+P0=-2E+2D-2C+2B+P0 P7=2B-P6=2B+2E-2D+2C-2B-P0=2E-2D+2C-P0 P8=2C-P7=2C-2E+2D-2C+P0=-2E+2D+P0 P9=2D-P8=2D+2E-2D-P0=2E-P0 P10=2E-P9=2E-2E+P0=P0 Question 6. For the leapfrog puzzle problems in both Exploratory Challenge 1 and Problem 5, we are given an odd number of points to leapfrog over. What if we", "$$\\mathbb{Q}$$, you can find torsion points of order $$m-1=1,2,3,4,5,6,7,8,9,10$$, and $$12$$.", "## Algebra and Trigonometry 10th Edition We know that the distance between the numbers is: $$74+48 = 122$$", "subtract the $$x$$-coordinates, we get $$5-1=4$$. These points are 4 units apart.", "send the point at infinity to $(1:1:1)$.", "$$5-1=4$$. These points are 4 units apart.", "notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$. after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$. Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta. Solving, we get $a=5$ and $b=-1$, so the distance is $a^2 + b^2 = \\boxed{026}$.", "+0 # coordinates 0 84 1 On a Cartesian coordinate plane, points (1,2) and (7,-4) are opposite vertices of a square. What is the area of the square? Feb 23, 2021 The distance between $(1,2)$ and $(7,-4)$ is $\\sqrt{6^2+6^2}=6\\sqrt{2}$ by the distance formula. This is the length of the diagonal of the square. Recall that the diagonal is equal to $\\sqrt{2}\\cdot\\text{side length}$. Therefore, $\\text{side length}=6$, and $\\text{Area}=6^2=\\boxed{36}$.", "# Solve this following Question: Find the distance of the point $(3,4,5)$ from the plane $\\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}-5 \\hat{\\mathrm{j}}+3 \\hat{\\mathrm{k}})=13$. Solution:", "Question 1: Calculate the distance between the points $(6, -4)$ and $(3,2)$ correct to $2$ decimal places. $(6, -4)$ and $(3, 2)$ Distance $= \\sqrt{(3-6)^2+(2-(-4))^2} = \\sqrt{9+36} = \\sqrt{45} = 6.71$ $\\\\$ Question 2: Find the distance between the points $(-2, -2)$ and $(1,0)$ correct to $3$ significant figures. $(-2, -2)$ and $(1,0)$ Distance $= \\sqrt{(1-(-2))^2+(0-(-2))^2} = \\sqrt{9+4} = \\sqrt{13} =3.61$ $\\\\$ Question 3: Show that the points $P(7, 3), Q(6, 3 + \\sqrt{3}) \\ and \\ R (5, 3)$ form an equilateral triangle. $P(7, 3), Q(6, 3 + \\sqrt{3}) \\ and \\ R (5, 3)$ $PQ = \\sqrt{(6-7)^2+(3 + \\sqrt{3}-3)^2} = \\sqrt{1+3} = \\sqrt{4} = 2$ $PR = \\sqrt{(5-7)^2+(3-3)^2} = \\sqrt{4+0} = \\sqrt{4} = 2$ $QR = \\sqrt{(5-6)^2+(3-(3 + \\sqrt{3})^2} = \\sqrt{1+3} = \\sqrt{4} = 2$ Therefore three sides $PQ, PR \\ and \\ QR$ are equal which makes it an equilateral triangle. $\\\\$ Question 4: The circle with center $(x, y)$ passes though the points $(3, 11), (14, 0) \\ and \\ (12, 8)$ . Find the values of $x \\ and \\ y$. Distance of the points from the center are equal. Therefore $\\sqrt{(3-x)^2+(11-y)^2}=\\sqrt{(12-x)^2+(8-y)^2}$ $9x^2-6x+121+y^2-22y=144+x^2-24x+64+y^2-16y$ $18x-6y=78$ $3x-y=13$ … … … … i) $\\sqrt{(3-x)^2+(11-y)^2}=\\sqrt{(14-x)^2+(0-y)^2}$ $9x^2-6x+121+y^2-22y=196+x^2-28x+y^2$ $22x-22y=66$ $x-y=3$… … … … ii) Solving i) and ii), we get $x = 5$ and $y = 2$. Therefore the", "You are given four points $$(1, 3), (-3, 5), (-5, -3),$$ and $$(7, -1).$$ Suppose two line segments are to constructed so that each point is used exactly once. What is the greatest possible distance between the midpoints of two such line segments?", "## Calculus (3rd Edition) 1) d = $2\\sqrt 2$ 2) d = $3$ 3) d = $\\sqrt {13}$ 4) d = $\\sqrt {37}$ The distance between point (x_{1}, y_{1}) and the point (x_{2}, y_{2}) is: $d = \\sqrt {(x_2-x_1)^{2}+(y_2-y_1)^{2}}$ 1) Points (1,4) and (3,2): $d = \\sqrt {(3-1)^{2}+(2-4)^{2}}$ $d= 2\\sqrt2$ 2) Points (2,4) and (2,1): $d = \\sqrt {(2-2)^{2}+(1-4)^{2}}$ $d= 3$ 3) Points (-2,3) and (0,0): $d = \\sqrt {(0-(-2))^{2}+(0-3)^{2}}$ $d= \\sqrt{13}$ 4) Points (-2,3) and (-3,-3): $d = \\sqrt {(-3-(-2))^{2}+(-3-3)^{2}}$ $d= \\sqrt{37}$", "passes through points (4, -5) and (4, 4). $$\\textrm{line passing through the points }( 4, -(5) )\\textrm{ and }( 4, 4 )$$", "$|b|=1$. $b^2+2a+1=0\\rightarrow b^2=-1-2a$. This is a circle in the complex plane centered at $(-1,0)$ with radius $2$ since $|a|=1$. The maximum distance from the origin is $3$ at $(-3,0)$ and similarly the minimum distance is $1$ at $(1,0)$. So $1\\le |b^2|\\le 3\\rightarrow 1\\le |b|\\le \\sqrt{3}$. Both solutions give the same lower bound, $1$. So the range is $\\sqrt{3}-1=\\boxed{\\textbf{(C) }\\sqrt{3}-1}$.", "prefer, you can keep $\\mathbb R^n$ and place a point numbered $(n+1)$ at $$(-t,-t,-t, \\ldots, -t)$$ for a special value of $t > 0$ that makes all the distances $\\sqrt 2.$ I get $$n t^2 + 2 t - 1 = 0$$ or, with $t>0,$ $$t = \\frac{\\sqrt {n+1} - 1}{n } = \\frac{1}{\\sqrt {n+1} + 1}$$", "# From A to B Geometry Level 1 Given that $$A = (-3, 6)$$ and $$B = (2, 18 )$$, what is the distance between points $$A$$ and $$B$$? ×", "anonymous one year ago hhhhhhhhhhheeeeeeeellllllllllllllllllllllllllllllpppppppppppppppppppppppppppppp!!!!!! 1. ali2x2 ? 2. anonymous 3. mathstudent55 First, can you read off the coordinates of points A and B? Each point is an ordered pair. 4. anonymous okay 5. mathstudent55 After you know the coordinates of the points, use them in the distance formula below. The distance, $$d$$, between points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is $$d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$", "You might look for the point $p$ of the line such that the segment frpm $p$ to $(3,2,-7)$ is orthogonal to the line. Then the distance from $(3,2,-7)$ to the line is equal to the distance from $(3,2,-7)$ to $p$. – José Carlos Santos May 4 '18 at 23:02", "Four points are arranged on a plane such that the distance between every pair is either $a$ or $b$. If $b>a$, find all possible values of $\\frac{b}{a}$." ]

[ "$a$ and $b$ are acute angles. Prove that $\\frac{a}{b}$ is $\\sin ^2 a – \\cos ^2 a$ \\$\\frac{2 \\", "B+\\sin C}}$$ By simplifying we get $$m=\\frac{\\cos B+\\cos C-1}{\\sin B-\\sin C}$$ Hence the angle of the line joining I & D with side BC is given as $$\\tan \\theta=|m|$$ $$\\implies \\theta=\\tan^{-1}\\left|\\frac{\\cos B+\\cos C-1}{\\sin B-\\sin C}\\right|$$", "are three points. Then the angle between $BA$and $BC$is A) ${{\\tan }^{-1}}\\left( \\frac{2}{3} \\right)$ B) ${{\\tan }^{-1}}\\left( \\frac{3}{2} \\right)$ C) ${{\\tan }^{-1}}\\left( \\frac{1}{3} \\right)$ D) ${{\\tan }^{-1}}\\left( \\frac{1}{2} \\right)$", "x = \\frac {\\sin 2x} 2.$$ There will be either only cos's or only sin's left. Convert them into double-angles with \\begin{align} \\cos^2 x &= \\frac {1 + \\cos 2x} 2 \\\\ \\sin^2 x &= \\frac {1 - \\cos 2x} 2. \\end{align}", "corresponding sides. Then $$\\sin A =\\frac{\\sinh a}{\\sinh c}\\qquad \\text{and}\\qquad \\cos A = \\frac{\\tanh b}{\\tanh c}$$", "the law of sines to get the angles, to wit, $$\\frac{\\sin A}{a}=\\frac{\\sin B}{b}=\\frac{\\sin C}{c}$$ So we complete the problem with $$A=\\sin^{-1}\\left( \\frac{a}{c}\\sin C \\right)\\\\ B=\\sin^{-1}\\left( \\frac{b}{c}\\sin C \\right)$$", "Proving $\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$ Prove $$\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$$ Can someone help? I separated $\\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves. 2 Answers Use the formulas $$\\sin a+\\sin b=2\\sin\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right),$$ $$\\cos a+\\cos b=2\\cos\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right).$$ Therefore, we have $$\\frac{\\sin a+\\sin b}{\\cos a+\\cos b}=\\frac{\\sin\\displaystyle\\left(\\frac{a+b}{2}\\right)}{\\cos\\displaystyle\\left(\\frac{a+b}{2}\\right)}=\\tan\\left(\\frac{a+b}{2}\\right).$$ If $$a,b$$ are acute angles one can proceed as follows: Draw a parallelogram with all four sides of length $$1,$$ as follows. The first side has one endpoint at $$(0,0)$$ and the angle between it and the $$x$$-axis is $$a$$. Its other endpoint is therefore at $$(\\cos a,\\sin a).$$ The second side begins at that latter point and its angle with the $$x$$-axis is $$b.$$ Its other endpoint is therefore $$(\\sin a + \\sin b, \\cos a+\\cos b).$$ The tangent of the angle from the origin to this last point is therefore $$(\\sin a+\\sin b)/(\\cos a + \\cos b).$$ But the geometric symmetry of the parallelogram shows that that angle is $$(a+b)/2.$$", "- b^2$$ 44. If $$\\sin A = \\frac{1}{2}$$ and $$\\sin B = \\frac{1}{3},$$ find the value of $$\\sin(A + B)$$ and $$\\sin(2A + 2B)$$ 45. If $$\\cos A = \\frac{11}{61}$$ and $$\\sin B = \\frac{4}{5},$$ find the value of $$\\sin^2 \\frac{A - B}{2}$$ and $$cos^2\\frac{A + B}{2},$$ the angle of $$A$$ and $$B$$ being positive acute angles. 46. Given $$\\sec A = \\frac{5}{4},$$ find $$\\tan\\frac{A}{2}$$ and $$\\tan A.$$ 47. If $$\\cos A = .3,$$ find the value of $$\\tan \\frac{A}{2},$$ and explain the resulting ambiguity. 48. If $$\\sin A + \\sin B = x$$ and $$\\cos A + \\cos B = y,$$ find the value of $$\\tan \\frac{A - B}{2}$$ Prove that 1. $$(\\cos A + \\cos B)^2 + (\\sin A - \\sin B)^2 = 4\\cos^2 \\frac{A + B}{2}$$ 2. $$(\\cos A + \\cos B)^2 + (\\sin A + \\sin B)^2 = 4\\cos^2 \\frac{A - B}{2}$$ 3. $$(\\cos A - \\cos B)^2 + (\\sin A - \\sin B)^2 = 4\\sin^2 \\frac{A - B}{2}$$ 4. $$\\sin^2\\left(\\frac{\\pi}{8} + \\frac{A}{2}\\right) - \\sin^2\\left(\\frac{\\pi}{8} -\\frac{A}{2}\\right) = \\frac{1}{\\sqrt{2}}\\sin A$$ 5. $$(\\tan 4A + \\tan 2A)(1 - \\tan^23A\\tan^2A) = 2\\tan 3A\\sec^2A$$ 6. $$\\left(1 + \\tan \\frac{A}{2} - \\sec\\frac{A}{2}\\right)\\left(1 + \\tan \\frac{A}{2} + \\sec\\frac{A}{2}\\right) = \\sin A\\sec^2\\frac{A}{2}$$ 7. $$\\frac{1 + \\sin A - \\cos A}{1 + \\sin A + \\cos A} = \\tan \\frac{A}{2}$$ 8. $$\\frac{1", "\\frac{7\\pi}{12}$$ Take tangent of both sides: $\\tan(\\frac{7\\pi}{12})$ as a sum of two angles and use the double angle relation (again) to find this. Equate the above two and solve for x 7. Oct 9, 2012 ### SammyS Staff Emeritus Beware of possible extraneous solutions.", "apply the formula of $tan(A + B)$ or $tan(A - B)$ and put the angles that are given in question so we can prove it.", "# geometry or algebra Geometry Level 2 $$a,b,c$$ denote sides opposite to angles $$\\alpha, \\beta, \\gamma$$. Given that $$\\frac { 1 }{ b+c } +\\frac { 1 }{ a+c } =\\frac { 3 }{ a+b+c }$$, then find angle $$\\gamma$$. × Problem Loading... Note Loading... Set Loading...", "find $a*b$ if $|a| = 4$, $|b|=8$, and the angle between $a$ and $b$ is [pi/-9] radians. $a*b=$ 2. $\\cos \\left( \\phi \\right) = \\frac{{a \\cdot b}}{{\\left\\| a \\right\\|\\left\\| b \\right\\|}}", "If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P. Question: If $\\angle A$ and $\\angle P$ are acute angles such that $\\tan A=\\tan P$, then show that $\\angle A=\\angle P$. Solution: Given: $\\tan A=\\tan P$ To show: $\\angle A=\\angle P$ Consider two right angled triangles $\\mathrm{ABC}$ and $\\mathrm{PQR}$ such that $\\tan A=\\tan P$ Therefore we have, $\\tan A=\\frac{B C}{A B}$ and $\\tan P=\\frac{Q R}{P Q}$ Since it is given that $\\tan A=\\tan P$ Therefore, $\\frac{B C}{A B}=\\frac{Q R}{P Q}$ Now by interchanging position of AB and QR by cross multiplication We get, $\\frac{B C}{Q R}=\\frac{A B}{P Q}$ Let $\\frac{B C}{Q R}=\\frac{A B}{P Q}=k$ (say)....(1) Now by cross multiplication $B C=k Q R$ and $A B=k P Q \\ldots \\ldots(2)$ Now by using Pythagoras theorem in triangles ABC and PQR We have, $A C^{2}=A B^{2}+B C^{2}$ and $P R^{2}=P Q^{2}+Q R^{2}$ Therefore $A C=\\sqrt{A B^{2}+B C^{2}}$ and $P R=\\sqrt{P Q^{2}+Q R^{2}}$ Now $\\frac{A C}{P R}=\\frac{\\sqrt{A B^{2}+B C^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now using equation (2) We get, $\\frac{A C}{P R}=\\frac{\\sqrt{(k P Q)^{2}+(k Q R)^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ $\\frac{A C}{P R}=\\frac{\\sqrt{k^{2} P Q^{2}+k^{2} Q R^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now by taking $k^{2}$ common We get, $\\frac{A C}{P R}=\\frac{\\sqrt{k^{2}\\left(P Q^{2}+Q R^{2}\\right)}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Therefore, $\\frac{A C}{P R}=\\frac{k \\sqrt{\\left(P Q^{2}+Q R^{2}\\right)}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now $\\sqrt{P", "sketch the angles, you will see we can calculate $Cos\\left(90+\\frac{A}{2}\\right)=-Sin\\left(\\frac{A}{2}\\right)$ and to find this you can use $Sin^2B=\\frac{1}{2}(1-Cos2B)\\Rightarrow\\ Sin^2\\left(\\frac{A}{2}\\right)=\\frac{1}{2}(1-CosA)$ etc", "# geometry or algebra Geometry Level 2 $$a,b,c$$ denote sides opposite to angles $$\\alpha, \\beta, \\gamma$$. Given that $$\\frac { 1 }{ b+c } +\\frac { 1 }{ a+c } =\\frac { 3 }{ a+b+c }$$, then find angle $$\\gamma$$. ×", "# 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $(vi)\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}= \\sec A+\\tan A$ M manish We need to prove - $\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}= \\sec A+\\tan A$ Taking LHS; By rationalising the denominator, we get; $\\\\= \\sqrt{\\frac{1+\\sin A}{1-\\sin A}\\times \\frac{1+\\sin A}{1+\\sin A}}\\\\\\\\ = \\sqrt{\\frac{(1+\\sin A)^2}{1-\\sin^2A}}\\\\\\\\ =\\frac{1+\\sin A}{\\cos A}\\\\\\\\ = \\sec A + \\tan A\\\\\\\\ = RHS$ Hence proved. Exams Articles Questions", "Q6. If $$∠\\ A$$ and $$∠\\ B$$ are acute angles such that cos A = cos B, then show that $$∠ \\ A \\ = \\ ∠ \\ B$$. In $$∆ \\ ABC$$, $$cosA \\ = \\ \\frac{AC}{AB} \\ = \\ \\frac{B}{H}$$ and, $$cosB \\ = \\ \\frac{BC}{AB}$$ But given that, cos A = cos B => $$\\frac{AC}{AB} \\ = \\ \\frac{BC}{AB}$$ => $$AC \\ = \\ BC$$ ∴ $$∠ \\ A \\ = \\ ∠ \\ B$$ [∵ Angles opposite to equal sides are equal]", "would the angles between $x$ and $a$ and $y$ and $a$ be equal to each other? If $a = \\left(1, 1, -1\\right)$, $x = \\left(2, 0, 1\\right)$, and $y = \\left(2, 1, 2\\right)$ what are the cosines of the two angles? c) Show that the inner product of $Px$ and $Py$ is the same as the inner products of $x$ with $Py$ and $y$ with $Px$ and explain why this is. What is the angle between the vectors $Px$ and $Py$? Answer: a) We have $P = aa^T/a^Ta$ so that $Px = \\left(aa^T/a^Ta\\right)x = \\frac{1}{a^Ta}\\left(aa^Tx\\right) = \\frac{a^Tx}{a^Ta}a$ and $Py = \\left(aa^T/a^Ta\\right)y = \\frac{1}{a^Ta}\\left(aa^Ty\\right) = \\frac{a^Ty}{a^Ta}a$ The inner product of $x$ with $Py$ is then $x^T\\left(Py\\right) = x^T\\left(a^Ty/a^Ta\\right)a$ $= \\left(a^Ty/a^Ta\\right)x^Ta = \\left(a^Tyx^Ta\\right)/a^Ta$ $= \\left(a^Tya^Tx\\right)/a^Ta = \\left(a^Txa^Ty\\right)/a^Ta$ and the inner product of $y$ with $Px$ is then $y^T\\left(Px\\right) = y^T\\left(a^Tx/a^Ta\\right)a$ $= \\left(a^Tx/a^Ta\\right)y^Ta = \\left(a^Txy^Ta\\right)/a^Ta$ $= \\left(a^Txa^Ty\\right)/a^Ta$ So the inner products of $x$ with $Py$ and $y$ with $Px$ are equal. b) Since $x$ and $y$ are arbitrary vectors, in general they would not make the same angle with respect to $a$. For example, consider the case when $a = \\left(1, 1, -1\\right)$, $x = \\left(2, 0, 1\\right)$, and $y = \\left(2, 1, 2\\right)$. If $\\theta_1$ is the angle between $x$ and $a$ then $\\cos \\theta_1 = a^Tx/\\left(\\|a\\|\\|x\\|\\right)$. Similarly, if $\\theta_2$", "- e^{i(a+b)} + e^{i(a-b)} - e^{i(-a+b)} + e^{i(-a-b)}}{4i}$$ With some cancellations on the numerator, this becomes $$\\sin{a}\\cos{b} - \\sin{b}\\cos{a} =\\frac{e^{i(a-b)} + e^{-i(a-b)}}{2i}$$ but since $$\\sin{x} = \\frac{e^{ix} - e^{-ix}}{2i}$$ we conclude $$\\sin{a}\\cos{b} - \\sin{b}\\cos{a} = \\sin{(a-b)}$$ I searched for an alternate way to deriving sum of angles relations, and by congruence came up with this alternate derivation. Perhaps it is acceptable. Take a coordinate pair say $(x,y)$ with angle $A$. We have the definitions $\\sin(A)= y/(x^2+y^2)^.5$, and $\\cos(A)= x/(x^2+y^2)^.5$. Now rotate, counterclockwise, the coordinate pair $(x,y)$ by an angle $B$ such that $\\angle A + \\angle B = \\angle C$ After rotation $(x,y)$ becomes $(x\\cos B-y\\sin B, x\\sin B+y\\cos B)$. (I'm leaving out a derivation for rotation since it is everywhere, See Rotate the graph of a function?). Since the angle of the rotated pair $(x,y)$ is $\\angle C$ the following equations follow: $$\\cos(C)=\\frac{x\\cos B-y\\sin B}{((x\\cos B-y\\sin B)^2+(x\\sin B+y\\cos B)^2)^.5}$$ Notice that: $$(x\\cos B-y\\sin B)^2+(x\\sin B+y\\cos B)^2)= (x^2+y^2)$$ Hence, $$\\cos(C)=\\frac{x\\cos B-y\\sin B}{(x^2+y^2)^.5}$$ And $$\\cos(A+B)= {\\cos(A)\\cos(B)}-{\\sin(A)\\sin(B)}$$ Other Relations-------------- $$\\sin(C)=\\frac{x\\sin B+y\\cos B}{(x^2+y^2)^.5} = {\\cos(A)\\sin(B)}+{\\sin(A)\\cos(B)}$$ ## $$\\sin(A+B)= {\\cos(A)\\sin(B)}+{\\sin(A)\\cos(B)}$$ We also have $$A\\tan(y/x)= \\angle A$$ and $$A\\tan\\frac{x\\sin B+y\\cos B}{x\\cos B-y\\sin B}= \\angle C$$ $$\\angle B +A\\tan(y/x)= A\\tan\\frac{x\\sin B+y\\cos B}{x\\cos B-y\\sin B}$$ $$\\angle B +A\\tan(y/x)= A\\tan\\frac{\\tan B+y/x}{1-(y/x)\\tan B}$$ Or, for example, by setting $Tan B = u$ and $y/x = v$ $$A\\tan(u) +A\\tan(v)=", "satisfy the equality $\\mathbf{\\frac{\\sin^2A+\\sin^2B+\\sin^2C}{\\cos^2A+\\cos^2B+\\cos^2C} = 2}$ is right angled." ]

[ "# In Triangle Abc, Prove the Following: a − B a + B = Tan ( a − B 2 ) Tan ( a + B 2 ) - Mathematics In triangle ABC, prove the following: $\\frac{a - b}{a + b} = \\frac{\\tan \\left( \\frac{A - B}{2} \\right)}{\\tan \\left( \\frac{A + B}{2} \\right)}$ #### Solution $\\text{ Assume }\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = k$ Consider the LHS of the equation $\\frac{a - b}{a + b} = \\frac{\\tan \\left( \\frac{A - B}{2} \\right)}{\\tan \\left( \\frac{A + B}{2} \\right)}$ $LHS = \\frac{a - b}{a + b}$ $= \\frac{k\\left( \\sin A - \\sin B \\right)}{k\\left( \\sin A + \\sin B \\right)}$ $\\because \\sin A - \\sin B = 2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right), \\sin A + \\sin B = 2\\sin\\left( \\frac{A + B}{2} \\right)\\cos\\left( \\frac{A - B}{2} \\right)$ $\\therefore LHS = \\frac{2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}{2\\sin\\left( \\frac{A + B}{2} \\right)\\cos\\left( \\frac{A - B}{2} \\right)}$ $= \\frac{\\tan\\left( \\frac{A - B}{2} \\right)}{\\tan\\left( \\frac{A + B}{2} \\right)} = RHS$ $\\text{ Hence proved } .$ Concept: Sine and Cosine Formulae and Their Applications Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 10 Sine and cosine formulae and their applications Exercise 10.1 | Q", "If sin α + sin β = a and cos α - cos β = b, then the value of $\\tan \\left( {\\frac{{\\alpha - \\beta }}{2}} \\right)$ will be 21 views closed If sin α + sin β = a and cos α - cos β = b, then the value of $\\tan \\left( {\\frac{{\\alpha - \\beta }}{2}} \\right)$ will be 1. a + b 2. a - b 3. $- \\frac{a}{b}$ 4. $- \\frac{b}{a}$ by (53.1k points) selected Correct Answer - Option 4 : $- \\frac{b}{a}$ Concept: $\\sin c + \\sin D = 2\\sin \\left( {\\frac{{C + D}}{2}} \\right).\\cos \\frac{{C - D}}{2} \\to \\left( 1 \\right)$ $\\cos C - \\cos D = 2\\sin \\left( {\\frac{{C + D}}{2}} \\right).\\sin \\left( {\\frac{{D - C}}{2}} \\right) \\to \\left( 2 \\right)$ Calculation: Given, sin α + sin β = a and cos α – cos β = b Using (1) & (2) sin α + sin β = a $2\\sin \\left( {\\frac{{\\alpha + \\beta }}{2}} \\right).\\cos \\left( {\\frac{{\\alpha - \\beta }}{2}} \\right) = a$ -----(1) cos α – cos β = b $2\\sin \\left( {\\frac{{\\alpha + \\beta }}{2}} \\right).\\sin \\left( {\\frac{{\\beta - \\alpha }}{2}} \\right) = b$ ----- (2) On dividing (2) by (1) we get- $\\frac{{2\\sin \\left( {\\frac{{\\alpha + \\beta }}{2}} \\right).\\sin \\left( {\\frac{{\\beta - \\alpha }}{2}} \\right)}}{{2\\sin \\left( {\\frac{{\\alpha", "5x}=\\frac{-2\\sin \\frac{3x+5x}{2}\\sin \\frac{3x-5x}{2}}{2\\sin \\frac{3x+5x}{2}\\cos \\frac{3x-5x}{2}} \\\\ & =\\frac{-2\\sin \\frac{8x}{2}\\sin \\frac{-2x}{2}}{2\\sin \\frac{8x}{2}\\cos \\frac{-2x}{2}} \\\\ & =\\frac{-2\\sin 4x\\sin \\left( -x \\right)}{2\\sin 4x\\cos \\left( -x \\right)} \\\\ & =-\\tan \\left( -x \\right) \\end{align} Thus, the result will be $\\tan x$. Hence, verified. Also, consider the identity: $\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=-\\tan x$ And use the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=\\frac{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}}{2\\sin \\frac{3x+x}{2}\\cos \\frac{3x-x}{2}} \\\\ & =\\frac{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}}{2\\sin \\frac{4x}{2}\\cos \\frac{2x}{2}} \\\\ & =\\frac{-2\\sin 2x\\sin x}{2\\sin 2x\\cos x} \\\\ & =-\\tan x \\end{align} Hence verified. Also, consider the identity: $\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=-\\cot 2x$ By using the sum-to product formula $\\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2}$ to numerator and $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=\\frac{2\\sin \\frac{3x-x}{2}\\cos \\frac{3x+x}{2}}{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}} \\\\ & =\\frac{2\\sin \\frac{2x}{2}\\cos \\frac{4x}{2}}{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}} \\\\ & =-\\frac{\\cos 2x}{\\sin 2x} \\\\ & =-\\cot 2x \\end{align} Hence verified. And also consider the identity: $\\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2}$ Also, use the sum-to product formula", "\\tan v}{1 \\mp \\tan u \\tan v}$ Double Angle Formulas: • $\\sin (2 u) =2 \\sin u \\cos u$ • $\\cos (2 u) =\\cos ^{2} u-\\sin ^{2} u$ • $\\cos (2 u) =2 \\cos ^{2} u-1$ • $\\cos (2 u) =1-2 \\sin ^{2} u$ • $\\tan (2 u) =\\frac{2 \\tan u}{1-\\tan ^{2} u}$ Half Angle Formulas: • $\\sin^{2} u=\\frac{1-\\cos (2 u)}{2}$ • $\\cos^{2} u=\\frac{1+\\cos (2 u)}{2}$ • $\\tan^{2} u=\\frac{1-\\cos (2 u)}{1+\\cos (2 u)}$ Sum to product formulas: • $\\sin u+\\sin v=2 \\sin \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\sin u-\\sin v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ • $\\cos u+\\cos v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\cos u-\\cos v=-2 \\sin \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ Product to Sum Formulas: • $\\sin u \\sin v=\\frac{1}{2}[\\cos (u-v)-\\cos (u+v)]$ • $\\cos u \\cos v=\\frac{1}{2}[\\cos (u-v)+\\cos (u+v)]$ • $\\sin u \\cos v=\\frac{1}{2}[\\sin (u+v)+\\sin (u-v)]$ • $\\cos u \\sin v=\\frac{1}{2}[\\sin (u+v)-\\sin (u-v)]$", "# In Triangle Abc, Prove the Following: C a − B = Tan ( a 2 ) + Tan ( B 2 ) Tan ( a 2 ) − Tan ( B 2 ) - Mathematics In triangle ABC, prove the following: $\\frac{c}{a - b} = \\frac{\\tan\\left( \\frac{A}{2} \\right) + \\tan \\left( \\frac{B}{2} \\right)}{\\tan \\left( \\frac{A}{2} \\right) - \\tan \\left( \\frac{B}{2} \\right)}$ #### Solution Let $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = k$ ...(1) We need to prove: $\\frac{c}{a - b} = \\frac{\\tan\\left( \\frac{A}{2} \\right) + \\tan \\left( \\frac{B}{2} \\right)}{\\tan \\left( \\frac{A}{2} \\right) - \\tan \\left( \\frac{B}{2} \\right)}$ Consider $LHS = \\frac{c}{a - b}$ $= \\frac{k\\sin C}{k\\left( \\sin A - \\sin B \\right)} \\left( \\text{ using } \\left( 1 \\right) \\right)$ $= \\frac{2\\sin\\frac{C}{2}\\cos\\frac{C}{2}}{2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}$ $= \\frac{\\sin\\left( \\frac{\\pi - \\left( A + B \\right)}{2} \\right)\\cos\\frac{C}{2}}{\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)} \\left( \\because A + B + C = \\pi \\right)$ $= \\frac{\\cos\\frac{C}{2}\\cos\\left( \\frac{A + B}{2} \\right)}{\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}$ $= \\frac{\\cos\\frac{C}{2}}{\\sin\\left( \\frac{A - B}{2} \\right)} . . . \\left( 2 \\right)$ $RHS = \\frac{\\tan\\frac{A}{2} + \\tan\\frac{B}{2}}{\\tan\\frac{A}{2} - \\tan\\frac{B}{2}}$ $= \\frac{\\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}} + \\frac{\\sin\\frac{B}{2}}{\\cos\\frac{B}{2}}}{\\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}} - \\frac{\\sin\\frac{B}{2}}{\\cos\\frac{B}{2}}}$ $= \\frac{\\sin\\frac{A}{2}\\cos\\frac{B}{2} + \\sin\\frac{B}{2}\\cos\\frac{A}{2}}{\\sin\\frac{A}{2}\\cos\\frac{B}{2} - \\sin\\frac{B}{2}\\cos\\frac{A}{2}}$ $= \\frac{\\sin\\left( \\frac{A + B}{2} \\right)}{\\sin\\left( \\frac{A - B}{2} \\right)}$ $= \\frac{\\sin\\left(", "- \\sin a \\sin b}$ Divide throughout by $\\cos a \\cos b$ tan(a+b)=(((sinacosb)/(cosacosb))+((sinbcosa)/(cosacosb)))/((cosacosb)/(cosacosb)-(sinasinb)/(cosacosb) Simplifyng we get $\\tan \\left(a + b\\right) = \\frac{\\tan a + \\tan b}{1 - \\tan a \\tan b}$ Here we have $\\tan a = \\frac{1}{3}$ and $\\tan b = \\frac{1}{5}$ so $\\tan \\left(a + b\\right) = \\frac{\\frac{1}{3} + \\frac{1}{5}}{1 - \\left(\\frac{1}{3} \\cdot \\frac{1}{5}\\right)} = \\frac{\\frac{8}{15}}{\\frac{14}{15}} = \\frac{8}{14} = \\frac{4}{7}$ ${\\tan}^{-} 1 \\left(\\frac{4}{7}\\right) = {\\tan}^{-} 1 \\left(\\frac{1}{3}\\right) + {\\tan}^{-} 1 \\left(\\frac{1}{5}\\right)$", "FORMULA – In a ∆ABC , if a+b+c = 2s, then it’s area can given by: $$\\Delta = \\sqrt{s(s-a)(s-b)(s-c)}$$ ### CIRCUMCIRCLE OF A TRIANGLE In any triangle ABC, a).$$R = \\frac{a}{2sinA}= \\frac{b}{2sinB} =\\frac{ c}{2sinC}$$ b). $$R = \\frac{abc}{4\\Delta}$$ Everywhere means area ### INCIRCLE OF A TRIANGLE In any triangle ABC, a).$$r = \\frac{\\Delta}{s}$$ b). $$r = (s-a)tan\\left(\\frac{A}{2}\\right)= (s-b)tan\\left(\\frac{B}{2}\\right)= (s-c)tan\\left(\\frac{C}{2}\\right)$$ c).\\begin{align*} r& = \\frac{a.sin\\left(\\frac{B}{2}\\right).sin\\left(\\frac{C}{2}\\right)}{cos\\left(\\frac{A}{2}\\right)}\\\\ r &= \\frac{b.sin\\left(\\frac{A}{2}\\right).sin\\left(\\frac{C}{2}\\right)}{cos\\left(\\frac{B}{2}\\right)}\\\\ r& = \\frac{c.sin\\left(\\frac{B}{2}\\right).sin\\left(\\frac{A}{2}\\right)}{cos\\left(\\frac{C}{2}\\right)} \\end{align*} d).$$r= 4Rsin\\left(\\frac{A}{2}\\right).sin\\left(\\frac{B}{2}\\right).sin\\left(\\frac{C}{2}\\right)$$ ### EXCRIBED CIRCLE OF A TRIANGLE In any triangle ABC a). \\begin{align*} r_1& =\\frac{\\Delta}{s-a}\\\\ r_2&= \\frac{\\Delta}{s-b}\\\\ r_3&= \\frac{\\Delta}{s-c} \\end{align*} b). \\begin{align*} r1& = s.tan\\left(\\frac{A}{2}\\right)\\\\ r2& = s.tan\\left(\\frac{B}{2}\\right)\\\\ r3& = s.tan\\left(\\frac{C}{2}\\right) \\end{align*} c). \\begin{align*} r_1 = \\frac{a.cos\\left(\\frac{B}{2}\\right).cos\\left(\\frac{C}{2}\\right)}{cos\\left(\\frac{A}{2}\\right)}\\\\ r_2 = \\frac{b.cos\\left(\\frac{C}{2}\\right).cos\\left(\\frac{A}{2}\\right)}{cos\\left(\\frac{B}{2}\\right)}\\\\ r_3 = \\frac{c.cos\\left(\\frac{A}{2}\\right).cos\\left(\\frac{B}{2}\\right)}{cos\\left(\\frac{C}{2}\\right)} \\end{align*} d). \\begin{align*} r_1& = 4R.sin\\left(\\frac{A}{2}\\right).cos\\left(\\frac{B}{2}\\right).cos\\left(\\frac{C}{2}\\right)\\\\ r_2&= 4R.sin\\left(\\frac{B}{2}\\right).cos\\left(\\frac{A}{2}\\right).cos\\left(\\frac{C}{2}\\right)\\\\ r_3&= 4R.sin\\left(\\frac{C}{2}\\right).cos\\left(\\frac{A}{2}\\right).cos\\left(\\frac{B}{2}\\right) \\end{align*} ### ORTHOCENTRE OF A TRIANGLE The triangle formed by joining the points F, E, and D is called the pedal triangle. In any triangle ABC, a). The distance of the orthocentre from the vertices of the triangle ABC is- Distance from $A = 2RcosA$ Distance from $B = 2RcosB$ Distance from $C = 2RcosC$ b). Distance of the orthocentre from the sides of the triangle ABC is- $2RcosBcosC$ , $2RcosCcosA$ , $2RcosAcosB$ #### POINTS TO BE NOTED 1). The circumradius of a padel triangle is $$\\frac{R}{2}$$ 2). Area of a pedal triangle is", "opposite divided by hypotenuse). – André Nicolas Sep 23 '12 at 6:37 5). Use $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$ and note that $a+b=1$ 6). $\\sin A\\cos A=\\frac{\\sin A}{\\cos A}.\\cos^2 A=\\tan A \\frac{1}{1+\\tan^2 A}=\\frac{\\sqrt{2}-1}{1+(\\sqrt{2}-1)^2}=\\frac{\\sqrt{2}-1}{2\\sqrt{2}(\\sqrt{2}-1)}=\\frac{\\sqrt{2}}{4}$ -", "= \\left(\\frac{1 + \\tan A}{1 - \\tan A}\\right)^2$ L.H.S. $= \\frac{1 + \\sin 2A}{1 - \\sin 2A} = \\frac{\\sin^2A + \\cos^2A + 2\\sin A\\cos A}{\\sin^2A + \\cos^2A - 2\\sin A\\cos A}$ $= \\left(\\frac{\\sin A + \\cos A}{\\sin A - \\cos A}\\right)^2$ Dividing numerator and denominator by $\\cos^2A,$ we get $= \\left(\\frac{1 + \\tan A}{1 - \\tan A}\\right)^2 =$ R.H.S. 50. We have to prove that $\\frac{1}{\\sin 10^\\circ} - \\frac{\\sqrt{3}}{\\cos 10^\\circ} = 4$ L.H.S. $= \\frac{1}{\\sin 10^\\circ} - \\frac{\\sqrt{3}}{\\cos 10^\\circ}$", "$\\cos (2A + 2B) = \\frac{3}{4}$ and $\\cos (2B + 2C) = \\frac{1}{9}$. Substituting the identity $\\cos (2A + 2B) = \\cos(2C)$, we get: $\\cos (2C) = \\frac{3}{4}$ and $\\cos (2A) = \\frac{1}{9}$. Since the third expression simplifies to the expression $\\frac{3}{2} + \\frac{\\cos (2A + 2C)}{2}$, taking inverse cosine and using the angles in angle addition formula yields the answer, $\\frac{111 - 4\\sqrt{35}}{72}$, giving us the answer $\\boxed{222}$. ### Solution 5 We will use the sum to product formula to simply these equations. Recall $$2\\sin{\\frac{\\alpha+\\beta}{2}}\\sin{\\frac{\\alpha-\\beta}{2}} = \\cos{\\alpha}+\\cos{\\beta}.$$ Using this, let's rewrite the first equation: $$\\cos^2(A) + \\cos^2(B) + 2 \\sin(A) \\sin(B) \\cos(C) = \\frac{15}{8}$$ $$\\cos^2(A) + \\cos^2(B) + (\\cos(A+B)+\\cos(A-B))\\cos(C).$$ Now, note that $\\cos(C)=-\\cos(A+B)$. $$\\cos^2(A) + \\cos^2(B) + (\\cos(A+B)+\\cos(A-B))(-\\cos(A+B))$$ $$\\cos^2(A) + \\cos^2(B) - \\cos(A+B)\\cos(A-B)+cos^2(A+B)=\\frac{15}{8}.$$ We apply the sum to product formula again. $$\\cos^2(A) + \\cos^2(B) - \\frac{\\cos(2A)+\\cos(2B)}{2}+cos^2(A+B)=\\frac{15}{8}.$$ Now, recall that $\\cos(2\\alpha)=2\\cos^2(\\alpha)-1$. We apply this and simplify our expression to get: $$\\cos^2(A+B)=\\frac{7}{8}$$ $$\\cos^2(C)=\\frac{7}{8}.$$ Analogously, $$\\cos^2(A)=\\frac{5}{9}.$$ $$\\cos^2(A+C)=\\frac{p-q\\sqrt{r}}{s}-1.$$ We can find this value easily by angle sum formula. After a few calculations, we get $\\frac{111 - 4\\sqrt{35}}{72}$, giving us the answer $\\boxed{222}$. ~superagh ### Solution 6 According to LOC $a^2+b^2-2ab\\cos{\\angle{c}}=c^2$, we can write it into $\\sin^2{\\angle{A}}+\\sin^2{\\angle{B}}-2\\sin{\\angle{A}}\\sin{\\angle{B}}\\cos{\\angle{C}}=\\sin^2{\\angle{C}}$. $\\sin^2{\\angle{A}}+\\sin^2{\\angle{B}}-2\\sin{\\angle{A}}\\sin{\\angle{B}}\\cos{\\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\\frac{15}{8}+sin^2C$ We can simplify to $2=sin^2C+\\frac{15}{8}$. Similarly, we can generalize $2=sin^2A+\\frac{14}{9}$. After solving, we can get that $sinA=\\frac{2}{3}; cosA=\\frac{\\sqrt{5}}{3}; sinC=\\frac{\\sqrt{2}}{4}; cosC=\\frac{\\sqrt{14}}{4}$ Assume the value we", "= 3\\sqrt{3}$ Thus, A.M. of $\\tan A, \\tan B, \\tan C =$ G.M. of $\\tan A, \\tan B, \\tan C$ $\\Rightarrow \\tan A = \\tan B = \\tan C$ 10. L.H.S. $= (a + b + c)\\tan\\frac{C}{2} = 2R(\\sin A + \\sin B + \\sin C)\\frac{\\sin\\frac{C}{2}}{\\cos\\frac{C}{2}}$ $= 2R\\left(2\\sin\\frac{A + B}{2}\\cos\\frac{A - B}{2} + 2\\sin\\frac{C}{2}\\cos\\frac{C}{2}\\right)\\frac{\\sin\\frac{C}{2}}{\\cos\\frac{C}{2}}$ $= 2R\\left(2\\cos\\frac{C}{2}\\cos\\frac{A - B}{2} + 2\\sin\\frac{C}{2}\\cos\\frac{C}{2}\\right)\\frac{\\sin\\frac{C}{2}}{\\cos\\frac{C}{2}}$ $= 2R\\left(2\\sin\\frac{C}{2}\\cos\\frac{A - B}{2} + 2\\sin^2\\frac{C}{2}\\right)$ $= 2R\\left(2\\cos\\frac{A + B}{2}\\cos\\frac{A - B}{2} + 2\\sin^2\\frac{C}{2}\\right)$ $= 2R\\left(\\cos A + \\cos B + 2\\sin^2\\frac{C}{2}\\right)$ R.H.S $= a\\cot\\frac{A}{2} + b\\cot\\frac{B}{2} - c\\cot\\frac{C}{2}$ $= 2R\\left(\\sin A\\cot\\frac{A}{2} = \\sin B\\cot\\frac{B}{2} - \\sin C\\cot\\frac{C}{2}\\right)$ $= 2R\\left(2\\cos^2\\frac{A}{2} + 2\\cos^2\\frac{B}{2} - 2\\cos^2\\frac{C}{2}\\right)$ $= 2R\\left(2\\cos^2\\frac{A}{2} + 2\\cos^2\\frac{B}{2} - 2 + 2\\sin^2\\frac{C}{2}\\right)$ $= 2R\\left(\\cos A + \\cos B + 2\\sin^2\\frac{C}{2}\\right)$ Thus, L.H.S. = R.H.S. 11. $\\sin^2\\theta = \\frac{1 - \\cos2\\theta}{2} \\Rightarrow \\sin^4\\theta = \\frac{(1 - \\cos2\\theta)^2}{4}$ Also, for a triangle $\\cos 2A + \\cos 2B + \\cos 2C = -1 -4\\cos A\\cos B\\cos C$ and $\\cos^22A + \\cos^2B + \\cos^2C = 1 + 2\\cos 2A\\cos 2B\\cos 2C$ L.H.S. $= \\frac{(1 - \\cos2A)^2}{4} + \\frac{(1 - \\cos2B)^2}{4} + \\frac{(1 - \\cos 2C)^2}{4}$ $= \\frac{1}{4}[3 - 2(\\cos2A +\\cos 2B + \\cos 2C) + \\cos^22A + \\cos^22B + \\cos^22C]$ $= \\frac{1}{4}[3 - 2(-1 - 4\\cos A\\cos B\\cos C) + 1 + 2\\cos 2A\\cos 2B\\cos 2C]$ $= \\frac{3}{2} + 2\\cos A\\cos B\\cos", "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "$$\\left|\\,\\frac{\\sin(x)-x}{\\tan(x)-x}+\\frac12\\,\\right|\\le\\epsilon\\tag{6}$$ Thus, $$\\lim_{x\\to0}\\,\\frac{\\sin(x)-x}{\\tan(x)-x}=-\\frac12\\tag{7}$$ Furthermore, \\begin{align} \\frac{\\tan(x)-\\sin(x)}{x^3} &=\\tan(x)(1-\\cos(x))\\frac1{x^3}\\\\ &=\\frac{\\sin(x)}{\\cos(x)}\\frac{\\sin^2(x)}{1+\\cos(x)}\\frac1{x^3}\\\\ &=\\frac1{\\cos(x)(1+\\cos(x))}\\left(\\frac{\\sin(x)}{x}\\right)^3\\tag{8} \\end{align} Therefore, $$\\lim_{x\\to0}\\frac{\\tan(x)-\\sin(x)}{x^3}=\\frac12\\tag{9}$$ Combining $(7)$ and $(9)$ yield $$\\lim_{x\\to0}\\frac{x-\\sin(x)}{x^3}=\\frac16\\tag{10}$$ Additionally, $$\\frac{\\sin(A)-\\sin(B)}{\\sin(A-B)} =\\frac{\\cos\\left(\\frac{A+B}{2}\\right)}{\\cos\\left(\\frac{A-B}{2}\\right)} =1-\\frac{2\\sin\\left(\\frac{A}{2}\\right)\\sin\\left(\\frac{B}{2}\\right)}{\\cos\\left(\\frac{A-B}{2}\\right)}\\tag{11}$$ Finishing Up: \\begin{align} &x\\sin(\\sin(x))-\\sin^2(x)\\\\ &=[\\color{#C00000}{(x-\\sin(x))+\\sin(x)}][\\color{#00A000}{(\\sin(\\sin(x))-\\sin(x))+\\sin(x)}]-\\sin^2(x)\\\\ &=\\color{#C00000}{(x-\\sin(x))}\\color{#00A000}{(\\sin(\\sin(x))-\\sin(x))}\\\\ &+\\color{#C00000}{(x-\\sin(x))}\\color{#00A000}{\\sin(x)}\\\\ &+\\color{#C00000}{\\sin(x)}\\color{#00A000}{(\\sin(\\sin(x))-\\sin(x))}\\\\ &=(x-\\sin(x))(\\sin(\\sin(x))-\\sin(x))+\\sin(x)(x-2\\sin(x)+\\sin(\\sin(x)))\\tag{12} \\end{align} Using $(10)$, we get that \\begin{align} &\\lim_{x\\to0}\\frac{(x-\\sin(x))(\\sin(\\sin(x))-\\sin(x))}{x^6}\\\\ &=\\lim_{x\\to0}\\frac{x-\\sin(x)}{x^3}\\lim_{x\\to0}\\frac{\\sin(\\sin(x))-\\sin(x)}{\\sin^3(x)}\\lim_{x\\to0}\\left(\\frac{\\sin(x)}{x}\\right)^3\\\\ &=\\frac16\\cdot\\frac{-1}6\\cdot1\\\\ &=-\\frac1{36}\\tag{13} \\end{align} and with $(10)$ and $(11)$, we have \\begin{align} &\\lim_{x\\to0}\\frac{\\sin(x)(x-2\\sin(x)+\\sin(\\sin(x)))}{x^6}\\\\ &=\\lim_{x\\to0}\\frac{\\sin(x)}{x}\\lim_{x\\to0}\\frac{x-2\\sin(x)+\\sin(\\sin(x))}{x^5}\\\\ &=\\lim_{x\\to0}\\frac{(x-\\sin(x))-(\\sin(x)-\\sin(\\sin(x))}{x^5}\\\\ &=\\lim_{x\\to0}\\frac{(x-\\sin(x))-\\sin(x-\\sin(x))\\left(1-\\frac{2\\sin\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{\\sin(x)}{2}\\right)}{\\cos\\left(\\frac{x-\\sin(x)}{2}\\right)}\\right)}{x^5}\\\\ &=\\lim_{x\\to0}\\frac{(x-\\sin(x))-\\sin(x-\\sin(x))+\\sin(x-\\sin(x))\\frac{2\\sin\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{\\sin(x)}{2}\\right)}{\\cos\\left(\\frac{x-\\sin(x)}{2}\\right)}}{x^5}\\\\ &=\\lim_{x\\to0}\\frac{\\sin(x-\\sin(x))}{x^3}\\frac{2\\sin\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{\\sin(x)}{2}\\right)}{x^2}\\\\[6pt] &=\\frac16\\cdot\\frac12\\\\[6pt] &=\\frac1{12}\\tag{14} \\end{align} Adding $(13)$ and $(14)$ gives $$\\color{#C00000}{\\lim_{x\\to0}\\frac{x\\sin(\\sin(x))-\\sin^2(x)}{x^6}=\\frac1{18}}\\tag{15}$$ Added Explanation for the Derivation of $(6)$ The explanation below works for $x\\gt0$ and $x\\lt0$. Just reverse the red inequalities. Assume that $x\\color{#C00000}{\\gt}0$ and $|x|\\lt\\pi/2$. Then $\\tan(x)-2\\tan(x/2)\\color{#C00000}{\\gt}0$. $(3)$ is equivalent to \\begin{align} &(-1/2-\\epsilon)(\\tan(x)-2\\tan(x/2))\\\\[4pt] \\color{#C00000}{\\le}&\\sin(x)-2\\sin(x/2)\\\\[4pt] \\color{#C00000}{\\le}&(-1/2+\\epsilon)(\\tan(x)-2\\tan(x/2))\\tag{16} \\end{align} for all $|x|\\lt\\delta$. Thus, for $k\\ge0$, \\begin{align} &(-1/2-\\epsilon)(2^k\\tan(x/2^k)-2^{k+1}\\tan(x/2^{k+1}))\\\\[4pt] \\color{#C00000}{\\le}&2^k\\sin(x/2^k)-2^{k+1}\\sin(x/2^{k+1})\\\\[4pt] \\color{#C00000}{\\le}&(-1/2+\\epsilon)(2^k\\tan(x/2^k)-2^{k+1}\\tan(x/2^{k+1}))\\tag{17} \\end{align} Summing $(17)$ from $k=0$ to $\\infty$ yields \\begin{align} &(-1/2-\\epsilon)\\left(\\tan(x)-\\lim_{k\\to\\infty}2^k\\tan(x/2^k)\\right)\\\\[4pt] \\color{#C00000}{\\le}&\\sin(x)-\\lim_{k\\to\\infty}2^k\\sin(x/2^k)\\\\[4pt] \\color{#C00000}{\\le}&(-1/2+\\epsilon)\\left(\\tan(x)-\\lim_{k\\to\\infty}2^k\\tan(x/2^k)\\right)\\tag{18} \\end{align} Since $\\lim\\limits_{k\\to\\infty}2^k\\tan(x/2^k)=\\lim\\limits_{k\\to\\infty}2^k\\sin(x/2^k)=x$, $(18)$ says \\begin{align} &(-1/2-\\epsilon)(\\tan(x)-x)\\\\[4pt] \\color{#C00000}{\\le}&\\sin(x)-x\\\\[4pt] \\color{#C00000}{\\le}&(-1/2+\\epsilon)(\\tan(x)-x))\\tag{19} \\end{align} which, since $\\epsilon$ is arbitrary is equivalent to $(6)$. - (+1) Side note: it is easier to reinvent L'Hopitale's rule than come up with this solution. I bet that authors of the book didn't bother much about methods avaliable to students when they were suggesting this problem – Norbert Jul 7 '13 at 14:48 @Norbert It is rather naive to say that (essentially, sloppy or careless) about G. H. Hardy. – Andrés E. Caicedo Jul 7 '13 at 15:03 You like challenge, don't you?", "# If sin α + sin β = a and cos α − cos β = b, Question: If $\\sin \\alpha+\\sin \\beta=a$ and $\\cos \\alpha-\\cos \\beta=b$, then $\\tan \\frac{\\alpha-\\beta}{2}=$ (a) $-\\frac{a}{b}$ (b) $-\\frac{b}{a}$ (c) $\\sqrt{a^{2}+b^{2}}$ (d) None of these Solution: (b) $-\\frac{b}{a}$ Given: sin α + sin β = a .....(i) cos α − cos β = b .....(ii) Dividing (i) by (ii): $\\Rightarrow \\frac{\\sin \\alpha+\\sin B}{\\cos \\alpha-\\cos B}=\\frac{a}{b}$ $\\Rightarrow \\frac{2 \\sin \\left(\\frac{\\alpha+\\beta}{2}\\right) \\cos \\left(\\frac{\\alpha-\\beta}{2}\\right)}{-2 \\sin \\left(\\frac{\\alpha+\\beta}{2}\\right) \\sin \\left(\\frac{\\alpha-\\beta}{2}\\right)}=\\frac{a}{b}$ $\\left[\\because \\sin A+\\sin B=2 \\sin \\left(\\frac{A+B}{2}\\right) \\cos \\left(\\frac{A-B}{2}\\right)\\right.$ and $\\left.\\cos A+\\cos B=-2 \\sin \\left(\\frac{A+B}{2}\\right) \\sin \\left(\\frac{A-B}{2}\\right)\\right]$ $\\Rightarrow \\frac{\\sin \\left(\\frac{\\alpha+\\beta}{2}\\right) \\cos \\left(\\frac{\\alpha-\\beta}{2}\\right)}{-\\sin \\left(\\frac{\\alpha+\\beta}{2}\\right) \\sin \\left(\\frac{\\alpha-\\beta}{2}\\right)}=\\frac{a}{b}$ $\\Rightarrow \\cot \\left(\\frac{\\alpha-\\beta}{2}\\right)=-\\frac{a}{b}$ $\\Rightarrow \\frac{1}{\\cot \\left(\\frac{\\alpha-\\beta}{2}\\right)}=\\frac{1}{-\\frac{a}{b}}$ $\\Rightarrow \\tan \\left(\\frac{\\alpha-\\beta}{2}\\right)=-\\frac{b}{a}$", "Solve that equation Question: $\\tan ^{-1}\\left(\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\right),-\\frac{\\pi}{4}<x<\\frac{\\pi}{4}$ Solution: Let $y=\\tan ^{-1}\\left[\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\right]$ $=\\tan ^{-1}\\left[\\sqrt{\\frac{2 \\sin ^{2} x / 2}{2 \\cos ^{2} x / 2}}\\right]\\left[\\begin{array}{r}\\because 1-\\cos x=2 \\sin ^{2} x / 2 \\\\ 1+\\cos x=2 \\cos ^{2} x / 2\\end{array}\\right]$ $=\\tan ^{-1}\\left[\\frac{\\sin x / 2}{\\cos x / 2}\\right]=\\tan ^{-1}\\left[\\tan \\frac{x}{2}\\right]$ Thus, $y=\\frac{x}{2}$ Differentiating both sides w.r.t. $x$ $\\frac{d y}{d x}=\\frac{1}{2} \\frac{d}{d x}(x)=\\frac{1}{2} \\cdot 1=\\frac{1}{2}$ Hence, $\\frac{d y}{d x}=\\frac{1}{2}$", "+ \\tan B + \\tan C = \\tan A\\tan B\\tan C]$ $= 2.a\\tan B\\tan C\\sec A =$ R.H.S. 30. We have to prove that $(b - c)\\cos\\frac{A}{2} = a\\sin\\frac{B - C}{2}$ $\\frac{b - c}{a} = \\frac{\\sin B - \\sin C}{\\sin A} = \\frac{2\\cos\\frac{B + C}{2}\\sin \\frac{B - C}{2}}{2\\sin\\frac{A}{2}\\cos\\frac{A}{2}}$ $= \\frac{2\\sin\\frac{A}{2}\\sin\\frac{B - C}{2}}{2\\sin\\frac{A}{2}\\cos\\frac{A}{2}} = \\frac{\\sin\\frac{B - C}{2}}{\\cos\\frac{A}{3}}$ 31. We have to prove that $\\tan\\left(\\frac{A}{2} + B\\right) = \\frac{c + b}{c - b}\\tan \\frac{A}{2}$ $\\frac{c + b}{c - b} = \\frac{\\sin C + \\sin B}{\\sin C - \\sin B}$ $= \\frac{2\\sin\\frac{B + C}{2}\\cos\\frac{C - B}{2}}{2\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2}}$ $= \\frac{\\tan\\frac{B + C}{2}}{\\tan\\frac{C - B}{2}} = \\frac{\\cot\\frac{A}{2}}{\\tan\\frac{\\pi - B - A - B}{2}}$ $= \\frac{\\tan\\left(\\frac{A}{2} + B\\right)}{\\tan\\frac{A}{3}}$ 32. We have to prove that $\\tan\\frac{A - B}{2} = \\frac{a - b}{a + b}\\cot\\frac{C}{2}$ $\\frac{a - b}{a + b} = \\frac{\\sin A - \\sin B}{\\sin A + \\sin B} = \\frac{2\\cos\\frac{A + B}{2}\\sin\\frac{A - B}{2}}{2\\sin\\frac{A + B}{2}\\cos\\frac{A - B}{2}}$ $= \\frac{\\tan\\frac{A - B}{2}}{\\tan \\frac{A + B}{2}} = \\frac{\\tan\\frac{A - B}{2}}{\\cot\\frac{C}{2}}$ 33. L.H.S. $= (b + c)\\cos A + (c + a)\\cos B + (a + b)\\cos C$ $= (a\\cos B + b\\cos A) + (b\\cos C + c\\cos B) + (a\\cos C + c\\cos A)$ $= c + a + b =$ R.H.S. 34. First term of L.H.S. $= \\frac{\\cos^2B - \\cos^2C}{b", "x^2)(1 - y^2) = 4xyz$$ Dividing both sides with $$(1 - x^2)(1 - y^2)(1 - z^2),$$ we get $$\\frac{x}{1 - x^2} + \\frac{y}{1 - y^2} + \\frac{z}{1 - z^2} = \\frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}$$ L.H.S. $$= \\frac{x}{1 - x^2} + \\frac{y}{1 - y^2} + \\frac{z}{1 - z^2} = \\frac{1}{2}[\\tan 2A + \\tan 2B + \\tan 2C]$$ $$= \\frac{1}{2}\\tan2A\\tan2B\\tan2C = \\frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}$$ 56. L.H.S. $$= (\\cos A + \\cos B) + (\\cos C + \\cos D)$$ $$= 2\\cos\\frac{A + B}{2}\\cos\\frac{A - B}{2} + 2\\cos\\frac{C + D}{2}\\cos\\frac{C - D}{2}$$ $$= 2\\cos\\frac{A + B}{2}\\cos\\frac{A - B}{2} + 2\\cos\\left(\\pi - \\frac{A + B}{2}\\right)\\cos\\frac{C - D}{2}$$ $$= 2\\cos\\frac{A + B}{2}\\cos\\frac{A - B}{2} - 2\\cos\\frac{A + B}{2}\\cos\\frac{C - D}{2}$$ $$= 2\\cos\\frac{A + B}{2}\\left[\\cos\\frac{A - B}{2} - \\cos\\frac{C - D}{2}\\right]$$ $$= 2\\cos\\frac{A + B}{2}.2\\sin\\frac{A - B + C - D}{4}\\sin\\frac{C - D - A + B}{4}$$ $$= 4\\cos\\frac{A + B}{2}\\sin\\frac{A + C - (B + C)}{4}\\sin\\frac{B + C - (A + D)}{4}$$ $$= 4\\cos\\frac{A + B}{2}\\sin\\frac{A + C -(2\\pi - A - C)}{4}\\sin\\frac{B + C - (2\\pi - B - C)}{4}$$ $$= 4\\cos\\frac{A + B}{2}\\sin\\frac{A + C - \\pi}{2}\\sin\\frac{B + C - \\pi}{2}$$ $$= 4\\cos\\frac{A + B}{2}\\cos\\frac{B + C}{2}\\cos\\frac{C + A}{2}$$ 57. L.H.S. $$= \\cos^2S + \\cos^2(S - A) + \\cos^2(S - B) + \\cos^2(S - C)$$", "\\cos a \\sin^2 b & = \\cos b \\sin a \\sin b \\cos C + \\sin b \\sin C \\sin a \\cot A. \\end$ The result follows on dividing by $\\sin a\\sin b$. Similar techniques with the other two cosine rules give CT3 and CT5. The other three equations follow by applying rules 1, 3 and 5 to the polar triangle. Half-angle and half-side formulae With $2s=\\left(a+b+c\\right)$ and $2S=\\left(A+B+C\\right)$, ::$\\begin & \\sinA=\\leftfrac\\right&\\qquad &\\sina=\\leftfrac\\right\\\\ex& \\cosA=\\leftfrac\\right&\\qquad &\\cosa=\\leftfrac\\right\\\\ex& \\tanA=\\leftfrac\\right&\\qquad &\\tana=\\leftfrac\\right\\end$ Another twelve identities follow by cyclic permutation. The proof (Todhunter, Art.49) of the first formula starts from the identity 2sin2(''A''/2) = 1–cos''A'', using the cosine rule to express ''A'' in terms of the sides and replacing the sum of two cosines by a product. (See sum-to-product identities.) The second formula starts from the identity 2cos2(''A''/2) = 1+cos''A'', the third is a quotient and the remainder follow by applying the results to the polar triangle. Delambre (or Gauss) analogies ::$\\begin &\\\\ \\frac =\\frac &\\qquad\\qquad & \\frac =\\frac \\\\ex\\frac =\\frac &\\qquad & \\frac =\\frac \\end$ Another eight identities follow by cyclic permutation. Proved by expanding the numerators and using the half angle formulae. (Todhunter, Art.54 and Delambre) Napier's analogies ::$\\begin &&\\\\2exdisplaystyle =\\frac \\cot &\\qquad & =\\frac \\tan \\\\ex=\\frac \\cot &\\qquad & =\\frac \\tan \\end$ Another eight identities follow by cyclic permutation. These identities", "2. $2\\sin B\\cos A=\\sin \\left( A+B \\right)-\\sin \\left( A-B \\right)$ 3. $2\\cos A\\cos B=\\cos \\left( A+B \\right)+\\cos \\left( A-B \\right)$ 4. $2\\sin A\\sin B=\\cos \\left( A-B \\right)-\\cos \\left( A+B \\right)$ 5. $\\sin C+\\sin D=2\\sin \\left( \\frac{C+1}{2} \\right)\\cos \\left( \\frac{C-D}{2} \\right)$ 6. $\\sin C-\\sin D=2\\cos \\left( \\frac{C+D}{2} \\right)\\sin \\left( \\frac{C-D}{2} \\right)$ 7. $\\cos C-\\cos D=2\\cos \\left( \\frac{C+D}{2} \\right)\\cos \\left( \\frac{C-D}{2} \\right)$ 8. $\\cos C-\\cos D=2\\sin \\left( \\frac{C+D}{2} \\right)\\sin \\left( \\frac{D-C}{2} \\right)$ ### T-Ratio Of Multiple Angles 1. $\\sin 2A=2\\sin A\\cos A=\\frac{2\\tan A}{1+{{\\tan }^{2}}A}$ 2. $\\cos 2A=2{{\\cos }^{2}}A-1=1-2{{\\sin }^{2}}A=\\frac{1-{{\\tan }^{2}}A}{1+{{\\tan }^{2}}A}={{\\cos }^{2}}A-{{\\sin }^{2}}A$ 3. $\\tan 2A=\\frac{2\\tan A}{1-{{\\tan }^{2}}A}$ 4. $\\sin 3A=3\\sin A-4{{\\sin }^{3}}A$ 5. $\\cos 3A=4{{\\cos }^{3}}A-3\\cos A$ 6. $\\tan 3A=\\frac{3\\tan A-{{\\tan }^{3}}A}{1-3{{\\tan }^{2}}A}$ Remember: $1+\\cos 2A=2{{\\cos }^{2}}A$ $1-\\cos 2A=2{{\\sin }^{2}}A$ $\\frac{1-\\cos 2A}{1+\\cos 2A}={{\\tan }^{2}}A$ #### Values Of T-Ratios At Some Useful Angles 1. $\\sin 75{}^\\circ =\\frac{\\sqrt{3}+1}{2\\sqrt{2}}=\\cos 15{}^\\circ$ 2. $\\cos 75{}^\\circ =\\frac{\\sqrt{3}-1}{2\\sqrt{2}}=\\sin 15{}^\\circ$ 3. $\\tan 75{}^\\circ =2+\\sqrt{3}=\\cot 15{}^\\circ$ 4. $\\cot 75{}^\\circ =2-\\sqrt{3}=\\tan 15{}^\\circ$ 5. $\\sin 9{}^\\circ =\\frac{\\sqrt{3+\\sqrt{5}}-\\sqrt{5-\\sqrt{5}}}{4}=\\cos 81{}^\\circ$ 6. $\\cos 9{}^\\circ =\\frac{\\sqrt{3+\\sqrt{5}}-\\sqrt{5-\\sqrt{5}}}{4}=\\sin 81{}^\\circ$ Important Results: $\\sin \\theta \\sin \\left( 60-\\theta \\right)\\sin \\left( 60+\\theta \\right)=\\frac{1}{4}\\sin \\theta$ $\\cos \\theta \\cos \\left( 60-\\theta \\right)\\cos \\left( 60+\\theta \\right)=\\frac{1}{4}\\cos \\theta$ $\\tan \\theta \\tan \\left( 60-\\theta \\right)\\tan \\left( 60+\\theta \\right)=\\tan 3\\theta$ Example: If $A+B=225{}^\\circ$ $\\left( \\frac{\\cot A}{1+\\cot A} \\right)\\left( \\frac{\\cot B}{1+\\cot B} \\right)=?$. Solution: $A+B=225=180+45{}^\\circ$ $\\tan \\left( A+B \\right)=\\tan \\left( \\bar{\\wedge }+45{}^\\circ \\right)$ $\\tan \\left( A+B", "\\frac{2\\pi}7 + \\cos \\frac{4\\pi}7 + \\cos \\frac{6\\pi}7\\right) = -1,$$ from which $$\\cos \\frac{2\\pi}7 + \\cos \\frac{4\\pi}7 + \\cos \\frac{6\\pi}7 = -\\frac12.$$ Therefore, we get $a = -\\left(-\\frac12\\right) = \\frac12.$ 2. Solve for $b:$ By Vieta's Formulas, we have $b = \\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 + \\cos \\frac{2\\pi}7 \\cos \\frac{6\\pi}7 + \\cos \\frac{4\\pi}7 \\cos \\frac{6\\pi}7.$ Note that $\\cos \\alpha \\cos \\beta = \\frac{ \\cos \\left(\\alpha + \\beta\\right) + \\cos \\left(\\alpha - \\beta\\right) }{2}$ for all $\\alpha$ and $\\beta.$ Therefore, we get $$b=\\frac{\\cos \\frac{6\\pi}7 + \\cos \\frac{2\\pi}7}2 + \\frac{\\cos \\frac{4\\pi}7 + \\cos \\frac{4\\pi}7}2 + \\frac{\\cos \\frac{6\\pi}7 + \\cos \\frac{2\\pi}7}2=\\cos \\frac{2\\pi}7 + \\cos \\frac{4\\pi}7 + \\cos \\frac{6\\pi}7=-\\frac12.$$ 3. Solve for $c:$ By Vieta's Formulas, we have $c = -\\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7.$ We multiply both sides by $8 \\sin{\\frac{2\\pi}{7}},$ then repeatedly apply the angle addition formula for sine: \\begin{align*} c \\cdot 8 \\sin{\\frac{2\\pi}{7}} &= -8 \\sin{\\frac{2\\pi}{7}} \\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7 \\\\ &= -4 \\sin \\frac{4\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7 \\\\ &= -2 \\sin \\frac{8\\pi}7 \\cos \\frac{8\\pi}7 \\\\ &= -\\sin \\frac{16\\pi}7 \\\\ &= -\\sin \\frac{2\\pi}7. \\end{align*} Therefore, we get $c = -\\frac18.$ Finally, the answer is $abc=\\frac12\\cdot\\left(-\\frac12\\right)\\cdot\\left(-\\frac18\\right)=\\boxed{\\textbf{(D) }\\frac{1}{32}}.$ ~Tucker (Solution) ~MRENTHUSIASM (Reformatting) ## Solution 4 (Product-to-Sum Identity) Note that the sum of the roots of unity equal zero, so the sum of their real parts equal zero, and" ]

[ "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "# In Triangle Abc, Prove the Following: a − B a + B = Tan ( a − B 2 ) Tan ( a + B 2 ) - Mathematics In triangle ABC, prove the following: $\\frac{a - b}{a + b} = \\frac{\\tan \\left( \\frac{A - B}{2} \\right)}{\\tan \\left( \\frac{A + B}{2} \\right)}$ #### Solution $\\text{ Assume }\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = k$ Consider the LHS of the equation $\\frac{a - b}{a + b} = \\frac{\\tan \\left( \\frac{A - B}{2} \\right)}{\\tan \\left( \\frac{A + B}{2} \\right)}$ $LHS = \\frac{a - b}{a + b}$ $= \\frac{k\\left( \\sin A - \\sin B \\right)}{k\\left( \\sin A + \\sin B \\right)}$ $\\because \\sin A - \\sin B = 2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right), \\sin A + \\sin B = 2\\sin\\left( \\frac{A + B}{2} \\right)\\cos\\left( \\frac{A - B}{2} \\right)$ $\\therefore LHS = \\frac{2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}{2\\sin\\left( \\frac{A + B}{2} \\right)\\cos\\left( \\frac{A - B}{2} \\right)}$ $= \\frac{\\tan\\left( \\frac{A - B}{2} \\right)}{\\tan\\left( \\frac{A + B}{2} \\right)} = RHS$ $\\text{ Hence proved } .$ Concept: Sine and Cosine Formulae and Their Applications Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 10 Sine and cosine formulae and their applications Exercise 10.1 | Q", "proved as follows. As per law of sines, $\\frac{a}{\\sin A}$ = $\\frac{b}{\\sin B}$. Let this ratio be k. Then, a = k*sin (A) and b = k*sin (B) Therefore, $\\left [ \\frac{(a-b)}{(a+b)} \\right ]$ = $\\frac{[k*\\sin (A)-k*\\sin (B)]}{[k*\\sin (A)+k*\\sin (B)]}$ = $\\frac{[\\sin (A)-\\sin (B)]}{[\\sin (A)+\\sin (B)]}$ As per the formula for sum and difference of two sine ratios, sin (A) - sin (B) = 2 sin $\\frac{1}{2}$ (A – B)*cos $\\frac{1}{2}$ (A + B) and sin (A) + sin (B) = 2 sin $\\frac{1}{2}$ (A + B)*cos $\\frac{1}{2}$ (A - B) Therefore, $\\left [ \\frac{(a-b)}{(a+b)} \\right ]$ = $\\frac{[2\\sin \\frac{1}{2}(A-B)*\\cos \\frac{1}{2}(A+B)]}{2\\sin \\frac{1}{2}(A+B)*\\cos \\frac{1}{2}(A-B)}$ =$\\frac{[\\tan \\frac{1}{2}(A-B)]}{[\\tan \\frac{1}{2}(A+B)]}$ Thus the law of tangent is proved. This law is more convenient than the law of cosine to solve a triangle when only two sides and the included angle is known. It is explained below. The above formula can be rewritten as, $\\tan$ $\\frac{1}{2}$ (A – B) = $[\\frac{(a – b)}{(a + b)}]$ * $\\tan$ $\\frac{1}{2}$ (A + B) = $[\\frac{(a – b)}{(a + b)}]$ * $\\cot$ $\\frac{1}{2}$ (C)Since we know all the items on the right, (A – B) can be figured out. We also know (A + B), since (A + B) = 180o – (C) By solving we can find the angles A and B. The remaining side c", "Ucale # Sum of sines / cosine in terms of products Values of sin(A+B) and cos(A+B) in terms of their product is given by 1. $\\displaystyle \\sin { A } +\\sin { B } =2\\sin { \\left( \\frac { A+B }{ 2 } \\right) } \\cos { \\left( \\frac { A-B }{ 2 } \\right) }$ 2. $\\displaystyle \\sin { A } -\\sin { B } =2\\cos { \\left( \\frac { A+B }{ 2 } \\right) } \\sin { \\left( \\frac { A-B }{ 2 } \\right) }$ 3. $\\displaystyle \\cos { A } +\\cos { B } =2\\cos { \\left( \\frac { A-B }{ 2 } \\right) } \\cos { \\left( \\frac { A-B }{ 2 } \\right) }$ 4. $\\displaystyle \\cos { A } -\\cos { B } =2\\cos { \\left( \\frac { A+B }{ 2 } \\right) } \\cos { \\left( \\frac { B-A }{ 2 } \\right) }$ 5. $\\displaystyle \\tan { A } +\\tan { B } =\\frac { \\sin { \\left( A+B \\right) } }{ \\cos { A } \\cos { B } } \\quad where\\quad A,B\\neq n\\pi +\\frac { \\pi }{ 2 }$ 6. $\\displaystyle \\tan { A } -\\tan { B } =\\frac { \\sin { \\left( A-B \\right) } }{ \\cos { A } \\cos { B } }", "- b^2$$ 44. If $$\\sin A = \\frac{1}{2}$$ and $$\\sin B = \\frac{1}{3},$$ find the value of $$\\sin(A + B)$$ and $$\\sin(2A + 2B)$$ 45. If $$\\cos A = \\frac{11}{61}$$ and $$\\sin B = \\frac{4}{5},$$ find the value of $$\\sin^2 \\frac{A - B}{2}$$ and $$cos^2\\frac{A + B}{2},$$ the angle of $$A$$ and $$B$$ being positive acute angles. 46. Given $$\\sec A = \\frac{5}{4},$$ find $$\\tan\\frac{A}{2}$$ and $$\\tan A.$$ 47. If $$\\cos A = .3,$$ find the value of $$\\tan \\frac{A}{2},$$ and explain the resulting ambiguity. 48. If $$\\sin A + \\sin B = x$$ and $$\\cos A + \\cos B = y,$$ find the value of $$\\tan \\frac{A - B}{2}$$ Prove that 1. $$(\\cos A + \\cos B)^2 + (\\sin A - \\sin B)^2 = 4\\cos^2 \\frac{A + B}{2}$$ 2. $$(\\cos A + \\cos B)^2 + (\\sin A + \\sin B)^2 = 4\\cos^2 \\frac{A - B}{2}$$ 3. $$(\\cos A - \\cos B)^2 + (\\sin A - \\sin B)^2 = 4\\sin^2 \\frac{A - B}{2}$$ 4. $$\\sin^2\\left(\\frac{\\pi}{8} + \\frac{A}{2}\\right) - \\sin^2\\left(\\frac{\\pi}{8} -\\frac{A}{2}\\right) = \\frac{1}{\\sqrt{2}}\\sin A$$ 5. $$(\\tan 4A + \\tan 2A)(1 - \\tan^23A\\tan^2A) = 2\\tan 3A\\sec^2A$$ 6. $$\\left(1 + \\tan \\frac{A}{2} - \\sec\\frac{A}{2}\\right)\\left(1 + \\tan \\frac{A}{2} + \\sec\\frac{A}{2}\\right) = \\sin A\\sec^2\\frac{A}{2}$$ 7. $$\\frac{1 + \\sin A - \\cos A}{1 + \\sin A + \\cos A} = \\tan \\frac{A}{2}$$ 8. $$\\frac{1", "• # question_answer Let A, B and C are the angles of a plain triangle and $\\tan \\frac{A}{2}=\\frac{1}{3},\\,\\,\\tan \\frac{B}{2}=\\frac{2}{3}$. Then $\\tan \\frac{C}{2}$ is equal to [Orissa JEE 2003] A) 7/9 B) 2/9 C) 1/3 D) 2/3 $A+B+C=\\pi$ $\\therefore \\,\\,\\,\\tan \\left( \\frac{A+B}{2} \\right)=\\tan \\left( \\frac{\\pi }{2}-\\frac{C}{2} \\right)$ Þ $\\frac{\\tan \\frac{A}{2}+\\tan \\frac{B}{2}}{1-\\tan \\frac{A}{2}.\\tan \\frac{B}{2}}=\\cot \\frac{C}{2}\\Rightarrow \\frac{\\frac{1}{3}+\\frac{2}{3}}{1-\\frac{1}{3}.\\frac{2}{3}}=\\frac{9}{7}=\\cot \\frac{C}{2}$ \\ $\\tan \\frac{C}{2}=\\frac{7}{9}$.", "\\tan v}{1 \\mp \\tan u \\tan v}$ Double Angle Formulas: • $\\sin (2 u) =2 \\sin u \\cos u$ • $\\cos (2 u) =\\cos ^{2} u-\\sin ^{2} u$ • $\\cos (2 u) =2 \\cos ^{2} u-1$ • $\\cos (2 u) =1-2 \\sin ^{2} u$ • $\\tan (2 u) =\\frac{2 \\tan u}{1-\\tan ^{2} u}$ Half Angle Formulas: • $\\sin^{2} u=\\frac{1-\\cos (2 u)}{2}$ • $\\cos^{2} u=\\frac{1+\\cos (2 u)}{2}$ • $\\tan^{2} u=\\frac{1-\\cos (2 u)}{1+\\cos (2 u)}$ Sum to product formulas: • $\\sin u+\\sin v=2 \\sin \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\sin u-\\sin v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ • $\\cos u+\\cos v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\cos u-\\cos v=-2 \\sin \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ Product to Sum Formulas: • $\\sin u \\sin v=\\frac{1}{2}[\\cos (u-v)-\\cos (u+v)]$ • $\\cos u \\cos v=\\frac{1}{2}[\\cos (u-v)+\\cos (u+v)]$ • $\\sin u \\cos v=\\frac{1}{2}[\\sin (u+v)+\\sin (u-v)]$ • $\\cos u \\sin v=\\frac{1}{2}[\\sin (u+v)-\\sin (u-v)]$", "# Trigonometry Compund Angles Problem I am studying for my trigonometry examination. But I cannot figure out how to this sum from the chapter \"Compound and Multiple Angles\". I am in the eleventh standard. The sum goes like this : $\\sin x + \\sin y = a$ and $\\cos x+\\cos y=b$. Find $\\tan \\frac{x+y}{2}$ Any help will be appreciated. - Recall the two (well, basically one) identities for the sine of a sum and the sine of a difference. We have $$\\sin(a+b)=\\sin a\\cos b+\\cos a\\sin b,$$ and $$\\sin(a-b)=\\sin a\\cos b-\\cos a\\sin b.$$ Adding we get $$\\sin(a+b)+\\sin(a-b)=2\\sin a \\cos b.$$ Put $x=a+b$ and $y=a-b$. Then $a=\\frac{x+y}{2}$ and $b=\\frac{x-y}{2}$. It follows that $$\\sin x+\\sin y=2\\sin\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x-y}{2}\\right).\\tag{1}$$ Use the addition/subtraction laws for cosine to get a similar expression for $\\cos x+\\cos y$, and it will be almost over. Remark: The identity (1) may already be part of your list of standard identities, and so may the corresponding identity about $\\cos x+\\cos y$. In that case, the calculation of $\\tan\\left(\\frac{x+y}{2}\\right)$ in terms of $a$ and $b$ can be done in one line.", "Proving $\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$ Prove $$\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$$ Can someone help? I separated $\\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves. 2 Answers Use the formulas $$\\sin a+\\sin b=2\\sin\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right),$$ $$\\cos a+\\cos b=2\\cos\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right).$$ Therefore, we have $$\\frac{\\sin a+\\sin b}{\\cos a+\\cos b}=\\frac{\\sin\\displaystyle\\left(\\frac{a+b}{2}\\right)}{\\cos\\displaystyle\\left(\\frac{a+b}{2}\\right)}=\\tan\\left(\\frac{a+b}{2}\\right).$$ If $$a,b$$ are acute angles one can proceed as follows: Draw a parallelogram with all four sides of length $$1,$$ as follows. The first side has one endpoint at $$(0,0)$$ and the angle between it and the $$x$$-axis is $$a$$. Its other endpoint is therefore at $$(\\cos a,\\sin a).$$ The second side begins at that latter point and its angle with the $$x$$-axis is $$b.$$ Its other endpoint is therefore $$(\\sin a + \\sin b, \\cos a+\\cos b).$$ The tangent of the angle from the origin to this last point is therefore $$(\\sin a+\\sin b)/(\\cos a + \\cos b).$$ But the geometric symmetry of the parallelogram shows that that angle is $$(a+b)/2.$$", "the cosine of the angle between them, namely $$\\cos a$$, from which we obtain immediately $\\cos a = \\cos b \\cos c + \\sin b \\sin c \\cos A. \\label{3.5.9} \\tag{3.5.9}$ To obtain the sine formula, we isolate $$\\cos A$$ from this equation, square both sides, and write $$1 − \\sin^2 A$$ for $$\\cos^2 A$$. Thus, $(\\sin b \\sin c \\cos A )^2 = (\\cos a - \\cos b \\cos c )^2 , \\label{3.5.10} \\tag{3.5.10}$ and when we have carried out these operations we obtain $\\sin^2 A = \\frac{\\sin^2 b \\sin^2 c - \\cos^2 a - \\cos^2 b \\cos^2 c + 2\\cos a \\cos b \\cos c}{\\sin^2 b \\sin^2 c} . \\label{3.5.11} \\tag{3.5.11}$ In the numerator, write $$1 − \\cos^2 b$$ for $$\\sin^2 b$$ and $$1 − \\cos^2 c$$ for $$\\sin^2 c$$, and divide both sides by $$\\sin^2 a$$. This results in $\\frac{\\sin^2 A}{\\sin^2 a} = \\frac{1-\\cos^2 a - \\cos^2 b - \\cos^2 c + 2\\cos a \\cos b \\cos c}{\\sin^2 a \\sin^2 b \\sin^2 c}. \\label{3.5.12} \\tag{3.5.12}$ At this stage the reader may feel that we are becoming bogged down in heavier and heavier algebra and getting nowhere. But, after a careful look at equation $$\\ref{3.5.12}$$, it may be noted with some delight that the next line is: Therefore $\\frac{\\sin A}{\\sin a} = \\frac{\\sin B}{\\sin b} =", "# In Triangle Abc, Prove the Following: C a − B = Tan ( a 2 ) + Tan ( B 2 ) Tan ( a 2 ) − Tan ( B 2 ) - Mathematics In triangle ABC, prove the following: $\\frac{c}{a - b} = \\frac{\\tan\\left( \\frac{A}{2} \\right) + \\tan \\left( \\frac{B}{2} \\right)}{\\tan \\left( \\frac{A}{2} \\right) - \\tan \\left( \\frac{B}{2} \\right)}$ #### Solution Let $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = k$ ...(1) We need to prove: $\\frac{c}{a - b} = \\frac{\\tan\\left( \\frac{A}{2} \\right) + \\tan \\left( \\frac{B}{2} \\right)}{\\tan \\left( \\frac{A}{2} \\right) - \\tan \\left( \\frac{B}{2} \\right)}$ Consider $LHS = \\frac{c}{a - b}$ $= \\frac{k\\sin C}{k\\left( \\sin A - \\sin B \\right)} \\left( \\text{ using } \\left( 1 \\right) \\right)$ $= \\frac{2\\sin\\frac{C}{2}\\cos\\frac{C}{2}}{2\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}$ $= \\frac{\\sin\\left( \\frac{\\pi - \\left( A + B \\right)}{2} \\right)\\cos\\frac{C}{2}}{\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)} \\left( \\because A + B + C = \\pi \\right)$ $= \\frac{\\cos\\frac{C}{2}\\cos\\left( \\frac{A + B}{2} \\right)}{\\sin\\left( \\frac{A - B}{2} \\right)\\cos\\left( \\frac{A + B}{2} \\right)}$ $= \\frac{\\cos\\frac{C}{2}}{\\sin\\left( \\frac{A - B}{2} \\right)} . . . \\left( 2 \\right)$ $RHS = \\frac{\\tan\\frac{A}{2} + \\tan\\frac{B}{2}}{\\tan\\frac{A}{2} - \\tan\\frac{B}{2}}$ $= \\frac{\\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}} + \\frac{\\sin\\frac{B}{2}}{\\cos\\frac{B}{2}}}{\\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}} - \\frac{\\sin\\frac{B}{2}}{\\cos\\frac{B}{2}}}$ $= \\frac{\\sin\\frac{A}{2}\\cos\\frac{B}{2} + \\sin\\frac{B}{2}\\cos\\frac{A}{2}}{\\sin\\frac{A}{2}\\cos\\frac{B}{2} - \\sin\\frac{B}{2}\\cos\\frac{A}{2}}$ $= \\frac{\\sin\\left( \\frac{A + B}{2} \\right)}{\\sin\\left( \\frac{A - B}{2} \\right)}$ $= \\frac{\\sin\\left(", "can be easily solved. Integrals of this sort can be done using some useful identities that are immediate corollaries of the angle-sum formulas $\\cos A \\cos B-\\sin A \\sin B=\\cos (A+B)$ and $\\sin A \\cos B+\\cos A \\sin B=\\sin (A+B),$ in conjunction with $\\sin (-A)=-\\sin A$ and $\\cos (-A)=\\cos A.$ We have $\\cos (A+B)=\\cos A \\cos B -\\sin A \\sin B\\;,$ and $\\; \\cos (A-B)=\\cos A \\cos B+\\sin A \\sin B.$ Adding these , we have $$\\cos (A+B)+\\cos (A-B)=2\\cos A \\cos B.$$ Subtracting them , we have $$\\cos (A+B)-\\cos (A-B)=-\\sin A \\sin B.$$ Similarly from the formulas for $\\sin (A\\pm B)$ we have $$\\sin (A+B)+\\sin (A-B)=2\\sin A \\cos B$$ and $$\\sin (A+B)-\\sin (A-B)=2\\cos A \\sin B.$$ From the angle-sum formulas, with $A=B$ and from $\\cos^2A+\\sin^2A=1,$ we have $$\\sin^2 A=(1-\\cos 2A)/2$$ and $$\\cos^2 A=(1+\\cos 2A)/2.$$ In particular $\\sin^2 A \\cos 4A=\\frac {1}{2}(1-\\cos 2A)\\cos 4A=\\frac {1}{2}(\\cos 4A-\\cos 2A\\cos 4A)=\\frac {1}{2}(\\cos 4A-\\frac {1}{2}(\\cos 6A +\\cos 2A)).$ For a lot more of this, see Trigonometry, by Hobson. (Dover Press re-print.) Use following trigonometric identities to solve this question, $$\\sin^2x=(\\frac{1-\\cos2x}{2})$$ & $$\\cos A\\cos B=\\frac{\\cos(A+B)+\\cos(A-B)}{2}$$. Given integration $$I=\\int \\sin^2x\\cos 4x\\ dx$$ $$put\\ \\sin^2x=\\frac{1-\\cos 2x}{2}$$ $$I=\\int \\Big(\\frac{1-\\cos 2x}{2}\\Big)\\cos 4x\\ dx$$ $$\\Rightarrow I= \\frac{1}{2}\\int \\Big(\\cos 4x-\\cos 2x \\cdot \\cos 4x\\Big)dx$$ $$\\Rightarrow I=\\frac{1}{2}\\int\\Big(\\cos 4x-\\frac{\\cos(2x+4x)+\\cos(2x-4x)}{2}\\Big)dx$$ $$\\Rightarrow I=\\frac{1}{2}\\int\\Big(\\cos 4x-\\frac{\\cos 6x+cos(-2x)} {2}\\Big)dx$$ $$\\Rightarrow I=\\frac{1}{2}\\int\\Big(\\cos 4x-\\frac{\\cos 6x+\\cos2x} {2}\\Big)dx\\ \\ \\ \\", "there exists only three distinct relations between the angles and the sides. For instance the system $$\\begin{eqnarray} \\frac{a}{\\sin A} &=&\\frac{b}{\\sin B} \\\\ \\frac{a}{\\sin A} &=&\\frac{c}{\\sin C}\\tag{1} \\\\ A+B+C &=&\\pi; \\end{eqnarray}$$ or this equivalent (yours (I),(II),(III)) $$\\begin{eqnarray} a^{2} &=&b^{2}+c^{2}-2bc\\cos A \\\\ b^{2} &=&c^{2}+a^{2}-2ac\\cos B \\tag{2}\\\\ c^{2} &=&a^{2}+b^{2}-2ab\\cos C \\end{eqnarray}$$ are two of them. Another is $$\\begin{eqnarray} \\frac{\\tan \\frac{A+B}{2}}{\\tan \\frac{A-B}{2}} &=&\\frac{a+b}{a-b} \\\\ \\frac{\\tan \\frac{B+C}{2}}{\\tan \\frac{B-C}{2}} &=&\\frac{b+c}{b-c} \\tag{3}\\\\ \\frac{\\tan \\frac{C+A}{2}}{\\tan \\frac{C-A}{2}} &=&\\frac{c+a}{c-a}, \\end{eqnarray}$$ from which one can derive $$$$\\frac{\\tan \\frac{A+B}{2}}{\\tan \\frac{A-B}{2}}+\\frac{\\tan \\frac{B+C}{2}}{\\tan \\frac{B-C}{2}}+\\frac{\\tan \\frac{C+A}{2}}{\\tan \\frac{C-A}{2}}=\\frac{a+b}{a-b}+% \\frac{b+c}{b-c}+\\frac{c+a}{c-a}.\\tag{4}$$$$ Also from $$\\begin{eqnarray} \\tan \\frac{A-B}{2} &=&\\frac{a-b}{a+b}\\cot \\frac{C}{2} \\\\ \\tan \\frac{B-C}{2} &=&\\frac{b-c}{b+c}\\cot \\frac{A}{2}\\tag{5} \\\\ \\tan \\frac{C-A}{2} &=&\\frac{c-a}{c+a}\\cot \\frac{B}{2}, \\end{eqnarray}$$ follows $$\\begin{eqnarray} \\tan \\frac{A-B}{2} \\tan \\frac{C}{2}+\\tan \\frac{B-C}{2} \\tan \\frac{A}{2}+\\tan \\frac{C-A}{2}\\tan \\frac{B}{2} &=&\\frac{a-b}{a+b}+\\frac{b-c}{b+c}+\\frac{c-a}{c+a}. \\; (6)\\end{eqnarray}$$ As for a reference I looked at my old trigonometry text book Compêndio de Trigonometria (in Portuguese) by J. Jorge Calado. The system $\\left( 3\\right)$ is the law of tangents (Wikipedia) and $\\left( 5\\right)$ can be deduct from $\\left( 3\\right)$ by applying the relation $A+B+C=\\pi$ to get \\begin{eqnarray} \\tan \\frac{A+B}{2} &=&\\tan \\left( \\frac{\\pi }{2}-\\frac{C}{2}\\right) =\\cot \\frac{C}{2}=\\left( \\tan \\frac{C}{2}\\right) ^{-1}\\tag{7} \\end{eqnarray} and similar to $\\tan \\frac{B+C}{2}$ and $\\tan \\frac{C+A}{2}$. -", "# 25) Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\\tan ^{-1} \\sqrt 2$ Let a be the semi-vertical angle of cone Let r , h , l are the radius , height , slent height of cone Now, $r = l\\sin a \\ and \\ h=l\\cos a$ we know that Volume of cone (V) = $\\frac{1}{3}\\pi r^2 h = \\frac{1}{3}\\pi (l\\sin a)^2(l\\cos a) = \\frac{\\pi l^3\\sin^2 a\\cos a}{3}$ Now, $\\frac{dV}{da}= \\frac{\\pi l^3}{3}\\left ( 2\\sin a\\cos a.\\cos a+\\sin^2a.(-\\sin a)\\right )= \\frac{\\pi l^3}{3}\\left ( 2\\sin a\\cos^2a-\\sin^3a \\right )$ $\\frac{dV}{da}=0\\\\ \\frac{\\pi l^3}{3}\\left ( 2\\sin a\\cos^2a-\\sin^3a \\right ) = 0\\\\ 2\\sin a\\cos^2a-\\sin^3a= 0\\\\ 2\\sin a\\cos^2a=\\sin^3a\\\\ \\tan^2 a = 2\\\\ a = \\tan^{-1}\\sqrt 2$ Now, $\\frac{d^2V}{da^2}= \\frac{\\pi l^3}{3}\\left ( 2\\cos a\\cos^2a+2\\cos a(-2\\cos a\\sin a+3\\sin^2a\\cos a) \\right )$ Now, at $a= \\tan ^{-1}\\sqrt 2$ $\\frac{d^2V}{dx^2}< 0$ Therefore, $a= \\tan ^{-1}\\sqrt 2$ is the point of maxima Hence proved Exams Articles Questions", "# Consider the angle sum formulas: sin(A+B)=sinA cosB+cosA sinB cos(A+B)=cosA cosB−sinA sinB show that tan(90^0+θ)=− 1/tanθ. and use this identity to prove that lines y = mx and y = m_1x are perpendicular iff mm_1=−1? Jan 8, 2017 Given $\\sin \\left(A + B\\right) = \\sin A \\cos B + \\cos A \\sin B \\ldots . \\left[1\\right]$ $\\cos \\left(A + B\\right) = \\cos A \\cos B - \\sin A \\sin B \\ldots . \\left[2\\right]$ So $\\tan \\left(A + B\\right) = \\sin \\frac{A + B}{\\cos} \\left(A + B\\right)$ $= \\frac{\\frac{\\sin A \\cos B}{\\cos A \\cos B} + \\frac{\\cos A \\sin B}{\\cos A \\cos B}}{\\frac{\\cos A \\cos B}{\\cos A \\cos B} - \\frac{\\sin A \\sin B}{\\cos A \\cos B}}$ $\\implies \\tan \\left(A + B\\right) = \\frac{\\tan A + \\tan B}{1 - \\tan A \\tan B} \\ldots . . \\left[3\\right]$ $\\implies \\tan \\left(A + B\\right) = \\frac{\\tan \\frac{A}{\\tan} A + \\tan \\frac{B}{\\tan} A}{\\frac{1}{\\tan} A - \\tan B}$ $\\implies \\tan \\left(A + B\\right) = \\frac{1 + \\tan B \\cdot \\cot A}{\\cot A - \\tan B} \\ldots . . \\left[4\\right]$ This is an identity so it is valid for real values of $A \\mathmr{and} B$.So putting $A = {90}^{\\circ} \\mathmr{and} B = \\theta$ in [4] we get $\\implies \\tan \\left({90}^{\\circ} + \\theta\\right) = \\frac{1 + \\tan \\theta \\cdot \\cot 90}{\\cot 90 - \\tan \\theta}$", "for Double Angle: S.no. Formulae 1. sin 2A = 2 sin A∙cos A 2. cos 2A = cos2A – sin2A=2cos2A – 1= 1 – 2sin2A 3. tan 2A = $$\\frac{2tanA}{1- {tan}^{2}A}$$ ## Trigonometric Formulas for Triple Angle: S.no. Formulae 1. Sin 3A = 3sinA – 4sin3A 2. Cos 3A = 4cos3A – 3cosA 3. Tan3A = $$\\frac{3tanA – {tan}^{3}A}{1- 3{tan}^{2}A}$$ 4. Cot3A = $$\\frac{{cot}^{3}A – 3cotA}{3{cot}^{2}A – 1}$$ ## Trigonometric product conversion: S.no. Formulae 1. $$\\frac{cos(A – B) – cos(A+B)}{2}$$ = sinA.sinB 2. $$\\frac{cos(A – B) + cos(A+B)}{2}$$ = cosA.cosB 3. $$\\frac{sin(A + B) + sin(A – B)}{2}$$ = sinA.cosB 4. $$\\frac{cos(A + B) + sin(A – B)}{2}$$ = cosA.sinB ## Extended product rule: S.no. Formulae 1. sinA.sinB = $$\\frac{cos\\frac{(A + B)}{2}.cos\\frac{(A-B)}{2}}{2}$$ 2. cosA.cosB = $$\\frac{cos\\frac{(A + B)}{2}.sin\\frac{(A-B)}{2}}{2}$$ 3. sinA.cosB = $$\\frac{sin\\frac{(A + B)}{2}.cos\\frac{(A-B)}{2}}{2}$$ 4. cosA.sinB = $$\\frac{sin\\frac{(A + B)}{2}.sin\\frac{(A-B)}{2}}{2}$$ ## Trigonometric Half Angle Identities: S.no. Formulae 1. $$\\sin\\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos\\: x}{2}}$$ 2. $$\\cos\\frac{x}{2}=\\pm \\sqrt{\\frac{1+\\cos\\: x}{2}}$$ 3. $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$ 4. $$\\tan(\\frac{x}{2}) = \\sqrt{\\frac{1-\\cos(x)}{1+\\cos(x)}}$$ Also, $$\\tan(\\frac{x}{2}) =\\frac{1-\\cos(x)}{\\sin(x)}$$ Trigonometric formulas will make you easy to solve questions. Keep a bookmark of this site to learn and revise it regularly. Class 10th trigonometric formula is here with complete evaluation. ## Trigonometric ratio of some specific angles: The trigonometric ratio of 0°, 30°, 45°, 60°, 90° is mentioned below. take a look", "to $\\tan (A + B) = \\frac{\\tan A + \\tan B}{1 - \\tan A\\tan B}$ We have shown that addition formula is true when angles involved are acute angles. The same proof can be applied to prove the results for all values of $A$ and $B$. Consider $A' = 90^{\\circ} + A \\therefore `\\sin A' = \\cos A$ and $\\cos A' = \\sin A$ $\\sin(A' + B) = \\cos (A + B) = \\cos A\\cos B - \\sin A\\sin B = \\sin A'\\cos B + \\cos A'\\sin B$ Similarly $\\cos(A' + B) = -\\sin(A + B) = -\\sin A\\cos B - \\sin B\\cos A = \\cos A'\\cos B - \\sin A'\\sin B$ We can prove it again for $B' = 90^{\\circ} + B$ and so on by increasing the values of $A$ and $B$. Then we can again increase values by $90^{\\circ}$ and proceeding this way we see that the formula holds true for all values of $A$ and $B$. ## 8.2. The Subtraction Formula¶ $\\sin(A - B) = \\sin A\\cos B - \\sin B\\cos A$ $\\cos(A - B) = \\cos A\\cos B + \\sin A\\sin B$ $\\tan(A - B) = \\frac{\\tan A - \\tan B}{1 + \\tan A\\tan B}$ Conside the diagram above. The angle $MOP$ is $A - B.$ We take a point $P,$ and draw", "2. $2\\sin B\\cos A=\\sin \\left( A+B \\right)-\\sin \\left( A-B \\right)$ 3. $2\\cos A\\cos B=\\cos \\left( A+B \\right)+\\cos \\left( A-B \\right)$ 4. $2\\sin A\\sin B=\\cos \\left( A-B \\right)-\\cos \\left( A+B \\right)$ 5. $\\sin C+\\sin D=2\\sin \\left( \\frac{C+1}{2} \\right)\\cos \\left( \\frac{C-D}{2} \\right)$ 6. $\\sin C-\\sin D=2\\cos \\left( \\frac{C+D}{2} \\right)\\sin \\left( \\frac{C-D}{2} \\right)$ 7. $\\cos C-\\cos D=2\\cos \\left( \\frac{C+D}{2} \\right)\\cos \\left( \\frac{C-D}{2} \\right)$ 8. $\\cos C-\\cos D=2\\sin \\left( \\frac{C+D}{2} \\right)\\sin \\left( \\frac{D-C}{2} \\right)$ ### T-Ratio Of Multiple Angles 1. $\\sin 2A=2\\sin A\\cos A=\\frac{2\\tan A}{1+{{\\tan }^{2}}A}$ 2. $\\cos 2A=2{{\\cos }^{2}}A-1=1-2{{\\sin }^{2}}A=\\frac{1-{{\\tan }^{2}}A}{1+{{\\tan }^{2}}A}={{\\cos }^{2}}A-{{\\sin }^{2}}A$ 3. $\\tan 2A=\\frac{2\\tan A}{1-{{\\tan }^{2}}A}$ 4. $\\sin 3A=3\\sin A-4{{\\sin }^{3}}A$ 5. $\\cos 3A=4{{\\cos }^{3}}A-3\\cos A$ 6. $\\tan 3A=\\frac{3\\tan A-{{\\tan }^{3}}A}{1-3{{\\tan }^{2}}A}$ Remember: $1+\\cos 2A=2{{\\cos }^{2}}A$ $1-\\cos 2A=2{{\\sin }^{2}}A$ $\\frac{1-\\cos 2A}{1+\\cos 2A}={{\\tan }^{2}}A$ #### Values Of T-Ratios At Some Useful Angles 1. $\\sin 75{}^\\circ =\\frac{\\sqrt{3}+1}{2\\sqrt{2}}=\\cos 15{}^\\circ$ 2. $\\cos 75{}^\\circ =\\frac{\\sqrt{3}-1}{2\\sqrt{2}}=\\sin 15{}^\\circ$ 3. $\\tan 75{}^\\circ =2+\\sqrt{3}=\\cot 15{}^\\circ$ 4. $\\cot 75{}^\\circ =2-\\sqrt{3}=\\tan 15{}^\\circ$ 5. $\\sin 9{}^\\circ =\\frac{\\sqrt{3+\\sqrt{5}}-\\sqrt{5-\\sqrt{5}}}{4}=\\cos 81{}^\\circ$ 6. $\\cos 9{}^\\circ =\\frac{\\sqrt{3+\\sqrt{5}}-\\sqrt{5-\\sqrt{5}}}{4}=\\sin 81{}^\\circ$ Important Results: $\\sin \\theta \\sin \\left( 60-\\theta \\right)\\sin \\left( 60+\\theta \\right)=\\frac{1}{4}\\sin \\theta$ $\\cos \\theta \\cos \\left( 60-\\theta \\right)\\cos \\left( 60+\\theta \\right)=\\frac{1}{4}\\cos \\theta$ $\\tan \\theta \\tan \\left( 60-\\theta \\right)\\tan \\left( 60+\\theta \\right)=\\tan 3\\theta$ Example: If $A+B=225{}^\\circ$ $\\left( \\frac{\\cot A}{1+\\cot A} \\right)\\left( \\frac{\\cot B}{1+\\cot B} \\right)=?$. Solution: $A+B=225=180+45{}^\\circ$ $\\tan \\left( A+B \\right)=\\tan \\left( \\bar{\\wedge }+45{}^\\circ \\right)$ $\\tan \\left( A+B", "product is $(v_1, v_2) = |v_1||v_2|cos(\\phi) = a \\sin x + b \\cos x$ where $\\phi$ is the angle between $v_1$ and $v_2$, but $|v_1| = \\sqrt{a^2 + b^2}, |v_2| = 1$. • Don't see any reason for a down vote. Canceling it by a +1. Nice answer – Shailesh Jun 21 '16 at 11:36 Find the extrema of $$a\\cos(x)+b\\sin(x).$$ By canceling the derivative, $$-a\\sin(z)+b\\cos(z)=0$$ or$$\\tan(z)=\\frac ba.$$ Then, with $$\\cos(z)=\\pm\\frac1{\\sqrt{\\tan^2(z)+1}}=\\pm\\frac a{\\sqrt{a^2+b^2}},\\\\\\sin(z)=\\pm\\frac{\\tan(z)}{\\sqrt{\\tan^2(z)+1}}=\\pm\\frac b{\\sqrt{a^2+b^2}},$$ you obtain $$a\\cos(z)+b\\sin(z)=\\pm\\sqrt{a^2+b^2}.$$ Then for all $x$, $$-\\sqrt{a^2+b^2}\\le a\\cos(x)+b\\sin(x)\\le\\sqrt{a^2+b^2}.$$ $$a\\sin{x}+b\\cos{x}=\\sqrt{a^2+b^2}(\\frac{a}{\\sqrt{a^2+b^2}}\\sin{x}+\\frac{b}{\\sqrt{a^2+b^2}}\\cos{x}$$ now let $$\\sin{\\alpha}=\\frac{b}{\\sqrt{a^2+b^2}}$$ then $$\\cos{\\alpha}=\\frac{a}{\\sqrt{a^2+b^2}}$$ thus $$a\\cos{x}+b\\sin{x}=\\sqrt{a^2+b^2}\\sin({\\alpha+x})<\\sqrt{a^2+b^2}$$", "# Proving $\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$ Prove $$\\tan\\left( \\frac {a+b}{2}\\right) = \\frac{\\sin a+\\sin b} {\\cos a + \\cos b }$$ Can someone help? I separated $\\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves. Use the formulas $$\\sin a+\\sin b=2\\sin\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right),$$ $$\\cos a+\\cos b=2\\cos\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right).$$ $$\\frac{\\sin a+\\sin b}{\\cos a+\\cos b}=\\frac{\\sin\\displaystyle\\left(\\frac{a+b}{2}\\right)}{\\cos\\displaystyle\\left(\\frac{a+b}{2}\\right)}=\\tan\\left(\\frac{a+b}{2}\\right).$$" ]

[ "are $2 x = \\frac{\\pi}{2} + 2 \\pi k$ and $2 x = \\frac{3 \\pi}{2} + 2 \\pi k$, (when the cosine is $0$) or $x = \\frac{\\pi}{4} + \\pi k$ and $x = \\frac{3 \\pi}{4} + \\pi k$ Checking the sign of $y ' = \\cos \\left(2 x\\right)$, we will find maximum values at the first set of critical numbers and minimum values at the second.", "also be done like this : Since $\\cos x = 1-\\frac{x^2}{2!}+\\frac{x^4}{4!}-\\frac{x^6}{6!}+...$ then $\\cos 2x = 1-\\frac{4x^2}{2!}+\\frac{16x^4}{4!}-\\frac{64x^6}{6!}+...$ so $\\frac{1-\\cos 2x}{x^2} = \\frac{4}{2!}-\\frac{16x^2}{4!}+\\frac{64x^4}{6!}+...$ taking the limit as $x \\rightarrow 0$ on both sides, you get $\\frac{4}{2!}=4/2=2$.", "Comment Share Q) Using principal value, calculate the following : $$\\sin^{-1} \\bigg( \\sin \\large\\frac{3\\pi}{5} \\bigg).$$ $Sin^{-1}sin(\\pi-\\frac{2\\pi}{5})=Sin^{-1}sin\\frac{2\\pi}{5}=\\frac{2\\pi}{5}$", "# Write the value of $$\\cos^{-1} \\bigg( \\cos\\large\\frac{7\\pi}{6} \\bigg).$$ $Cos^{-1}Cos(\\pi+\\frac{\\pi}{6})=Cos^{-1}Cos(\\pi-\\frac{\\pi}{6})$ $=\\frac{5\\pi}{6}$", "is obtainable, viz. $$x=\\frac{\\pi}{6}$$. Rewrite the equation as $$\\;\\cos(2x)=\\cos \\bigl(\\frac\\pi 2-x\\bigr)$$. Then use that $$\\cos\\alpha=\\cos\\beta\\iff \\alpha\\equiv\\pm\\beta \\pmod{2\\pi}.$$", "# Using principal value, calculate the following : $$\\sin^{-1} \\bigg( \\sin \\large\\frac{3\\pi}{5} \\bigg).$$ Ans: $Sin^{-1}sin(\\pi-\\frac{2\\pi}{5})=Sin^{-1}sin\\frac{2\\pi}{5}=\\frac{2\\pi}{5}$", "2x + 3) 25. Evaluate $\\int { { a }^{ 3{ log }_{ a }x } } dx$ 26. Evaluate $\\int { \\frac { 2{ cos }^{ 2 }x-cos2x }{ { sin }^{ 2 }x } }$ 27. 5 x 3 = 15 28. Evaluate $\\int { \\frac { { ax }^{ 2 }+bx+v }{ \\sqrt { x } } dx }$ 29. Evaluate $\\int { \\sqrt { 2x+1dx } }$ 30. Evaluate $\\int { \\frac { dx }{ { \\left( 2x+3 \\right) }^{ 2 } } }$ 31. Evaluate $\\int { \\frac { { { x }^{ 4 }+{ x }^{ 4 }+1 } }{ { x }^{ 2 }-x+1 } }$ 32. Evaluate $\\int { \\frac { cos2x-cos2\\alpha }{ cosx-cos\\alpha } } dx$ 33. 2 x 5 = 10 34. Evaluate $\\int { \\frac { { 3x }^{ 2 }+2x+1 }{ x } dx }$ 35. If f'(x) = a sin x + b cos x and f'(0) = 4, f(0) = 3, f$\\left( \\frac { \\pi }{ 2 } \\right)$ = 5, find f(x).", "Given that $x = \\pi^{\\pi^{\\pi}}$, evaluate $$\\frac{1}{\\frac{1}{4}(\\sin^4{x}+\\cos^4{x})-\\frac{1}{6}(\\sin^6{x}+\\cos^6{x})}$$", "\\sin x - 2a &= 0 \\\\ 4y^3 +\\frac{1}{2}\\sin 2y +a &=0 \\end{array} \\right.$$ Compute the value of $\\cos(x+2y)$ Let $f$ be a real-valued function such that $f(x) + 2f(\\frac{2002}{x}) = 3x$ for all $x > 0$. Find $f(2)$.", "\\pi^2 \\int_0^{\\infty} \\frac{dx}{e^x+1}\\\\ +i \\,PV \\int_0^{2 \\pi} dy \\frac{y^2}{e^{i y}+1}-i \\pi^3=0$$ where $PV$ denotes the Cauchy principal value. Note that $$PV \\int_0^{2 \\pi} dy \\frac{y^2}{e^{i y}+1} = \\frac12 \\int_0^{2 \\pi} dy \\: y^2 -i \\frac12 \\, PV \\int_0^{2 \\pi} dy \\frac{y^2 \\sin{y}}{1+\\cos{y}}$$ Equating imaginary parts, we see that $$-4 \\pi \\int_0^{\\infty} dx \\frac{x}{e^x+1} + \\frac12 \\frac{(2 \\pi)^3}{3} - \\pi^3=0$$ or $$\\int_0^{\\infty} dx \\frac{x}{e^x+1} = \\frac13 \\pi^2-\\frac14\\pi^2 = \\frac{\\pi^2}{12}$$", "# Evaluate :$\\sin(n\\pi+(-1)^n\\large\\frac{\\pi}{4})$ where $n$ is an integer, $\\begin{array}{1 1}(A)\\;\\large\\frac{1}{\\sqrt 2}&(B)\\;\\large\\frac{1}{2}\\\\(C)\\;1&(D)\\;0\\end{array}$", "109 views Iam getting Max = $\\frac{9}{8}$ and min = $0$ maximum match but minimum doesn't match the answer which has been given max=1.125 //9/8=1.125 min = -2 how did u get max? Yes, it has to be $-2$, i calculated min at $x=2\\pi$, which is wrong. $\\color{olive}{f(x) = \\sin(x) + \\cos(2x)}$ $\\color{olive}{f'(x) = \\cos(x) - 2\\sin(2x)}$ $\\color{olive}{f'(x) = \\cos(x) - 4\\sin(x)\\cos(x)}$ $\\color{olive}{f'(x) = \\cos(x)(1- 4\\sin(x))}$ For minima and maxima, $f'(x) = 0$ $f'(x) = \\cos(x)(1- 4\\sin(x)) = 0$ $\\cos(x) = 0$ (or) $1 - 4\\sin(x) = 0$ $x = \\frac{\\pi}{2},\\frac{3\\pi}{2}$ (or) $\\sin(x) = \\frac{1}{4} \\Rightarrow x = \\sin^{-1}(\\frac{1}{4})$ Now, the proper method for finding minima and maxima is to check whether $f''(x) \\gt 0$ or $f''(x) \\lt 0$, but here as we are required to find both minima and maxima, i will directly put values of x and find out. When $x = \\frac{\\pi}{2}$, $f(x) = 1 + (-1) = 0$ When $x = \\frac{3\\pi}{2}$, $f(x) = \\sin(\\frac{3\\pi}{2}) + \\cos(3\\pi) = -1 + (-1) = -2$ $\\sin(x) = \\frac{1}{4}$, $\\cos(x) = \\frac{\\sqrt{15}}{4}$, $\\sin(2x) = 2\\sin(x)\\cos(x) = 2*\\frac{1}{4}*\\frac{\\sqrt{15}}{4} = \\frac{\\sqrt{15}}{8}$ and $\\cos(2x) = \\frac{\\sqrt{64-15}}{8} = \\frac{7}{8}$ When, $x = \\sin^{-1}(\\frac{1}{4})$, $f(x) = \\sin(x) + \\cos(2x) = \\frac{1}{4} + \\frac{7}{8} = \\frac{2}{8} + \\frac{7}{8} = \\frac{9}{8}$ Hence, $\\color{navy}{f(x)_{max} = \\frac{9}{8}}$ and $\\color{navy}{f(x)_{min} = -2}$", "are $\\frac{\\sqrt2}}{2}$ and $\\frac{\\sqrt{3}}{2}$, respectively. Therefore: $cos(\\frac{5\\pi}{12}) + cos(\\frac{\\pi}{12}) = 2(\\frac{ \\sqrt{2}}{2}) (\\frac{\\sqrt{3} }{2})$. Finally, this becomes: $\\frac{ \\sqrt{2}\\sqrt{3} }{2} = \\frac{ \\sqrt{6}}{2}$.", "Trigonometry 7th Edition $\\frac{\\pi}{2}$. Let $\\cos^{-1}(0)=x$. Then, $\\cos x=0$. Needless to say, $cos(\\frac{\\pi}{2})=0$. Therefore, x= $\\frac{\\pi}{2}$.", "\\frac{\\pi}{2})$$ Thus, $$\\frac{1}{\\cos{x}} - \\cos{x} = f(x) \\gt x^2 \\mbox{ for } x \\in (0, \\frac{\\pi}{2})$$ Done. Note that for $x\\in\\left(0,\\frac{\\pi}{2}\\right)$ $$\\tan x \\sin x>\\left(x+\\frac{x^3}{3}\\right)\\left(x-\\frac{x^3}{6}\\right)=x^2+\\frac{x^4}{6}-\\frac{x^6}{18}>x^2\\iff3-x^2>0\\iff x < \\sqrt 3$$", "... ### Computing $\\int_0^{\\pi\\over2} \\frac{dx}{1+\\sin^2(x)}$? How would you compute$$\\int_0^{\\pi\\over2} \\frac{dx}{1+\\sin^2(x)}\\, \\, ?$$", "Use $\\sin 2A= 2\\sin A \\cos A$ and $\\cos 2A= \\cos^2 A- \\sin^2 A$ to get $$\\frac{\\sin A \\cos A}{\\cos^2 A- \\sin^2 A}=\\frac{\\sin 2A}{2\\cos 2A}=\\frac12\\tan{2A},$$ which is equivalent to $$\\frac{ \\tan A} {1 - \\tan^2 A} .$$", "$0< k < \\frac{n}{2}$ we get the quadratic factors $$( x - (\\cos \\frac{2 k \\pi}{n} + i \\sin \\frac {2 k \\pi}{n}) )(x-(\\cos \\frac{2 (n-k) \\pi}{n} + i \\sin \\frac {2 (n-k) \\pi}{n}))=\\\\=x^2 - 2\\cos\\frac{2 k \\pi}{n}\\, x + 1$$ Thus we get the real factorization of $x^n-1$: $$x^n-1 =(x-1)\\prod_{k=1}^{\\frac{n-1}{2}} (x^2 - 2\\cos\\frac{2 k \\pi}{n}\\, x + 1 )$$ if $n$ odd and $$x^n-1 =(x-1)(x+1)\\prod_{k=1}^{\\frac{n}{2}-1} (x^2 - 2\\cos\\frac{2 k \\pi}{n}\\, x + 1 )$$ if $n$ is even.", "2\\cos \\frac{8\\pi}{9} = 2 - 2\\cos \\frac{\\pi}{9} = 4\\sin^2 \\frac{\\pi}{18}$ 3) $2 + 2\\cos \\frac{14\\pi}{9} = 2 - 2\\cos \\frac{5\\pi}{9} = 4\\sin^2 \\frac{5\\pi}{18}$", "+0 # precalculus, 0 199 2 +132 Find all values of $$x$$ such that $$0\\leq x\\leq 2\\pi$$ and $$\\sin(x)-\\cos(x) = \\dfrac{1}{\\sqrt{2}}$$. been stuck on this for a while now, help would be appreciated! Nov 10, 2021 #1 0 $\\sin x - \\cos x = \\frac{1}{\\sqrt{2}}$ $1 - \\sin 2x = \\frac{1}{2}$ $\\sin 2x = \\frac{1}{2}$ $2x = \\frac{\\pi}{6}$ or $2x = \\frac{5 \\pi}{6}$ $x = \\frac{\\pi}{12}$ or $x = \\frac{5 \\pi}{12}$ Nov 10, 2021 #2 +117437 0 Thanks for answering guest but you need to put your latex in a latex box for it to render properly. Nov 12, 2021" ]

[ "5 & 6\\\\ 2 & 1 & 4 & 5 & 6 & 3 \\end{pmatrix}$ has two cycles, namely: (4) \\begin{align} \\quad (12) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 2 & 1 & 3 & 4 & 5 & 6 \\end{pmatrix} , \\quad (3456) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 2 & 4 & 5 & 6 & 3 \\end{pmatrix} \\end{align} We can also compose cycles. Let $\\alpha = (12)$ and $\\beta = (3456)$ and consider the composition $(\\alpha \\circ \\beta)(x) = \\alpha(\\beta(x))$ for all $x \\in \\{ 1, 2, 3, 4, 5, 6 \\}$. We have that $(\\alpha(\\beta(1)) = \\alpha(1) = 2$, $(\\alpha(\\beta(2)) = \\alpha(2) = 1$, $\\alpha(\\beta(3)) = \\alpha(4) = 4$, $\\alpha(\\beta(4)) = \\alpha(5) = 5$, $\\alpha(\\beta(5)) = \\alpha(6) = 6$, and $\\alpha(\\beta(6)) = \\alpha(3) = 3$, and so: (5) \\begin{align} \\quad \\alpha \\circ \\beta = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 2 & 1 & 4 & 5 & 6 & 3 \\end{pmatrix} = \\sigma \\end{align}", "$$\\overline{AB}$$ (the $$x$$-axis). Then $$P^\\prime_x=P_x$$ and $$P^\\prime_y=-P_y$$, giving the equations \\begin{align} (\\beta\\,c+\\gamma\\,b \\cos A)(\\alpha^\\prime+\\beta^\\prime+\\gamma^\\prime)&=(\\beta^\\prime\\,c+\\gamma^\\prime\\,b\\cos A)(\\alpha+\\beta+\\gamma) \\\\[4pt] \\gamma\\,(\\alpha^\\prime+\\beta^\\prime+\\gamma^\\prime)&= - \\gamma^\\prime\\,(\\alpha+\\beta+\\gamma) \\end{align}\\tag{3} Solving the system for, say, $$\\alpha^\\prime$$ and $$\\beta^\\prime$$ gives \\begin{align} \\alpha^\\prime &= -\\frac{\\gamma^\\prime}{\\gamma\\,c} \\left(\\;\\alpha\\,c + 2 \\gamma\\,(c - b \\cos A)\\;\\right) = -\\frac{\\gamma^\\prime}{\\gamma\\,c} \\left(\\;\\alpha\\,c + 2 \\gamma\\,a \\cos B\\;\\right) \\\\[4pt] \\beta^\\prime &= -\\frac{\\gamma^\\prime}{\\gamma\\,c}\\left(\\;\\beta\\,c + 2 \\gamma\\,b \\cos A\\;\\right) \\end{align}\\tag{4} from which we can deduce barycentric coordinates, in a smattering of variants, \\begin{align} \\alpha^\\prime:\\beta^\\prime:\\gamma^\\prime\\quad&=\\quad \\alpha\\,c + 2 \\gamma\\,a \\cos B\\;:\\; \\beta\\,c + 2 \\gamma\\,b \\cos A\\;:\\; -\\gamma\\,c \\\\[8pt] \\quad&=\\quad \\alpha+\\frac{2\\gamma\\,a\\cos B}{c}:\\beta+\\frac{2\\gamma\\,b \\cos A}{c} : \\gamma - 2\\gamma \\\\[8pt] \\quad&=\\quad \\alpha+2\\gamma\\,\\sin A\\cos B \\csc C:\\beta+2\\gamma\\,\\cos A\\sin B\\csc C : \\gamma - 2\\gamma \\\\[8pt] \\quad&=\\quad \\alpha+\\frac{2\\gamma\\,S_B}{c^2}:\\beta+\\frac{2\\gamma\\,S_A}{c^2} : \\gamma - 2\\gamma \\end{align} \\tag{5} As a sanity check: • Any point on the $$x$$-axis has $$\\gamma=0$$; reflection fixes the such a point, and we see from $$(5)$$ that, indeed $$\\alpha^\\prime : \\beta^\\prime : 0 = \\alpha:\\beta:0$$. • The reflection of $$C$$, which has barycentric coordinates $$(0,0,1)$$, should have Cartesian coordinates $$(b\\cos A,-b\\sin A)$$; from the first form in $$(5)$$, \\begin{align} \\frac{\\alpha^\\prime A + \\beta^\\prime B + \\gamma^\\prime C}{\\alpha^\\prime+\\beta^\\prime+\\gamma^\\prime} &= \\frac{A\\cdot 2a\\cos B + B\\cdot 2b\\cos A -C\\cdot c}{2a\\cos B+2b\\cos A-c} \\\\[8pt] &=\\frac{(2bc\\cos A-bc \\cos A, -bc\\sin A)}{2c-c} \\\\[8pt] &= (b\\cos A,-b\\sin A) \\end{align} \\tag{6} as expected. Without loss of generaliy let $$A=(x_1,y_1), B=(0,0),", "721 + (-7\\alpha + 15) p^2 x_0^3 \\\\ 722 & \\qquad {} + (-11\\alpha - 2) x_0^2 + (-3\\alpha - 2) p^3 x_0 - y^2 723 \\\\ 724 &= 0,\\\\ 725 p &= 2. 726\\end{align*} 727To blow up at $(x_0,y,p)$, we introduce projective coordinates 728$(x_1,y_1,p_1)$ with $x_{0} y_1 = x_{1} y$, $x_{0} p_{1} = x_{1} p$, and 729$y p_1 = y_{1} p$. We look in all three affine covers and check for regularity. 730 731\\begin{description} 732\\item[$x_{1} = 1$] We have $y = x_{0} y_{1}$, $p = x_{0} p_{1}$. We get 733 $g_2(x_{0},y_{1},p_{1}) = 0$, $x_{0} p_{1} = 2$, where 734 \\begin{align*} 735 g_2(x_{0},y_{1},p_{1}) &= x_{0}^{-2}g_{1}(x_{0},x_{0}y_{1},x_{0}p_{1}) \\\\ 736 &= -x_0^4 + (-3\\alpha + 5) p_1 x_0^4 + (5\\alpha - 47) x_0^2 737 + (-7\\alpha + 15) p_1^2 x_0^3 \\\\ 738 & \\qquad{} + (-11\\alpha - 2) + (-3\\alpha - 2) p_1^3 x_0^2 - y_1^2 \\,. 739 \\end{align*} 740 In the reduction we have either $x_{0} = 0$ or $p_1 = 0$. 741 \\begin{description} 742 \\item[$x_{0} = 0$] $(y_{1} + \\alpha + 1)^2 = 0$. 743 This is a new component which we denote $B$. It has genus~0 and 744 multiplicity~2. We check regularity along~$B$ at 745 $(x_{0}, y_{1} + \\alpha + 1, p_{1}-t, 2)$, with $t$ in $\\Z_2\\unr$, and 746 find that $B$ is nowhere regular. 747 \\item[$p_{1} =", "+ 1)\\\\ =&\\ 31 * 33 \\end{align} or \\begin{align} 2^{10} - 1 =&\\ 1111111111_2 = 1023 \\\\=&\\ 1100000000_2 + 0011000000_2 + 0000110000_2 + 0000001100_2 + 0000000011_2 \\\\ =&\\ 3*2^8 + 3*2^6 + 3*2^4 + 3*2^2 + 3*2^0 = 1023\\\\ =&\\ 3*(2^8 + 2^6 + 2^4 + 2^2 + 2^0) = 1023 \\\\ =&\\ 3*341 = 1023 \\end{align} If $n$ is composite (i.e. non-prime) then the $1$'s can be grouped into sets which will make $2^n - 1$ non-prime. Basically, assume that $n$ is non-prime such that $n = \\alpha \\beta$, then $2^n - 1$ will be divisible by $2^\\alpha - 1$ and $2^\\beta - 1$.", "\\label{eq:A1},\\\\ A_2^{\\prime \\sigma} &= g_2^A t \\frac{ -\\pi \\alpha_A' }{\\Gamma(\\alpha_A + 1)} \\frac{1 - e^{-i\\pi\\alpha_A}}{2 \\sin \\pi \\alpha_A} \\left(r_2 s\\right)^{\\alpha_A - 1} ,\\\\ A_3^\\sigma &= 0,\\\\ A_4^\\sigma &= g_4^V \\frac{ -\\pi \\alpha_V' }{\\Gamma(\\alpha_V)} \\frac{1 - e^{-i\\pi\\alpha_V}}{2 \\sin \\pi \\alpha_V} \\left(r_4 s\\right)^{\\alpha_V - 1} \\label{eq:A4}, \\end{align} where for $\\sigma = s$ ($\\sigma = v$) we have $V = \\omega$ and $A = h$ ($V = \\rho$ and $A = b$). We introduced the scaling parameters $r_{i=1,2,4}$, which are fitted to the data. Finally we use linear Regge trajectories \\begin{align} \\alpha_V(t) &= \\alpha_{V0}+ \\alpha_V' t = 1 + 0.9(t - m_\\rho^2),\\\\ \\alpha_A(t) &= \\alpha_{A0}+ \\alpha_A' t = 0 + 0.7(t - m_\\pi^2). \\end{align} We have fitted 8 parameters to the available data. The result is \\begin{align} \\label{eq:param1} g_1^\\rho &= \\phantom{+} 3.434 \\pm 0.083 \\mbox{ GeV}^{-2}, & g_2^b &= \\phantom{+} 4.946 \\pm 1.491 \\mbox{ GeV}^{-2}, & g_4^\\rho &= -1.397 \\pm 0.085 \\mbox{ GeV}^{-1},\\\\ g_1^\\omega &= \\phantom{+} 0.116 \\pm 0.074 \\mbox{ GeV}^{-2},& g^h_2 & = (2/3) g^b_2, & g_4^\\omega &= -3.346 \\pm 0.087 \\mbox{ GeV}^{-1},\\\\ r_1 &= \\phantom{+}3.001 \\pm 0.087 \\mbox{ GeV}^{-1}, & r_2 &= \\phantom{+}6.204 \\pm 2.484 \\mbox{ GeV}^{-1}, & r_4 &= \\phantom{+}1.974 \\pm 0.101 \\mbox{ GeV}^{-1}. \\end{align} The results is presented below and compared to the data from [Dewire71] and [Braunschweig70]. The dashed (dotted-dashed) line indicates the isoscalar (isovector) contribution.", "y) \\mid ≤ M < 1 \\\\ \\end{align} Then this implies that: (6) \\begin{align} \\quad \\mid \\alpha - x_{n+1} \\mid + \\mid \\beta - y_{n+1} \\mid \\: ≤ \\: M \\mid \\alpha - x_n \\mid + \\mid \\beta - y_n \\mid) \\end{align} Thus if the initial approximation $(x_0, y_0) \\in D$ then the successive approximations $(x_n, y_n) \\in D$ for $n = 0, 1, 2, ...$. We will use the $1$-norm (defined as $\\| x \\|_1 = \\sum_{i=1}^{n} \\mid x_i \\mid$) for our error. Let $e_{n+1} = \\begin{bmatrix} \\alpha - x_{n+1}\\\\ \\beta - y_{n+1} \\end{bmatrix}$. We have that: (7) \\begin{align} \\quad \\| e_{n+1} \\|_1 = \\mid \\alpha - x_{n+1} \\mid + \\mid \\beta - y_{n+1} \\mid \\end{align} Inductively we see that $\\| e_{n+1} \\|_1 ≤ M^{n+1} \\| e_0 \\|_1$: (8) \\begin{align} \\quad \\quad \\| e_{n+1} \\|_1 \\: ≤ \\: M(\\mid \\alpha - x_n \\mid + \\mid \\beta - y_n \\mid) \\: ≤ \\: M^2 (\\mid \\alpha - x_{n-1} \\mid + \\mid \\beta - y_{n-1} \\mid ) \\: ≤ \\: ... \\: ≤ \\: M^{n+1} (\\mid \\alpha - x_0 \\mid + \\mid \\beta - y_0 \\mid) = M^{n+1} \\| e_0 \\|_1 \\end{align} Therefore for $e_n = \\begin{bmatrix} \\alpha - x_n\\\\ \\beta - y_n \\end{bmatrix}$ we have that $\\| e_n \\|_1 = \\mid \\alpha - x_n \\mid + \\mid \\beta", "- 7(\\alpha\\beta^{5} + \\beta\\gamma^{5} + \\gamma\\alpha^{5})\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,+ 14(\\alpha^{2}\\beta^{3} + \\beta^{2}\\gamma^{3} + \\gamma^{2}\\alpha^{3}) + 8\\tag{15}\\end{align} In the same manner we can calculate $D_{5} = -g_{2}^{6}D'_{5}$ where $$D'_{5} = \\begin{vmatrix}\\beta & \\alpha & 0 & \\gamma & 0 & 1\\\\ 1 & \\beta & \\alpha & 0 & \\gamma & 0\\\\ 0 & 1 & \\beta & \\alpha & 0 & 0\\\\ 0 & 0 & 1 & \\beta & \\alpha & \\gamma\\\\ \\gamma & 0 & 0 & 1 & \\beta & 0\\\\ 0 & \\gamma & 0 & 0 & 1 & \\alpha\\end{vmatrix}$$ and \\begin{align}D'_{5}&= (\\alpha\\beta^{5} + \\beta\\gamma^{5} + \\gamma\\alpha^{5}) - 4(\\alpha^{2}\\beta^{3} + \\beta^{2}\\gamma^{3} + \\gamma^{2}\\alpha^{3})\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,+ 3(\\alpha^{3}\\beta + \\beta^{3}\\gamma + \\gamma^{3}\\alpha) - 8\\tag{16}\\end{align} Since we know the value of $D'$ in a closed form as $$D' = \\frac{D}{g_{2}^{7}} = -\\frac{f^{8}(-q^{7})}{q^{14}f^{8}(-q^{49})}\\tag{17}$$ the real issue is to evaluate $D'_{5}$ in a closed form. For this we try to find relation between $D'$ and $D'_{5}$. We use a new set of variables $y_{1}, y_{2}, y_{3}$ defined by $$y_{1} = \\alpha^{3}\\beta, y_{2} = \\beta^{3}\\gamma, y_{3} = \\gamma^{3}\\alpha\\tag{18}$$ and then apply equation $(10)$ to evaluate the following expressions \\begin{align}\\alpha^{2}\\beta^{3}&= y_{1}y_{2} = - y_{1} - 1\\tag{19a}\\\\ \\beta^{2}\\gamma^{3} &= y_{2}y_{3} = - y_{2} - 1\\tag{19b}\\\\ \\gamma^{2}\\alpha^{3} &= y_{3}y_{1} = - y_{3} - 1\\tag{19c}\\end{align} $$y_{1}y_{2}y_{3} = 1\\tag{20}$$ $$\\alpha\\beta^{5} = y_{1} - y_{2} + 1, \\beta\\gamma^{5}", "we try to find real numbers $$\\alpha$$, $$\\beta$$, and $$\\gamma$$ such that \\begin{align*} x^2 + 2 & = \\alpha(x^2 + x) + \\beta(x^2+1) + \\gamma(x+1) \\end{align*} Rewritting the right-hand side by collecting like terms, we obtain \\begin{align*} x^2 + 2 & = (\\alpha + \\beta) x^2 + (\\alpha+\\gamma)x + (\\beta+\\gamma) \\end{align*} Comparing the coefficients of the different powers of $$x$$ on both sides, we get that \\begin{align*} \\alpha + \\beta & = 1 \\\\ \\alpha + \\gamma & = 0 \\\\ \\beta + \\gamma & = 2 \\end{align*} The first equation comes from the coefficients of $$x^2$$, the second from those of $$x$$, and the third from the constant terms. This system has a unique solution with $$\\alpha = -\\dfrac{1}{2}$$, $$\\beta = \\dfrac{3}{2}$$, and $$\\gamma = \\dfrac{1}{2}$$. In other words, $x^2 + 2 = -\\dfrac{1}{2}(x^2+x) + \\dfrac{3}{2}(x^2+1) + \\dfrac{1}{2}(x+1).$ One can now simplify the right-hand side to see that equality holds. 5. We show that $$\\operatorname{span}\\left(\\left\\{\\begin{bmatrix} 1\\\\1\\end{bmatrix}, \\begin{bmatrix} 1 \\\\-1\\end{bmatrix}\\right\\}\\right)=\\mathbb{R}^2$$. What we need to show is that every $$\\displaystyle\\begin{bmatrix} x\\\\y\\end{bmatrix} \\in \\mathbb{R}^2$$ can be written as a linear combination of $$\\begin{bmatrix} 1\\\\1\\end{bmatrix}$$ and $$\\begin{bmatrix} 1 \\\\-1\\end{bmatrix}$$. In other words, we want to determine if there exist scalars $$\\alpha$$ and $$\\beta$$ such that $$\\displaystyle\\begin{bmatrix} x\\\\y\\end{bmatrix} = \\alpha\\begin{bmatrix} 1\\\\1\\end{bmatrix}+ \\beta\\begin{bmatrix} 1\\\\-1\\end{bmatrix}$$, or equivalently, $$\\displaystyle\\begin{bmatrix} x\\\\y\\end{bmatrix} = \\begin{bmatrix} \\alpha + \\beta\\\\\\alpha", "cubic expression in the form: \\begin{align} ax^3 + bx^2 + cx + d \\\\ \\end{align} If we call the roots $\\alpha$, $\\beta$ and $\\gamma$ then: \\begin{align} a(x-\\alpha)(x-\\beta)(x-\\gamma) = ax^3 + bx^2 + cx + d \\\\ \\end{align} Where: \\begin{align} \\alpha + \\beta + \\gamma = \\frac{-b}{a} \\\\ \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha= \\frac{c}{a} \\\\ \\alpha\\beta\\gamma= \\frac{-d}{a} \\\\ \\end{align} In this specific case, a = 1. That lot should help you! Also please notice that the 'a' I use to represent the coefficient of x cubed is not the same as the 'a' you have been given in the question. Extra Hint (Only use if you are still stuck): \\begin{align} \\alpha\\beta\\gamma = \\frac{a}{k} \\times a \\times ak = a^3 \\end{align} Last edited: Aug 12, 2011", "& \\alpha \\beta \\pm a \\alpha \\\\ \\alpha \\beta \\pm a \\alpha & \\beta^2 + \\alpha^2\\end{bmatrix}\\end{aligned} \\quad\\quad\\quad(11) so our equations are now \\begin{aligned}1 = \\alpha^2 + a^2 \\end{aligned} \\quad\\quad\\quad(12) \\begin{aligned}1 = \\beta^2 + \\alpha^2\\end{aligned} \\quad\\quad\\quad(13) \\begin{aligned}0 = \\alpha (\\beta \\pm a )\\end{aligned} \\quad\\quad\\quad(14) There are two cases to distinguish here. The first is the more trivial $\\alpha = 0$ case, for which we find $a^2 = \\beta^2 = 1$. For the other case we have \\begin{aligned}\\beta = \\mp a\\end{aligned} \\quad\\quad\\quad(15) Again, writing out in full to check, this reduces our system of equations \\begin{aligned}\\begin{bmatrix}1 & 0 \\\\ 0 & 1 \\\\ \\end{bmatrix}=\\begin{bmatrix}\\alpha & a \\\\ \\mp a & \\pm \\alpha \\\\ \\end{bmatrix}\\begin{bmatrix}\\alpha & \\mp a \\\\ a & \\pm \\alpha \\\\ \\end{bmatrix}= \\begin{bmatrix}\\alpha^2 + a^2 & 0 \\\\ 0 & a^2 + \\alpha^2\\end{bmatrix}\\end{aligned} \\quad\\quad\\quad(16) We have now only one constraint left, and have reduced things to a single degree of freedom \\begin{aligned}1 = \\alpha^2 + a^2 \\end{aligned} \\quad\\quad\\quad(17) Or \\begin{aligned}\\alpha = (1 - a^2)^{1/2}\\end{aligned} \\quad\\quad\\quad(18) We’ve already used $\\pm$ to distinguish the roots of $\\alpha = \\pm b$, so here lets imply that this square root can take either positive or negative values, but that we are treating the sign of this the same whereever seen. Our transformation, employing $a$ as the free variable is now known to", "# Solve the following Question: Let $\\alpha$ and $\\beta$ be the roots of the equation, $5 x^{2}+6 x-2=0$. If $S_{n}=\\alpha^{n}+\\beta^{n}, n=1,2,3, \\ldots$, then : 1. (1) $6 S_{6}+5 S_{5}=2 S_{4}$ 2. (2) $6 S_{6}+5 S_{5}+2 S_{4}=0$ 3. (3) $5 S_{6}+6 S_{5}=2 S_{4}$ 4. (4) $5 S_{6}+6 S_{5}+2 S_{4}=0$ Correct Option: , 3 Solution: Since, $\\alpha$ and $\\beta$ are the roots of the equaton $5 x^{2}+6 x-2=0$ Then, $5 \\alpha^{2}+6 \\alpha-2=0,5 \\beta^{2}+6 \\beta-2=0$ $5 \\alpha^{2}+6 \\alpha=2$ $5 S_{6}+6 S_{5}=5\\left(\\alpha^{6}+\\beta^{6}\\right)+6\\left(\\alpha^{5}+\\beta^{5}\\right)$ $=\\left(5 \\alpha^{4}+6 \\alpha^{5}\\right)+\\left(5 \\beta^{6}+6 \\beta^{5}\\right)$ $=\\alpha^{4}\\left(5 \\alpha^{2}+6 \\alpha\\right)+\\beta^{4}\\left(5 \\beta^{2}+6 \\beta\\right)$ $=2\\left(\\alpha^{4}+\\beta^{4}\\right)=2 S_{4}$", "$b$ and $c$, we will be finished. Note that $$\\gamma+\\delta=\\alpha^3+\\beta^3+\\frac{\\alpha^3+\\beta^3}{\\alpha^3\\beta^3}.$$ Now $\\alpha^3+\\beta^3=(\\alpha+\\beta)^3 -3\\alpha\\beta(\\alpha+\\beta)$. This can be easily expressed in terms of $b$ and $c$, and therefore so can $\\gamma+\\delta$. We leave dealing with $\\gamma\\delta$ to you. $(x-\\alpha)(x - \\beta) = x^2 + bx + c \\implies \\alpha + \\beta = -b,\\quad \\alpha\\beta = c$ If $r$ is a root of the equation $x^2 + bx + c = 0$, Then \\begin{align} r^2 + br + c &= 0 \\\\ r^2 &= -br - c \\\\ r^3 &= -br^2 - cr \\\\ &= -b(-br - c) - cr \\\\ &= b^2r + bc - cr\\\\ &= (b^2-c)r + bc \\end{align} It follows that \\begin{align} \\alpha^3 + \\beta^3 &= ((b^2-c)\\alpha + bc) + ((b^2-c)\\beta + bc) \\\\ &= (b^2-c)(\\alpha+\\beta) + 2bc \\\\ &= (b^2-c)(-b) + 2bc \\\\ &= -b^3 + bc + 2bc \\\\ &= -b^3 + 3bc \\end{align} and \\begin{align} \\alpha^3 \\beta^3 &= (\\alpha\\beta)^3 \\\\ &= c^3 \\end{align} So the polynomial is $(x-\\alpha^3)(x - \\beta^3) = x^2 + (b^3 - 3bc)x + c^3$ Just to check $(x-2)(x+3) = x^2+x-6$ So $\\quad b=1 \\quad \\text{and}\\quad c=-6$ $b^3 - 3bc = 1 + 18 = 19$ $c^3 = -216$ $(x-8)(x+27) = x^2 +19x - 216$", "cases where we know $$\\alpha$$, $$\\beta$$, and $$\\gamma$$, and where $$f$$ is easy to compute. Then combine such equations to solve for $$A$$, $$B$$, and $$C$$. The book has the new $$f$$ value listed first, then derives the values of the variables which satisfy that equation, but for clarity I will begin by defining values for the variables, then wave my hand and get the function $$f$$ which satisfies the recurrence under those variables. The first case (in the book) has, \\begin{align} \\alpha &= 1\\\\ \\beta &= 0\\\\ \\gamma &= 0\\\\ \\end{align} So that for $$n=2^m+l$$, \\begin{align} f(n) = 2^m \\end{align} Plugging in the values we get, \\begin{align} f(n) &= A(n) \\alpha + B(n) \\beta + C(n) \\gamma\\\\ 2^m &= A(n) \\end{align} Next we make, \\begin{align} \\alpha &= 1,\\\\ \\beta &= -1,\\\\ \\gamma &= -1. \\end{align} So that, \\begin{align} f(n) = 1. \\end{align} Plugging in the values we get, \\begin{align} f(n) &= A(n) \\alpha + B(n) \\beta + C(n) \\gamma\\\\ 1 &= A(n)-B(n)-C(n) \\end{align} And finally we take, \\begin{align} \\alpha &= 1\\\\ \\beta &= 0\\\\ \\gamma &= 1 \\end{align} So that, \\begin{align} f(n) = n \\end{align} Plugging in the values we get, \\begin{align} f(n) &= A(n) \\alpha + B(n) \\beta + C(n) \\gamma\\\\ n&= A(n) + C(n) \\end{align} Now we have, \\begin{align} A(n) &= 2^m\\\\ 1 &= A(n)", "1. Picked from choosing small random coefficients $x^3+a_1x+a_0$ and $x^2+b_1x+b_0$. \\begin{align} f_\\alpha(x) &= x^3+x-17, \\alpha\\approx 2.44176\\\\ f_\\beta(x) &= x^2+10x-20, \\beta = -3+\\sqrt 5\\\\ f_{\\alpha\\beta}(x) &= x^6+140 x^4+27200 x^3+400 x^2+68000 x-2312000 \\end{align} The other two roots of $f_{\\alpha}(x)$ are complex. Now \\begin{align} h(f_{\\alpha\\beta}) &= \\log(2312000)\\approx 14.6536\\\\ h(f_\\alpha) &= \\log(17)\\\\ h(f_\\beta) &= \\log(20)\\\\ nh(f_\\alpha) + mh(f_\\beta) &\\approx 14.6536 \\end{align} So $h(f_{\\alpha\\beta}) \\approx nh(f_\\alpha) + mh(f_\\beta)$. The \"extra\" term contributes: $$C(m,n) = \\log(\\binom{6}{3} (3+1)^{2/2} (2+1)^{3/2}) \\approx 6.02995$$ Example 2. Another one, picking random coefficients in $[-2^{16},2^{16}]$: \\begin{align} f_{\\alpha}(x) &= 26249 + 49279 x + x^3, \\alpha \\approx -0.532658\\\\ f_{\\beta}(x) &= -55643 + 47743 x + x^2, \\beta = \\frac{1}{2} \\left(-\\sqrt{2279616621}-47743\\right)\\\\ f_{\\alpha\\beta}(x) &= -118701754103013160002707 - 191207722325905327988897 x + 7518736182133771609 x^2 - 2856759510952394846 x^3 + 112331743403465 x^4 + x^6 \\end{align} giving heights: \\begin{align} h(f_\\alpha) &= \\log(49279)\\\\ h(f_\\beta) &= \\log(55643)\\\\ nh(f_\\alpha)+mh(f_\\beta) &\\approx 54.3906\\\\ h(f_{\\alpha\\beta}) &= \\log(191207722325905327988897)\\approx 53.6076 \\end{align} where $C(m,n)$ is still $\\approx 6.02995$. Example 3. An example where $h(f_{\\alpha\\beta}) > nh(f_\\alpha) + mh(f_\\beta)$. \\begin{align} f_\\alpha(x) &= -33909+19672 x+x^3\\\\ f_\\beta(x) &= -41717+40527 x+x^2\\\\ f_{\\alpha\\beta}(x) &= -83477449753997709876453+47047233759121009092744 x+673477622777766976 x^2+2257259257212603840 x^3+32311676318536 x^4+x^6 \\end{align} giving heights \\begin{align} h(f_\\alpha) &= \\log(33909)\\\\ h(f_\\beta) &= \\log(41717)\\\\ h(f_{\\alpha\\beta}) &= \\log(83477449753997709876453)\\\\ h(f_{\\alpha\\beta}) - nh(f_\\alpha) - mh(f_\\beta) &\\approx 7.10543*10^{-15} > 0 \\end{align}", "bi, \\beta^{\\prime} = c + di, \\gamma^{\\prime} = -(c - di), delta^{\\prime} = a - bi$ Then, $\\alpha\\delta - \\beta\\gamma = x^2 + y^2 + z^2 + t^2$ $\\alpha^{\\prime}\\delta^{\\prime} = beta^{\\prime}\\gamma^{\\prime} = a^2 + b^2 + c^2 + d^2$ $\\alpha\\alpha^{\\prime} + \\beta\\gamma^{\\prime} = (ax - by - cz - dt) + i(bx + ay + dz - ct)$ $\\gamma\\beta^{\\prime} + \\delta\\delta^{\\prime} = \\overline{\\alpha\\alpha^{\\prime} + \\beta\\gamma^{\\prime}}$ $\\therefore (\\alpha\\alpha^{\\prime} + \\beta\\gamma^{\\prime})(\\gamma\\beta^{\\prime} + \\delta\\delta^{\\prime}) = (ax - by - cz - dt)^2 + (bx + ay + dz - ct)^2$ Further $\\alpha\\beta^{\\prime} + \\beta\\delta^{\\prime} = (cx - dy + az + bt) + i(dx + cy -bz + at)$ $\\gamma\\alpha^{\\prime} + \\delta\\gamma^{\\prime} = -(cx - dy + az + by) + i(dz + cy - bz + at)$ i.e. $-(\\alpha\\beta^{\\prime} + \\beta\\delta^{\\prime})(\\gamma\\alpha^{\\prime} + \\delta\\gamma^{\\prime}) = (cx - dy + az + bt)^2 + (dx + cy - bz + at)^2$ Substituting the obtained expressions into the original identity, we find $(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + t^2) =$ $(ax - by -cz -dt)^2 + (bx + ay -dz + ct)^2 + (cx + dy + az -bt)^2 + (dx - cy + bz + at)^2$ 32. Expanding the expression $$(\\cos\\theta + i \\sin\\theta)^n$$ using binomial expansion, using De Moivre’s formula and comparing real and imaginary", "with small norm, yielding a linear system, then pinpoint possible generators. First we factors primes smaller than the Minkowski bound, primes not shown below are inert. \\begin{aligned}(3)&=(3,\\alpha)(3,\\alpha-1)(3,\\alpha+1) \\\\ (7)&=(7,\\alpha)(7,\\alpha^2-3)\\\\ (11)&=(11,\\alpha-1)(11,\\alpha^2+\\alpha+1)\\\\ (13)&=(13,\\alpha+3)(13,\\alpha^2-3\\alpha-6)\\\\ (17)&=(17,\\alpha-2)(17,\\alpha^2+2\\alpha-2)\\\\ (19)&=(19,\\alpha+7)(19,\\alpha^2-7\\alpha+3)\\\\ (23)&=(23,\\alpha-8)(23,\\alpha^2+8\\alpha+6)\\\\ (29)&=(29,\\alpha+4)(29,\\alpha+6)(29,\\alpha-10)\\\\ \\end{aligned} Let $$\\mathfrak{p}_1=(3,\\alpha),\\mathfrak{p}_2=(3,\\alpha-1),\\cdots,\\mathfrak{p}_{17}=(29,\\alpha+6),\\mathfrak{p}_{18}=(29,\\alpha-10)$$ which are named in their order of appearance. Now we find elements with small norm: $$N(\\alpha) = -3\\times 7 \\qquad N(\\alpha+1)=-3^2 \\qquad N(\\alpha-1)=-3\\times 11 \\qquad N(\\alpha+3)=-3\\times 13$$ $$N(\\alpha+4)=3\\times 29 \\qquad N(\\alpha^2+\\alpha-2)=-3^3\\times 11 \\qquad N(\\alpha^2-\\alpha-2)=3^3\\times 17$$ $$N(\\alpha^2-\\alpha+1)=3\\times 13\\times 19 \\qquad N(\\alpha^2-2\\alpha-2)=3\\times 13\\times 23 \\qquad N(3\\alpha-1)=-23\\times 29$$ $$N(4\\alpha-1)=-3^2\\times 13^2 \\qquad N(4\\alpha+7)=3\\times 7\\times 11 \\qquad N(4\\alpha+9)=3\\times 17\\times 19$$ Now we factor then into $$\\mathfrak{p}_n$$. For example, to factor $$(\\alpha)$$, we have three possibilities: $$(\\alpha) = \\mathfrak{p}_1\\mathfrak{p}_4 \\qquad \\mathfrak{p}_2\\mathfrak{p}_4 \\qquad \\mathfrak{p}_3\\mathfrak{p}_4$$ I rule out the latter two. If $$(\\alpha) = \\mathfrak{p}_2\\mathfrak{p}_4 = (3,\\alpha+2)(7,\\alpha-7)$$, then $$(\\alpha+2)(\\alpha-7)/\\alpha \\in \\mathcal{O}_K$$, but this number is not integral. If $$(\\alpha) = \\mathfrak{p}_3 \\mathfrak{p}_4 = (3,\\alpha-2)(7,\\alpha-7)$$, then $$(\\alpha-2)(\\alpha-7)/\\alpha \\in \\mathcal{O}_K$$, this is neither integral as well. Hence $$(\\alpha) = \\mathfrak{p}_1\\mathfrak{p}_4$$. Similarily, we can factor other ideals, giving $$(\\alpha) = \\mathfrak{p}_1\\mathfrak{p}_4 \\qquad (\\alpha+1) = \\mathfrak{p}_3^2 \\qquad (\\alpha-1) = \\mathfrak{p}_2 \\mathfrak{p}_6 \\qquad (\\alpha+3) = \\mathfrak{p}_1 \\mathfrak{p}_8$$ $$(\\alpha+4) = \\mathfrak{p}_3 \\mathfrak{p}_{16} \\qquad (\\alpha^2+\\alpha-2) = \\mathfrak{p}_2^3 \\mathfrak{p}_6 \\qquad (\\alpha^2-\\alpha-2) = \\mathfrak{p}_3^3 \\mathfrak{p}_{10}$$ $$(\\alpha^2-\\alpha+1) = \\mathfrak{p}_3 \\mathfrak{p}_8 \\mathfrak{p}_{12} \\qquad (\\alpha^2-2\\alpha-2) = \\mathfrak{p}_2 \\mathfrak{p}_8 \\mathfrak{p}_{14} \\qquad (3\\alpha-1) = \\mathfrak{p}_{14} \\mathfrak{p}_{18}$$ $$(4\\alpha-1) = \\mathfrak{p}_2^2 \\mathfrak{p}_8^2 \\qquad (4\\alpha+7)", "is $$2\\alpha+\\beta=5,\\quad 3\\beta+\\alpha=0.$$ From the second one, we get $\\alpha=-3\\beta$.then from the first one we obtain $$2(-3\\beta)+\\beta=5\\implies \\beta=-1$$ and finally $\\alpha=-3\\beta=-3(-1)=3$. P.S. In the solution on your book, the system is transformed into equivalent ones by replacing one equation with sum of that equation with a multiple of the other one. See Reduced Row-Echelon Form in your pdf. $$\\begin{bmatrix} 2&1&5\\\\1&3&0\\\\ \\end{bmatrix} \\overset{}{\\to} \\begin{bmatrix} 1&-2&5\\\\1&3&0\\\\ \\end{bmatrix} \\overset{}{\\to} \\begin{bmatrix} 1&-2&5\\\\0&5&-5\\\\ \\end{bmatrix}\\\\ \\overset{}{\\to} \\begin{bmatrix} 1&-2&5\\\\0&1&-1\\\\ \\end{bmatrix} \\overset{}{\\to} \\begin{bmatrix} 1&0&3\\\\0&1&-1\\\\ \\end{bmatrix}.$$", "+ 2}(f_{4}^{f_{1}^{f_{2}(3).2 + 2}(f_{2}^{f_{1}^{2}(3)}(f_{\\omega + 2}(3))) + 1}(f_{\\omega^2.2 + \\omega + 2}^{2}(3))).2 + f_{3}(f_{\\omega}^{f_{\\omega}^{2}(3) + 2}(f_{\\omega + 1}(3)))$$ corrected using Version 2 code $$\\beta(9,3) = f_{\\varphi(1,0)}(3)$$ $$\\beta(27,3) = f_{\\varphi(1,0,0)}(3) = f_{\\Gamma_0}(3)$$ $$\\beta(81,3) = f_{\\varphi(1,0,0,0)}(3) = f_{SVO}(3)$$ by definition Base v = 4 $$\\beta(0,4) = 0$$ $$\\beta(1,4) = 4$$ $$\\beta(2,4) = f_{2}(4) = 64$$ $$\\beta(e,4) = \\beta(2.71828182845905,4) = f_{1}^{2}(f_{2}^{3}(4)) + f_{1}^{3}(f_{2}(4)) + f_{1}^{3}(4) + 2$$ $$\\beta(3,4) = f_{1}^{4}(f_{3}(4)) + f_{1}^{2}(4) + 12$$ $$\\beta(\\pi,4) = \\beta(3.14159265358979,4) = f_{1}^{3}(f_{2}^{f_{1}^{2}(4)}(f_{3}(4))) + 1$$ $$\\beta(4,4) = f_{2}^{f_{1}^{4}(f_{2}^{f_{2}^{f_{1}^{f_{1}^{f_{1}^{f_{1}^{3}(4) + f_{1}^{2}(4) + 15}(f_{2}(4)) + f_{1}^{f_{1}^{3}(4) + f_{1}^{2}(4) + 6}(f_{2}(4)) + 6}(f_{2}^{2}(4)) + 2}(f_{2}^{3}(4))}(f_{3}(4)) + 3}(f_{3}^{2}(4)))}(f_{3}^{3}(4))$$ $$\\beta(5,4) = f_{f_{1}^{3}(4) + 2}(f_{\\omega^2.2 + 2}^{3}(4)) + 7$$ $$\\beta(16,4) = f_{\\varphi(1,0)}(4)$$ $$\\beta(64,4) = f_{\\varphi(1,0,0)}(4) = f_{\\Gamma_0}(4)$$ $$\\beta(256,4) = f_{\\varphi(1,0,0,0)}(4)$$ $$\\beta(1024,4) = f_{\\varphi(1,0,0,0,0)}(4) = f_{SVO}(4)$$ by definition $$\\beta(2.99046262281061,3)$$ $$= f_{1}^{f_{1}^{3}(f_{2}(3)) + f_{1}^{2}(f_{2}(3)) + f_{1}^{2}(3) + 3}(f_{2}^{2}(3)) + f_{1}^{f_{1}^{3}(f_{2}(3)) + f_{1}^{2}(f_{2}(3)) + f_{1}^{2}(3) + 2}(f_{2}^{2}(3)) + f_{1}^{f_{1}^{3}(f_{2}(3)) + 1}(f_{2}^{2}(3)) + 1$$ $$<$$ Googol $$= 10^{100}$$ $$< \\beta(2.99046262281062,3) = f_{1}^{f_{1}^{3}(f_{2}(3)) + f_{1}^{2}(f_{2}(3)) + f_{1}^{2}(3) + 4}(f_{2}^{2}(3))$$ ## Granularity Examples of this Function The Beta Function uses the full depth of the Real Numbers to enable almost every ordinal and big number to be described. In these examples, corrected using Version 3 code, big numbers can be accessed around a large ordinal by slightly changing the precision of the input real number. $$\\beta(12.02,3) =", "3&1&5&7&6&4&0&2\\cr 4&6&2&0&1&3&7&5\\cr 5&7&3&1&0&2&6&4\\cr 6&4&0&2&3&1&5&7\\cr 7&5&1&3&2&0&4&6\\cr}.$$ Similarly we get that $$L_{\\alpha+1}=\\pmatrix{ 0&3&4&7&1&2&5&6\\cr 1&2&5&6&0&3&4&7\\cr 2&1&6&5&3&0&7&4\\cr 3&0&7&4&2&1&6&5\\cr 4&7&0&3&5&6&1&2\\cr 5&6&1&2&4&7&0&3\\cr 6&5&2&1&7&4&3&0\\cr 7&4&3&0&6&5&2&1\\cr}.$$ The general theory tells us that $L_\\alpha$ and $L_{\\alpha+1}$ are MOLS. We observe that, in accordance with the earlier part of the solution, they both have the required extra property. We also observe that the two MOLS share the same set of columns. This is built in to the method of construction. - Alternatively we could arrange the other diagonal to have entries $L(x,\\gamma-x)=L(x,\\gamma+x)$ for any non-zero \"pairwise sum $\\gamma\\in F$\". Also note that this is specific to characteristic two. In other characteristics we might place zero in the middle and pair up $x$ and $-x$. – Jyrki Lahtonen Jul 19 '12 at 17:01", "= 3$. Compute the following: a. $E(X)$ and $V(X)$ b. $P(X\\leq 6)$ c. $P(1.8\\leq X \\leq 6)$ d. $P(X\\geq 3)$. #### Solution Let $X$ denote the lifetime (in hundreds of hours) of vaccume tube. Given that $X\\sim W(\\alpha,\\beta)$, where $\\alpha =2$ and $\\beta=3$. The probability density function of $X$ is \\begin{aligned} f(x;\\alpha, \\beta)&=\\frac{\\alpha}{\\beta} \\big(\\dfrac{x}{\\beta}\\big)^{\\alpha-1}e^{-\\big(\\dfrac{x}{\\beta}\\big)^\\alpha};\\; x>0,\\alpha,\\beta>0. \\end{aligned} The distribution function of $X$ is \\begin{aligned} F(x) &= 1- e^{-\\big(x/\\beta\\big)^\\alpha}. \\end{aligned} a. Mean and variance of $X$ \\begin{aligned} E(X) &= \\beta \\Gamma (\\dfrac{1}{\\alpha}+1)\\\\ &=3\\Gamma(\\dfrac{1}{2}+1)\\\\ &=3\\Gamma(3/2)\\\\ &=3\\times\\dfrac{1}{2}\\Gamma(1/2)\\\\ &=\\dfrac{3}{2}\\times\\sqrt{\\pi}\\\\ &=\\dfrac{3}{2}\\times1.7725\\\\ &=2.6588 \\end{aligned} \\begin{aligned} V(X) &= \\beta^2 \\bigg[\\Gamma (\\dfrac{2}{\\alpha}+1) -\\bigg(\\Gamma (\\dfrac{1}{\\alpha}+1) \\bigg)^2\\bigg]\\\\ &=3^2 \\bigg[\\Gamma (\\dfrac{2}{2}+1) -\\bigg(\\Gamma (\\dfrac{1}{2}+1) \\bigg)^2\\bigg]\\\\ &=9\\bigg[\\Gamma(2)-\\big(\\Gamma(3/2)\\big)^2\\bigg]\\\\ &=9\\bigg[1-\\bigg(\\frac{1}{2}\\Gamma(1/2)\\bigg)^2\\bigg]\\\\ &=9\\bigg[1-\\bigg(\\frac{\\sqrt{\\pi}}{2}\\bigg)^2\\bigg]\\\\ &=9\\bigg[1-\\bigg(\\frac{\\sqrt{3.1416}}{2}\\bigg)^2\\bigg]\\\\ &=1.931846 \\end{aligned} b. $P(X\\leq 6)$ \\begin{aligned} P(X\\leq 6) &=F(6)\\\\ &= 1-e^{-(6/3)^{2}}\\\\ &= 1-e^{-(2)^{2}}\\\\ &= 1-e^{-(4)}\\\\ &=1-0.0183\\\\ &=0.9817 \\end{aligned} c. $P(1.8\\leq X \\leq 6)$ \\begin{aligned} P(1.8 \\leq X\\leq 6) &=F(6)-F(1.8)\\\\ &= \\bigg[1-e^{-(6/3)^{2}}\\bigg] -\\bigg[1-e^{-(1.8/3)^{2}}\\bigg]\\\\ &= e^{-(0.6)^{2}}-e^{-(2)^{2}}\\\\ &= e^{-(0.36)}-e^{-(4)}\\\\ &=0.6977-0.0183\\\\ &=0.6794 \\end{aligned} d. $P(X\\geq 3)$ \\begin{aligned} P(X\\geq 3) &=1-P(X< 3)\\\\ &= 1-F(3)\\\\ &= 1-\\bigg[1-e^{-(3/3)^{2}}\\bigg]\\\\ &= e^{-(1)^{2}}\\\\ &=0.3679 \\end{aligned} ## Weibull Distribution Example 2 Assume that the life of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with $\\alpha = 300$ hours and $\\beta = 0.5$. Calculate the probability that a. a disk lasts at least 600 hours, b. a disk fails before 500 hours. #### Solution Let $X$ denote the life" ]

[ "24 '18 at 12:34 • @italy John Doe's comment above this is the reason why. Do you see now? – Kevin Apr 24 '18 at 15:24 Note that $$64r^2S^4+4=4(16r^2S^4+1)$$ so $$\\sqrt{64r^2S^4+4}=\\sqrt{4(16r^2S^4+1)}=2\\sqrt{16r^2S^4+1}$$ Finally, \\begin{align} \\frac{-8rS^2\\pm\\sqrt{64r^2S^4+4}}{2} &=\\frac{-8rS^2\\pm2\\sqrt{16r^2S^4+1}}{2}\\\\[6px] &=\\frac{2(-4rS^2\\pm\\sqrt{16r^2S^4+1}\\,)}{2}\\\\[6px] &=-4rS^2\\pm\\sqrt{16r^2S^4+1} \\end{align} Your original equation apparently was $$x^2+8rS^2x-1=0$$ Whenever you have a factor2$that can be extracted from the coefficient of the degree$1$term, the simplification above can be performed: if the equation is$ax^2+2\\beta x+c=0, the quadratic formula yields \\begin{align} \\frac{-2\\beta\\pm\\sqrt{4\\beta^2-4ac}}{2a} &=\\frac{-2\\beta\\pm\\sqrt{4(\\beta^2-ac)}\\,}{2a}\\\\[6px] &=\\frac{-2\\beta\\pm2\\sqrt{\\beta^2-ac}}{2a}\\\\[6px] &=\\frac{-\\beta\\pm\\sqrt{\\beta^2-ac}}{a} \\end{align} In your casea=1$,$\\beta=-4rS^2$and$c=-1$. It's not necessary to remember one more formula, just to recall that the simplification can be done. We have$\\frac{-8rS^2}{2}=-4rS^2$and$\\frac{\\sqrt{4X}}{2}=\\sqrt{X}$. • This answer could be helped by doing that last simplification over an extra step ($2=\\sqrt 4\\$ for example). That seems to be where OP's confusion lies. – John Doe Apr 24 '18 at 11:19 • @JohnDoe: seeing his comment, the OP believes in the distributivity of the square root over addition ! – Yves Daoust Apr 24 '18 at 11:58 • @YvesDaoust Ohh right I see, well thats a completely different issue! – John Doe Apr 24 '18 at 12:31", "14.4 & & 138.03 & 1.7 & 2 \\\\ \\eta & 0^{-} & 0^{+} & 3 & & 548.8 & 1.5 & 2 \\\\ \\rho & 1^{-} & 1^{+} & 0.9 & 6.1 & 769 & 1.85 & 4 \\\\ \\omega & 1^{-} & 0^{-} & 24.5 & 0 & 782.6 & 1.85 & 4 \\\\ \\delta & 0^{+} & 1^{-} & 2.488 & & 983 & 2.0 & 2 \\\\ \\sigma^{a} & 0^{+} & 0^{+} & 8.9437 & & 550 & 1.9 & 2 \\\\ & & & (18.3773) & & (720) & (2.0) & (2) \\\\ \\hline \\end{array}$$$$ At each vertex is used a form factor like this: $$$$\\mathcal{F}_{\\alpha}\\left[\\left(\\boldsymbol{q}^{\\prime}-\\boldsymbol{q}\\right)^{2}\\right]=\\left[\\frac{\\Lambda_{\\alpha}^{2}-m_{\\alpha}^{2}}{\\Lambda_{\\alpha}^{2}+\\left(\\boldsymbol{q}^{\\prime}-\\boldsymbol{q}\\right)^{2}}\\right]^{n_{\\alpha}}$$$$ Lastly i use the following equation to compute the Cross section $$$$\\sigma=\\frac{1}{64 \\pi^{2} s} \\frac{\\mathrm{p}_{f}}{\\mathrm{p}_{i}} \\int\\left|\\mathcal{M}_{f i}\\right|^{2} \\mathrm{~d} \\Omega$$$$ Where the cross section must include a factor of $$1/2$$ because each final state may only be counted once in integrating over the angles To my mind all of this should yield reasonable results. Not exact since this is a tree level treatment of the scattering but my results as i mentioned before do not even come close to the experimental results. I know that for Moller Scattering (Elastic electron scattering) tree level treatment is not enough since the integral is diverging. Is this one case where tree", "For case 1, \\begin{align}(1)&\\implies \\frac 1n+\\frac{1}{2n}=\\alpha+\\frac 13\\\\&\\implies \\alpha=\\frac{9-2n}{6n}\\\\&\\implies0\\le \\frac{9-2n}{6n}\\lt \\frac 12\\\\&\\implies (n,\\alpha)=(2,5/12),(3,1/6),(4,1/24)\\end{align} I think that you can do for case 2 similarly. • Sir, could you please show me how you went from $0\\le \\frac{9-2n}{6n}\\lt \\frac 12$ to $(n,\\alpha)=(2,5/12),(3,1/6),(4,1/24)$? – user342209 Jun 10 '16 at 15:09 • @user342209: From $0\\le (9-2n)/(6n)\\lt 1/2$, we have $n=2,3,4$. For each $n$, you can get $\\alpha$ from $\\alpha=(9-2n)/(6n)$. – mathlove Jun 10 '16 at 15:11 • Thank you Sir. Actually I had meant to ask how you got the values of $n$. – user342209 Jun 10 '16 at 15:12 • Did you perhaps use the Wavy Curve method, and then just consider the Integral values in the intervals for the values of $n$? – user342209 Jun 10 '16 at 15:13 • @user342209: Since $n\\not=0$, you can multiply it by $6n^2\\ (\\gt 0)$ to get $0\\le n(9-2n)\\lt 3n^2$, i.e. $n(2n-9)\\le 0$ and $n(5n-9)\\gt 0$. I think that you can continue from here. – mathlove Jun 10 '16 at 16:05 We start by rewriting $\\left\\lfloor{x}\\right\\rfloor = k$ and $\\{x\\} = d$, with $k \\in \\Bbb{N}$ and $d \\in [0, 1)$, where we exclude negative integers thanks to a comment on the OP. Now there are two cases: either $d \\in [0, 0.5)$ or $d \\in [0.5, 1)$. Notice that if $d \\in [0.5, 1)$", "5 & 6\\\\ 2 & 1 & 4 & 5 & 6 & 3 \\end{pmatrix}$ has two cycles, namely: (4) \\begin{align} \\quad (12) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 2 & 1 & 3 & 4 & 5 & 6 \\end{pmatrix} , \\quad (3456) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 2 & 4 & 5 & 6 & 3 \\end{pmatrix} \\end{align} We can also compose cycles. Let $\\alpha = (12)$ and $\\beta = (3456)$ and consider the composition $(\\alpha \\circ \\beta)(x) = \\alpha(\\beta(x))$ for all $x \\in \\{ 1, 2, 3, 4, 5, 6 \\}$. We have that $(\\alpha(\\beta(1)) = \\alpha(1) = 2$, $(\\alpha(\\beta(2)) = \\alpha(2) = 1$, $\\alpha(\\beta(3)) = \\alpha(4) = 4$, $\\alpha(\\beta(4)) = \\alpha(5) = 5$, $\\alpha(\\beta(5)) = \\alpha(6) = 6$, and $\\alpha(\\beta(6)) = \\alpha(3) = 3$, and so: (5) \\begin{align} \\quad \\alpha \\circ \\beta = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 2 & 1 & 4 & 5 & 6 & 3 \\end{pmatrix} = \\sigma \\end{align}", "is added to their sum, the result is positive. ... 9 As Lucian pointed out this follows immediately from the properties of cyclotomic polynomials. We have the factorization (into polynomials irreducible over \\Bbb{Q}):$$ x^n-1=\\prod_{d\\mid n}\\Phi_d(x). $$Your observation follows from the following as factorization:$$ a^n-b^n=b^n[\\left(\\frac ab\\right)^n-1]=b^n\\prod_{d\\mid n}\\Phi_d(\\frac ab)= \\prod_{d\\mid ... 8 Let $$f(x)=2x^3-3x^2-12x+1$$ Now, $f(0)>0$ and $f(1)<0 \\implies 0< \\alpha <1$ $f(-2)<0$ and $f(-1)>0 \\implies -2< \\beta <-1$ $f(3)<0$ and $f(4)>0 \\implies 3< \\gamma <4$ $\\therefore \\lfloor\\alpha\\rfloor + \\lfloor\\beta\\rfloor + \\lfloor\\gamma\\rfloor = 0+(-2)+3=1$ 8 $$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$ Set $t=x^2,z=x^2+x+1$. $\\Longrightarrow$ \\begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\\\&=(x^2-x-1)(4x^2-x-1)\\end{align} $$\\boxed{\\color{red}{x_{1,2}=\\frac{1}{2}\\pm\\frac{\\sqrt5}{2},\\;x_{3,4}=\\frac{1}{8}\\pm \\frac{\\sqrt{17}}{8}}}$$ 7 Set $A=x^2,B=x^2+x+1$. Then, \\begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\\\&=(2A-B)(5A-B)\\\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\\\&=(x^2-x-1)(4x^2-x-1)\\end{align} 7 Localise the roots between two consecutive integers and use the Intermediate value theorem. Let $p(x)=2x^3-3x^2-12x+1$. As $p'(x)=6x^2-6x-12=6(x+1)(x-2)$, $p(x)$ has a local maximum at $x=-1$ and a local minimum at $x=2$: $$p(-1)=8,\\quad p(2)=-19.$$ Thus we know there are three real roots: $x1<-1$, $-1< x_2 <2$, $x_3>2$. Now a table of the ... 7 For a simple approach consider the function $y=x^4+4x^3-8x^2=x^2\\cdot\\left((x+2)^2-12\\right)$ - the intersections with the line $y=-p$ will give the roots of the original. Since this is just a horizontal line in the normal $x,y$ plane, a quick sketch will show that the number of real roots is governed by the relationship of $p$ to the local minima/maxima of ... 7 The route you take is fruitful. $p\\left(x\\right)=\\left(x-2\\right)\\left(x-8\\right)q\\left(x\\right)$ leads to: $$\\left(x-4\\right)q\\left(2x\\right)=2\\left(x-2\\right)q\\left(x\\right)$$ Then", "# On $\\big(\\tfrac{1+\\sqrt{5}}{2}\\big)^{12}=\\small 161+72\\sqrt{5}$ and $\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small3/4} \\sqrt[4]{161+72\\sqrt{5}\\,x}}$ (This summarizes scattered results from here, here, here and elsewhere. See also this older post.) I. Cubic Define $\\beta= \\tfrac{\\Gamma\\big(\\tfrac56\\big)}{\\Gamma\\big(\\tfrac13\\big)\\sqrt{\\pi}}= \\frac{1}{48^{1/4}\\,K(k_3)}$. Then we have the nice evaluations, \\begin{aligned}\\frac{3}{5^{5/6}} &=\\,_2F_1\\big(\\tfrac{1}{3},\\tfrac{1}{3};\\tfrac{5}{6};-4\\big)\\\\ &=\\beta\\,\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[3]{x^2+4x^3}}\\\\[1.7mm] &=\\beta\\,\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small2/3} \\sqrt[3]{\\color{blue}{9+4\\sqrt{5}}\\,x}}\\\\[1.7mm] &=2^{1/3}\\,\\beta\\,\\int_0^\\infty\\frac{dx}{\\sqrt[3]{9+\\cosh x}} \\end{aligned}\\tag1 and, \\begin{aligned}\\frac{4}{7} &=\\,_2F_1\\big(\\tfrac{1}{3},\\tfrac{1}{3};\\tfrac{5}{6};-27\\big)\\\\ &=\\beta\\,\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[3]{x^2+27x^3}}\\\\[1.7mm] &=\\beta\\,\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small2/3} \\sqrt[3]{\\color{blue}{55+12\\sqrt{21}}\\,x}}\\\\[1.7mm] &=2^{1/3}\\,\\beta\\,\\int_0^\\infty\\frac{dx}{\\sqrt[3]{55+\\cosh x}} \\end{aligned}\\tag2 Note the powers of fundamental units, $$U_{5}^6 = \\big(\\tfrac{1+\\sqrt{5}}{2}\\big)^6=\\color{blue}{9+4\\sqrt{5}}$$ $$U_{21}^3 = \\big(\\tfrac{5+\\sqrt{21}}{2}\\big)^3=\\color{blue}{55+12\\sqrt{21}}$$ Those two instances can't be coincidence. II. Quartic Define $\\gamma= \\tfrac{\\sqrt{2\\pi}}{\\Gamma^2\\big(\\tfrac14\\big)}= \\frac{1}{2\\sqrt2\\,K(k_1)}=\\frac1{2L}$ with lemniscate constant $L$. Then we have the nice, \\begin{aligned}\\frac{2}{3^{3/4}} &=\\,_2F_1\\big(\\tfrac{1}{4},\\tfrac{1}{4};\\tfrac{3}{4};-3\\big)\\\\ &=\\gamma\\,\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[4]{x^3+3x^4}}\\\\[1.7mm] &\\overset{\\color{red}?}=\\gamma\\,\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small3/4} \\sqrt[4]{\\color{blue}{7+4\\sqrt{3}}\\,x}}\\\\[1.7mm] &=2^{1/4}\\,\\gamma\\,\\int_0^\\infty\\frac{dx}{\\sqrt[4]{7+\\cosh x}} \\end{aligned}\\tag3 and, \\begin{aligned}\\frac{3}{5}&=\\,_2F_1\\big(\\tfrac{1}{4},\\tfrac{1}{4};\\tfrac{3}{4};-80\\big)\\\\ &=\\gamma\\,\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[4]{x^3+80x^4}}\\\\[1.7mm] &\\overset{\\color{red}?}=\\gamma\\,\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small3/4} \\sqrt[4]{\\color{blue}{161+72\\sqrt{5}}\\,x}}\\\\[1.7mm] &=2^{1/4}\\,\\gamma\\,\\int_0^\\infty\\frac{dx}{\\sqrt[4]{161+\\cosh x}} \\end{aligned}\\tag4 with $a=161$ given by Noam Elkies in this comment. (For $4$th roots, I just assumed the equality using the blue radicals based on the ones for cube roots.) Note again the powers of fundamental units, $$U_{3}^2 = \\big(2+\\sqrt3\\big)^2=\\color{blue}{7+4\\sqrt{3}}$$ $$U_{5}^{12} = \\big(\\tfrac{1+\\sqrt{5}}{2}\\big)^{12}=\\color{blue}{161+72\\sqrt{5}}$$ Just like for the cube roots version, these can't be coincidence. Questions: Is it true these observations can be explained by, let $b=2a+1$, then, $$\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[3]{x^2+ax^3}}=\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small2/3} \\sqrt[3]{b+\\sqrt{b^2-1}\\,x}}=2^{1/3}\\int_0^\\infty\\frac{dx}{\\sqrt[3]{b+\\cosh x}}$$ $$\\int_0^1 \\frac{dx}{\\sqrt{1-x}\\,\\sqrt[4]{x^3+ax^4}}=\\int_{-1}^1\\frac{dx}{\\left(1-x^2\\right)^{\\small3/4} \\sqrt[4]{b+\\sqrt{b^2-1}\\,x}}=2^{1/4}\\int_0^\\infty\\frac{dx}{\\sqrt[4]{b+\\cosh x}}$$ • Hello, I really admire your work (especially on your blog), and I really want to someday be as much of an expert as you are. May I ask, where were you educated/where did you learn so much amazing number theory?", "solutions. Here $n=3$ so, let's take $m=0,1,2$. $m=0\\implies \\cos\\frac{\\pi}{3}+i\\sin \\frac{\\pi}{3}=\\frac{1+i\\sqrt3}{2}$ $m=1\\implies \\cos \\pi+i\\sin\\pi=-1$ $m=2\\implies \\cos\\frac{5\\pi}{3}+i\\sin \\frac{5\\pi}{3}=\\frac{1-i\\sqrt3}{2}$ - I think the second factor in your first line should be $x^2-x+1$. – Daryl Sep 8 '12 at 12:32 @Daryl, thanks for identifying the lapsus calami. – lab bhattacharjee Sep 8 '12 at 12:34 Hi, where did you got $x^3+1=(x+1)(x^2-x+1)$? I mean also the site, because I found $x^n-1=(x-1)(x^{n-1}+ \\dots +1)$ - planetmath.org/encyclopedia/PrimeFactorsOfXn1.html. – alvoutila Sep 8 '12 at 19:02 $x^3+y^3=(x)^3+(y)^3=(x+y)^3-3xy(x+y)=(x+y)((x+y)^2-3xy)=(x+y)(x^2+y^2-xy)$ – lab bhattacharjee Sep 9 '12 at 3:39 Let $x=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^3 = -1$. Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$ We can separate the real and imaginary parts of this equation: \\begin{align} a^3 -3ab^2 & = -1 \\\\ 3a^2b - b^3 & = 0 \\end{align} Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution $$x =-1.$$ So suppose $b\\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \\pm a\\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$a^3 - 9a^3 = -1$$ so $a = \\frac12.$ Since $b = \\pm a\\sqrt3$, we have $b=\\pm\\frac{\\sqrt3}2$. This gives us the other two solutions, namely $$x=\\frac12 \\pm\\!\\frac{\\sqrt3}2i.$$ - Set $\\displaystyle x=re^{i \\theta}$. So $\\displaystyle r^3e^{i3\\theta}= x^3= -1= e^{i \\pi}$, hence $r^3=1$ and $3", "\\frac{ y_{12} }{ 6 } + \\frac{ 2y_{13} }{ 3 } - \\frac{ y_{21} }{ 6 } - \\frac{ y_{22} }{ 6 } + \\frac{ y_{23} }{ 6 } - \\beta_3 - \\mu \\\\ \\alpha_2 & = -\\frac{ y_{11} }{ 6 } - \\frac{ y_{12} }{ 6 } + \\frac{ y_{13} }{ 3 } + \\frac{ y_{21} }{ 6 } + \\frac{ y_{22} }{ 6 } + \\frac{ 2 y_{23} }{ 3 }- \\beta_3 - \\mu \\\\ \\beta_1 & = \\frac{ y_{11} }{ 2 } - \\frac{ y_{13} }{ 2 } + \\frac{ y_{21} }{ 2 } - \\frac{ y_{23} }{ 2 } + \\beta_3 \\\\ \\beta_2 & = \\frac{ y_{12}}{ 2 } - \\frac{ y_{13} }{ 2 } + \\frac{ y_{22} }{ 2 } - \\frac{ y_{23} }{ 2 } + \\beta_3 \\end{align} ## c Give a set of basis vectors for $\\mathcal N (X)$. We need to find the basis of vectors $a$ such that $X a = 0.$ \\begin{align} a_1 + a_2 + a_3 + a_4 + a_5 + a_6 & = 0 \\\\ a_1 + a_2 + a_3 & = 0 \\\\ a_4 + a_5 + a_6 & = 0 \\\\ a_1 + a_4 & = 0 \\\\ a_2 + a_5 & = 0 \\\\ a_3 + a_6 & = 0 \\end{align} Fix $a_5$", "and smaller than $\\frac43 \\beta$ at the same time. So you can pick a number between $\\frac43$ and $2$ say $\\frac53$ then discuss the following $2$ cases : 1. $|x|>|y|^{\\frac53}$, then $x^3 y^2 < x^{3 + \\frac65} = x^{4.2}$, and $x^4 + y^6 > x^4$ thus $\\frac{x^3y^2}{x^4 + y^6} < x^{0.2}$. 2. $|x| \\leqslant |y|^{\\frac53}$, then $x^3 y^2 \\leqslant y^{5 + 2} = y^{7}$, and $x^4 + y^6 \\leqslant y^6$ thus $\\frac{x^3y^2}{x^4 + y^6} \\leqslant y$. So in summary $\\frac{x^3y^2}{x^4 + y^6} \\leqslant \\max{\\{x^{0.2},y\\}} < \\left(x^2 + y^2\\right)^{0.1}$, thus converges to $0$. if $(x,y)$ approaches $(0,0)$ while $x^3=y^2$, this doesn't tend to 0, so... (sorry, I read $\\frac{x^2y^3}{x^4+y^6}$.) You may want to try $X^2=x^4$ and $Y^2=y^6$, and use polar coordinates $X=r\\cos\\theta$, $Y=r\\sin\\theta$. You should get a $r^{\\alpha}$ estimate, which readily gives the answer...", "/\\partial {{\\beta }_{1}}=0\\,\\!$: ${{x}_{i1}}\\ln {{t}_{1}}+{{x}_{i1}}\\ln {{t}_{2}}+{{x}_{i1}}\\ln {{t}_{3}}+{{x}_{i1}}\\ln {{t}_{4}}-4{{\\beta }_{1}}=0\\,\\!$ Thus: \\begin{align} {{{\\hat{\\beta }}}_{1}} & = & \\frac{1}{4}(-\\ln {{t}_{1}}+\\ln {{t}_{2}}-\\ln {{t}_{3}}+\\ln {{t}_{4}}) \\\\ & = & \\frac{1}{4}(-3.2958+3.2189-3.912+4.0073) \\\\ & = & 0.0046 \\end{align}\\,\\! Setting $\\partial \\Lambda /\\partial {{\\beta }_{2}}=0\\,\\!$: ${{x}_{i2}}\\ln {{t}_{1}}+{{x}_{i2}}\\ln {{t}_{2}}+{{x}_{i3}}\\ln {{t}_{3}}+{{x}_{i4}}\\ln {{t}_{4}}-4{{\\beta }_{2}}=0\\,\\!$ Thus: \\begin{align} {{{\\hat{\\beta }}}_{2}} & = & \\frac{1}{4}(-\\ln {{t}_{1}}-\\ln {{t}_{2}}+\\ln {{t}_{3}}+\\ln {{t}_{4}}) \\\\ & = & \\frac{1}{4}(-3.2958-3.2189+3.912+4.0073) \\\\ & = & 0.3512 \\end{align}\\,\\! Knowing ${{\\hat{\\beta }}_{0}}\\,\\!$, ${{\\hat{\\beta }}_{1}}\\,\\!$ and ${{\\hat{\\beta }}_{2}}\\,\\!$, ${{\\hat{\\sigma }}^{\\prime }}\\,\\!$ can now be obtained. Setting $\\partial \\Lambda /\\partial {\\sigma }'=0\\,\\!$: $-\\frac{4}{{{\\sigma }'}}+\\frac{1}{{{({\\sigma }')}^{3}}}\\underset{i=1}{\\overset{4}{\\mathop \\sum }}\\,{{[\\ln ({{t}_{i}})-(3.6085+0.0046{{x}_{i1}}+0.3512{{x}_{i2}})]}^{2}}=0\\,\\!$ Thus: \\begin{align} {{{\\hat{\\sigma }}}^{\\prime }} & = & \\frac{1}{2}\\sqrt{\\underset{i=1}{\\overset{4}{\\mathop \\sum }}\\,{{[\\ln ({{t}_{i}})-(3.6085+0.0046{{x}_{i1}}+0.3512{{x}_{i2}})]}^{2}}} \\\\ & = & 0.043 \\end{align}\\,\\! Once the estimates have been calculated, the likelihood ratio test can be carried out for the two factors. ### Likelihood Ratio Test The likelihood ratio test for factor $A\\,\\!$ is conducted by using the likelihood value corresponding to the full model and the likelihood value when $A\\,\\!$ is not included in the model. The likelihood value corresponding to the full model (in this case $\\mu _{i}^{\\prime }={{\\beta }_{0}}+{{\\beta }_{1}}{{x}_{i1}}+{{\\beta }_{2}}{{x}_{i2}}\\,\\!$) is: \\begin{align} L & = & \\underset{i=1}{\\overset{4}{\\mathop \\prod }}\\,\\left[ \\frac{1}{{{t}_{i}}{{{\\hat{\\sigma }}}^{\\prime }}\\sqrt{2\\pi }}{{e}^{-\\frac{1}{2}{{\\left( \\frac{\\ln ({{t}_{i}})-({{{\\hat{\\beta }}}_{0}}+{{{\\hat{\\beta }}}_{1}}{{x}_{i1}}+{{{\\hat{\\beta }}}_{2}}{{x}_{i2}})}{{{{\\hat{\\sigma }}}^{\\prime }}} \\right)}^{2}}}} \\right] \\\\ & = & 0.000537311 \\end{align}\\,\\! The corresponding logarithmic value is $\\ln (L)=\\ln (0.000537311)=-7.529\\,\\!$.", "$4$ and $6$ is that these are the only values of $n$ such that $p_n(x)$ has degree at most two. $\\sin\\pi/5$ is still pretty special, as the roots of $p_5(x)$ can be expressed in terms of radicals. This will be true for infinitely many, but not all $n$; the smallest $n$ for which $\\sin \\pi/n$ cannot be expressed in terms of radicals is $n=11$. A nice way to see these polynomials is via deMoivre's formula. Write $\\sin x = (u-1/u)/2i$ where $u=e^{ix}$. For $n=3$ the binomial theorem gives \\begin{align*} (2i)^3\\sin^3 x &=(u-1/u)^3\\\\ &=(u^3-3u+3/u-1/u^3)\\\\ &=2i(\\sin 3x -3\\sin x) \\end{align*} giving $\\sin 3x=3\\sin x- 4\\sin^3x$. Taking $x=\\pi/3$ and $z=\\sin \\pi/3$, we get $$0= 3z-4z^3,$$ giving $p_3(z)$. You can get $p_5$ similarly. • The Chebyshev polynomials are fascinating, thank you for introducing me to them. – electronpusher Dec 26 '16 at 1:06 You can use the addition and subtraction formula together with a table of sine and cosine values to calculate exponentially more values. $$\\sin(\\alpha + \\beta) = \\sin(\\alpha)\\cos(\\beta) + \\sin(\\beta)\\cos(\\alpha)$$ together with $$\\sin\\left(\\sum_{k=1}^n \\frac{d_k\\pi}{2^k}\\right), \\text{where } d_k\\in\\{0,1\\}$$ Factoring out $\\pi$, and knowing the binary number system can represent all numbers ( with sufficient number of digits provided ), we can accomplish what you ask for, using a small table of preset values to calculate any sine or cosine up to", "# ISI2014-DCG-31 111 views For real $\\alpha$, the value of $\\int_{\\alpha}^{\\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is 1. $\\alpha$ 2. $[\\alpha]$ 3. $1$ 4. $\\dfrac{[\\alpha] + [\\alpha +1]}{2}$ in Calculus retagged 0 B? 2 Case 1:$\\alpha$>=0 Let α=1.23, α+1=2.23, =$\\int_{1.23}^{2} x dx$ +$\\int_{2}^{2.23} x dx$ =$\\int_{1.23}^{2} 1$ + $\\int_{2}^{2.23} 2 dx$ ($\\because$ floor between $1.23, 2 = 1$, similarly floor of 2 and 2.23 is 2. ) = 0.77 + 2*(0.23) =0.77+0.46=1.23 ($\\alpha$) Case 2:$\\alpha$<0 Let α=-0.2, α+1=0.8, = $\\int_{-0.2}^{0} x dx$+ $\\int_{0}^{0.8} x dx$ = $\\int_{-0.2}^{0} -1 dx$ + $\\int_{0}^{0.8} 0 dx$ = -1(0-(-0.2) + 2*(0) =0.2 ($\\alpha$) In both case, $\\int_{\\alpha}^{\\alpha +1} x dx$ = $\\alpha$ Hope, it helps. text might be small! ## Related questions 1 182 views The integral $\\int _0^{\\frac{\\pi}{2}} \\frac{\\sin^{50} x}{\\sin^{50}x +\\cos^{50}x} dx$ equals $\\frac{3 \\pi}{4}$ $\\frac{\\pi}{3}$ $\\frac{\\pi}{4}$ none of these The value of the definite integral $\\int_0^{\\pi} \\mid \\frac{1}{2} + \\cos x \\mid dx$ is $\\frac{\\pi}{6} + \\sqrt{3}$ $\\frac{\\pi}{6} - \\sqrt{3}$ $0$ $\\frac{1}{2}$ The value of the integral $\\displaystyle{}\\int_{-1}^1 \\dfrac{x^2}{1+x^2} \\sin x \\sin 3x \\sin 5x dx$ is $0$ $\\frac{1}{2}$ $– \\frac{1}{2}$ $1$ If $A(t)$ is the area of the region bounded by the curve $y=e^{-\\mid x \\mid}$ and the portion of the $x$-axis between $-t$ and $t$, then $\\underset{t \\to \\infty}{\\lim} A(t)$", "12:17 • @JyrkiLahtonen: It is an exercise in Cohen's Combinatorial Group Theory, somewhere in the first couple of chapters. I didn't remember all the details when I started writing, all I had in mind was \"use that n+1 and n are rel. prime\" :). – user641 Sep 22 '11 at 12:21 In this answer I give credit to Jyrki Lahtonen for the answer he posted. There are holes in his post, so I sensed the need for a step by step answer (firstly to convince myself, but also other people in doubt), and so here it is. $\\bbox[5px,border:2px solid]{\\begin{array}{cc}a^3=b^{-1}a^2b&(\\alpha)\\\\b^3=a^{-1}b^2a &(\\beta)\\end{array}}$ $b^{6}=b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)=a^{-1}b^4a\\quad(\\alpha)$ $b^{12}=b^6b^6=(a^{-1}b^4a)(a^{-1}b^4a)=a^{-1}b^8a\\quad(\\alpha)$ $b^{18}=b^{12}b^6=(a^{-1}b^8a)(a^{-1}b^4a)=a^{-1}b^{12}a\\quad(\\alpha)$ $b^{27}=b^{18}b^6b^3=(a^{-1}b^{12}a)(a^{-1}b^4a)(a^{-1}b^2a)=a^{-1}b^{18}a\\quad(\\alpha)$ $b^{27}=a^{-1}b^{18}a=a^{-2}b^{12}a^2=a^{-3}b^8a^3=\\ (\\beta)\\ =(b^{-1}a^2b)^{-1}b^8(b^{-1}a^2b)=(b^{-1}a^{-2}b)b^8(b^{-1}b^{-1}a^2b)=b^{-1}a^{-2}b^8a^2b=b^{-1}(a^{-1}(a^{-1}b^8a)a)b=b^{-1}(b^{18})b=b^{18}$ $b^{27}=b^{18}\\Rightarrow b^9=1$ $b^9=b^3b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)(a^{-1}b^2a)=(a^{-1}b^6a)=1\\Rightarrow b^6=1\\quad (\\beta)$ $b^3=b^9b^{-6}=1=a^{-1}b^2a\\Rightarrow b^2=1$ $b=b^3b^{-2}=1$ and then $a^3=b^{-1}a^2b=a^2\\Rightarrow a=1\\quad (\\alpha)$. $\\bbox[5px,border:2px solid]{a=b=1}$ Another way. $$b^2a=ab^3,$$ which gives $$b^2ab^{-2}=ab$$ and we obtain: $$b^2a^3b^{-2}=(ab)^3$$ and from here $$b^2(b^{-1}a^2b)b^{-2}=(ab)^3$$ or $$ba^2b^{-1}=(ab)^3.$$ Also, $$b^2a^2b^{-2}=(ab)^2,$$ which gives $$a^2=b^{-2}(ab)^2b^2$$ and we obtain: $$b\\left(b^{-2}(ab)^2b^2\\right)b^{-1}=(ab)^3$$ or $$b^{-1}(ab)^2b=(ab)^3$$ or $$(ab)^2=(ba)^3.$$ Similarly, $$(ba)^2=(ab)^3,$$ which gives $$(ab)^2=(ba)(ab)^3$$ or $$ba^2b=e.$$ Similarly, $$ab^2a=e.$$ Now, since $$a^{-1}b^2a=b^3,$$ we obtain $$a^2b^3=e.$$ In another hand, $$ab^2a=e$$ gives $$b^2=a^{-2}$$ and $$a^2b^2=e.$$ Id est, $$e=a^2b^3=eb=b,$$ $$a=e$$ and we are done!", "one is ok.) My attempt for small x: \\begin{align} I(x) &= \\frac{1}{2\\pi}\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} e^{x \\cos^2(\\theta)}d\\theta\\\\ &= \\frac{1}{2\\pi}\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} d\\theta [1 + x\\cos^2{\\theta} + \\order{x^2}]\\\\ &= \\frac{1}{2\\pi}\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} d\\theta [1 + \\frac{x}{2}(1+\\cos{2\\theta}) + \\order{x^2}]\\\\ &\\approx \\frac{1}{2\\pi} [\\pi + \\frac{x}{2}(\\pi)] \\end{align} Therefore, \\begin{align} &\\boxed{ I(x)\\approx \\frac{1}{2} + \\frac{x}{4} \\text{for small $x$}} \\end{align} — niagarajohn Nov 18 '18 at 14:53 Thank you for responding. I mean that $\\mathfrak p_v$ is the prime ideal of $K$ corresponding to the place $v$, $(\\mathfrak p_v, L/K)$ is the Frobenius element, and this automorphism is raised to the power of $ord_v(x)$, which is the normalized valuation of $K$ (so that $\\ord_v(\\mathfrak p_v) = 1$). — D_S Feb 10 '15 at 0:51 Nope, because $-7 =21$. In your $G$, the element $7$ has order $4$. The elements of order $7$ in $\\mathbb{Z}_{28}$ are $4, 8, 12, 16, 20, 24$. In general, in $\\mathbb{Z}_{n}$ the oder of $m$ is $\\ord(m)=\\frac{n}{\\gcd (m,n)}$. — N. S. Aug 27 '12 at 15:39 In this one, \\ord is not used in math mode: Hi, welcome to Math.SE. I just edited your post to make a few things clearer. Note that you can use \\pmod{number here}. Also, it usually helps to split things into separate lines to make things easier to read. I'm not sure what you were trying to do with \\ind and \\ord", "## e Show that $\\alpha_1 - \\alpha_2$ is estimable and give its least squares estimator. Take $\\lambda^T = [0, 1,-1, 0, 0, 0]$ such that $\\lambda^T b = \\alpha_1 - \\alpha_2.$ Now we need to find $a$ such that $a^T X = \\lambda ^T.$ $[a_1, a_2, a_3, a_4, a_5, a_6] \\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 1 & 0 \\\\ 1 & 1 & 0 & 0 & 0 & 1 \\\\ 1 & 0 & 1 & 1 & 0 & 0 \\\\ 1 & 0 & 1 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 0 & 1 \\end{bmatrix} = [0, 1, -1, 0, 0, 0]$ \\begin{align} a_1 + a_2 + a_3 + a_4 + a_5 + a_6 & = 0 \\\\ a_1 + a_2 + a_3 & = 1 \\\\ a_4 + a_5 + a_6 & = -1 \\\\ a_1 + a_4 & = 0 \\\\ a_2 + a_5 & = 0 \\\\ a_3 + a_6 & = 0. \\end{align} Now fix $a_5$ and $a_6$. \\begin{align} a_2 & = -a_5 \\\\ a_3 & = -a_6 \\\\ a_4 & = -1 + a_5 + a_6 \\\\ a_1 & = 1 - a_5 - a_6", "but we can use our knowledge of combinations to solve this. $\\editable{B}=\\editable{C}$B=C because of of the symmetry. $\\editable{B}$B also equals the value of $\\nCr{9}{4}$9C4 and $\\editable{C}$C$=$=$\\nCr{9}{5}$9C5, but we also know that $\\nCr{9}{4}=\\nCr{9}{5}$9C4=9C5 (confirming what we already knew from symmetry that the values will be the same). $\\editable{B}$B $=$= $\\nCr{9}{4}$9C4 $=126$=126 Thus both $\\editable{B}$B and $\\editable{C}=126$C=126. ## Pascal's triangle and binomial distributions $(a+b)^0$(a+b)0 $=$= $1$1 $(a+b)^1$(a+b)1 $=$= $a+b$a+b $(a+b)^2$(a+b)2 $=$= $a^2+2ab+b^2$a2+2ab+b2 $(a+b)^3$(a+b)3 $=$= $a^3+3a^2b+3ab^2+b^3$a3+3a2b+3ab2+b3 $(a+b)^4$(a+b)4 $=$= $a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4 $(a+b)^5$(a+b)5 $=$= $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5 Consider the distributions above of $(a+b)^n$(a+b)n. Particularly note the following patterns. • For each distribution to the power $n$n, there are $n+1$n+1 elements. • For each term, the sum of the exponents is $n$n. • Powers of $a$a decrease from left to right, from $n$n down to $0$0. • Powers of $b$b increase from left to right, from $0$0 up to $n$n • The coefficients start at $1$1, end at $1$1, and are the terms of the relevant row from Pascals triangle. #### Worked example ##### Example 2 What are the coefficients for the distribution of $(a+1)^7$(a+1)7, and then write out the full distribution. So we can see that we will have $n+1=8$n+1=8 terms. We can refer to the relevant row in Pascal's triangle, specifically this row This shows us that the coefficients will be $1,7,21,35,35,21,7,1$1,7,21,35,35,21,7,1. Thus the full", "$~ a_0[3a_2\\alpha +6a_2 (n + m ) ] + a_2[\\alpha(3a_0 - 11a_2) + (-22n a_2 +6m a_0 - 10a_2 m) + a_2\\sigma^2]$ $~ a_0[5ma_2 + 5na_2 + 4n m a_2 ] + a_2[ -15ma_2 - 25na_2 -12 n m a_2 + 5ma_0 ]$ $~ -[ 2n(n-1)a_2^2 +2m(m-1)a_2^2 ] + [ -8n(n-1)a_2^2 +4m(m-1)a_0 a_2 ]$ $~x^8$ : $~ a_2[3a_2\\alpha +6a_2 (n + m ) ] + a_2[5ma_2 + 5na_2 + 4n m a_2 ] + [ 2n(n-1)a_2^2 +2m(m-1)a_2^2 ]$ After simplification: $~x^0$ : $~ \\alpha(10a_0^2) + \\sigma^2(- 2a_0^2) -20n a_0a_2 + 10ma_0^2$ $~x^2$ : $~\\alpha(- 11a_0^2 +20 a_0a_2) + \\sigma^2(a_0^2 - 4 a_0a_2) + 60n a_0a_2 -20na_2^2- 25m a_0^2 + 20m a_0a_2 + 8n m a_0a_2$$- [ 8n(n-1)a_2^2 + 2m(m-1)a_0^2 ]$ $~x^4$ : $~ \\alpha(10a_2^2 - 22 a_0a_2 +3a_0^2) - \\sigma^2(2a_2^2- 2a_0a_2 ) - 47n a_0a_2+ 60n a_2^2 - 50m a_0a_2 +11m a_0^2+ 10ma_2^2-12 n m a_0a_2 + 8n m a_2^2$ $~ + 16n(n-1)a_2^2 - 4m(m-1)a_0 a_2 + 2m(m-1)a_0^2$ $~x^6$ : $~ \\alpha(6a_0a_2 - 11a_2^2) + \\sigma^2(a_2^2) + 11 n a_0a_2 +22 m a_0a_2 - 47n a_2^2 - 25m a_2^2 -12 n m a_2^2 + 4n m a_0a_2$ $~ - 10 n(n-1)a_2^2 -2m(m-1)a_2^2 +4m(m-1)a_0 a_2$ $~x^8$ : $~ a_2[3a_2\\alpha +6a_2 (n + m ) ] + a_2[5ma_2 + 5na_2 + 4n m a_2 ] + [ 2n(n-1)a_2^2 +2m(m-1)a_2^2 ]$", "x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi - x_1 - x_2 - x_3 - x_4.$$ We have $$x_1, x_2, x_3, x_4, x_5 > 0; \\ x_1 + x_2+x_3+x_4+x_5=\\pi.$$ Note that $$\\big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\big)^5 = \\frac{5\\sqrt{5}-11}{2}$$. We have \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin x_5} {\\sin (x_1+x_2) \\sin (x_2 + x_3) \\sin (x_3 + x_4) \\sin (x_4+x_5) \\sin (x_5+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\Big)^5\\\\ =\\ &\\frac{(\\sin x_1)^2(\\sin x_2)^2\\sin (2x_1 + 2x_2)}{(\\sin (x_1+x_2))^2\\sin 2x_2 (\\sin (x_1 + 2x_2))^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &\\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &-\\frac{5\\sqrt{5}-11}{16}\\frac{ (\\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\\sqrt{5}+1)(\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3)^2} {y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\\\ =\\ &0. \\end{align} • This is interesting +1. Since the general Jacobian matrix I computed was singular, maybe there are multiple points where the inequalities are attained. – dezdichado Jun 19 '19 at 19:45 • When $n = 4,$ it must be that $2x_1+x_2", "at the $(0,0)$, $a=16$, and $(8,6)$ is a point on the ellipse. Hence $$\\frac{8^2}{16^2}+\\frac{6^2}{b^2} = 1.$$ Now from this, we can solve for $b$ to get the final equation: ... 2 Given a parametric curve $(x(t), y(t))$, the (signed) curvature is given by \\begin{equation*} \\kappa(t) = \\frac{x' y'' - y' x''}{((x')^2 + (y')^2)^{3/2}}. \\end{equation*} The ellipse $9(x-1)^2 + y^2 = 9$ can be rewritten in standard form as $(x-1)^2 + \\frac{y^{2}}{9} = 1$. Then parametrize it as \\begin{align*} x(t) & = 1 + \\cos{t}\\\\ y(t) & = ... 2 Did you try a less clever method? Put $x=\\alpha x' + \\beta y'$ and $y=-\\beta x' + \\alpha y'$ Then the equation $5x^2 + 5y^2 - 6xy - 8 = 0$ becomes $(5(\\alpha^2+\\beta^2)+6\\alpha\\beta)x'^2+(5(\\alpha^2+\\beta^2)-6\\alpha\\beta)y'^2-6x'y'(\\alpha^2-\\beta^2)-8=0$ If you want to remove any cross product, then $\\alpha^2-\\beta^2=0$ EDITED after abel comment ... 2 Hint I did what you tell $(x=p+q , y=q-p)$ and the raw result I obtained for the expression is $$4 p^2+32 p+8 q^2-32 q+92=0$$ I let you the task of grouping but it seems that you made some mistakes. I am sure that you can take from here and fix the problem. 2 the tangent to the parabola at $(4, 4)$ has slope $1/2$ so the radius has slope $-2.$ let the center of the circle touching the parabola $y^2", "Gluing Lemma. Also: (2) \\begin{align} \\quad H_0(s) = H(s, 0) = \\left\\{\\begin{matrix} x & 0 \\leq s \\leq 0\\\\ \\alpha (2s) & 0 \\leq s \\leq \\frac{1}{2}\\\\ \\alpha^{-1}(2s- 1) & \\frac{1}{2} \\leq s \\leq 1\\\\ x & 1\\leq s \\leq 1 \\\\ \\end{matrix}\\right. = \\alpha\\alpha^{-1} \\end{align} (3) \\begin{align} \\quad H_1(s) = H(s, 1) = \\left\\{\\begin{matrix} x & 0 \\leq s \\leq \\frac{1}{2} \\\\ \\alpha (2s - 1) & \\frac{1}{2} \\leq s \\leq \\frac{1}{2}\\\\ \\alpha^{-1}(2s) & \\frac{1}{2} \\leq s \\leq \\frac{1}{2}\\\\ x & \\frac{1}{2} \\leq s \\leq 1 \\\\ \\end{matrix}\\right. = c_x \\end{align} • Furthermore: (4) \\begin{align} \\quad H_t(0) = x \\alpha\\alpha^{-1}(s) = c_x(s) \\quad \\forall t \\in I, \\forall s \\in I \\end{align} (5) \\begin{align} \\quad H_t(1) = x \\alpha\\alpha^{-1}(s) = c_x(s) \\quad \\forall t \\in I, \\forall s \\in I \\end{align} • So indeed $\\alpha\\alpha^{-1} \\simeq_{\\{0, 1\\}} c_x$. So $[\\alpha\\alpha^{-1}] = [c_x$, i.e.: (6)" ]

[ "# Where's the other 4 equations? Algebra Level 5 Let $$a, b, c, x, y, z$$ be positive real numbers such that \\begin{align*} abx + bcy + acz &= xyz - 2abc \\\\ axz + bxy + cyz &= xyz. \\\\ \\end{align*} How many solutions are there to this system? If there are an infinite number of solutions, input $$-1$$ as your answer. If there are no solutions to the system, input $$0$$. ×", "Algebra Level 5 Three real numbers $$x, y,$$ and $$z$$ such that $$x < y < z$$ satisfy the following three equations: \\begin{align} x + y + z &= 4 \\\\ x^2 + y^2 + z^2 &= 10 \\\\ x^3 + y^3 + z^3 &= 22. \\end{align} Find the value of $$z.$$ This problem is not original. ×", "Four Equations in Four Unknowns Algebra Level 2 \\begin{aligned} x + y + z + w &= 10 \\\\ x^2 + y^2 + z^2 + w^2 &= 30 \\\\ x^3 + y^3 + z^3 + w^3 &= 100\\\\ xyzw &= 24 \\end{aligned} What is a solution of the system of equations above? ×", "# Are you smarter than me? Algebra Level 5 Find the number of solutions $$\\left(x,y,z\\right)$$ to the system of equations \\begin{align} x+2y+4z&=9\\\\ 4yz+2zx+xy&=13\\\\ xyz&=13 \\end{align} such that at least two of $$x,y,z$$ are integers. ×", "# Danny's system of equations Algebra Level 3 If $$x,y,$$ and $$z$$ are real numbers that satisfy the following system of equations: \\begin{align} x^2 +6y & = -14 \\\\ y^2 + 12z &= -63 \\\\ z^2 - 4x &= 28, \\end{align} what is the value of $$\\lvert x + y + z \\rvert$$? This problem is posed by Danny H. ×", "# Complex Equations Algebra Level 4 \\begin{align} (2-3i)z + (5+i)w &= 19-4i\\\\ (1-i)z + (2+i)w &= 9-i \\end{align} Let $$z$$ and $$w$$ be the complex numbers satisfying the equations above. Compute $$|z+w|^2$$. Note: $$i^2 = -1$$, and $$|z|$$ is the absolute value of $$z$$. ×", "+0 # system 0 37 2 If w, x, y, and z are real numbers satisfying: w+x+y = -2, w+x+z = 4, w+y+z = 19 x+y+z = 16 what is wx + yz? Jun 26, 2022 ### 1+0 Answers #1 +2222 +1 Adding everything gives us: $$3(w+x+y+z) = 37$$ This means $$w + x + y + z = {37 \\over 3}$$. Now, you can use this to solve for w, x, y, and z, and compute the answer. Can you take it from here? Jun 27, 2022", "# Weird Equations Algebra Level 5 \\begin{align*} x^4+2y^3 - x & = -\\frac{1}{4} + 3\\sqrt{3} \\\\ y^4 + 2x^3 - y & = -\\frac{1}{4} - 3\\sqrt{3} \\end{align*} Find the number of solutions in real numbers to the system of equations above. ×", "x-1+2-3=-1 \\\\ & x-2=-1 \\\\ & x=1 \\end{align} Therefore, $x=1,y=-1,z=2,w=-3$ Hence, the provided statement makes sense.", "+0 # The real numbers x and y are such that 0 94 1 The real numbers x and y are such that \\begin{align*} x + y &= 4, \\\\ x^2 + y^2 &= 22, \\\\ x^4 &= y^4 - 176 \\sqrt{7}. \\end{align*} Find x-y Jun 13, 2020", "of information into equations, it turned out that $x$ was the smallest number, $y$ the middle number, and $z$ the largest number. This is not necessary, however, as we could have written down the same system of equations with the right hand sides (12, 17, 19) in any order. Solution: 2 Reasoning to produce special equations Instead of three equations, each involving two unknons, there is another way to set up equations to find the solution. Suppose again that $x,y,$ and $z$ are the three whole numbers we are looking for and this time we arrange them so that $x \\leq y \\leq z$. We know that \\begin{align} 2(x+y+z) &=(x+y)+(x+z)+(y+z)\\\\ &=12+17+19\\\\ &=48 \\end{align} We know that $x + y$ is the smallest of the three sums followed by $x + z$ followed by $y + z$ because we assumed $x \\leq y \\leq z$. So we conclude that $$x + y + z = 24$$ Now we are given that $x + y = 12$. Subtracting this from the equation $x + y + z = 24$ gives us $z = 12$. Similarly from $x + z = 17$ we can subtract this from $x + y + z = 24$ to deduce that $y = 7$. Finally we get $x = 5$ by subtracting $y+ z =", "x \\end{align} Note, that there a two branches of $W$ in the range $[-\\tfrac 1e, 0[$. Thus the above equation will give two solutions...", "Systems Consider the linear system \\begin{aligned} 2x+y+z&=5,\\\\ x-3y+2z&=1.\\end{aligned} Which of the following best describes the set of solutions to this system? # Linear Systems For what real-number values of $$a$$ will the following system of equations have no solutions? \\begin{aligned} ax+2y&=4\\\\ 2x-ay&=6.\\end{aligned} ×", "## Find all The Solutions Hii ! - Find all the real solutions positive ( if its exist ) For This system : $\\left\\{ \\begin{array}{rcl} x^3+y^3+z^3 = x+y+z \\\\ x^2+y^2+z^2 = xyz \\end{array}\\right.$", "# RMO 2015! Algebra Level 4 \\begin{cases} \\begin{align} (1 + x)\\big(1 + x^2\\big)\\big(1 + x^4\\big) &= 1 + y^7 \\\\\\\\ (1 + y)\\big(1 + y^2\\big)\\big(1 + y^4\\big) &= 1 + x^7 \\end{align} \\end{cases} Find the number of ordered pairs of real numbers $$(x, y)$$ for which the above system of equations is satisfied. ×", "Complex Equations \\begin{aligned} (2-3i)z + (5+i)w &= 19-4i\\\\ (1-i)z + (2+i)w &= 9-i \\end{aligned} Let $z$ and $w$ be the complex numbers satisfying the equations above. Compute $|z+w|^2$. Note: $i^2 = -1$, and $|z|$ is the absolute value of $z$. ×", "How many $$(x,y)$$ are there? Algebra Level 4 How many real numbered ordered pairs $$(x, y)$$ satisfy the following inequalities: \\begin{align} (x+2y)^2 &\\leq (2x-3)(4y+3)\\\\ (x-2y)^2 &\\leq (5-2x)(4y-5)? \\end{align} ×", "Let's assume that $W$ is analytic in $z$ so that we can write \\begin{align} W(x,y,z) = \\sum_{k=0}^\\infty w_k(x,y) z^k \\end{align} Notice that we might have tried to include negative powers of $z$ in this expansion, but those would be prohibited by the boundary condition. Moreover, the boundary condition $W(x,y,0) = V_0$ implies that $w_0(x,y) = V_0$. Laplace's equation $\\nabla^2W = 0$ then gives \\begin{align} \\sum_{k=0}^\\infty \\Big[(\\partial_x^2 + \\partial_y^2) w_k(x,y)+(k+2)(k+1)w_{k+2}(x,y)\\Big]z^k =0 \\end{align} Setting the coefficient of each power of $z$ in the series to $0$ gives us constraints on the functions $w_k$ for $k\\geq 0$; \\begin{align} (\\partial_x^2 + \\partial_y^2) w_k(x,y) = -(k+2)(k+1)w_{k+2}(x,y) \\end{align} For $k=2$, we obtain \\begin{align} 0=(\\partial_x^2+\\partial_y^2)w_0 = -2 w_2 \\end{align} which gives $w_2=0$. Applying this repeatedly shows that $w_k = 0$ for all even $k$. For odd $k$, we obtain an infinite tower of equations that couples all $w_k$ for odd $k$ to $w_1$. A particularly simple solution to this tower is obtained by setting $w_k = 0$ for all $k\\geq 3$ in which chase $w_1$ satisfies the two-dimensional Laplace equation. We could, for example, set $w_1 = C$ for some constant $C$. This would give the solution that exhibits translation invariance in $x$ and $y$ mentioned by Michael Brown in the comments. • Umm... I believe your Laplace equation is wrong. The $k(k-1)w_k$ term should", "# Symmetric System of Equations Algebra Level 4 Find the number of real solutions of the system of equations. $\\left\\{ \\begin{array}{l} x^2-y^2=z,\\\\ y^2-z^2=x,\\\\ z^2-x^2=y. \\end{array} \\right.$ ×", "# Difference between revisions of \"2014 UMO Problems/Problem 5\" ## Problem Find all positive real numbers $x, y$, and $z$ that satisfy both of the following equations. \\begin{align*} xyz & = 1\\\\ x^2 + y^2 + z^2 & = 4x\\sqrt{yz}- 2yz \\end{align*} ## Solution By AM-GM $$x^2+y^2+z^2 + 2yz\\ge x^2 + 4yz\\ge 4x\\sqrt{yz}$$ Hence, the second equation implies that $y=z$ and $x^2=4yz\\implies x=2y=2z$. Now we plug it into the first equation to get $(x,y,z) = \\left(\\sqrt[3]{4}, \\frac{\\sqrt[3]4}{2}, \\frac{\\sqrt[3]4}{2}\\right)$" ]

[ "xyz; \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\quad \\end{align} (b) Prove that \\begin{align} \\frac{x}{{1 - {x^2}}} + \\frac{y}{{1 - {y^2}}} + \\frac{z}{{1 - {z^2}}} \\ge \\frac{{3\\sqrt 3 }}{2}\\\\ {\\text{if}}\\;0 < x,\\;y,\\;z < 1\\;{\\rm{and}}\\;xy + yz + zx = 1. \\qquad \\qquad \\qquad \\qquad \\qquad \\end{align} Solution: Both problems can be solved by trigonometric substitutions. (a) Let there be there is an acute triangle $$ABC$$ with $$\\tan A = x,\\;\\tan B = y$$, and $$\\tan C = z$$. Note that \\begin{align}\\frac{{\\tan A}}{{\\sqrt {1 + {{\\tan }^2}A} }} = \\frac{{\\tan A}}{{\\sec A}} = \\sin A.\\end{align} The desired inequality becomes \\begin{align}\\sin A + \\sin B + \\sin C \\le \\frac{{3\\sqrt 3 }}{2}\\end{align} which we’ve covered earlier. (b) From the given condition, we can assume that there is an acute triangle $$ABC$$ such that \\begin{align}\\tan \\frac{A}{2} = x, \\qquad \\tan \\frac{B}{2} = y, \\qquad \\tan \\frac{C}{2} = z.\\end{align} By the double-angle formulas, it suffices to prove that \\begin{align}\\tan A + \\tan B + \\tan C \\ge 3\\sqrt 3 \\end{align} which is another inequality we’ve encountered already. Example - 16 Compute the sums $\\left( \\begin{matrix} n \\\\ 1 \\\\ \\end{matrix} \\right)\\sin a+\\left( \\begin{matrix} n \\\\ 2 \\\\ \\end{matrix} \\right)\\sin 2a+....+\\left( \\begin{matrix} n \\\\ n \\\\ \\end{matrix} \\right)\\sin na$ and $\\left( \\begin{matrix} n \\\\ 1 \\\\ \\end{matrix} \\right)\\cos a+\\left( \\begin{matrix} n", "} \\right)}^5}}}\\,\\,d\\theta } } \\\\&\\qquad\\qquad\\qquad\\qquad= \\int {\\frac{{{{\\sin }^7}\\theta \\cos \\theta }}{{{{\\cos }^{10}}\\theta }}\\,\\,d\\theta } \\\\&\\qquad\\qquad\\qquad\\qquad= \\int {{{\\tan }^7}\\theta {{\\sec }^2}\\theta \\,d\\theta } \\end{align} Notice now that the simple substitution $$\\tan \\theta = y$$ will reduce $$I$$ to a simple integrable form: \\begin{align}&\\qquad \\quad\\;\\;\\tan \\theta = y\\\\ &\\Rightarrow \\quad {\\sec ^2}\\theta d\\theta = dy\\end{align} \\begin{align} &I = \\int {{{\\tan }^7}\\theta } {\\sec ^2}\\theta d\\theta = \\int {{y^7}dy} \\\\&\\qquad\\qquad\\qquad\\qquad\\;= \\frac{{{y^8}}}{8} + C \\\\ &\\qquad\\qquad\\qquad\\qquad\\;= \\frac{{{{\\tan }^8}\\theta }}{8} + C \\\\ &\\qquad\\qquad\\qquad\\qquad\\;= \\frac{{{x^8}}}{{8{{\\left( {1 - {x^2}} \\right)}^4}}} + C\\;\\;\\;\\;\\;\\;\\;\\;\\left( {∵ x = \\sin \\theta } \\right) \\end{align} There is one last point to be observed. The answers obtained by methods (a) and (b) might seem to be different from each other. However, verify that they are not! The answer obtained by (a) can be converted into the answer obtained by (b) by just a simple splitting of the constant of integration. Indefinite Integration grade 11 | Questions Set 1 Indefinite Integration Indefinite Integration grade 11 | Questions Set 2 Indefinite Integration", "$2\\theta - \\dfrac{5\\pi}{6} = n\\pi$. Hence, $\\theta = \\dfrac{\\pi(6n+5)}{12}$. As topsquark pointed out, tangent has a period of $\\pi$, but it did not seem obvious to me that meant you will find solutions every $\\dfrac{\\pi}{2}$. Edit: Now that I think about it, topsquark's hint was more straightforward than I thought. You set $\\theta + \\dfrac{\\pi}{6} = \\pi-\\theta$. Instead, you needed to add $n\\pi$ to one side. For example: $\\theta + \\dfrac{\\pi}{6} = \\pi-\\theta + n\\pi$. That would give you the same answer I got using the difference of angles formula for sine. 4. ## Re: all solutions of equation involving tan Originally Posted by Ant Hi, Let $0 \\leq \\theta \\leq \\pi$ Find all solutions to: $\\tan(\\theta + \\dfrac{\\pi}{6})=\\tan(\\pi-\\theta)$ Equating the two arguements and solving we find $\\theta = \\dfrac{5\\pi}{12}$. How do I find the other solution? Thanks First of all, remember that \\displaystyle \\begin{align*} \\tan{ \\left( \\alpha + \\beta \\right) } \\equiv \\frac{\\tan{( \\alpha )} + \\tan{ ( \\beta ) }}{ 1 - \\tan{( \\alpha )} \\tan{ (\\beta )}} \\end{align*} and also \\displaystyle \\begin{align*} \\tan{ ( \\pi - \\theta )} \\equiv -\\tan{(\\theta ) } \\end{align*}, so that means \\displaystyle \\begin{align*} \\tan{ \\left( \\theta + \\frac{\\pi}{6} \\right) } &= \\tan{ \\left( \\pi - \\theta \\right) } \\\\ \\frac{\\tan{ \\left( \\theta \\right)} + \\tan{ \\left( \\frac{\\pi}{6} \\right) } }{1", "For example: $a=b=c=0$ or $a=0$ and $b=2\\pi-c$. – robjohn Jul 25 '20 at 12:03 HINT $A+B+C = 180$ $A+B = 180 - C$ We'll apply tangent function: $\\tan (A+B) = \\tan (180 - C)$ We'll consider the identity: $\\tan(x+y) = \\frac{\\tan x + \\tan y}{1-\\tan x\\tan y}$ $\\frac{\\tan A + \\tan B}{1-\\tan A\\tan B} = \\frac{\\tan 180 - \\tan C}{1+\\tan 180\\tan C}$ But $\\tan 180 = 0$, therefore, we'll get: $\\frac{\\tan A + \\tan }{1-\\tan A\\tan B}$ = $\\frac{0 - \\tan C}{1+0}$ $\\frac{\\tan A + \\tan B}{1-\\tan A\\tan B} = -\\tan C$ We'll multiply by $(1-\\tan A\\tan B)$: $\\tan A + \\tan B = -\\tan C +\\tan A\\tan B\\tan C$ Hence $\\tan A + \\tan B+ \\tan C = \\tan A\\tan B\\tan C$ Here is a geometric proof, for the case that all three angles are acute: $QRUV$ are collinear because $B+90^\\circ+(90^\\circ-B)=180^\\circ$. $STV$ are collinear because $A+B+C=180^\\circ$, so $\\angle QSV=\\angle UTV=C$. Similar triangles $\\triangle PQR\\sim\\triangle TRS$ and $\\triangle RTU \\sim \\triangle SRQ$ give $\\displaystyle \\frac{QP}{RQ} = \\frac{RT}{SR} = \\frac{TU}{RQ}$, and therefore $TU=QP=1$. Then, \\begin{align}& \\tan A + \\tan B + \\tan C = QR+RU+UV = QV \\\\ &= QP \\frac{QR}{QP}\\, \\frac{QS}{QR} \\, \\frac{QV}{QS} = 1 \\cdot \\tan(A) \\tan(B) \\tan(C) \\end{align} When one of the angles is obtuse, let it (without loss of generality) be $C$. Then a similar", "b^2 \\\\ 2^2 & = b^2 \\\\ b & = 2 \\end{align*} Therefore the equation is $$y=2 \\times 2^{x} + 1$$. This is a tangent graph and so we use $$y = a \\tan \\theta + q$$. To find $$q$$ we note that the graph has been shifted down by 2 units (the point $$(180^\\circ;-2)$$ is given). Therefore $$q = -2$$. To find $$a$$ we substitute $$(135^\\circ;-1)$$ into the equation: \\begin{align*} y & = a \\tan \\theta - 2 \\\\ -1 & = a \\tan (135^\\circ) - 2 \\\\ 1 & = -a \\\\ a & = -1 \\end{align*} Therefore the equation is $$y=-\\tan \\theta - 2$$. State the coordinates at $$A,~B,~C$$ and $$D$$. $$A( 90^{\\circ};1 ),~B(90 ^{\\circ};-1 ),~C( 180^{\\circ}; 2) \\text{ and } D(360 ^{\\circ};1)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. 0 What is the amplitude of $$f(x)$$. 1 Evaluate: $$f(180^{\\circ}) - g(180^{\\circ})$$ . \\begin{align*} f(180^{\\circ}) - g(180^{\\circ}) &= 2 - (-3)\\\\ &=5 \\end{align*} State the coordinates at $$A,~B,~C$$ and $$D$$. $$A( 90^{\\circ}; 1),~B(270 ^{\\circ}; 3),~C(270 ^{\\circ}; 1) \\text{ and } D(360 ^{\\circ}; 2)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. 2 What is the amplitude of $$g(x)$$. 3 Evaluate: $$g(180^{\\circ}) - f(180^{\\circ})$$ . \\begin{align*} f(180^{\\circ}) - g(180^{\\circ}) &= 4 - 2 \\\\ &=2 \\end{align*} $$y=2^{x}$$ and $$y=-2^{x}$$ are sketched", "# 2009 AMC 12A Problems/Problem 25 ## Problem The first two terms of a sequence are $a_1 = 1$ and $a_2 = \\frac {1}{\\sqrt3}$. For $n\\ge1$, $$a_{n + 2} = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$? $\\textbf{(A)}\\ 0\\qquad \\textbf{(B)}\\ 2 - \\sqrt3\\qquad \\textbf{(C)}\\ \\frac {1}{\\sqrt3}\\qquad \\textbf{(D)}\\ 1\\qquad \\textbf{(E)}\\ 2 + \\sqrt3$ ## Solution Consider another sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ such that $a_n = \\tan{\\theta_n}$, and $0 \\leq \\theta_n < 180$. The given recurrence becomes \\begin{align*} a_{n + 2} & = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\\\ \\tan{\\theta_{n + 2}} & = \\frac {\\tan{\\theta_n} + \\tan{\\theta_{n + 1}}}{1 - \\tan{\\theta_n}\\tan{\\theta_{n + 1}}} \\\\ \\tan{\\theta_{n + 2}} & = \\tan(\\theta_{n + 1} + \\theta_n) \\end{align*} It follows that $\\theta_{n + 2} \\equiv \\theta_{n + 1} + \\theta_{n} \\pmod{180}$. Since $\\theta_1 = 45, \\theta_2 = 30$, all terms in the sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ will be a multiple of $15$. Now consider another sequence $\\{b_1, b_2, b_3...\\}$ such that $b_n = \\theta_n/15$, and $0 \\leq b_n < 12$. The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \\equiv b_{n + 1} + b_n \\pmod{12}$. As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the", "# 2013 AMC 12B Problems/Problem 20 ## Problem For $135^\\circ < x < 180^\\circ$, points $P=(\\cos x, \\cos^2 x), Q=(\\cot x, \\cot^2 x), R=(\\sin x, \\sin^2 x)$ and $S =(\\tan x, \\tan^2 x)$ are the vertices of a trapezoid. What is $\\sin(2x)$? $\\textbf{(A)}\\ 2-2\\sqrt{2}\\qquad\\textbf{(B)}\\3\\sqrt{3}-6\\qquad\\textbf{(C)}\\ 3\\sqrt{2}-5\\qquad\\textbf{(D}}\\ -\\frac{3}{4}\\qquad\\textbf{(E)}\\ 1-\\sqrt{3}$ (Error compiling LaTeX. ! Undefined control sequence.) ## Solution Let $f,g,h,j$ be $\\sin, \\cos, \\tan, \\cot$ (not respectively). Then we have four points $(f,f^2),(g,g^2),(h,h^2),(j,j^2)$, and the two lines connecting two pairs of them must be parallel (as we are dealing with a trapezoid). WLOG, we take the line connecting the first two and the line connecting the last two to be parallel, so that $\\frac{g^2-f^2}{g-f} = \\frac{j^2-h^2}{j-h}$, or $g+f = j+h$. Now, we must find how to match up $\\sin, \\cos, \\tan, \\cot$ to $f,g,h,j$ so that the above equation has a solution. The three possible ways are $\\sin x+\\cos x = \\tan x+\\cot x$, $\\sin x+\\tan x = \\cos x+\\cot x$, and $\\sin x+\\cot x = \\cos x+\\tan x$. On the interval $135^\\circ < x < 180^\\circ$, we have $0<\\sin x<\\frac{1}{\\sqrt{2}},-1< \\cos x<-\\frac{1}{\\sqrt{2}},-1< \\tan x<0,-\\infty< \\cot x<-1$, so the first two of those possible ways do not work because the LHS and the RHS have different ranges. Thus, it must be the last possible way, $\\sin x+\\cot x = \\cos", "\\int \\frac1{(x^2+6x+10)^2}\\, dx =\\int \\frac1{\\big((x+3)^2+1\\big)^2}\\, dx\\amp = \\int \\frac1{(u^2+1)^2}\\,du.\\notag\\\\ \\end{align} Now make the substitution $$u=\\tan(\\theta)\\text{,}$$ $$du=\\sec^2(\\theta) \\, d\\theta\\text{:}$$ \\begin{align} \\amp = \\int \\frac1{(\\tan^2(\\theta) +1)^2}\\sec^2(\\theta) \\, d\\theta\\notag\\\\ \\amp = \\int\\frac 1{(\\sec^2(\\theta) )^2}\\sec^2(\\theta) \\, d\\theta\\notag\\\\ \\amp = \\int \\cos^2(\\theta) \\, d\\theta.\\notag\\\\ \\end{align} Applying a power reducing formula, we have \\begin{align} \\amp = \\int \\left(\\frac12 +\\frac12\\cos(2\\theta)\\right)\\, d\\theta\\notag\\\\ \\amp = \\frac12\\theta + \\frac14\\sin(2\\theta) + C\\text{.}\\label{eq_extrigsub7}\\tag{6.4.2} \\end{align} We need to return to the variable $$x\\text{.}$$ As $$u=\\tan(\\theta)\\text{,}$$ $$\\theta = \\tan^{-1}(u)\\text{.}$$ Using the identity $$\\sin(2\\theta) = 2\\sin(\\theta) \\cos(\\theta)$$ and using the reference triangle found in Key Idea 6.4.3(b), we have \\begin{equation*} \\frac14\\sin(2\\theta) = \\frac12\\frac u{\\sqrt{u^2+1}}\\cdot\\frac 1{\\sqrt{u^2+1}} = \\frac12\\frac u{u^2+1}\\text{.} \\end{equation*} Finally, we return to $$x$$ with the substitution $$u=x+3\\text{.}$$ We start with the expression in Equation (6.4.2): \\begin{align*} \\frac12\\theta + \\frac14\\sin(2\\theta) + C \\amp = \\frac12\\tan^{-1}(u) + \\frac12\\frac{u}{u^2+1}+C\\\\ \\amp = \\frac12\\tan^{-1}(x+3) + \\frac{x+3}{2(x^2+6x+10)}+C\\text{.} \\end{align*} Stating our final result in one line, \\begin{equation*} \\int\\frac1{(x^2+6x+10)^2}\\, dx=\\frac12\\tan^{-1}(x+3) + \\frac{x+3}{2(x^2+6x+10)}+C\\text{.} \\end{equation*} Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of $$x$$ to one in terms of $$\\theta\\text{,}$$ then converting back to $$x$$) and then evaluate using the original bounds. It is much more straightforward, though, to", "& =\\int\\frac{\\overbrace{-\\frac{1}{2}dt}^{xdx}}{\\sqrt{t}}\\\\ & =\\int-\\frac{1}{2}t^{-1/2}dt\\\\ & =\\frac{-1}{2}(2t^{1/2})+C\\\\ & =-\\sqrt{1-x^{2}}+C.&&\\small{(t=1-x^{2})}\\end{aligned} Finally \\begin{aligned} \\int\\arcsin x\\,dx & =x\\arcsin x-\\left(-\\sqrt{1-x^{2}}+C\\right)\\\\ & =x\\arcsin x+\\sqrt{1-x^{2}}+C_{1},\\end{aligned} where $$C_{1}=-C$$. Example 6 Evaluate $${\\displaystyle \\int x\\tan^{-1}x\\,dx}$$. [Notice that $$\\tan^{-1}x$$ is the same as $$\\arctan x$$] Solution 6 The LIATE strategy suggests that we let $u=\\tan^{-1}x,\\qquad dv=x\\,dx.$ Notice that because it is easy to differentiate $$\\tan^{-1}x$$ but hard to integrate, and also because its derivative is algebraic; it is the best choice for $$u$$. So $du=\\frac{1}{1+x^{2}}dx,\\qquad\\qquad v=\\frac{1}{2}x^{2}.$ and \\begin{aligned} \\int x\\tan^{-1}x\\,dx & =\\overbrace{\\frac{1}{2}x^{2}\\tan^{-1}x}^{uv}-\\int\\overbrace{\\frac{x^{2}}{2(1+x^{2})}dx}^{vdu}\\\\ & =\\frac{1}{2}x^{2}\\tan^{-1}x-\\frac{1}{2}\\int\\frac{(1+x^{2})-1}{1+x^{2}}dx\\\\ & =\\frac{1}{2}x^{2}\\tan^{-1}x-\\frac{1}{2}\\int dx+\\frac{1}{2}\\int\\frac{dx}{1+x^{2}}\\\\ & =\\frac{1}{2}x^{2}\\tan^{-1}x-\\frac{1}{2}x+\\frac{1}{2}\\tan^{-1}x+C\\\\ & =\\frac{1}{2}(x^{2}+1)\\tan^{-1}x-\\frac{1}{2}x+C.\\end{aligned} The same result is obtained more quickly if we take $v=\\frac{1}{2}x^{2}+C_{1}.$ Then integration by parts gives $\\int x\\tan^{-1}x\\,dx=\\left(\\frac{1}{2}x^{2}+C_{1}\\right)\\tan^{-1}x-\\int\\frac{\\frac{1}{2}x^{2}+C_{1}}{1+x^{2}}dx.$ Now taking $$C_{1}=1/2$$ simplifies the last integral and leads to the following result: \\begin{aligned} \\int x\\tan^{-1}x & =\\frac{1}{2}\\left(x^{2}+1\\right)\\tan^{-1}x-\\frac{1}{2}\\int\\overset{1}{\\cancel{\\frac{x^{2}+1}{1+x^{2}}}}dx\\\\ & =\\frac{1}{2}\\left(x^{2}+1\\right)\\tan^{-1}x-\\frac{1}{2}x+C.\\end{aligned} Example 7 Evaluate $${\\displaystyle \\int\\tan^{-1}\\sqrt{x}\\:dx}$$. Solution 7 Let $u=\\tan^{-1}\\sqrt{x},\\qquad\\text{and}\\qquad dv=dx.$ Then $u=\\frac{1}{1+(\\sqrt{x})^{2}}\\frac{1}{2\\sqrt{x}}dx,\\qquad\\qquad v=x+C_{1}.$ and $\\int\\tan^{-1}\\sqrt{x}\\,dx=(x+C_{1})\\tan^{-1}\\sqrt{x}-\\int(x+C_{1})\\frac{1}{1+x}\\frac{1}{2\\sqrt{x}}dx.$ Taking $$C_{1}=1$$ simplifies the second integral. So \\begin{aligned} \\int\\tan^{-1}\\sqrt{x}dx & =(x+1)\\tan^{-1}\\sqrt{x}-\\int\\frac{1}{2\\sqrt{x}}dx\\\\ & =(x+1)\\tan^{-1}\\sqrt{x}-\\sqrt{x}+C.\\end{aligned} Example 8 Evaluate $${\\displaystyle \\int\\frac{x}{\\cos^{2}x}dx}$$. Solution 8 Because we can easily integrate $$\\frac{1}{\\cos^{2}x}=\\sec^{2}x$$, we let $u=x,\\qquad\\qquad dv=\\frac{1}{\\cos^{2}x}dx=\\sec^{2}x\\ dx.$ Notice that this choice for $$u$$ agrees with the LIATE strategy. So $u=x\\quad\\Rightarrow\\quad du=dx$ $dv=\\sec^{2}x\\ dx\\quad\\Rightarrow\\quad v=\\tan x,$ and $\\int\\frac{x}{\\cos^{2}x}dx=\\underbrace{x\\tan x}_{uv}-\\int\\underbrace{\\tan x}_{v}\\underbrace{dx}_{du}.$ We know $\\int\\tan xdx=\\int\\frac{\\sin x}{\\cos x}dx=-\\int\\frac{d(\\cos x)}{\\cos x}=-\\ln|\\cos x|+C,$ so \\begin{aligned} \\int\\frac{x}{\\cos^{2}x}dx & =x\\tan x-(-\\ln|\\cos x|+C)\\\\ & =x\\tan x+\\ln|\\cos x|+C_{1},\\end{aligned} with", "&\\tan(A - B) + \\tan(B - C) + \\tan(C - A) \\\\ &\\quad = \\tan((A - C) + (C - B)) + \\tan(B - C) + \\tan(C - A) \\\\ &\\quad = \\frac{\\tan(A - C) + \\tan(C - B)}{1 - \\tan(A - C)\\tan(C - B)} + \\tan(B - C) + \\tan(C - A). \\end{align*} Since all we need to know is whether this is $0$, we can put it over a single denominator, which we can then disregard, and save some algebra. This is $0$ if and only if \\begin{align*} 0 &= \\tan(A - C) + \\tan(C - B) - \\tan(C - B) - \\tan(A - C)\\\\ &\\quad + [\\tan(A - C) + \\tan(C - B)]\\tan(A - C)\\tan(C - B) \\\\ &= \\tan(A - C)\\tan(C - B) [\\tan(A - C) + \\tan(C - B)]. \\end{align*} So we see that $\\tan(A - C) = 0$, or $\\tan(C - B) = 0$, or $\\tan(A - C) = - \\tan(C- B)$. Since $A$, $B$ and $C$ are angles of a triangle, we are restricted in the range that they can take, and we find that $A = C$, or $C = B$, or $A - C = - C + B$. This last possibility is equivalent to $A = B$. We have established that the equation holds if and only if", "B < {90^\\circ}$$ and $$A > B$$. If $$\\tan \\,(A + B) = \\sqrt 3 \\,\\,\\rm{and}\\,\\,\\tan \\left( {A - B} \\right)\\, = \\frac{1}{\\sqrt 3}$$ , find the values of $$A$$ and $$B$$. Solution: We have: $\\tan \\left( {A + B} \\right) = \\sqrt 3 \\Rightarrow A + B = {60^ \\circ}$ $\\tan \\left( {A - B} \\right) = \\frac{1}{{\\sqrt 3 }} \\Rightarrow A - B = {30^\\circ }$ Solving these two equations gives $$\\boxed{A = 45^\\circ }$$ and $$\\boxed{B = 15^\\circ }$$. Think: Why is the fact that $${0^ \\circ } < A + B < {90^ \\circ }\\,\\,\\rm{and}\\,\\,A > B$$ significant? Did we make any use of this fact? Challenge 2: If $$\\sin (A - B) = \\frac{1}{2},\\cos (A + B) = \\frac{1}{2}$$ and $${0^\\circ } < A + B \\leq 90^\\circ ,A > B,$$ find the values of $$A$$ and $$B$$. Example 4: If \\begin{align}\\frac{{\\sin {{60}^ \\circ} + \\cos {{30}^ \\circ }}}{{1 + \\cos {{60}^ \\circ } + \\sin {{30}^ 0 }}} = \\sin \\theta \\end{align} Find the value of $$\\theta$$ . Solution: We have: \\begin{align}\\frac{{\\frac{{\\sqrt 3 }}{2} + \\frac{{\\sqrt 3 }}{2}}}{{1 + \\frac{1}{2} + \\frac{1}{2}}} &= \\sin \\theta \\\\ \\Rightarrow \\frac{{\\sqrt 3 }}{2} &= \\sin \\theta \\\\ \\\\\\Rightarrow &\\boxed {\\theta = {60^ \\circ}}\\end{align} Example 5: Later on, we will study a lot of trigonometric identities", "$$au^2 + bv^2 \\equiv 0 \\pmod{|c|} \\implies au^2 \\equiv -bv^2 \\pmod{|c|} \\tag{4}\\label{eq4A}$$ From \\eqref{eq2A}, we have $$aux + bvy \\equiv 0 \\pmod{|c|} \\implies aux \\equiv -bvy \\pmod{|c|} \\tag{5}\\label{eq5A}$$ Multiplying both sides of \\eqref{eq5A} by $$u$$ and using \\eqref{eq4A} gives \\begin{aligned} (au^2)x & \\equiv -buvy \\pmod{|c|} \\\\ (-bv^2)x & \\equiv -buvy \\pmod{|c|} \\\\ vx & \\equiv uy \\pmod{|c|} \\end{aligned}\\tag{6}\\label{eq6A} In the last line, I used that $$b$$ and $$v$$ are relatively prime to $$c$$. Next, multiply both sides of \\eqref{eq6A} by $$bvy$$, plus use \\eqref{eq4A} and \\eqref{eq5A}, to get \\begin{aligned} (bv^2)xy & \\equiv buvy^2 \\pmod{|c|} \\\\ (-au^2)xy & \\equiv buvy^2 \\pmod{|c|} \\\\ (-au)xy & \\equiv bvy^2 \\pmod{|c|} \\\\ -a(uy)x & \\equiv bvy^2 \\pmod{|c|} \\\\ -a(vx)x & \\equiv bvy^2 \\pmod{|c|} \\\\ -ax^2 & \\equiv by^2 \\pmod{|c|} \\\\ ax^2 + by^2 & \\equiv 0 \\pmod{|c|} \\end{aligned}\\tag{7}\\label{eq7A} Similarly, you can show that $$by^2 + cz^2 \\equiv 0 \\pmod{|a|}$$ and $$ax^2 + cz^2 \\equiv 0 \\pmod{|b|}$$. As $$a,b,c$$ are relatively prime to each other, you can put these together to get that \\eqref{eq3A} holds. Note I did not use anywhere that $$a,b$$ and $$c$$ are square-free integers, so I'm note sure why this was stated as the question's result holds without that requirement. • Excellent, thank you for your thourough answer – PerelMan Oct 1 at 2:00", "the first half of the 1800s (e.g. one was in Crelle’s Journal [= Journal für die reine und angewandte Mathematik]), but I don’t have the exact references with me now. - Starting with $\\tan(\\pi/3)=\\sqrt{3}$ and $\\tan(\\pi/4)=1$ and using $$\\tan(x/2)=\\frac{\\sqrt{1+\\tan^2(x)}-1}{\\tan(x)}\\tag{1}$$ and $$\\tan(x+y)=\\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}\\tag{2}$$ and $$(\\cos(x),\\sin(x))=\\frac{(1,\\tan(x))}{\\sqrt{1+\\tan^2(x)}}\\tag{3}$$ we can construct the sine and cosine of all rational multiples of $\\pi$ where the denominator is a power of $2$, or $3$ times a power of $2$. For example, $x=\\pi/4$ with $(1)$ gives $$\\tan(\\pi/8)=\\sqrt{2}-1\\tag{4}$$ then $(2)$ and $(4)$ yields \\begin{align} \\tan(3\\pi/8) &=\\tan(\\pi/4+\\pi/8)\\\\ &=\\frac{1+\\tan(\\pi/8)}{1-\\tan(\\pi/8)}\\\\ &=\\sqrt{2}+1\\tag{5} \\end{align} Then $(3)$ and $(5)$ give $$(\\cos(3\\pi/8),\\sin(3\\pi/8))=\\frac{(1,\\sqrt{2}+1)}{\\sqrt{4+2\\sqrt{2}}}\\tag{6}$$ -", "$$\\cos(2\\theta) = {\\cos}^2 \\theta-{\\sin}^2 \\theta = \\cos^2 \\theta - (1-{\\cos}^2 \\theta) = 2 \\cos^2 \\theta - 1$$ Similarly, to derive the double-angle formula for tangent, replacing $$\\alpha=\\beta=\\theta$$ in the sum formula gives \\begin{align*} \\tan(\\alpha+\\beta)&= \\dfrac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\[4pt] \\tan(\\theta+\\theta)&= \\dfrac{\\tan \\theta+\\tan \\theta}{1-\\tan \\theta \\tan \\theta}\\\\[4pt] \\tan(2\\theta)&= \\dfrac{2\\tan \\theta}{1-{\\tan}^2 \\theta} \\end{align*} DOUBLE-ANGLE FORMULAS \\begin{align} \\sin(2\\theta)&= 2 \\sin \\theta \\cos \\theta\\\\[4pt] \\cos(2\\theta)&= \\begin{cases}{\\cos}^2 \\theta-{\\sin}^2 \\theta \\\\ 1-2 {\\sin}^2 \\theta \\\\ 2{\\cos}^2 \\theta-1 \\end{cases}\\\\[4pt] \\tan(2\\theta)&= \\dfrac{2 \\tan \\theta}{1-{\\tan}^2\\theta} \\end{align} Example $$\\PageIndex{1}$$: Use Double-Angle Formulas to Evaluate a Trigonometric Expression Given that $$\\tan \\theta=−\\dfrac{3}{4}$$ and $$\\theta$$ is in quadrant II, find the following: a. $$\\sin(2\\theta)$$ $$\\qquad$$ b. $$\\cos(2\\theta)$$ $$\\qquad$$ c. $$\\tan(2\\theta)$$ Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $$\\tan \\theta=−\\dfrac{3}{4}$$, such that $$\\theta$$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $$\\theta$$ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: $$c^2 = {(-4)}^2+{(3)}^2 = 16+9 = 25 \\rightarrow c= 5$$ Therefore, $${ \\begin{matrix} \\boxed { \\begin{array}{llr} x &=& -4 \\\\ y &=& 3\\\\ r &=& 5 \\end{array}} \\end{matrix} }$$.", "Curious Cheetah # Double u Substitution: ∫tan³θ dθ ## Introduction $$u$$ substitution is an important technique in integration. In the problem explored in this article, we look at a moderately advanced example in which $$u$$ substitution is performed twice; it also involves trigonometric and logarithmic functions, as well as several other key issues in integration. ## Problem $\\tag{1}\\int \\tan^3\\theta\\,d\\theta$ ## Background Information In order to complete this problem, we need two related pieces of information: \\begin{align} \\tag{2a} d(\\tan \\theta) = \\frac{1}{\\cos^2 \\theta}\\,d\\theta \\\\ \\tag{2b} \\cos^2 \\theta = \\frac{1}{1+\\tan^2 \\theta} \\end{align} Note that (2a) is a rephrasing of the standard formula $$d(\\tan \\theta) = \\sec^2 \\theta\\,d\\theta$$. The calculus texts with which Iʼm familiar tend to simply present that and the formulae for $$d(\\cot \\theta)$$, $$d(\\sec \\theta)$$, $$d(\\csc \\theta)$$ as formulae to memorize. However, itʼs important that students do actually work these out; indeed, the derivatives of all six of the main trigonometric functions are based on the simple pair: $$d(\\sin \\theta) = \\cos \\theta\\,d\\theta; d(\\cos \\theta) = -\\sin \\theta\\,d\\theta$$. Memorize that pair, and everything else can be generated as needed. In this case, we need $$d(\\tan \\theta)$$. This is derived as follows: \\begin{align} \\tag{3a} d(\\tan \\theta) &= d\\left(\\frac{\\sin \\theta}{\\cos \\theta}\\right)\\\\ \\tag{3b} &= \\frac{\\cos \\theta\\cdot d(\\sin \\theta) - \\sin \\theta\\cdot d(\\cos \\theta)}{\\cos^2 \\theta}\\\\ \\tag{3c} &= \\frac{\\cos \\theta\\cdot\\cos \\theta\\,d\\theta - \\sin", "$f(x) \\equiv 0 \\pmod {7^3}$. We attempt to find a solution to $f(x) \\equiv 0 \\pmod 7$. We have that: (7) \\begin{align} \\quad f(0) &= -13 \\equiv 1 \\pmod 7 \\\\ \\quad f(1) &= -8 \\equiv 6 \\pmod 7 \\\\ \\quad f(2) &= 21 \\equiv 0 \\pmod 7 \\\\ \\quad f(3) &= 116 \\equiv 4 \\pmod 7 \\\\ \\quad f(4) &= 343 \\equiv 0 \\pmod 7 \\\\ \\quad f(5) &= 792 \\equiv 1 \\pmod 7\\\\ \\quad f(6) &= 1577 \\equiv 2 \\pmod 7 \\end{align} Let $s = 2$ and $t = 4$. Then $s$ and $t$ are solutions to $f(x) \\equiv 0 \\pmod p$. Now the derivative of $f(x)$ is: (8) $$f'(x) = 4x^3 + 3x^2 + 4x + 1$$ Plugging in $s$ and $t$ into $f'$ gives us: (9) \\begin{align} \\quad f'(s) = f'(2) = 53 \\equiv 4 \\not \\equiv 0 \\pmod 7 \\\\ \\quad f'(t) = f'(4) = 321 \\equiv 6 \\not \\equiv 0 \\pmod 7 \\end{align} So Hensel's Lemma can be applied to both $s$ and $t$. First set $a_1 = 2$. From above we have that $f'(2) = 4$. So $[f'(2)]^{-1} = [4]^{-1} = 2 \\pmod 7$. Hence: (10) \\begin{align} \\quad a_2 & \\equiv a_1 - f(a_1)[f'(2)]^{-1} \\pmod {7^2} \\\\ & \\equiv 2 - f(2) \\cdot 2 \\pmod {7^2} \\\\ & \\equiv 2 -", "$y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \\ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\\Delta \\le 0$. Then, $$\\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \\le 0$$ as desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$. Then $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $a:b:c = 8:5:5$. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle. $\\blacksquare$ ## Solution 2 Let $A$, $B$, $C$ be the three angles of the triangle, then $r/AP = \\tan{(A/2)}$, $r/BQ=\\tan{(B/2)}$, and $r/CR=\\tan{(C/2)}$. Without loss of generality, assume $A \\leq B \\leq C$, then $CR=min{(AP,BQ,CR)}$. Denote $x=\\tan{(A/2)}$, $x=\\tan{(B/2)}$, $z=\\tan{(C/2)}$, then we have, $$5(x+y+z)- 3z=6 \\qquad \\qquad (1)$$ Meanwhile, $$z = \\tan{(C/2)} = \\tan{(90- (A+B)/2))} = \\frac{1-xy}{x+y}$$ Substitute $z$ in $(1)$, we get $$5x^{2}+8xy+5y^{2} - 6x - 6y+2=0 \\qquad \\qquad (2)$$ Treating (2) as a quadratic equation for $x$, the discriminant is: $$\\Delta = (8y-6)^2 - 4*5(5y^2-6y+2)=-36y^2+24y-4 = -(6y-2)^2 \\leq 0$$ For $x$, $y$ to", "ad+bc & \\equiv 0 \\mod p\\\\ bd & \\equiv B^2 \\mod p. \\end{align} From our first equation we deduce that $c\\equiv-a\\mod p$. Multiplying the third equation by $d$ and substituting the fourth into the resultant equation yields \\begin{align} a(d^2-B^2)\\equiv 0\\mod p. \\end{align} Now, since $p$ is prime, we have two possible cases, $d^\\equiv B^2\\mod p$ or $a\\equiv 0 \\mod p$. (It is possible for both cases to be fulfilled simultaneously, but we need only concentrate on having one of them satisfied). #### Case 1: $d^2\\equiv B^2\\mod p$ In this case we have $d\\equiv\\pm B\\mod p$ and so $b\\equiv d\\mod p$. Plugging this into $ac+b+d\\equiv A\\mod p$ from before we get \\begin{align} a^2&\\equiv 2d-A\\mod p\\\\ &\\equiv \\pm 2B-A\\mod p. \\end{align} This means, that if $\\pm 2B-A$ is a quadratic residue $\\mod p$, then $g(x)$ has the factorisation \\begin{align} g(x)=(x^2+ax+d)(x^2-ax+d) \\end{align} where $d\\equiv\\pm B\\mod p$ and $a^2\\equiv 2d-A\\mod p$. ##### Example Let $p=7$. In our polynomial with $A=-4$ and $B=12$ we see that $-2B-A\\equiv 1\\mod 7$, and $1$ is a quadratic residue $\\mod 7$, so $f(x)$ can be factored as \\begin{align} f(x)=(x^2+x+2)(x^2+6x+2)\\mod 7. \\end{align} If $\\pm 2B-A$ is not a residue $\\mod p$, then we must be in case 2. #### Case 2: $a\\equiv 0\\mod p$ In this case we also have $c\\equiv 0\\mod p$, and so we are now", "{n^2} &= \\left( {m + n} \\right)\\left( {m - n} \\right)\\\\&= 2\\tan \\theta \\times 2\\sin \\theta \\\\&= 4\\tan \\theta \\sin \\theta\\end{align} And, \\begin{align}&\\sqrt {mn} = \\sqrt {{{\\tan }^2}\\theta - {{\\sin }^2}\\theta } \\\\ \\qquad\\;&\\qquad\\;= \\sqrt {\\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }} - {{\\sin }^2}\\theta } \\\\ \\qquad\\;&\\qquad\\;= \\sin \\theta \\sqrt {\\frac{1}{{{{\\cos }^2}\\theta }} - 1} \\\\\\qquad\\; &\\qquad\\;= \\frac{{\\sin \\theta }}{{\\cos \\theta }} \\times \\sqrt {1 - {{\\cos }^2}\\theta } \\\\ \\qquad\\; &\\qquad\\;= \\tan \\theta \\sin \\theta \\end{align} Thus, \\begin{align} \\boxed {\\frac{{{m^2} - {n^2}}}{{\\sqrt {mn} }} = 4} \\end {align} Example 4 : If $$\\tan A = n \\,\\tan B$$ and $$\\sin A = m\\,\\sin B$$, find the value of $${\\cos ^2}A$$ in terms of $$m$$ and $$n$$. Solution: We need to eliminate $$B$$ from these two relations. We have : $\\sin B = \\frac{{\\sin A}}{m},\\tan B = \\frac{{\\tan A}}{n}$ $\\Rightarrow {\\cos ^2}B = 1 - \\frac{{{{\\sin }^2}A}}{{{m^2}}},{\\sec ^2}B = 1 + \\frac{{{{\\tan }^2}A}}{{{n^2}}}$ Taking the product, we have : \\begin{align}&{\\cos ^2}B{\\sec ^2}B = \\left( {1 - \\frac{{{{\\sin }^2}A}}{{{m^2}}}} \\right)\\left( {1 + \\frac{{{{\\tan }^2}A}}{{{n^2}}}} \\right)\\\\ &\\Rightarrow 1 = 1 - \\frac{{{{\\sin }^2}A}}{{{m^2}}} + \\frac{{{{\\tan }^2}A}}{{{n^2}}} - \\frac{{{{\\sin }^2}A\\;{{\\tan }^2}A}}{{{m^2}{n^2}}}\\\\ &\\Rightarrow {n^2}{\\sin ^2}A - {m^2}{\\tan ^2}A + {\\sin ^2}A\\;{\\tan ^2}A = 0\\\\ &\\Rightarrow {n^2} - {m^2}{\\sec ^2}A + {\\tan ^2}A = 0\\\\ &\\Rightarrow {n^2} - {m^2}{\\sec ^2}A + {\\sec ^2}A -", "mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Dec 12 '19 at 10:28 • Welcome to Mathematics Stack Exchange. $x\\equiv-2\\bmod4$ and $x\\equiv-2\\bmod5\\implies x\\equiv-2\\bmod 20$. Now solve that and $x\\equiv9\\bmod 13$ – J. W. Tanner Dec 12 '19 at 10:31 • I added an answer showing how to use the constant-case CRT optimization mentioned by @J.W.T This is the easiest approach in cases like this. – Bill Dubuque Dec 12 '19 at 16:52 We have , $$x\\equiv2 \\mod 4 \\implies \\color{#d05}{x=4a+2}\\\\$$ \\begin{align}x&\\equiv3 \\mod 5 \\\\4a+2&\\equiv3\\mod5\\\\ 4a&\\equiv 1\\mod5 \\\\ a &\\equiv 4 \\mod 5\\implies\\color{#2cd}{a = 5b+4}\\end{align} \\begin{align}x&\\equiv9 \\mod 13 \\\\ 4(5b+4)+2&\\equiv 9\\mod13 \\\\ 20b+18&\\equiv9\\mod13 \\\\ 20b&\\equiv 4\\mod 13 \\implies \\color{#2c0}{b = 13c+8}\\end{align} So , $$x = 4a+2 = 20b+18=260c + 178$$ • To fill the two gaps: $\\ a = 5b+4\\,$ so $\\,x = 4a+2 = 4(5b+4) = 20b+18.\\$ $\\bmod 13\\!:\\ 9\\equiv x\\equiv 20b+18 \\equiv -6b+18\\iff 6b\\equiv 9\\iff b\\equiv 9/6\\equiv 3/2\\equiv 16/2\\equiv 8.\\$ This method of solving the congruences in pairs works generally, see here for a proof. – Bill Dubuque Dec 12 '19 at 17:05 • @BillDubuque I Left that for the O.P to fill in . On a different note , why was the answer downvoted ?" ]

[ "b^2 \\\\ 2^2 & = b^2 \\\\ b & = 2 \\end{align*} Therefore the equation is $$y=2 \\times 2^{x} + 1$$. This is a tangent graph and so we use $$y = a \\tan \\theta + q$$. To find $$q$$ we note that the graph has been shifted down by 2 units (the point $$(180^\\circ;-2)$$ is given). Therefore $$q = -2$$. To find $$a$$ we substitute $$(135^\\circ;-1)$$ into the equation: \\begin{align*} y & = a \\tan \\theta - 2 \\\\ -1 & = a \\tan (135^\\circ) - 2 \\\\ 1 & = -a \\\\ a & = -1 \\end{align*} Therefore the equation is $$y=-\\tan \\theta - 2$$. State the coordinates at $$A,~B,~C$$ and $$D$$. $$A( 90^{\\circ};1 ),~B(90 ^{\\circ};-1 ),~C( 180^{\\circ}; 2) \\text{ and } D(360 ^{\\circ};1)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. 0 What is the amplitude of $$f(x)$$. 1 Evaluate: $$f(180^{\\circ}) - g(180^{\\circ})$$ . \\begin{align*} f(180^{\\circ}) - g(180^{\\circ}) &= 2 - (-3)\\\\ &=5 \\end{align*} State the coordinates at $$A,~B,~C$$ and $$D$$. $$A( 90^{\\circ}; 1),~B(270 ^{\\circ}; 3),~C(270 ^{\\circ}; 1) \\text{ and } D(360 ^{\\circ}; 2)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. 2 What is the amplitude of $$g(x)$$. 3 Evaluate: $$g(180^{\\circ}) - f(180^{\\circ})$$ . \\begin{align*} f(180^{\\circ}) - g(180^{\\circ}) &= 4 - 2 \\\\ &=2 \\end{align*} $$y=2^{x}$$ and $$y=-2^{x}$$ are sketched", "# 2013 AMC 12B Problems/Problem 20 ## Problem For $135^\\circ < x < 180^\\circ$, points $P=(\\cos x, \\cos^2 x), Q=(\\cot x, \\cot^2 x), R=(\\sin x, \\sin^2 x)$ and $S =(\\tan x, \\tan^2 x)$ are the vertices of a trapezoid. What is $\\sin(2x)$? $\\textbf{(A)}\\ 2-2\\sqrt{2}\\qquad\\textbf{(B)}\\3\\sqrt{3}-6\\qquad\\textbf{(C)}\\ 3\\sqrt{2}-5\\qquad\\textbf{(D}}\\ -\\frac{3}{4}\\qquad\\textbf{(E)}\\ 1-\\sqrt{3}$ (Error compiling LaTeX. ! Undefined control sequence.) ## Solution Let $f,g,h,j$ be $\\sin, \\cos, \\tan, \\cot$ (not respectively). Then we have four points $(f,f^2),(g,g^2),(h,h^2),(j,j^2)$, and the two lines connecting two pairs of them must be parallel (as we are dealing with a trapezoid). WLOG, we take the line connecting the first two and the line connecting the last two to be parallel, so that $\\frac{g^2-f^2}{g-f} = \\frac{j^2-h^2}{j-h}$, or $g+f = j+h$. Now, we must find how to match up $\\sin, \\cos, \\tan, \\cot$ to $f,g,h,j$ so that the above equation has a solution. The three possible ways are $\\sin x+\\cos x = \\tan x+\\cot x$, $\\sin x+\\tan x = \\cos x+\\cot x$, and $\\sin x+\\cot x = \\cos x+\\tan x$. On the interval $135^\\circ < x < 180^\\circ$, we have $0<\\sin x<\\frac{1}{\\sqrt{2}},-1< \\cos x<-\\frac{1}{\\sqrt{2}},-1< \\tan x<0,-\\infty< \\cot x<-1$, so the first two of those possible ways do not work because the LHS and the RHS have different ranges. Thus, it must be the last possible way, $\\sin x+\\cot x = \\cos", "xyz; \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\quad \\end{align} (b) Prove that \\begin{align} \\frac{x}{{1 - {x^2}}} + \\frac{y}{{1 - {y^2}}} + \\frac{z}{{1 - {z^2}}} \\ge \\frac{{3\\sqrt 3 }}{2}\\\\ {\\text{if}}\\;0 < x,\\;y,\\;z < 1\\;{\\rm{and}}\\;xy + yz + zx = 1. \\qquad \\qquad \\qquad \\qquad \\qquad \\end{align} Solution: Both problems can be solved by trigonometric substitutions. (a) Let there be there is an acute triangle $$ABC$$ with $$\\tan A = x,\\;\\tan B = y$$, and $$\\tan C = z$$. Note that \\begin{align}\\frac{{\\tan A}}{{\\sqrt {1 + {{\\tan }^2}A} }} = \\frac{{\\tan A}}{{\\sec A}} = \\sin A.\\end{align} The desired inequality becomes \\begin{align}\\sin A + \\sin B + \\sin C \\le \\frac{{3\\sqrt 3 }}{2}\\end{align} which we’ve covered earlier. (b) From the given condition, we can assume that there is an acute triangle $$ABC$$ such that \\begin{align}\\tan \\frac{A}{2} = x, \\qquad \\tan \\frac{B}{2} = y, \\qquad \\tan \\frac{C}{2} = z.\\end{align} By the double-angle formulas, it suffices to prove that \\begin{align}\\tan A + \\tan B + \\tan C \\ge 3\\sqrt 3 \\end{align} which is another inequality we’ve encountered already. Example - 16 Compute the sums $\\left( \\begin{matrix} n \\\\ 1 \\\\ \\end{matrix} \\right)\\sin a+\\left( \\begin{matrix} n \\\\ 2 \\\\ \\end{matrix} \\right)\\sin 2a+....+\\left( \\begin{matrix} n \\\\ n \\\\ \\end{matrix} \\right)\\sin na$ and $\\left( \\begin{matrix} n \\\\ 1 \\\\ \\end{matrix} \\right)\\cos a+\\left( \\begin{matrix} n", "B < {90^\\circ}$$ and $$A > B$$. If $$\\tan \\,(A + B) = \\sqrt 3 \\,\\,\\rm{and}\\,\\,\\tan \\left( {A - B} \\right)\\, = \\frac{1}{\\sqrt 3}$$ , find the values of $$A$$ and $$B$$. Solution: We have: $\\tan \\left( {A + B} \\right) = \\sqrt 3 \\Rightarrow A + B = {60^ \\circ}$ $\\tan \\left( {A - B} \\right) = \\frac{1}{{\\sqrt 3 }} \\Rightarrow A - B = {30^\\circ }$ Solving these two equations gives $$\\boxed{A = 45^\\circ }$$ and $$\\boxed{B = 15^\\circ }$$. Think: Why is the fact that $${0^ \\circ } < A + B < {90^ \\circ }\\,\\,\\rm{and}\\,\\,A > B$$ significant? Did we make any use of this fact? Challenge 2: If $$\\sin (A - B) = \\frac{1}{2},\\cos (A + B) = \\frac{1}{2}$$ and $${0^\\circ } < A + B \\leq 90^\\circ ,A > B,$$ find the values of $$A$$ and $$B$$. Example 4: If \\begin{align}\\frac{{\\sin {{60}^ \\circ} + \\cos {{30}^ \\circ }}}{{1 + \\cos {{60}^ \\circ } + \\sin {{30}^ 0 }}} = \\sin \\theta \\end{align} Find the value of $$\\theta$$ . Solution: We have: \\begin{align}\\frac{{\\frac{{\\sqrt 3 }}{2} + \\frac{{\\sqrt 3 }}{2}}}{{1 + \\frac{1}{2} + \\frac{1}{2}}} &= \\sin \\theta \\\\ \\Rightarrow \\frac{{\\sqrt 3 }}{2} &= \\sin \\theta \\\\ \\\\\\Rightarrow &\\boxed {\\theta = {60^ \\circ}}\\end{align} Example 5: Later on, we will study a lot of trigonometric identities", "following conditions are equivalent: \\begin{align*} \\tan x + \\tan y &= 0\\\\ \\tan x &= -\\tan y\\\\ \\tan x &= \\tan(-y)\\\\ x &= -y + k\\pi\\\\ x+y&= k\\pi \\end{align*} If we use $y=2x$ we get \\begin{align*} 3x &= k\\pi\\\\ x &= k\\frac\\pi3 \\end{align*} In line 4 you divided both side with $\\tan x$ assuming that $\\tan x\\neq0$ ($1$) So, if $\\tan x\\neq0$, you are on right track. $x=60^\\circ$ ($2$) But, if $\\tan x=0$, you didn't consider this one. So, $\\tan x=0\\implies x=0^\\circ$ $$\\tan x+\\tan2x=\\dfrac{\\sin(x+2x)}{\\cos x\\cos2x}$$ So, we need $\\sin3x=0\\implies3x=n\\pi$ where $n$ is any integer By the double angle formula we get $$\\tan(2x)+\\tan(x)=\\frac{2\\tan(x)}{1-\\tan^2(x)}+\\tan(x)=\\frac{3-\\tan^2(x)}{1-\\tan^2(x)}\\tan(x),$$ so that $\\tan(x)=0$ is certainly a solution. But $\\tan(x)=\\pm\\sqrt3$ as well, so that the initial claim is false.", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "x & = 0\\\\ x^2 &= x &&\\text{(adding }x\\text{ to both sides)}\\\\ \\frac{x^2}{x} &= \\frac{x}{x} &&\\text{(divide by }x\\text{, which assumes }x\\neq 0)\\\\ x &= 1. \\end{align*} This assumes $x\\ne0$. Check this out: \\begin{align*} x^2 - x & = 0\\\\ x(x-1) &= 0 &&\\text{(factor)}\\\\ x=0 \\quad &\\text{or} \\quad (x-1)=0 && \\text{(by some rule in } \\mathbb{R} \\text{ that says } ab=0 \\implies a=0 \\lor b=0 \\text{)}\\\\ x=0 &\\lor \\ x=1.\\\\ \\end{align*} Another good example is shown when solving a trigonometric equation. There are four solutions to the equation: $\\sin x \\tan x = \\sin x$, \\begin{align} \\sin x \\tan x &= \\sin x \\\\ \\sin x \\tan x - \\sin x &= 0 \\\\ \\sin x (\\tan x - 1) &= 0 \\end{align} $\\sin x = 0$ or $\\tan x = 1$. The solution set is $\\{0^\\circ, 45^\\circ, 180^\\circ, 225^\\circ\\}$. However, trying to solve the equation by dividing each side by $\\sin x$ would lead to just $\\tan x = 1$, which would only give you $x = 45^\\circ$ or $x = 225^\\circ$. The other two solutions would not appear. The missing solutions are the ones that make the divisor ($\\sin x$) equal zero. For this reason we avoid dividing by a variable expression. If you do it correctly, you most certainly can divide both sides of an equation", "\\frac{x}{x} &&\\text{(divide by }x\\text{, which assumes }x\\neq 0)\\\\ x &= 1. \\end{align*} This assumes $x\\ne0$. Check this out: \\begin{align*} x^2 – x & = 0\\\\ x(x-1) &= 0 &&\\text{(factor)}\\\\ x=0 \\quad &\\text{or} \\quad (x-1)=0 && \\text{(by some rule in } \\mathbb{R} \\text{ that says } ab=0 \\implies a=0 \\lor b=0 \\text{)}\\\\ x=0 &\\lor \\ x=1.\\\\ \\end{align*} Another good example is shown when solving a trigonometric equation. There are four solutions to the equation: $\\sin x \\tan x = \\sin x$, \\begin{align} \\sin x \\tan x &= \\sin x \\\\ \\sin x \\tan x – \\sin x &= 0 \\\\ \\sin x (\\tan x – 1) &= 0 \\end{align} $\\sin x = 0$ or $\\tan x = 1$. The solution set is $\\{0^\\circ, 45^\\circ, 180^\\circ, 225^\\circ\\}$. However, trying to solve the equation by dividing each side by $\\sin x$ would lead to just $\\tan x = 1$, which would only give you $x = 45^\\circ$ or $x = 225^\\circ$. The other two solutions would not appear. The missing solutions are the ones that make the divisor ($\\sin x$) equal zero. For this reason we avoid dividing by a variable expression. If you do it correctly, you most certainly can divide both sides of an equation by a variable. As long as $x \\ne 0$, $\\dfrac 1x$ exists; and, as", "the first half of the 1800s (e.g. one was in Crelle’s Journal [= Journal für die reine und angewandte Mathematik]), but I don’t have the exact references with me now. - Starting with $\\tan(\\pi/3)=\\sqrt{3}$ and $\\tan(\\pi/4)=1$ and using $$\\tan(x/2)=\\frac{\\sqrt{1+\\tan^2(x)}-1}{\\tan(x)}\\tag{1}$$ and $$\\tan(x+y)=\\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}\\tag{2}$$ and $$(\\cos(x),\\sin(x))=\\frac{(1,\\tan(x))}{\\sqrt{1+\\tan^2(x)}}\\tag{3}$$ we can construct the sine and cosine of all rational multiples of $\\pi$ where the denominator is a power of $2$, or $3$ times a power of $2$. For example, $x=\\pi/4$ with $(1)$ gives $$\\tan(\\pi/8)=\\sqrt{2}-1\\tag{4}$$ then $(2)$ and $(4)$ yields \\begin{align} \\tan(3\\pi/8) &=\\tan(\\pi/4+\\pi/8)\\\\ &=\\frac{1+\\tan(\\pi/8)}{1-\\tan(\\pi/8)}\\\\ &=\\sqrt{2}+1\\tag{5} \\end{align} Then $(3)$ and $(5)$ give $$(\\cos(3\\pi/8),\\sin(3\\pi/8))=\\frac{(1,\\sqrt{2}+1)}{\\sqrt{4+2\\sqrt{2}}}\\tag{6}$$ -", "For example: $a=b=c=0$ or $a=0$ and $b=2\\pi-c$. – robjohn Jul 25 '20 at 12:03 HINT $A+B+C = 180$ $A+B = 180 - C$ We'll apply tangent function: $\\tan (A+B) = \\tan (180 - C)$ We'll consider the identity: $\\tan(x+y) = \\frac{\\tan x + \\tan y}{1-\\tan x\\tan y}$ $\\frac{\\tan A + \\tan B}{1-\\tan A\\tan B} = \\frac{\\tan 180 - \\tan C}{1+\\tan 180\\tan C}$ But $\\tan 180 = 0$, therefore, we'll get: $\\frac{\\tan A + \\tan }{1-\\tan A\\tan B}$ = $\\frac{0 - \\tan C}{1+0}$ $\\frac{\\tan A + \\tan B}{1-\\tan A\\tan B} = -\\tan C$ We'll multiply by $(1-\\tan A\\tan B)$: $\\tan A + \\tan B = -\\tan C +\\tan A\\tan B\\tan C$ Hence $\\tan A + \\tan B+ \\tan C = \\tan A\\tan B\\tan C$ Here is a geometric proof, for the case that all three angles are acute: $QRUV$ are collinear because $B+90^\\circ+(90^\\circ-B)=180^\\circ$. $STV$ are collinear because $A+B+C=180^\\circ$, so $\\angle QSV=\\angle UTV=C$. Similar triangles $\\triangle PQR\\sim\\triangle TRS$ and $\\triangle RTU \\sim \\triangle SRQ$ give $\\displaystyle \\frac{QP}{RQ} = \\frac{RT}{SR} = \\frac{TU}{RQ}$, and therefore $TU=QP=1$. Then, \\begin{align}& \\tan A + \\tan B + \\tan C = QR+RU+UV = QV \\\\ &= QP \\frac{QR}{QP}\\, \\frac{QS}{QR} \\, \\frac{QV}{QS} = 1 \\cdot \\tan(A) \\tan(B) \\tan(C) \\end{align} When one of the angles is obtuse, let it (without loss of generality) be $C$. Then a similar", "# 2009 AMC 12A Problems/Problem 25 ## Problem The first two terms of a sequence are $a_1 = 1$ and $a_2 = \\frac {1}{\\sqrt3}$. For $n\\ge1$, $$a_{n + 2} = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$? $\\textbf{(A)}\\ 0\\qquad \\textbf{(B)}\\ 2 - \\sqrt3\\qquad \\textbf{(C)}\\ \\frac {1}{\\sqrt3}\\qquad \\textbf{(D)}\\ 1\\qquad \\textbf{(E)}\\ 2 + \\sqrt3$ ## Solution Consider another sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ such that $a_n = \\tan{\\theta_n}$, and $0 \\leq \\theta_n < 180$. The given recurrence becomes \\begin{align*} a_{n + 2} & = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\\\ \\tan{\\theta_{n + 2}} & = \\frac {\\tan{\\theta_n} + \\tan{\\theta_{n + 1}}}{1 - \\tan{\\theta_n}\\tan{\\theta_{n + 1}}} \\\\ \\tan{\\theta_{n + 2}} & = \\tan(\\theta_{n + 1} + \\theta_n) \\end{align*} It follows that $\\theta_{n + 2} \\equiv \\theta_{n + 1} + \\theta_{n} \\pmod{180}$. Since $\\theta_1 = 45, \\theta_2 = 30$, all terms in the sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ will be a multiple of $15$. Now consider another sequence $\\{b_1, b_2, b_3...\\}$ such that $b_n = \\theta_n/15$, and $0 \\leq b_n < 12$. The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \\equiv b_{n + 1} + b_n \\pmod{12}$. As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the", "- {{89}^\\circ }} \\right)\\\\&\\qquad\\qquad\\qquad = \\tan {1^\\circ }\\cot {1^\\circ }\\;\\;= 1\\end{align} Similarly, take the product of the second and second last terms: $\\tan {2^\\circ }\\tan {88^\\circ } = \\tan {2^\\circ }\\cot {2^\\circ } = 1$ Thus, we can group the terms in $$E$$ into pairs, each of whose product is 1. One term will still be left: $$\\tan {45^\\circ }$$. But the value of this term is also 1. For this, please go through the trigonometric ratios of specific angles. Thus, $\\boxed {E=1}$ Challenge 2: Find the value of the following expression: $E = \\tan {48^\\circ }\\tan {23^\\circ }\\tan {42^\\circ }\\tan {67^\\circ }$ Tip: Pair up complementary angles. Example 4: If $$\\tan 2A = \\cot \\left( {A - {{21}^\\circ }} \\right)$$, where $$2A$$ is an acute angle, find $$A$$. Solution: We have, \\begin{align}\\tan 2A &= \\tan \\left( {{{90}^\\circ } - \\left( {A - {{21}^\\circ }} \\right)} \\right)\\\\ \\qquad\\quad&= \\tan \\left( {{{111}^\\circ } - A} \\right)\\\\ \\Rightarrow 2A &= {111^\\circ } - A\\end{align} $\\Rightarrow \\boxed{A = 37^\\circ }$ Challenge 3: If $$\\sin 3A = \\cos \\left( {A - 26^\\circ } \\right)$$, where $$3A$$ is an acute angle, find the value of $$A$$. Tip: Use a similar approach as in example-4. Example 5 : In any right $$\\Delta ABC$$, prove that $\\sin \\left( {\\frac{{B + C}}{2}} \\right) = \\cos \\frac{A}{2}$", "$$\\text{cosec}^2 A = 1 + \\cot^2 A$$ Sometimes you are required to manipulate the identity to suit the question. For example, we may need to make sin2 A the subject: \\begin{align} \\sin^2 A + \\cos^2 A & = 1 \\\\ \\sin^2 A & = 1 - \\cos^2 A \\end{align} ## Compound Angle Formula (alternatively Addition & Subtraction Formula) $$\\sin (A \\pm B) = \\sin A \\cos B \\pm \\cos A \\sin B$$ $$\\cos (A \\pm B) = \\cos A \\cos B \\mp \\sin A \\sin B$$ $$\\tan (A \\pm B) = {\\tan A \\pm \\tan B \\over 1\\mp \\tan A \\tan B}$$ Some popular ways to utilize the formula: \\begin{align} \\sin 3A & = \\sin (2A + A) \\\\ & = \\sin 2A \\cos A + \\cos 2A \\sin A \\\\ \\\\ \\cos (15^\\circ) & = \\cos (60^\\circ - 45^\\circ) \\\\ & = \\cos 60^\\circ \\cos 45^\\circ + \\sin 60^\\circ \\sin 45^\\circ \\end{align} ## Double Angle Formula $$\\sin 2A = 2\\sin A \\cos A$$ $$\\cos 2A = \\cos^2 A - \\sin^2 A = 2\\cos^2 A - 1 = 1 - 2\\sin^2 A$$ $$\\tan 2A = {2\\tan A \\over 1 - \\tan^2 A}$$ Sometimes you are required to change the subject of the formula to suit the question's requirement. The Half Angle formulas can be derived from the Double", "v=\\tan x,$ and $\\int\\sec^{3}x\\,dx=\\overbrace{\\sec x\\ \\tan x}^{uv}-\\int\\overbrace{\\tan^{2}x\\,\\sec x\\,dx}^{vdu}.$ Because $$1+\\tan^{2}x=\\sec^{2}x$$, we have \\begin{aligned} \\int\\sec^{3}x\\,dx & =\\sec x\\,\\tan x-\\int(\\sec^{2}x-1)\\sec x\\,dx\\\\ & =\\sec x\\,\\tan x-\\int\\sec^{3}x\\,dx-\\int\\sec x\\,dx,\\end{aligned} or $2\\int\\sec^{3}x\\,dx=\\sec x\\,\\tan x-\\int\\sec x\\,dx$ Previously we showed that $\\int\\sec x\\,dx=\\ln|\\sec x+\\tan x|+K.$ Therefore, $\\int\\sec^{3}x\\,dx=\\frac{1}{2}\\sec x\\:\\tan x+\\frac{1}{2}\\ln|\\sec x+\\tan x|+C.$ 1. Reference: Kasube, Herbert E. “A technique for integration by parts.” The American Mathematical Monthly 90.3 (1983): 210-211.↩︎", "$y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \\ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\\Delta \\le 0$. Then, $$\\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \\le 0$$ as desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$. Then $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $a:b:c = 8:5:5$. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle. $\\blacksquare$ ## Solution 2 Let $A$, $B$, $C$ be the three angles of the triangle, then $r/AP = \\tan{(A/2)}$, $r/BQ=\\tan{(B/2)}$, and $r/CR=\\tan{(C/2)}$. Without loss of generality, assume $A \\leq B \\leq C$, then $CR=min{(AP,BQ,CR)}$. Denote $x=\\tan{(A/2)}$, $x=\\tan{(B/2)}$, $z=\\tan{(C/2)}$, then we have, $$5(x+y+z)- 3z=6 \\qquad \\qquad (1)$$ Meanwhile, $$z = \\tan{(C/2)} = \\tan{(90- (A+B)/2))} = \\frac{1-xy}{x+y}$$ Substitute $z$ in $(1)$, we get $$5x^{2}+8xy+5y^{2} - 6x - 6y+2=0 \\qquad \\qquad (2)$$ Treating (2) as a quadratic equation for $x$, the discriminant is: $$\\Delta = (8y-6)^2 - 4*5(5y^2-6y+2)=-36y^2+24y-4 = -(6y-2)^2 \\leq 0$$ For $x$, $y$ to", "$$n \\in Z$$ ## Chapter 3 Ex.3.4 Question 8 Find the principal and general solutions of the following equation: $${\\sec ^2}2x = 1 - \\tan 2x$$ ### Solution \\begin{align}{\\sec ^2}2x &= 1 - \\tan 2x\\\\1 - {\\tan ^2}2x &= 1 - \\tan 2x\\\\{\\tan ^2}2x - \\tan 2x &= 0\\\\\\tan 2x\\left( {\\tan 2x - 1} \\right) &= 0\\end{align} Now, $$\\tan 2x = 0$$ or $$\\tan 2x - 1 = 0$$ Therefore, \\begin{align}\\tan 2x &= 0\\\\2x &= n\\pi \\quad n \\in Z\\\\x &= \\frac{{n\\pi }}{2}\\end{align} Or, \\begin{align}\\tan 2x - 1 &= 0\\\\\\tan 2x &= 1\\\\&= - \\tan \\frac{\\pi }{4}\\\\&= \\tan \\left( {\\pi - \\frac{\\pi }{4}} \\right)\\\\&= \\tan \\frac{{3\\pi }}{4}\\\\2x &= n\\pi + \\frac{{3\\pi }}{4} \\qquad n \\in Z\\\\x &= \\frac{{n\\pi }}{2} + \\frac{{3\\pi }}{8}\\end{align} Therefore, $$x = \\frac{{n\\pi }}{2}$$ or $$\\left( \\frac{n\\pi }{2}+\\frac{3\\pi }{8} \\right)$$, where $$n\\in Z$$ ## Chapter 3 Ex.3.4 Question 9 Find the principal and general solutions of the following equation: $$\\sin x + \\sin 3x + \\sin 5x = 0$$ ### Solution \\begin{align} & \\sin x+\\sin 3x+\\sin 5x=0 \\\\ & 2\\sin \\left( \\frac{5x+x}{2} \\right)\\cos \\left( \\frac{5x-x}{2} \\right)+\\sin 3x=0\\\\& \\quad \\qquad \\left[ \\because \\sin A+\\sin B=2\\sin \\left( \\frac{A+B}{2} \\right)\\cos \\left( \\frac{A-B}{2} \\right) \\right] \\\\ & 2\\sin 3x\\cos 2x+\\sin 3x=0 \\\\ & \\sin 3x\\left( 2\\cos 2x+1 \\right)=0 \\\\ \\end{align} Now, $$\\sin 3x = 0$$ or $$2\\cos 2x +", "$2\\theta - \\dfrac{5\\pi}{6} = n\\pi$. Hence, $\\theta = \\dfrac{\\pi(6n+5)}{12}$. As topsquark pointed out, tangent has a period of $\\pi$, but it did not seem obvious to me that meant you will find solutions every $\\dfrac{\\pi}{2}$. Edit: Now that I think about it, topsquark's hint was more straightforward than I thought. You set $\\theta + \\dfrac{\\pi}{6} = \\pi-\\theta$. Instead, you needed to add $n\\pi$ to one side. For example: $\\theta + \\dfrac{\\pi}{6} = \\pi-\\theta + n\\pi$. That would give you the same answer I got using the difference of angles formula for sine. 4. ## Re: all solutions of equation involving tan Originally Posted by Ant Hi, Let $0 \\leq \\theta \\leq \\pi$ Find all solutions to: $\\tan(\\theta + \\dfrac{\\pi}{6})=\\tan(\\pi-\\theta)$ Equating the two arguements and solving we find $\\theta = \\dfrac{5\\pi}{12}$. How do I find the other solution? Thanks First of all, remember that \\displaystyle \\begin{align*} \\tan{ \\left( \\alpha + \\beta \\right) } \\equiv \\frac{\\tan{( \\alpha )} + \\tan{ ( \\beta ) }}{ 1 - \\tan{( \\alpha )} \\tan{ (\\beta )}} \\end{align*} and also \\displaystyle \\begin{align*} \\tan{ ( \\pi - \\theta )} \\equiv -\\tan{(\\theta ) } \\end{align*}, so that means \\displaystyle \\begin{align*} \\tan{ \\left( \\theta + \\frac{\\pi}{6} \\right) } &= \\tan{ \\left( \\pi - \\theta \\right) } \\\\ \\frac{\\tan{ \\left( \\theta \\right)} + \\tan{ \\left( \\frac{\\pi}{6} \\right) } }{1", "function, and drop them into the equation I just wrote down. Voila, you're done. 7. x=2+5sinT/cosT (x-2)/5sinT=1/cosT; y=1+3(1/cosT) (y-1)/3=1/cosT; (x-2)/5sinT=(y-1)/3 3(x-2)/5sinT+1=y I don't know if I screwed that up or not... I'm a little confused about using the tan^2 identity :/ would be nice to impress the teacher, lol. 8. What Ackbeet was suggesting was what I erased -- I ended up getting two possible equations when I tried to solve for y, and I thought I was missing something. I'll just show you the beginning part: Solve each equation for the corresponding trig function: \\begin{aligned} tan\\, t &= \\frac{x - 2}{5} \\\\ sec\\, t &= \\frac{y - 1}{3} \\end{aligned} Use the pythagorean identity $1 + tan^2\\, t = sec^2\\, t$ that Ackbeet mentioned and plug in : \\begin{aligned} 1 + tan^2\\, t &= sec^2\\, t \\\\ 1 + \\left( \\frac{x - 2}{5} \\right)^2 &= \\left( \\frac{y - 1}{3} \\right)^2 \\\\ \\end{aligned} EDIT: Duh, this is an equation for a hyperbola. Who says we need to solve for y? I'm such an idiot. So just put it in standard form: \\begin{aligned} 1 + \\left( \\frac{x - 2}{5} \\right)^2 &= \\left( \\frac{y - 1}{3} \\right)^2 \\\\ \\frac{(y - 1)^2}{9} - \\frac{(x - 2)^2}{25} &= 1 \\end{aligned} 9. Hello, wiseguy! $\\begin{array}{cccc}x &=& 2 + 5 \\tan t & [1]\\\\ y", "+ 17} tangent \\tan text insert \\text{some text} there exists \\exists there exists only one \\exists! therefore \\therefore theta \\theta triangle \\triangle ABC union \\cup vector arrow \\vec{AB} Examples! Example: Addition or Subtraction \\begin{align*} \\\\ 3.47 \\\\ \\underline{-0.92} \\end{align*} Example: Addition or Subtraction in Equations \\begin{array}{ccccccc} 2.2x&+&5&=1.2x&-4 \\\\ -1.2x & & & -1.2x & \\end{array} Example: Systems of Equations \\begin{displaymath} \\left\\{ \\begin{array}{lr} x + y = 50\\\\ x + 43 = 2y \\end{array} \\right. \\end{displaymath} Example: The Matrix* \\begin{bmatrix} a &b & c \\\\ d & e & f\\\\ g & h & i \\end{bmatrix} Example: The Determinant* \\begin{vmatrix} a & b & c \\\\ d & e &f \\\\ g & h & i \\end{vmatrix} Example: Augmented Matrix** \\left[\\begin{array}{ccc|c} a&b&c&d\\\\ e&f&g&h \\end{array} \\right] Example: 6 Trig Identities \\begin{align*} \\sin x &= a & \\csc x &= b \\\\ \\cos x &= c & \\sec x &= d \\\\ \\tan x &= e & \\cot x &= f \\end{align*} Example: The Align Function* \\begin{align*} 10x &= 2x + 4 \\\\ 10x - 2x &= 4 \\\\ 8x &= 4 \\\\ x &= 2 \\end{align*} *This expression does not require dollar signs. **The number of c's indicates the number of columns. \"\\\\\" indicates the start of a new row. This expression does not require dollar signs.", "$$\\cos(2\\theta) = {\\cos}^2 \\theta-{\\sin}^2 \\theta = \\cos^2 \\theta - (1-{\\cos}^2 \\theta) = 2 \\cos^2 \\theta - 1$$ Similarly, to derive the double-angle formula for tangent, replacing $$\\alpha=\\beta=\\theta$$ in the sum formula gives \\begin{align*} \\tan(\\alpha+\\beta)&= \\dfrac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\[4pt] \\tan(\\theta+\\theta)&= \\dfrac{\\tan \\theta+\\tan \\theta}{1-\\tan \\theta \\tan \\theta}\\\\[4pt] \\tan(2\\theta)&= \\dfrac{2\\tan \\theta}{1-{\\tan}^2 \\theta} \\end{align*} DOUBLE-ANGLE FORMULAS \\begin{align} \\sin(2\\theta)&= 2 \\sin \\theta \\cos \\theta\\\\[4pt] \\cos(2\\theta)&= \\begin{cases}{\\cos}^2 \\theta-{\\sin}^2 \\theta \\\\ 1-2 {\\sin}^2 \\theta \\\\ 2{\\cos}^2 \\theta-1 \\end{cases}\\\\[4pt] \\tan(2\\theta)&= \\dfrac{2 \\tan \\theta}{1-{\\tan}^2\\theta} \\end{align} Example $$\\PageIndex{1}$$: Use Double-Angle Formulas to Evaluate a Trigonometric Expression Given that $$\\tan \\theta=−\\dfrac{3}{4}$$ and $$\\theta$$ is in quadrant II, find the following: a. $$\\sin(2\\theta)$$ $$\\qquad$$ b. $$\\cos(2\\theta)$$ $$\\qquad$$ c. $$\\tan(2\\theta)$$ Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $$\\tan \\theta=−\\dfrac{3}{4}$$, such that $$\\theta$$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $$\\theta$$ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: $$c^2 = {(-4)}^2+{(3)}^2 = 16+9 = 25 \\rightarrow c= 5$$ Therefore, $${ \\begin{matrix} \\boxed { \\begin{array}{llr} x &=& -4 \\\\ y &=& 3\\\\ r &=& 5 \\end{array}} \\end{matrix} }$$." ]

[ "30^\\circ \\sin 60^\\circ,$$ as desired.", "$A = {\\sin}^{-} 1 \\left(\\frac{9 \\sin {150}^{\\circ}}{10}\\right) = {26}^{\\circ} 45 '$ and therefore $C = {180}^{\\circ} - {150}^{\\circ} - {26}^{\\circ} 45 ' = {3}^{\\circ} 15 '$", "So, I'll use the expression in the next calculation instead of an approximation. $\\sin 12^{\\circ}=\\frac{10\\sin 18^{\\circ}}{x}$ $x\\sin12^{\\circ}=10\\sin 18^{\\circ}$ $x=\\frac{10\\sin 18^{\\circ}}{\\sin 12^{\\circ}}\\approx 14.86$ ft. Great notes like always.", "+0 0 220 1 Simplify $\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$ Mar 31, 2020 I think $$\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$$equals to $$\\frac{\\sqrt{3}}{4}$$. I have heard that you can use caculator to symplifiy this! But I still think im right.", "for a surprise if he wants it to \"come out to $$18.9^\\circ$$\" 9. Mar 16, 2005 ### Novus Dakota but how do you do it 1.10/1.52 =.723 sin 20=.723 then im stuck.... 10. Mar 16, 2005 ### Data Look for the $$\\sin^{-1}$$ button on your calculator. Apply it as in the equation I put above. 11. Mar 17, 2005 ### K.J.Healey first calculate sin(20) then multiply that by 1.52 then divide by 1.10 then take the arcsin/inverse sin of that number make sure its in degrees and yorue done. 28.2", "$2\\cos^2(15^\\circ)-1$ 2. $2\\sin(25^\\circ)\\cos(25^\\circ)$ 3. $1-2\\sin^2(35^\\circ)$ 4. $\\cos^2(60^\\circ)-\\sin^2(60^\\circ)$ 5. $2\\sin(125^\\circ)\\cos(125^\\circ)$ 6. $1-2\\sin^2(32^\\circ)$", "solve this system, the solution is $$d=\\frac{1}{1+2\\frac{\\sin(36^\\circ)}{1-\\sin(36^\\circ)}}\\approx0.2596$$", "\\pm {42.0038}^{\\circ} + {360}^{\\circ} k \\quad$integer $k$", "$y$ from $(1)$ in $(2)$ and simplifying (cancel off $x^2$), we get $$4\\cos^{2}72^{\\circ} + 2\\cos72^{\\circ} - 1 = 0$$ $$\\cos72^{\\circ} = \\sin18^{\\circ} = \\frac{\\sqrt{5} - 1}{4}$$", "Use an appropriate local linear approximation to estimate the value of $\\sin(46^{\\circ})$. $\\sin(46^{\\circ}) \\approx$", "could use $60 + 15 = 75$ and $\\sin \\left(A + B\\right)$ to find $\\sin {75}^{\\circ}$. Although if the question had been to find $\\sin {75}^{\\circ} , I ' d p r o b a b l y u s e$30^@$\\mathmr{and}$45^@#", "= 4\\sqrt 2 \\sin \\left( {\\pi t + {{45}^ \\circ }} \\right)$", "6^\\circ$ -", "equal to the right side, which is $\\sin \\left( 30{}^\\circ +60{}^\\circ \\right)$ or $\\sin 90{}^\\circ$ and equal to $\\sin 30{}^\\circ \\cos 60{}^\\circ +\\cos 30{}^\\circ \\sin 60{}^\\circ$. Hence, they are equal.", "-24, 0)|= 7 \\cdot 7 \\cdot \\sin(\\vec u, \\vec v)$ $12 \\cdot \\sqrt{13}= 49 \\cdot \\sin(\\vec u, \\vec v)$ ... and now solve for $\\sin(\\vec u, \\vec v)$", "dividing. Now I see that you first divide by 2 then divide by sin 5 Attached Files: • Screenshot (13).png File size: 67.7 KB Views: 99 11. Apr 27, 2015 Orodruin Staff Emeritus So from the last line of that solution it is quite apparent that there is an arithmetic error. If it would be correct, then $\\sin(5^\\circ) = 4.41/5.06$, which obviously is not the case (it is 0.087, not 0.87). The answer is off by a factor 10. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Have something to add? Draft saved Draft deleted", "Volume 23 , Issue 3 1987 Going from one place to another can be a difficult problem. Have you ever wondered how your calculator or computer finds quantities such as $\\sin(63^\\circ)$ or $\\tan(17^\\circ)$? In a recent edition of Mathematical Spectrum (the English equivalent of Parabola) the following famous problem was discussed: Q.720 When the initial digit of a whole number $x$ is deleted, the number decreases by a factor of 13. Find all possible values of $x$. Q.708 A four digit number $abcd$ has the property that $a+b = c\\times d$ and also $a\\times b = c+d$. Find all possibilities.", "i \\sin 330^{\\circ} ) = \\dfrac{3\\sqrt{3}}{14}- \\dfrac{3}{14} i$$ .", "to this problem is $$\\{114.3^\\circ,335.7^\\circ\\}$$", "\\sin (40^\\circ )$ $x = 4.40025...$ Therefore $$YZ = 4.4cm\\,(to\\,1\\,d.p.)$$" ]

[ "4. . \\scriptsize \\begin{align*}\\cos {{50}^\\circ} & =\\cos ({{25}^\\circ}+{{25}^\\circ})\\\\=2{{\\cos }^{2}}{{25}^\\circ}-1\\\\\\therefore 2{{\\cos }^{2}}{{25}^\\circ} & =\\cos {{50}^\\circ}+1\\\\\\therefore {{\\cos }^{2}}{{25}^\\circ} & =\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}\\\\\\therefore \\cos {{25}^\\circ} & =\\sqrt{{\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}}}\\quad \\quad \\cos {{25}^\\circ} \\gt 0\\therefore {{\\cos }^{2}}{{25}^\\circ} \\gt 0\\\\&=\\sqrt{{\\displaystyle \\frac{{\\cos ({{{90}}^\\circ}-{{{40}}^\\circ})+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{\\sin {{{40}}^\\circ}+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{b+1}}{2}}}\\end{align*} 5. . \\scriptsize \\begin{align*}\\cos {{80}^\\circ}&=\\cos ({{40}^\\circ}+{{40}^\\circ})\\\\&=1-2{{\\sin }^{2}}{{40}^\\circ}\\\\&=1-2{{b}^{2}}\\end{align*} 6. . \\scriptsize \\begin{align*}\\cos (-{{800}^\\circ})&=\\cos (-{{80}^\\circ}-2\\cdot {{360}^\\circ})\\\\&=\\cos (-{{80}^\\circ})\\\\&=\\cos {{80}^\\circ}\\\\&=1-2{{b}^{2}}\\quad \\quad \\text{From e}\\text{.}\\end{align*} 2. . \\scriptsize \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{1}{{\\tan 2A}}\\\\&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{{\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-(1-2{{{\\sin }}^{2}}A)}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{2{{{\\sin }}^{2}}A}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{\\sin A}}{{\\cos A}}\\\\&=\\tan A=\\text{RHS}\\end{align*} 3. $\\scriptsize 13\\cos \\theta =-5$ and $\\scriptsize {{180}^\\circ}\\le \\theta \\le {{360}^\\circ}$ 1. . \\scriptsize \\begin{align*}\\cos \\theta & =-\\displaystyle \\frac{5}{{13}}\\\\\\cos 2\\theta & =2{{\\cos }^{2}}\\theta -1\\\\&=2{{\\left( {-\\displaystyle \\frac{5}{{13}}} \\right)}^{2}}-1\\\\&=2\\left( {\\displaystyle \\frac{{25}}{{169}}} \\right)-1\\\\&=\\displaystyle \\frac{{50}}{{169}}-1\\\\&=\\displaystyle \\frac{{50-169}}{{169}}\\\\&=-\\displaystyle \\frac{{119}}{{169}}\\end{align*} 2. . \\scriptsize \\begin{align*}\\sin 2\\theta &=2\\sin \\theta \\cos \\theta \\\\&=2\\times \\left( {-\\displaystyle \\frac{{12}}{{13}}\\times -\\displaystyle \\frac{5}{{13}}} \\right)\\\\&=2\\times \\left( {\\displaystyle \\frac{{60}}{{169}}} \\right)\\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} 3. . \\scriptsize \\begin{align*}\\tan 2\\theta &=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\cos 2\\theta }}\\\\&=\\displaystyle \\frac{{\\displaystyle \\frac{{120}}{{169}}}}{{-\\displaystyle \\frac{{119}}{{169}}}}\\\\&=-\\displaystyle \\frac{{120}}{{169}}\\times \\displaystyle \\frac{{169}}{{119}}\\\\&=-\\displaystyle \\frac{{120}}{{119}}\\end{align*} 4. . \\scriptsize \\begin{align*}\\sin ({{180}^\\circ}-2\\theta )&=\\sin 2\\theta \\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} Back to Exercise 1.2 ### Unit 1: Assessment 1. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\sin {{75}^\\circ}+\\sin {{15}^\\circ}\\\\&=\\sin ({{45}^\\circ}+{{30}^\\circ})+\\sin ({{45}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{45}^\\circ}\\cos {{30}^\\circ}+\\cos {{45}^\\circ}\\sin {{30}^\\circ}+\\sin {{45}^\\circ}\\cos {{30}^\\circ}-\\cos {{45}^\\circ}\\sin {{30}^\\circ}\\\\&=2\\sin {{45}^\\circ}\\cos {{30}^\\circ}\\\\&=2\\times \\displaystyle \\frac{1}{{\\sqrt{2}}}\\times \\displaystyle \\frac{{\\sqrt{3}}}{2}\\\\&=\\displaystyle \\frac{{\\sqrt{3}}}{{\\sqrt{2}}}\\\\&=\\displaystyle \\frac{{\\sqrt{6}}}{2}=\\text{RHS}\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\sin \\theta }}-\\displaystyle \\frac{{\\cos 2\\theta }}{{\\cos \\theta }}\\\\&=\\displaystyle \\frac{{2\\sin \\theta", "\\cos 48{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 12{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\sin 6{}^\\circ \\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ If we observe the above equation carefully then we can say that the equation can be arranged such that it will take the form of sin2x. Therefore we will multiply and divide the equation by $2\\cos 6{}^\\circ$ therefore we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\sin 6{}^\\circ \\times \\dfrac{2\\cos 6{}^\\circ }{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{2\\times \\sin 6{}^\\circ \\times \\cos 6{}^\\circ }{2\\times \\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ To proceed further in the solution we should know the formula given below, Formula: $\\sin 2x=2\\times \\sin x\\times \\cos x$ ……………………………… (i) By using above formula we can write the above equation as, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin \\left( 2\\times 6{}^\\circ \\right)}{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin 12{}^\\circ }{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin 12{}^\\circ \\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times", "of length $\\,3,\\, 4,\\, 6\\,$. • Note that $\\,6 < 3 + 4\\,$. By SSS, the triangle is unique. • Use the Law of Cosines with the longest side: $$\\begin{gather} \\cssId{sb36}{6^2 = 3^2 + 4^2 - 2\\cdot 3\\cdot 4\\cdot\\cos F}\\cr\\cr \\cssId{sb37}{\\cos F \\approx -0.4583} \\cr\\cr \\cssId{sb38}{F = \\arccos(\\cos F) \\approx 117.3^\\circ} \\end{gather}$$ • Use the Law of Sines with the shortest side: $$\\begin{gather} \\cssId{sb40}{\\frac{\\sin 117.3^\\circ}{6} = \\frac{\\sin D}{3}}\\cr\\cr \\cssId{sb41}{\\sin D = \\frac{3\\sin 117.3^\\circ}{6} \\approx 0.4443}\\cr\\cr \\cssId{sb42}{D = \\arcsin(0.4443) \\approx 26.38^\\circ} \\end{gather}$$ • $\\,E = 180^\\circ - 117.3^\\circ - 26.38^\\circ = 36.32^\\circ$ • Summary: ANGLES OPPOSITE SIDES $D = 26.38^\\circ$ $d = 3$ $E = 36.32^\\circ$ $e = 4$ $F = 117.3^\\circ$ $f = 6$ This is the correct triangle! Note: At left, the corrections would be: $$\\begin{gather} \\cssId{sb56}{F = 180^\\circ - \\arcsin(0.8886) \\approx 117.3^\\circ}\\cr\\cr \\cssId{sb57}{E = 180^\\circ - 26.38^\\circ - 117.3^\\circ = 36.32^\\circ} \\end{gather}$$ ## Congruence Theorems CONGRUENCE THEOREMS A unique triangle is determined by each of these configurations: SAS side-angle-side (two sides and an included angle) SSS side-side-side (three sides) AAS/SAA angle-angle-side (or, side-angle-angle) (two angles and a non-included side) ASA angle-side-angle (two angles and an included side) If $$\\begin{gather} \\cssId{sb65}{e > 0}\\cr \\cssId{sb66}{f > 0}\\cr \\cssId{sb67}{0^\\circ < D < 180^\\circ} \\end{gather}$$ then there exists a unique triangle having the two sides and included angle. Name the", "of $EF=FA,\\,$ that is, $\\displaystyle \\cot(18^{\\circ})-\\cot(24^{\\circ}) = 2\\frac{\\sin(18^{\\circ})}{\\sin(48^{\\circ})}\\,$. We use the fact that $s=\\sin(18^{\\circ}),\\,$ being equal to $\\displaystyle \\frac{\\varphi - 1}{2},\\,$ has the property $\\displaystyle 2 s^2 = \\frac{1}{2} - s.\\,$ Thus \\displaystyle\\begin{align} \\cot(18^{\\circ})-\\cot(24^{\\circ}) &= \\frac{\\sin(6^{\\circ})}{s\\sin(24^{\\circ})} = \\frac{2\\sin(6^{\\circ})\\cos(24^{\\circ})}{s\\sin(48^{\\circ})}\\\\ &=\\frac{\\sin(30^{\\circ})-s}{s\\sin(48^{\\circ})} = \\frac{2 s^2}{s\\sin(48^{\\circ})} = \\frac{2 s}{\\sin(48^{\\circ})}. \\end{align} and we are done. ### Acknowledgment The problem has been posted at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group by Konstantin Knop. Konstantin has requested not to leave out a similar problem: I do leave it as an exercise to the reader. It's the same circle, though. Grégoire Nicollier has come up with a direct proof (P.S.) of $BI\\perp AC.$ Konstantin has also added an extra point to the circle on observed that, be symmetry, points C, B, E, O, N are also concyclic. Grégoire Nicollier has later added another proof (Solution 2), a variant of which has been placed on a separate page.", "and \\begin{align*}b = 30^\\circ\\end{align*}. \\begin{align*}\\cos (45^\\circ - 30^\\circ) & = \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{6} + \\sqrt{2}}{4}\\end{align*} #### Example C Find the exact value of \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*}, in radians. Solution: \\begin{align*}\\cos \\frac{5 \\pi}{12} = \\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right )\\end{align*}, notice that \\begin{align*}\\frac{\\pi}{4} = \\frac{3 \\pi}{12}\\end{align*} and \\begin{align*}\\frac{\\pi}{6} = \\frac{2 \\pi}{12}\\end{align*} \\begin{align*}\\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right ) & = \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} \\\\ \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ & = \\frac{\\sqrt{6} - \\sqrt{2}}{4}\\end{align*} ### Vocabulary Cosine Sum Formula: The cosine sum formula relates the cosine of a sum of two arguments to a set of sine and cosines functions, each containing one argument. Cosine Difference Formula: The cosine difference formula relates the cosine of a difference of two arguments to a set of sine and cosines functions, each containing one argument. ### Guided Practice 1. Find the exact value for \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*} 2. Find the exact value for \\begin{align*}\\cos \\frac{7 \\pi}{12}\\end{align*} 3. Find the exact value for \\begin{align*}\\cos 345^\\circ\\end{align*} Solutions: 1. \\begin{align*}\\cos \\frac{5 \\pi}{12} & = \\cos \\left (\\frac{2 \\pi}{12} + \\frac{3", "5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d = \\sqrt{34 - 30\\cos 57^\\circ}$ 4. $d \\approx 4.202$ 5. $\\displaystyle \\frac{\\sin 57^\\circ}{d} = \\frac{\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} = \\frac{\\sin E}{3}$ 6. $\\displaystyle\\sin E = \\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} \\approx 0.5987$ 7. $\\displaystyle E = \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\approx 36.78^\\circ$ 8. $\\displaystyle\\,F = 180^\\circ - 57^\\circ - E \\ =\\ 180^\\circ - 57^\\circ - \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\ \\approx\\ 86.22^\\circ$ It's just scary looking. (But, every ‘$\\,=\\,$’ is strictly correct.) Potential Remedy #3 with Comments Another potential remedy is to use ‘$\\,\\approx\\,$’ for everything that is truly an approximation. In this case, pretty much everything after the first occurrence of an approximation gets ‘$\\,\\approx\\,$’: 1. $d^2 = 3^2 + 5^2 - 2\\cdot 3\\cdot 5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d \\approx 4.202$ 4. $\\displaystyle\\frac{\\sin 57^\\circ}{4.202} \\approx \\frac{\\sin E}{3}$ 5. $\\displaystyle\\sin E \\approx \\frac{3\\sin 57^\\circ}{4.202} \\approx 0.5988$ 6. $E \\approx \\arcsin(0.5988) \\approx 36.78^\\circ$ 7. $\\,F \\approx 180^\\circ - 57^\\circ - 36.78^\\circ = 86.22^\\circ$ I don't like this at all. It makes it ‘sound like’ (say) the Law of Sines only sort of works. Potential Remedy #4 with Comments One final potential remedy is to use an ellipsis ‘$\\,\\ldots\\,$’ to indicate missing digits of a number. In this case,", "30^{\\circ}$$ $$a = \\displaystyle \\frac {b\\sin(130^{\\circ})}{\\sin(20^{\\circ})} = 22.398$$ $$c = \\displaystyle \\frac {b\\sin(30^{\\circ})}{\\sin(20^{\\circ})} = 14.619$$ 2. $$A = 180 ^{\\circ} - 120 ^{\\circ} - 30 ^{\\circ} = 30 ^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(120^{\\circ})}{\\sin30^{\\circ})} = 27.713$$ $$c = \\displaystyle \\frac {a\\sin(30^{\\circ})}{sin(30^{\\circ})} = 16$$ 3. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(30^{\\circ})} = 17.851$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 107.76^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 42.24^{\\circ}$$ 4. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(108^{\\circ})} = 20.438^{\\circ}$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 27.73^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 44.27^{\\circ}$$ 5. $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 38.44^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 61.73^{\\circ}$$ $$C = \\arccos\\left(\\displaystyle \\frac {a^2 + b^2 - c^2}{2ab}\\right) = 79.84^{\\circ}$$ 6. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 66.67^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 66.67^{\\circ} = 73.33^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(73.33^{\\circ})}{\\sin(40^{\\circ})} = 10.432$$ 7. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 32.39^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 32.39^{\\circ} = 107.61^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(107.61^{\\circ})}{\\sin(40^{\\circ})} = 17.794$$ ### Exercise $$\\PageIndex{15}$$ 1. A ladder $$10$$ meters long is leaning against the side of a building. It makes an angle of $$75^{\\circ}$$ with the ground. How far up", "}$ $\\displaystyle \\text{iv) } \\text{LHS } =\\sin 23^{\\circ} + \\sin 37^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{23^{\\circ}+37^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{23^{\\circ}-37^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 30^{\\circ} \\cos (-7^{\\circ})$ $\\displaystyle = 2 \\times \\frac{1}{2} \\cos 7^{\\circ}$ $\\displaystyle = \\cos 7^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{v) } \\text{LHS } =\\sin 105^{\\circ} + \\cos 105^{\\circ}$ $\\displaystyle =\\sin 105^{\\circ} + \\cos (90^{\\circ}+ 15^{\\circ})$ $\\displaystyle =\\sin 105^{\\circ} - \\sin 15^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{105^{\\circ}-15^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{105^{\\circ}+15^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 45^{\\circ} \\cos 60^{\\circ}$ $\\displaystyle = 2 \\times \\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}$ $\\displaystyle = \\frac{1}{\\sqrt{2}} = \\cos 45^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{vi) } \\text{LHS } =\\sin 40^{\\circ} + \\sin 20^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{40^{\\circ}+20^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{20^{\\circ}-40^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 30^{\\circ} \\cos (-10^{\\circ})$ $\\displaystyle = 2 \\times \\frac{1}{2} \\cos 10^{\\circ}$ $\\displaystyle = \\cos 10^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\\\$ Question 3: Prove that: $\\displaystyle \\text{i) } \\cos 55^{\\circ} + \\cos 65^{\\circ} + \\cos 175^{\\circ} = 0 \\hspace{1.0cm} \\text{ii) } \\sin 50^{\\circ} - \\sin 70^{\\circ} + \\sin 10^{\\circ} = 0$ $\\displaystyle \\text{iii) } \\cos 80^{\\circ} + \\cos 40^{\\circ} - \\cos 20^{\\circ} = 0 \\hspace{1.0cm} \\text{v) } \\cos 20^{\\circ} + \\cos 100^{\\circ} + \\cos 140^{\\circ} = 0$ $\\displaystyle \\text{v) } \\sin \\frac{5\\pi}{18} - \\cos \\frac{4\\pi}{9} = \\sqrt{3} \\sin \\frac{\\pi}{9}", "the problem gives us: We don’t seem to know much about the side $PR$, whereas parts of $PQ$ and $QR$ are known. Hence, in calculating the area of $\\Delta PQR$, it seems like using the formula $\\text{Area of } \\Delta PQR = \\displaystyle\\frac{1}{2} |PQ||QR|\\sin PQR = \\frac{1}{2} |PQ||QR| \\sin 60^\\circ$ could work. Let’s label the diagram further to give more info on $PQ$ and $QR$: We have 2 unknown quantities we would like to find: $|PQ|$ and $|QR|$. We can use the sine rule to reduce that to just finding $|PQ|$: \\begin{aligned} \\displaystyle\\frac{|PQ|}{\\sin 45^\\circ} &= \\frac{|QR|}{\\sin 75^\\circ}, \\\\ |QR| &= \\frac{|PQ|}{\\sin 45^\\circ} \\sin 75^\\circ = \\sqrt{2}|PQ| \\sin 75^\\circ. \\end{aligned} From the diagram, we know that $|PQ| = 2 + |AP|$. We can use the sine rule again to find the value of $|AP|$: \\begin{aligned} \\displaystyle\\frac{|AP|}{\\sin 45^\\circ} &= \\frac{|AF|}{\\sin 75^\\circ}, \\\\ |AP| &= \\frac{1}{\\sin 75^\\circ} \\sin 45^\\circ = \\frac{1}{\\sqrt{2}\\sin 75^\\circ}. \\end{aligned} Hence, we have \\begin{aligned} \\text{Area} &= \\displaystyle\\frac{1}{2}|PQ| \\cdot \\sqrt{2} |PQ| \\sin 75^\\circ \\cdot \\sin 60^\\circ \\\\ &= \\frac{\\sqrt{2}}{2}\\frac{\\sqrt{3}}{2} \\sin 75^\\circ |PQ|^2 \\\\ &= \\frac{\\sqrt{6}}{4} \\sin 75^\\circ \\left( 2 + \\frac{1}{\\sqrt{2}\\sin 75^\\circ} \\right)^2 \\\\ &= \\frac{\\sqrt{6}}{4}\\sin 75^\\circ \\left( 4 + \\frac{4}{\\sqrt{2}\\sin 75^\\circ} + \\frac{1}{2\\sin^2 75^\\circ} \\right) \\\\ &= \\sqrt{6}\\sin 75^\\circ + \\sqrt{3} + \\frac{\\sqrt{6}}{8 \\sin 75^\\circ}. \\end{aligned} It remains to get rid of the $\\sin 75^\\circ$ term. $\\sin 75^\\circ", "= \\frac{\\cos 1^\\circ}{\\sin 1^\\circ}. (6)$$ Now we notice that we can multiply both sides by $\\sin 1^\\circ$. Doing so gives us $$\\sin 1^\\circ(2 (\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ) + 1) = \\cos 1^\\circ. (7)$$ Now, simplifying $\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas. Doing so, gives us $$(\\sin 1^\\circ)(2 \\sin 2^\\circ)+(\\sin 1^\\circ)(2\\sin 4^\\circ) +(\\sin 1^\\circ)(2\\sin 6^\\circ) + (\\sin 1^\\circ)(2\\sin 8^\\circ) +(\\sin 1^\\circ)(2 \\sin 10^\\circ) ... + (\\sin 1^\\circ)(2\\sin 88^\\circ) + \\sin 1^\\circ = \\cos 1^\\circ. (8)$$ Now we can use our product of sines to sum formula and see if we can find a pattern. We will start with expanding $(\\sin 1^\\circ)(2 \\sin 2^\\circ)$. Because the format of our formula for the product of sines is $\\sin \\alpha + \\sin \\beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\\sin 1^\\circ)(\\sin 2^\\circ)$. Our formula for the product of sines is $\\sin \\alpha + \\sin \\beta = \\frac{1}{2}[\\cos(\\alpha - \\beta) - \\cos(\\alpha + \\beta)]$. Plugging in our values into the formula and simplifying gives us $2\\left(\\frac{1}{2}(\\cos 1^\\circ", "= \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{(ii) } \\text{LHS } = \\sin \\frac{8 \\pi}{3} \\cos \\frac{23 \\pi}{6} + \\cos \\frac{13 \\pi}{3} \\sin \\frac{35 \\pi}{6}$ $\\displaystyle = \\sin \\Big( 3\\pi - \\frac{\\pi}{3} \\Big) \\cos \\Big( 4\\pi - \\frac{\\pi}{6} \\Big) + \\cos \\Big( 4\\pi + \\frac{\\pi}{3} \\Big) \\sin \\Big( 6\\pi - \\frac{\\pi}{6} \\Big)$ $\\displaystyle = \\sin \\frac{\\pi}{3} \\cos \\frac{\\pi}{6} + \\cos \\frac{\\pi}{3} \\Big( - \\sin \\frac{\\pi}{6} \\Big)$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\Big( - \\frac{1}{2} \\Big)$ $\\displaystyle = \\frac{3}{4} - \\frac{1}{4} = \\frac{2}{4} = \\frac{1}{2}$ $\\displaystyle \\text{(iii) } \\text{LHS } = \\cos 24^{\\circ} + \\cos 55^{\\circ} + \\cos 125^{\\circ} + \\cos 204^{\\circ} + \\cos 300^{\\circ}$ $\\displaystyle = ( \\cos 24^{\\circ} + \\cos 204^{\\circ}) + ( \\cos 55^{\\circ} + \\cos 125^{\\circ}) + \\cos 300^{\\circ}$ $\\displaystyle = ( \\cos 24^{\\circ} + \\cos (\\pi + 24^{\\circ}) ) + ( \\cos 55^{\\circ} + \\cos (\\pi - 55^{\\circ})) + \\cos (2\\pi - 60^{\\circ})$ $\\displaystyle = \\cos 24^{\\circ} - \\cos 24^{\\circ} + \\cos 55^{\\circ} - \\cos 55^{\\circ} + \\cos 60^{\\circ}$ $\\displaystyle = \\cos 60^{\\circ} = \\frac{1}{2} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{(iv) } \\text{LHS } = \\tan (-225^{\\circ}) \\cot(-405^{\\circ}) - \\tan (-765^{\\circ}) \\cot (675^{\\circ})$ $\\displaystyle = -\\tan 225^{\\circ} (-\\cot 405^{\\circ} ) + \\tan 765^{\\circ} \\cot 675^{\\circ}$ $\\displaystyle = \\tan \\Big( \\pi + \\frac{\\pi}{4} \\Big) \\cot \\Big( 2\\pi + \\frac{\\pi}{4} \\Big) +", "1. ## prove $\\sin 47° + \\sin 61° - \\sin 11° - \\sin 25° = \\cos 7°$. 2. ## Sum to Product First, note that 47-11=61-25=36, so rearrange the left hand side to get $(\\sin 47^\\circ - \\sin 11^\\circ) + (\\sin 61^\\circ - \\sin 25^\\circ)$. From this, we use the sum-to-product formulas, specifically the formula $\\sin A - \\sin B = 2\\cos \\left ( \\frac{A+B}{2} \\right ) \\sin \\left ( \\frac{A-B}{2} \\right )$. Thus we get $\\sin 47^\\circ + \\sin 61^\\circ - \\sin 11^\\circ - \\sin 25^\\circ = (\\sin 47^\\circ - \\sin 11^\\circ) + (\\sin 61^\\circ - \\sin 25^\\circ)$ $\\,\\,= 2 \\cos 29^\\circ \\sin 18^\\circ + 2 \\cos 43^\\circ \\sin 18^\\circ$ $\\,\\,= 2(\\cos 29^\\circ + \\cos 43^\\circ)\\sin 18^\\circ$ We use the sum-to-product formulas again, this time using $\\cos A + \\cos B = 2\\cos \\left ( \\frac{A+B}{2} \\right ) \\cos \\left ( \\frac{A-B}{2} \\right )$, so $\\sin 47^\\circ + \\sin 61^\\circ - \\sin 11^\\circ - \\sin 25^\\circ = 2(\\cos 29^\\circ + \\cos 43^\\circ)\\sin 18^\\circ$ $\\,\\,= 2(2\\cos 36^\\circ \\cos (-7^\\circ))\\sin 18^\\circ$ $\\,\\,= 4\\cos 7^\\circ \\cos 36^\\circ \\sin 18^\\circ$, as $\\cos(-x)=\\cos x$. Working from the geometry of the pentagon, one can find the sine and cosine for angles that are multiples of 18 degrees, and thus find that: $\\cos 36^\\circ = \\frac{1 + \\sqrt{5}}{4}$, $\\sin 18^\\circ = \\frac{-1 + \\sqrt{5}}{4}$,", "50^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ (\\cos 10^{\\circ} \\cos 50^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ (2 \\cos 50^{\\circ} \\cos 10^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\Big( \\cos (50^{\\circ}+10^{\\circ}) + \\cos (50^{\\circ}-10^{\\circ}) \\Big) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\cos 60^{\\circ} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\frac{1}{2} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{4} \\cos 40^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} (2\\cos 40^{\\circ} \\cos 70^{\\circ})$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} ( \\cos (70^{\\circ}+40^{\\circ}) + \\cos (70^{\\circ}-40^{\\circ}) )$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} \\cos 110 + \\frac{\\sqrt{3}}{8} \\cos 30^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos 110 + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos (180 - 70) + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} - \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{3}{16} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{ii) } \\text{LHS } = \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} ( 2 \\cos 160^{\\circ} \\cos 40^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} [ \\cos ( 160^{\\circ}+40^{\\circ})", "& = 180^\\circ\\\\ m \\angle A & = 57^\\circ\\end{align*} Now, use the Law of Sines to set up ratios for \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*}. \\begin{align*}\\frac{\\sin 57^\\circ}{a} = \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\end{align*} \\begin{align*}\\frac{\\sin 57^\\circ}{a} & = \\frac{\\sin 38^\\circ}{12} && \\quad \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\\\ a \\cdot \\sin 38^\\circ & = 12 \\cdot \\sin 57^\\circ && b \\cdot \\sin 38^\\circ = 12 \\cdot \\sin 85^\\circ\\\\ a & = \\frac{12 \\cdot \\sin 57^\\circ}{\\sin 38^\\circ} \\approx 16.4 && \\qquad \\quad \\ \\ b = \\frac{12 \\cdot \\sin 85^\\circ}{\\sin 38^\\circ} \\approx 19.4\\end{align*} Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth. Solution: Set up the ratio for \\begin{align*}\\angle B\\end{align*} using Law of Sines. \\begin{align*}\\frac{\\sin 95^\\circ}{27} & = \\frac{\\sin B}{16}\\\\ 27 \\cdot \\sin B & = 16 \\cdot \\sin 95^\\circ\\\\ \\sin B & = \\frac{16 \\cdot \\sin 95^\\circ}{27} \\rightarrow \\sin^{-1} \\left ( \\frac{16 \\cdot \\sin 95^\\circ}{27} \\right ) = 36.2^\\circ\\end{align*} To find \\begin{align*}m \\angle C\\end{align*} use the Triangle Sum Theorem. \\begin{align*}m \\angle C + 95 ^\\circ + 36.2^\\circ = 180^\\circ \\rightarrow m \\angle C = 48.8^\\circ\\end{align*} To find \\begin{align*}c\\end{align*}, use the Law of Sines again. \\begin{align*}\\frac{\\sin 95^\\circ}{27} = \\frac{\\sin 48.8^\\circ}{c}\\end{align*} \\begin{align*}c \\cdot \\sin 95^\\circ & = 27 \\cdot \\sin 48.8^\\circ\\\\ c & = \\frac{27 \\cdot \\sin 48.8^\\circ}{\\sin 95^\\circ} \\approx 20.4\\end{align*} ## Law of Cosines Law", "\\cos x }{2\\cos ^2 x-1}=\\dfrac{\\sin (2x)}{ \\cos (2x)}=\\tan (2x)$$ 58) $$(\\sin^2x-1)^2=\\cos(2x)+\\sin^4x$$ 59) $$\\sin(3x)=3\\sin x \\cos^2x-\\sin^3x$$ \\begin{align*} \\sin (x+2x) &= \\sin x \\cos (2x)+\\sin (2x) \\cos x\\\\ &= \\sin x(\\cos ^2 x - \\sin ^2 x)+2\\sin x \\cos x \\cos x\\\\ &= \\sin x \\cos ^2 x-\\sin ^3 x + 2\\sin x\\cos ^2 x\\\\ &= 3\\sin x\\cos ^2 x - \\sin ^3 x \\end{align*} 60) $$\\cos(3x)=\\cos^3x-3\\sin^2x\\cos x$$ 61) $$\\dfrac{1+\\cos (2t)}{\\sin (2t)-\\cos t}=\\dfrac{2\\cos t}{2\\sin t-1}$$ \\begin{align*} \\dfrac{1+\\cos (2t)}{\\sin (2t)-\\cos t} &= \\dfrac{1+2\\cos ^2t-1}{2\\sin t\\cos t-\\cos t}\\\\ &= \\dfrac{2\\cos ^2t}{\\cos t(2\\sin t-1)}\\\\ &= \\dfrac{2\\cos t}{2\\sin t-1} \\end{align*} 62) $$\\sin(16x)=16 \\sin x \\cos x \\cos(2x)\\cos(4x)\\cos(8x)$$ 63) $$\\cos(16x)=(\\cos^2(4x)-\\sin^2(4x)-\\sin(8x))(\\cos^2(4x)-\\sin^2(4x)+\\sin(8x))$$ \\begin{align*} (\\cos^2(4x)-\\sin^2(4x)-\\sin(8x))(\\cos^2(4x)-\\sin^2(4x)+\\sin(8x)) &= (\\cos(8x)-\\sin(8x))(\\cos(8x)+\\sin(8x))\\\\ &= \\cos ^2 (8x)-\\sin ^2 (8x)\\\\ &= \\cos(16x) \\end{align*} ## 7.4: Sum-to-Product and Product-to-Sum Formulas From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. The product-to-sum formulas can rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. We can also derive the sum-to-product identities from the product-to-sum identities using substitution. The sum-to-product formulas are used to rewrite sum or difference as products of sines and cosines. #### Verbal 1) Starting with the product to sum formula $$\\sin \\alpha \\cos \\beta =\\dfrac{1}{2} \\left[\\sin(\\alpha +\\beta )+\\sin(\\alpha -\\beta ) \\right]$$$,$ explain", "deliver each package, at what time must she leave? 1. ASA 2. AAS 3. neither 4. ASA 5. AAS 6. AAS 1. Student answers will vary but they should notice that in both cases you know or can find an angle and the side across from it. 1. \\begin{align*}\\frac{\\sin 11.7^\\circ}{a} = \\frac{\\sin 144.5^\\circ}{16}, a = 5.6\\end{align*} 2. \\begin{align*}\\frac{\\sin 41.3^\\circ}{214.9} = \\frac{\\sin 39.7^\\circ}{d}, d = 208.0\\end{align*} 3. not enough information 4. \\begin{align*}\\frac{\\sin 40.3^\\circ}{l} = \\frac{\\sin 123.5^\\circ}{6.3}, l = 4.9\\end{align*} 5. \\begin{align*}\\frac{\\sin 9^\\circ}{o} = \\frac{\\sin 31^\\circ}{15}, o = 4.6\\end{align*} 6. \\begin{align*}\\frac{\\sin 127^\\circ}{q} = \\frac{\\sin 21.8^\\circ}{3.62}, q = 7.8\\end{align*} 2. \\begin{align*}\\angle{G} = 180^\\circ - 62.1^\\circ - 21.3^\\circ = 96.6^\\circ\\end{align*} \\begin{align*}\\frac{\\sin 96.6^\\circ}{g} = \\frac{\\sin 21.3^\\circ}{108}, g = 295.3 && \\frac{\\sin 62.1^\\circ}{h} = \\frac{\\sin 21.3^\\circ}{108}, h = 262.8\\end{align*} 3. \\begin{align*}\\frac{\\sin A}{a} & = \\frac{\\sin B}{b} && \\text{Law of Sines} \\\\ a (\\sin B) & = b (\\sin A) && \\text{Cross multiply} \\\\ \\frac{a}{b} & = \\frac{\\sin A}{\\sin B} && \\text{Divide by}\\ b(\\sin B)\\end{align*} 1. \\begin{align*}\\tan 54^\\circ = \\frac{h}{7.15} \\rightarrow h = 9.8, \\cos 67^\\circ = \\frac{9.8}{x} \\rightarrow x = 25.2\\end{align*} 2. The angle we are finding is the one at the far left side of the triangle. \\begin{align*}8.9^2 = 11.2^2 + 12.6^2 - 2 \\cdot 11.2 \\cdot 12.6 \\cos A \\rightarrow A = 43.4^\\circ, \\frac{\\sin 43.4^\\circ}{x} = \\frac{\\sin 31}{11.2} \\rightarrow x = 14.9.\\end{align*}", "7^o \\cos 14^o \\cos 28^o \\cos 56^o}{2 \\sin 7^o }$ $=$ $\\frac{2 \\sin 14^o \\cos 14^o \\cos 28^o \\cos 56^o}{4 \\sin 7^o }$ $=$ $\\frac{2 \\sin 28^o \\cos 28^o \\cos 56^o}{8 \\sin 7^o }$ $=$ $\\frac{2 \\sin 56^o \\cos 56^o}{16 \\sin 7^o }$ $=$ $\\frac{\\sin 112^o}{16 \\sin 7^o }$ $=$ $\\frac{\\sin (180^o - 68^o)}{16 \\sin (90^o - 83^o) }$ $=$ $\\frac{\\sin 68^o}{16 \\cos 83^o }$ $=$ RHS. Hence proved. $\\\\$ Question 34: Prove that $\\cos$ $\\frac{2\\pi}{15}$ $\\cos$ $\\frac{4\\pi}{15}$ $\\cos$ $\\frac{8\\pi}{15}$ $\\cos$ $\\frac{16\\pi}{15}$ $=$ $\\frac{1}{16}$ LHS = $\\cos$ $\\frac{2\\pi}{15}$ $\\cos$ $\\frac{4\\pi}{15}$ $\\cos$ $\\frac{8\\pi}{15}$ $\\cos$ $\\frac{16\\pi}{15}$ $= \\cos 24^o \\cos 48^o \\cos 96^o \\cos 192^o$ $=$ $\\frac{2 \\sin 24^o \\cos 24^o \\cos 48^o \\cos 96^o \\cos 192^o}{2 \\sin 24^o}$ $=$ $\\frac{2 \\sin 48^o \\cos 48^o \\cos 96^o \\cos 192^o}{4 \\sin 24^o}$ $=$ $\\frac{2 \\sin 96^o \\cos 96^o \\cos 192^o}{8 \\sin 24^o}$ $=$ $\\frac{2 \\sin 192^o \\cos 192^o}{16 \\sin 24^o}$ $=$ $\\frac{\\sin 384^o}{16 \\sin 24^o}$ $=$ $\\frac{\\sin (360^o+24^o)}{16 \\sin 24^o}$ $=$ $\\frac{\\sin 24^o}{16 \\sin 24^o}$ $=$ $\\frac{1}{16}$ $=$ RHS. Hence proved. $\\\\$ Question 35: Prove that $\\cos$ $\\frac{\\pi}{5}$ $\\cos$ $\\frac{2\\pi}{5}$ $\\cos$ $\\frac{4\\pi}{5}$ $\\cos$ $\\frac{8\\pi}{5}$ $= -$ $\\frac{1}{16}$ LHS = $\\cos$ $\\frac{\\pi}{5}$ $\\cos$ $\\frac{2\\pi}{5}$ $\\cos$ $\\frac{4\\pi}{5}$ $\\cos$ $\\frac{8\\pi}{5}$ $= \\cos 12^o \\cos 24^o \\cos 48^o \\cos 96^o$ $=$ $\\frac{2 \\sin 12^o \\cos 12^o \\cos 24^o \\cos 48^o \\cos 96^o}{2 \\sin 12^o}$ $=$", "\\frac{\\cos A - \\cos 3A + \\cos 3A - \\cos 9A + \\cos 9A - \\cos 17A}{\\sin 3A - \\sin A + \\sin 9A - \\sin 3A + \\sin 17A - \\sin 9A}$ $= \\frac{\\cos A - \\cos 17A}{\\sin 17A - \\sin A}$ $= \\frac{2\\sin 8A\\sin 9A}{2\\cos 9A\\sin 8A} = \\tan 9A$ 44. We have to prove that, $\\frac{\\cos 2A\\cos 3A - \\cos 2A\\cos 7A + \\cos A\\cos 10A}{\\sin 4A\\sin 3A - \\sin 2A\\sin 5A + \\sin 4A\\sin 7A} =\\cot 6A\\cot 5A$ L.H.S. $= \\frac{\\cos 5A + \\cos A - \\cos 9A - \\cos 5A + \\cos 11A + \\cos 9A}{\\cos A - \\cos 7A - \\cos 3A + cos 7A + \\cos 3A - \\cos 11A}$ $= \\frac{\\cos A + \\cos 11A}{\\cos A - \\cos 11A}$ $= \\frac{\\cos 6A\\cos 5A}{\\sin 6A\\sin 5A} = \\cot 6A\\cot 5A$ 45. We have to prove that, $\\cos(36^\\circ - A)\\cos(36^\\circ + A) + \\cos(54^\\circ + A)\\cos(54^\\circ - A) = \\cos 2A$ L.H.S. $= \\frac{1}{2}(\\cos (72^\\circ)+ \\cos 2A) + \\frac{1}{2}(\\cos 108^\\circ + \\cos 2A)$ $= \\frac{1}{2}(\\sin 18^\\circ + - \\sin 18^\\circ + 2\\cos 2A)[\\because \\sin 18^\\circ = \\cos(90^\\circ - 18^\\circ) = \\cos 72^\\circ$ and $\\cos 108^\\circ = \\cos(90^\\circ + 18^\\circ) = -\\sin 18^\\circ]$ $= \\cos 2A$ 46. We have to prove that $\\cos A\\sin(B - C) + \\cos B\\sin(C - A) + \\cos", "is $$a\\big(\\!\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ$$ $$+r^{5}\\cos300^\\circ+r^{6}\\cos360^\\circ+\\cdots\\big).$$ Because the angles repeat after 6 terms, this sum is equal to $$aS(1+r^{6}+r^{12}+r^{18}+\\cdots)=\\frac{aS}{1-r^{6}},$$ where $$S=\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ+ r^{5}\\cos300^\\circ.$$ Similarly, the $y$-coordinate of $P$ will be $\\displaystyle{\\frac{aT}{1-r^{6}}}$, where $$T=\\sin0^\\circ+r\\sin60^\\circ + r^{2}\\sin120^\\circ+r^{3}\\sin180^\\circ+r^{4}\\sin240^\\circ+ r^{5}\\sin300^\\circ.$$In this case $r=\\frac12$ and $a = 5$, so $$S=1+\\frac14-\\frac18-\\frac18-\\frac1{32}+\\frac1{64}=\\frac{63}{64},$$ $$T=0+\\frac{\\sqrt3}4+\\frac{\\sqrt3}8+0-\\frac{\\sqrt3}{32}-\\frac{\\sqrt3}{64}=\\frac{21\\sqrt3}{64},$$and the coordinates of $P$ are $$\\left(\\frac{5S}{1-\\frac1{64}}, \\frac{5T}{1-\\frac1{64}}\\right)=\\left(5,\\frac{5\\sqrt3}{3}\\right).$$Thus the square of the distance from the origin to $P$ is $25+\\frac{25}3=\\frac{100}3$. The requested sum is $100+3=103$. ## Solution 5 (Official MAA 2) Let point $O$ be the origin in the complex plane. Point $P$ is the complex sum $5(1+z+z^2+\\cdots) = \\frac{5}{1-z}$, where $z=\\frac{1+i\\sqrt3}4$. The distance squared is$${OP}^2=\\left|\\frac5{1-\\frac{1+i\\sqrt3}4}\\right|^{2}= \\frac{(4\\cdot5)^2}{\\left|4-(1+i\\sqrt3)\\right|^2}=\\frac{400}{9+3}=\\frac{100}3.$$ -avn", "= \\frac{\\sin B}{6.1}, B & = 84.8^\\circ \\\\ 180^\\circ - 32.6^\\circ - 84.8^\\circ, C & = 62.6^\\circ\\end{align*} (d) No solution (e) Both cases need angle \\begin{align*}C\\end{align*} and side \\begin{align*}c\\end{align*}. \\begin{align*}\\text{Case}\\ 1: C & = 180^\\circ - 47^\\circ - 77.2^\\circ = 55.8^\\circ, \\frac{\\sin 55.8^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 50.9 \\\\ \\text{Case}\\ 2: C & = 180^\\circ - 47^\\circ - 102.8^\\circ = 30.2^\\circ,\\frac{\\sin 30.2^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 30.95\\end{align*} 2. To find the area of the rhombus, use the formula \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} and then multiply that by 2. We first need to find one of the angles that are opposite the given diagonal (they are both the same measurement). We will call it angle \\begin{align*}A\\end{align*}. \\begin{align*}21.5^2 = 12^2 + 12^2 - 2(12)(12) \\sin A, A = 127.2^\\circ\\end{align*}, which means the other two angles are both \\begin{align*}52.8^\\circ (360^\\circ - 127.2^\\circ - 127.2^\\circ\\end{align*} and then divide by 2). \\begin{align*}K = 2 \\left (\\frac{1}{2}(12)(12) \\sin 127.2^\\circ \\right ) = 114.7\\end{align*} 3. Divide the pentagon into three triangles, drawing segments from \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{5}\\end{align*}, called \\begin{align*}x\\end{align*} below, and \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{4}\\end{align*}, called \\begin{align*}y\\end{align*} below. With these three triangles, only the middle triangle needs us to find two sides and the angle between them (called \\begin{align*}\\angle{Z}\\end{align*} below) to use \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} (the outer" ]

[ "4. . \\scriptsize \\begin{align*}\\cos {{50}^\\circ} & =\\cos ({{25}^\\circ}+{{25}^\\circ})\\\\=2{{\\cos }^{2}}{{25}^\\circ}-1\\\\\\therefore 2{{\\cos }^{2}}{{25}^\\circ} & =\\cos {{50}^\\circ}+1\\\\\\therefore {{\\cos }^{2}}{{25}^\\circ} & =\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}\\\\\\therefore \\cos {{25}^\\circ} & =\\sqrt{{\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}}}\\quad \\quad \\cos {{25}^\\circ} \\gt 0\\therefore {{\\cos }^{2}}{{25}^\\circ} \\gt 0\\\\&=\\sqrt{{\\displaystyle \\frac{{\\cos ({{{90}}^\\circ}-{{{40}}^\\circ})+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{\\sin {{{40}}^\\circ}+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{b+1}}{2}}}\\end{align*} 5. . \\scriptsize \\begin{align*}\\cos {{80}^\\circ}&=\\cos ({{40}^\\circ}+{{40}^\\circ})\\\\&=1-2{{\\sin }^{2}}{{40}^\\circ}\\\\&=1-2{{b}^{2}}\\end{align*} 6. . \\scriptsize \\begin{align*}\\cos (-{{800}^\\circ})&=\\cos (-{{80}^\\circ}-2\\cdot {{360}^\\circ})\\\\&=\\cos (-{{80}^\\circ})\\\\&=\\cos {{80}^\\circ}\\\\&=1-2{{b}^{2}}\\quad \\quad \\text{From e}\\text{.}\\end{align*} 2. . \\scriptsize \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{1}{{\\tan 2A}}\\\\&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{{\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-(1-2{{{\\sin }}^{2}}A)}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{2{{{\\sin }}^{2}}A}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{\\sin A}}{{\\cos A}}\\\\&=\\tan A=\\text{RHS}\\end{align*} 3. $\\scriptsize 13\\cos \\theta =-5$ and $\\scriptsize {{180}^\\circ}\\le \\theta \\le {{360}^\\circ}$ 1. . \\scriptsize \\begin{align*}\\cos \\theta & =-\\displaystyle \\frac{5}{{13}}\\\\\\cos 2\\theta & =2{{\\cos }^{2}}\\theta -1\\\\&=2{{\\left( {-\\displaystyle \\frac{5}{{13}}} \\right)}^{2}}-1\\\\&=2\\left( {\\displaystyle \\frac{{25}}{{169}}} \\right)-1\\\\&=\\displaystyle \\frac{{50}}{{169}}-1\\\\&=\\displaystyle \\frac{{50-169}}{{169}}\\\\&=-\\displaystyle \\frac{{119}}{{169}}\\end{align*} 2. . \\scriptsize \\begin{align*}\\sin 2\\theta &=2\\sin \\theta \\cos \\theta \\\\&=2\\times \\left( {-\\displaystyle \\frac{{12}}{{13}}\\times -\\displaystyle \\frac{5}{{13}}} \\right)\\\\&=2\\times \\left( {\\displaystyle \\frac{{60}}{{169}}} \\right)\\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} 3. . \\scriptsize \\begin{align*}\\tan 2\\theta &=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\cos 2\\theta }}\\\\&=\\displaystyle \\frac{{\\displaystyle \\frac{{120}}{{169}}}}{{-\\displaystyle \\frac{{119}}{{169}}}}\\\\&=-\\displaystyle \\frac{{120}}{{169}}\\times \\displaystyle \\frac{{169}}{{119}}\\\\&=-\\displaystyle \\frac{{120}}{{119}}\\end{align*} 4. . \\scriptsize \\begin{align*}\\sin ({{180}^\\circ}-2\\theta )&=\\sin 2\\theta \\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} Back to Exercise 1.2 ### Unit 1: Assessment 1. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\sin {{75}^\\circ}+\\sin {{15}^\\circ}\\\\&=\\sin ({{45}^\\circ}+{{30}^\\circ})+\\sin ({{45}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{45}^\\circ}\\cos {{30}^\\circ}+\\cos {{45}^\\circ}\\sin {{30}^\\circ}+\\sin {{45}^\\circ}\\cos {{30}^\\circ}-\\cos {{45}^\\circ}\\sin {{30}^\\circ}\\\\&=2\\sin {{45}^\\circ}\\cos {{30}^\\circ}\\\\&=2\\times \\displaystyle \\frac{1}{{\\sqrt{2}}}\\times \\displaystyle \\frac{{\\sqrt{3}}}{2}\\\\&=\\displaystyle \\frac{{\\sqrt{3}}}{{\\sqrt{2}}}\\\\&=\\displaystyle \\frac{{\\sqrt{6}}}{2}=\\text{RHS}\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\sin \\theta }}-\\displaystyle \\frac{{\\cos 2\\theta }}{{\\cos \\theta }}\\\\&=\\displaystyle \\frac{{2\\sin \\theta", "\\cos 48{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 12{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\sin 6{}^\\circ \\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ If we observe the above equation carefully then we can say that the equation can be arranged such that it will take the form of sin2x. Therefore we will multiply and divide the equation by $2\\cos 6{}^\\circ$ therefore we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\sin 6{}^\\circ \\times \\dfrac{2\\cos 6{}^\\circ }{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{2\\times \\sin 6{}^\\circ \\times \\cos 6{}^\\circ }{2\\times \\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ To proceed further in the solution we should know the formula given below, Formula: $\\sin 2x=2\\times \\sin x\\times \\cos x$ ……………………………… (i) By using above formula we can write the above equation as, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin \\left( 2\\times 6{}^\\circ \\right)}{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin 12{}^\\circ }{2\\cos 6{}^\\circ }\\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times \\cos 48{}^\\circ$ On rearranging the above equation we will get, $\\therefore$ L.H.S. (Left Hand Side) $=\\dfrac{\\sin 12{}^\\circ \\times \\cos 12{}^\\circ \\times \\cos 24{}^\\circ \\times", "of length $\\,3,\\, 4,\\, 6\\,$. • Note that $\\,6 < 3 + 4\\,$. By SSS, the triangle is unique. • Use the Law of Cosines with the longest side: $$\\begin{gather} \\cssId{sb36}{6^2 = 3^2 + 4^2 - 2\\cdot 3\\cdot 4\\cdot\\cos F}\\cr\\cr \\cssId{sb37}{\\cos F \\approx -0.4583} \\cr\\cr \\cssId{sb38}{F = \\arccos(\\cos F) \\approx 117.3^\\circ} \\end{gather}$$ • Use the Law of Sines with the shortest side: $$\\begin{gather} \\cssId{sb40}{\\frac{\\sin 117.3^\\circ}{6} = \\frac{\\sin D}{3}}\\cr\\cr \\cssId{sb41}{\\sin D = \\frac{3\\sin 117.3^\\circ}{6} \\approx 0.4443}\\cr\\cr \\cssId{sb42}{D = \\arcsin(0.4443) \\approx 26.38^\\circ} \\end{gather}$$ • $\\,E = 180^\\circ - 117.3^\\circ - 26.38^\\circ = 36.32^\\circ$ • Summary: ANGLES OPPOSITE SIDES $D = 26.38^\\circ$ $d = 3$ $E = 36.32^\\circ$ $e = 4$ $F = 117.3^\\circ$ $f = 6$ This is the correct triangle! Note: At left, the corrections would be: $$\\begin{gather} \\cssId{sb56}{F = 180^\\circ - \\arcsin(0.8886) \\approx 117.3^\\circ}\\cr\\cr \\cssId{sb57}{E = 180^\\circ - 26.38^\\circ - 117.3^\\circ = 36.32^\\circ} \\end{gather}$$ ## Congruence Theorems CONGRUENCE THEOREMS A unique triangle is determined by each of these configurations: SAS side-angle-side (two sides and an included angle) SSS side-side-side (three sides) AAS/SAA angle-angle-side (or, side-angle-angle) (two angles and a non-included side) ASA angle-side-angle (two angles and an included side) If $$\\begin{gather} \\cssId{sb65}{e > 0}\\cr \\cssId{sb66}{f > 0}\\cr \\cssId{sb67}{0^\\circ < D < 180^\\circ} \\end{gather}$$ then there exists a unique triangle having the two sides and included angle. Name the", "and \\begin{align*}b = 30^\\circ\\end{align*}. \\begin{align*}\\cos (45^\\circ - 30^\\circ) & = \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ \\cos 15^\\circ & = \\frac{\\sqrt{6} + \\sqrt{2}}{4}\\end{align*} #### Example C Find the exact value of \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*}, in radians. Solution: \\begin{align*}\\cos \\frac{5 \\pi}{12} = \\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right )\\end{align*}, notice that \\begin{align*}\\frac{\\pi}{4} = \\frac{3 \\pi}{12}\\end{align*} and \\begin{align*}\\frac{\\pi}{6} = \\frac{2 \\pi}{12}\\end{align*} \\begin{align*}\\cos \\left (\\frac{\\pi}{4} + \\frac{\\pi}{6} \\right ) & = \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} \\\\ \\cos \\frac{\\pi}{4} \\cos \\frac{\\pi}{6} - \\sin \\frac{\\pi}{4} \\sin \\frac{\\pi}{6} & = \\frac{\\sqrt{2}}{2} \\times \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{2}}{2} \\times \\frac{1}{2} \\\\ & = \\frac{\\sqrt{6} - \\sqrt{2}}{4}\\end{align*} ### Vocabulary Cosine Sum Formula: The cosine sum formula relates the cosine of a sum of two arguments to a set of sine and cosines functions, each containing one argument. Cosine Difference Formula: The cosine difference formula relates the cosine of a difference of two arguments to a set of sine and cosines functions, each containing one argument. ### Guided Practice 1. Find the exact value for \\begin{align*}\\cos \\frac{5 \\pi}{12}\\end{align*} 2. Find the exact value for \\begin{align*}\\cos \\frac{7 \\pi}{12}\\end{align*} 3. Find the exact value for \\begin{align*}\\cos 345^\\circ\\end{align*} Solutions: 1. \\begin{align*}\\cos \\frac{5 \\pi}{12} & = \\cos \\left (\\frac{2 \\pi}{12} + \\frac{3", "5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d = \\sqrt{34 - 30\\cos 57^\\circ}$ 4. $d \\approx 4.202$ 5. $\\displaystyle \\frac{\\sin 57^\\circ}{d} = \\frac{\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} = \\frac{\\sin E}{3}$ 6. $\\displaystyle\\sin E = \\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}} \\approx 0.5987$ 7. $\\displaystyle E = \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\approx 36.78^\\circ$ 8. $\\displaystyle\\,F = 180^\\circ - 57^\\circ - E \\ =\\ 180^\\circ - 57^\\circ - \\arcsin\\left(\\frac{3\\sin 57^\\circ}{\\sqrt{34 - 30\\cos 57^\\circ}}\\right) \\ \\approx\\ 86.22^\\circ$ It's just scary looking. (But, every ‘$\\,=\\,$’ is strictly correct.) Potential Remedy #3 with Comments Another potential remedy is to use ‘$\\,\\approx\\,$’ for everything that is truly an approximation. In this case, pretty much everything after the first occurrence of an approximation gets ‘$\\,\\approx\\,$’: 1. $d^2 = 3^2 + 5^2 - 2\\cdot 3\\cdot 5\\cdot\\cos 57^\\circ$ 2. $d^2 = 34 - 30\\cos 57^\\circ$ 3. $d \\approx 4.202$ 4. $\\displaystyle\\frac{\\sin 57^\\circ}{4.202} \\approx \\frac{\\sin E}{3}$ 5. $\\displaystyle\\sin E \\approx \\frac{3\\sin 57^\\circ}{4.202} \\approx 0.5988$ 6. $E \\approx \\arcsin(0.5988) \\approx 36.78^\\circ$ 7. $\\,F \\approx 180^\\circ - 57^\\circ - 36.78^\\circ = 86.22^\\circ$ I don't like this at all. It makes it ‘sound like’ (say) the Law of Sines only sort of works. Potential Remedy #4 with Comments One final potential remedy is to use an ellipsis ‘$\\,\\ldots\\,$’ to indicate missing digits of a number. In this case,", "30^{\\circ}$$ $$a = \\displaystyle \\frac {b\\sin(130^{\\circ})}{\\sin(20^{\\circ})} = 22.398$$ $$c = \\displaystyle \\frac {b\\sin(30^{\\circ})}{\\sin(20^{\\circ})} = 14.619$$ 2. $$A = 180 ^{\\circ} - 120 ^{\\circ} - 30 ^{\\circ} = 30 ^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(120^{\\circ})}{\\sin30^{\\circ})} = 27.713$$ $$c = \\displaystyle \\frac {a\\sin(30^{\\circ})}{sin(30^{\\circ})} = 16$$ 3. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(30^{\\circ})} = 17.851$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 107.76^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 42.24^{\\circ}$$ 4. $$c = \\sqrt{a^2 + b^2 - 2ab\\cos(108^{\\circ})} = 20.438^{\\circ}$$ $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 27.73^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 44.27^{\\circ}$$ 5. $$A = \\arccos\\left(\\displaystyle \\frac {c^2 + b^2 - a^2}{2cb}\\right) = 38.44^{\\circ}$$ $$B = \\arccos\\left(\\displaystyle \\frac {c^2 + a^2 - b^2}{2ca}\\right) = 61.73^{\\circ}$$ $$C = \\arccos\\left(\\displaystyle \\frac {a^2 + b^2 - c^2}{2ab}\\right) = 79.84^{\\circ}$$ 6. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 66.67^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 66.67^{\\circ} = 73.33^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(73.33^{\\circ})}{\\sin(40^{\\circ})} = 10.432$$ 7. $$C = \\arcsin\\left(\\displaystyle \\frac {c\\sin(40^{\\circ})}{a}\\right) = 32.39^{\\circ}$$ $$B = 180^{\\circ} - 40^{\\circ} - 32.39^{\\circ} = 107.61^{\\circ}$$ $$b = \\displaystyle \\frac {a\\sin(107.61^{\\circ})}{\\sin(40^{\\circ})} = 17.794$$ ### Exercise $$\\PageIndex{15}$$ 1. A ladder $$10$$ meters long is leaning against the side of a building. It makes an angle of $$75^{\\circ}$$ with the ground. How far up", "of $EF=FA,\\,$ that is, $\\displaystyle \\cot(18^{\\circ})-\\cot(24^{\\circ}) = 2\\frac{\\sin(18^{\\circ})}{\\sin(48^{\\circ})}\\,$. We use the fact that $s=\\sin(18^{\\circ}),\\,$ being equal to $\\displaystyle \\frac{\\varphi - 1}{2},\\,$ has the property $\\displaystyle 2 s^2 = \\frac{1}{2} - s.\\,$ Thus \\displaystyle\\begin{align} \\cot(18^{\\circ})-\\cot(24^{\\circ}) &= \\frac{\\sin(6^{\\circ})}{s\\sin(24^{\\circ})} = \\frac{2\\sin(6^{\\circ})\\cos(24^{\\circ})}{s\\sin(48^{\\circ})}\\\\ &=\\frac{\\sin(30^{\\circ})-s}{s\\sin(48^{\\circ})} = \\frac{2 s^2}{s\\sin(48^{\\circ})} = \\frac{2 s}{\\sin(48^{\\circ})}. \\end{align} and we are done. ### Acknowledgment The problem has been posted at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group by Konstantin Knop. Konstantin has requested not to leave out a similar problem: I do leave it as an exercise to the reader. It's the same circle, though. Grégoire Nicollier has come up with a direct proof (P.S.) of $BI\\perp AC.$ Konstantin has also added an extra point to the circle on observed that, be symmetry, points C, B, E, O, N are also concyclic. Grégoire Nicollier has later added another proof (Solution 2), a variant of which has been placed on a separate page.", "}$ $\\displaystyle \\text{iv) } \\text{LHS } =\\sin 23^{\\circ} + \\sin 37^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{23^{\\circ}+37^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{23^{\\circ}-37^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 30^{\\circ} \\cos (-7^{\\circ})$ $\\displaystyle = 2 \\times \\frac{1}{2} \\cos 7^{\\circ}$ $\\displaystyle = \\cos 7^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{v) } \\text{LHS } =\\sin 105^{\\circ} + \\cos 105^{\\circ}$ $\\displaystyle =\\sin 105^{\\circ} + \\cos (90^{\\circ}+ 15^{\\circ})$ $\\displaystyle =\\sin 105^{\\circ} - \\sin 15^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{105^{\\circ}-15^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{105^{\\circ}+15^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 45^{\\circ} \\cos 60^{\\circ}$ $\\displaystyle = 2 \\times \\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}$ $\\displaystyle = \\frac{1}{\\sqrt{2}} = \\cos 45^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{vi) } \\text{LHS } =\\sin 40^{\\circ} + \\sin 20^{\\circ}$ $\\displaystyle =2 \\sin \\Big( \\frac{40^{\\circ}+20^{\\circ}}{2} \\Big) \\cos \\Big( \\frac{20^{\\circ}-40^{\\circ}}{2} \\Big)$ $\\displaystyle = 2 \\sin 30^{\\circ} \\cos (-10^{\\circ})$ $\\displaystyle = 2 \\times \\frac{1}{2} \\cos 10^{\\circ}$ $\\displaystyle = \\cos 10^{\\circ} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\\\$ Question 3: Prove that: $\\displaystyle \\text{i) } \\cos 55^{\\circ} + \\cos 65^{\\circ} + \\cos 175^{\\circ} = 0 \\hspace{1.0cm} \\text{ii) } \\sin 50^{\\circ} - \\sin 70^{\\circ} + \\sin 10^{\\circ} = 0$ $\\displaystyle \\text{iii) } \\cos 80^{\\circ} + \\cos 40^{\\circ} - \\cos 20^{\\circ} = 0 \\hspace{1.0cm} \\text{v) } \\cos 20^{\\circ} + \\cos 100^{\\circ} + \\cos 140^{\\circ} = 0$ $\\displaystyle \\text{v) } \\sin \\frac{5\\pi}{18} - \\cos \\frac{4\\pi}{9} = \\sqrt{3} \\sin \\frac{\\pi}{9}", "= \\frac{\\cos 1^\\circ}{\\sin 1^\\circ}. (6)$$ Now we notice that we can multiply both sides by $\\sin 1^\\circ$. Doing so gives us $$\\sin 1^\\circ(2 (\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ) + 1) = \\cos 1^\\circ. (7)$$ Now, simplifying $\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas. Doing so, gives us $$(\\sin 1^\\circ)(2 \\sin 2^\\circ)+(\\sin 1^\\circ)(2\\sin 4^\\circ) +(\\sin 1^\\circ)(2\\sin 6^\\circ) + (\\sin 1^\\circ)(2\\sin 8^\\circ) +(\\sin 1^\\circ)(2 \\sin 10^\\circ) ... + (\\sin 1^\\circ)(2\\sin 88^\\circ) + \\sin 1^\\circ = \\cos 1^\\circ. (8)$$ Now we can use our product of sines to sum formula and see if we can find a pattern. We will start with expanding $(\\sin 1^\\circ)(2 \\sin 2^\\circ)$. Because the format of our formula for the product of sines is $\\sin \\alpha + \\sin \\beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\\sin 1^\\circ)(\\sin 2^\\circ)$. Our formula for the product of sines is $\\sin \\alpha + \\sin \\beta = \\frac{1}{2}[\\cos(\\alpha - \\beta) - \\cos(\\alpha + \\beta)]$. Plugging in our values into the formula and simplifying gives us $2\\left(\\frac{1}{2}(\\cos 1^\\circ", "the problem gives us: We don’t seem to know much about the side $PR$, whereas parts of $PQ$ and $QR$ are known. Hence, in calculating the area of $\\Delta PQR$, it seems like using the formula $\\text{Area of } \\Delta PQR = \\displaystyle\\frac{1}{2} |PQ||QR|\\sin PQR = \\frac{1}{2} |PQ||QR| \\sin 60^\\circ$ could work. Let’s label the diagram further to give more info on $PQ$ and $QR$: We have 2 unknown quantities we would like to find: $|PQ|$ and $|QR|$. We can use the sine rule to reduce that to just finding $|PQ|$: \\begin{aligned} \\displaystyle\\frac{|PQ|}{\\sin 45^\\circ} &= \\frac{|QR|}{\\sin 75^\\circ}, \\\\ |QR| &= \\frac{|PQ|}{\\sin 45^\\circ} \\sin 75^\\circ = \\sqrt{2}|PQ| \\sin 75^\\circ. \\end{aligned} From the diagram, we know that $|PQ| = 2 + |AP|$. We can use the sine rule again to find the value of $|AP|$: \\begin{aligned} \\displaystyle\\frac{|AP|}{\\sin 45^\\circ} &= \\frac{|AF|}{\\sin 75^\\circ}, \\\\ |AP| &= \\frac{1}{\\sin 75^\\circ} \\sin 45^\\circ = \\frac{1}{\\sqrt{2}\\sin 75^\\circ}. \\end{aligned} Hence, we have \\begin{aligned} \\text{Area} &= \\displaystyle\\frac{1}{2}|PQ| \\cdot \\sqrt{2} |PQ| \\sin 75^\\circ \\cdot \\sin 60^\\circ \\\\ &= \\frac{\\sqrt{2}}{2}\\frac{\\sqrt{3}}{2} \\sin 75^\\circ |PQ|^2 \\\\ &= \\frac{\\sqrt{6}}{4} \\sin 75^\\circ \\left( 2 + \\frac{1}{\\sqrt{2}\\sin 75^\\circ} \\right)^2 \\\\ &= \\frac{\\sqrt{6}}{4}\\sin 75^\\circ \\left( 4 + \\frac{4}{\\sqrt{2}\\sin 75^\\circ} + \\frac{1}{2\\sin^2 75^\\circ} \\right) \\\\ &= \\sqrt{6}\\sin 75^\\circ + \\sqrt{3} + \\frac{\\sqrt{6}}{8 \\sin 75^\\circ}. \\end{aligned} It remains to get rid of the $\\sin 75^\\circ$ term. $\\sin 75^\\circ", "50^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\Big[ (\\cos 10^{\\circ} \\cos 50^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ (2 \\cos 50^{\\circ} \\cos 10^{\\circ}) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ \\Big( \\cos (50^{\\circ}+10^{\\circ}) + \\cos (50^{\\circ}-10^{\\circ}) \\Big) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\cos 60^{\\circ} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{4} \\Big[ ( \\frac{1}{2} + \\cos 40^{\\circ} ) \\cos 70^{\\circ} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{4} \\cos 40^{\\circ} \\cos 70^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} (2\\cos 40^{\\circ} \\cos 70^{\\circ})$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} ( \\cos (70^{\\circ}+40^{\\circ}) + \\cos (70^{\\circ}-40^{\\circ}) )$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{8} \\cos 110 + \\frac{\\sqrt{3}}{8} \\cos 30^{\\circ}$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos 110 + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} + \\cos (180 - 70) + \\cos 30 \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\cos 70^{\\circ} - \\cos 70^{\\circ} + \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{\\sqrt{3}}{8} \\Big[ \\frac{\\sqrt{3}}{2} \\Big]$ $\\displaystyle = \\frac{3}{16} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{ii) } \\text{LHS } = \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ}$ $\\displaystyle = \\frac{1}{2} \\Big[ 2 \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} ( 2 \\cos 160^{\\circ} \\cos 40^{\\circ}) \\Big]$ $\\displaystyle = \\frac{1}{2} \\Big[ \\cos 80^{\\circ} [ \\cos ( 160^{\\circ}+40^{\\circ})", "= \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{(ii) } \\text{LHS } = \\sin \\frac{8 \\pi}{3} \\cos \\frac{23 \\pi}{6} + \\cos \\frac{13 \\pi}{3} \\sin \\frac{35 \\pi}{6}$ $\\displaystyle = \\sin \\Big( 3\\pi - \\frac{\\pi}{3} \\Big) \\cos \\Big( 4\\pi - \\frac{\\pi}{6} \\Big) + \\cos \\Big( 4\\pi + \\frac{\\pi}{3} \\Big) \\sin \\Big( 6\\pi - \\frac{\\pi}{6} \\Big)$ $\\displaystyle = \\sin \\frac{\\pi}{3} \\cos \\frac{\\pi}{6} + \\cos \\frac{\\pi}{3} \\Big( - \\sin \\frac{\\pi}{6} \\Big)$ $\\displaystyle = \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\Big( - \\frac{1}{2} \\Big)$ $\\displaystyle = \\frac{3}{4} - \\frac{1}{4} = \\frac{2}{4} = \\frac{1}{2}$ $\\displaystyle \\text{(iii) } \\text{LHS } = \\cos 24^{\\circ} + \\cos 55^{\\circ} + \\cos 125^{\\circ} + \\cos 204^{\\circ} + \\cos 300^{\\circ}$ $\\displaystyle = ( \\cos 24^{\\circ} + \\cos 204^{\\circ}) + ( \\cos 55^{\\circ} + \\cos 125^{\\circ}) + \\cos 300^{\\circ}$ $\\displaystyle = ( \\cos 24^{\\circ} + \\cos (\\pi + 24^{\\circ}) ) + ( \\cos 55^{\\circ} + \\cos (\\pi - 55^{\\circ})) + \\cos (2\\pi - 60^{\\circ})$ $\\displaystyle = \\cos 24^{\\circ} - \\cos 24^{\\circ} + \\cos 55^{\\circ} - \\cos 55^{\\circ} + \\cos 60^{\\circ}$ $\\displaystyle = \\cos 60^{\\circ} = \\frac{1}{2} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\text{(iv) } \\text{LHS } = \\tan (-225^{\\circ}) \\cot(-405^{\\circ}) - \\tan (-765^{\\circ}) \\cot (675^{\\circ})$ $\\displaystyle = -\\tan 225^{\\circ} (-\\cot 405^{\\circ} ) + \\tan 765^{\\circ} \\cot 675^{\\circ}$ $\\displaystyle = \\tan \\Big( \\pi + \\frac{\\pi}{4} \\Big) \\cot \\Big( 2\\pi + \\frac{\\pi}{4} \\Big) +", "1. ## prove $\\sin 47° + \\sin 61° - \\sin 11° - \\sin 25° = \\cos 7°$. 2. ## Sum to Product First, note that 47-11=61-25=36, so rearrange the left hand side to get $(\\sin 47^\\circ - \\sin 11^\\circ) + (\\sin 61^\\circ - \\sin 25^\\circ)$. From this, we use the sum-to-product formulas, specifically the formula $\\sin A - \\sin B = 2\\cos \\left ( \\frac{A+B}{2} \\right ) \\sin \\left ( \\frac{A-B}{2} \\right )$. Thus we get $\\sin 47^\\circ + \\sin 61^\\circ - \\sin 11^\\circ - \\sin 25^\\circ = (\\sin 47^\\circ - \\sin 11^\\circ) + (\\sin 61^\\circ - \\sin 25^\\circ)$ $\\,\\,= 2 \\cos 29^\\circ \\sin 18^\\circ + 2 \\cos 43^\\circ \\sin 18^\\circ$ $\\,\\,= 2(\\cos 29^\\circ + \\cos 43^\\circ)\\sin 18^\\circ$ We use the sum-to-product formulas again, this time using $\\cos A + \\cos B = 2\\cos \\left ( \\frac{A+B}{2} \\right ) \\cos \\left ( \\frac{A-B}{2} \\right )$, so $\\sin 47^\\circ + \\sin 61^\\circ - \\sin 11^\\circ - \\sin 25^\\circ = 2(\\cos 29^\\circ + \\cos 43^\\circ)\\sin 18^\\circ$ $\\,\\,= 2(2\\cos 36^\\circ \\cos (-7^\\circ))\\sin 18^\\circ$ $\\,\\,= 4\\cos 7^\\circ \\cos 36^\\circ \\sin 18^\\circ$, as $\\cos(-x)=\\cos x$. Working from the geometry of the pentagon, one can find the sine and cosine for angles that are multiples of 18 degrees, and thus find that: $\\cos 36^\\circ = \\frac{1 + \\sqrt{5}}{4}$, $\\sin 18^\\circ = \\frac{-1 + \\sqrt{5}}{4}$,", "& = 180^\\circ\\\\ m \\angle A & = 57^\\circ\\end{align*} Now, use the Law of Sines to set up ratios for \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*}. \\begin{align*}\\frac{\\sin 57^\\circ}{a} = \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\end{align*} \\begin{align*}\\frac{\\sin 57^\\circ}{a} & = \\frac{\\sin 38^\\circ}{12} && \\quad \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\\\ a \\cdot \\sin 38^\\circ & = 12 \\cdot \\sin 57^\\circ && b \\cdot \\sin 38^\\circ = 12 \\cdot \\sin 85^\\circ\\\\ a & = \\frac{12 \\cdot \\sin 57^\\circ}{\\sin 38^\\circ} \\approx 16.4 && \\qquad \\quad \\ \\ b = \\frac{12 \\cdot \\sin 85^\\circ}{\\sin 38^\\circ} \\approx 19.4\\end{align*} Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth. Solution: Set up the ratio for \\begin{align*}\\angle B\\end{align*} using Law of Sines. \\begin{align*}\\frac{\\sin 95^\\circ}{27} & = \\frac{\\sin B}{16}\\\\ 27 \\cdot \\sin B & = 16 \\cdot \\sin 95^\\circ\\\\ \\sin B & = \\frac{16 \\cdot \\sin 95^\\circ}{27} \\rightarrow \\sin^{-1} \\left ( \\frac{16 \\cdot \\sin 95^\\circ}{27} \\right ) = 36.2^\\circ\\end{align*} To find \\begin{align*}m \\angle C\\end{align*} use the Triangle Sum Theorem. \\begin{align*}m \\angle C + 95 ^\\circ + 36.2^\\circ = 180^\\circ \\rightarrow m \\angle C = 48.8^\\circ\\end{align*} To find \\begin{align*}c\\end{align*}, use the Law of Sines again. \\begin{align*}\\frac{\\sin 95^\\circ}{27} = \\frac{\\sin 48.8^\\circ}{c}\\end{align*} \\begin{align*}c \\cdot \\sin 95^\\circ & = 27 \\cdot \\sin 48.8^\\circ\\\\ c & = \\frac{27 \\cdot \\sin 48.8^\\circ}{\\sin 95^\\circ} \\approx 20.4\\end{align*} ## Law of Cosines Law", "{3}{17}}\\right)^{\\circ }} }$，${\\displaystyle \\mathbf {\\left({\\frac {360}{17}}\\right)^{\\circ }} }$：正十七邊形 ${\\displaystyle \\operatorname {cos} {2\\pi \\over 17}={\\frac {-1+{\\sqrt {17}}+{\\sqrt {34-2{\\sqrt {17}}}}+2{\\sqrt {17+3{\\sqrt {17}}-{\\sqrt {34-2{\\sqrt {17}}}}-2{\\sqrt {34+2{\\sqrt {17}}}}}}}{16}}}$ ### 22.5°：正八邊形 ${\\displaystyle \\sin {\\frac {\\pi }{8}}=\\sin 22.5^{\\circ }={\\tfrac {1}{2}}({\\sqrt {2-{\\sqrt {2}}}})}$ ${\\displaystyle \\cos {\\frac {\\pi }{8}}=\\cos 22.5^{\\circ }={\\tfrac {1}{2}}({\\sqrt {2+{\\sqrt {2}}}})\\,}$ ${\\displaystyle \\tan {\\frac {\\pi }{8}}=\\tan 22.5^{\\circ }={\\sqrt {2}}-1\\,}$ ### 24°：12°的二倍 ${\\displaystyle \\sin {\\frac {2\\pi }{15}}=\\sin 24^{\\circ }={\\tfrac {1}{8}}\\left[{\\sqrt {3}}({\\sqrt {5}}+1)-{\\sqrt {2}}{\\sqrt {5-{\\sqrt {5}}}}\\right]\\,}$ ${\\displaystyle \\cos {\\frac {2\\pi }{15}}=\\cos 24^{\\circ }={\\tfrac {1}{8}}\\left({\\sqrt {6}}{\\sqrt {5-{\\sqrt {5}}}}+{\\sqrt {5}}+1\\right)\\,}$ ${\\displaystyle \\tan {\\frac {2\\pi }{15}}=\\tan 24^{\\circ }={\\tfrac {1}{2}}\\left[{\\sqrt {2(25+11{\\sqrt {5}})}}-{\\sqrt {3}}(3+{\\sqrt {5}})\\right]\\,}$ ### 180/7°，${\\displaystyle \\mathbf {\\left(25{\\frac {5}{7}}\\right)^{\\circ }} }$，${\\displaystyle \\mathbf {\\left({\\frac {180}{7}}\\right)^{\\circ }} }$：正七邊形 ${\\displaystyle \\cos {\\frac {\\pi }{7}}=\\cos {\\frac {180}{7}}^{\\circ }=\\cos 25{\\frac {5}{7}}^{\\circ }={\\frac {1}{6}}+{\\frac {1-{\\sqrt {3}}i}{24}}{\\sqrt[{3}]{28-84{\\sqrt {3}}i}}+{\\frac {1+{\\sqrt {3}}i}{24}}{\\sqrt[{3}]{28-84{\\sqrt {3}}i}}}$ ### 27°：12°与15°的和 ${\\displaystyle \\sin {\\frac {3\\pi }{20}}=\\sin 27^{\\circ }={\\tfrac {1}{8}}\\left[2{\\sqrt {5+{\\sqrt {5}}}}-{\\sqrt {2}}\\;({\\sqrt {5}}-1)\\right]\\,}$ ${\\displaystyle \\cos {\\frac {3\\pi }{20}}=\\cos 27^{\\circ }={\\tfrac {1}{8}}\\left[2{\\sqrt {5+{\\sqrt {5}}}}+{\\sqrt {2}}\\;\\left({\\sqrt {5}}-1\\right)\\right]\\,}$ ${\\displaystyle \\tan {\\frac {3\\pi }{20}}=\\tan 27^{\\circ }={\\sqrt {5}}-1-{\\sqrt {5-2{\\sqrt {5}}}}\\,}$ ### 30°：正六邊形 ${\\displaystyle \\sin {\\frac {\\pi }{6}}=\\sin 30^{\\circ }={\\tfrac {1}{2}}\\,}$ ${\\displaystyle \\cos {\\frac {\\pi }{6}}=\\cos 30^{\\circ }={\\tfrac {1}{2}}{\\sqrt {3}}\\,}$ ${\\displaystyle \\tan {\\frac {\\pi }{6}}=\\tan 30^{\\circ }={\\tfrac {1}{3}}{\\sqrt {3}}\\,}$ ### 33°：15°与18°的和 ${\\displaystyle \\sin {\\frac {11\\pi }{60}}=\\sin 33^{\\circ }={\\tfrac {1}{4}}{\\sqrt {8-{\\sqrt {3}}-{\\sqrt {15}}+{\\sqrt {10-2{\\sqrt {5}}}}}}\\,}$ ${\\displaystyle \\cos {\\frac {11\\pi }{60}}=\\cos 33^{\\circ }={\\tfrac {1}{4}}{\\sqrt {8+{\\sqrt {3}}+{\\sqrt {15}}-{\\sqrt {10-2{\\sqrt {5}}}}}}\\,}$ ${\\displaystyle \\tan {\\frac", "\\cos x }{2\\cos ^2 x-1}=\\dfrac{\\sin (2x)}{ \\cos (2x)}=\\tan (2x)$$ 58) $$(\\sin^2x-1)^2=\\cos(2x)+\\sin^4x$$ 59) $$\\sin(3x)=3\\sin x \\cos^2x-\\sin^3x$$ \\begin{align*} \\sin (x+2x) &= \\sin x \\cos (2x)+\\sin (2x) \\cos x\\\\ &= \\sin x(\\cos ^2 x - \\sin ^2 x)+2\\sin x \\cos x \\cos x\\\\ &= \\sin x \\cos ^2 x-\\sin ^3 x + 2\\sin x\\cos ^2 x\\\\ &= 3\\sin x\\cos ^2 x - \\sin ^3 x \\end{align*} 60) $$\\cos(3x)=\\cos^3x-3\\sin^2x\\cos x$$ 61) $$\\dfrac{1+\\cos (2t)}{\\sin (2t)-\\cos t}=\\dfrac{2\\cos t}{2\\sin t-1}$$ \\begin{align*} \\dfrac{1+\\cos (2t)}{\\sin (2t)-\\cos t} &= \\dfrac{1+2\\cos ^2t-1}{2\\sin t\\cos t-\\cos t}\\\\ &= \\dfrac{2\\cos ^2t}{\\cos t(2\\sin t-1)}\\\\ &= \\dfrac{2\\cos t}{2\\sin t-1} \\end{align*} 62) $$\\sin(16x)=16 \\sin x \\cos x \\cos(2x)\\cos(4x)\\cos(8x)$$ 63) $$\\cos(16x)=(\\cos^2(4x)-\\sin^2(4x)-\\sin(8x))(\\cos^2(4x)-\\sin^2(4x)+\\sin(8x))$$ \\begin{align*} (\\cos^2(4x)-\\sin^2(4x)-\\sin(8x))(\\cos^2(4x)-\\sin^2(4x)+\\sin(8x)) &= (\\cos(8x)-\\sin(8x))(\\cos(8x)+\\sin(8x))\\\\ &= \\cos ^2 (8x)-\\sin ^2 (8x)\\\\ &= \\cos(16x) \\end{align*} ## 7.4: Sum-to-Product and Product-to-Sum Formulas From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. The product-to-sum formulas can rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. We can also derive the sum-to-product identities from the product-to-sum identities using substitution. The sum-to-product formulas are used to rewrite sum or difference as products of sines and cosines. #### Verbal 1) Starting with the product to sum formula $$\\sin \\alpha \\cos \\beta =\\dfrac{1}{2} \\left[\\sin(\\alpha +\\beta )+\\sin(\\alpha -\\beta ) \\right]$$$,$ explain", "small right-angled triangle next to the angle $18^\\circ$ Then, you will see that $\\angle ACD=90^\\circ-18^\\circ=72^\\circ$ (vert opp. angles) $\\angle BAC=180^\\circ-72^\\circ=108^\\circ$ (supplementary angles in trapezium) By sine rule, $\\displaystyle \\frac{\\sin\\angle ABC}{30}=\\frac{\\sin 108^\\circ}{40.9}$ $\\sin\\angle ABC=0.697596$ $\\angle ABC=44.23^\\circ$ $\\angle ACB=180^\\circ-44.23^\\circ-108^\\circ=27.77^\\circ=27.8^\\circ$ (1 d.p.) (b) By Sine Rule, $\\displaystyle\\frac{AB}{\\sin\\angle ACB}=\\frac{30}{\\sin 44.23^\\circ}$ $AB=\\frac{30}{\\sin 44.23^\\circ}\\times\\sin 27.77^\\circ=20.04=20.0 m$ (shown) (c) $\\angle BCD=\\angle ACD-\\angle ACB=72^\\circ-27.77^\\circ=44.23^\\circ$ By Cosine Rule, $BD^2=40.9^2+50^2-2(40.9)(50)\\cos 44.23^\\circ=1242.139$ $BD=35.24=35.2 m$ (d) $\\displaystyle\\frac{\\sin\\angle BDC}{40.9}=\\frac{\\sin 44.23^\\circ}{35.24}$ $\\sin\\angle BDC=0.80957$ $\\angle BDC=54.05^\\circ$ angle of depression = $90^\\circ-54.05^\\circ=35.95^\\circ=36.0^\\circ$ (1 d.p.) (e) Let X be the point where the man is at the shortest distance from D. Draw a right-angle triangle XDC. $\\displaystyle\\cos 72^\\circ=\\frac{XC}{50}$ $XC=50\\cos 72^\\circ=15.5 m$ ## Rectangle Maths Tuition Solution: (a) $\\angle SPC=\\angle SRQ=90^\\circ$ $\\angle PCS=\\angle SQR$ (given) Thus $\\triangle PCS$ is similar to $\\triangle RQS$ (AAA) (b)(I) $\\angle KRQ=90^\\circ /2=45^\\circ$ (KR bisects $\\angle QRS$) Thus $\\angle QKR=180^\\circ-90^\\circ-45^\\circ=45^\\circ$ Hence $\\triangle KQR$ is an isosceles triangle. So $KQ=QR=PS=\\frac{1}{2}PQ$ Hence, $\\begin{array}{rcl}PK&=&PQ-KQ\\\\ &=&PQ-\\frac{1}{2}PQ\\\\ &=&\\frac{1}{2}PQ\\\\ &=&\\frac{1}{2}(2PS)\\\\ &=&PS \\end{array}$ (shown) (ii) By similar triangles, $\\displaystyle\\frac{PC}{PS}=\\frac{RQ}{RS}=\\frac{PS}{PQ}=\\frac{PS}{2PS}=\\frac{1}{2}$ Thus $PC=\\frac{1}{2}PS=\\frac{1}{2} PK$ Hence $\\begin{array}{rcl}CK&=&PK-PC\\\\ &=&PK-\\frac{1}{2}PK\\\\ &=&\\frac{1}{2}PK\\\\ &=&PC \\end{array}$ (shown) ## Tips on attempting Geometrical Proof questions (E Maths Tuition) Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths) 1) Draw extended lines and additional lines. (using pencil) Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z”", "\\Big[ \\cos \\frac{2\\pi}{3} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ \\cos \\Big( \\frac{\\pi}{2} + \\frac{\\pi}{6} \\Big) + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\sin \\frac{\\pi}{6} + \\cos 2x \\Big]$ $\\displaystyle = 2 \\cos x \\Big[ - \\frac{1}{2} + \\cos 2x \\Big]$ $\\displaystyle = - \\cos x + 2 \\cos x \\cos 2x$ $\\displaystyle = - \\cos x + [ \\cos (2x + x) + \\cos ( 2x - x) ]$ $\\displaystyle = - \\cos x + \\cos 3x + \\cos x$ $\\displaystyle = \\cos 3x =$ RHS $\\displaystyle \\\\$ Question 5: Prove that: $\\displaystyle \\text{ i) } \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos 50^{\\circ} \\cos 70^{\\circ} = \\frac{3}{16} \\hspace{1.0cm} \\text{ ii) } \\cos 40^{\\circ} \\cos 80^{\\circ} \\cos 160^{\\circ} = - \\frac{1}{8}$ $\\displaystyle \\text{ iii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 80^{\\circ} = \\frac{\\sqrt{3}}{8} \\hspace{1.0cm} \\text{ iv) } \\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ} = \\frac{1}{8}$ $\\displaystyle \\text{ v) } \\tan 20^{\\circ} \\tan 40^{\\circ} \\tan 60^{\\circ} \\tan 80^{\\circ} = 3 \\hspace{1.0cm} \\text{ vi) } \\tan 20^{\\circ} \\tan 30^{\\circ} \\tan 40^{\\circ} \\tan 80^{\\circ} = 1$ $\\displaystyle \\text{ vii) } \\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 60^{\\circ} \\sin 70^{\\circ} = \\frac{\\sqrt{3}}{16} \\hspace{1.0cm} \\text{ viii) } \\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 60^{\\circ} \\sin 80^{\\circ} = \\frac{3}{16}$ $\\displaystyle \\text{i) } \\text{LHS } = \\cos 10^{\\circ} \\cos 30^{\\circ} \\cos", "to $\\mathbb{R}^2$. You obtained $(a, 2b^2)$ for this derivative, which is a map from $\\mathbb{R}^2$ to $\\mathbb{R}$. You should instead find $$D(\\pi_1, s \\circ \\pi_2)(a,b) = \\begin{pmatrix} 1 & 0 \\\\ 0 & 2b \\end{pmatrix}.$$ You didn't write out what $Dp$ is either. You should find that $$Dp(a,b) = (b,a).$$ Putting this all together, you have that \\begin{align*} Df(a,b) & = D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b)) \\circ Dp((\\pi_{1}, s\\circ \\pi_{2})(a,b)) \\circ D(\\pi_1, s \\circ \\pi_2)(a,b) \\\\ & = D(\\sin)(ab^2) \\circ Dp(a,b^2) \\circ D(\\pi_1, s \\circ \\pi_2)(a,b) \\\\ & = \\cos(ab^2) (b^2, a) \\begin{pmatrix} 1 & 0 \\\\ 0 & 2b \\end{pmatrix} \\\\ & = \\cos(ab^2) (b^2, 2ab) \\\\ & = (b^2 \\cos(ab^2), 2ab \\cos(ab^2)), \\end{align*} as desired. - So this confirms it. Thanks Henry. – roo Aug 9 '12 at 22:49 But this leads me to another issue which I will edit my post with momentarily. – roo Aug 9 '12 at 23:00 Sorry that took so long. I hope my question is still clear. – roo Aug 9 '12 at 23:38 This is a typo in the book. You can see in the Remark after the chain rule theorem $2$-$2$ that that composition symbol was supposed to be a multiplication dot. - I interpreted that remark in the following way: $Df$ is a linear transformation satisfying the equation", "= \\frac{\\sin B}{6.1}, B & = 84.8^\\circ \\\\ 180^\\circ - 32.6^\\circ - 84.8^\\circ, C & = 62.6^\\circ\\end{align*} (d) No solution (e) Both cases need angle \\begin{align*}C\\end{align*} and side \\begin{align*}c\\end{align*}. \\begin{align*}\\text{Case}\\ 1: C & = 180^\\circ - 47^\\circ - 77.2^\\circ = 55.8^\\circ, \\frac{\\sin 55.8^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 50.9 \\\\ \\text{Case}\\ 2: C & = 180^\\circ - 47^\\circ - 102.8^\\circ = 30.2^\\circ,\\frac{\\sin 30.2^\\circ}{c} = \\frac{\\sin 47^\\circ}{45}, c = 30.95\\end{align*} 2. To find the area of the rhombus, use the formula \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} and then multiply that by 2. We first need to find one of the angles that are opposite the given diagonal (they are both the same measurement). We will call it angle \\begin{align*}A\\end{align*}. \\begin{align*}21.5^2 = 12^2 + 12^2 - 2(12)(12) \\sin A, A = 127.2^\\circ\\end{align*}, which means the other two angles are both \\begin{align*}52.8^\\circ (360^\\circ - 127.2^\\circ - 127.2^\\circ\\end{align*} and then divide by 2). \\begin{align*}K = 2 \\left (\\frac{1}{2}(12)(12) \\sin 127.2^\\circ \\right ) = 114.7\\end{align*} 3. Divide the pentagon into three triangles, drawing segments from \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{5}\\end{align*}, called \\begin{align*}x\\end{align*} below, and \\begin{align*}\\angle{2}\\end{align*} to \\begin{align*}\\angle{4}\\end{align*}, called \\begin{align*}y\\end{align*} below. With these three triangles, only the middle triangle needs us to find two sides and the angle between them (called \\begin{align*}\\angle{Z}\\end{align*} below) to use \\begin{align*}K = \\frac{1}{2} \\ bc \\sin A\\end{align*} (the outer" ]

[ "simplify tan 67.5?...", "+ 4$ 4. Simplify And that's it", "{{25}^{\\circ }}-\\tan {{20}^{\\circ }}\\ \\ and\\ \\ \\tan {{24}^{\\circ }}\\tan {{21}^{\\circ }}=1-\\tan {{24}^{\\circ }}-\\tan {{21}^{\\circ }}\\$", "or $30^\\circ$ it gives $0.5$, and for close to $180^\\circ$ or $0^\\circ$ it gives close to $0$. - it is simple and robust. i never think this. thank you very much. – niro Sep 1 '11 at 14:50", "mentioned above, you shouldn't say that $\\tan(90^\\circ)$ is infinite but you should really say that it's undefined (careful, though, if a curious student enters $\\tan(\\pi/2)$ into a certain free online calculator, they'll get something like $1.63317787 \\times 10^{16}$). • Using the double angle prove that $\\sin(90^\\circ)=1$ • Using the Pythagorean identity, prove $\\cos(90^\\circ)=0$. • By evaluating $\\tan(x)$ using the definition at $x=89^\\circ,\\, 89.9^\\circ,\\, 89.99^\\circ,\\dots$ and at $x=91^\\circ,\\, 90.1^\\circ,\\,90.01^\\circ,\\dots$ try to explain a little more about they sense in which $\\tan(90^\\circ)$ is undefined. Now try to explain it using the unit circle.", "1 L 8 { [ { 4 2T 1... ##### Simplify $4 \\times(8-3 \\times 2) \\div 2$ A. 20 B. 5 C. 4 D. 16 Simplify $4 \\times(8-3 \\times 2) \\div 2$ A. 20 B. 5 C. 4 D. 16...", "+0 0 220 1 Simplify $\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$ Mar 31, 2020 I think $$\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$$equals to $$\\frac{\\sqrt{3}}{4}$$. I have heard that you can use caculator to symplifiy this! But I still think im right.", "\\dfrac{ v_y }{ v_x } \\bigg ) \\ = \\ tan^{-1} \\bigg ( \\dfrac{ 9.8 }{ 6.25 } \\bigg ) \\ = \\ 57.47^{ \\circ }$", "# Just simplify everything, right? Geometry Level 4 In the domain $[ 0^ \\circ, 360 ^ \\circ ],$ how many solutions are there to $\\frac {\\tan 2x- \\tan x}{1+\\tan 2x\\tan x}=1?$ ×", "$$\\tan 54^\\circ = \\tan (30^\\circ + 24^\\circ) = \\frac{\\sin(30^\\circ+24^\\circ)}{\\cos(30^\\circ+24^\\circ)}$$", "# How do you multiply and simplify (1+tanx)^2? ${\\left(1 + \\tan x\\right)}^{2} = \\left(1 + \\tan x\\right) \\left(1 + \\tan x\\right)$---> FOIL $= 1 + \\tan x + \\tan x + {\\tan}^{2} x$ $= 1 + 2 \\tan x + {\\tan}^{2} x$", "7)^9(21x^2 - 126x + 173)$ - Your simplify routine seems to add a few too many minus signs. – Thomas Andrews Feb 15 '13 at 19:03", "Apr 21 '13 at 20:56 Hint: $z=x+iy \\to tan\\phi=\\frac{y}{x}$", "# $\\large\\frac{\\tan 80^{\\circ} -\\tan 10^{\\circ}}{\\tan 70^{\\circ}}$ is equal to $\\begin {array} {1 1} (a)\\;0 & \\quad (b)\\;1 \\\\ (c)\\;2 & \\quad (d)\\;3 \\end {array}$ (c) 2", "1. Simplification What does $-2 = x \\tan 45 ^\\circ - \\frac {9.8 \\times x^2}{2 \\times 14^2}$ give for the value of $x$? I did this but am not getting the correct value of $x$ so can someone show me the value of $x$ that you are obtaining? 2. Hello, There are two possible values for x... You can simplify tan(45) into 1. So it yields -784=392x-9.8x² Edit : x=4.19 or x=-1.91 3. Originally Posted by Moo Hello, There are two possible values for x... You can simplify tan(45) into 1. So it yields -784=392x-9.8x² Edit : x=4.19 or x=-1.91 Ermm i did get -1.91 but got 41.9 for the other value Edit : @ Air, to check ur answers u can simply replace the value of x u obtained back in the formula to check if it holds 4. Originally Posted by Kai Ermm i did get -1.91 but got 41.9 for the other value Edit : @ Air, to check ur answers u can simply replace the value of x u obtained back in the formula to check if it holds Yes, the point didn't want to got at its correct place Thanks", "1. $t^0$ 2. $t^-7$ 3. $1/t^4$ 4. -4 Simplify: $(x^10)/(8x^5).$ 1. $(x^5)/8$ 2. $(1)/(8x^5)$ 3. $8x^5$ 4. $(8)/(x^5)$ Simplify $x^2 * x^4$. 1. $x^8$ 2. $x^6$ 3. $8x^2$ 4. 6x 5. $x^2$ Simplify $(2r)^2$. 1. $2^2 + r^2$ 2. $4r$ 3. $4r^2$ 4. $16$", "$sin(140^\\circ) = 0.6428$, $cos(140^\\circ) = -0.7660$, and $tan(140^\\circ) = -0.8391$.", "# Trigonometric Sum with a Twist Algebra Level pending Find the exact value of the following sum $\\frac{1}{\\sqrt{3} - \\tan(20^\\circ)} +\\frac{1}{\\sqrt{3} + \\tan(40^\\circ)}+\\frac{1}{\\sqrt{3} - \\tan(80^\\circ)}.$ ×", "\\circ (1~3~2) = (1~3~2)$ which is equivalent to (3 2 1). -Dan • November 30th 2013, 05:17 AM Stormey Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1). Thanks Dan! • November 30th 2013, 05:44 AM topsquark", "by $2(\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ))$. However, if we write these roots as $r_1,...,r_6$, then we also have $$\\sum_i r_i^2 = \\left(\\sum_i r_i\\right)^2 - 2\\sum_{i < j} r_ir_j$$ But $\\sum_i r_i$ is the coefficient of $x^5$ in the above equation, namely zero, and $\\sum_{i < j} r_ir_j$ is the coefficient of $x^4$, namely $-33$. So you get $$2(\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ)) = -2\\times-33$$ So we conclude that $$\\tan^2(20^\\circ) + \\tan^2(40^\\circ) + \\tan^2(80^\\circ) = 33$$" ]

[ "{{25}^{\\circ }}-\\tan {{20}^{\\circ }}\\ \\ and\\ \\ \\tan {{24}^{\\circ }}\\tan {{21}^{\\circ }}=1-\\tan {{24}^{\\circ }}-\\tan {{21}^{\\circ }}\\$", "– 2m\\tan{25^{\\circ}} &= \\tan{25^{\\circ}} – 2 \\\\ m + 2m\\tan{25^{\\circ}} &= 2 – \\tan{25^{\\circ}} \\\\ m(1 + 2\\tan{25^{\\circ}}) &= 2 – \\tan{25^{\\circ}} \\\\ m &= \\frac{2 – \\tan{25^{\\circ}}}{1 + 2\\tan{25^{\\circ}}} \\\\ m &= 0.8 \\color{red} \\cdots (1) \\\\ \\frac{2-m}{1+2m} &= -\\tan{25^{\\circ}} \\\\ 2-m &= -\\tan{25^{\\circ}} – 2m\\tan{25^{\\circ}} \\\\ -m + 2m\\tan{25^{\\circ}} &= -\\tan{25^{\\circ}} – 2 \\\\ 2m\\tan{25^{\\circ}} – m &= \\tan{25^{\\circ}} + 2 \\\\ m(2\\tan{25^{\\circ}} – 1) &= \\tan{25^{\\circ}} + 2 \\\\ m &= \\frac{\\tan{25^{\\circ}} + 2}{2\\tan{25^{\\circ}} – 1} \\\\ m &= 36.6 \\color{red} \\cdots (2) \\\\ \\therefore m &= 0.8 \\text{ or } m = 36.6 &\\color{red} \\text{by (1) and (2)} \\\\ \\end{aligned} \\\\", "=0$: Ref angle: $\\scriptsize \\theta ={{0}^\\circ}$ General solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos \\theta =1$: Ref angle: $\\scriptsize \\theta ={{0}^\\circ}$ General solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ Overall general solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ 4. . \\scriptsize \\begin{align*}5\\tan 2x-1 & ={{\\tan }^{2}}2x+5\\\\\\therefore {{\\tan }^{2}}2x-5\\tan 2x+6 & =0\\\\\\therefore (\\tan 2x-3)(\\tan 2x-2) & =0\\\\\\therefore \\tan 2x=3\\text{ } & \\text{or }\\tan 2x=2\\end{align*} $\\scriptsize \\tan 2x=3$: Ref angle: $\\scriptsize 2x={{71.6}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & ={{71.57}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{35.8}^\\circ}+k{{.90}^\\circ},k\\in \\mathbb{Z}\\end{align*} $\\scriptsize \\tan 2x=2$: Ref angle: $\\scriptsize 2x={{63.4}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & ={{63.4}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{31.71}^\\circ}+k{{.90}^\\circ},k\\in \\mathbb{Z}\\end{align*} 2. . 1. . \\scriptsize \\begin{align*}\\text{LHS}&=\\sin x\\cos 3x-\\cos x\\sin 3x\\\\&=\\sin x\\cos (2x+x)-\\cos x\\sin (2x+x)\\\\&=\\sin x\\left( {\\cos 2x\\cos x-\\sin 2x\\sin x} \\right)-\\cos x\\left( {\\sin 2x\\cos x+\\cos 2x\\sin x} \\right)\\\\&=\\sin x\\left( {\\left( {2{{{\\cos }}^{2}}x-1} \\right)\\cos x-2\\sin x\\cos x\\sin x} \\right)-\\cos x\\left( {2\\sin x\\cos x\\cos x+(2{{{\\cos }}^{2}}x-1)\\sin x} \\right)\\\\&=\\sin x\\left( {2{{{\\cos }}^{3}}x-\\cos x-2{{{\\sin }}^{2}}x\\cos x} \\right)-\\cos x\\left( {2\\sin x{{{\\cos }}^{2}}x+2\\sin x{{{\\cos }}^{2}}x-\\sin x} \\right)\\\\&=2\\sin x{{\\cos }^{3}}x-\\sin x\\cos x-2{{\\sin }^{3}}x\\cos x-2\\sin x{{\\cos }^{3}}x-2\\sin x{{\\cos }^{3}}x+\\sin x\\cos x\\\\&=-2\\sin x{{\\cos }^{3}}x-2{{\\sin }^{3}}x\\cos x\\\\&=-2\\sin x\\cos x({{\\cos }^{2}}x+{{\\sin }^{2}}x)\\\\&=-\\sin 2x\\end{align*} Therefore: \\scriptsize \\begin{align*}-\\sin 2x & =\\tan {{140}^\\circ}\\\\\\therefore \\sin 2x & =-\\tan {{140}^\\circ}\\\\ & =0.839\\end{align*} Ref angle: $\\scriptsize 2x={{57.05}^\\circ}$ Sine is positive in the first and second quadrants. $\\scriptsize 2x={{180}^\\circ}-{{57.05}^\\circ}={{122.95}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & =57.05+k{{.360}^\\circ}\\text{ or 2}x={{122.95}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{28.525}^\\circ}+k{{.180}^\\circ}\\text{", "# $\\large\\frac{\\tan 80^{\\circ} -\\tan 10^{\\circ}}{\\tan 70^{\\circ}}$ is equal to $\\begin {array} {1 1} (a)\\;0 & \\quad (b)\\;1 \\\\ (c)\\;2 & \\quad (d)\\;3 \\end {array}$ (c) 2", "Question # Prove that $tan {15^ \\circ } + \\tan {30^ \\circ } + \\tan {15^ \\circ }\\tan {30^ \\circ } = 1$ Verified 163.5k+ views Hint: Here use trigonometric identities and formulas on the LHS part of the equation and simplify using basic trigonometric angles to prove LHS=RHS. We know that $tan {45^ \\circ } = 1$ We can write $tan {45^ \\circ } = \\tan ({30^ \\circ } + {15^ \\circ })$ We also know that $tan(A + B) = \\dfrac{{\\tan A + \\tan B}}{{1 - \\tan A\\tan B}}$ By using this we can write $\\Rightarrow \\tan ({30^ \\circ } + {15^ \\circ }) = \\dfrac{{\\tan {{30}^ \\circ } + \\tan {{15}^ \\circ }}}{{1 - \\tan {{30}^ \\circ }\\tan {{15}^ \\circ }}} = \\tan {45^ \\circ } = 1$ By solving above equation $\\Rightarrow \\tan {30^ \\circ } + \\tan {15^ \\circ } = 1 - \\tan {30^ \\circ }\\tan {15^ \\circ }$ By rearranging the equation we get the result $\\Rightarrow \\tan {30^ \\circ } + \\tan {15^ \\circ } + \\tan {30^ \\circ }\\tan {15^ \\circ } = 1$ Hence Proved Note: - In such a type of question always try to apply the formula of $tan(A + B)$ or $tan(A - B)$ and put the angles that are given in question so we", "## tan A = tan B + tan C + tan D $\\tan \\, 84^\\circ \\; = \\; tan \\, 78^\\circ \\; + \\; tan \\, 72^\\circ \\; + \\; tan \\, 60^\\circ$ $\\tan \\, 80^\\circ \\; = \\; tan \\, 70^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 50^\\circ$ Find other 4-tuples $(a, b, c, d)$ of distinct integers between $0^\\circ$ and $90^\\circ$ that satisfy the relation $tan \\, A^\\circ\\; = \\; tan \\, B^\\circ \\; + \\; tan \\, C^\\circ \\; + \\; tan \\, D^\\circ$ Other solutions: $\\tan \\, 72^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 36^\\circ \\; + \\; tan \\, 6^\\circ$ $\\tan \\, 70^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 40^\\circ \\; + \\; tan \\, 10^\\circ$ $\\tan \\, 72^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 24^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 36^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 72^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 36^\\circ$ How many such equations can you produce if we allow repeated angles? Hint: there are 49 such equations Determine", "\\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 20{}^\\circ +\\tan (90{}^\\circ -20{}^\\circ )\\tan 50{}^\\circ \\tan 20{}^\\circ$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 50{}^\\circ +\\cot 20{}^\\circ \\tan 50{}^\\circ \\tan 20{}^\\circ$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 20{}^\\circ +\\frac{1}{{\\tan 20{}^\\circ }}(\\tan 50{}^\\circ \\tan 20{}^\\circ )$ $\\displaystyle \\therefore \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =2\\tan 50{}^\\circ +\\tan 20{}^\\circ$ (iv) $\\displaystyle \\ \\ \\tan 3A=\\tan (2A+A)$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\tan 3A=\\frac{{\\tan 2A+\\tan A}}{{1-\\tan 2A\\tan A}}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\tan 3A-\\tan 3A\\tan 2A\\tan A=\\tan 2A+\\tan A$ $\\displaystyle \\therefore \\ \\ \\ \\ \\tan 3A-\\tan 2A-\\tan A=\\tan 3A\\tan 2A\\tan A$ 8. (i) If $\\displaystyle A+B=45{}^\\circ,$ prove that $\\displaystyle (1+\\tan A)(1+\\tan B)=2.$ (ii) If $\\displaystyle A+B=45{}^\\circ,$ prove that $\\displaystyle (\\cot A - 1)(\\cot B - 1)=2.$ (iii) If $\\displaystyle A-B=45{}^\\circ,$ prove that $\\displaystyle (1+\\tan A)(1+\\tan B)=2\\tan A.$ Show/Hide Solution (i) $\\displaystyle \\ A+B=45{}^\\circ$ $\\displaystyle \\therefore \\ \\ \\ \\tan (A+B)=\\tan 45{}^\\circ =1$ $\\displaystyle \\therefore \\ \\ \\ \\frac{{\\tan A+\\tan B}}{{1-\\tan A\\tan B}}=1$ $\\displaystyle \\begin{array}{l}\\therefore \\ \\ \\ \\tan A+\\tan B=1-\\tan A\\tan B\\\\\\\\\\therefore \\ \\ \\ \\tan A+\\tan B+\\tan A\\tan B=1\\\\\\\\\\therefore \\ \\ \\ 1+\\tan A+\\tan B+\\tan A\\tan B=2\\\\\\\\\\therefore \\ \\ \\ \\left( {1+\\tan", "value of the ratio for that angle. $\\cos 45° = \\frac{1}{\\sqrt{2}}$ $$\\tan 45° =$$ Remember to first identify the hypotenuse, opposite and adjacent sides for the given angle. Then write down the correct fraction for each ratio. You can confirm your answer by using your calculator to find the value of the ratio for that angle. $\\tan 45° = \\frac{1}{1} = 1$ Evaluate the following without using a calculator. Select the closest answer from the list provided. $$\\sin{60^\\circ} - \\tan{60^\\circ}$$ $\\begin{array}{c c c c c } 0 &-\\dfrac{1}{2} &\\dfrac{2}{\\sqrt{3}} &-\\dfrac{\\sqrt{3}}{2} &-\\dfrac{2}{\\sqrt{3}} \\end{array}$ \\begin{align*} \\sin{60^\\circ} - \\tan{60^\\circ} & = \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{3}}{1} \\\\ & = \\frac{\\sqrt{3} - 2\\sqrt{3}}{2} \\\\ & = -\\frac{\\sqrt{3}}{2} \\end{align*} $$\\tan{30^\\circ} - \\cos{30^\\circ}$$ $\\begin{array}{c c c c c } 0 &-\\dfrac{1}{2\\sqrt{3}} &\\dfrac{\\sqrt{3}}{2} &-\\dfrac{2}{\\sqrt{3}} &-\\dfrac{\\sqrt{3}}{2} \\end{array}$ \\begin{align*} \\tan{30^\\circ} - \\cos{30^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{2 \\sqrt{3}} \\\\ & = -\\frac{1}{2\\sqrt{3}} \\end{align*} $$\\tan{60^\\circ} - \\sin{60^\\circ} - \\tan{60^\\circ}$$ $\\begin{array}{c c c c c} -\\dfrac{\\sqrt{3}}{2} &-\\dfrac{\\sqrt{3}}{1} &-\\dfrac{1}{2} &-\\dfrac{1}{1} &-\\dfrac{1}{\\sqrt{2}} \\end{array}$ \\begin{align*} \\tan{60^\\circ} - \\sin{60^\\circ} - \\tan{60^\\circ} & = \\frac{\\sqrt{3}}{1} - \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{3}}{1} \\\\ & = \\frac{2\\sqrt{3} - \\sqrt{3} - 2\\sqrt{3}}{2} \\\\ & = -\\frac{\\sqrt{3}}{2} \\end{align*} $$\\sin{30^\\circ} \\times \\sin{30^\\circ} \\times \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} \\dfrac{1}{2} &\\dfrac{1}{2\\sqrt{3}} &\\dfrac{1}{8} &\\dfrac{1}{4} &\\dfrac{\\sqrt{3}}{4\\sqrt{2}} \\end{array}$ \\begin{align*} \\sin{30^\\circ} \\times \\sin{30^\\circ} \\times \\sin{30^\\circ} & = \\frac{1}{2} \\times \\frac{1}{2} \\times", "let $\\arctan x = 30^{\\circ}$, or in other words $x = \\tan(30^{\\circ}) = 1/\\sqrt3$.", "angle $30^\\circ$, and calculating $\\tan(\\theta) = O/A$ $\\tan(30^\\circ) = O/600$ $600\\tan(30^\\circ) = O$ $346.4 = O$ -", "cm, then $\\angle B = \\underline{\\hspace{2.0cm}}$ OR Two triangles are similar if their corresponding sides are $\\underline{\\hspace{2.0cm}}$ $\\underline{\\angle B = 90^{\\circ}}$ We have to first check if the $\\triangle ABC$ is a right angled triangle. If it is a right angled triangle, then Pythagoras theorem should hold $\\Rightarrow 12^2 = ( 6\\sqrt{3})^2 + (6)^2$ $\\Rightarrow 144 = 108 + 36$ $\\Rightarrow 144 = 144$. Therefore the $\\triangle ABC$ is a right angled triangle. $\\therefore \\angle B = 90^{\\circ}$ OR Two triangles are similar if there corresponding sides are $\\underline{proportional}$. If $\\triangle ABC$ and $\\triangle PQO$ are similar, then $\\frac{AB}{PQ}$ $=$ $\\frac{BC}{QR}$ $=$ $\\frac{AC}{PR}$ $\\\\$ Question 14: The value of $( \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 89^{\\circ})$ is equal to $\\underline{\\hspace{2.0cm}}$ $\\underline{\\hspace{0.5cm} 1 \\hspace{0.5cm} }$ $\\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 89^{\\circ}$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\tan 46^{\\circ} \\ldots\\tan 89^{\\circ}$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\tan (90^{\\circ} - 44^{\\circ}) \\ldots \\tan (90^{\\circ} - 1^{\\circ})$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\cot 44^{\\circ} \\ldots \\cot 1^{\\circ}$ $= \\tan 45^{\\circ} = 1$ $\\\\$ Question 15: In Fig. 3, the angles of depressions from the observing positions $O_1$ and $O_2$ respectively of the object $A$ are $\\underline{\\hspace{2.0cm}}$ $\\underline{30^{\\circ} \\& 45^{\\circ}}$ Therefore the angle of depression are", "}}=\\tan 62{}^\\circ .$ (ii) $\\displaystyle \\tan 50{}^\\circ =\\tan 40{}^\\circ +2\\tan 10{}^\\circ .$ (iii) $\\displaystyle \\tan 70{}^\\circ =2\\tan 50{}^\\circ +\\tan 20{}^\\circ .$ (iv) $\\displaystyle \\tan 3A-\\tan 2A-\\tan A=\\tan 3A\\tan 2A\\tan A.$ Show/Hide Solution (i) $\\displaystyle \\ \\ \\ \\ \\frac{{\\cos 17{}^\\circ +\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ -\\sin 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{\\displaystyle \\frac{{\\cos 17{}^\\circ }}{{\\cos 17{}^\\circ }}+\\displaystyle \\frac{{\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ }}}}{{\\displaystyle \\frac{{\\cos 17{}^\\circ }}{{\\cos 17{}^\\circ }}-\\displaystyle \\frac{{\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ }}}}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{1+\\tan 17{}^\\circ }}{{1+\\tan 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{\\tan 45{}^\\circ +\\tan 17{}^\\circ }}{{1+\\tan 45{}^\\circ \\tan 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\tan (45{}^\\circ +17{}^\\circ )$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\tan 62{}^\\circ$ (ii) $\\displaystyle \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\tan (40{}^\\circ +10{}^\\circ )$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\frac{{\\tan 40{}^\\circ +\\tan 10{}^\\circ }}{{1-\\tan 40{}^\\circ \\tan 10{}^\\circ }}$ $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ -\\tan 50{}^\\circ \\tan 40{}^\\circ \\tan 10{}^\\circ =\\tan 40{}^\\circ +\\tan 10{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\tan 40{}^\\circ +\\tan 10{}^\\circ +\\tan 50{}^\\circ \\tan 40{}^\\circ \\tan 10{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "= \\dfrac{x}{ \\cot \\beta - \\cot \\alpha}$ Solution Express $75^{\\circ}$ as a sum of two special angles. $75^{\\circ} = 30^{\\circ} + 45^{\\circ}$ Hence $\\tan 75^{\\circ} = \\tan (30^{\\circ} + 45^{\\circ}) = \\dfrac{\\tan 30^{\\circ}+\\tan 45^{\\circ}}{1 - \\tan 30^{\\circ}\\tan 45^{\\circ}}$ Knowing that $\\tan 30^{\\circ} = \\dfrac{\\sqrt{3}}{3}$ and $\\tan 45^{\\circ} = 1$, evaluate: $\\tan 75^{\\circ} = \\dfrac{\\dfrac{\\sqrt{3}}{3}+1}{1-\\dfrac{\\sqrt{3}}{3}} = 2+\\sqrt{3}$ Solution $\\alpha$ is an angle in standard position and therefore: $\\quad \\sin \\alpha = \\dfrac{y}{-2} = \\dfrac{1}{5}$ Solve the above equation for $y$ to obtain: $\\quad y = - 2/5$ Solution Since $\\sec \\alpha \\gt 0$, $\\cos \\alpha = \\dfrac{1}{\\sec \\alpha} \\gt 0$ $\\cos \\alpha = \\sqrt {1-\\sin^2 \\alpha} = \\dfrac{3\\sqrt{15}}{16}$ Hence: $\\cot \\alpha = \\dfrac{ \\cos \\alpha}{ \\sin \\alpha} = \\dfrac{ \\dfrac{3\\sqrt{15}}{16}}{ - 11/16} = -\\dfrac{3\\sqrt{15}}{11}$ Solution The $x$ coordinate of the given point $(2,k)$ is positive and therefore $\\quad \\cos \\beta = \\dfrac{2}{\\sqrt{2^2+k^2}}$ is always positive. The signs of $\\tan \\beta = \\dfrac{k}{2}$ and $\\sin \\beta = \\dfrac{k}{\\sqrt{2^2+k^2}}$ depends on $k$, hence A) and B) are not right answers. Since $\\quad \\cos \\beta \\gt 0$, $\\sec \\beta \\gt 0$, hence D) is not the right answer. $\\sin \\beta + \\cos \\beta = 1$ is valid for $k = 0$ only and therefore E) is not the right answer. The identity $\\cos^2 \\beta + \\sin^2 \\beta = 1$ gives $\\cos \\beta", "60^o \\cot 78^o$ $\\displaystyle = \\cot 60^o = \\frac{1}{\\sqrt{3}}$ $\\displaystyle \\text{ix) } \\tan 5^o \\tan 25^o \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\tan (90^o-85^o) \\tan (90^o-65^o) \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\cot 85^o \\cot 65^o \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\tan 30^o = \\frac{1}{\\sqrt{3}}$ $\\displaystyle \\\\$ Question 4: Express each of the following in terms of trigonometric ratios of angles between $\\displaystyle 0$ and $\\displaystyle 45^o$. $\\displaystyle \\text{i) }\\sin 85^o + \\mathrm{cosec} 85^o = \\sin (90^o-5^o) + \\mathrm{cosec} (90^o-5^o) = \\cos 5^o + \\sec 5^o$ $\\displaystyle \\text{ii) } \\mathrm{cosec} 69^o + \\cot 69^o = \\mathrm{cosec} (90^o-21^o) + \\cot (90^o-21^o) = \\sec 21^o + \\tan 21^o$ $\\displaystyle \\\\$ Question 5: Prove that: $\\displaystyle \\text{i) }\\tan 1^o \\tan 2^o \\tan 3^o \\cdots \\tan 89^o = 1$ $\\displaystyle \\text{ii) }\\cos 1^o \\cos 2^o \\cos 3^o \\cdots \\cos 180^o = 0$ i) $\\displaystyle \\text{LHS} = \\tan 1^o \\tan 2^o \\tan 3^o \\cdots \\tan 89^o$ $\\displaystyle = \\tan (90^o-89^o) \\tan (90^o-88^o) \\tan (90^o-87^o) \\cdots \\tan 87^o \\tan 88^o \\tan 89^o$ $\\displaystyle = \\cot 89^o \\cot 88^o \\cot 87^o \\cdots \\tan 87^o \\tan 88^o \\tan 89^o$ $\\displaystyle = 1 \\times 1 \\times 1 \\cdots \\times 1 = 1 = RHS$ Hence Proved. ii) $\\displaystyle \\text{LHS} = \\cos 1^o \\cos 2^o \\cos 3^o \\cdots \\cos 180^o$ $\\displaystyle", "to $\\tan (A + B) = \\frac{\\tan A + \\tan B}{1 - \\tan A\\tan B}$ We have shown that addition formula is true when angles involved are acute angles. The same proof can be applied to prove the results for all values of $A$ and $B$. Consider $A' = 90^{\\circ} + A \\therefore `\\sin A' = \\cos A$ and $\\cos A' = \\sin A$ $\\sin(A' + B) = \\cos (A + B) = \\cos A\\cos B - \\sin A\\sin B = \\sin A'\\cos B + \\cos A'\\sin B$ Similarly $\\cos(A' + B) = -\\sin(A + B) = -\\sin A\\cos B - \\sin B\\cos A = \\cos A'\\cos B - \\sin A'\\sin B$ We can prove it again for $B' = 90^{\\circ} + B$ and so on by increasing the values of $A$ and $B$. Then we can again increase values by $90^{\\circ}$ and proceeding this way we see that the formula holds true for all values of $A$ and $B$. ## 8.2. The Subtraction Formula¶ $\\sin(A - B) = \\sin A\\cos B - \\sin B\\cos A$ $\\cos(A - B) = \\cos A\\cos B + \\sin A\\sin B$ $\\tan(A - B) = \\frac{\\tan A - \\tan B}{1 + \\tan A\\tan B}$ Conside the diagram above. The angle $MOP$ is $A - B.$ We take a point $P,$ and draw", "DBE)-\\tan^2(\\alpha) &=& \\left(1-\\tan^2(\\angle DBE)\\tan^2(\\alpha) \\right)\\tan^2(\\angle DBE) \\\\ -\\tan^2(\\alpha) &=& -\\tan^4(\\angle DBE)\\tan^2(\\alpha) \\\\ \\tan^4(\\angle DBE)\\tan^2(\\alpha) -\\tan^2(\\alpha) &=& 0 \\\\ \\underbrace{ \\tan^2(\\alpha)}_{\\neq 0}\\underbrace{(\\tan^4(\\angle DBE)-1)}_{=0} &=& 0 \\\\ \\tan^4(\\angle DBE)-1 &=& 0 \\\\ \\tan^4(\\angle DBE) &=& 1 \\\\ \\tan (\\angle DBE) &=& 1 \\\\ \\angle DBE &=& \\arctan(1) \\\\ \\mathbf{\\angle DBE} &=& \\mathbf{45^\\circ} \\\\ \\hline \\end{array}$$ $$2.\\ \\mathbf{ AP}:$$ $$\\begin{array}{|rcll|} \\hline \\cot (\\angle DBE) &=& b \\quad | \\quad \\angle DBE = 45^\\circ \\\\ \\cot (45^\\circ) &=& b \\\\ 1 &=& b \\\\\\\\ \\cot (\\angle CBE)=\\cot (\\angle DBE-\\alpha)=\\cot (45^\\circ-\\alpha) = b+d&=&1+b \\\\ \\cot (\\angle DBC)=\\cot(\\alpha) = b+2d=1+2d \\\\\\\\ \\cot (\\angle DBE) &=& 1 \\\\ \\cot (45^\\circ-\\alpha) &=& 1+d \\\\ \\cot (\\alpha) &=& 1+2d \\\\\\\\ 2(1+d) &=& 1+(1+2d) \\\\ 2\\cot (\\angle 45^\\circ-\\alpha) &=& 1+ \\cot (\\alpha) \\\\ && \\boxed{ \\text{Formula}\\\\ \\cot(x-y)=\\dfrac{\\cot(x)\\cot(y)+1}{\\cot(y)-\\cot(x)}\\\\ \\cot(45^\\circ-\\alpha)=\\dfrac{\\cot(45^\\circ)\\cot(\\alpha)+1}{\\cot(\\alpha)-\\cot(45^\\circ)}\\\\ =\\cot(45^\\circ-\\alpha)=\\dfrac{1*\\cot(\\alpha)+1}{\\cot(\\alpha)-1}\\\\ =\\cot(45^\\circ-\\alpha)=\\dfrac{\\cot(\\alpha)+1}{\\cot(\\alpha)-1} } \\\\\\\\ 2\\left(\\dfrac{\\cot(\\alpha)+1}{\\cot(\\alpha)-1}\\right) &=& 1+ \\cot (\\alpha) \\\\ (1+ \\cot (\\alpha))(\\cot(\\alpha)-1)&=& 2 (\\cot(\\alpha)+1)\\\\ (1+ \\cot (\\alpha))((\\cot(\\alpha)-1)-2 (\\cot(\\alpha)+1)&=& 0\\\\ (1+ \\underbrace{\\cot(\\alpha)}_{\\neq-1} ) ( \\underbrace{\\cot(\\alpha)-3}_{=0} ) &=& 0\\\\ \\cot(\\alpha)-3 &=& 0 \\\\ \\mathbf{\\cot(\\alpha)} &=& \\mathbf{3} \\\\ \\hline \\end{array}$$ $$\\mathbf{BD=\\ ?}$$ $$\\begin{array}{|rcll|} \\hline \\cos(\\angle DBE) &=& \\dfrac{BD}{BE} \\\\ \\cos(45^\\circ) &=& \\dfrac{BD}{10} \\\\ BD &=& 10\\cos(45^\\circ) \\\\ BD &=& 10\\dfrac{\\sqrt{2}}{2} \\\\ \\mathbf{BD} &=& \\mathbf{5 \\sqrt{2} } \\\\ \\hline \\end{array}$$ $$\\mathbf{AC=\\ ?}$$ $$\\begin{array}{|rcll|} \\hline \\cot(\\alpha)&=& \\dfrac{BD}{DC}\\times \\dfrac{2}{2} \\\\\\\\ \\cot(\\alpha)&=& \\dfrac{2BD}{2DC} \\quad | \\quad 2DC = AC \\\\\\\\ \\cot(\\alpha)&=& \\dfrac{2BD}{AC} \\\\\\\\ AC&=& \\dfrac{2BD}{\\cot(\\alpha)} \\quad", "Go to content These formulae are often refered to as the addition formulae: \\sin(A \\pm B) = \\sin A \\cos B \\pm \\cos A \\sin B \\cos(A \\pm B) = \\cos A \\cos B \\mp \\sin A \\sin B \\tan(A \\pm B) = \\displaystyle \\frac{\\tan A \\pm \\tan B}{1 \\mp \\tan A \\tan B}", "-1/\\tan 15^\\circ$$ Now, $\\tan 2x = 2\\tan x / (1-\\tan^2 x)$ Put $x=15^\\circ$ in the above equation and calculate $\\tan 15^\\circ$ from here and put in $$\\tan 105^\\circ = -1/\\tan15^\\circ$$ Do the same thing for your other question. • Thanks @Micah, you have edited my response in such a beautiful fashion. I am new here, so, i dont know how to write mathematical equations here. – th ie Jul 20 '15 at 20:16", "10^\\circ\\sin\\alpha\\right)\\;,$$ $$\\left(4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ\\right)\\sin\\alpha=\\sqrt{3}\\cos\\alpha\\;,$$ $$\\tan\\alpha=\\cfrac{\\sqrt{3}}{4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ}\\;.$$ Hence, $$\\angle ADB=\\alpha=\\arctan\\left(\\cfrac{\\sqrt{3}}{4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ}\\right)\\simeq\\\\\\simeq 77,3361794^\\circ.$$ • He he, it is like my picture. – Aqua Aug 29 '20 at 18:23 • You are right, I do not have a program to draw, so I have used your drawing. I am sorry. Aug 29 '20 at 22:01 • If you don't have a program, but want to post an answer, you can take a picture of a real drawing. 😊 Aug 30 '20 at 4:46 • But I am not good at drawing. Aug 30 '20 at 6:49", "+ 40)tan 30 Cross multiplication. x = x tan 30 + 40 tan 30 x- x tan 30 = 40 tan 30 x = $\\frac{40tan30}{1-tan30}$ ≈ 54.6 ft." ]

[ "{{25}^{\\circ }}-\\tan {{20}^{\\circ }}\\ \\ and\\ \\ \\tan {{24}^{\\circ }}\\tan {{21}^{\\circ }}=1-\\tan {{24}^{\\circ }}-\\tan {{21}^{\\circ }}\\$", "1. ## Double-angle Formulae Having a bit of trouble with this one. Can anyone help me out? Many thanks. Q. In the triangle pqr (see attachment), $\\displaystyle |\\angle qrp|=90^{\\circ}$ & |rp| = h. s is a point on [qr] such that $\\displaystyle |\\angle spq|=2B$ & $\\displaystyle |\\angle rps|=45^{\\circ}-B$, $\\displaystyle 0^{\\circ}<B<45^{\\circ}$ . If $\\displaystyle |sr|=h\\tan(45-B)$, show that $\\displaystyle |qs|=2h \\tan B$. Attempt: $\\displaystyle \\frac{|qr|}{h}=\\tan(2B+45-B)$ $\\displaystyle |qr|=h\\tan(45+B)$ $\\displaystyle |qr|-|sr|=h\\tan(45+B)-h\\tan(45-B)$ $\\displaystyle |qs|=\\frac{h\\tan45+h\\tan B}{1-\\tan45\\tan B}-(\\frac{h\\tan45-h\\tan B}{1+\\tan45\\tan B})$ $\\displaystyle |qs|=\\frac{2h\\tan B}{1-\\tan^245\\tan B}$ 2. ## Re: Double-angle Formulae Hello, GrigOrig99! By the way, don't you know that $\\displaystyle \\tan45^o = 1\\,?$ Having a bit of trouble with this one. No wonder . . . the problem is stated incorrectly. $\\displaystyle \\text{In }\\Delta PQR,\\,\\angle QRP=90^o\\,\\text{ and }\\,RP = h.\\;S\\text{ is a point on }QR$ $\\displaystyle \\text{ such that }\\angle SPQ= 2B\\text{ and }\\angle RPS =45^o-B,\\;0^o<B<45^o$ $\\displaystyle \\text{If }SR=a=h\\tan(45-B),\\,\\text{ show that: }\\:x=QS = 2h\\tan ({\\color{red}2}B)$. Code: o P * * | *2B * | * * 45-B| * * | h * * | * * | * * | Q o * * * * * * * o * * * * * * * o R x S a Note that $\\displaystyle \\angle QPR \\:=\\:2B + 45^o-B \\:=\\:45^o+B$ In $\\displaystyle \\Delta QPR\\!:\\;\\tan(45+B) \\,=\\,\\frac{x+a}{h} \\quad\\Rightarrow\\quad x+a \\:=\\:h\\tan(45+B)$ .[1] In $\\displaystyle \\Delta SPR\\!:\\;\\tan(45-B) \\,=\\,\\frac{a}{h} \\quad\\Rightarrow\\quad", "=0$: Ref angle: $\\scriptsize \\theta ={{0}^\\circ}$ General solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos \\theta =1$: Ref angle: $\\scriptsize \\theta ={{0}^\\circ}$ General solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ Overall general solution: $\\scriptsize \\theta ={{0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ 4. . \\scriptsize \\begin{align*}5\\tan 2x-1 & ={{\\tan }^{2}}2x+5\\\\\\therefore {{\\tan }^{2}}2x-5\\tan 2x+6 & =0\\\\\\therefore (\\tan 2x-3)(\\tan 2x-2) & =0\\\\\\therefore \\tan 2x=3\\text{ } & \\text{or }\\tan 2x=2\\end{align*} $\\scriptsize \\tan 2x=3$: Ref angle: $\\scriptsize 2x={{71.6}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & ={{71.57}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{35.8}^\\circ}+k{{.90}^\\circ},k\\in \\mathbb{Z}\\end{align*} $\\scriptsize \\tan 2x=2$: Ref angle: $\\scriptsize 2x={{63.4}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & ={{63.4}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{31.71}^\\circ}+k{{.90}^\\circ},k\\in \\mathbb{Z}\\end{align*} 2. . 1. . \\scriptsize \\begin{align*}\\text{LHS}&=\\sin x\\cos 3x-\\cos x\\sin 3x\\\\&=\\sin x\\cos (2x+x)-\\cos x\\sin (2x+x)\\\\&=\\sin x\\left( {\\cos 2x\\cos x-\\sin 2x\\sin x} \\right)-\\cos x\\left( {\\sin 2x\\cos x+\\cos 2x\\sin x} \\right)\\\\&=\\sin x\\left( {\\left( {2{{{\\cos }}^{2}}x-1} \\right)\\cos x-2\\sin x\\cos x\\sin x} \\right)-\\cos x\\left( {2\\sin x\\cos x\\cos x+(2{{{\\cos }}^{2}}x-1)\\sin x} \\right)\\\\&=\\sin x\\left( {2{{{\\cos }}^{3}}x-\\cos x-2{{{\\sin }}^{2}}x\\cos x} \\right)-\\cos x\\left( {2\\sin x{{{\\cos }}^{2}}x+2\\sin x{{{\\cos }}^{2}}x-\\sin x} \\right)\\\\&=2\\sin x{{\\cos }^{3}}x-\\sin x\\cos x-2{{\\sin }^{3}}x\\cos x-2\\sin x{{\\cos }^{3}}x-2\\sin x{{\\cos }^{3}}x+\\sin x\\cos x\\\\&=-2\\sin x{{\\cos }^{3}}x-2{{\\sin }^{3}}x\\cos x\\\\&=-2\\sin x\\cos x({{\\cos }^{2}}x+{{\\sin }^{2}}x)\\\\&=-\\sin 2x\\end{align*} Therefore: \\scriptsize \\begin{align*}-\\sin 2x & =\\tan {{140}^\\circ}\\\\\\therefore \\sin 2x & =-\\tan {{140}^\\circ}\\\\ & =0.839\\end{align*} Ref angle: $\\scriptsize 2x={{57.05}^\\circ}$ Sine is positive in the first and second quadrants. $\\scriptsize 2x={{180}^\\circ}-{{57.05}^\\circ}={{122.95}^\\circ}$ General solution: \\scriptsize \\begin{align*}2x & =57.05+k{{.360}^\\circ}\\text{ or 2}x={{122.95}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore x & ={{28.525}^\\circ}+k{{.180}^\\circ}\\text{", "Question # Prove that $tan {15^ \\circ } + \\tan {30^ \\circ } + \\tan {15^ \\circ }\\tan {30^ \\circ } = 1$ Verified 163.5k+ views Hint: Here use trigonometric identities and formulas on the LHS part of the equation and simplify using basic trigonometric angles to prove LHS=RHS. We know that $tan {45^ \\circ } = 1$ We can write $tan {45^ \\circ } = \\tan ({30^ \\circ } + {15^ \\circ })$ We also know that $tan(A + B) = \\dfrac{{\\tan A + \\tan B}}{{1 - \\tan A\\tan B}}$ By using this we can write $\\Rightarrow \\tan ({30^ \\circ } + {15^ \\circ }) = \\dfrac{{\\tan {{30}^ \\circ } + \\tan {{15}^ \\circ }}}{{1 - \\tan {{30}^ \\circ }\\tan {{15}^ \\circ }}} = \\tan {45^ \\circ } = 1$ By solving above equation $\\Rightarrow \\tan {30^ \\circ } + \\tan {15^ \\circ } = 1 - \\tan {30^ \\circ }\\tan {15^ \\circ }$ By rearranging the equation we get the result $\\Rightarrow \\tan {30^ \\circ } + \\tan {15^ \\circ } + \\tan {30^ \\circ }\\tan {15^ \\circ } = 1$ Hence Proved Note: - In such a type of question always try to apply the formula of $tan(A + B)$ or $tan(A - B)$ and put the angles that are given in question so we", "10^\\circ\\sin\\alpha\\right)\\;,$$ $$\\left(4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ\\right)\\sin\\alpha=\\sqrt{3}\\cos\\alpha\\;,$$ $$\\tan\\alpha=\\cfrac{\\sqrt{3}}{4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ}\\;.$$ Hence, $$\\angle ADB=\\alpha=\\arctan\\left(\\cfrac{\\sqrt{3}}{4\\sin 10^\\circ-\\sqrt{3}\\tan 10^\\circ}\\right)\\simeq\\\\\\simeq 77,3361794^\\circ.$$ • He he, it is like my picture. – Aqua Aug 29 '20 at 18:23 • You are right, I do not have a program to draw, so I have used your drawing. I am sorry. Aug 29 '20 at 22:01 • If you don't have a program, but want to post an answer, you can take a picture of a real drawing. 😊 Aug 30 '20 at 4:46 • But I am not good at drawing. Aug 30 '20 at 6:49", "# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 28 The expression $\\frac{\\tan 50{}^\\circ -\\tan 20{}^\\circ }{1+\\tan 50{}^\\circ \\tan 20{}^\\circ }$ is written as $\\tan 30{}^\\circ$ and the exact value of $\\tan 30{}^\\circ$ is $\\frac{1}{\\sqrt{3}}$. #### Work Step by Step Use the difference formula of tangent and rewrite the expression as the difference of angles to obtain the tangent of the angle as, \\begin{align} & \\tan \\left( 50{}^\\circ -20{}^\\circ \\right)=\\frac{\\tan 50{}^\\circ -\\tan 20{}^\\circ }{1+\\tan 50{}^\\circ \\tan 20{}^\\circ } \\\\ & \\tan \\left( 30{}^\\circ \\right)=\\frac{\\tan 50{}^\\circ -\\tan 20{}^\\circ }{1+\\tan 50{}^\\circ \\tan 20{}^\\circ } \\end{align} Therefore, the expression $\\frac{\\tan 50{}^\\circ -\\tan 20{}^\\circ }{1+\\tan 50{}^\\circ \\tan 20{}^\\circ }$ is equivalent to $\\tan 30{}^\\circ$. From the knowledge of trigonometric ratios defined for tangent of an angle, the exact value of $\\tan 30{}^\\circ$ is $\\frac{1}{\\sqrt{3}}$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "\\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 20{}^\\circ +\\tan (90{}^\\circ -20{}^\\circ )\\tan 50{}^\\circ \\tan 20{}^\\circ$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 50{}^\\circ +\\cot 20{}^\\circ \\tan 50{}^\\circ \\tan 20{}^\\circ$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =\\tan 50{}^\\circ +\\tan 20{}^\\circ +\\frac{1}{{\\tan 20{}^\\circ }}(\\tan 50{}^\\circ \\tan 20{}^\\circ )$ $\\displaystyle \\therefore \\ \\ \\ \\ \\ \\ \\tan 70{}^\\circ =2\\tan 50{}^\\circ +\\tan 20{}^\\circ$ (iv) $\\displaystyle \\ \\ \\tan 3A=\\tan (2A+A)$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\tan 3A=\\frac{{\\tan 2A+\\tan A}}{{1-\\tan 2A\\tan A}}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\tan 3A-\\tan 3A\\tan 2A\\tan A=\\tan 2A+\\tan A$ $\\displaystyle \\therefore \\ \\ \\ \\ \\tan 3A-\\tan 2A-\\tan A=\\tan 3A\\tan 2A\\tan A$ 8. (i) If $\\displaystyle A+B=45{}^\\circ,$ prove that $\\displaystyle (1+\\tan A)(1+\\tan B)=2.$ (ii) If $\\displaystyle A+B=45{}^\\circ,$ prove that $\\displaystyle (\\cot A - 1)(\\cot B - 1)=2.$ (iii) If $\\displaystyle A-B=45{}^\\circ,$ prove that $\\displaystyle (1+\\tan A)(1+\\tan B)=2\\tan A.$ Show/Hide Solution (i) $\\displaystyle \\ A+B=45{}^\\circ$ $\\displaystyle \\therefore \\ \\ \\ \\tan (A+B)=\\tan 45{}^\\circ =1$ $\\displaystyle \\therefore \\ \\ \\ \\frac{{\\tan A+\\tan B}}{{1-\\tan A\\tan B}}=1$ $\\displaystyle \\begin{array}{l}\\therefore \\ \\ \\ \\tan A+\\tan B=1-\\tan A\\tan B\\\\\\\\\\therefore \\ \\ \\ \\tan A+\\tan B+\\tan A\\tan B=1\\\\\\\\\\therefore \\ \\ \\ 1+\\tan A+\\tan B+\\tan A\\tan B=2\\\\\\\\\\therefore \\ \\ \\ \\left( {1+\\tan", "60^o \\cot 78^o$ $\\displaystyle = \\cot 60^o = \\frac{1}{\\sqrt{3}}$ $\\displaystyle \\text{ix) } \\tan 5^o \\tan 25^o \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\tan (90^o-85^o) \\tan (90^o-65^o) \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\cot 85^o \\cot 65^o \\tan 30^o \\tan 65^o \\tan 85^o$ $\\displaystyle = \\tan 30^o = \\frac{1}{\\sqrt{3}}$ $\\displaystyle \\\\$ Question 4: Express each of the following in terms of trigonometric ratios of angles between $\\displaystyle 0$ and $\\displaystyle 45^o$. $\\displaystyle \\text{i) }\\sin 85^o + \\mathrm{cosec} 85^o = \\sin (90^o-5^o) + \\mathrm{cosec} (90^o-5^o) = \\cos 5^o + \\sec 5^o$ $\\displaystyle \\text{ii) } \\mathrm{cosec} 69^o + \\cot 69^o = \\mathrm{cosec} (90^o-21^o) + \\cot (90^o-21^o) = \\sec 21^o + \\tan 21^o$ $\\displaystyle \\\\$ Question 5: Prove that: $\\displaystyle \\text{i) }\\tan 1^o \\tan 2^o \\tan 3^o \\cdots \\tan 89^o = 1$ $\\displaystyle \\text{ii) }\\cos 1^o \\cos 2^o \\cos 3^o \\cdots \\cos 180^o = 0$ i) $\\displaystyle \\text{LHS} = \\tan 1^o \\tan 2^o \\tan 3^o \\cdots \\tan 89^o$ $\\displaystyle = \\tan (90^o-89^o) \\tan (90^o-88^o) \\tan (90^o-87^o) \\cdots \\tan 87^o \\tan 88^o \\tan 89^o$ $\\displaystyle = \\cot 89^o \\cot 88^o \\cot 87^o \\cdots \\tan 87^o \\tan 88^o \\tan 89^o$ $\\displaystyle = 1 \\times 1 \\times 1 \\cdots \\times 1 = 1 = RHS$ Hence Proved. ii) $\\displaystyle \\text{LHS} = \\cos 1^o \\cos 2^o \\cos 3^o \\cdots \\cos 180^o$ $\\displaystyle", "# The relationship between tan(x) and square roots Please note: I am working in DEGREES I think the easiest way to illustrate my point is by showing some examples: • $\\tan(0^\\circ) = \\sqrt 0 = 0$ • $\\tan(22.5^\\circ) = \\sqrt 2 -1$ • $3 \\cdot \\tan(30 ^\\circ) =\\sqrt 3$ • $\\tan(45 ^\\circ) =\\sqrt 1 = 1$ • $\\tan(60 ^\\circ) =\\sqrt 3$ • $\\tan(75 ^\\circ) = 2 + \\sqrt 3$ Ok, so there are some nice examples that have whole numbers, but then there are some that are less \"pretty\": • $\\tan(54.73561... ^\\circ) = \\sqrt 2$ • $\\tan(65.90515... ^\\circ) = \\sqrt 5$ ... and so on ... These aren't very impressive, because once can easily generate these irrational numbers in order to get the square root of something by simply using the arctan(x) function. (arctan($\\sqrt 2$) = 54.73561... Let's look at this useless equation: $\\tan( \\arctan(\\sqrt x)) = \\sqrt x$ * SHOCKER * So, I was wondering, is there possibly a way to represent this without using arctan($\\sqrt x$) by the use of an infinite series - for example: $\\tan(\\sum_{n=0}^\\infty$ something in terms of x ) = $\\sqrt x$ I look forward to any answers or responses I may get :) Kind regards Joshua :) # EDIT: An interesting \"discovery\" (not really) I'm putting \"discovery\" in inverted commas because", "cm, then $\\angle B = \\underline{\\hspace{2.0cm}}$ OR Two triangles are similar if their corresponding sides are $\\underline{\\hspace{2.0cm}}$ $\\underline{\\angle B = 90^{\\circ}}$ We have to first check if the $\\triangle ABC$ is a right angled triangle. If it is a right angled triangle, then Pythagoras theorem should hold $\\Rightarrow 12^2 = ( 6\\sqrt{3})^2 + (6)^2$ $\\Rightarrow 144 = 108 + 36$ $\\Rightarrow 144 = 144$. Therefore the $\\triangle ABC$ is a right angled triangle. $\\therefore \\angle B = 90^{\\circ}$ OR Two triangles are similar if there corresponding sides are $\\underline{proportional}$. If $\\triangle ABC$ and $\\triangle PQO$ are similar, then $\\frac{AB}{PQ}$ $=$ $\\frac{BC}{QR}$ $=$ $\\frac{AC}{PR}$ $\\\\$ Question 14: The value of $( \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 89^{\\circ})$ is equal to $\\underline{\\hspace{2.0cm}}$ $\\underline{\\hspace{0.5cm} 1 \\hspace{0.5cm} }$ $\\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 89^{\\circ}$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\tan 46^{\\circ} \\ldots\\tan 89^{\\circ}$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\tan (90^{\\circ} - 44^{\\circ}) \\ldots \\tan (90^{\\circ} - 1^{\\circ})$ $= \\tan 1^{\\circ} \\tan 2^{\\circ} \\ldots \\tan 44^{\\circ} \\tan 45^{\\circ} \\cot 44^{\\circ} \\ldots \\cot 1^{\\circ}$ $= \\tan 45^{\\circ} = 1$ $\\\\$ Question 15: In Fig. 3, the angles of depressions from the observing positions $O_1$ and $O_2$ respectively of the object $A$ are $\\underline{\\hspace{2.0cm}}$ $\\underline{30^{\\circ} \\& 45^{\\circ}}$ Therefore the angle of depression are", "– 2m\\tan{25^{\\circ}} &= \\tan{25^{\\circ}} – 2 \\\\ m + 2m\\tan{25^{\\circ}} &= 2 – \\tan{25^{\\circ}} \\\\ m(1 + 2\\tan{25^{\\circ}}) &= 2 – \\tan{25^{\\circ}} \\\\ m &= \\frac{2 – \\tan{25^{\\circ}}}{1 + 2\\tan{25^{\\circ}}} \\\\ m &= 0.8 \\color{red} \\cdots (1) \\\\ \\frac{2-m}{1+2m} &= -\\tan{25^{\\circ}} \\\\ 2-m &= -\\tan{25^{\\circ}} – 2m\\tan{25^{\\circ}} \\\\ -m + 2m\\tan{25^{\\circ}} &= -\\tan{25^{\\circ}} – 2 \\\\ 2m\\tan{25^{\\circ}} – m &= \\tan{25^{\\circ}} + 2 \\\\ m(2\\tan{25^{\\circ}} – 1) &= \\tan{25^{\\circ}} + 2 \\\\ m &= \\frac{\\tan{25^{\\circ}} + 2}{2\\tan{25^{\\circ}} – 1} \\\\ m &= 36.6 \\color{red} \\cdots (2) \\\\ \\therefore m &= 0.8 \\text{ or } m = 36.6 &\\color{red} \\text{by (1) and (2)} \\\\ \\end{aligned} \\\\", "-1/\\tan 15^\\circ$$ Now, $\\tan 2x = 2\\tan x / (1-\\tan^2 x)$ Put $x=15^\\circ$ in the above equation and calculate $\\tan 15^\\circ$ from here and put in $$\\tan 105^\\circ = -1/\\tan15^\\circ$$ Do the same thing for your other question. • Thanks @Micah, you have edited my response in such a beautiful fashion. I am new here, so, i dont know how to write mathematical equations here. – th ie Jul 20 '15 at 20:16", "\\ \\ \\ \\ \\ \\ \\ \\ \\ =-x\\\\\\\\\\ \\ \\ \\ \\tan 125{}^\\circ =\\tan (90{}^\\circ +35{}^\\circ )\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\cot 35{}^\\circ \\end{array}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\frac{1}{{\\tan 35{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\frac{1}{x}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\frac{{\\tan 145{}^\\circ -\\tan 125{}^\\circ }}{{1+\\tan 145{}^\\circ \\tan 125{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ =\\ \\ \\frac{{-x-\\left(\\displaystyle {-\\frac{1}{x}} \\right)}}{{1-x\\left(\\displaystyle {-\\frac{1}{x}} \\right)}}$ $\\displaystyle \\ \\ \\ \\ =\\ \\ \\frac{{\\displaystyle \\frac{{1-{{x}^{2}}}}{x}}}{2}$ $\\displaystyle \\ \\ \\ \\ =\\ \\ \\frac{{1-{{x}^{2}}}}{{2x}}$ 5. Prove that (i) $\\displaystyle \\sin (A+B)\\sin (A-B)={{\\sin }^{2}}A-{{\\sin }^{2}}B.$ (ii) $\\displaystyle \\sin (A+B)\\sin (A-B)={{\\cos }^{2}}B-{{\\cos }^{2}}A.$ (iii) $\\displaystyle \\cos (A+B)\\cos (A-B)={{\\cos }^{2}}A-{{\\sin }^{2}}B.$ (iv) $\\displaystyle \\cos (A+B)\\cos (A-B)={{\\cos }^{2}}B-{{\\sin }^{2}}A.$ Show/Hide Solution (i)$\\displaystyle \\ \\ \\ \\sin (A+B)\\sin (A-B)$ $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\ =(\\sin A\\cos B+\\cos A\\sin B)(\\sin A\\cos B-\\cos A\\sin B)\\\\\\\\ \\ \\ \\ \\ \\ ={{\\sin }^{2}}A{{\\cos }^{2}}B-{{\\cos }^{2}}A{{\\sin }^{2}}B\\\\\\\\ \\ \\ \\ \\ \\ ={{\\sin }^{2}}A(1-{{\\sin }^{2}}B)-(1-{{\\sin }^{2}}A){{\\sin }^{2}}B\\\\\\\\ \\ \\ \\ \\ \\ ={{\\sin }^{2}}A-{{\\sin }^{2}}A{{\\sin }^{2}}B-{{\\sin }^{2}}B+{{\\sin }^{2}}A{{\\sin }^{2}}B\\\\\\\\ \\ \\ \\ \\ \\ ={{\\sin }^{2}}A-{{\\sin }^{2}}B\\end{array}$ (ii)$\\displaystyle \\ \\", "## tan A = tan B + tan C + tan D $\\tan \\, 84^\\circ \\; = \\; tan \\, 78^\\circ \\; + \\; tan \\, 72^\\circ \\; + \\; tan \\, 60^\\circ$ $\\tan \\, 80^\\circ \\; = \\; tan \\, 70^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 50^\\circ$ Find other 4-tuples $(a, b, c, d)$ of distinct integers between $0^\\circ$ and $90^\\circ$ that satisfy the relation $tan \\, A^\\circ\\; = \\; tan \\, B^\\circ \\; + \\; tan \\, C^\\circ \\; + \\; tan \\, D^\\circ$ Other solutions: $\\tan \\, 72^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 36^\\circ \\; + \\; tan \\, 6^\\circ$ $\\tan \\, 70^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 40^\\circ \\; + \\; tan \\, 10^\\circ$ $\\tan \\, 72^\\circ \\; = \\; tan \\, 60^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 24^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 66^\\circ \\; + \\; tan \\, 60^\\circ \\; + \\; tan \\, 36^\\circ$ $\\tan \\, 78^\\circ \\; = \\; tan \\, 72^\\circ \\; + \\; tan \\, 42^\\circ \\; + \\; tan \\, 36^\\circ$ How many such equations can you produce if we allow repeated angles? Hint: there are 49 such equations Determine", "\\cos {25}^{\\circ} + h \\sin {25}^{\\circ}$ $\\implies \\sec \\alpha + \\tan \\alpha = \\cos {25}^{\\circ} + \\sin {25}^{\\circ} = 1.3289 \\ldots \\left(1\\right)$ Now $1.3289 \\left(\\sec \\alpha - \\tan \\alpha\\right) = {\\sec}^{2} \\alpha - {\\tan}^{2} \\alpha = 1$ $\\implies \\sec \\alpha - \\tan \\alpha = \\frac{1}{1.3289}$ $\\implies \\sec \\alpha - \\tan \\alpha = 0.7525 \\ldots \\left(2\\right)$ Adding (1) and (2) we get $2 \\sec \\alpha = 2.0814$ $\\implies \\sec \\alpha = 1.0407$ $= \\cos \\alpha = 0.9609$ $\\implies \\alpha = {\\cos}^{-} 1 \\left(0.9609\\right) = {16.077}^{\\circ}$ And $\\beta = {90}^{\\circ} - \\alpha = {73.923}^{\\circ}$ Part-(a) By the given condition of part (a) of the question Area of the upper triangle = Area of the lower triangle $\\implies \\frac{1}{2} \\times h \\times h \\tan \\alpha = \\frac{1}{2} h \\cos {25}^{\\circ} \\times h \\sin {25}^{\\circ}$ $\\tan \\alpha = \\frac{1}{2} \\times 2 \\cos {25}^{\\circ} \\times \\sin {25}^{\\circ} = \\frac{1}{2} \\times \\sin {50}^{\\circ} = 0.3830$ $\\implies \\alpha = {\\tan}^{-} 1 \\left(0.3830\\right) = {20.958}^{\\circ}$ Then $\\beta = {90}^{\\circ} - {20.95}^{\\circ} = {69.042}^{\\circ}$", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "angle $30^\\circ$, and calculating $\\tan(\\theta) = O/A$ $\\tan(30^\\circ) = O/600$ $600\\tan(30^\\circ) = O$ $346.4 = O$ -", "}}=\\tan 62{}^\\circ .$ (ii) $\\displaystyle \\tan 50{}^\\circ =\\tan 40{}^\\circ +2\\tan 10{}^\\circ .$ (iii) $\\displaystyle \\tan 70{}^\\circ =2\\tan 50{}^\\circ +\\tan 20{}^\\circ .$ (iv) $\\displaystyle \\tan 3A-\\tan 2A-\\tan A=\\tan 3A\\tan 2A\\tan A.$ Show/Hide Solution (i) $\\displaystyle \\ \\ \\ \\ \\frac{{\\cos 17{}^\\circ +\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ -\\sin 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{\\displaystyle \\frac{{\\cos 17{}^\\circ }}{{\\cos 17{}^\\circ }}+\\displaystyle \\frac{{\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ }}}}{{\\displaystyle \\frac{{\\cos 17{}^\\circ }}{{\\cos 17{}^\\circ }}-\\displaystyle \\frac{{\\sin 17{}^\\circ }}{{\\cos 17{}^\\circ }}}}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{1+\\tan 17{}^\\circ }}{{1+\\tan 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\frac{{\\tan 45{}^\\circ +\\tan 17{}^\\circ }}{{1+\\tan 45{}^\\circ \\tan 17{}^\\circ }}$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\tan (45{}^\\circ +17{}^\\circ )$ $\\displaystyle \\ \\ \\ \\ \\ \\ =\\tan 62{}^\\circ$ (ii) $\\displaystyle \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\tan (40{}^\\circ +10{}^\\circ )$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\frac{{\\tan 40{}^\\circ +\\tan 10{}^\\circ }}{{1-\\tan 40{}^\\circ \\tan 10{}^\\circ }}$ $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ -\\tan 50{}^\\circ \\tan 40{}^\\circ \\tan 10{}^\\circ =\\tan 40{}^\\circ +\\tan 10{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\tan 50{}^\\circ =\\tan 40{}^\\circ +\\tan 10{}^\\circ +\\tan 50{}^\\circ \\tan 40{}^\\circ \\tan 10{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "# Quintuple Angle Formulas/Tangent ## Theorem $\\tan 5 \\theta = \\dfrac {\\tan^5 \\theta - 10 \\tan^3 \\theta + 5 \\tan \\theta} {1 - 10 \\tan^2 \\theta + 5 \\tan^4 \\theta}$ where $\\tan$ denotes tangent. ## Proof $\\displaystyle \\tan 5 \\theta$ $=$ $\\displaystyle \\frac {\\sin 5 \\theta} {\\cos 5 \\theta}$ Tangent is Sine divided by Cosine $\\displaystyle$ $=$ $\\displaystyle \\frac {5 \\sin \\theta - 20 \\sin^3 \\theta + 16 \\sin^5 \\theta} {16 \\cos^5 \\theta - 20 \\cos^3 \\theta + 5 \\cos \\theta}$ Quintuple Angle Formulas $\\displaystyle$ $=$ $\\displaystyle \\frac {5 \\frac {\\tan \\theta} {\\cos^4 \\theta} - 20 \\frac {\\tan^3 \\theta} {\\cos^2 \\theta} + 16 \\tan^5 \\theta} {16 \\cos \\theta - \\frac {20} {\\cos^2 \\theta} + \\frac 5 {\\cos^4 \\theta} }$ dividing top and bottom by $\\cos^5 \\theta$ $\\displaystyle$ $=$ $\\displaystyle \\frac {5 \\tan \\theta \\sec^4 \\theta - 20 \\tan^3 \\theta \\sec^2 \\theta + 16 \\tan^5 \\theta} {16 - 20 \\sec^2 \\theta + 5 \\sec^4 \\theta}$ Secant is Reciprocal of Cosine $\\displaystyle$ $=$ $\\displaystyle \\frac {5 \\tan \\theta \\paren {1 + \\tan^2 \\theta}^2 - 20 \\tan^3 \\theta \\paren {1 + \\tan^2 \\theta} + 16 \\tan^5 \\theta} {16 - 20 \\paren {1 + \\tan^2 \\theta} + 5 \\paren {1 + \\tan^2 \\theta}^2}$ Difference of Squares of Secant and Tangent $\\displaystyle$ $=$ $\\displaystyle \\frac {\\tan^5 \\theta - 10 \\tan^3 \\theta + 5", "(Take $\\displaystyle \\pi = \\frac{22}{7}$ ) [4] (a) Given $\\displaystyle \\sin^2 34^{\\circ} + \\sin^2 56^{\\circ} + 2 \\tan 18^{\\circ} \\tan 72^{\\circ} - \\cot ^2 30^{\\circ}$ $\\displaystyle = \\sin^2 34^{\\circ} + \\sin^2 (90^{\\circ} - 34^{\\circ}) + 2 \\tan 18^{\\circ} \\tan (90^{\\circ}-18^{\\circ}) - \\cot ^2 30^{\\circ}$ $\\displaystyle = \\sin^2 34^{\\circ} + \\cos^2 34^{\\circ} + 2 \\tan 18^{\\circ} \\cot 18^{\\circ} - (\\sqrt{3})^2$ $\\displaystyle = 1+ 2 \\tan 18^{\\circ} \\times \\frac{1}{\\tan 18^{\\circ}}-3$ $\\displaystyle = 1+2-3 = 0$ (b) Let $\\displaystyle f(x)=x^3+10x^2-37x+26$ Putting $\\displaystyle x =1$ , we get $\\displaystyle f(1)=1+10-37+26=0$ By factor theorem $\\displaystyle x-1$ is actor of $\\displaystyle f(x)$ $\\displaystyle \\begin{array}{r l l } x-1 ) & \\overline {x^3+10x^2-37x+26} & (x^2+11x-26 \\\\ (-) & \\underline {x^3-x^2} & \\\\ & \\hspace{1.0cm} 11x^2-37x+26 & \\\\ (-) &\\hspace{1.0cm} \\underline{11x^2-37x} & \\\\ & \\hspace{2.0cm} {-26x+26} & \\\\ (-) & \\hspace{2.0cm} \\underline{ -26x+26} & \\\\ & \\hspace{3.0cm} \\times & \\\\ \\end{array}$ On dividing $\\displaystyle x^{3}+10x^{2}-37x+26$ by $\\displaystyle (x-1)$ , we get $\\displaystyle x^{2}+11x-26$ as the quotient and remainder $\\displaystyle = 0$ Therefore the other factor of $\\displaystyle f(x)$ are the factor of $\\displaystyle x^{2}+11x-26$ Now, $\\displaystyle x^{2}+11x-26$ $\\displaystyle = x^{2}+13x-2x-26$ $\\displaystyle = x(x+13)-2(x+13)$ $\\displaystyle = (x+13)(x-2)$ Hence $\\displaystyle x^{3}+10x^{2}-37x+26=(x-1)(x-2)(x+13)$ (c) Area of rectangle $\\displaystyle ABCD= 14 \\times 7 = 98 \\text{ cm }$ $\\displaystyle \\text{Area of quarter circle } BFEC=\\frac{1}{4}\\pi (7^{2})=\\frac{49}{4}\\pi$ $\\displaystyle \\text{Area of semicircle }" ]

[ "than $C$, so we can discard this, and $D=(2,1).$ The value of $s$ we seek is $1$. #### Approach 2: using the angle in a semicircle If $A$,$B$ and $D$ are points on a circle with $AB$ a diameter of that circle, then $\\widehat{ADB}$ is a right angle. The converse is also true; if $\\widehat{ADB}=\\dfrac{\\pi}{2}$, then $AB$ has to be a diameter of the circle. Thus $D$ is on the circle with diameter $AB$. The midpoint of $AB$ is $(3,3)$ and the length of $AB$ is $2\\sqrt{5}$, and so the circle has centre $(3,3)$ and radius $\\sqrt{5}$. Thus the equation of the circle is $(x-3)^2+(y-3)^2=5$. The point $D$ is on this, so $(2s-3)^2 +(s-3)^2 = 5$, or $5s^2-18s+13=0$. The solution then proceeds as before. #### Approach 3: using the scalar product We know that the lines $DA$ and $DB$ have to be perpendicular. Thus $(\\mathbf{d}-\\mathbf{a})\\cdot(\\mathbf{d}-\\mathbf{b})=0,$ where $\\mathbf{d}$ is the position vector of $D$, and where $\\cdot$ represents the scalar product. Now $\\mathbf{d} = \\begin{pmatrix}2s\\\\s\\end{pmatrix},$ so $\\mathbf{d}-\\mathbf{a} =\\begin{pmatrix}2s-2\\\\s-5\\end{pmatrix} , \\mathbf{d}-\\mathbf{b}= \\begin{pmatrix}2s-4\\\\s-1\\end{pmatrix}.$ Thus we have \\begin{align*} (\\mathbf{d}-\\mathbf{a})\\cdot(\\mathbf{d}-\\mathbf{b})&=\\begin{pmatrix}2s-2\\\\s-5\\end{pmatrix} \\cdot \\begin{pmatrix}2s-4\\\\s-1\\end{pmatrix}\\\\ &=(2s-2)(2s-4)+(s-5)(s-1)\\\\ &=5s^2-18s+13=0, \\end{align*} as before, and so $\\mathbf{d}=\\begin{pmatrix}2\\\\1\\end{pmatrix}, s = 1.$", "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "1. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 4 \\\\ 1 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 3 During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): $1.2$ miles north, $6.1$ miles $38$ degrees east of south, $4.0$ miles at $89$ degrees east of north, and $6.5$ miles at $31$ degrees east of north. Find the distance between starting and ending positions (O'Hanian 1985). Problem 4 Find $k$ so that these two vectors are perpendicular. $\\begin{pmatrix} k \\\\ 1 \\end{pmatrix} \\qquad \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$ Problem 5 Describe the set of vectors in $\\mathbb{R}^3$ orthogonal to this one. $\\begin{pmatrix} 1 \\\\ 3 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 1. Find the angle between the diagonal of the unit square in $\\mathbb{R}^2$ and one of the axes. 2. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^3$ and one of the axes. 3. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^n$ and one of the axes. 4. What is the limit, as $n$ goes to $\\infty$,", "+0 # Pls help me +1 32 1 +57 The vectors $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}$ satisfy $\\|\\mathbf{a}\\| = \\|\\mathbf{b}\\| = 1,$ $\\|\\mathbf{c}\\| = 2,$ and $\\mathbf{a} \\times (\\mathbf{a} \\times \\mathbf{c}) + \\mathbf{b} = \\mathbf{0}.$If $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{c},$ then find all possible values of $\\theta,$ in degrees. Apr 25, 2021", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "+0 # Let $\\theta$ be the angle between $\\bold{a}$ and $\\bold{b}$, where $\\bold{a}$ and $\\bold{b}$ are the vectors defined in part (a). Find \\$\\cos 0 171 2 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$. Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\cos(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{\\vec{a}\\cdot \\vec{b}}{|\\vec{a}|\\cdot |\\vec{b}|} } \\quad | \\quad |\\vec{a}| = \\sqrt{2^2+(-6)^2+5^2}=\\sqrt{65},\\ |\\vec{b}| = \\sqrt{(-1)^2+(-2)^2+0^2}=\\sqrt{5} \\\\\\\\ \\cos(\\theta) &=& \\dfrac{\\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix}\\cdot \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix} } {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{2\\cdot (-1)+(-6)\\cdot(-2)+5\\cdot 0} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{10} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {65\\cdot 5 } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {325 } \\\\\\\\ \\mathbf{\\cos^2(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{4}{13}} \\\\ \\hline \\end{array}$$ Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix}", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "Math Relative position of lines Angle between two lines # Angle between two lines The angle between two lines is calculated similarly as the angle between two vectors. ! ### Remember In contrast to vectors, the angle between two lines always discribes the angle less than 90°. Two lines are given: $\\text{g: } \\vec{x} = \\vec{u} + r \\cdot \\vec{a}$ $\\text{h: } \\vec{x} = \\vec{v} + s \\cdot \\vec{b}$ For determining the angle the direction vectors of the lines are used: $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\gamma = \\cos^{-1}\\left(\\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}\\right)$ i ### Method 1. Calculate the scalar product of the direction vectors 2. Calculate the magnitude of the direction vectors 3. Insert results into the formula ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Calculate scalar product $\\vec{a}\\cdot\\vec{b}$ $=\\begin{pmatrix}9 \\\\ 8 \\\\ 7 \\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ 1 \\\\ 2\\end{pmatrix}$ $=9\\cdot1+8\\cdot1+7\\cdot2$ $=31$ 2. #### Calculate magnitude of the direction vectors $|\\vec{a}|=\\sqrt{9^2+8^2+7^2}$ $=\\sqrt{194}$ $|\\vec{b}|=\\sqrt{1^2+1^2+2^2}$ $=\\sqrt{6}$ 3. #### Insert results into the formula $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\cos(\\gamma) = \\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}$ $\\gamma = \\cos^{-1}\\left(\\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}\\right)$ $\\approx24.68°$", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "vector 6i+aj+bk has magnitude 21 and perpendicular to 3i+2j+2k .Find the possible values of a and b ,showing all necessary working. (a)Either: $$\\vec{PR}=2\\vec{PQ}=2(q-p)$$ $$\\vec{OR}=p+2q-2p=2q-p$$ $$\\vec{OR}=\\vec{PQ}=q-p$$ $$\\vec{OR}=\\vec{OQ}+\\vec{QR}=q+q-p=2q-p$$ (b)$$6^{2}+a^{2}+b^{2}=21^{2}$$ 18+2a+2b=0 $$a^{2}+\\left ( -a-9 \\right )^{2}=405$$ $$\\left ( 2 \\right )\\left ( a^{2}+9a-162 \\right )\\left (= 0 \\right )$$ a=9 or -18 b=-18 0r 9 Question The position vectors of points A and B relative to an origin O are a and b respectively. The position vectors of points C and D relative to O are 3a and 2b respectively. It is given that $$a=\\begin{pmatrix}2\\\\ 1\\\\ 2\\end{pmatrix}$$ and $$b=\\begin{pmatrix}4\\\\ 0\\\\ 6\\end{pmatrix}$$ (i) Find the unit vector in the direction of $$\\vec{CD}$$ (ii) The point E is the mid-point of CD. Find angle EOD. (i)$$\\vec{CD}=-3\\begin{pmatrix}2\\\\ 1\\\\ 2\\end{pmatrix}+2\\begin{pmatrix}4\\\\ 0\\\\ 6\\end{pmatrix}=\\begin{pmatrix}2\\\\ -3\\\\ 6\\end{pmatrix}$$ $$Unit vector =\\frac{1}{7}\\begin{pmatrix}2\\\\ -3\\\\ 6\\end{pmatrix}$$ (ii)$$\\vec{OE}=\\begin{pmatrix}6\\\\ 3\\\\ 6\\end{pmatrix}+\\begin{pmatrix}1\\\\-1\\tfrac{1}{2}\\\\ 3\\end{pmatrix}=\\begin{pmatrix}7\\\\ 1\\tfrac{1}{2}\\\\ 9\\end{pmatrix}$$ $$\\vec{OE}.\\vec{OD}=56+0+108=164$$ $$\\left | \\vec{OE} \\right |=\\sqrt{132.25}=11.5$$ $$\\left | \\vec{OD} \\right |=\\sqrt{208}$$ $$164=\\sqrt{132.25}\\times \\sqrt{208}\\times \\cos \\Theta$$ $$\\Theta =8.6^{\\circ}$$ Question The diagram shows a pyramid OABCD in which the vertical edge OD is 3 units in length. The point E is the centre of the horizontal rectangular base OABC. The sides OA and AB have lengths of 6 units and 4 units respectively. The unit vectors i, j and k are parallel t0 $$\\vec{OA}$$,$$\\vec{OC}$$ and $$\\vec{OD}$$ respectively. (i) Express each of the", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "= \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}$ Now, using the formula for the angle between two vectors: $\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ 4. (Original post by The_Big_E) A general point on the line L has coordinates: $\\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}$ Call this point C, with position vector $\\vec{OC}$ The angle between vectors $\\vec{AB}$ and $\\vec{BC}$ is $90^{\\circ}$ So first we have to find $\\vec{BC}$. $\\vec{BC} = \\vec{OC} - \\vec{OB}$ $\\vec{BC} = \\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}-\\begin{pmatrix} 1\\\\ 0\\\\ -2 \\end{pmatrix}$ $\\vec{BC} = \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}$ Now, using the formula for the angle between two vectors: $\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ Thank you, I get", "Find $\\mathbf{a} \\cdot \\mathbf{b}$ if $|\\mathbf{a}| = 6$, $|\\mathbf{b}| = 6$, and the angle between $\\mathbf{a}$ and $\\mathbf{b}$ is $\\pi/2$.", "## Angle Between Two Lines The angle between two lines $\\vec{r}_1= \\vec{r}_{10}+ \\vec{u} t, \\: \\vec{r}_2= \\vec{r}_{20}+ \\vec{v} t$ is the angle between their direction vectors. We can find the angle $\\theta$ using the formula $cos(\\theta ) =\\frac{ \\vec{u} \\cdot \\vec{v}}{\\| \\vec{u} \\| \\| \\vec{v} \\| }$ . Example: Let $\\vec{r}_1= \\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix} + t \\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix}, \\: \\vec{r}_2= \\begin{pmatrix}0\\\\0\\\\4\\end{pmatrix} + s \\begin{pmatrix}2\\\\-1\\\\2\\end{pmatrix}$ . The direction bvectors are $\\vec{u}= \\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix}, \\: \\vec{v}= \\begin{pmatrix}2\\\\-1\\\\2\\end{pmatrix}$ . $cos \\theta=\\frac{\\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix} \\cdot \\begin{pmatrix}2\\\\-1\\\\2\\end{pmatrix}}{\\| \\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix} \\| \\| \\begin{pmatrix}2\\\\-1\\\\2\\end{pmatrix} \\|}=\\frac{1 \\times 2 + 1 \\times -1 +2 \\times 2}{\\sqrt{1^2+1^2+2^2} \\sqrt{2^2+1^2+(-2)^2}}=\\frac{5}{\\sqrt{6} \\sqrt{9}}=0.6804$ $\\theta=cos^{-1}(0.6804)=47.1^o$ Refresh", "\\mathbf{E}_y \\cdot \\mathbf{e}_c & \\mathbf{E}_z \\cdot \\mathbf{e}_c \\end{bmatrix} = \\begin{bmatrix} a_x & a_y & a_z \\\\ b_x & b_y & b_z \\\\ c_x & c_y & c_z \\end{bmatrix}$$ where $\\mathbf{E}_i$ are unit vectors along the reference coordinate axes and $\\mathbf{e}_j$ are unit vectors along the axes of the ellipsoid. Then, the Euler angles are given by \\begin{align} \\theta & = \\cos^{-1}(R_{33}) \\\\ \\phi & = \\begin{cases} \\tan^{-1}(R_{21}/R_{11}) & \\quad \\text{for} \\quad \\theta \\sim 0 \\\\ \\tan^{-1}(-R_{13}/R_{23}) & \\quad \\text{otherwise} \\end{cases}\\\\ \\psi & = \\begin{cases} 0 & \\quad \\text{for} \\quad \\theta \\sim 0 \\\\ \\tan^{-1}(R_{31}/R_{32}) & \\quad \\text{otherwise} \\end{cases} \\end{align} The Euler angles can be converted into direction cosines using $\\mathbf{R} = \\mathbf{D} \\mathbf{C} \\mathbf{B}$ where $$\\mathbf{D} = \\begin{bmatrix} \\cos\\phi & -\\sin\\phi & 0 \\\\ \\sin\\phi & \\cos\\phi & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} ~,~~ \\mathbf{C} = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & \\cos\\theta & -\\sin\\theta \\\\ 0 & \\sin\\theta & \\cos\\theta \\end{bmatrix} ~,~~ \\mathbf{B} = \\begin{bmatrix} \\cos\\psi & -\\sin\\psi & 0 \\\\ \\sin\\psi & \\cos\\psi & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}$$ We can easily check that these transformations produce orthogonal rotation matrices. The R function that we use to do the Euler angle computations is given below. Euler angles computed using the above function are identical to those produced", "Since $(21,26)=1$, there exists $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$. Remember that finding $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$ is equivalent to solving the diophantine equation $21x+26y=1$. \\begin{align*} 26&=21\\cdot 1+5\\\\ 21&=5\\cdot 4+1 \\end{align*} and \\begin{align*} 1&=21-5\\cdot 4\\\\ &=21-(26-21)4\\\\ &=5\\cdot 21+26(-4) \\end{align*} Thus, we find $21^{-1}\\equiv 5\\mod 26$. \\begin{align*} A^{-1}&=\\begin{pmatrix} 5\\cdot 8 & -5\\cdot 3\\\\ -5\\cdot 7 & 5\\cdot 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 40 & -15\\\\ -35 & 10 \\end{pmatrix}\\\\ &\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$. Then $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=A\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\ cx+dy \\end{pmatrix}$$ This gives a linear map or a linear transformation from vectors to vectors. We introduce the following theorem without proof. Theorem. Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ and set $D=ad-bc$ as before. Then the following are equivalent: 1. $(D,N)=1$. 2. $A$ has an inverse matrix. 3. if $x$ and $y$ are not both $0$ in $\\mathbb{Z}/N\\mathbb{Z}$, then $A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4. $A$ gives a one-to-one correspondence of $(\\mathbb{Z}/N\\mathbb{Z})^2$ with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the", "1. ## Proving a quadrilateral is a rectangle? I'm having a little trouble with a question I'm doing. I'll write it below. The vertices of a quadrilateral are C(-2,4), D(3,9), E(6,6), and F(1,1). a) Show that the quadrilateral is a rectangle. First I calculated the length of its four sides. CD = square root of 50 DE = square root of 18 EF = square root of 50 CF = square root of 18 Ok, so far so good. Now I have to show that it has a right angle, which would require two adjacent sides being perpendicular. mCD = 1, but mDE = -6/0? Did I mess up somewhere? Any help/advice is appreciated. 2. Originally Posted by mathdonkey mCD = 1, but mDE = -6/0? Now, what are you doing here? If you are familiair with vectors, try to express each side as a vector. For an example, $\\bar{CD}=\\begin{pmatrix}3-(-2) \\\\ 9-4\\end{pmatrix}=\\begin{pmatrix}5 \\\\ 5\\end{pmatrix}$ and $\\bar{DE}=\\begin{pmatrix}6-3 \\\\ 6-9\\end{pmatrix}=\\begin{pmatrix}3 \\\\ -3\\end{pmatrix}$ For $\\bar{CD}$ and $\\bar{DE}$ to perpendicular we must have that the scalar product is zero. $\\bar{CD} \\cdot \\bar{DE}=\\begin{pmatrix}5 \\\\ 5\\end{pmatrix} \\cdot \\begin{pmatrix}3 \\\\ -3\\end{pmatrix}= 5(3)+5(-3)=0$, so the vectors are perpendicular. 3. Originally Posted by Aliquantus Now, what are you doing here? If you are familiair with vectors, try to express each side as a vector. For an example,", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "# Area of the triangle formed by three vector lines I've had made several failed attempts at this question, but I'll show you my most recent one. Question: Show that the lines with equations $$\\mathbf r_1 = \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\\\ \\end{pmatrix} + \\lambda\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\\\ \\end{pmatrix}$$ $$\\mathbf r_2 = \\begin{pmatrix} -2 \\\\ 3 \\\\ -1 \\\\ \\end{pmatrix} + \\mu\\begin{pmatrix} 1 \\\\ -1 \\\\ 0 \\\\ \\end{pmatrix}$$ $$\\mathbf r_3 = \\begin{pmatrix} 2 \\\\ -1 \\\\ -1 \\\\ \\end{pmatrix} + t\\begin{pmatrix} -1 \\\\ 2 \\\\ 3 \\\\ \\end{pmatrix}$$ form a triangle, and find its area. I've 5.53 and 6.02 square units in my solution. For the sake of brevity, I'll outline my method as my actual attempt is several pages long. 1. Find co-ordinates/position vectors for the vertices of the triangle using: $$r_1 = r_2$$ $$r_1 = r_3$$ $$r_2 = r_3$$ 2. Using position vectors of vertices to find vector equations of the three lines. 3. Calculate the magnitude of two of the lines. 4. Calculate the dot product of these two lines. 5. Find the angle between these two lines using $$cos\\theta = \\frac{a \\cdot b}{\\lvert a\\rvert \\lvert b\\rvert}$$ Now the area of the triangle is $$Area = \\frac{1}{2}\\lvert a\\rvert \\lvert b\\rvert sin\\theta$$ However, this has given me 6.02 and 5.53", "$\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ 4. (Original post by The_Big_E) A general point on the line L has coordinates: $\\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}$ Call this point C, with position vector $\\vec{OC}$ The angle between vectors $\\vec{AB}$ and $\\vec{BC}$ is $90^{\\circ}$ So first we have to find $\\vec{BC}$. $\\vec{BC} = \\vec{OC} - \\vec{OB}$ $\\vec{BC} = \\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}-\\begin{pmatrix} 1\\\\ 0\\\\ -2 \\end{pmatrix}$ $\\vec{BC} = \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}$ Now, using the formula for the angle between two vectors: $\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ Thank you, I get it now TSR Support Team We have a brilliant team of more than 60 Support Team" ]

[ "$(\\mathbf{a}+\\mathbf{b})\\cdot(\\mathbf{c}+\\mathbf{d}) = \\mathbf{a}\\cdot\\mathbf{c} + \\mathbf{a}\\cdot\\mathbf{d} + \\mathbf{b}\\cdot\\mathbf{c} + \\mathbf{b}\\cdot\\mathbf{d}$ ## Mixed associative law The dot product has a homogeneity $\\alpha (\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot \\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Bilinear Often the mixed associative law and distributive law are combined to show that the dot product is bilinear: $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Cauchy-Schwarz Inequality $|\\mathbf{a}\\cdot\\mathbf{b}|\\leq|\\mathbf{a}||\\mathbf{b}|$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Fast calculation of Cosine The definition of the dot-product is used extensively in computer graphics since it speeds up computations in many circumstances by avoiding inefficient trigonometric functions. Additionally, working with normalized vectors gives the cosine between these two vectors by taking the dot product of them: $\\cos\\theta = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}||\\mathbf{b}|} = \\hat{\\mathbf{a}}\\cdot\\hat{\\mathbf{b}}$ ### Making perpendicular vectors Since two vectors are perpendicular when $$\\mathbf{a}\\cdot\\mathbf{b} = 0$$, we can easily construct a perpendicular vector by zeroing all components except 2, flipping those two and reversing the sign of one of them. For example with $$\\mathbf{a} = (\\mathbf{a}_1, \\mathbf{a}_2, \\dots, \\mathbf{a}_n)$$ the vectors $$\\mathbf{a}' = (-\\mathbf{a}_2, \\mathbf{a}_1, 0, \\dots, 0)$$ or $$\\mathbf{a}'' = (0, -\\mathbf{a}_3, \\mathbf{a}_2, 0, \\dots, 0)$$, etc are all perpendcular to $$\\mathbf{a}$$,", "then we have that the columns of B act on the columns of A. Now that we see that matrix multiplication can fill our world with dot products, let’s say a bit more about what the dot product gives us. For two vectors u and v, it turns out that ${\\bf u}\\cdot{\\bf v}= |{\\bf u}||{\\bf v}|\\cos \\theta$, where |vector| means its length, and $\\theta$ is the angle between the two vectors. (From here we can confirm what I said earlier about ${\\bf u}\\cdot{\\bf u}$ being its length squared (since the angle between a vector and itself is 0 and the cosine of 0 is 1).) So what? This means that our product of canoes and waterfalls above is actually a bunch of lengths and angles: $\\left[\\begin{array}{ccc} \\mbox{---} \\ {\\bf r}_1\\ \\mbox{---} \\\\ \\mbox{---} \\ {\\bf r}_2\\ \\mbox{---} \\\\ \\mbox{---} \\ {\\bf r}_3\\ \\mbox{---} \\\\ \\end{array}\\right] \\left[\\begin{array}{ccc} \\vline & \\vline & \\vline \\\\ {\\bf c}_1 & {\\bf c}_2 & {\\bf c}_3 \\\\ \\vline & \\vline & \\vline \\\\ \\end{array}\\right] = \\left[\\begin{array}{ccc} {\\bf r}_1 \\cdot {\\bf c}_1 & {\\bf r}_1 \\cdot {\\bf c}_2 & {\\bf r}_1 \\cdot {\\bf c}_3 \\\\ {\\bf r}_2 \\cdot {\\bf c}_1 & {\\bf r}_2 \\cdot {\\bf c}_2 & {\\bf r}_2 \\cdot {\\bf c}_3 \\\\ {\\bf r}_3 \\cdot {\\bf c}_1 & {\\bf r}_3 \\cdot {\\bf c}_2 &", "Limited access Let: $$||{\\bf u}|| =6$$ $$||{\\bf u} + {\\bf v}||=6\\cdot \\sqrt{2}$$ ...and let's assume that ${\\bf u}$ makes an angle of $45 ^{\\circ}$ with a positive $x$-axis, while ${\\bf u} + {\\bf v}$ makes an angle of $90 ^{\\circ}$ with the positive $x$-axis. What is the component form of ${\\bf v}$? A", "If A= <5 ,1 ,-2 > and B= <4 ,-2 ,3 >, what is A*B -||A|| ||B||? May 12, 2017 $12 - \\sqrt{870}$ Explanation: For two vectors $A$ and $B$ of the form $A = < {A}_{x} , {A}_{y} , {A}_{z} >$ and $B = < {B}_{x} , {B}_{y} , {B}_{z} > ,$ the dot product $A \\cdot B$ is given by $\\left({a}_{x} \\cdot {b}_{x}\\right) + \\left({a}_{y} \\cdot {b}_{y}\\right) + \\left({a}_{z} \\cdot {b}_{z}\\right)$. For $A = < 5 , 1 , - 2 >$ and $B = < 4 , - 2 , 3 >$, we have: $A \\cdot B = \\left(5 \\cdot 4\\right) + \\left(1 \\cdot - 2\\right) + \\left(- 2 \\cdot 3\\right) = 20 - 2 - 6 = \\textcolor{b l u e}{12}$ The next part is the product of the magnitudes of vectors $A$ and $B$. The magnitude of a vector $A = < {A}_{x} , {A}_{y} , {A}_{z} >$ is given by: $| A | = \\sqrt{{\\left({A}_{x}\\right)}^{2} + {\\left({A}_{y}\\right)}^{2} + {\\left({A}_{z}\\right)}^{2}}$ For the given vectors $A$ and $B$: $| A | = \\sqrt{{\\left(5\\right)}^{2} + {\\left(1\\right)}^{2} + {\\left(- 2\\right)}^{2}} = \\sqrt{25 + 1 + 4} = \\sqrt{30}$ $| B | = \\sqrt{{\\left(4\\right)}^{2} + {\\left(- 2\\right)}^{2} + {\\left(3\\right)}^{2}} = \\sqrt{16 + 4 + 9} = \\sqrt{29}$ We then have $| A | \\cdot | B |", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "can be defined on vector spaces over other fields such as the set of complex numbers, $\\mathbb{C}$. Definition: If $V$ is a vector space over the field $\\mathbb{R}$ and $(\\cdot, \\cdot)$ is an inner product of $V$ then $(V, (\\cdot, \\cdot))$ is said to be an Inner Product Space. Perhaps the most familiar inner product space for the reader is the set $\\mathbb{R}^n$ with the inner product $(\\cdot, \\cdot)$ defined for all vectors $\\mathbf{x} = (x_1, x_2, ..., x_n), \\mathbf{y} = (y_1, y_2, ..., y_n) \\in \\mathbb{R}^n$ by: (1) \\begin{align} (\\mathbf{x}, \\mathbf{y}) = x_1y_1 + x_2y_2 + ... + x_ny_n = \\sum_{k=1}^{n} x_ky_k \\end{align} This inner product is common denoted $\\mathbf{x} \\cdot \\mathbf{y}$ as opposed to the notation $(\\mathbf{x}, \\mathbf{y})$. If we consider the $\\mathbb{R}^3$ and the vectors $\\mathbf{x} = (1, 2, 3)$ and $\\mathbf{y} = (2, -1, 4)$ then the inner product between these two vectors is: (2) \\begin{align} \\quad \\mathbf{x} \\cdot \\mathbf{y} = (1, 2, 3) \\cdot (2, -1, 4) = (1)(2) + (2)(-1) + (3)(4) = 12 \\end{align} It is very easy to verify that this is indeed an inner product on $\\mathbb{R}^n$ and it is given a special name - the Euclidean inner product on $\\mathbb{R}^n$ or simply, the dot product.", "# If A= <6 ,-5 ,3 > and B= <5 ,4 ,-9 >, what is A*B -||A|| ||B||? Mar 28, 2016 $A \\cdot B - | A | \\cdot | B | = - 17 - 92.4124 = - 109.4124$ #### Explanation: Where $A \\cdot B$ represents the dot product between vectors $A$ and $B$. This is given by: $A \\cdot B = \\left({a}_{1} \\cdot {b}_{1}\\right) + \\left({a}_{2} \\cdot {b}_{2}\\right) + \\left({a}_{3} \\cdot {b}_{3}\\right)$. Hence for $A = < 6 , - 5 , 3 >$ and $B = < 5 , 4 , - 9 >$ A*B=(6*5)+(-5)* 4+(3*(-9) or $A \\cdot B = 30 - 20 - 27 = - 17$ $| A |$ represents the magnitude of vector $A$ and is given by $\\sqrt{{a}_{1}^{2} + {a}_{2}^{2} + {a}_{3}^{2}}$. Hence $| A | \\cdot | B |$ = $\\left(\\sqrt{{6}^{2} + {\\left(- 5\\right)}^{2} + {3}^{2}}\\right) \\cdot \\left(\\sqrt{{5}^{2} + {4}^{2} + {\\left(- 9\\right)}^{2}}\\right)$ = $\\left(\\sqrt{36 + 25 + 9}\\right) \\cdot \\left(\\sqrt{25 + 16 + 81}\\right)$ = $\\sqrt{70} \\cdot \\sqrt{122} = 8.3666 \\cdot 11.0454 = 92.4124$ Hence $A \\cdot B - | A | \\cdot | B | = - 17 - 92.4124 = - 109.4124$", "$$|\\mathbf{x} \\cdot \\mathbf{z} - \\mathbf{y} \\cdot \\mathbf{w} | \\leq |\\mathbf{x} \\cdot \\mathbf{z} - \\mathbf{z} \\cdot \\mathbf{w} + \\mathbf{z} \\cdot \\mathbf{w} - \\mathbf{y} \\cdot \\mathbf{w} |$$ $$\\leq |\\mathbf{z}| \\cdot |\\mathbf{x} - \\mathbf{w}| + |\\mathbf{w}| \\cdot |\\mathbf{z} - \\mathbf{y}|$$ $$\\leq ||\\mathbf{z}|| \\cdot ||\\mathbf{x} - \\mathbf{w}|| + ||\\mathbf{w}|| \\cdot ||\\mathbf{z} - \\mathbf{y}||$$ $$\\leq ||\\mathbf{x} - \\mathbf{w}|| + ||\\mathbf{z} - \\mathbf{y}||$$ Any input would be appreciated. - Everything seems good, but for part $c$, what do you mean by the absolute value of a vector? I see this in your second line of equations, and in the third, you pass to vector norms. This is the only advice I have; otherwise, great job! – Clayton Feb 15 '13 at 4:06 It's easier to work with everything at once. For (a), since you've noticed that $x\\cdot y - x\\cdot z=||x||(||y||cost-||z||cosk)$, you could use the triangle inequality: $|||x||(||y||cost-||z||cosk|<2(||y||+||z||)<2(3+4)=14$. It looks good to me otherwise. – josh Feb 15 '13 at 4:07 a) follows from $|x \\cdot y - x \\cdot z| = |x \\cdot (y -z)| \\leq \\|x\\| \\|y-z\\| \\leq \\|x\\| (\\|y\\|+\\|z\\|) = 2 (3+4)$. The first inequality in your answer for b) is incorrect. First do the triangle inequality, then upper bound using the fact that $|x \\cdot y| < 1$ when $x,y \\in B$.", "two vectors being perpendicular is that the angle between them is $90^{\\circ}$. Once we leave two and three dimensions (or jump to the space of polynomials, as we’ll do soon), the concept of an angle isn’t as helpful. Another way to define perpindicular is through an operation known as the dot productSuppose we take two 2D vectors, $\\mathbf{x}$ and $\\mathbf{y}$. Each vector will have coordinates: $\\mathbf{x} = (x_{1},x_{2})$ and $\\mathbf{y} = (y_{1}, y_{2})$. The dot product is a special type of multiplication defined on vectors, and denoted $\\cdot$: $$\\mathbf{x}\\cdot\\mathbf{y} = x_{1}y_{1} + x_{2}y_{2}$$ The dot product can be described in words as the sum of the component-wise multiplication of the coordinates of the two vectors. Now, we can say that two vectors are perpendicular if their dot product is 0. That is, $\\mathbf{x}$ and $\\mathbf{y}$ are perpindicular if $\\mathbf{x}\\cdot\\mathbf{y} = 0$. (This article gives a nice overview of how we move from the algebraic to the geometric definition of perpendicular and orthogonal.) Remember, perpendicular doesn’t make sense once we get into higher dimensions. Orthogonal is a more general notion of vectors being perpendicular, and is defined for two vectors (of any length) as their dot product equalling zero. From Dot Product to Inner Product The dot product is used on vectors, and defines another type of product that", "to take the dot product of two $$N$$-bit binary vectors, $$\\vec{a} \\cdot \\vec{b} = \\sum_i^N a_i b_i$$. Since each $$a_i$$ and $$b_i$$ are ±1, their product is 1 if $$a_i = b_i$$ and -1 otherwise. So the dot product is the total count of how many bit places match, minus how many places don’t. Because the two counts must sum to $$N$$, we have $$\\vec{a} \\cdot \\vec{b} = 2 \\left(\\text{# places where a = b}\\right) - N$$.In notation: \\begin{align*} \\vec{a} \\cdot \\vec{b} &= \\sum_{i=1}^N a_i b_i \\\\\\\\ &= \\sum_{i=1}^N \\mathbf{1}_{a_i = b_i} + -\\mathbf{1}_{a_1 \\neq b_i} \\\\\\\\ &= \\sum_{i=1}^N \\mathbf{1}_{a_i = b_i} -\\sum_{i=1}^N \\mathbf{1}_{a_1 \\neq b_i} \\\\\\\\ &= 2 \\sum_{i=1}^N \\mathbf{1}_{a_i = b_i} - N, \\end{align*} where $\\mathbf{1}$ is an indicator function and the last equality holds because the equality condition partitions the bits. Since we’re encoding these vectors as 0/1 bit vectors, $$a \\text{ XNOR } b$$ is precisely 1 where $$a$$ matches $$b$$ and 0 where it doesn’t, so we can compute the dot product as $$a \\cdot b = 2 \\text{ popcount}\\left(a\\text{ XNOR } b\\right) - N$$. This takes just a few instructions and is very SIMD-friendly. Since matrix multiplication, convolution, and most other important operations for neural networks are made up of dot products, this makes the forward pass of a binary neural network", "vector in $\\mathbb{R}^{n}$? For that we get \\begin{aligned}\\mathbf{x} \\hat{\\mathbf{n}} \\mathbf{x} &=(\\mathbf{x} \\cdot \\hat{\\mathbf{n}} \\hat{\\mathbf{n}} + \\mathbf{x} \\wedge \\hat{\\mathbf{n}} \\hat{\\mathbf{n}} ) \\hat{\\mathbf{n}} \\mathbf{x} \\\\ &=(\\mathbf{x} \\cdot \\hat{\\mathbf{n}} + \\mathbf{x} \\wedge \\hat{\\mathbf{n}} ) (\\mathbf{x} \\cdot \\hat{\\mathbf{n}} \\hat{\\mathbf{n}} + \\mathbf{x} \\wedge \\hat{\\mathbf{n}} \\hat{\\mathbf{n}} ) \\\\ &=((\\mathbf{x} \\cdot \\hat{\\mathbf{n}})^2 + (\\mathbf{x} \\wedge \\hat{\\mathbf{n}})^2) \\hat{\\mathbf{n}}+ 2 (\\mathbf{x} \\cdot \\hat{\\mathbf{n}}) (\\mathbf{x} \\wedge \\hat{\\mathbf{n}}) \\hat{\\mathbf{n}}\\end{aligned} We can read 2.5 off of this result by inspection for the $\\hat{\\mathbf{n}} = \\mathbf{e}_1$ case. It is now straightforward to show that the product $(\\mathbf{x} \\hat{\\mathbf{n}})^p \\hat{\\mathbf{n}}$ is a vector for integer $p \\ge 2$. We’ve covered the $p=2$ case, justifying an assumption that this product has the following form \\begin{aligned}(\\mathbf{x} \\hat{\\mathbf{n}})^{p-1} \\hat{\\mathbf{n}} = a \\hat{\\mathbf{n}} + b (\\mathbf{x} \\wedge \\hat{\\mathbf{n}}) \\hat{\\mathbf{n}},\\end{aligned} \\hspace{\\stretch{1}}(2.6) for scalars $a$ and $b$. The induction test becomes \\begin{aligned}(\\mathbf{x} \\hat{\\mathbf{n}})^{p} \\hat{\\mathbf{n}} &= (\\mathbf{x} \\hat{\\mathbf{n}})^{p-1} (\\mathbf{x} \\hat{\\mathbf{n}}) \\hat{\\mathbf{n}} \\\\ &= (\\mathbf{x} \\hat{\\mathbf{n}})^{p-1} \\mathbf{x} \\\\ &= (a + b (\\mathbf{x} \\wedge \\hat{\\mathbf{n}}) ) ((\\mathbf{x} \\cdot \\hat{\\mathbf{n}} )\\hat{\\mathbf{n}} + (\\mathbf{x} \\wedge \\hat{\\mathbf{n}}) \\hat{\\mathbf{n}}) \\\\ &= ( a(\\mathbf{x} \\cdot \\hat{\\mathbf{n}} )^2 - b (\\mathbf{x} \\wedge \\hat{\\mathbf{n}})^2 ) \\hat{\\mathbf{n}}+ ( a + b(\\mathbf{x} \\cdot \\hat{\\mathbf{n}} ) ) (\\mathbf{x} \\wedge \\hat{\\mathbf{n}}) \\hat{\\mathbf{n}}.\\end{aligned} Again we have a vector split nicely into projective and rejective components, so for any integer power of $p$ our iterator 2.4 employing the geometric product is a mapping", "# arrow_back Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. 8 views Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. It suffices to show that $$\\mathbf{i} \\cdot \\mathbf{j}=\\mathbf{i} \\cdot \\mathbf{k}=\\mathbf{j} \\cdot \\mathbf{k}=\\mathbf{0}$$ because it will follow automatically from the symmetry property of the dot product that $$\\mathbf{j} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{j}=\\mathbf{0}$$ Although the orthogonality of the vectors in $S$ is evident geometrically from Figure $3.2 .2$ it is confirmed algebraically by the computations \\begin{aligned} &\\mathbf{i} \\cdot \\mathbf{j}=(1,0,0) \\cdot(0,1,0)=0 \\\\ &\\mathbf{i} \\cdot \\mathbf{k}=(1,0,0) \\cdot(0,0,1)=0 \\\\ &\\mathbf{j} \\cdot \\mathbf{k}=(0,1,0) \\cdot(0,0,1)=0 \\end{aligned} by Platinum (106,844 points) selected by ## Related questions Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$.Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors.Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. ... close Notice: Undefined index: avatar", "\\cdot \\mathbf{A}_\\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum. Momentum Now move on to (16). For the factors other than $\\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a \\mathbf{a} (\\mathbf{k} \\wedge \\mathbf{c}) }}\\right\\rangle&=\\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) - \\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) \\\\ &=\\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}) - \\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}),\\end{aligned} and the other \\begin{aligned}\\sigma_a \\left\\langle{{ \\sigma_a (\\mathbf{k} \\wedge \\mathbf{c}) \\mathbf{a} }}\\right\\rangle&=\\sigma_a (\\mathbf{k} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{c}) - \\sigma_a (\\mathbf{c} \\cdot \\sigma_a) (\\mathbf{a} \\cdot \\mathbf{k}) \\\\ &=\\mathbf{k} (\\mathbf{a} \\cdot \\mathbf{c}) - \\mathbf{c} (\\mathbf{a} \\cdot \\mathbf{k}).\\end{aligned} The momentum $\\mathbf{P} = \\int T(\\gamma^0) \\wedge \\gamma_0$ in this volume follows by computation of \\begin{aligned}&\\sigma_a \\left\\langle{{ \\sigma_a\\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} - i \\mathbf{k} \\wedge {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\right) \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} + i \\mathbf{k} \\wedge \\mathbf{A}_\\mathbf{k} \\right) }}\\right\\rangle \\\\ &= i \\mathbf{A}_\\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{k} \\right) - i \\mathbf{k} \\left( \\left( -\\frac{1}{{c}} {{(\\dot{\\mathbf{A}}_{\\mathbf{k}})}}^{*} + i \\mathbf{k} {{\\phi_{\\mathbf{k}}}}^{*} \\right) \\cdot \\mathbf{A}_\\mathbf{k} \\right) \\\\ &- i \\mathbf{k} \\left( \\left( \\frac{1}{{c}} \\dot{\\mathbf{A}}_\\mathbf{k} + i \\mathbf{k} \\phi_\\mathbf{k} \\right) \\cdot {{\\mathbf{A}_\\mathbf{k}}}^{*} \\right) + i {{\\mathbf{A}_{\\mathbf{k}}}}^{*} \\left( \\left(", "for any basis, not necessarily the standard basis. Then you will get components with respect to the given basis -- if they all coincide, then the vectors also coincide. It's not a rigorous proof but maybe helps to make the statement intuitive. Here is a proof. If $$\\mathbf{A}\\cdot\\mathbf{C} =\\mathbf{B}\\cdot\\mathbf{C}$$ for all $$\\mathbf{C}$$, then $$(\\mathbf{A}-\\mathbf{B})\\cdot\\mathbf{C} = 0$$ for all $$\\mathbf{C}$$. In particular we can choose $$\\mathbf{C} = \\mathbf{A}-\\mathbf{B}$$ so that $$(\\mathbf{A}-\\mathbf{B})\\cdot(\\mathbf{A}-\\mathbf{B})=0$$. Since the dot product is positive definite, $$\\mathbf{v} \\cdot \\mathbf{v} = 0$$ only if $$\\mathbf{v} = 0$$. We conclude $$\\mathbf{A} - \\mathbf{B} = 0$$, so $$\\mathbf{A} = \\mathbf{B}$$.", "|| \\mathbf{x} |||| \\mathbf{y}|| - || \\mathbf{x} |||| \\mathbf{z}|| > -6 -8=-14.$$ This gives, $$-14 <|| \\mathbf{x} |||| \\mathbf{y}|| cos (\\theta) - || \\mathbf{x} |||| \\mathbf{z}|| cos (\\alpha) < 14$$ or, $$\\left | || \\mathbf{x} |||| \\mathbf{y}|| cos (\\theta) - || \\mathbf{x} |||| \\mathbf{z}|| cos (\\alpha) \\right | < 14,$$ $$| \\mathbf{x} \\cdot \\mathbf{y} - \\mathbf{x} \\cdot \\mathbf{z}| < 14.$$ Is this correct? If it is, is my explanation of the extreme values rigorous enough? (b) Set $B:= \\left \\{ \\mathbf{x} \\in \\mathbb{R}^n : ||\\mathbf{x}|| <1 \\right \\}.$ If $\\mathbf{x}, \\mathbf{y}, \\mathbf{z} \\in B$, show that $$\\mathbf{w} = \\frac{(\\mathbf{x} \\cdot \\mathbf{y})\\mathbf{z} + (\\mathbf{x} \\cdot \\mathbf{z})\\mathbf{y} + (\\mathbf{z} \\cdot \\mathbf{y})\\mathbf{x}}{3}$$ belongs to B. Using the same reasoning as before, and defining the dot products as the product of the lengths of the vectors and the cosine of the angle between the vectors, I get $$|| \\mathbf{w} || = \\left \\| \\frac{(\\mathbf{x} \\cdot \\mathbf{y})\\mathbf{z} + (\\mathbf{x} \\cdot \\mathbf{z})\\mathbf{y} + (\\mathbf{z} \\cdot \\mathbf{y})\\mathbf{x}}{3} \\right \\| < \\left \\| \\frac{\\mathbf{z} + \\mathbf{y} + \\mathbf{x}}{3} \\right \\| < \\frac{||\\mathbf{z}||}{3} + \\frac{||\\mathbf{y}||}{3} + \\frac{||\\mathbf{x}||}{3} < 1$$ So, $\\mathbf{w} \\in B$. (c) Let B be as above, and let $\\mathbf{z}$, $\\mathbf{w} \\in B$. Prove that $$|\\mathbf{x} \\cdot \\mathbf{z} - \\mathbf{y} \\cdot \\mathbf{w} | \\leq \\left \\| \\mathbf{y}-\\mathbf{z} \\right \\| + \\left \\| \\mathbf{x}-\\mathbf{w} \\right \\|$$", "and thus the dot product of $V_1$ and $V_2$ is $-1 + 2 (1/2) = 0$, as before. If we use an orthonormal basis, then we do not have any such cross-terms. In general, one must define how to compute the dot product in some basis and then how to compute the dot product in other bases will be determined by the condition that the dot product be independent of the choice of basis. If we don't satisfy this property then what we've defined isn't a true inner product. An inner product is a lot less unique than one might first suspect. As you have essentially noted: Theorem: Let $V$ be a finite-dimensional linear space over the field $\\mathcal{F}$ of real or complex numbers with basis $B=\\{ b_1,b_2,\\cdots,b_n\\}$. Then there exists an inner product $(\\cdot,\\cdot)_{B}$ on $V$ with respect to which $B$ is an orthonormal basis of $V$. Proof: Assume $\\mathcal{F}=\\mathbb{R}$ (the proof is essentially the same for $\\mathcal{F}=\\mathbb{C}$.) Define a linear map $L : \\mathbb{R}^{n}\\rightarrow V$ by $$L(\\alpha_1,\\alpha_2,\\cdots,\\alpha_n)=\\alpha_1b_1+\\alpha_2b_2+\\cdots\\alpha_nb_n.$$ Then $L$ is a linear bijection from $\\mathbb{R}^{n}$ onto $V$. It is easy to verify that the following defines an inner product $(\\cdot,\\cdot)_{B}$ on $V$: $$(x,y)_{B}=(L^{-1}x,L^{-1}y)_{\\mathbb{R}^{n}}.$$ If $e_1=(1,0,0,\\cdots,1)$, $e_2=(0,1,0,\\cdots,0)$, $\\cdots$, $e_n=(0,0,\\cdots,1)$ are the standard basis elements of $\\mathbb{R}^{n}$, then $L^{-1}b_k = e_k$ and, therefore, $(b_l,b_m)_{B}=(e_l,e_m)_{\\mathbb{R}^{n}}=\\delta_{l,m}$, which makes $B$ an", "Inner Product As we continue to study vectors, we will be looking at the concept of distance and angles between vectors. In order to do so, we need to define an inner product on a vector space. Over the powers of real numbers, we were able to use the dot product as an inner product; however, this is not the only possibility. As such, we will look at examples with dot products and with a different inner product. Over $$\\mathbb{R}^{3}$$ Let $$\\mathbf{u}=(1,1,2)$$ and $$\\mathbf{v}=(-1,3,0)$$ in the vector space $$\\mathbb{R}^{3}$$, then find the following • $$||\\mathbf{u}||$$, • $$d(\\mathbf{u},\\mathbf{v})$$, • $$\\mathbf{u}\\cdot \\mathbf{v}$$ and • the angle between $$\\mathbf{u}$$ and $$\\mathbf{v}$$. $$||\\mathbf{u}||$$ We first note that the magnitude of a vector in a vector space with a defined inner product is given by \\begin{align*} ||\\mathbf{u}||=\\sqrt{\\mathbf{u}\\cdot \\mathbf{u}}. \\end{align*} As such, we just need to find the dot product of $$\\mathbf{u}$$ with itself. This is then found at \\begin{align*} ||\\mathbf{u}||&=\\sqrt{\\mathbf{u}\\cdot \\mathbf{u}} \\\\ &=\\sqrt{(1,1,2) \\cdot (1,1,2)} \\\\ &=\\sqrt{1*1+1*1+2*2} \\\\ &=\\sqrt{6}. \\end{align*} $$d(\\mathbf{u},\\mathbf{v})$$ In order to find the distance between the two vectors, we need to find \\begin{align*} d(\\mathbf{u},\\mathbf{v})&=||\\mathbf{u}-\\mathbf{v}|| \\\\ &=\\sqrt{((1,1,2)-(-1,3,0))\\cdot((1,1,2-(-1,3,0))} \\\\ &=\\sqrt{(2,-2,2) \\cdot (2,-2,2)} \\\\ &=\\sqrt{2*2+(-2)(-2)+2*2} \\\\ &=\\sqrt{12}. \\end{align*} $$\\mathbf{u} \\cdot \\mathbf{v}$$ In order to find the dot product of the two vectors, we need only use the definition to find that \\begin{align*} \\mathbf{u}", "## Proof of Integral Identity for Dot Product of Normal With Product of a Scalar With a Vector Over a Surface Theorem If $\\phi$ is a differentiable scalar function and $\\mathbf{f}$ is a differentiable vector field then $\\int \\int_S \\phi \\mathbf{F} \\cdot \\mathbf{n} dS = \\int \\int \\int_V \\mathbf{F} \\cdot (\\mathbf{\\nabla} \\phi) dV +\\int \\int \\int_V \\phi \\mathbf{\\nabla} \\cdot \\mathbf{F} dV$ Proof Apply the Divergence Theorem to the vector field $\\phi \\mathbf{F}$ to obtain $\\int \\int_S \\phi \\mathbf{F} \\cdot \\mathbf{n} dS = \\int \\int \\int_V \\mathbf{\\nabla} \\cdot (\\phi \\mathbf{F}) dV$ Now use the identity $\\mathbf{\\nabla} \\cdot (\\phi \\mathbf{F}) = \\phi \\mathbf{\\nabla} \\cdot \\mathbf{F} + \\mathbf{F} \\cdot (\\mathbf{\\nabla} \\phi)$ to obtain \\begin{aligned} \\int \\int_S \\phi \\mathbf{F} \\cdot \\mathbf{n} dS &= \\int \\int \\int_V \\phi \\mathbf{\\nabla} \\cdot \\mathbf{F} + \\mathbf{F} \\cdot (\\mathbf{\\nabla} \\phi) dV \\\\ &= \\int \\int \\int_V \\phi \\mathbf{\\nabla} \\cdot \\mathbf{F} dV+ \\int \\int \\int_V \\mathbf{F} \\cdot (\\mathbf{\\nabla} \\phi) dV \\end{aligned}", "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "Finally in the second line we see things like $$\\mathbf x \\mathbf x^T$$. This a matrix. For example $$M=\\mathbf a \\mathbf b^T$$ is a matrix whose components are given by products of the components of the vectors: $$M_{ij}=a_i b_j$$. There's no conflict with having both $$\\mathbf x \\mathbf x^T$$ and $$\\mathbf x^T \\mathbf x$$ at the same time. They just mean different things. Note that dot product multiplication commutes so $$\\mathbf x\\cdot \\mathbf{\\hat n} = \\mathbf{\\hat n}\\cdot \\mathbf x$$. With that in mind you can write $$(\\mathbf x \\cdot \\mathbf{\\hat n})^2 = (\\mathbf x \\cdot \\mathbf{\\hat n})(\\mathbf x \\cdot \\mathbf{\\hat n}) = (\\mathbf{\\hat n}\\cdot \\mathbf x)(\\mathbf x \\cdot \\mathbf{\\hat n})=(\\mathbf{\\hat n}^T \\mathbf x)(\\mathbf x^T \\mathbf{\\hat n})$$, having switched between the various notational conventions ;). But the parentheses don't really matter here, and you could just as well imagine forming the matrix $$\\mathbf x \\mathbf x^T$$ first, and then multiplying it on both sides by $$\\mathbf{\\hat n}$$ later, instead of taking the dot products first and multiplying the resulting scalars second. The notation in this text sample was really bad, so don't feel discouraged. $$\\vec u=(\\vec x^T\\,\\vec n)\\,\\vec n$$ \\begin{align*} & \\text{\\vec u~ is also}\\\\ &\\vec{u}=\\vec{n}\\,\\vec{n}^T\\,\\vec x= \\left[ \\begin {array}{ccc} {n_{{x}}}^{2}&n_{{x}}n_{{y}}&n_{{x}}n_{{z} }\\\\ n_{{x}}n_{{y}}&{n_{{y}}}^{2}&n_{{y}}n_{{z}} \\\\ n_{{x}}n_{{z}}&n_{{y}}n_{{z}}&{n_{{z}}}^{2} \\end {array} \\right]\\vec x =\\boldsymbol A\\,\\vec{x}\\\\ &\\text{with}\\\\ &\\boldsymbol A^T=A\\\\ &\\boldsymbol A^T\\,A=A\\,A=\\vec{n}\\,\\vec{n}^T\\,\\vec{n}\\,\\vec{n}^T=\\boldsymbol A \\end{align*} where" ]

[ "than $C$, so we can discard this, and $D=(2,1).$ The value of $s$ we seek is $1$. #### Approach 2: using the angle in a semicircle If $A$,$B$ and $D$ are points on a circle with $AB$ a diameter of that circle, then $\\widehat{ADB}$ is a right angle. The converse is also true; if $\\widehat{ADB}=\\dfrac{\\pi}{2}$, then $AB$ has to be a diameter of the circle. Thus $D$ is on the circle with diameter $AB$. The midpoint of $AB$ is $(3,3)$ and the length of $AB$ is $2\\sqrt{5}$, and so the circle has centre $(3,3)$ and radius $\\sqrt{5}$. Thus the equation of the circle is $(x-3)^2+(y-3)^2=5$. The point $D$ is on this, so $(2s-3)^2 +(s-3)^2 = 5$, or $5s^2-18s+13=0$. The solution then proceeds as before. #### Approach 3: using the scalar product We know that the lines $DA$ and $DB$ have to be perpendicular. Thus $(\\mathbf{d}-\\mathbf{a})\\cdot(\\mathbf{d}-\\mathbf{b})=0,$ where $\\mathbf{d}$ is the position vector of $D$, and where $\\cdot$ represents the scalar product. Now $\\mathbf{d} = \\begin{pmatrix}2s\\\\s\\end{pmatrix},$ so $\\mathbf{d}-\\mathbf{a} =\\begin{pmatrix}2s-2\\\\s-5\\end{pmatrix} , \\mathbf{d}-\\mathbf{b}= \\begin{pmatrix}2s-4\\\\s-1\\end{pmatrix}.$ Thus we have \\begin{align*} (\\mathbf{d}-\\mathbf{a})\\cdot(\\mathbf{d}-\\mathbf{b})&=\\begin{pmatrix}2s-2\\\\s-5\\end{pmatrix} \\cdot \\begin{pmatrix}2s-4\\\\s-1\\end{pmatrix}\\\\ &=(2s-2)(2s-4)+(s-5)(s-1)\\\\ &=5s^2-18s+13=0, \\end{align*} as before, and so $\\mathbf{d}=\\begin{pmatrix}2\\\\1\\end{pmatrix}, s = 1.$", "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "\\begin{pmatrix}x_1 + x_2 \\\\ y_1 + y_2 \\\\ z_1 + z_2\\end{pmatrix}$ $a \\begin{pmatrix}x \\\\ y \\\\ z\\end{pmatrix} = \\begin{pmatrix}ax \\\\ ay \\\\ az\\end{pmatrix}$ ## Magnitude We can find the magnitude (length) of a vector by using Pythagoras’ theorem: $\\lVert \\mathbf{v} \\rVert = \\lVert \\begin{pmatrix}x \\\\ y \\\\ z\\end{pmatrix} \\rVert = \\sqrt{x^2 + y^2 + z^2}$ ## Normalisation It often makes it easier to work with vectors if they have a length of $$1$$. We can scale a vector to have a length of $$1$$ (i.e. normalise the vector) by dividing by its length: $\\hat{\\mathbf{v}} = \\frac{\\mathbf{v}}{\\lVert \\mathbf{v} \\rVert}$ ## Dot product The dot product is one method of multiplying vectors, producing a scalar result. The process is very simple - we just multiply element-by-element and add them all together: $\\mathbf{v_1} \\cdot \\mathbf{v_2} = \\begin{pmatrix}x_1 \\\\ y_1 \\\\ z_1\\end{pmatrix} \\cdot \\begin{pmatrix}x_2 \\\\ y_2 \\\\ z_2\\end{pmatrix} = (x_1 \\cdot x_2) + (y_1 \\cdot y_2) + (z_1 \\cdot z_2)$ The dot product is useful for finding the angle between two vectors, since for any two vectors $$\\mathbf{v_1}$$ and $$\\mathbf{v_2}$$ the following rule holds (where $$\\theta$$ is the angle between the vectors): $\\mathbf{v_1} \\cdot \\mathbf{v_2} = \\lVert \\mathbf{v_1} \\rVert \\lVert \\mathbf{v_2} \\rVert \\cos \\theta$ This has the nice side-effect that the dot product is $$0$$ when the two vectors are perpendicular. ##", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "equal to ${\\displaystyle \\mathbf {a} ={\\begin{pmatrix}a_{1}\\\\a_{2}\\\\a_{3}\\\\a_{4}\\end{pmatrix}}.}$ This can be written in terms of the four standard basis vectors (e1, e2, e3, e4), given by ${\\displaystyle \\mathbf {e} _{1}={\\begin{pmatrix}1\\\\0\\\\0\\\\0\\end{pmatrix}};\\mathbf {e} _{2}={\\begin{pmatrix}0\\\\1\\\\0\\\\0\\end{pmatrix}};\\mathbf {e} _{3}={\\begin{pmatrix}0\\\\0\\\\1\\\\0\\end{pmatrix}};\\mathbf {e} _{4}={\\begin{pmatrix}0\\\\0\\\\0\\\\1\\end{pmatrix}},}$ so the general vector a is ${\\displaystyle \\mathbf {a} =a_{1}\\mathbf {e} _{1}+a_{2}\\mathbf {e} _{2}+a_{3}\\mathbf {e} _{3}+a_{4}\\mathbf {e} _{4}.}$ Vectors add, subtract and scale as in three dimensions. The dot product of Euclidean three-dimensional space generalizes to four dimensions as ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}+a_{4}b_{4}.}$ It can be used to calculate the norm or length of a vector, ${\\displaystyle \\left|\\mathbf {a} \\right|={\\sqrt {\\mathbf {a} \\cdot \\mathbf {a} }}={\\sqrt {{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}+{a_{4}}^{2}}},}$ and calculate or define the angle between two vectors as ${\\displaystyle \\theta =\\arccos {\\frac {\\mathbf {a} \\cdot \\mathbf {b} }{\\left|\\mathbf {a} \\right|\\left|\\mathbf {b} \\right|}}.}$ Minkowski spacetime is four-dimensional space with geometry defined by a nondegenerate pairing different from the dot product: ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}-a_{4}b_{4}.}$ As an example, the distance squared between the points (0,0,0,0) and (1,1,1,0) is 3 in both the Euclidean and Minkowskian 4-spaces, while the distance squared between (0,0,0,0) and (1,1,1,1) is 4 in Euclidean space and 2 in Minkowski space; increasing ${\\displaystyle b_{4}}$ actually decreases the metric distance. This leads to many of the well known apparent \"paradoxes\" of relativity. The cross product is not defined in", "Since $(21,26)=1$, there exists $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$. Remember that finding $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$ is equivalent to solving the diophantine equation $21x+26y=1$. \\begin{align*} 26&=21\\cdot 1+5\\\\ 21&=5\\cdot 4+1 \\end{align*} and \\begin{align*} 1&=21-5\\cdot 4\\\\ &=21-(26-21)4\\\\ &=5\\cdot 21+26(-4) \\end{align*} Thus, we find $21^{-1}\\equiv 5\\mod 26$. \\begin{align*} A^{-1}&=\\begin{pmatrix} 5\\cdot 8 & -5\\cdot 3\\\\ -5\\cdot 7 & 5\\cdot 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 40 & -15\\\\ -35 & 10 \\end{pmatrix}\\\\ &\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$. Then $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=A\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\ cx+dy \\end{pmatrix}$$ This gives a linear map or a linear transformation from vectors to vectors. We introduce the following theorem without proof. Theorem. Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ and set $D=ad-bc$ as before. Then the following are equivalent: 1. $(D,N)=1$. 2. $A$ has an inverse matrix. 3. if $x$ and $y$ are not both $0$ in $\\mathbb{Z}/N\\mathbb{Z}$, then $A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4. $A$ gives a one-to-one correspondence of $(\\mathbb{Z}/N\\mathbb{Z})^2$ with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the", "is given by \\begin{align}x'&=x\\cos\\theta-y\\sin\\theta \\\\ y'&=x\\sin\\theta+y\\cos\\theta, \\end{align} or, in matrix form, $$\\begin{bmatrix}x'\\\\y'\\end{bmatrix}\\begin{bmatrix}\\cos\\theta&-\\sin\\theta\\\\\\sin\\theta&\\cos\\theta\\end{bmatrix}\\begin{bmatrix}x\\\\y\\end{bmatrix}.$$ (If a positive angle instead represents a clockwise rotation, change the sign of the $\\sin\\theta$ terms.) If you don’t have the angle $\\theta$ between the blue and green vectors (call them $\\mathbf b$ and $\\mathbf g$) handy, you can recover it from them via the dot product: $$\\mathbf b\\cdot\\mathbf g = \\|\\mathbf b\\|\\,\\|\\mathbf g\\|\\cos\\theta$$ and so $$\\cos\\theta={\\mathbf b\\cdot\\mathbf g\\over\\|\\mathbf b\\|\\,\\|\\mathbf g\\|}.$$ You don’t have to compute the angle $\\theta$ itself to derive the rotation. Once you have $\\cos\\theta$, you can get $\\sin\\theta$ via the identity $\\cos^2\\theta+\\sin^2\\theta=1$, though you have to take a bit of care to get the sign right. In your situation we can simplify things a bit. Assuming that “up” corresponds to the positive $y$-direction, $\\mathbf b\\cdot\\mathbf g=(0,y_b)\\cdot(x_g,y_g)=y_by_g$, but $\\|\\mathbf b\\|=y_b=\\|\\mathbf g\\|$, so $$\\cos\\theta={y_g\\over y_b}={y_g\\over\\sqrt{x_g^2+y_g^2}}$$ and with a bit of thought you should be able to convince yourself that $$\\sin\\theta=-{x_g\\over y_b}=-{x_g\\over\\sqrt{x_g^2+y_g^2}}.$$ The rotation matrix is therefore $$R=\\frac1{\\sqrt{x_g^2+y_g^2}}\\begin{bmatrix}y_g&x_g\\\\-x_g&y_g\\end{bmatrix}$$ which represents the transformation \\begin{align}x'&={y_gx+x_gy\\over\\sqrt{x_g^2+y_g^2}}\\\\y'&={-x_gx+y_gy\\over\\sqrt{x_g^2+y_g^2}}.\\end{align} Of course, if $\\mathbf b$ and $\\mathbf g$ are unit vectors, this becomes even simpler since there’s no need to normalize. Note that we could’ve instead derived this matrix directly from the vector $\\mathbf g$ by taking advantage of the fact that the columns of a transformation matrix are the images of", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "outer product is a row vector multiplied on the left by a column vector: $\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$ where $\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi:[1] $\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p", "+0 # Let $\\theta$ be the angle between $\\bold{a}$ and $\\bold{b}$, where $\\bold{a}$ and $\\bold{b}$ are the vectors defined in part (a). Find \\$\\cos 0 171 2 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$. Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix} \\quad \\text{and} \\quad b = \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix}$$. Find $$\\cos^2 \\theta$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\cos(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{\\vec{a}\\cdot \\vec{b}}{|\\vec{a}|\\cdot |\\vec{b}|} } \\quad | \\quad |\\vec{a}| = \\sqrt{2^2+(-6)^2+5^2}=\\sqrt{65},\\ |\\vec{b}| = \\sqrt{(-1)^2+(-2)^2+0^2}=\\sqrt{5} \\\\\\\\ \\cos(\\theta) &=& \\dfrac{\\begin{pmatrix} 2 \\\\ -6 \\\\ 5 \\end{pmatrix}\\cdot \\begin{pmatrix} -1 \\\\ -2 \\\\ 0 \\end{pmatrix} } {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{2\\cdot (-1)+(-6)\\cdot(-2)+5\\cdot 0} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos(\\theta) &=& \\dfrac{10} {\\sqrt{65}\\cdot \\sqrt{5} } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {65\\cdot 5 } \\\\\\\\ \\cos^2(\\theta) &=& \\dfrac{100} {325 } \\\\\\\\ \\mathbf{\\cos^2(\\theta)} &\\mathbf{=}& \\mathbf{\\dfrac{4}{13}} \\\\ \\hline \\end{array}$$ Feb 28, 2019 #1 +23071 +3 Let $$\\theta$$ be the angle between a and b, where a and b are the vectors defined in $$a = \\begin{pmatrix}", "# Inner and Outer product in Linear Algebra Let’s take two coordinate vectors, $\\textbf v$ and $\\textbf u$. Both of them can be represented as column matrices (why not as row matrices?) of order $m \\times 1$ and $n \\times 1$. The inner product of these two coordinate vectors is, $\\textbf v^T \\cdot \\textbf u$. $\\textbf v = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3\\\\...\\\\a_m \\end{pmatrix}$ $\\textbf v^T = \\begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \\end{pmatrix}$ $\\textbf u = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3\\\\...\\\\ b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = \\begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 \\\\ b_2 \\\\b_3 \\\\ ... \\\\b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$ Notice that it only works if $m = n$ and it always results in a matrix of order $1 \\times 1$. The outer product of these two vectors is $\\textbf v \\cdot \\textbf u^T$. $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\\\ ... \\\\ a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \\end{pmatrix}$ $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\\\a_2 b_1 &", "\\end{pmatrix}^k = \\begin{pmatrix} A_{11}^k & 0 & \\cdots & 0 \\\\ 0 & A_{22}^k & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & A_{nn}^k \\end{pmatrix}$$ meaning its easy to raise a diagonal matrix to a power. When raising an arbitrary matrix (not necessarily a diagonal matrix) to a power, it is often helpful to exploit this property by diagonalizing the matrix first. The inner and outer products Given two column vectors a and b, the Euclidean inner product and outer product are the simplest special cases of the matrix product, by transposing the column vectors into row vectors.[5] The inner product is a column vector multiplied on the left by a row vector: $$\\mathbf{a}\\cdot \\mathbf{b} = \\mathbf{a}^\\mathrm{T}\\mathbf{b}$$ More explicitly, $$\\mathbf{a}\\cdot \\mathbf{b} = \\begin{pmatrix}a_1 & a_2 & \\cdots & a_n\\end{pmatrix} \\begin{pmatrix}b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n\\end{pmatrix} = a_1b_1+a_2b_2+\\cdots+a_nb_n = \\sum_{i=1}^n a_ib_i$$ The outer product is a row vector multiplied on the left by a column vector: $$\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$$ where $$\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots", "x}+{\\mathbf y}\\neq {\\mathbf 0} \\}\\\\ & = \\min\\{ w({\\mathbf z}) : {\\mathbf z} \\neq {\\mathbf 0} \\}\\text{.} \\end{align*} ### SubsectionLinear Codes From Example 8.16, it is now easy to check that the minimum nonzero weight is $$3\\text{;}$$ hence, the code does indeed detect and correct all single errors. We have now reduced the problem of finding “good” codes to that of generating group codes. One easy way to generate group codes is to employ a bit of matrix theory. Define the inner product of two binary $$n$$-tuples to be \\begin{equation*} {\\mathbf x} \\cdot {\\mathbf y} = x_1 y_1 + \\cdots + x_n y_n\\text{,} \\end{equation*} where $${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^\\transpose$$ and $${\\mathbf y} = (y_1, y_2, \\ldots, y_n)^\\transpose$$ are column vectors. 5 For example, if $${\\mathbf x} = (011001)^\\transpose$$ and $${\\mathbf y} = (110101)^\\transpose\\text{,}$$ then $${\\mathbf x} \\cdot {\\mathbf y} = 0\\text{.}$$ We can also look at an inner product as the product of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^\\transpose {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}\\text{.} \\end{align*} Since we will be working with matrices, we will write binary", "{J} =\\mathbf {ji} -\\mathbf {ij} ={\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}}}$ is a 90° anticlockwise rotation operator in 2d. It can be left-dotted with a vector r = xi + yj to produce the vector, ${\\displaystyle (\\mathbf {ji} -\\mathbf {ij} )\\cdot (x\\mathbf {i} +y\\mathbf {j} )=x\\mathbf {ji} \\cdot \\mathbf {i} -x\\mathbf {ij} \\cdot \\mathbf {i} +y\\mathbf {ji} \\cdot \\mathbf {j} -y\\mathbf {ij} \\cdot \\mathbf {j} =-y\\mathbf {i} +x\\mathbf {j} ,}$ in summary ${\\displaystyle \\mathbf {J} \\cdot \\mathbf {r} =\\mathbf {r} _{\\mathrm {rot} }}$ or in matrix notation ${\\displaystyle {\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}}{\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\begin{pmatrix}-y\\\\x\\end{pmatrix}}.}$ For any angle θ, the 2d rotation dyadic for a rotation anti-clockwise in the plane is ${\\displaystyle \\mathbf {R} =\\mathbf {I} \\cos \\theta +\\mathbf {J} \\sin \\theta =(\\mathbf {ii} +\\mathbf {jj} )\\cos \\theta +(\\mathbf {ji} -\\mathbf {ij} )\\sin \\theta ={\\begin{pmatrix}\\cos \\theta &-\\sin \\theta \\\\\\sin \\theta &\\;\\cos \\theta \\end{pmatrix}}}$ where I and J are as above, and the rotation of any 2d vector a = axi + ayj is ${\\displaystyle \\mathbf {a} _{\\mathrm {rot} }=\\mathbf {R} \\cdot \\mathbf {a} }$ #### 3d rotations A general 3d rotation of a vector a, about an axis in the direction of a unit vector ω and anticlockwise through angle θ, can be performed using Rodrigues' rotation formula in the dyadic form ${\\displaystyle \\mathbf {a} _{\\mathrm {rot} }=\\mathbf {R} \\cdot \\mathbf {a} \\,,}$ ${\\displaystyle \\mathbf {R} =\\mathbf {I} \\cos", "Math Relative position of lines Angle between two lines # Angle between two lines The angle between two lines is calculated similarly as the angle between two vectors. ! ### Remember In contrast to vectors, the angle between two lines always discribes the angle less than 90°. Two lines are given: $\\text{g: } \\vec{x} = \\vec{u} + r \\cdot \\vec{a}$ $\\text{h: } \\vec{x} = \\vec{v} + s \\cdot \\vec{b}$ For determining the angle the direction vectors of the lines are used: $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\gamma = \\cos^{-1}\\left(\\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}\\right)$ i ### Method 1. Calculate the scalar product of the direction vectors 2. Calculate the magnitude of the direction vectors 3. Insert results into the formula ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Calculate scalar product $\\vec{a}\\cdot\\vec{b}$ $=\\begin{pmatrix}9 \\\\ 8 \\\\ 7 \\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ 1 \\\\ 2\\end{pmatrix}$ $=9\\cdot1+8\\cdot1+7\\cdot2$ $=31$ 2. #### Calculate magnitude of the direction vectors $|\\vec{a}|=\\sqrt{9^2+8^2+7^2}$ $=\\sqrt{194}$ $|\\vec{b}|=\\sqrt{1^2+1^2+2^2}$ $=\\sqrt{6}$ 3. #### Insert results into the formula $\\cos(\\gamma) = \\frac{|\\vec{a}\\cdot\\vec{b}|}{|\\vec{a}|\\cdot|\\vec{b}|}$ $\\cos(\\gamma) = \\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}$ $\\gamma = \\cos^{-1}\\left(\\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}\\right)$ $\\approx24.68°$", "= \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}$ Now, using the formula for the angle between two vectors: $\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ 4. (Original post by The_Big_E) A general point on the line L has coordinates: $\\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}$ Call this point C, with position vector $\\vec{OC}$ The angle between vectors $\\vec{AB}$ and $\\vec{BC}$ is $90^{\\circ}$ So first we have to find $\\vec{BC}$. $\\vec{BC} = \\vec{OC} - \\vec{OB}$ $\\vec{BC} = \\begin{pmatrix} 3-3x\\\\ -4+5x\\\\ 2+2x \\end{pmatrix}-\\begin{pmatrix} 1\\\\ 0\\\\ -2 \\end{pmatrix}$ $\\vec{BC} = \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}$ Now, using the formula for the angle between two vectors: $\\cos \\theta = \\frac{\\vec{AB}\\cdot \\vec{BC}}{\\left | \\vec{AB} \\right |\\left | \\vec{BC} \\right |}$ Given that the angle between the vectors is 90, cos90 = 0. $\\vec{AB}\\cdot \\vec{BC} = \\begin{pmatrix} -2\\\\ 4\\\\ -4 \\end{pmatrix}\\cdot \\begin{pmatrix} 2-3x\\\\ -4+5x\\\\ 4+2x \\end{pmatrix}=0$ $18x-36=0$ $x=2$ So the point C on line L is where $x=2$. $C\\begin{pmatrix} 3-3(2)\\\\ -4+5(2)\\\\ 2+2(2) \\end{pmatrix}$ Therefore, $C\\begin{pmatrix} -3\\\\ 6\\\\ 6 \\end{pmatrix}$ Thank you, I get", "\\cdot (b_2\\textbf{j}) + (a_2\\textbf{j}) \\cdot (b_1\\textbf{i}) +(a_2\\textbf{j}) \\cdot (b_2\\textbf{j}) \\\\ &= a_1b_1(\\textbf{i} \\cdot \\textbf{i}) + a_1b_2(\\textbf{i} \\cdot \\textbf{j}) + a_2b_1(\\textbf{j} \\cdot \\textbf{i}) + a_2b_2(\\textbf{j} \\cdot \\textbf{j}) \\end{align*} Given the properties of the scalar product and the fact both $\\textbf{i}$ and $\\textbf{j}$ have magnitude 1, we get following definition #### Scalar product of vectors in terms of components If $\\textbf{a} = a_1\\textbf{i} + a_2\\textbf{j} + a_3\\textbf{k}$ and $\\textbf{b} = b_1\\textbf{i} + b_2\\textbf{j} + b_3\\textbf{k}$, then $$\\textbf{a} \\cdot \\textbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$ In column notation, if $\\textbf{a} = \\begin{pmatrix}a_1 \\ a_2 \\ a_3\\end{pmatrix}$ and $\\textbf{b} = \\begin{pmatrix}b_1 \\ b_2 \\ b_3\\end{pmatrix}$, then $\\textbf{a} \\cdot \\textbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ ## 7.2 Finding the angle between two vectors Rearranging the definition of the scalar product gives following fact #### Angle between two vectors The angle $\\theta$ between any two non-zero vectors $\\textbf{a}$ and $\\textbf{b}$ is given by $$\\cos\\theta = \\frac{\\textbf{a} \\cdot \\textbf{b}}{|\\textbf{a}||\\textbf{b}|}$$", "see from the image below this projection is nothing but $|\\mathbf a| \\cos \\theta$ where $\\theta$ is the angle between $\\mathbf a$ and $\\mathbf b$. Thus $\\mathbf a\\cdot \\mathbf b = |\\mathbf a| \\cos \\theta \\cdot |\\mathbf b|$. The dot product holds some notable properties: 1. $\\mathbf a \\cdot \\mathbf b = \\mathbf b \\cdot \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\cdot \\mathbf b = \\alpha \\cdot (\\mathbf a \\cdot \\mathbf b)$ 3. $(\\mathbf a + \\mathbf b)\\cdot \\mathbf c = \\mathbf a \\cdot \\mathbf c + \\mathbf b \\cdot \\mathbf c$ I.e. it is a commutative function which is linear with respect to both arguments. Let's denote the unit vectors as $$\\mathbf e_x = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, \\mathbf e_y = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}, \\mathbf e_z = \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}.$$ With this notation we can write the vector $\\mathbf r = (x;y;z)$ as $r = x \\cdot \\mathbf e_x + y \\cdot \\mathbf e_y + z \\cdot \\mathbf e_z$. And since for unit vectors $$\\mathbf e_x\\cdot \\mathbf e_x = \\mathbf e_y\\cdot \\mathbf e_y = \\mathbf e_z\\cdot \\mathbf e_z = 1,\\\\ \\mathbf e_x\\cdot \\mathbf e_y = \\mathbf e_y\\cdot \\mathbf e_z = \\mathbf e_z\\cdot \\mathbf e_x = 0$$ we can see that in terms of coordinates for $\\mathbf" ]

[ "we have$\\tan (a-b)=\\frac{\\tan a - \\tan b}{1+\\tan a \\tan b} \\$, hence $$\\arctan a -\\arctan b = \\arctan \\frac{a-b}{1+ab}$$ and this gives the desired result. (By the way, I wonder why someone would intentionally write the answer in such a complicated form!)", "$f'(x)=\\frac1{1+x^2},$ where $f(x)=\\arctan x;$ and $f'(x)=1,$ where $f(x)=x.$ Plus, chain rule is involved here.", "+ \\frac{1}{4} \\ln | \\frac{x - 1}{x + 1} | - \\arctan \\frac{x}{2} + C$.", "$x=0$ OR $x=4$ #### Example Find all critical points of $f(x)=in(x+1)-arctan(x)$ $f'(x)=\\frac{1}{x+1} -\\frac{1}{x^{2}+1}$ $=\\frac{(x^{2}+1)-(x+1)}{(x^{2}+1)}$ $=\\frac{x^{2}-x}{(x+1)(x^{2}+1)}$ $=\\frac{x(x-1)}{(x+1)(x^{2}+1)}$ Thus, $f(x)$ has critical points at - $x=1 \\Rightarrow f'(x)$ undef $x=0 \\Rightarrow f'(x)=0$ $x=1 \\Rightarrow f'(x)=0$", "$arctan(\\frac{1}{\\sqrt{3}}) = \\frac{\\pi}{6} > \\pi (\\frac{1}{\\sqrt{3}})^4 = \\frac{\\pi}{9}$", "Find the first partial derivatives of $f(x,y,z) = z \\ \\arctan(\\frac{y}{x})$ at the point (4, 4, -3). A. $\\frac{\\partial f}{\\partial x}(4, 4, -3) =$ B. $\\frac{\\partial f}{\\partial y}(4, 4, -3) =$ C. $\\frac{\\partial f}{\\partial z}(4, 4, -3) =$", "1, \\Rightarrow x_2 = \\arctan 1 + \\pi k = \\frac{\\pi}{4} + \\pi k, k \\in \\mathbb{Z}.$ $x_1 = \\frac{\\pi}{2} + \\pi n; \\;x_2 = \\frac{\\pi}{4} + \\pi k; \\;n, k \\in \\mathbb{Z}.$", "Now let's try $$x=\\sqrt{\\arctan \\theta}$$,$$dx=\\frac{1}{(1+\\theta^2)2\\sqrt{\\arctan \\theta}}$$ $$\\frac{\\tan^4(x^2)}{8x} + \\frac{\\tan^6(x^2)}{12x} + \\frac{1}{2} \\int \\frac{(1+\\theta^2)\\theta^4}{2\\arctan^{\\frac{3}{2}}\\theta (1+\\theta^2)}d\\theta$$ $$\\frac{\\tan^4(x^2)}{8x} + \\frac{\\tan^6(x^2)}{12x} + \\frac{1}{4} \\int \\frac{\\theta^4}{\\arctan^{\\frac{3}{2}}\\theta} d\\theta$$", "2. You're expected to know that the derivative of $\\arctan x$ equals $\\frac1{1+x^2}.$ By the Chain Rule, we have that the derivative of $\\arctan y^9$ equals $\\frac1{1+y^{18}}\\cdot(y^9)'.$", "$$x^{4}$$ $$\\frac{\\arctan (x^{4})}{4}+C$$", "xdx=\\frac{1}{(b+1)^2+1}$$ and hence $$I(a,b)=\\int_a^b\\frac{1}{(s+1)^2+1}ds=\\arctan (b+1)-\\arctan(a+1).$$", "(2x+\\arctan \\frac {B}{A})}$$", "I can't believe I didn't think of doing that before. In case anyone else is wondering, the answer turns out to be-- $\\theta = arctan(\\frac{2v^2 \\frac{+}{} v^2 \\sqrt{4- \\frac{4g^2x^2}{v^4} - \\frac {8gy}{v^2}}}{2gx})$ Thanks again.", "Problem 10: If $y = \\arcsin{\\sqrt{1-x}} + \\arccos{\\sqrt{x}}$, then $\\frac{dy}{dx}$ equals: (a) $\\frac{1}{\\sqrt{x(1-x)}}$ (b) $\\frac{1}{x(1+x)}$ (c) $\\frac{-1}{\\sqrt{x(1-x)}}$ (d) none Regards, Nalin Pithwa", "{dx}{1 + x^2} = \\arctan x + C$ 8. Ack, my mistake. I made that question more complicated than it had to be by forgetting that $\\int\\frac{dx}{1+x^2} = arctan(x) + C$ -_-", "x = \\frac{b}{d} \\arctan\\left(\\frac{x}{d} \\right) + \\frac{a}{2} \\log\\left(x^2+d^2\\right)$$", "$\\frac{3 \\pi}{4}$ radians $\\tan \\left(\\pi - \\frac{\\pi}{4}\\right) = \\tan \\left(3 \\frac{\\pi}{4}\\right)$ = -1. Hence $\\arctan \\left(- 1\\right) = \\frac{3 \\pi}{4}$ radians", "u}{\\partial y}= \\pm\\frac{-y^2-x^2}{x^2+y^2}=\\mp 1.$$ Hence, we need $u(x,y)=-\\arctan\\frac{y}{x}$.", "It thus suffices to consider $$h(x) = \\arctan{x} - \\frac{x}{(1 + x^2)},$$ and show that $$h(x) \\ge 0$$ for $$x \\ge 0$$. But $$h(0) = 0$$, and $$h'(x) = \\frac{2x^2}{(1 + x^2)^2} \\ge 0$$ for all $$x$$. This completes the proof.", "Find the first partial derivatives of $f(x,y,z) = z \\ \\arctan(\\frac{y}{x})$ at the point (5, 5, 2). A. $\\frac{\\partial f}{\\partial x}(5, 5, 2) =$ B. $\\frac{\\partial f}{\\partial y}(5, 5, 2) =$ C. $\\frac{\\partial f}{\\partial z}(5, 5, 2) =$ Your overall score for this problem is" ]

[ "# Proving a curious formula of $\\pi$? I have recently come across this statement without proof. $$\\pi = 128 \\arctan\\frac{1}{40} -4\\arctan\\frac{1}{239} -16\\arctan\\frac{1}{515} -32\\arctan\\frac{1}{4030} -64\\arctan\\frac{1}{32060}$$ I'd put down my approach but, to be frank, I've gotten nowhere with this. How to go about this ? Use the fact that $\\arctan(1/a)$ is the argument of $a+i$, and that the arguments of complex numbers add up when you multiply them. According to WolframAlpha, we have \\begin{aligned} &\\frac{(40+i)^{128}}{(239+i)^4 (515+i)^{16} (4030+i)^{32} (32060+i)^{64}} = \\\\ &-1 / 37403944359352749280528518983232679702 \\\\ & 01985315749502348525466597837636105197 \\\\ & 87830439618227322115549670041854205583 \\\\ & 78215314658650047572142913167759891935 \\\\ & 23573829633433227264657819301199042671 \\\\ & 95356826263444502459300305177919563475 \\\\ & 022474784673838736016384 . \\end{aligned} Since this is a negative number, it has argument $\\pi$, and this must agree with your sum, except that they might differ by $2\\pi n$ for some integer $n$, since arguments are not uniquely defined. But just by numerical evaluation one should be convinced that your sum is a least close enough to $\\pi$ to rule out all the options except $n=0$. Q.E.D. • Can you explain why the argument has tobe$\\pi$ because it is a negative number ? Mar 29 '16 at 14:42 • (Strictly speaking, $\\pi$ is an argument of a negative real number. Also $-\\pi$, $3\\pi$, etc. are possible arguments.) Mar 29 '16 at 14:51", "If we divide both of these by $\\cos A \\cos B \\cos C \\cos D$, then we have $$\\tan(A + B + C + D) = \\frac {1 - \\sum \\tan A \\tan B + \\tan A \\tan B \\tan C \\tan D}{\\sum \\tan A - \\sum \\tan A \\tan B \\tan C},$$ which makes for more direct, less error-prone computations. Substitution gives the desired answer. ### Solution 3: Complex Numbers Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\\arctan\\frac{1}{n}$, is the argument of $n+i$. The sum of these angles is then just the argument of the product $$(3+i)(4+i)(5+i)(n+i)$$ and expansion give us $(48n-46)+(48+46n)i$. Since the argument of this complex number is $\\frac{\\pi}{4}$, its real and imaginary parts must be equal. So we set them equal and expand the product to get: $48n - 46 = 48 + 46n.$ Therefore, $n$ equals $\\boxed{047}$.", "principal argument of $1+2i$, and $\\arctan \\frac 12$ is the principal argument of $1+\\frac12 i$. Therefore their sum is an argument of $(1+2i)(1+\\frac12 i) = \\frac52 i$, so it must be $\\pi/2+2\\pi n$ for some $n$. But by considering the range of $\\arctan$ we can see that $n=0$ is the only possibility, so $$\\arctan 2 + \\arctan\\frac12 = \\frac\\pi2$$ - In fact arctanx=pi/2-arctan1/x, there are few ways of proving this, one way is the right-angled triangle way suggested by user84413 who just commented on your post, or you could tan both sides and carefully show they are both equivalent. -", "= \\arctan \\frac{3}{5} \\approx 0.54$. Notice that $\\varphi + \\alpha = \\arctan \\frac{5}{3} + \\arctan \\frac{3}{5} = \\frac{1}{2}\\pi$, as expected. Now, the argument of a complex number is by definition the angle made with the positive real axis. So to find the argument of $-3+5\\mathrm i$ you stand at the origin looking at the positive reals, e.g. at number $1 = 1+0\\mathrm i$. To face the number $-3+5\\mathrm i$ you need to turn anti-clockwise to the positive imaginary axis, and then continue on through the angle $\\alpha$. Hence, $\\arg(-3+5\\mathrm i) = \\frac{1}{2}\\pi + \\alpha = \\frac{1}{2}\\pi+\\arctan\\frac{3}{5} \\approx 2.11$", "four. We want to prove $4 \\arctan (1/5) - \\arctan (1/239) = \\frac{\\pi}{4}.$ How do we prove that? Well, the left-hand side may be written as $\\arg[(5+i)^4 (239-i)]$ That's great because we just need to compute the complex number which is the product of powers. Now, $(5+i)^2 = 5^2 + 10 i -1 = 24+10i = 2(12+5i)$ and $\\eq{ (5+i)^4 &= [(5+i)^2]^2 = 2^2 (12+5i)^2 =\\\\ &= 4 (144+120i - 25) = 4(119+120i). }$ Now, multiply it by the last factor,$\\eq{ (5+i)^4 (239-i) &= 4(119+120i)(239-i) = \\\\ &= 4(119\\times 239+120)+\\\\ &+4(120\\times 239i-119i) }$ What you may immediately see is that the real part and the imaginary part of the number coincide because $120+119=239$. So the complex number arising from the product has the form $N(1+i)$ where $N$ is a positive integer so it's clear that the argument is $\\pi/4$ which we wanted to prove. There are billions of known identities in mathematics and many of them may be proven in a similar indisputable way. But our discussion is about the hypothetical identities for which we can't do it – or we're lazy to do it. Is the numerical evidence enough to make us feel certain? In most cases we know, it is. But there may exist examples in which it is not. Playing with $j$-functions Just a month", "# What is the principal argument of $-5-5i$? When i calculated it by $\\tan^{-1}\\dfrac{y}{x}$, I got $\\dfrac{\\pi}{4}$, then i added $\\pi$ to make it in the right quadrant, so my final answer is $\\dfrac{5\\pi}{4}$ However, the correct answer is $-\\dfrac{3\\pi}{4}$...why is that? • The \"principal argument\" is often taken to be normalized to lie in $(-\\pi,\\pi]$. Then instead of adding $\\pi$, you should have subtracted $\\pi$. – Daniel Fischer Jan 12 '14 at 15:32 • Now thats a good answer...thanks! :) – Mohdak Jan 12 '14 at 15:37 Using the definition of atan2, $$\\arg(-5-5i)=\\arctan\\left(\\frac{-5}{-5}\\right)-\\pi=\\arctan1-\\pi=\\frac\\pi4-\\pi$$ • Your answer is absolutely correct except the notation that you need to use $\\mathrm{Arg}$. – NasuSama Jan 12 '14 at 15:44", "- \\frac{1}{2} \\arcsin(0.87)$$. From the inequality $$0 < \\frac{1}{2}\\arcsin(0.87) < \\frac{\\pi}{4}$$, we first multiply through by $$-1$$ and then add $$\\frac{\\pi}{2}$$ to get $$\\frac{\\pi}{2} > \\frac{\\pi}{2} -\\frac{1}{2} \\arcsin(0.87) > \\frac{\\pi}{4}$$, or $$\\frac{\\pi}{4} < \\frac{\\pi}{2} -\\frac{1}{2} \\arcsin(0.87) < \\frac{\\pi}{2}$$. Proceeding with the usual arguments, we find the only solutions which lie in $$[0,2\\pi)$$ correspond to $$k = 0$$ and $$k=1$$, namely $$x =\\frac{\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$ and $$x = \\frac{3\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$. All told, we have found four solutions to $$\\sin(2x) = 0.87$$ in $$[0,2\\pi)$$: $$x =\\frac{1}{2} \\arcsin(0.87)$$, $$x=\\frac{1}{2} \\arcsin(0.87) + \\pi$$, $$x =\\frac{\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$ and $$x = \\frac{3\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$. By graphing $$y = \\sin(2x)$$ and $$y = 0.87$$, we confirm our results. \\hspace{0.75in} \\includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq06.jpg} \\\\ \\$y = \\tan\\left(\\frac{x}{2}\\right)\\) and \\boldmath $$y = -3$$ & Each of the problems in Example \\ref{TrigEqnEx1} featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page \\pageref{trigeqnstrategy1}. We repeat here the advice given when solving systems of nonlinear equations in section \\ref{NonLinear} -- when it comes to solving equations involving the trigonometric functions, it helps to just try something. Next, we focus on solving inequalities involving", "# How do you find the exact value of cos[arc tan(-2/3)]? Jun 12, 2015 The exact value is apparently $\\frac{3}{\\sqrt{13}}$. However, you'll have to use a calculator to find $\\arctan \\left(- \\frac{2}{3}\\right)$. It's an irrational radian value that you'll get to be $- 33.69006752598 \\ldots e t {c}^{o}$. Wolfram Alpha simply gives it as: $- \\frac{180}{\\pi} \\cdot \\arctan \\left(- \\frac{2}{3}\\right)$ which doesn't really give a satisfactory exact answer (such as $\\frac{2}{\\sqrt{2}}$). It's just the conversion of $\\arctan \\left(- \\frac{2}{3}\\right)$ to degrees... If you just take this exact result and take the $\\cos$ of it, you'll get $\\frac{3}{\\sqrt{13}}$, but it probably won't be obvious at all that it's equal to that from just looking at it. $\\sqrt{13} = 3.605551275 \\ldots e t c$. $\\frac{3}{\\sqrt{13}} \\approx 0.83205 \\ldots e t c$ Jun 14, 2015 Find $\\cos \\left(\\arctan \\left(- \\frac{2}{3}\\right)\\right)$ #### Explanation: $\\tan x = - \\frac{2}{3}$ Calculator gives; $x = - 33.69$ and $x = - 33.69 + 180 = 146.31$ Therefor, $\\cos x = \\pm 0.83$", "# Euler's Approximation of pi. I recently stumbled across the formula: $$\\pi=20\\arctan\\frac{1}{7}+8\\arctan\\frac{3}{79}$$ developed by Euler, for approximating pi. I evaluated it to several thousand decimal places and up to that point, it accurately represented pi. When does this equation break down in its representation of pi? - It doesn't. It's an exact identity. – Qiaochu Yuan Jan 11 '12 at 4:02 A derivation of the identity and related ones can be found here – t.b. Jan 11 '12 at 4:43 Quick proof: The complex number $(7 + i)^{20} (79 + 3 i)^8 = -149011611938476562500000000000000$ lies on the negative real axis, so $20\\arctan\\frac{1}{7}+8\\arctan\\frac{3}{79} = \\pi + 2\\pi n$. Consideration of the sizes of the angles involved shows that $n$ must be zero. – Hans Lundmark Jan 11 '12 at 7:46 See this as well. – J. M. Jan 12 '12 at 9:17 $-149011611938476562500000000000000 = 2^{14}5^{40}$, just for info. – Tito Piezas III Oct 5 '13 at 2:55 That formula doesn't break down. It can be proven that the right-hand side is equal to $\\pi$. One reason it comes up in connection with approximation is that the arctangent terms can be approximated well by adding terms from the Taylor expansion of the arctangent function, $$\\arctan(x)=x-\\frac{x^3}{3}+\\frac{x^5}{5}-\\frac{x^7}{7}+\\frac{x^9}{9}-\\cdots.$$ This is valid for $|x|\\leq 1$, but the series converges more rapidly the closer $x$", "function. $\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239,", "function. $\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239,", "$\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239, 2*n+1);", "do not need to evaluate this $$\\arctan$$. Recall that the argument of the product is the sum of arguments of the factors and the latter are very easy to evaluate. $$\\frac\\pi4+\\frac\\pi6=\\frac {5\\pi}{12}.$$ You should recognize that the argument of $$z=1+i$$ is $$\\frac\\pi4.$$ You should also recognize that the argument of $$w = \\sqrt3 + i$$ is $$\\arctan(1/\\sqrt3),$$ which you should realize is $$\\frac\\pi2 - \\frac\\pi3 = \\frac\\pi6.$$ So $$zw$$ is the product of a number whose argument is $$\\frac\\pi4$$ and a number whose argument is $$\\frac\\pi6.$$ The argument of $$zw$$ is therefore $$\\frac\\pi4 + \\frac\\pi6.$$ I will let you finish from there! In my opinion, any attempt to evaluate $$\\arctan(2+\\sqrt3)$$ in this context by any other method than the above is a waste of time.", "the measurement from brim to brim, and a 30 cubit cord could completely encircle the main body of the vessel below the lip. ### The Indiana bill The \"Indiana Pi Bill\" of 1897, which never passed out of the committee of the Indiana General Assembly in the U.S., has been claimed to imply a number of different values for pi, although the closest it comes to explicitly asserting one is the wording \"the ratio of the diameter and circumference is as five-fourths to four\", which would make π = 16/5 = 3.2, a discrepancy of nearly 2 percent. ## Development of efficient formulae ### Machin-like formulae For fast calculations, one may use formulae such as Machin's: $\\frac{\\pi}{4} = 4 \\arctan\\frac{1}{5} - \\arctan\\frac{1}{239}$ together with the Taylor series expansion of the function arctan(x). This formula is most easily verified using polar coordinates of complex numbers, producing: $(5+i)^4\\cdot(-239+i)=-114244-114244i.$ Another example is: $\\frac{\\pi}{4} = \\arctan\\frac{1}{2} + \\arctan\\frac{1}{3}$ which is verified as above as producing a 45° vector: $(2+i)\\cdot(3+i) = 5+5i.$ Formulae of this kind are known as Machin-like formulae. ### Other classical formulae Other formulae that have been used to compute estimates of pi include: \\begin{align} \\pi \\approxeq 768 \\sqrt{2 - \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2+1}}}}}}}}} \\approxeq 3.141590463236763. \\end{align} $\\pi =", "$\\beta=\\arctan(1/x)$, then $\\tan\\beta=1/x$, that is $$x=\\cot\\beta=\\tan\\left(\\frac{\\pi}{2}-\\beta\\right)$$ Therefore $$\\arctan x=\\arctan\\tan\\left(\\frac{\\pi}{2}-\\beta\\right)=\\frac{\\pi}{2}-\\beta$$ by the hypothesis that $x>0$, so that $0<\\arctan(1/x)<\\pi/2$. One may also use complex numbers: We are multiplying two complex numbers with argument $\\frac{1}{x}$ and $x$. So, we desire to show that $\\arg((1 + ix)(x + i)) = \\frac{\\pi}{2}$ We expand the product to get $(x^2 + 1)i$ — since there is no real part and the imaginary part is $> 0$, the argument is $\\frac{\\pi}{2}$ Assume $x>0$, then \\begin{align}&\\tan^{-1} x +\\tan^{-1} \\dfrac1x \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\dfrac1{\\tan\\tan^{-1} x} \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\cot \\tan^{-1} x \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\tan(\\dfrac{\\pi}2- \\tan^{-1}x) \\left(\\because\\text{for $x>0$,}\\;\\dfrac{\\pi}2- \\tan^{-1}x\\in\\left(0,\\dfrac{\\pi}2\\right)\\right) You want $y+z$ for $y,z$ satisfying $\\sin y / \\cos y=\\cos z/\\sin z$. Since $\\sin y=\\cos(\\pi/2-y)$ it follows that $z=\\pi/2-y$.", "$f'(x)=\\frac1{1+x^2},$ where $f(x)=\\arctan x;$ and $f'(x)=1,$ where $f(x)=x.$ Plus, chain rule is involved here.", "10 2 11 × 12 1 13 - 14 1/X 15 X<>Y 16 - 17 × 18 DSE 00 19 GTO 00 20 END Example Make sure we're at the beginning of the program: RTN Calculate $$\\arctan \\frac{1}{2}$$: 2 1/X 53 R/S STO 01 4.636476090008061162142562314612144e-1 Calculate $$\\arctan \\frac{1}{3}$$: 3 1/X 33 R/S 3.217505543966421934014046143586614e-1 Calculate $$\\pi=4\\left(\\arctan \\frac{1}{2} + \\arctan \\frac{1}{3}\\right)$$: RCL 01 + 4 × 3.141592653589793238462643383279503 Which is correct to all places. (07-23-2022 11:31 PM)Valentin Albillo Wrote: it converges very fast (order 6), the general term is extremely simple to program needing just 6 steps, and this makes for speedy looping and very short running times The code in the loop of this program isn't much more complicated but instead of using 1000000 terms to get 33 correct digits we only need 53 + 33 = 86 terms to get 34 correct digits of $$\\pi$$. And there are ways to even reduce this further e.g. by using Machin's formula: \\begin{align} \\frac{\\pi}{4}=4 \\arctan \\frac{1}{5} - \\arctan \\frac{1}{239} \\end{align} References 07-24-2022, 06:39 PM (This post was last modified: 07-24-2022 07:25 PM by pauln.) Post: #25 pauln Member Posts: 84 Joined: May 2022 RE: Pi Approximation Day In particular, in Python it is trivial to compute $$\\pi$$ (with no library, just integer arithmetic) and other constants with any number of decimals. For", "# Tag Info 7 As you know $$\\matrix{ {0 \\le \\sin (x) \\le 1,} & {0 \\le x \\le \\pi } \\cr }$$ and hence $$0 \\le \\sin(x) + 2\\sin(2x) + 3\\sin(3x) + 4\\sin(4x) \\le 10$$ and we can have your equality only when all of the $\\sin$'s are equal to $1$, i.e, $$\\matrix{ {\\sin (x) = 1} \\hfill & \\to \\hfill & {x = {\\pi \\over 2}} \\hfill \\cr {\\sin (2x) = ... 5 Notice that we can write$$ \\frac{2n}{n^4 - n^2 + 1} = \\frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$In view of the addition formula for the tangent, we find that$$ \\arctan \\bigg( \\frac{x-y}{1+xy} \\bigg) = \\arctan x - \\arctan y $$for any x > y > 0. Thus we have$$ \\arctan\\bigg( \\frac{2n}{n^4 - n^2 + 1} \\bigg) = \\arctan(n^2+n) - ... 5 The statemnet $\\sqrt{\\cos^2(x)}=|\\cos(x)|$ is true for all real $x$. Wolfram|Alpha, on the other hand, has the often annoying habit of considering all complex numbers by default. So, it's not willing to apply the identity $$\\sqrt{x^2}=|x|$$ unless it knows that $x$ is real. Given that $\\cos(x)$ is real for all real $x$, the identity you cite is true for any ... 4 $\\sin \\theta + \\cos \\theta = \\sqrt 2 \\sin (\\theta + \\frac{\\pi}{4})$ Sine is positive in the first and second quadrants. So", "D\\}$. $$\\newcommand{\\Ln}{\\operatorname{Ln}}$$ In this case you should write $$\\Ln(x+i)$$ for the main branch of the complex logarithm. Note that $$\\Ln(x+i)=\\ln(|x+i|)+i·\\arg(x+i)=\\tfrac12·\\ln(1+x^2)+i·(\\tfrac\\pi2-\\arctan(x)+k·2\\pi).$$ As such, your solution further transforms to \\begin{align} &=\\frac12\\ln|x|-\\frac{1-i}4\\Ln(x+1-i)-\\frac{1+i}4\\Ln(x+1+i)+C_2 \\\\ &=\\frac12\\ln|x|-\\frac12\\Re\\Bigl((1+i)(\\ln|x+1+i|+i\\arg(x+1+i)\\Bigr)+C_2 \\\\ &=\\frac12\\ln|x|-\\frac14\\ln((x+1)^2+1)+\\frac12\\left(\\frac\\pi2-\\arctan(x+1)\\right)+C_2 \\end{align} You need to think of $\\ln(y)$ as a function that solves the equation $y = e^x$ for x. It becomes readily apparent that there are an infinite amount of possible solutions for a complex x, as euler's formula states: $$e^{ix} = \\cos(x) + i\\sin(x)$$ This implies that any multiple of the imaginary unit will have its real part plus a full cycle: ($n \\in \\mathbb{Z}$) $$x = ln|y| + 2i n\\pi$$ For any negative y, you need to rotate in the complex plane by 180 degrees, this means an odd multiple of pi: $$y \\in \\mathbb{R^-} \\Rightarrow x = ln|y| + (2n-1)i\\pi \\Rightarrow |y|e^{(2n-1)i\\pi} = -|y|$$ Taking any integral on the negative side of the $x^{-1}$ function from a to b, where $a < b$, results in the following: $$(ln|b| + (2n-1)i\\pi) - (ln|a| + (2n-1)i\\pi)$$ $$\\Rightarrow ln|b| - ln|a|$$ Peculiar occurrences happen when taking the integral over the interval where it is discontinuous (at zero): $$\\int_{-a}^a x^{-1}dx, a \\in \\mathbb{R^+} = (ln(a) + 2in\\pi) - (ln(a) + i(2n-1)\\pi)$$ $$= 2in\\pi - 2in\\pi + i\\pi = i\\pi$$ This answer makes no sense, and is", "at 18:28 The argument of $(2+i)$ is $\\arctan(1/2)$. When you raise this number to the 4th power, you have $(-7+24i)$, and this number's argument is $\\arctan(-24/7)+\\pi$. Since raising a complex number to the 4th power multiplies its argument by $4$, we have that $$4\\arctan(1/2)=\\arctan(-24/7)+\\pi\\text{.}$$ Apply $\\tan$ to both sides, and $$\\tan(4\\arctan(1/2))=-{\\frac{24}{7}}\\text{.}$$ Therefore $t=1/2$ is a solution to $\\tan(4\\arctan(t))=-{\\frac{24}{7}}\\text{.}$ Let $2+i=re^{i\\theta}$ so that $(2+i)^4=r^4e^{4i\\theta}$ We have $$2+i=r\\cos \\theta +ir\\sin \\theta$$ and $$(2+i)^4=r^4\\cos 4\\theta+ir^4\\sin 4\\theta$$ From these we can compute $\\tan \\theta$ and $\\tan 4\\theta$. To complete the question you will need to compute $\\tan 4\\theta$ in two ways - first by using $\\tan \\theta =\\frac 12$ and the expression for $\\tan 4\\theta$ in terms of $\\tan \\theta$, and secondly by computing $(2+i)^4$ directly." ]

[ "# Proving a curious formula of $\\pi$? I have recently come across this statement without proof. $$\\pi = 128 \\arctan\\frac{1}{40} -4\\arctan\\frac{1}{239} -16\\arctan\\frac{1}{515} -32\\arctan\\frac{1}{4030} -64\\arctan\\frac{1}{32060}$$ I'd put down my approach but, to be frank, I've gotten nowhere with this. How to go about this ? Use the fact that $\\arctan(1/a)$ is the argument of $a+i$, and that the arguments of complex numbers add up when you multiply them. According to WolframAlpha, we have \\begin{aligned} &\\frac{(40+i)^{128}}{(239+i)^4 (515+i)^{16} (4030+i)^{32} (32060+i)^{64}} = \\\\ &-1 / 37403944359352749280528518983232679702 \\\\ & 01985315749502348525466597837636105197 \\\\ & 87830439618227322115549670041854205583 \\\\ & 78215314658650047572142913167759891935 \\\\ & 23573829633433227264657819301199042671 \\\\ & 95356826263444502459300305177919563475 \\\\ & 022474784673838736016384 . \\end{aligned} Since this is a negative number, it has argument $\\pi$, and this must agree with your sum, except that they might differ by $2\\pi n$ for some integer $n$, since arguments are not uniquely defined. But just by numerical evaluation one should be convinced that your sum is a least close enough to $\\pi$ to rule out all the options except $n=0$. Q.E.D. • Can you explain why the argument has tobe$\\pi$ because it is a negative number ? Mar 29 '16 at 14:42 • (Strictly speaking, $\\pi$ is an argument of a negative real number. Also $-\\pi$, $3\\pi$, etc. are possible arguments.) Mar 29 '16 at 14:51", "= \\arctan \\frac{3}{5} \\approx 0.54$. Notice that $\\varphi + \\alpha = \\arctan \\frac{5}{3} + \\arctan \\frac{3}{5} = \\frac{1}{2}\\pi$, as expected. Now, the argument of a complex number is by definition the angle made with the positive real axis. So to find the argument of $-3+5\\mathrm i$ you stand at the origin looking at the positive reals, e.g. at number $1 = 1+0\\mathrm i$. To face the number $-3+5\\mathrm i$ you need to turn anti-clockwise to the positive imaginary axis, and then continue on through the angle $\\alpha$. Hence, $\\arg(-3+5\\mathrm i) = \\frac{1}{2}\\pi + \\alpha = \\frac{1}{2}\\pi+\\arctan\\frac{3}{5} \\approx 2.11$", "# How do you find the exact value of cos[arc tan(-2/3)]? Jun 12, 2015 The exact value is apparently $\\frac{3}{\\sqrt{13}}$. However, you'll have to use a calculator to find $\\arctan \\left(- \\frac{2}{3}\\right)$. It's an irrational radian value that you'll get to be $- 33.69006752598 \\ldots e t {c}^{o}$. Wolfram Alpha simply gives it as: $- \\frac{180}{\\pi} \\cdot \\arctan \\left(- \\frac{2}{3}\\right)$ which doesn't really give a satisfactory exact answer (such as $\\frac{2}{\\sqrt{2}}$). It's just the conversion of $\\arctan \\left(- \\frac{2}{3}\\right)$ to degrees... If you just take this exact result and take the $\\cos$ of it, you'll get $\\frac{3}{\\sqrt{13}}$, but it probably won't be obvious at all that it's equal to that from just looking at it. $\\sqrt{13} = 3.605551275 \\ldots e t c$. $\\frac{3}{\\sqrt{13}} \\approx 0.83205 \\ldots e t c$ Jun 14, 2015 Find $\\cos \\left(\\arctan \\left(- \\frac{2}{3}\\right)\\right)$ #### Explanation: $\\tan x = - \\frac{2}{3}$ Calculator gives; $x = - 33.69$ and $x = - 33.69 + 180 = 146.31$ Therefor, $\\cos x = \\pm 0.83$", "principal argument of $1+2i$, and $\\arctan \\frac 12$ is the principal argument of $1+\\frac12 i$. Therefore their sum is an argument of $(1+2i)(1+\\frac12 i) = \\frac52 i$, so it must be $\\pi/2+2\\pi n$ for some $n$. But by considering the range of $\\arctan$ we can see that $n=0$ is the only possibility, so $$\\arctan 2 + \\arctan\\frac12 = \\frac\\pi2$$ - In fact arctanx=pi/2-arctan1/x, there are few ways of proving this, one way is the right-angled triangle way suggested by user84413 who just commented on your post, or you could tan both sides and carefully show they are both equivalent. -", "$f'(x)=\\frac1{1+x^2},$ where $f(x)=\\arctan x;$ and $f'(x)=1,$ where $f(x)=x.$ Plus, chain rule is involved here.", "10 2 11 × 12 1 13 - 14 1/X 15 X<>Y 16 - 17 × 18 DSE 00 19 GTO 00 20 END Example Make sure we're at the beginning of the program: RTN Calculate $$\\arctan \\frac{1}{2}$$: 2 1/X 53 R/S STO 01 4.636476090008061162142562314612144e-1 Calculate $$\\arctan \\frac{1}{3}$$: 3 1/X 33 R/S 3.217505543966421934014046143586614e-1 Calculate $$\\pi=4\\left(\\arctan \\frac{1}{2} + \\arctan \\frac{1}{3}\\right)$$: RCL 01 + 4 × 3.141592653589793238462643383279503 Which is correct to all places. (07-23-2022 11:31 PM)Valentin Albillo Wrote: it converges very fast (order 6), the general term is extremely simple to program needing just 6 steps, and this makes for speedy looping and very short running times The code in the loop of this program isn't much more complicated but instead of using 1000000 terms to get 33 correct digits we only need 53 + 33 = 86 terms to get 34 correct digits of $$\\pi$$. And there are ways to even reduce this further e.g. by using Machin's formula: \\begin{align} \\frac{\\pi}{4}=4 \\arctan \\frac{1}{5} - \\arctan \\frac{1}{239} \\end{align} References 07-24-2022, 06:39 PM (This post was last modified: 07-24-2022 07:25 PM by pauln.) Post: #25 pauln Member Posts: 84 Joined: May 2022 RE: Pi Approximation Day In particular, in Python it is trivial to compute $$\\pi$$ (with no library, just integer arithmetic) and other constants with any number of decimals. For", "$\\pi$!) The only other way to get $\\pi$ in the \"answer\" would be to recognize that the $\\arctan(x)$ can equal $\\theta + n\\cdot\\pi$ (where $n = 0, \\pm1,2,3,...$). So the $\\arctan(1/3) = 0.322 +n\\cdot\\pi = (0.102 + n)\\cdot\\pi$ (where $n = 0, \\pm1,2,3,...$). Not very satisfying. You can use Newton-Rapson with $\\tan\\left(x\\right) - 1/3 = 0$ and the 'starting point' $x = \\pi/6$: \\begin{align} &\\texttt{Clear[i, x];} \\\\ &\\texttt{x = Pi/6;} \\\\ & \\\\ &\\texttt{For[i = 0, i < 5, ++i,} \\\\ &\\texttt{x -= (Tan[x] - 1/3)/Sec[x]^2; Print[N[x]]]} \\end{align} \\begin{align} &\\texttt{0.340586, 0.321873, 0.321751, 0.321751}\\,,\\ \\color{red}{0.321751} \\end{align} For the evaluation of $\\tan^{-1}(x)$, you can also use Padé approximants such as $$\\frac{ x+\\frac{4 x^3}{15}} { 1+\\frac{3 x^2}{5}}\\tag 1$$ $$\\frac{x+\\frac{11 x^3}{21} } {1+\\frac{6 x^2}{7}+\\frac{3 x^4}{35} } \\tag 2$$ $$\\frac{x+\\frac{7 x^3}{9}+\\frac{64 x^5}{945}} {1+\\frac{10 x^2}{9}+\\frac{5 x^4}{21} } \\tag 3$$ $$\\frac{x+\\frac{34 x^3}{33}+\\frac{x^5}{5} } { 1+\\frac{15 x^2}{11}+\\frac{5 x^4}{11}+\\frac{5 x^6}{231}} \\tag 4$$ Using the above formulae, you would get $\\frac{139}{432}$, $\\frac{250}{777}$, $\\frac{20806}{64665}$, $\\frac{19593}{60895}$ which are $\\approx 0.3217592593$, $\\approx 0.3217503218$, $\\approx 0.3217505606$, $\\approx 0.3217505542$ while the \"exact\" value would be $\\approx 0.3217505544$. Don't forget you can construct it. A geometric solution is also a solution.", "function. $\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239,", "function. $\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239,", "- \\frac{1}{2} \\arcsin(0.87)$$. From the inequality $$0 < \\frac{1}{2}\\arcsin(0.87) < \\frac{\\pi}{4}$$, we first multiply through by $$-1$$ and then add $$\\frac{\\pi}{2}$$ to get $$\\frac{\\pi}{2} > \\frac{\\pi}{2} -\\frac{1}{2} \\arcsin(0.87) > \\frac{\\pi}{4}$$, or $$\\frac{\\pi}{4} < \\frac{\\pi}{2} -\\frac{1}{2} \\arcsin(0.87) < \\frac{\\pi}{2}$$. Proceeding with the usual arguments, we find the only solutions which lie in $$[0,2\\pi)$$ correspond to $$k = 0$$ and $$k=1$$, namely $$x =\\frac{\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$ and $$x = \\frac{3\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$. All told, we have found four solutions to $$\\sin(2x) = 0.87$$ in $$[0,2\\pi)$$: $$x =\\frac{1}{2} \\arcsin(0.87)$$, $$x=\\frac{1}{2} \\arcsin(0.87) + \\pi$$, $$x =\\frac{\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$ and $$x = \\frac{3\\pi}{2} - \\frac{1}{2}\\arcsin(0.87)$$. By graphing $$y = \\sin(2x)$$ and $$y = 0.87$$, we confirm our results. \\hspace{0.75in} \\includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq06.jpg} \\\\ \\$y = \\tan\\left(\\frac{x}{2}\\right)\\) and \\boldmath $$y = -3$$ & Each of the problems in Example \\ref{TrigEqnEx1} featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page \\pageref{trigeqnstrategy1}. We repeat here the advice given when solving systems of nonlinear equations in section \\ref{NonLinear} -- when it comes to solving equations involving the trigonometric functions, it helps to just try something. Next, we focus on solving inequalities involving", "$\\arctan(x) = \\sum_{n=0}^\\infty \\frac{2^{2n} (n!)^2}{(2n + 1)!} \\frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.$ This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for $$\\pi$$, first discovered by John Machin in 1706, $\\frac{\\pi}{4} = 4 \\arctan \\frac{1}{5} – \\arctan \\frac{1}{239}$ Here are the implementations for this formula. C++ double pi_summation_fast(int order) { using boost::math::factorial; double sum = 0.0; for (unsigned int n=0; n<order; ++n) { double f = factorial<double>(n); double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1); double A = pow(25./26., n+1)/pow(5., 2*n+1); double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1); sum += common*( 4*A - B ); } return 4*sum; } Python def pi_summation_fast(N): sum = 0.0 for n in range(0,N): f = factorial(n) common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1) A = math.pow(25/26, n+1)/math.pow(5, 2*n+1) B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1) sum = sum + common*( 4*A - B ) return 4*sum; JavaScript function pi_summation_fast(N) { let sum = 0.0; for (let n=0; n<N; ++n) { const f = factorial(n); const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1); const A = pow(25/26, n+1)/pow(5, 2*n+1); const B = pow(239*239 / (239*239 + 1), n+1)/pow(239, 2*n+1);", "# Computational efficiency of Machin-like formulae From what I have read, it appears that the most efficient methods of calculating $\\pi$ are Machin-like formulae. And it is known that certain formulas are more efficient than others. Are there any theoretical properties (rates of convergence, etc) of these formulas that make them more efficient than others? Like, why is $$\\frac{\\pi}{4} = 183\\arctan\\frac{1}{239} + 32\\arctan\\frac{1}{1023} - 68\\arctan\\frac{1}{5832} + 12\\arctan\\frac{1}{110443} - 12\\arctan\\frac{1}{4841182} - 100\\arctan\\frac{1}{6826318}$$ more efficient than $$\\frac{\\pi}{4} = \\arctan\\frac{1}{2} + \\arctan\\frac{1}{3}$$ (as claimed by Wikipedia) Or is this a purely empirical result? I presume that the calculation of arctan is done using the Maclaurin series. - Many of the recent record computations use algorithms that are not at all Machin-like. In Machin-like methods, it is basically Maclaurin series, with speedup techniques. Note that if we have a \"long\" Machin-like formulas, like the $6$-term one you quote, the individual terms can be computed in parallel. – André Nicolas Jun 4 '12 at 22:27 In the Maclaurin series of $\\arctan(x)$, the $n$'th term is $x^{2n-1}/(2n-1)$. When $|x|$ is small, the important part here is the $x^{2n-1}$. So to get accuracy $\\epsilon$, you need $2n-1 \\approx \\log (\\epsilon)/\\log (x)$. Roughly speaking, in a Machin-like formula $\\sum_j a_j \\arctan(1/b_j)$ the number of terms you need to calculate is inversely proportional to the log of", "# Partial Derivative of arctan Given that $$f(x,y)=\\tan^{-1}\\left(\\frac{x+y}{1-xy}\\right)$$ Find $f_x(x,y)$ My attempt, \\begin{aligned} f_x(x,y)&=\\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\\cdot\\frac{1}{1+\\left(\\frac{x+y}{1-xy}\\right)^2}\\\\ &=\\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\\\ &=\\frac{1+y^2}{1+y^2+x^2+x^2y^2} \\end{aligned} But the given answer is $\\frac{1}{x^2+1}$. How? Hint: $$1+x^2+y^2+x^2y^2=(1+x^2)+y^2(1+x^2)=?$$ • Thanks ! I got it. – Mathxx Mar 27 '18 at 16:00 Because $$\\frac{1+y^2}{1+y^2+x^2+x^2y^2}$$ and $$\\frac{1}{x^2+1}$$ are the same thing. • When $y \\ne \\pm i$ – Paul Sinclair Mar 27 '18 at 17:24 Alternatively, denote: $x=\\tan a; y=\\tan b$. Then: $$z=\\arctan \\frac{\\tan a+\\tan b}{1-\\tan a\\tan b}=a+b.$$ So: $$z_x=a_x=(\\arctan x)'=\\frac{1}{1+x^2}.$$", "# Euler's Approximation of pi. I recently stumbled across the formula: $$\\pi=20\\arctan\\frac{1}{7}+8\\arctan\\frac{3}{79}$$ developed by Euler, for approximating pi. I evaluated it to several thousand decimal places and up to that point, it accurately represented pi. When does this equation break down in its representation of pi? - It doesn't. It's an exact identity. – Qiaochu Yuan Jan 11 '12 at 4:02 A derivation of the identity and related ones can be found here – t.b. Jan 11 '12 at 4:43 Quick proof: The complex number $(7 + i)^{20} (79 + 3 i)^8 = -149011611938476562500000000000000$ lies on the negative real axis, so $20\\arctan\\frac{1}{7}+8\\arctan\\frac{3}{79} = \\pi + 2\\pi n$. Consideration of the sizes of the angles involved shows that $n$ must be zero. – Hans Lundmark Jan 11 '12 at 7:46 See this as well. – J. M. Jan 12 '12 at 9:17 $-149011611938476562500000000000000 = 2^{14}5^{40}$, just for info. – Tito Piezas III Oct 5 '13 at 2:55 That formula doesn't break down. It can be proven that the right-hand side is equal to $\\pi$. One reason it comes up in connection with approximation is that the arctangent terms can be approximated well by adding terms from the Taylor expansion of the arctangent function, $$\\arctan(x)=x-\\frac{x^3}{3}+\\frac{x^5}{5}-\\frac{x^7}{7}+\\frac{x^9}{9}-\\cdots.$$ This is valid for $|x|\\leq 1$, but the series converges more rapidly the closer $x$", "# Proving that $\\cos(\\frac{\\arctan(\\frac{11}{2})}{3}) = \\frac{2}{\\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \\sqrt{3}-2$ and $x= -\\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\\sqrt 5 \\sin \\left( \\frac{\\arctan(\\frac{2}{11})}{3}+\\pi/3 \\right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\\cos(\\frac{\\arctan(\\frac{11}{2})}{3}) = \\frac{2}{\\sqrt{5}}$ without using either of these results. • I would like to apologize in advance in case I come across as rambling and being unhelpful. – cpiegore Feb 11 '16 at 4:10 • You might try other methods of finding the roots, such as using some online root finder. (This is because maybe you'll get some third answer that contains pieces of the other two). I assume you've looked for unusual trig identities that help. – Nate 8 Feb 11 '16 at 4:30 • Are you allowed to use multiple angle formulae? – David Feb 11 '16 at 4:34 •", "So this gives us the identity $2 \\arctan \\frac12 = \\arctan \\frac43$. When $x = \\frac43$, $\\frac{1-x}{1+x} = -\\frac17$, so we have $$\\frac\\pi4 = \\arctan \\frac43 - \\arctan \\frac17 = 2\\arctan \\frac12 - \\arctan \\frac17.$$ If we take $(x,y) = (41,29)$, we get the Pythagorean triple $(20, 21, 29)$, which has $q=5$ and $p=2$, this gives us the identity $2 \\arctan \\frac25 = \\arctan \\frac{20}{21}$. When $x = \\frac{20}{21}$, $\\frac{1-x}{1+x} = \\frac{1}{41}$, so $$\\frac\\pi 4 = \\arctan \\frac{20}{21} + \\arctan \\frac1{41} = 2 \\arctan \\frac25 + \\arctan \\frac{1}{41}.$$ Since $\\frac25 \\approx \\frac13$, we could try to write $\\arctan \\frac25 = \\arctan \\frac13 + \\arctan y$, and get $y = \\frac1{17}$. So we can rewrite the identity as $$\\frac\\pi4 = 2 \\arctan \\frac13 + 2\\arctan \\frac1{17} + \\arctan \\frac1{41}.$$ We get the infamous formula with $239$ in it by taking $x=239$ and $y = 169$ as our Pell solution, which gives us the Pythagorean triple $119^2 + 120^2 = 169^2$ and $(p,q) = (5,12)$. So $2 \\arctan \\frac{5}{12} = \\arctan \\frac{120}{119}$; when $x = \\frac{120}{119}$, $\\frac{1-x}{1+x} = -\\frac{1}{239}$, so we get the identity $$\\frac\\pi4 = 2\\arctan \\frac{5}{12} - \\arctan \\frac{1}{239}.$$ Here, we get extra lucky, because $5$ and $12$ form a Pythagorean triple themselves, so we can write $\\arctan \\frac{5}{12} = 2 \\arctan \\frac15$, giving you the identity $(1)$.", "x$ near $x=0$ up to fifth order and use this (together with the fact that $\\sin\\frac{\\pi}{6}=\\frac{1}{2}$) to obtain a rational approximation to the value of $\\pi$. I thought this was an interesting bit of a question, but relatively few of them attempted it (which, I guess, goes to show that examiners and students have different ideas about what constitutes an “interesting” exam question). Anyway, the fifth-order Taylor-MacLaurin expansion for $\\arcsin x$ is $x = x + \\frac12\\cdot\\frac{x^3}{3} + \\frac12\\cdot\\frac34\\cdot\\frac{x^5}{5}$ and so by setting $x=\\frac12$ and multiplying the result by 6 we should get a rational approximation to $\\pi$. As it happens, we get $\\frac{2009}{640} = 3.1390625$, which is clearly getting there but isn’t as good as 3.14, 22/7 or 355/133. And although we can get a much better approximation by calculating more terms of the Taylor-MacLaurin series, it’s clear that this isn’t a particularly efficient way of doing things. Fortunately, there are many better ways of calculating rational or decimal approximations to $\\pi$. In 1706, the English mathematician John Machin (c.1686-1751) published the identity $\\frac{\\pi}{4} = 4\\arctan\\frac{1}{5} - \\arctan\\frac{1}{239}$ and used it (together with the MacLaurin series for $\\arctan x$) to calculate the first hundred decimal places of $\\pi$. The Indian mathematician Srinivasa Ramanujan (1887-1920) discovered the formula $\\frac{1}{\\pi} = \\frac{2\\sqrt{2}}{9801} \\sum^\\infty_{k=0} \\frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$ which converges extremely rapidly.", "If we divide both of these by $\\cos A \\cos B \\cos C \\cos D$, then we have $$\\tan(A + B + C + D) = \\frac {1 - \\sum \\tan A \\tan B + \\tan A \\tan B \\tan C \\tan D}{\\sum \\tan A - \\sum \\tan A \\tan B \\tan C},$$ which makes for more direct, less error-prone computations. Substitution gives the desired answer. ### Solution 3: Complex Numbers Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\\arctan\\frac{1}{n}$, is the argument of $n+i$. The sum of these angles is then just the argument of the product $$(3+i)(4+i)(5+i)(n+i)$$ and expansion give us $(48n-46)+(48+46n)i$. Since the argument of this complex number is $\\frac{\\pi}{4}$, its real and imaginary parts must be equal. So we set them equal and expand the product to get: $48n - 46 = 48 + 46n.$ Therefore, $n$ equals $\\boxed{047}$.", "6:18 We can also use $$\\arctan(u) \\pm \\arctan(v) = \\arctan\\left(\\frac{u \\pm v}{1 \\mp uv}\\right)$$ to obtain in four steps $$\\frac{\\frac15 - \\frac1{239}}{1 + \\frac1{5\\cdot 239}}=\\frac{239-5}{5\\cdot 239+1}=\\frac{234}{5\\cdot 239+1}=\\frac9{46} \\to$$ $$\\to \\frac{\\frac15 + \\frac9{46}}{1 - \\frac15\\frac9{46}}= \\frac7{17} \\\\\\to \\frac{\\frac15 + \\frac7{17}}{1 - \\frac15\\frac7{17}}= \\frac2{3} \\\\\\to \\frac{\\frac15 + \\frac2{3}}{1 - \\frac15\\frac2{3}}= 1$$ • Very nice. My best compliments. :-) I'm amazed how you remember all these equalities. – Sebastiano Nov 16 '19 at 23:22 • @Sebastiano Thanks! I've encounered it recently in another different context. It is a very useful identity indeed. – user Nov 16 '19 at 23:26 A slightly faster variant of the same computation using the identity $$\\tan^{-1} u \\pm \\tan^{-1} v = \\tan^{-1} \\frac{u \\pm v}{1 \\mp u v}$$ can be performed by observing that in the special case $$u = v$$ $$2\\tan^{-1} u = \\tan^{-1} \\frac{2u}{1-u^2}.$$ Consequently, we iterate $$g(u) = 2u/(1-u^2)$$ twice for $$u = 1/5$$ to obtain $$4 \\tan^{-1} \\frac{1}{5} = \\tan^{-1} g(g(\\tfrac{1}{5})) = \\tan^{-1} \\frac{120}{119}.$$ Now we apply the original formula to obtain $$4 \\tan^{-1} \\frac{1}{5} - \\tan^{-1} \\frac{1}{239} = \\tan^{-1} \\frac{\\frac{120}{119} - \\frac{1}{239}}{1 + \\frac{120}{(119)(239)}} = \\tan^{-1} 1 = \\frac{\\pi}{4}.$$ In all, we used three steps instead of four. It is also worth noting that when $$u, v \\in \\mathbb Q$$, we can write $$\\tan^{-1} \\frac{p}{q} \\pm \\tan^{-1} \\frac{r}{s} = \\tan^{-1} \\frac{ps", "$\\beta=\\arctan(1/x)$, then $\\tan\\beta=1/x$, that is $$x=\\cot\\beta=\\tan\\left(\\frac{\\pi}{2}-\\beta\\right)$$ Therefore $$\\arctan x=\\arctan\\tan\\left(\\frac{\\pi}{2}-\\beta\\right)=\\frac{\\pi}{2}-\\beta$$ by the hypothesis that $x>0$, so that $0<\\arctan(1/x)<\\pi/2$. One may also use complex numbers: We are multiplying two complex numbers with argument $\\frac{1}{x}$ and $x$. So, we desire to show that $\\arg((1 + ix)(x + i)) = \\frac{\\pi}{2}$ We expand the product to get $(x^2 + 1)i$ — since there is no real part and the imaginary part is $> 0$, the argument is $\\frac{\\pi}{2}$ Assume $x>0$, then \\begin{align}&\\tan^{-1} x +\\tan^{-1} \\dfrac1x \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\dfrac1{\\tan\\tan^{-1} x} \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\cot \\tan^{-1} x \\\\\\\\=&\\tan^{-1} x +\\tan^{-1}\\tan(\\dfrac{\\pi}2- \\tan^{-1}x) \\left(\\because\\text{for $x>0$,}\\;\\dfrac{\\pi}2- \\tan^{-1}x\\in\\left(0,\\dfrac{\\pi}2\\right)\\right) You want $y+z$ for $y,z$ satisfying $\\sin y / \\cos y=\\cos z/\\sin z$. Since $\\sin y=\\cos(\\pi/2-y)$ it follows that $z=\\pi/2-y$." ]

[ "# Difference between revisions of \"1996 AHSME Problems/Problem 27\" ## Problem Consider two solid spherical balls, one centered at $(0, 0,\\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$", "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $\\left(0, 0,\\frac{21}{2}\\right)$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution 1 The two equations of the balls are $$x^2 + y^2 + \\left(z - \\frac{21}{2}\\right)^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the second ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in \\{-2, -1, 0, 1, 2\\}$. This gives $5$ points. Thus, there are $\\boxed{13}$ possible points, giving answer $\\boxed{D}$.", "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $(0, 0,\\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution The two equations of the balls are $$x^2 + y^2 + (z - \\frac{21}{2})^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the seocnd ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in {-2, -1, 0, 1, 2\\}$ (Error compiling LaTeX. ! Missing } inserted.). This gives $5$ points. Thus, there are", "# What is the intersection of two balls in $\\mathbb{R}^3$? I know that the What is the intersection of two balls in $\\mathbb{R}^2$ is either 0,1 or 2 points (since a ball in $\\mathbb{R}^2$ is just a circle I can drew this). What is the intersection of two balls in $\\mathbb{R}^3$ ? is it at max 3 points ? I would also like to know about the more general case of $n>1$ - Two spheres will intersect in a circle. Possibly with zero or imaginary radius. – Ｊ. Ｍ. Apr 28 '12 at 9:50 A ball is often taken to mean the interior of a sphere, perhaps including the edge, which would make their intersections a lens – Henry Apr 28 '12 at 9:59", "with open balls of radius $\\frac\\epsilon2$.", "the center of the ball and the point $x$, instead of $[0,x)$. But that case is proven similarly.", "# What is the intersection of two balls in $\\mathbb{R}^3$? I know that the What is the intersection of two balls in $\\mathbb{R}^2$ is either 0,1 or 2 points (since a ball in $\\mathbb{R}^2$ is just a circle I can drew this). What is the intersection of two balls in $\\mathbb{R}^3$ ? is it at max 3 points ? I would also like to know about the more general case of $n>1$", "# What is the intersection of two balls in $\\mathbb{R}^3$? I know that the What is the intersection of two balls in $\\mathbb{R}^2$ is either 0,1 or 2 points (since a ball in $\\mathbb{R}^2$ is just a circle I can drew this). What is the intersection of two balls in $\\mathbb{R}^3$ ? is it at max 3 points ? I would also like to know about the more general case of $n>1$ - Two spheres will intersect in a circle. Possibly with zero or imaginary radius. – Ｊ. Ｍ. Apr 28 '12 at 9:50 A ball is often taken to mean the interior of a sphere, perhaps including the edge, which would make their intersections a lens – Henry Apr 28 '12 at 9:59 No it will be a circle or a point (or empty), see this wikimedia image: - It is also possible for the two spheres to not intersect, if e.g. the spheres are concentric or very far apart. – Ｊ. Ｍ. Apr 28 '12 at 9:57 Of course, I added that part. – Listing Apr 28 '12 at 9:58 Thank you both, the picture really help me see that! – Belgi Apr 28 '12 at 10:08", "is impossible. • I was 56 seconds behind $\\ldots$ – Christian Blatter Oct 25 '11 at 20:54 Assume $n$ points in the cube are each at distance at least $1$ from the others. Then the balls of radius $\\frac12$ centered at these points are disjoint. Their total volume is $n$ times the volume $\\frac43\\pi\\left(\\frac12\\right)^3=\\frac16\\pi$ of a ball of radius $\\frac12$. On the other hand every point in one of these balls is at distance at most $\\frac12$ from the cube of side $9$, in particular all the balls are included in a cube of side $10$. Hence $\\frac16\\pi n\\leqslant 10^3$, that is, $n\\leqslant1909$.", "form the whole set?? – Ishfaaq Feb 14 '14 at 20:00 • Just choose the balls with radius $n$ and centre $0$. Or the balls with rational centres and radius $1$. Or - there are a lot of ways. – Daniel Fischer Feb 14 '14 at 20:04 • Rational centres and radius $1$. That was my initial idea. Will try both. Thanks loads for the advice.. – Ishfaaq Feb 14 '14 at 20:06", "$B_2(\\rho)$ be the balls with radius $\\rho$ centered at $(-\\frac{d}{2},0)$ and $(\\frac{d}{2},0)$ respectively. Let $A(\\rho_1,\\rho_2)$ be the area of $B_1(\\rho_1) \\cap B_2(\\rho_2)$. The area of the your shaped area is simply $$\\frac12 \\left( A(r_{1'}, r_{2'} ) - A(r_{1},r_{2'}) - A(r_{1'},r_{2}) + A(r_1,r_2) \\right)$$ – achille hui Nov 27 '13 at 12:09 Can't be resolved in terms of calculus and that. -", "ball is described by $$\\rho \\leq 2R(\\cos\\theta \\sin\\phi \\sin\\alpha + \\cos\\phi \\cos\\alpha).$$ The bounds on the solid region of intersection are therefore $$0 \\leq \\theta \\leq 2\\pi,\\qquad 0 \\leq \\phi \\leq \\alpha,\\qquad 0 \\leq \\rho \\leq 2R(\\cos\\theta \\sin\\phi \\sin\\alpha + \\cos\\phi \\cos\\alpha).$$ If you really did mean \"half a cone with vertex angle $\\alpha$\" (rather than a \"cone with half-vertex angle $\\alpha$\", as I've assumed), this description will require modification (and more information: where is the center of the sphere on the half-cone?).", "which gives : $$\\frac{|f(x)-f(z)|}{|x-z|} Now take the segment which link $$x$$ to $$y$$, cover it by balls where $$f$$ is $$(L+\\delta)$$-Lipschitz on each ball, and use the triangle inequality on the centers of the balls and the intersections ( we can manage to have finite number of balls, each ball is centered on a point on the segment and intersect only two other balls, with intersection points on the segment, except for the ball centered on $$x$$ and the one centered on $$y$$). It suffices now to take $$\\delta$$ to $$0$$ to get the result. Now for the other direction, we see that $$J_{f}(x)(v) = \\lim \\frac{|f(x)-f(x+tv)|}{t}≤L$$, hence : $$|J_{f}(x)|≤L$$.", "equation $x^2 + y^2 + z^2 = 1$ from $x^2 + y^2 + \\left(z - \\frac{3}{2}\\right)^2 = 1$ gives $$-3z + \\frac{9}{4} = 0.$$ So $z = \\frac{3}{4}$. When $z = \\frac{3}{4}$ both equations, $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + \\left(z - \\frac{3}{2}\\right)^2 = 1$ simplify to $$x^2 + y^2 = \\frac{7}{16}.$$ So the intersection of these two spheres is the circle $x^2 + y^2 = \\frac{7}{16}$ contained in the plane where $z = \\frac{3}{4}$. This is pictured below: Below, the purple sphere has been removed from the picture in order to see more clearly where the two spheres meet (the shaded circle on the green sphere).", "}}^{2}}-\\frac{{132}^{2}}{{\\mathrm{\\pi }}^{2}}\\right)$ = $\\frac{1762}{\\mathrm{\\pi }}-\\frac{1322}{\\mathrm{\\pi }}$ = (176 - 132)(176 + 132) / $\\mathrm{\\pi }$ = (44 x 308) / (22/7) = 4312 sq.m 9 1502 Q: If a solid sphere of radius 10 cms is moulded into 8 spherical solid balls of equal radius, then surface area of each ball (in sq.cm) is ? A) 100 π B) 101/π C) 99 π/12 D) 54/13π", "ball. $$\\large \\frac{1}{c^2}\\frac{{\\gamma }_{\\mathrm{v}}}{{\\gamma }_{\\mathrm{v}}+1}=\\frac{1}{v^2}\\left(1-\\frac{1}{{\\gamma }_{\\mathrm{v}}}\\right)$$which I proved in the complete answer above and here.", "# placing balls inside ball [duplicate] Is it possible to put pairwise disjoint open 3d-balls with radii $\\frac{1}{2},\\frac{1}{3},\\frac{1}{4},\\dots$ inside a unit ball? not an original question, I found it somewhere in the internet once, but without any answer. ## marked as duplicate by Alex M., Jeremy Rickard, Watson, user91500, gebruikerJun 25 '16 at 8:56 • The answer is certainly yes. You can already do this in two dimensions, as discussed (with lovely images) here: math.stackexchange.com/questions/1491754/… – mjqxxxx May 27 '16 at 14:47 • I think the analogue for area in this case is volume so. Volume of unit ball in $\\Bbb R^3$ is ${4\\over3}\\pi$, and the volume of the spheres to be fitted is ${4\\over3}\\pi(\\zeta(3)-1)$. With something like 3.34 units of volume left over. – snulty May 27 '16 at 14:57 As pointed out by @mjqxxxx, this is possible. In fact, it is possible to pack two families of open balls of radii $\\frac12, \\frac13, \\ldots$ into unit sphere. The picture below illustrate one possible configuration (with $n$ up to $300$): $\\hspace1in$ In above picture, the two families are related to each other by a reflection with respect to origin. Let us concentrate on the families of brown balls. The brown ball with radius $\\frac12$ is centered at $(0,0,-\\frac12)$. For $n \\ge 3$, the brown ball with radius", "balls of the same color. $X$ has geometric distribution ($X\\sim G(p)$). We know if $Y\\sim G(q)$, then $E(Y) = \\frac{1}{q}$. Thus, $$\\bbox[5px,border:2px solid #C0A000]{E(X) = \\frac{1}{p} = \\frac{1}{\\frac{1}{2}\\times \\frac{2}{4} \\times \\frac{2}{5} + \\frac{1}{2} \\times \\frac{2}{4} \\times \\frac{3}{5}}}$$", "Use the open cover by balls of radius $\\varepsilon$, and use the centres plus intersection points.", "\\|X\\| \\cos \\theta$ where $\\theta$ is the angle between $X$ and $Z$. Noting that the area of a spherical cap of opening angle $\\theta$ is proportional to $1 - \\cos \\theta$, we see that the probability of $Y$ being in the ball is $\\min\\left(1, \\frac{1}{2} + \\frac{R^2 - \\|X\\|^2 - d^2}{4 d \\|X\\|}\\right)$. So you could select $X$ using a distribution proportional to this factor times the uniform density in the ball, then select $Z$ uniformly in the spherical cap $\\cos \\theta \\le \\min\\left(1, \\frac{R^2 - \\|X\\|^2 - d^2}{2d\\|X\\|}\\right)$ - Thanks so much for your answer! – SeptemberGrass Jan 5 '12 at 6:58" ]

[ "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $\\left(0, 0,\\frac{21}{2}\\right)$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution 1 The two equations of the balls are $$x^2 + y^2 + \\left(z - \\frac{21}{2}\\right)^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the second ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in \\{-2, -1, 0, 1, 2\\}$. This gives $5$ points. Thus, there are $\\boxed{13}$ possible points, giving answer $\\boxed{D}$.", "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $(0, 0,\\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution The two equations of the balls are $$x^2 + y^2 + (z - \\frac{21}{2})^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the seocnd ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in {-2, -1, 0, 1, 2\\}$ (Error compiling LaTeX. ! Missing } inserted.). This gives $5$ points. Thus, there are", "point in the second shell! This is what I want to show. More precisely, I’ll show that there are at least 30. So far I only know there are at most 30 thanks to Egan’s computer search. We can probably dream up a proof, but we haven’t gotten around to it. Why 30? Remember the standard description of the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice. Define a half-integer to be an integer plus $\\frac{1}{2}\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}$. The ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice consists of all 8-tuples of real numbers $\\left({x}_{0},{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6},{x}_{7}\\right) \\left(x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7\\right) $ such that • the ${x}_{i}x_i$ are either all integers or all half-integers, and • the sum of all the ${x}_{i}x_i$ is even. The shortest nonzero vectors in this lattice — the vectors in the first shell — have length $\\sqrt{2}\\sqrt\\left\\{2\\right\\}$, like this: $\\left(1,1,0,0,0,0,0,0\\right) \\left(1,1,0,0,0,0,0,0\\right) $ The second shortest nonzero vectors — the vectors in the second shell — have length $22$, like this: $\\left(1,1,1,1,0,0,0,0\\right) \\left(1,1,1,1,0,0,0,0\\right) $ So, we have the opportunity to find an expanded copy of ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice, dilated by a factor of $\\sqrt{2}\\sqrt\\left\\{2\\right\\}$, inside the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice. Our goal is to find 30 such copies containing our favorite point in the second shell. Let’s take this as our favorite: $\\left(2,0,0,0,0,0,0,0\\right) \\left(2,0,0,0,0,0,0,0\\right) $ And here’s the cool part: there’s one such copy for each way to give", "}[\\text{-}100, 100],\\text{ how many lattice points are on the line: }y \\:=\\:-\\tfrac{4}{3}x + \\tfrac{70}{3}$ We have: . $y \\:=\\:\\text{-}x - \\frac{x}{3} + 23 + \\frac{1}{3} \\quad\\Rightarrow\\quad y \\:=\\:\\text{-}x + 23 + \\frac{1-x}{3}$ Since $y$ is an integer, then $1-x$ must be a multiple of 3. . . That is: . $1 - x \\:=\\:3a\\:\\text{ for some integer }a$ We have: . $x \\:=\\:1-3a$ Since $x \\ge \\text{-}100\\!:\\;\\;1-3a \\,\\ge\\,\\text{-}100 \\quad\\Rightarrow\\quad \\text{-}3a\\,\\ge\\,\\text{-}101 \\quad\\Rightarrow\\quad a\\,\\le\\,33$ Since $x \\le 100\\!:\\;\\;1-3a\\,\\le\\,100 \\quad\\Rightarrow\\quad \\text{-}3a \\,\\le\\,99 \\quad\\Rightarrow\\quad a \\,\\ge\\,\\text{-}33$ . . Thus: . $a \\in [\\text{-}33,\\,33]$ Then $a$ (and hence, $x$) can take on 67 values. Therefore, there are $67$ lattice points on the line segment.", "all coordinates are $±\\frac{1}{2}\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}$ with either $11$ or $33$ minus signs are already roots, and by switching all the signs we get the half-integer roots with $55$ or $77$ minus signs. There are $135135$ non-zero elements of norm $00$, and these all correspond to lattice points in shell $22$, with $1616$ lattice points per element of the vector space. a) There are $7070$ elements of shape $\\left(1,1,1,1,0,0,0,0\\right)\\left\\left(1,1,1,1,0,0,0,0\\right\\right)$. We get $88$ lattice points by changing an even number of signs (including $00$). We get another $88$ lattice points by taking the complement and then changing an odd number of signs. b) There is $11$ element of shape $\\left(1,1,1,1,1,1,1,1\\right)\\left\\left(1,1,1,1,1,1,1,1\\right\\right)$. This corresponds to the $1616$ lattice points of shape $\\left(±2,0,0,0,0,0,0,0\\right)\\left\\left(\\pm2,0,0,0,0,0,0,0\\right\\right)$. c) There are $6464$ elements like ${\\left(±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2}\\right)}^{*}\\left\\left(\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac \\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}\\right\\right)^*$, with $11$ or $33$ minus signs. We get $88$ actual lattice points by replacing $±\\frac{1}{2}\\pm\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}$ by $\\mp \\frac{3}{2}\\mp\\frac\\left\\{3\\right\\}\\left\\{2\\right\\}$ in one coordinate, and another $88$ by changing the signs of all coordinates. This accounts for all $16\\cdot 135=216016\\cdot135=2160$ points in shell $22$. Isotropic: $\\begin{array}{|ll|}\\hline \\mathrm{shape}& \\mathrm{number}\\\\ \\left(1,1,1,1,1,1,1,1\\right)& 1\\\\ \\left(1,1,1,1,0,0,0,0\\right)& 70\\\\ {\\left(±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2}\\right)}^{*}& 64\\\\ \\mathrm{total}& 135\\\\ \\hline\\end{array}\\array\\left\\{\\arrayopts\\left\\{\\collayout\\left\\{left\\right\\}\\rowlines\\left\\{solid\\right\\}\\collines\\left\\{solid\\right\\}\\frame\\left\\{solid\\right\\}\\right\\} \\mathbf\\left\\{shape\\right\\}&\\mathbf\\left\\{number\\right\\}\\\\ \\left\\left(1,1,1,1,1,1,1,1\\right\\right)&1\\\\ \\left\\left(1,1,1,1,0,0,0,0\\right\\right)&70\\\\ \\left\\left(\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{ 1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\}\\right\\right)^*&64\\\\ \\mathbf\\left\\{total\\right\\}&\\mathbf\\left\\{135\\right\\} \\right\\}$ Anisotropic: $\\begin{array}{|ll|}\\hline \\mathrm{shape}& \\mathrm{number}\\\\ \\left(1,1,1,1,1,1,0,0\\right)& 28\\\\ \\left(1,1,0,0,0,0,0,0\\right)& 28\\\\ \\left(±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2},±\\frac{1}{2}\\right)& 64\\\\ \\mathrm{total}& 120\\\\ \\hline\\end{array}\\array\\left\\{\\arrayopts\\left\\{\\collayout\\left\\{left\\right\\}\\rowlines\\left\\{solid\\right\\}\\collines\\left\\{solid\\right\\}\\frame\\left\\{solid\\right\\}\\right\\} \\mathbf\\left\\{shape\\right\\}&\\mathbf\\left\\{number\\right\\}\\\\ \\left\\left(1,1,1,1,1,1,0,0\\right\\right)&28\\\\ \\left\\left(1,1,0,0,0,0,0,0\\right\\right)&28\\\\ \\left\\left(\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{ 1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\},\\pm\\tfrac\\left\\{1\\right\\}\\left\\{2\\right\\}\\right\\right)&64\\\\ \\mathbf\\left\\{total\\right\\}&\\mathbf\\left\\{120\\right\\} \\right\\}$ Since the quadratic form in ${𝔽}_{2}\\mathbb\\left\\{F\\right\\}_2$ comes from the quadratic form in", "centered around the origin, and has radius $r$. Also, w.l.o.g. $z_1$ has maximal modulus (although not necessarily uniquely so). Consider the disc $D^\\prime = D_{\\frac{|z_1|+r}{2}}(\\frac{z_1}{|z_1|}\\frac{|z_1|-r}{2})$. Note that for all $z\\not\\in D^\\prime$ we have $$|z|\\geq \\left|z-\\frac{z_1}{|z_1|}\\frac{|z_1|-r}{2}\\right| - \\left|\\frac{z_1}{|z_1|}\\frac{|z_1|-r}{2}\\right| \\geq \\frac{|z_1|+r}{2} - \\left|\\frac{|z_1|-r}{2}\\right| = |z_1|,$$ and equality holds only when $z,z_1$ have the same argument (but then, only for $z=z_1$). It follows that $z_1\\not\\in D^\\prime$, and is alone in having this property among $z_1,\\ldots,z_n$. • Thanks for your answer. I don't quite understand this bit: \"and equality holds only when $z, z_1$ have the same argument (but then, only for $z=z_1$)\". Why does that necessarily follow? Suppose $z_2$ is another point with $|z_1|=|z_2|$, then I don't see why $z_2$ is forced to have the same argument as $z_1$. – Prism Sep 23 '13 at 21:32 • @Prism, that's because for all $z,w\\in\\mathbb{C}$, $|z+w|=|z|+|w|$ only if $z,w$ are linearly dependent over $\\mathbb{R}$. – Jonathan Y. Sep 23 '13 at 22:16 • Ah right. That makes sense! (+1) – Prism Sep 24 '13 at 22:00 After a suitable Möbius transform we can assume that $\\partial D$ is the real axis and that the $z_k=x_k+i y_k$ $\\>(1\\leq k\\leq n)$ lie in the upper half plane, whereby $$y_k=1\\quad(1\\leq k\\leq r),\\qquad y_k>1\\quad(r+1\\leq k\\leq n)$$ for some $r\\geq1$. When $r=1$ the \"circle\" $y=1+\\epsilon$ for some small", "Example. The number of integer lattice points that satisfy the conditions $$-n\\leq x,y,z\\leq n,\\quad -s\\leq x+y+z\\leq s$$ for some $n$, $s\\in\\mathbb N$, is equal to $$\\frac{1}{2\\pi}\\int\\limits_{-\\pi}^{\\pi}\\left(\\frac{\\sin \\frac{2n+1}{2}t}{\\sin\\frac {t}{2}}\\right)^3\\frac{\\sin \\frac{2s+1}{2}t}{\\sin \\frac{t}{2}} dt.$$", "Difference between revisions of \"2011 AMC 12B Problems/Problem 19\" Problem A lattice point in an $xy$-coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx + 2$ passes through no lattice point with $0 < x \\leq 100$ for all $m$ such that $\\frac{1}{2} < m < a$. What is the maximum possible value of $a$? $\\textbf{(A)}\\ \\frac{51}{101} \\qquad \\textbf{(B)}\\ \\frac{50}{99} \\qquad \\textbf{(C)}\\ \\frac{51}{100} \\qquad \\textbf{(D)}\\ \\frac{52}{101} \\qquad \\textbf{(E)}\\ \\frac{13}{25}$ Solution It is very easy to see that the $+2$ in the graph does not impact whether it passes through lattice. We need to make sure that $m$ cannot be in the form of $\\frac{a}{b}$ for $1\\le b\\le 100$, otherwise the graph $y= mx$ passes through lattice point at $x = b$. We only need to worry about $\\frac{a}{b}$ very close to $\\frac{1}{2}$, $\\frac{m+1}{2m+1}$, $\\frac{m+1}{2m}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\\le b\\le 100$, the smallest is $\\frac{50}{99}$, so answer is $\\boxed{\\frac{50}{99}\\ \\(\\textbf{(B)}}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)", "at 3:01 • Also, the $L_1$ norm is not strict. – Mike Earnest Jan 5 at 15:34 $$\\frac{x+y}{|x|+|y|}=\\frac{|x|}{|x|+|y|}\\Big(\\frac{x}{|x|}\\Big)+\\frac{|y|}{|x|+|y|}\\Big(\\frac{y}{|y|}\\Big)$$ This shows that $$v:=(x+y)/(|x|+|y|)$$ is on the segment connecting $$x/|x|$$ to $$y/|y|$$. Since the unit sphere has no segments, $$|v|$$ cannot be $$1$$. By the triangle inequality, $$|v|\\le 1$$. Suppose $$V$$'s unit sphere contains no line segments, and $$x, y \\in V$$ such that $$\\|x + y\\| = \\|x\\| + \\|y\\|.$$ Let $$z$$ be the point on the line segment $$[0, x + y]$$ that you would expect to be distance $$\\|x\\|$$ from $$0$$ and distance $$\\|y\\|$$ from $$x + y$$. Working this out, you'll get $$z = \\frac{\\|x\\|(x + y)}{\\|x + y\\|}.$$ Note that $$z$$ lies on the spheres $$S[0; \\|x\\|]$$ and $$S[x + y; \\|y\\|]$$. Also note the same is true for $$x$$. That is, $$x$$ and $$z$$ lie in both spheres. Let's suppose they're different points. Since they both lie in $$S[0; \\|x\\|]$$, it follows from the convexity of the ball that $$\\frac{x + z}{2}$$ must lie in the open ball $$B(0; \\|x\\|)$$, which is to say $$\\left\\|\\frac{x + z}{2}\\right\\| < \\|x\\|$$. On the same token, we have $$\\frac{x + z}{2} \\in B(x + y, \\|y\\|)$$. Hence, $$\\|x + y\\| \\le \\left\\|\\frac{x + z}{2}\\right\\| + \\left\\|x + y - \\frac{x + z}{2}\\right\\| < \\|x\\| + \\|y\\| =", "wanted to describe the region the is bounded by the planes z=1, z=2, x=1, x=2 and y=0 and the parabaloid y=x^2, I would say the region is given by $$1 \\leq x \\leq 2$$ $$0 \\leq y \\leq x^2$$ $$1 \\leq z \\leq 2$$ It would make no sense to say $$f(x,y,z)= \\left\\{\\begin{array}{lr}1 \\leq x \\leq 2 \\\\ 0 \\leq y \\leq x^2 \\\\1 \\leq z \\leq 2 \\end{array}\\right.$$ You can't set a function equal to an inequality or set of inequalities, and you can't describe a 3D region by a function. Okay, then you have: $$\\rho = \\sqrt{x^2+y^2+z^2}$$ $$\\phi = \\cos^{-1}\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)$$ $$\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)$$ Last edited by a moderator: May 4, 2017 5. Sep 8, 2010 ### Telemachus Here it is: $$x>0$$ $$y>0$$ $$x^2+y^2\\leq{\\frac{z^2}{3}}\\forall{}z\\in{[\\frac{\\sqrt[ ]{3}}{2}},\\sqrt[ ]{3}]}$$ $$x^2+y^2+z^2\\leq{4}\\forall{z\\in{(\\sqrt[ ]{3},2]}}$$ $$x^2+y^2+z^2\\geq{1}$$ Sorry, it took some time to make the draw. Anyway, its limited by the planes zy and zx, and by the sphere x²+y²+z²=1.", "these last two results for $C2 \\land C3$. (Using numerical integration, we get the much lower values of 0.001497721920258410 and 0.003256122941383665, respectively.) A set of values of $a,b,c$ which satisfy $C4$ are $$\\left\\{\\frac{2}{3} (\\cos (\\alpha )+1),\\frac{2}{3} \\left(-\\frac{1}{2} \\sqrt{3} \\sin (\\alpha )-\\frac{\\cos (\\alpha )}{2}+1\\right),\\frac{2}{3} \\left(\\frac{1}{2} \\sqrt{3} \\sin (\\alpha )-\\frac{\\cos (\\alpha )}{2}+1\\right)\\right\\},$$ for $\\frac{\\pi}{3} \\leq \\alpha \\leq \\frac{5 \\pi}{3}$. So, I would like to obtain results of integration of the value 36 over $[0,1]^3$ of $C1 \\land C3$ and $C2 \\land C3$ for either specific values of $a,b,c$, satisfying $C4$, or even without particular values being specified. ### Counting the Number of Lattice Points in an $n$-Dimensional Sphere Math Overflow Recent Questions - Fri, 05/17/2019 - 13:03 Let $S_n(R)$ denote the number of lattice points in an $n$-dimensional \"sphere\" with radius $R$. For clarification, I am interested in lattice points found both strictly inside the sphere, and on its surface. I want to count exactly how many such points there are. This count corresponds to the amount of sets of integers that are solutions to the equation $$a_1^2+a_2^2+a_3^2+...+a_{n-1}^2+a_n^2= R^2$$ where the $a_i$'s are required to be integers, not necessarily positive, and sets that differ just by the order of the summands are also considered distinct, i.e for the case $n=3, R=5$, the following are both counted: $3^2+(-4)^2+0^2$ and $(-4)^2+0^2+3^2$. This", "the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice as the one giving the densest packing of spheres among all lattices in 8 dimensions. This packing was long suspected to be optimal even among those that do not arise from lattices — but this fact was proved only in 2016, by the young mathematician Maryna Viazovska [V]. We can also describe the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice more explicitly. In suitable coordinates, it consists of vectors for which: • the components are either all integers or all integers plus $\\frac{1}{2}\\textstyle\\left\\{\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}\\right\\}$, and • the components sum to an even number. This lattice consists of all integral linear combinations of the 8 rows of this matrix: $\\left(\\begin{array}{rrrrrrrr}1& -1& 0& 0& 0& 0& 0& 0\\\\ 0& 1& -1& 0& 0& 0& 0& 0\\\\ 0& 0& 1& -1& 0& 0& 0& 0\\\\ 0& 0& 0& 1& -1& 0& 0& 0\\\\ 0& 0& 0& 0& 1& -1& 0& 0\\\\ 0& 0& 0& 0& 0& 1& -1& 0\\\\ 0& 0& 0& 0& 0& 1& 1& 0\\\\ -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}\\end{array}\\right) \\left\\left( \\begin\\left\\{array\\right\\}\\left\\{rrrrrrrr\\right\\} 1&-1&0&0&0&0&0&0 \\\\ 0&1&-1&0&0&0&0&0 \\\\ 0&0&1&-1&0&0&0&0 \\\\ 0&0&0&1&-1&0&0&0 \\\\ 0&0&0&0&1&-1&0&0 \\\\ 0&0&0&0&0&1&-1&0 \\\\ 0&0&0&0&0&1&1&0 \\\\ -\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\} \\end\\left\\{array\\right\\} \\right\\right) $ The inner product of any row vector with itself is 2, while the inner product of distinct row vectors is either 0 or -1. Thus, any two of these", "or when all $x_i = \\frac{3}{4},$ with Euclidean minimum'' equal to $\\frac{7}{8}.$ Notice that with $\\vec x = \\left( \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4} \\right),$ the integer lattice points $\\vec y$ such that $q( \\vec x - \\vec y)=\\frac{7}{8}$ include $\\vec y = \\left( 0,0,0,0,0,0,0\\right)$ and all seven cyclic permutations (including the identity) of $\\vec y = \\left( 0,1,0,1,0,1,0\\right),$ another seven for $\\vec y = \\left( 1,0,0,0,0,0,0\\right),$ another seven for $\\vec y = \\left( 1,0,1,0,0,0,0\\right),$ finally seven for $\\vec y = \\left( 1,0,0,1,0,0,0\\right),$ a total of 29 lattice points on the ellipsoid, of 128 in the standard unit 7-cube. The Gram matrix for the form is $$Q \\; \\; = \\; \\; \\left( \\begin{array}{ccccccc} 1 & \\frac{1}{2} & 0 & 0 & 0 & 0 & \\frac{1}{2}\\\\\\ \\frac{1}{2} & 1 & \\frac{1}{2} & 0 & 0 & 0 & 0 \\\\\\ 0 & \\frac{1}{2} & 1 & \\frac{1}{2} & 0 & 0 & 0 \\\\\\ 0 & 0 & \\frac{1}{2} & 1 & \\frac{1}{2} & 0 & 0 \\\\\\ 0 & 0 & 0 & \\frac{1}{2} & 1 & \\frac{1}{2} & 0 \\\\\\ 0 & 0 & 0 & 0 & \\frac{1}{2} & 1 & \\frac{1}{2} \\\\\\ \\frac{1}{2} & 0 & 0 & 0 & 0 & \\frac{1}{2} & 1 \\end{array} \\right) ,$$ which has determinant", "by $\\{ \\frac{l}{2}, m, \\frac{l}{2}+m \\}$. That is, $L_l^* = L_l \\sqcup (L_l+m) \\sqcup (L_l+\\frac{l}{2}) \\sqcup (L_l+(\\frac{l}{2}+m))$ This gives us three lattices $\\begin{cases} M_1 &= L_l \\sqcup (L_l+m) = L \\\\ M_2 &= L_l \\sqcup (L_l+\\frac{l}{2}) \\\\ M_3 &= L_l \\sqcup (L_l+(\\frac{l}{2}+m)) \\end{cases}$ and all three of them are unimodular because $L_l \\subsetneq M_i \\subseteq M_i^* \\subsetneq L_l^*$ and $L_l$ is of index $4$ in $L_l^*$. Now, let’s assume the norm of $l$, that is, $l.l \\in 4 \\mathbb{Z}$. Then, either the norm of $\\frac{l}{2}$ is odd (but then the norm of $\\frac{l}{2}+m$ must be even), or the norm of $\\frac{l}{2}$ is even, in which case the norm of $\\frac{l}{2}+m$ is odd. That is, either $M_2$ or $M_3$ is an even unimodular lattice, the other one being an odd unimodular lattice. Let’s take for $l$ and $m$ the vectors $\\lambda = \\frac{1}{\\sqrt{2}} (1,1,\\dots,1) \\in L – 2L$ and $\\mu = \\sqrt{2}(1,0,\\dots,0) \\in L$, then $\\lambda.\\lambda = \\frac{1}{2}\\times 24 = 12 \\quad \\text{and} \\quad \\mu.\\lambda = 1$ Because $\\frac{\\lambda}{2}.\\frac{\\lambda}{2} = \\frac{12}{4}=3$ is odd, we have that $\\Lambda = L_{\\lambda} \\sqcup (L_{\\lambda} + (\\frac{\\lambda}{2} + \\mu))$ is an even unimodular lattice, which is the Leech lattice, and $\\Lambda_{odd} = L_{\\lambda} \\sqcup (L_{\\lambda} + \\frac{\\lambda}{2})$ is an odd unimodular lattice, called the odd Leech lattice. John Leech (1926-1992) – Photo Credit", "such that $\\frac{y}{x}<2^n$ thus $y<2^nx$. Aha, for $(\\Leftarrow)$, you've used the weak Archimedean property (equivalent to Archimedean property, for lattice ordered groups): Let $a,b\\!\\in\\!K_{\\geq0}$, $\\mathbb{N} a\\!\\leq\\!b$, and suppose that $a\\!\\neq\\!0$. Then $1\\!\\leq\\!2,\\frac{b}{a}$ and $2^\\mathbb{N}\\!\\leq\\!\\frac{b}{a}$, so by assumption, $2\\!=\\!1$, $\\rightarrow\\leftarrow$. Thank you very much! – Leon Jul 30 '12 at 21:03", "0, d(d+1)\\le 2k \\right\\}.$ We will in fact show that ${\\mathcal A}(\\alpha) \\le d+a\\left(k-\\frac{d(d+1)}{2}\\right)$ with $d=M_0+N$. (The next paragraph will show that $d(d+1)\\le 2k$.) To see this, observe that $\\Lambda$, not including the upper boundary, encloses all of the lattice points in the triangle with vertices $(0,0),(d-1,0)$, and $(0,d-1)$, and there are $d(d+1)/2$ of these. In addition, the line segment from $(M_0,N)$ to $(d,0)$, and the line segment from $(d,0)$ to $(M,0)$, include an additional $\\max\\{M_1-1,0\\}$ lattice points which are enclosed by $\\Lambda$ and not on the upper boundary. This gives a lower bound on the lattice point count $L(\\Lambda)$, from which it follows that $I(\\alpha)\\ge d(d+1)+2\\max\\{0,M_1-1\\}.$ In the inequality that we want to prove, $k=I(\\alpha)/2$, so it is enough to show that ${\\mathcal A}(\\alpha) \\le d+a\\max\\{0,M_1-1\\}.$ By our choice of $d$ and our previous calculation of ${\\mathcal A}(\\alpha)$, this inequality is $M_1-N\\le \\max\\{0,M_1-1\\}.$ This says that if $M_1=0$ then $N\\ge 0$, and if $M_1>0$ then $N>0$. This follows immediately from the definition of $N$, and so we have proved the desired inequality. The above calculation shows that an ECH generator of index $2k$ with maximum symplectic action has the form $e_{1,0}^d$ with action $d$, or $e_{1,0}^{d-1}e_{m,1}$ with $m=k-d(d+1)/2+1>0$ which has action $d+a(k-d(d+1)/2)$. The choice of $d$ that gives the maximum symplectic action depends on $a$.", "are already roots, and by switching all the signs we get the half-integer roots with $5$ or $7$ minus signs. There are $135$ non-zero elements of norm $0$, and these all correspond to lattice points in shell $2$, with $16$ lattice points per element of the vector space. a) There are $70$ elements of shape $\\left(1,1,1,1,0,0,0,0\\right)$. We get $8$ lattice points by changing an even number of signs (including $0$). We get another $8$ lattice points by taking the complement and then changing an odd number of signs. b) There is $1$ element of shape $\\left(1,1,1,1,1,1,1,1\\right)$. This corresponds to the $16$ lattice points of shape $\\left(\\pm2,0,0,0,0,0,0,0\\right)$. c) There are $64$ elements like $\\left(\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2}\\right)^*$, with $1$ or $3$ minus signs. We get $8$ actual lattice points by replacing $\\pm\\frac{1}{2}$ by $\\mp\\frac{3}{2}$ in one coordinate, and another $8$ by changing the signs of all coordinates. This accounts for all $16\\cdot135=2160$ points in shell $2$. Isotropic $\\array{\\arrayopts{\\collayout{left}\\rowlines{solid}\\collines{solid}\\frame{solid}} \\mathbf{shape}&\\mathbf{number}\\\\ \\left(1,1,1,1,1,1,1,1\\right)&1\\\\ \\left(1,1,1,1,0,0,0,0\\right)&70\\\\ \\left(\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2}\\right)^*&64\\\\ \\mathbf{total}&\\mathbf{135} }$ Anisotropic $\\array{\\arrayopts{\\collayout{left}\\rowlines{solid}\\collines{solid}\\frame{solid}} \\mathbf{shape}&\\mathbf{number}\\\\ \\left(1,1,1,1,1,1,0,0\\right)&28\\\\ \\left(1,1,0,0,0,0,0,0\\right)&28\\\\ \\left(\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2},\\pm\\frac{1}{2}\\right)&64\\\\ \\mathbf{total}&\\mathbf{120} }$ Since the quadratic form in $\\mathbb{F}_2$ comes from the quadratic form in Euclidean space, it is preserved by the Weyl group $W(E_8)$. In fact the homomorphism $W(E_8)\\rightarrow GO_8^+(2)$ is onto, although (contrary to what I said in an earlier comment) it is a double cover—the element of $W(E_8)$ that reverses", "$d_N^*$ are at times very “unforgiving” inasmuch as a single point in a suboptimal position can catastrophically reduce measure of the entire point distribution. In our case, regardless of how large $N$ is, the $d_N^*$ is typically determined by the four points closest to each pole, especially $t_0$ and $t_3$. However, what is also known about this lattice is that the largest Voronoi polygon is at the pole. Thus, in trying to maximize $d_N$ by separating the initial polar points in the sequence, actually makes the void at the pole even larger! Thus, we present an alternative to lattice 2 which is generally more preferable, as it does not exhibit such a large void near the poles. It is almost identical to lattice 2 but with two differences. Firstly, it uses $\\varepsilon =\\frac{7}{2}$ for $1 \\leq i \\leq N{-2}$. Secondly, in addition to these $N{-2}$ points, the first and final points are defined to be at located each pole. That is, $$t_0=(0,0); \\; t_{N{-1}} = (1,0); \\quad t_i = \\left( \\frac{i{+3.5}}{N{+6}}, \\frac{i}{\\phi} \\right) \\quad \\textrm{for }\\; 1 \\leq i \\leq N{-2} \\tag{4}$$ This lattice results in $d^*=3.313$, which for $N>90$ is even better than lattice #2. Thus for $N=100$, the values of the first coordinate would simply be: $$\\{ 0; \\; \\frac{3.5}{106},\\frac{4.5}{106},\\frac{5.5}{106},\\ldots, \\frac{99.5}{106},\\frac{100.5}{106},\\frac{101.5}{106} ; \\; 1\\}$$ Comparison For", "since due to symmetry all other missing coordinates would be found from permuting x, y and z (factor 6) or changing the sign of the coordinates (factor 8). Due to our new restriction your inequalities simplify to $$x^2+y^2\\geq \\frac{1}{16}$$ $$0\\leq x\\leq y\\leq z\\leq \\frac12$$ (everything else drops out) This can be written $$0\\leq x\\leq\\frac12$$ $$\\sqrt{\\frac{1}{16}-x^2}\\leq y\\leq\\frac12$$ $$x\\leq y\\leq \\frac12$$ $$y\\leq z\\leq \\frac12$$ We need to split into two cases so that we have to take into account only one of the middle two inequalities for y. (depending on the interval one inequality will dominate). $$0\\leq x\\leq \\frac{1}{\\sqrt{32}}$$ $$\\sqrt{\\frac{1}{16}-x^2}\\leq y\\leq \\frac12$$ $$y\\leq z\\leq \\frac12$$ and $$\\frac{1}{\\sqrt{32}}\\leq x\\leq \\frac12$$ $$x\\leq y\\leq \\frac12$$ $$y\\leq z\\leq \\frac12$$ Therefore $$V=48\\left(\\int_0^{\\frac{1}{\\sqrt{32}}}\\mathrm{d}x\\int_{\\sqrt{\\frac{1}{16}-x^2}}^{\\frac12}\\mathrm{d}y\\int_y^{\\frac12}\\mathrm{d}z+\\int_{\\frac{1}{\\sqrt{32}}}^{\\frac12}\\mathrm{d}x\\int_x^{\\frac12}\\mathrm{d}y\\int_y^{\\frac12}\\mathrm{d}z\\right)$$ Please check it for yourself, as initially I used slightly different numbers for myself. Last edited: Dec 10, 2009 17. Dec 10, 2009 ### Gerenuk I got good news for both of you! In one of sliders equations there is actually some truth. $$|\\overline{A}\\cap\\overline{B}\\cap\\overline{C}|$$ $$= |U| - (|A| + |B| + |C| - |A\\cap B| - |A \\cap C| - |B\\cap C| + |A\\cap B \\cap C|)$$ Therefore you need to calculate V=(Cube)-3(Cylinder)+3(Bicylinder)-(Tricylinder) (see http://mathworld.wolfram.com/SteinmetzSolid.html) So it seems sliders mistake was that he thought the bicylinder and the tricylinder have equal volume. 18. Dec 11, 2009 ### slider142 This statement is nowhere in", "5.431 Å) 32. Germanium (Ge) (a = 5.657 Å) 50. Gray Tin (Sn) (a = 6.491 Å) • TBD ## Diamond-like Two-particle Also known as zincblende. This is effectively ## Double-filled Diamond-like Two-particle The diamond lattice includes 8 \"tetrahedral holes\", with only 4 occupied in a 'normal' diamond structure. A two-particle lattice can be formed by filling all 8 internal holes with the 2nd particle-type, in which case the particles exist in a 1:2 ratio. This is often called a CaF2 lattice. ### Particle Positions There are 22 positions. In total there are 12 particles in the unit cell: • $8 \\, \\mathrm{corners} \\, \\times \\, \\frac{1}{8} = 1$ • $\\left(0,0,0\\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6 \\, \\mathrm{faces} \\, \\times \\, \\frac{1}{2} = 3$ • $\\left(0,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},0,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},0\\right),\\left(1,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},1,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},1\\right)$ • $4 \\, \\mathrm{principal \\,\\, tetrahedral \\,\\, sites} \\, \\times \\, 1 = 4$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+(0,0,0)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ • $4 \\, \\mathrm{tetrahedral \\,\\, holes} \\, \\times \\, 1 = 4$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-(0,0,0)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ ### Neighbors There are 12 particles in the unit cell: 4 of type A, and 8 of type B. The type A particles each connect to 8 particles of type B. The type B particles each connect (tetrahedrally) to 4 particles of type A. The lattice may also be thought of as having" ]

[ "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $\\left(0, 0,\\frac{21}{2}\\right)$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution 1 The two equations of the balls are $$x^2 + y^2 + \\left(z - \\frac{21}{2}\\right)^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the second ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in \\{-2, -1, 0, 1, 2\\}$. This gives $5$ points. Thus, there are $\\boxed{13}$ possible points, giving answer $\\boxed{D}$.", "# 1996 AHSME Problems/Problem 27 ## Problem Consider two solid spherical balls, one centered at $(0, 0,\\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$ ## Solution The two equations of the balls are $$x^2 + y^2 + (z - \\frac{21}{2})^2 \\le 36$$ $$x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5 \\pm 6$, and the seocnd ball goes from $1 \\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2 + y^2 \\le \\frac{23}{4}$$ $$x^2 + y^2 \\le \\frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|, |y| \\le 3$: For $x=\\pm 2$, we must have $y=0$. This gives $2$ points. For $x = \\pm 1$, we can have $y\\in \\{-1, 0, 1\\}$. This gives $2\\cdot 3 = 6$ points. For $x = 0$, we can have $y \\in {-2, -1, 0, 1, 2\\}$ (Error compiling LaTeX. ! Missing } inserted.). This gives $5$ points. Thus, there are", "# Difference between revisions of \"1996 AHSME Problems/Problem 27\" ## Problem Consider two solid spherical balls, one centered at $(0, 0,\\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls? $\\text{(A)}\\ 7\\qquad\\text{(B)}\\ 9\\qquad\\text{(C)}\\ 11\\qquad\\text{(D)}\\ 13\\qquad\\text{(E)}\\ 15$", "of one of these balls, then $$1\\left\\| {z_1 } \\right\\| = \\left\\| {z_2 } \\right\\| = \\frac{1}{2}\\left\\| {z_1 } \\right\\| + \\frac{1}{2}\\left\\| {z_2 } \\right\\| = \\left\\| {\\frac{1}{2}z_1 + \\frac{1}{2}z_2 } \\right\\|$$ so equality is attained in the inequality $$\\left\\| {z_{1 + } z_2 } \\right\\| \\leqslant \\left\\| {z_1 } \\right\\| + \\left\\| {z_2 } \\right\\|$$ despite the fact that neither z1 nor z2 is a nonnegative real multiple of the other. Furthermore, the presence of the corners leads to the existence of multiple norming functionals for some points z of the unit sphere of each of these spaces, that is, norm-one members z* of the dual space such that $$z^* z = \\left\\| z \\right\\|$$. To see why this would be, let Z be either of these two spaces and let z0 be one of the four corners of B z . Then there are infinitely many different straight lines that pass through z0 without intersecting the interior of the closed unit ball; let l1 and l2 be two of them. By Mazur’s separation theorem, there are members z 1 * and z 2 * of Z*, necessarily different, such that if j ∈ {1, 2}, then z j * z = 1 when zl j and z j * z ≤ 1 when zB z .", "R^n, ... 2 It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip \\mathbb{Z} with the usual metric, then B_{1/2}(0) = B_{1/3}(0), for instance. 6 In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the \"discrete metric\" d(x,y)=0 if x=y and 1 otherwise, all balls B_r(x) for r \\leq 1 are the same (they are just \\{ x \\}) while all balls B_r(x) for r>1 are also the same ... 2 We have:$$ \\int_{0}^{\\pi}\\frac{dx}{\\left(\\sin^2 x\\right)^{\\frac{1}{3}}}=\\frac{\\Gamma\\left(\\frac{1}{2}\\right)\\Gamma\\left(\\frac{1}{6}\\right)}{\\Gamma\\left(\\frac{2}{3}\\right)}$$hence:$$\\begin{eqnarray*} \\left|\\int_{\\pi}^{+\\infty}\\frac{dx}{x^2\\left(\\sin^2 x\\right)^{\\frac{1}{3}}}\\,dx\\right|&\\leq& ... 1 Assuming we are using the usual topology of $\\mathbb{R}$. a. $A_1 = \\left\\{ \\frac{1}{n\\pi} \\,|\\, n \\in \\mathbb{Z} -\\{0\\}\\right\\}$. So clearly $A_1$ is NOT open , since it has isolate points (actually all of its points are isolated points). It is also clear that $A_1$ is NOT closed, since $0$ is in its closure and $0\\notin A_1$. In other words, there is a ... 0 Hint. You have an Euler-type integral, you may then rewrite your initial integral in terms of the Appel hypergeometric function $$F_1(a,b_1,b_2,c;x,y)=\\dfrac{\\Gamma(c)}{\\Gamma(a)\\Gamma(c-a)}\\int_0^1\\frac{v^{a-1}(1-v)^{c-a-1}}{(1-xv)^{-b_1}(1-yv)^{-b_2}}~dv$$ where $\\Re c>\\Re a>0$ with the change of variable $$v=\\frac{u}{1-e^{-\\lambda x}}.$$ 0 I think a simpler and", "easy to characterize. Take a bunch of equal-sized spheres in 8 dimensions. Get as many of these spheres to touch a single sphere as you possibly can. Then, get as many to touch those spheres as you possibly can, and so on. Unlike in 3 dimensions, where there is ‘wiggle room’, you have no choice about how to proceed, except for an overall rotation and translation. The balls will inevitably be centered at points of the ${\\mathrm{E}}_{8} \\mathrm\\left\\{E\\right\\}_8$ lattice! More concretely, we can find coordinates in which the ${\\mathrm{E}}_{8} \\mathrm\\left\\{E\\right\\}_8$ lattice consists of vectors for which: 1) the components are either all integers or all integers plus $\\frac{1}{2} \\textstyle\\left\\{\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}\\right\\}$, and 2) the components sum to an even number. This lattice consists of all integral linear combinations of the 8 rows of this matrix: $\\left(\\begin{array}{rrrrrrrr}1& -1& 0& 0& 0& 0& 0& 0\\\\ 0& 1& -1& 0& 0& 0& 0& 0\\\\ 0& 0& 1& -1& 0& 0& 0& 0\\\\ 0& 0& 0& 1& -1& 0& 0& 0\\\\ 0& 0& 0& 0& 1& -1& 0& 0\\\\ 0& 0& 0& 0& 0& 1& -1& 0\\\\ 0& 0& 0& 0& 0& 1& 1& 0\\\\ -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}& -\\frac{1}{2}\\end{array}\\right) \\left\\left( \\begin\\left\\{array\\right\\}\\left\\{rrrrrrrr\\right\\} 1&-1&0&0&0&0&0&0 \\\\ 0&1&-1&0&0&0&0&0 \\\\ 0&0&1&-1&0&0&0&0 \\\\ 0&0&0&1&-1&0&0&0 \\\\ 0&0&0&0&1&-1&0&0 \\\\ 0&0&0&0&0&1&-1&0 \\\\ 0&0&0&0&0&1&1&0 \\\\ -\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}&-\\frac\\left\\{1\\right\\}\\left\\{2\\right\\} \\end\\left\\{array\\right\\} \\right\\right) $ The", "($$S$$) of absolutely separable states (those that can not be “entangled” by global unitary transformations) is defined by the additional inequality (eq. (1)) Halder $$begin{equation} x – z leq 2 sqrt{y (1-x-y-z)}. end{equation}$$ The maximal (inscribed) ball contained within $$S$$ is composed of the points for which $$begin{equation} x^2 +y^2 +z^2 +(1-x-y-z)^2 leq frac{1}{3}, end{equation}$$ while the minimal (circumscribed) ball containing $$S$$ is composed of the points for which $$begin{equation} x^2 +y^2 +z^2 +(1-x-y-z)^2 leq frac{3}{8}. end{equation}$$ So, let us pose the question of whether or not the John ellipsoids JohnEllipoids containing and contained within the set $$S$$ of absolutely separable two-qubit states are simply the inscribed and circumscribed 4-balls indicated above? (cf. 9and15D) Let us point out that in the answer to 4-ball, we report formulas for the Hilbert-Schmidt probabilities (relative volumes) of the inscribed and circumscribed two balls, that is, $$begin{equation} frac{35 pi }{23328 sqrt{3}} approx 0.00272132 end{equation}$$ and the considerably larger $$begin{equation} frac{35 sqrt{frac{1}{3} left(2692167889921345-919847607929856 sqrt{6}right)} pi}{27518828544} approx 0.0483353. end{equation}$$ Further, in AbsSepVol the formula for the Hilbert-Schmidt volume $$begin{equation} frac{29902415923}{497664}-frac{50274109}{512 sqrt{2}}-frac{3072529845 pi }{32768 sqrt{2}}+frac{1024176615 cos ^{-1}left(frac{1}{3}right)}{4096 sqrt{2}} approx 0.00365826 end{equation}$$ of the intermediate absolutely separable set $$S$$ has been given. Parallel analyses to those discussed above for two-qubit systems do not appear computationally feasible for the next higher-dimensional “qubit-qutrit” systems–for which the density matrices", "spheres are to be included. Had we wanted the spheres left out, we would have asked for the open region bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out.) Linh V. ### Problem 51 Find the center $C$ and the radius $a$ for the spheres. $$(x+2)^{2}+y^{2}+(z-2)^{2}=8$$ Linh V. ### Problem 52 Find the center $C$ and the radius $a$ for the spheres. $$(x-1)^{2}+\\left(y+\\frac{1}{2}\\right)^{2}+(z+3)^{2}=25$$ Linh V. ### Problem 53 Find the center $C$ and the radius $a$ for the spheres. $$(x-\\sqrt{2})^{2}+(y-\\sqrt{2})^{2}+(z+\\sqrt{2})^{2}=2$$ Linh V. ### Problem 54 Find the center $C$ and the radius $a$ for the spheres. $$x^{2}+\\left(y+\\frac{1}{3}\\right)^{2}+\\left(z-\\frac{1}{3}\\right)^{2}=\\frac{16}{9}$$ Linh V. ### Problem 55 Find the center $C$ and the radius $a$ for the spheres. $$x^{2}+y^{2}+z^{2}+4 x-4 z=0$$ Linh V. ### Problem 56 Find the center $C$ and the radius $a$ for the spheres. $$x^{2}+y^{2}+z^{2}-6 y+8 z=0$$ Linh V. ### Problem 57 Find the center $C$ and the radius $a$ for the spheres. $$2 x^{2}+2 y^{2}+2 z^{2}+x+y+z=9$$ Linh V. ### Problem 58 Find the center $C$ and the radius $a$ for the spheres. $$3 x^{2}+3 y^{2}+3 z^{2}+2 y-2 z=9$$ Linh V. ### Problem 59 Find the center $C$ and the radius $a$ for", "the $$2^d$$ heuristic obviously fails here. To be more precise, here's a smooth estimate with this flavor: $$N_{r,d} = (2 e^{1+\\chi} d /r^2)^{r^2} \\; ,$$ for any $$r \\leq \\sqrt{d}/2$$, where $$B(r)$$ is the ball of radius $$r$$ and the error parameter $$\\chi$$ satisfies $$-\\frac{r^2}{d} - \\frac{\\log(C r)}{r^2} \\leq \\chi \\leq \\sqrt{\\frac{C}{\\log(d/r^2)}} \\; ,$$ for some not very large constant $$C > 0$$ [*]. Notice that these bounds are quite tight for $$1 \\ll r \\ll \\sqrt{d}$$, since in this regime $$\\chi$$ is subconstant. (A smooth function in $$r$$ can't give a much better approximation when $$r$$ is constant since, e.g., $$N_{\\sqrt{2} - \\varepsilon, d} = 2d$$, but $$N_{\\sqrt{2}, d} \\approx d^2$$. But the binomial coefficient is quite accurate in this case for integer $$r^2$$.) This gives $$\\frac{N_{2r,d}}{N_{r,d}} \\approx (e d /r^2)^{3r^2}2^{-5r^2}$$ for $$1 \\ll r \\ll \\sqrt{d}$$. Geometrically, we can think of this strange behavior as a result of the integer lattice having short points that shouldn't be there.'' In particular, in $$d$$ dimensions, the ball of radius $$1$$ has volume much less than one, $$\\approx (C/d)^{d/2}$$, but the integer lattice manages to have $$2d+1$$ points inside this ball anyway. The hard case, $${\\bf r \\approx \\sqrt{d} }$$ When $$r \\ll \\sqrt{d}$$ or $$r \\gg \\sqrt{d}$$, there are nice smooth functions with closed formulas that approximate $$N_{r,d}$$", "these last two results for $C2 \\land C3$. (Using numerical integration, we get the much lower values of 0.001497721920258410 and 0.003256122941383665, respectively.) A set of values of $a,b,c$ which satisfy $C4$ are $$\\left\\{\\frac{2}{3} (\\cos (\\alpha )+1),\\frac{2}{3} \\left(-\\frac{1}{2} \\sqrt{3} \\sin (\\alpha )-\\frac{\\cos (\\alpha )}{2}+1\\right),\\frac{2}{3} \\left(\\frac{1}{2} \\sqrt{3} \\sin (\\alpha )-\\frac{\\cos (\\alpha )}{2}+1\\right)\\right\\},$$ for $\\frac{\\pi}{3} \\leq \\alpha \\leq \\frac{5 \\pi}{3}$. So, I would like to obtain results of integration of the value 36 over $[0,1]^3$ of $C1 \\land C3$ and $C2 \\land C3$ for either specific values of $a,b,c$, satisfying $C4$, or even without particular values being specified. ### Counting the Number of Lattice Points in an $n$-Dimensional Sphere Math Overflow Recent Questions - Fri, 05/17/2019 - 13:03 Let $S_n(R)$ denote the number of lattice points in an $n$-dimensional \"sphere\" with radius $R$. For clarification, I am interested in lattice points found both strictly inside the sphere, and on its surface. I want to count exactly how many such points there are. This count corresponds to the amount of sets of integers that are solutions to the equation $$a_1^2+a_2^2+a_3^2+...+a_{n-1}^2+a_n^2= R^2$$ where the $a_i$'s are required to be integers, not necessarily positive, and sets that differ just by the order of the summands are also considered distinct, i.e for the case $n=3, R=5$, the following are both counted: $3^2+(-4)^2+0^2$ and $(-4)^2+0^2+3^2$. This", "# Ratio of volumes of $n$-dimensional unit balls Prove that $$\\left( \\frac{nS_n}{S_{n-1}}\\right)^{1/n} \\le 2$$ where $S_i$ is the volume of the $i$th dimensional unit ball and $n\\ge 2$. I think we can use the fact that an $n$-dimensional unit ball is contained in a hypercube whose edge measures $2$ units, and the ball contains a hypercube whose edge measures $\\sqrt{2}$ units. So we have $\\sqrt{2}^n\\le S_n \\le 2^n$ and thus $$\\left( \\frac{nS_n}{S_{n-1}}\\right)^{1/n} \\le \\left( \\frac{n2^n}{\\sqrt{2}^{n-1}} \\right)^{1/n} = n^{1/n}2^{\\frac{n+1}{2n}} = n^{1/n}\\sqrt{2}\\sqrt{2}^{1/n}.$$ So if we take $n^{1/n} \\le 2$ and $\\sqrt{2}^{1/n}\\le \\sqrt{2}^{1/2}$, then an upper bound to my problem is $2^{1+3/4}$ which isn't good enough. Do we need another approach or do we get better bounds on $n^{1/n}$? • Since $3^{1 / 3} \\sqrt(2) \\sqrt{2}^{1 / n} > 2$, no estimate on $n^{1 / n}$ will work. – Travis Willse Jan 26 '18 at 12:55 • Your argument is very nice, but it only works if $n^{\\frac{1}{n}}2^{\\frac{n+1}{2n}} \\le 2 \\Leftrightarrow n \\le 2^{\\frac{n-1}{2}}$, which is true for all $n \\ge 7$, since the function on the right is exponential. So if you can find a better bound for lower $n$'s it's fine, but it might require an explicit formula?? – 57Jimmy Jan 26 '18 at 12:55 ## 2 Answers It is enough to show that $S_n\\leq 2\\,S_{n-1}$, which is trivial,", "# Math Help - q-ary code proof using spheres 1. ## q-ary code proof using spheres Show that there is no q-ary code of length 5 and distance 3 with more than $\\frac{q^5}{5q-4}$ codewords. You should prove this from first principles using spheres and without using known bounds on the number of codewords. This is my working so far: First prove that there is a q-ary code of length 5 and distance 3 with exactly $\\frac{q^5}{5q-4}$ codewords. consider the spheres of radius 1 centred on codewords. Each such sphere contains 1 + 5 = 6 words. Since the code has distance 3, no two such spheres intersect. Otehrwise we would have a pair of codewords at distance 2 or less. Since there are $\\frac{q^5}{5q-4}$ codewords, there are $\\frac{6q^5}{5q-4}$ distinct words in these spheres. However there are only $q^5$ distince q-ary words of length 5. therefore we have to show that $\\frac{6q^5}{5q-4}$ = $q^5$ for some q that is a natural non-zero number. i worked out q = 2 therefore such a code exists. but now im having problems proving that no q-ary code with MORE THAN $\\frac{q^5}{5q-4}$ codewords exists. also, with q = 2 the no of codewords = $\\frac{16}{3}$ can anyone tell me why this isnt a whole number? 2. There is some mistake in your argument.", "• # question_answer If $(2,\\,\\,3,\\,\\,5)$ are ends of the diameter of a sphere${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x$$-12y-2z+20=0$. Then coordinates of the other end are A) $(4,\\,\\,9,\\,\\,-3)$ B) $(4,\\,\\,3,\\,\\,5)$C) $(4,\\,\\,3,\\,\\,-3)$ D) $(4,\\,\\,-3,\\,\\,9)$ Solution : Let the co-ordinate of other ends are$(x,\\,\\,y,\\,\\,z)$. The centre of sphere is$C(3,\\,\\,6,\\,\\,1)$. Therefore,$\\frac{x+2}{2}=3\\Rightarrow x=4$ $\\frac{y+3}{2}=6\\Rightarrow y=9$and$\\frac{z+5}{2}=1\\Rightarrow z=-3$ You need to login to perform this action. You will be redirected in 3 sec", "point in the second shell! This is what I want to show. More precisely, I’ll show that there are at least 30. So far I only know there are at most 30 thanks to Egan’s computer search. We can probably dream up a proof, but we haven’t gotten around to it. Why 30? Remember the standard description of the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice. Define a half-integer to be an integer plus $\\frac{1}{2}\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}$. The ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice consists of all 8-tuples of real numbers $\\left({x}_{0},{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6},{x}_{7}\\right) \\left(x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7\\right) $ such that • the ${x}_{i}x_i$ are either all integers or all half-integers, and • the sum of all the ${x}_{i}x_i$ is even. The shortest nonzero vectors in this lattice — the vectors in the first shell — have length $\\sqrt{2}\\sqrt\\left\\{2\\right\\}$, like this: $\\left(1,1,0,0,0,0,0,0\\right) \\left(1,1,0,0,0,0,0,0\\right) $ The second shortest nonzero vectors — the vectors in the second shell — have length $22$, like this: $\\left(1,1,1,1,0,0,0,0\\right) \\left(1,1,1,1,0,0,0,0\\right) $ So, we have the opportunity to find an expanded copy of ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice, dilated by a factor of $\\sqrt{2}\\sqrt\\left\\{2\\right\\}$, inside the ${\\mathrm{E}}_{8}\\mathrm\\left\\{E\\right\\}_8$ lattice. Our goal is to find 30 such copies containing our favorite point in the second shell. Let’s take this as our favorite: $\\left(2,0,0,0,0,0,0,0\\right) \\left(2,0,0,0,0,0,0,0\\right) $ And here’s the cool part: there’s one such copy for each way to give", "5.431 Å) 32. Germanium (Ge) (a = 5.657 Å) 50. Gray Tin (Sn) (a = 6.491 Å) • TBD ## Diamond-like Two-particle Also known as zincblende. This is effectively ## Double-filled Diamond-like Two-particle The diamond lattice includes 8 \"tetrahedral holes\", with only 4 occupied in a 'normal' diamond structure. A two-particle lattice can be formed by filling all 8 internal holes with the 2nd particle-type, in which case the particles exist in a 1:2 ratio. This is often called a CaF2 lattice. ### Particle Positions There are 22 positions. In total there are 12 particles in the unit cell: • $8 \\, \\mathrm{corners} \\, \\times \\, \\frac{1}{8} = 1$ • $\\left(0,0,0\\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6 \\, \\mathrm{faces} \\, \\times \\, \\frac{1}{2} = 3$ • $\\left(0,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},0,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},0\\right),\\left(1,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},1,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},1\\right)$ • $4 \\, \\mathrm{principal \\,\\, tetrahedral \\,\\, sites} \\, \\times \\, 1 = 4$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+(0,0,0)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ • $4 \\, \\mathrm{tetrahedral \\,\\, holes} \\, \\times \\, 1 = 4$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-(0,0,0)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ ### Neighbors There are 12 particles in the unit cell: 4 of type A, and 8 of type B. The type A particles each connect to 8 particles of type B. The type B particles each connect (tetrahedrally) to 4 particles of type A. The lattice may also be thought of as having", "cubes of type B with a connected type A particle at the center. However the extended lattice is not BCC-like, since only every second \"BCC cube\" has a central particle of type A. ## Double Inter-penetrating Diamond Two-Particle This lattice is formed from two diamond lattices that inter-penetrate. It is often called the NaTl lattice. ### Particle Positions There are 36 positions. In total there are 16 particles in the unit cell, 8 of each type. #### Particle Type A • $8 \\, \\mathrm{corners} \\, \\times \\, \\frac{1}{8} = 1$ • $\\left(0,0,0\\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6 \\, \\mathrm{faces} \\, \\times \\, \\frac{1}{2} = 3$ • $\\left(0,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},0,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},0\\right),\\left(1,\\frac{1}{2},\\frac{1}{2}\\right),\\left(\\frac{1}{2},1,\\frac{1}{2}\\right),\\left(\\frac{1}{2},\\frac{1}{2},1\\right)$ • $4 \\, \\mathrm{internal \\,\\, tetrahedral \\,\\, sites} \\, \\times \\, 1 = 4$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+(0,0,0)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{1}{4},\\frac{1}{4},\\frac{1}{4}\\right)+\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ #### Particle Type B • $12 \\, \\mathrm{edges} \\, \\times \\, \\frac{1}{4} = 3$ • $\\left(\\frac{1}{2},0,0\\right) , \\left(0,\\frac{1}{2},0\\right) , \\left(0,0,\\frac{1}{2}\\right)$ • $\\left(1,0,\\frac{1}{2}\\right) , \\left(1,1,\\frac{1}{2}\\right) , \\left(0,1,\\frac{1}{2}\\right) , \\left(\\frac{1}{2},1,0\\right) , \\left(1,\\frac{1}{2},0\\right) , \\left(\\frac{1}{2},0,1\\right) , \\left(0,\\frac{1}{2},1\\right), \\left(1,\\frac{1}{2},1\\right) , \\left(\\frac{1}{2},1,1\\right)$ • $1 \\, \\mathrm{central} \\, \\times \\, 1 = 1$ • $\\left(\\frac{1}{2},\\frac{1}{2},\\frac{1}{2}\\right)$ • $4 \\, \\mathrm{internal \\,\\, tetrahedral \\,\\, sites} \\, \\times \\, 1 = 4$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-(0,0,0)=\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)=\\left(\\frac{3}{4},\\frac{1}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},0,\\frac{1}{2}\\right)=\\left(\\frac{1}{4},\\frac{3}{4},\\frac{1}{4}\\right)$ • $\\left(\\frac{3}{4},\\frac{3}{4},\\frac{3}{4}\\right)-\\left(\\frac{1}{2},\\frac{1}{2},0\\right)=\\left(\\frac{1}{4},\\frac{1}{4},\\frac{3}{4}\\right)$ ### Examples #### Atomics • Sodium Thallium crystal (NaTl) ## Cristobalite Cristobalite is a diamond-like lattice formed in silicon dioxide (SiO2).", "2pi i}oint _{C_{r}}{frac {f(zeta )}{zeta ^{k+1}}},dzeta } and Cr is the circle about 0 of radius r > 0. Suppose f is bounded: i.e. there exists a constant M such that {displaystyle |f(z)|leq M} for all z. We can estimate directly {displaystyle |a_{k}|leq {frac {1}{2pi }}oint _{C_{r}}{frac {|f(zeta )|}{|zeta |^{k+1}}},|dzeta |leq {frac {1}{2pi }}oint _{C_{r}}{frac {M}{r^{k+1}}},|dzeta |={frac {M}{2pi r^{k+1}}}oint _{C_{r}}|dzeta |={frac {M}{2pi r^{k+1}}}2pi r={frac {M}{r^{k}}},} where in the second inequality we have used the fact that {displaystyle |z|=r} on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem. Another proof uses the mean value property of harmonic functions. Proof[2] Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since f is bounded, the averages of it over the two balls are arbitrarily close, and so f assumes the same value at any two points. The proof can be adapted to the case where the harmonic function", "inequality by adding 1/4 to each side and made the middle of the inequality a sphere centered at (0,0,1/2). Here is my full attempt. The inequality represents the region on/between the spheres centered at (0,0,1/2) with radii 1 1/4 and 3 1/4. The region contains all of the points on the sphere with radius 1 1/4 and includes all of the points of the second sphere except the outermost points where r = 3 1/4. I think this might be right, but I just thought I'd check with you guys. Thanks! Well, if z=0, you have an annulus in the xy plane. What do you have if x = 0? y = 0? 3. Hello, statelyplump! $\\text{The inequality represents the region on/between the spheres centered}$ $\\text{at }(0,0,\\frac{1}{2})\\text{ with radii }1\\frac{1}{4}\\text{ and }3\\frac{1}{4}.$ $\\text{The region contains all of the points on the inner sphere}$ $\\text{but not the points on the outer sphere.}$ When you completed-the-square, you got $\\frac{13}{4}$ and $\\frac{5}{4}$ on the \"other side\". Recall the general equation: . $(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 \\:=\\:r^2$ What you have is the radius-squared. The radii are: . $\\dfrac{\\sqrt{13}}{2}$ and $\\dfrac{\\sqrt{5}}{2}$ 4. dwsmith, I think you must have misunderstood the question. statelyplump, in addition to what Soroban said, you are misusing some words. Strictly, a \"sphere\" is the surface of a \"ball\"", "the spheres. $$x^{2}+y^{2}+z^{2}-4 x+6 y-10 z=11$$ Linh V. ### Problem 60 Find the center $C$ and the radius $a$ for the spheres. $$(x-1)^{2}+(y-2)^{2}+(z+1)^{2}=103+2 x+4 y-2 z$$ Linh V. ### Problem 61 Find equations for the spheres whose centers and radii are given. $$\\begin{array}{l}\\text { (Radius) } \\\\\\sqrt{14} \\\\2 \\\\\\frac{4}{9} \\\\7\\end{array}$$ (Center) $$(1,2,3)$$ Linh V. ### Problem 62 Find equations for the spheres whose centers and radii are given. $$\\begin{array}{l}\\text { (Radius) } \\\\\\sqrt{14} \\\\2 \\\\\\frac{4}{9} \\\\7\\end{array}$$ (Center) $$(0,-1,5)$$ Linh V. ### Problem 63 Find equations for the spheres whose centers and radii are given. $$\\begin{array}{l}\\text { (Radius) } \\\\\\sqrt{14} \\\\2 \\\\\\frac{4}{9} \\\\7\\end{array}$$ (Center) $$\\left(-1, \\frac{1}{2},-\\frac{2}{3}\\right)$$ Linh V. ### Problem 64 Find equations for the spheres whose centers and radii are given. $$\\begin{array}{l}\\text { (Radius) } \\\\\\sqrt{14} \\\\2 \\\\\\frac{4}{9} \\\\7\\end{array}$$ (Center) $$(0,-7,0)$$ Linh V. ### Problem 65 Find a formula for the distance from the point $P(x, y, z)$ to the a. $x$ -axis. b. $y$ -axis. c. z-axis. Linh V. ### Problem 66 Find a formula for the distance from the point $P(x, y, z)$ to the a. $x y$ -plane. b. $y z$ -plane. c. $x z$ -plane. Linh V. ### Problem 67 Find the perimeter of the triangle with vertices $A(-1,2,1)$ $B(1,-1,3),$ and $C(3,4,5)$ Linh V. ### Problem 68 Show that the point $P(3,1,2)$ is equidistant", "is impossible. • I was 56 seconds behind $\\ldots$ – Christian Blatter Oct 25 '11 at 20:54 Assume $n$ points in the cube are each at distance at least $1$ from the others. Then the balls of radius $\\frac12$ centered at these points are disjoint. Their total volume is $n$ times the volume $\\frac43\\pi\\left(\\frac12\\right)^3=\\frac16\\pi$ of a ball of radius $\\frac12$. On the other hand every point in one of these balls is at distance at most $\\frac12$ from the cube of side $9$, in particular all the balls are included in a cube of side $10$. Hence $\\frac16\\pi n\\leqslant 10^3$, that is, $n\\leqslant1909$." ]

[ "other across the $$x$$-axis.", "$-P$, which is the reflection point with respect to the $x$-axis.", "Maybe using the matrix of map $\\sigma_a$ ? #### Klaas van Aarsen ##### MHB Seeker Staff member Ok!! Can we get a specific vector? Maybe using the matrix of map $\\sigma_a$ ? The question only asks for a fixed vector. We know now that we need a vector perpendicular to the line of reflection. The line has an angle a with respect to the positive x-axis. So $(\\cos a,\\,\\sin a)$ is a directional vector along the line. Can we find a vector perpendicular to it?", "### Home > PC3 > Chapter 9 > Lesson 9.2.2 > Problem9-115 9-115. ​Write the matrix associated with reflecting across the $x$-axis and then rotating $90^\\circ$ counterclockwise. Review your work from problem 9-114. Multiply the transformation matrices in the opposite order you did in problem 9-114.", "Thread: find the matrix for the transformation 1. find the matrix for the transformation the question is: Find the matrix for the transformation which first reflects across the main diagonal, then projects onto the line 2y+3^(1/2)=0 and then reflects about the line 3^(1/2)y=2x. am i supposed to start by finding the reflection across the main diagonal then multiplying it by the projected first line then multiplying that by the reflected second line? also, does anyone know how to reflect across the main diagonal in terms such as T(x,y) = (? , ?)o help is appreciated 2. Re: find the matrix for the transformation Originally Posted by bakinbacon am i supposed to start by finding the reflection across the main diagonal then multiplying it by the projected first line then multiplying that by the reflected second line? I think you have the right ideas, but your choice of terms is not quite right. What you want to do is to fix a basis (preferably the standard ones) and find the matrices associated with each linear transformation described above. Then you can compose all of these operations by multiplying the associated matrices. also, does anyone know how to reflect across the main diagonal in terms such as T(x,y) = (? , ?)o A linear transformation is determined by what it", "Symmetric with respect to the $y$-axis.", "2\\theta, \\sin 2\\theta)$ and $(0,1)$ gets reflected to $(\\sin 2\\theta, -\\cos 2\\theta)$. Another way is to observe that we can rotate an arbitrary mirror line onto the x-axis, then reflect across the x-axis, and rotate back. (The matrix product $[T]_{xy}$ can be seen as operating this way.) We took neither of these two approaches, because to justify them rigorously takes a bit of work, that is avoided by the pure linear algebra approach. Note also that $[T]_{uv}$ and $[T]_{xy}$ are orthogonal matrices, with determinant $-1$, as expected. \\section*{Reflection across a line of given direction vector} Suppose instead of being given an angle $\\theta$, we are given the unit direction vector $u$ to reflect the vector $w$. We can derive the matrix for the reflection directly, without involving any trigonometric functions. In the decomposition $\\vw = a \\vu + b\\vv$, we note that $b = \\vw \\cdot \\vv$. Therefore $\\vw' = (a\\vu + b\\vv) - 2b\\vv = \\vw - 2(\\vw \\cdot \\vv) \\vv\\,.$ (In fact, this is the formula used in the \\PMlinkescapetext{source code} to draw the diagram in this entry.) To derive the matrix with respect to $x,y$ coordinates, we resort to a trick: $\\vw' = I\\vw - 2\\vv (\\vw \\cdot \\vv) = I\\vw - 2\\vv (\\transpose{\\vv} \\vw) = I\\vw - 2(\\vv \\transpose{\\vv}) \\vw\\,.$ Therefore the matrix of", "cannot use a specific matrix. I hope this makes sense. This site uses Akismet to reduce spam. Learn how your comment data is processed. ##### Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis Let $F:\\R^2\\to \\R^2$ be the function that maps each vector in $\\R^2$ to its reflection with respect to $x$-axis. Determine... Close", "+1 vote The mirror image of the above test about the $\\text{X}$-axis is recategorized So, the correct answer is $(B).$", "### Home > PC > Chapter 12 > Lesson 12.2.2 > Problem12-51 12-51. Earlier we found the matrix associated with the composition of rotating 90° counterclockwise and then reflecting across the x-axis. Now find the matrix associated with reflecting across the x-axis and then rotating 90°. Homework Help ✎ From problem 12-45: A = matrix associated with 90º counterclockwise rotation B = martix associated with reflecting across the x-axis Multiply A and B in the opposite order you did in problem 12-45. That is, find BA.", "I) • For any matrix, A, it follows from the definition of I that I × A = A . ## Reflections . … … $\\left[ \\begin{matrix} -1&0 \\\\ 0&1 \\\\ \\end{matrix} \\right]$ … causes a reflection in the x direction (across the y-axis or on the y-axis) . … … $\\left[ \\begin{matrix} 1&0 \\\\ 0&-1 \\\\ \\end{matrix} \\right]$ … causes a reflection in the y direction (across the x-axis or on the x-axis)} . A reflection in the x-direction (across the y-axis or on the y-axis) could be written as: … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} -1&0 \\\\ 0&1 \\\\ \\end{matrix} \\right] \\, \\left[ \\begin{matrix} x \\\\ y \\\\ \\end{matrix} \\right]$ . so {by multiplying matrices} … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} -x \\\\ y \\\\ \\end{matrix} \\right]$ or … … $x' = -x$ … and … … $y' = y$ For example, if we start with the point $(5, \\; 3)$ under this transformation, the image is $(-5, \\; 3)$ . Compare this to starting with the standard cubic graph of $y = x^3$ under this transformation, the image is $y' = (-x')^3$ . ## Dilations . … … $\\left[ \\begin{matrix} a&0 \\\\ 0&1 \\\\ \\end{matrix} \\right]$ … causes a dilation by", "# Finding the equation of the line over which the linear transformation reflects I'm trying to find the equation of the line (y=kx) over which the linear transformation, given by the transformation matrix A (it represents reflection over the y-axis):\\begin{bmatrix}-1 & 0\\\\0 & 1\\end{bmatrix} reflects. I know how to find k: -2cosφ = tr(A) -2cosφ = 0 φ = arccos(0) = 90 k = tg(45) = 1 So now, so I've read, I have to find the eigenvector for the eigenvalue 1: A-λI = $\\begin{bmatrix}-2 & 0\\\\0 & 0\\end{bmatrix}$ How do I procede from here? I see that there is one free variable (y). - You get that $x = 0$ and $y$ is free so the equation of the line is $x = 0$.", "result here with a single linear transformation matrix, i.e. a reflection across the plane $x_1=....x_2$ then negative scaling, but I'd have to work out the specifics.", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "# Finding the eigenvectors of a reflection matrix How could I find the eigenvectors of the following matrix? $$\\left(\\begin{matrix} \\cos(\\theta) & \\sin(\\theta)\\\\ \\sin(\\theta)& -\\cos(\\theta)\\end{matrix}\\right)$$ I found the eigenvalues $1$ and $-1$ but I'm struggling with calculating the eigenvectors. Once you have the eigenvalues $\\lambda=\\pm1$, you can proceed to find corresponding eigenvectors by the usual method of solving the equation $A\\mathbf x=\\lambda\\mathbf x$, which amounts to finding the kernel of $A-\\lambda I$. It might be more instructive, however, to approach this geometrically. The matrix $A$ represents a reflection in some line through the origin. On that line, reflection is the identity map, so we expect to have $1$ as an eigenvalue with that line as its eigenspace. Reflection can be thought of as reversing the component of a vector that’s perpendicular to the mirror line, so the other eigenvalue must be $-1$, with eigenspace the orthogonal complement of that line. In $\\mathbb R^2$, this will be the perpendicular line. (We already knew that we’d have orthogonal eigenvectors because $A$ is symmetric, but it’s nice to verify that a different way.) What this means in terms of the computations is that once we’ve found an eigenvector for one eigenvalue, we can immediately write down an eigenvector for the other. This matches the eigenvalues you found by, I presume, finding the", "as areas. Example. Find a real matrix A such that reflects points across the line . The line makes an angle of with the positive x-axis. I can accomplish this transformation as follows: Thus, Contact information", "### Home > PC > Chapter 12 > Lesson 12.2.2 > Problem12-51 12-51. Earlier we found the matrix associated with the composition of rotating $90°$ counterclockwise and then reflecting across the $x$-axis. Now find the matrix associated with reflecting across the $x$-axis and then rotating $90°$. From problem 12-45: $A =$ matrix associated with $90º$ counterclockwise rotation $B =$ martix associated with reflecting across the $x$-axis $A = \\left[ \\begin{array} { c c } { 0 } & { 1 } \\\\ { - 1 } & { 0 } \\end{array} \\right] \\quad B = \\left[ \\begin{array} { c c } { 1 } & { 0 } \\\\ { 0 } & { - 1 } \\end{array} \\right]$ Multiply $A$ and $B$ in the opposite order you did in problem 12-45. That is, find $BA$.", "on the $y$-axis.", "the $y$-axis.", "Matrix for the reflection over the null space of a matrix I'm working on my latest linear algebra assignment and one question is as follows: In $\\mathbb R^3$ let R be the reflection over the null space of the matrix A = [4 4 5] Find the matrix which represents R using standard coordinates. I am familiar with the fact that the matrix for the orthogonal projection onto a subspace is given by [P] = A($A^TA$)$A^T$, where A is a matrix with columns that form a basis for the space in question. However, I am not familiar with the concept of a \"reflection over the null space\". How does this compare with the matrix for the orthogonal projection onto the null space? Is it a similar process? Thanks for any help. First of all, the formula should be $$P = B(B^TB)^{-1}B^T$$ where the columns of $B$ form of a basis of $ker(A)$. Think geometrically when solving it. Points are to be reflected in a plane which is the kernel of $A$ (see third item): • find a basis $v_1, v_2$ in $ker(A)$ and set up $B = (v_1 \\, v_2)$ • build the projector $P$ onto $ker(A)$ with above formula • geometrically the following happens to a point $x = (x_1 \\, x_2 \\, x_3)$ while reflecting in" ]

[ "projection onto the $x$-axis, \\item $T_3$: reflection about the line $y=x$, \\item $T_4$: reflection about the $x$-axis. \\end{itemize} You may use the applet that we built in class for this problem: \\url{https://sagecell.sagemath.org/?q=mtduir}. \\begin{parts} \\part[8] Write a matrix for each of the given transformations and label them correctly. \\begin{solutionorbox}[1.5in] \\end{solutionorbox} \\part[4] Find the matrix that first reflects about the $x$-axis and then rotates $90^\\circ$ (counterclockwise). \\begin{solutionorbox}[1.5in] \\end{solutionorbox} \\end{parts} \\end{questions} \\end{document}", "# Find the standard matrix of the transformation $T:\\mathbb{R}^2\\to \\mathbb{R}^2$ that corresponds to the reflection through the line Find the standard matrix of the transformation $T:\\mathbb{R}^2\\to \\mathbb{R}^2$ that corresponds to the reflection through the line $x_2=2x_1$ followed by reflection through the line $x_1=3x_2$ First you change coordinates to where your basis is given by $\\begin{pmatrix} 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix} 2 \\\\ -1\\end{pmatrix}$. This is done by multiplying by $$\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}^{-1} = \\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}.$$ Next, you reflect the second coordinate. This is done by multiplying by $$\\begin{pmatrix} 1 & 0 \\\\ 0 & -1\\end{pmatrix}.$$ Finally, you change your coordinates back by multiplying by $$\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}.$$ Thus, your transformation is multiplication by $$\\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix} = \\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 2 \\\\ -2 & 1\\end{pmatrix} = \\frac{1}{5}\\begin{pmatrix}-3 & 4 \\\\ 4 & 3 \\end{pmatrix}.$$ Now try to do the second reflection yourself. • You should explain to OP where the 1/5 factor comes from, as the determinant of the matrix. – Pieter Geerkens Apr 20 '13 at 13:04 • I assumed that the OP knows how", "# Matrix with linear transformation with reflection Find the matrix of the linear transformation A which is the reflection in the line $y = \\sqrt{2}x$ with respect to the standard basis in $\\mathbb{R^2}$. I Have no idea how to approach this problem... rotate the co-ordinate system anti-clockwise through $\\arctan \\sqrt{2}$ $$\\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ now reflect in $x'$-axis $$\\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix}$$ finally, return to original coordinates: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix}$$ altogether: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\\\ =\\frac 13 \\begin{pmatrix} -1 & 2\\sqrt{2} \\\\ 2\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ • why did you do arctan to rotate it? – user192898 Nov 16 '14 at 6:37 • this angle brings the x-axis to co-incide with the line $y=\\sqrt{2}x$ – David Holden Nov 16 '14 at 6:39 The", "# Linear transformation reflected across y-axis? We have that: The linear transformation matrix for a reflection across the line $$y = mx$$ is: $$\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}$$ So reflection across the x-axis would be: $$\\frac{1}{1 + (0)^2}\\begin{pmatrix}1-(0)^2&2(0)\\\\2(0)&(0)^2-1\\end{pmatrix} = \\begin{pmatrix}1 & 0 \\\\ 0&-1\\end{pmatrix}$$ But I’m having trouble thinking through what the matrix would be when we want to reflect every vector across the y-axis... First of all, reflection with respect to $$y$$ axis cannot be handled by formula : $$S=\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}\\tag{1}$$ for the reason that, exceptionally, there is no $$m$$ for which the $$y$$ axis has equation $$y=mx$$ (one could consider that $$m \\to \\infty$$ but it would be necessary to invoke a continuity of the symmetry operator...). Happily, there is an easy way to bypass this difficulty. Here is how. Let us remark that, in general, $$y=mx$$ can be written $$y=(\\tan \\alpha) \\ x$$ where $$\\alpha$$ is the (polar) angle made by the straight line with $$x$$ axis. Therefore we can identify : $$m=\\tan(\\alpha)\\tag{2}$$ If we plug (2) into (1), we recognize the so-called \"tangent half angle formulas\" (https://www.math24.net/weierstrass-substitution/) : $$S=\\begin{pmatrix}\\cos(2 \\alpha)&\\ \\ \\ \\sin(2 \\alpha)\\\\ \\sin(2 \\alpha)&-\\cos(2 \\alpha)\\end{pmatrix}\\tag{3}$$ which has strong similarities with a rotation matrix : it is an isometry matrix (its columns are unit length and are orthogonal). The difference with a rotation matrix", "= \\begin{pmatrix} x, -y \\end{pmatrix} \\label{9.2}$ Figure $$\\PageIndex{1}$$: Reflection across the x-axis Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} -x, y \\end{pmatrix} \\label{9.3}$ Figure $$\\PageIndex{2}$$: Reflection across the y-axis More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} = \\begin{pmatrix} -y, -x \\end{pmatrix} \\label{9.4}$ Figure $$\\PageIndex{3}$$: Reflection across the axis that is rotated 45° with with respect to x-axis. In two dimensions, the appropriate matrix to represent rotation by an angle $$\\theta$$ about the origin is $R(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.5}$ In three dimensions, rotations about the $$x$$, $$y$$ and $$z$$ axes acting on a vector $$\\begin{pmatrix} x, y, z \\end{pmatrix}$$ are represented by the following matrices. $R_{x}(\\theta) = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & \\cos\\theta & -\\sin\\theta \\\\ 0 & \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.6a}$ $R_{y}(\\theta) = \\begin{pmatrix} \\cos\\theta & 0 & -\\sin\\theta \\\\ 0 & 1 &", "B. $\\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ E. $\\begin{pmatrix} 0 &-1 \\\\-1 & 0 \\end{pmatrix}$ C. $\\begin{pmatrix} 1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ Solution The reflection across the line $y = mx$ can be represented by the matrix in form of $\\begin{pmatrix} m & 0 \\\\ 0 & m \\end{pmatrix}$. For the reflection across the line $y=\\dfrac13x$, the representation matrix is given by $T_1 = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix}$ For the reflection across the line $y=-3x$, the representation matrix is given by $T_1 = \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix}$ Hence, the representation matrix for $T$ can be expressed as follows. \\begin{aligned} T & = T_2 \\cdot T_1 \\\\ & = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix} \\cdot \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix} \\end{aligned} [collapse] Problem Number 21 A photocopier can make copies of images/writing of different sizes. A rectangular image is copied with certain settings. If this setting can be compared to the transformation process of the matrix $\\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix}$, then dilated in the origin by scale factor of $3$, then the area of ​​the rectangular image will be $\\cdots$ times the original area. A. $12$ C.", "have their w component increased after being multiplied by the projection matrix and when perspective divide is done this vertex will move closer to the origin. Perspective Divide $$\\begin{pmatrix} x \\\\ y \\\\ z \\\\ w \\end{pmatrix} \\mapsto \\begin{pmatrix} \\frac{x}{w} \\\\ \\frac{y}{w} \\\\ \\frac{z}{w} \\end{pmatrix}$$ # touch_appMake Your Own Projection Matrix The program below lets you specify the arguments ( l, r, t, b, n, f) for the projection matrix. Using the matrix you specified see where any vector is mapped to after projection and perspective divide, as well as seeing the effect of your matrix on the scenes below. tuneTune Projection Settings Use the sliders below to change the settings of the projection matrix & see how it affects the animation to the right. Changes to the matrix are used in computing the transformation of $$\\overrightarrow{a}$$ below. $$left =$$ -1.0 $$top =$$ 1.0 $$near =$$ 1.0 $$right =$$ 1.0 $$bottom =$$ -1.0 $$far =$$ 10.0 See how the vector $$\\overrightarrow{a}$$ is transformed after being transformed by the projection matrix and after perspective divide. Change the components of $$\\overrightarrow{a}$$ using the vector object below and press the blue button to calculate the final position of $$\\overrightarrow{a}$$. $$\\overrightarrow{a}$$ After Perspective Projection Transformation $$\\overrightarrow{a}$$ After Perspective Divide", "Free Version Easy Transformation Summary - f(-3x) ALGEBR-WEK1RV Given $f(x)$, which of the following transformations have been performed to obtain $f(-3x)$? Select ALL that apply. A Reflected across the $x-$axis B Reflected across the $y-$axis C Shifted left $3$ units D Shifted right $3$ units E Compressed horizontally by a factor of $3$ F Stretched horizontally by a factor of $3$", "& -1 \\end{pmatrix} = \\begin{pmatrix} x, -y \\end{pmatrix} \\label{9.2}$ Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} -x, y \\end{pmatrix} \\label{9.3}$ More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} = \\begin{pmatrix} -y, -x \\end{pmatrix} \\label{9.4}$ In two dimensions, the appropriate matrix to represent rotation by an angle $$\\theta$$ about the origin is $R(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.5}$ In three dimensions, rotations about the $$x$$, $$y$$ and $$z$$ axes acting on a vector $$\\begin{pmatrix} x, y, z \\end{pmatrix}$$ are represented by the following matrices. $R_{x}(\\theta) = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & \\cos\\theta & -\\sin\\theta \\\\ 0 & \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.6a}$ $R_{y}(\\theta) = \\begin{pmatrix} \\cos\\theta & 0 & -\\sin\\theta \\\\ 0 & 1 & 0 \\\\ \\sin\\theta & 0 & \\cos\\theta \\end{pmatrix} \\label{9.6b}$ $R_{z}(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta & 0 \\\\ \\sin\\theta & \\cos\\theta & 0 \\\\", "homography at all. Because the motion is planar, and let it be in XY plane, with Z axis perpendicular to the image plane, the rotation is: $$R=\\begin{pmatrix}\\cos\\alpha & -\\sin\\alpha & 0 \\\\ \\sin\\alpha & \\cos\\alpha & 0 \\\\ 0 & 0 & 1\\end{pmatrix}$$ The inhomogeneous translation vector is: $$C=\\begin{pmatrix}C_1 \\\\ C_2 \\\\ 0 \\end{pmatrix}$$ Then the derivation is as follows: $$x=K[I | 0]\\begin{pmatrix}X_1 \\\\ X_2 \\\\ X_3 \\\\ 1 \\end{pmatrix}$$ $$K^{-1}x=\\begin{pmatrix}X_1 \\\\ X_2 \\\\ X_3\\end{pmatrix}$$ $$x'=K'R\\bigg[I \\big| -\\begin{pmatrix}C_1 \\\\ C_2 \\\\ 0 \\end{pmatrix}\\bigg]\\begin{pmatrix}X_1 \\\\ X_2 \\\\ X_3 \\\\ 1 \\end{pmatrix} = K'R\\Bigg(\\begin{pmatrix}X_1 \\\\ X_2 \\\\ X_3\\end{pmatrix} - \\begin{pmatrix}C_1 \\\\ C_2 \\\\ 0 \\end{pmatrix}\\Bigg)= K'R\\Bigg(K^{-1}x - \\begin{pmatrix}C_1 \\\\ C_2 \\\\ 0 \\end{pmatrix}\\Bigg)$$ This is not a linear transformation, therefore is not homography. The same thing follows of course if a motion is simply a translation. If there is a rotation only, or change in camera parameters K, or both, then points will be related under homography. But if a camera center changes, it is no longer true.", "in the $$y$$ axis maps $$x$$ to $$-x$$. The appropriate matrix is very like the identity matrix but with a change in sign for the appropriate element. Reflection in the $$x$$ axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} x, -y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\end{pmatrix}$ Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is: $\\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -x \\\\ y \\end{pmatrix} \\label{9.3}$ More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -y \\\\ -x \\end{pmatrix} \\label{9.4}$ ## The $$n$$-fold Rotation Operation, $$C_n$$ The counterclockwise rotation about $$z$$ is: $C_n\\bf{u} = \\begin{pmatrix} \\cos{\\frac{2\\pi}{n}} & -\\sin{\\frac{2\\pi}{n}} & 0 \\\\ \\sin{\\frac{2\\pi}{n}} & \\cos{\\frac{2\\pi}{n}} & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x' \\\\ y' \\\\ z \\end{pmatrix}$ For clockwise", "& 0 & -(F+N)/(F-N) & -2.F.N/(F-N) \\\\ 0 & 0 & -1 & 0 \\end{pmatrix}$$ (TODO: need an image to show this transform) This is the canonical OpenGL perspective projection matrix: $$\\small\\begin{pmatrix} 2.N/(R-L) & 0 & (R+L)/(R-L) & 0 \\\\ 0 & 2.N/(T-B) & (T+B)/(T-B) & 0 \\\\ 0 & 0 & -(F+N)/(F-N) & -2.F.N/(F-N) \\\\ 0 & 0 & -1 & 0 \\end{pmatrix}$$", "$A$ is the transformation matrix for $V$. In fact wrt to $V$ we have: $v_1 = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}_{V}$. Hence: $$[A]_V[v_1]_V = \\begin{pmatrix} 1 & -1 & 1 \\\\ 1 & 0 &-1 \\\\ 0 & 1 & 0 \\end{pmatrix}_V\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}_{V} = \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}_{V} = [A(v_1)]_V$$ If you want to switch basis to the standard basis then you have to multiply the transition matrix from $V$ to the standard basis by $A$. And the transition matrix is $T = \\begin{pmatrix} 0 & 1 & a \\\\ 0 & 0 & b \\\\ 1 & 0 & c \\end{pmatrix}$. And indeed $T[A]_VT^{-1}v_1 = f(v_1)$, where the basis is the standard one, unless specified. • Awesome, that's the information I wanted. Thanks! – Staki42 Feb 9 '17 at 23:59", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "a single matrix • Lighting (done in eye-coordinates) • Projection Transformation • Orthographic • Perspective • glMatixMode(GL_PROJECTION) • gluPerspective • glOrtho • glViewport • Clipping Algorithms • Viewport Transformation • glViewport(); 3. Rasterization Stage Here all the primitives are rasterized (converted to pixels) 4. Fragment Operations Stage Additional per-pixel operations are performed here # Geometric Transformations Note: geometric transformations are applied via matrix multiplication, and matrix multiplication is not commutative, so order is important. Order is like this: $M_{final} = TransN\\ x\\ ...\\ x\\ Trans3\\ x\\ Trans2\\ x\\ Trans1$ ## Modelling Transformation Transforms the model into the world space • glScale $\\begin{pmatrix} x & 0 & 0 & 0 \\\\ 0 & y & 0 & 0 \\\\ 0 & 0 & z & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}$ • glRotate (CCW) $\\begin{pmatrix} cos(\\theta) & -sin(\\theta) & 0 & 0 \\\\ sin(\\theta) & cos(\\theta) & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}$ $\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & cos(\\theta) & -sin(\\theta) & 0 \\\\ 0 & sin(\\theta) & cos(\\theta) & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}$ $\\begin{pmatrix} cos(\\theta) & 0 & sin(\\theta) & 0 \\\\ 0 & 1 & 0 &", "The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially. To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed. In the end you obtain $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}\\mapsto \\frac1{110}\\begin{pmatrix} -\\sqrt{11} + 99 & -3 \\, \\sqrt{11} - 33 & 10 \\, \\sqrt{11} \\\\ -3 \\, \\sqrt{11} - 33 & -9 \\, \\sqrt{11} + 11 & 30 \\, \\sqrt{11} \\\\ -10 \\, \\sqrt{11} & -30 \\, \\sqrt{11} & -10 \\, \\sqrt{11} \\end{pmatrix}\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}- \\frac1{110}\\begin{pmatrix}", "# Reflection across a line? The linear transformation matrix for a reflection across the line $y = mx$ is: $$\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}$$ My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks • When you say \"completely new\", does that mean too new to know about bases and basis changes? – Daniel Fischer Oct 13 '13 at 21:20 • @DanielFischer Yup – dsd Oct 13 '13 at 21:23 You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be: The line $y = mx$ is parametrised by $t \\cdot \\begin{pmatrix}1\\\\m\\end{pmatrix}$. The line orthogonal to it is parametrised by $r \\cdot \\begin{pmatrix}-m\\\\1\\end{pmatrix}$. The line $y = mx$ shall be fixed, the line orthogonal to it shall be reflected, so you want a matrix $R$ with $$R \\begin{pmatrix}1 & -m\\\\ m & 1\\end{pmatrix} = \\begin{pmatrix}1 & m\\\\ m & -1\\end{pmatrix},$$ and that means \\begin{align} R &= \\begin{pmatrix}1 & m\\\\m&-1\\end{pmatrix} \\begin{pmatrix}1&-m\\\\m&1\\end{pmatrix}^{-1}\\\\ & = \\begin{pmatrix}1&m\\\\m&-1\\end{pmatrix}\\cdot \\frac{1}{1+m^2}\\begin{pmatrix}1&m\\\\-m&1\\end{pmatrix}\\\\ &= \\frac{1}{1+m^2} \\begin{pmatrix}1 - m^2 & 2m\\\\2m &m^2-1\\end{pmatrix}. \\end{align} Another way: to reflect along a line", "=$$\\begin{pmatrix}3\\\\-2\\\\ \\end{pmatrix}$$ The image of (2, -3) is (3, -2) The given matrix represent reflection on the line y = -x. The reflection matrix on y = -x is$$\\begin{pmatrix}0&-1\\\\-1&0\\\\ \\end{pmatrix}$$. The point P(-4, 6) represent in matrix form$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$. The image of point P$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0&-1\\\\-1&0\\\\ \\end{pmatrix}$$$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}-6\\\\4\\\\ \\end{pmatrix}$$ ∴ The image of P(-4, 6) is P'(-6, 4) Ans. Here, The matrix$$\\begin{bmatrix}0&-1\\\\1&0\\\\ \\end{bmatrix}$$ represents the rotation about origin through positive quarter turn. The point (a, b) write in matrix form is $$\\begin{pmatrix}a\\\\b\\\\ \\end{pmatrix}$$ The transformation matrix is$$\\begin{pmatrix}0&-1\\\\1&0\\\\ \\end{pmatrix}$$ Image of the object =$$\\begin{pmatrix}0&-1\\\\1&0\\\\ \\end{pmatrix}$$ $$\\begin{pmatrix}a\\\\b\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0 * a - 1 * b\\\\1 * a + 0 * b\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0 - b\\\\a + 0\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}-b\\\\a\\\\ \\end{pmatrix}$$ Hence, the image of (a, b) = (-b, a) Ans. Given vertices areA(0, 0), B(1, 0), C(1, 1) and D(0, 1) and the image of square areA'(0, 0), B'(3, 0), C'(4, 1) and D'(1, 1). Let the transformation matrix (M) =$$\\begin{pmatrix}a&b\\\\c&d\\\\ \\end{pmatrix}$$ Vertices of square write in matrix form $$\\begin{pmatrix}A&B&C&D\\\\0&1&1&0\\\\0&0&1&1\\\\ \\end{pmatrix}$$ and$$\\begin{pmatrix}A'&B'&C'&D'\\\\0&3&4&1\\\\0&0&1&1\\\\ \\end{pmatrix}$$ $$\\begin{pmatrix}a&b\\\\c&d\\\\ \\end{pmatrix}$$$$\\begin{pmatrix}0&1&1&0\\\\0&0&1&1\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0&3&4&1\\\\0&0&1&1\\\\ \\end{pmatrix}$$ Comparing the corresponding elements of equal matrix. a = 3, b = 1, c = 0 and d = 1 ∴ The required matrix =$$\\begin{pmatrix}3&1\\\\0&1\\\\ \\end{pmatrix}$$. or,$$\\begin{pmatrix}0 + 0&a +0&a + b&0 + b\\\\0 + 0&c + 0&c + d&0 +", "10:07 This is the correct answer $$\\begin{bmatrix} 0 & -1 \\\\ -1 & 0 \\end{bmatrix}$$ The first column is the reflection of $(1,0)$ and the second column is the reflection of $(0,1)$.", "described does not result in the black curve, but rather the red one. If you transform $\\ln(x)$ to $\\ln(x-3)$, the curve indeed shifts to the right $3$ units. However, when you know transform $\\ln(x-3)$ to $\\ln(-(x-3))$, the curve simply reflects across the line $x=3$ (since this line acts like the new $y$ axis for $\\ln(x-3)$), and does not create a new asymptote at $x=-3$ (i.e. the reflection of $\\ln(x-3)$ to $\\ln(-(x-3))$ does not reflect across the line $x=0$). – Dave May 21 '17 at 14:53 • That's weird, I know I'd be taking up your time, but may I please ask that you explain why the line x=3 becomes the 'new y-axis'? – Mathematician May 21 '17 at 14:56 • What I mean by \"like the new $y$ axis\" is just that the line $x=3$ is the reflection line for $\\ln(x-3)$ to $\\ln(-(x-3))$, much like how the $y$ axis (i.e. $x=0$) is the reflection line for $\\ln(x)$ to $\\ln(-x)$. – Dave May 21 '17 at 14:58 Elaborating on @Dave's comment, here's the sketch of what is going on. • You know that if you have a function $f(x)$, then $f(-x)$ reflects the function over the $y$-axis. • You know that if you have a function $f(x)$, then $f(x+3)$ shifts the function by $3$ to the left. • Starting" ]

[ "projection onto the $x$-axis, \\item $T_3$: reflection about the line $y=x$, \\item $T_4$: reflection about the $x$-axis. \\end{itemize} You may use the applet that we built in class for this problem: \\url{https://sagecell.sagemath.org/?q=mtduir}. \\begin{parts} \\part[8] Write a matrix for each of the given transformations and label them correctly. \\begin{solutionorbox}[1.5in] \\end{solutionorbox} \\part[4] Find the matrix that first reflects about the $x$-axis and then rotates $90^\\circ$ (counterclockwise). \\begin{solutionorbox}[1.5in] \\end{solutionorbox} \\end{parts} \\end{questions} \\end{document}", "# Matrix with linear transformation with reflection Find the matrix of the linear transformation A which is the reflection in the line $y = \\sqrt{2}x$ with respect to the standard basis in $\\mathbb{R^2}$. I Have no idea how to approach this problem... rotate the co-ordinate system anti-clockwise through $\\arctan \\sqrt{2}$ $$\\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ now reflect in $x'$-axis $$\\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix}$$ finally, return to original coordinates: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix}$$ altogether: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\\\ =\\frac 13 \\begin{pmatrix} -1 & 2\\sqrt{2} \\\\ 2\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ • why did you do arctan to rotate it? – user192898 Nov 16 '14 at 6:37 • this angle brings the x-axis to co-incide with the line $y=\\sqrt{2}x$ – David Holden Nov 16 '14 at 6:39 The", "# Find the standard matrix of the transformation $T:\\mathbb{R}^2\\to \\mathbb{R}^2$ that corresponds to the reflection through the line Find the standard matrix of the transformation $T:\\mathbb{R}^2\\to \\mathbb{R}^2$ that corresponds to the reflection through the line $x_2=2x_1$ followed by reflection through the line $x_1=3x_2$ First you change coordinates to where your basis is given by $\\begin{pmatrix} 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix} 2 \\\\ -1\\end{pmatrix}$. This is done by multiplying by $$\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}^{-1} = \\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}.$$ Next, you reflect the second coordinate. This is done by multiplying by $$\\begin{pmatrix} 1 & 0 \\\\ 0 & -1\\end{pmatrix}.$$ Finally, you change your coordinates back by multiplying by $$\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}.$$ Thus, your transformation is multiplication by $$\\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix} = \\frac{1}{5}\\begin{pmatrix} 1 & 2 \\\\ 2 & -1\\end{pmatrix}\\begin{pmatrix} 1 & 2 \\\\ -2 & 1\\end{pmatrix} = \\frac{1}{5}\\begin{pmatrix}-3 & 4 \\\\ 4 & 3 \\end{pmatrix}.$$ Now try to do the second reflection yourself. • You should explain to OP where the 1/5 factor comes from, as the determinant of the matrix. – Pieter Geerkens Apr 20 '13 at 13:04 • I assumed that the OP knows how", "= \\begin{pmatrix} x, -y \\end{pmatrix} \\label{9.2}$ Figure $$\\PageIndex{1}$$: Reflection across the x-axis Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} -x, y \\end{pmatrix} \\label{9.3}$ Figure $$\\PageIndex{2}$$: Reflection across the y-axis More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} = \\begin{pmatrix} -y, -x \\end{pmatrix} \\label{9.4}$ Figure $$\\PageIndex{3}$$: Reflection across the axis that is rotated 45° with with respect to x-axis. In two dimensions, the appropriate matrix to represent rotation by an angle $$\\theta$$ about the origin is $R(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.5}$ In three dimensions, rotations about the $$x$$, $$y$$ and $$z$$ axes acting on a vector $$\\begin{pmatrix} x, y, z \\end{pmatrix}$$ are represented by the following matrices. $R_{x}(\\theta) = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & \\cos\\theta & -\\sin\\theta \\\\ 0 & \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.6a}$ $R_{y}(\\theta) = \\begin{pmatrix} \\cos\\theta & 0 & -\\sin\\theta \\\\ 0 & 1 &", "# Linear transformation reflected across y-axis? We have that: The linear transformation matrix for a reflection across the line $$y = mx$$ is: $$\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}$$ So reflection across the x-axis would be: $$\\frac{1}{1 + (0)^2}\\begin{pmatrix}1-(0)^2&2(0)\\\\2(0)&(0)^2-1\\end{pmatrix} = \\begin{pmatrix}1 & 0 \\\\ 0&-1\\end{pmatrix}$$ But I’m having trouble thinking through what the matrix would be when we want to reflect every vector across the y-axis... First of all, reflection with respect to $$y$$ axis cannot be handled by formula : $$S=\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}\\tag{1}$$ for the reason that, exceptionally, there is no $$m$$ for which the $$y$$ axis has equation $$y=mx$$ (one could consider that $$m \\to \\infty$$ but it would be necessary to invoke a continuity of the symmetry operator...). Happily, there is an easy way to bypass this difficulty. Here is how. Let us remark that, in general, $$y=mx$$ can be written $$y=(\\tan \\alpha) \\ x$$ where $$\\alpha$$ is the (polar) angle made by the straight line with $$x$$ axis. Therefore we can identify : $$m=\\tan(\\alpha)\\tag{2}$$ If we plug (2) into (1), we recognize the so-called \"tangent half angle formulas\" (https://www.math24.net/weierstrass-substitution/) : $$S=\\begin{pmatrix}\\cos(2 \\alpha)&\\ \\ \\ \\sin(2 \\alpha)\\\\ \\sin(2 \\alpha)&-\\cos(2 \\alpha)\\end{pmatrix}\\tag{3}$$ which has strong similarities with a rotation matrix : it is an isometry matrix (its columns are unit length and are orthogonal). The difference with a rotation matrix", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "B. $\\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ E. $\\begin{pmatrix} 0 &-1 \\\\-1 & 0 \\end{pmatrix}$ C. $\\begin{pmatrix} 1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ Solution The reflection across the line $y = mx$ can be represented by the matrix in form of $\\begin{pmatrix} m & 0 \\\\ 0 & m \\end{pmatrix}$. For the reflection across the line $y=\\dfrac13x$, the representation matrix is given by $T_1 = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix}$ For the reflection across the line $y=-3x$, the representation matrix is given by $T_1 = \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix}$ Hence, the representation matrix for $T$ can be expressed as follows. \\begin{aligned} T & = T_2 \\cdot T_1 \\\\ & = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix} \\cdot \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix} \\end{aligned} [collapse] Problem Number 21 A photocopier can make copies of images/writing of different sizes. A rectangular image is copied with certain settings. If this setting can be compared to the transformation process of the matrix $\\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix}$, then dilated in the origin by scale factor of $3$, then the area of ​​the rectangular image will be $\\cdots$ times the original area. A. $12$ C.", "& -1 \\end{pmatrix} = \\begin{pmatrix} x, -y \\end{pmatrix} \\label{9.2}$ Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} -x, y \\end{pmatrix} \\label{9.3}$ More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} x, y \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} = \\begin{pmatrix} -y, -x \\end{pmatrix} \\label{9.4}$ In two dimensions, the appropriate matrix to represent rotation by an angle $$\\theta$$ about the origin is $R(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.5}$ In three dimensions, rotations about the $$x$$, $$y$$ and $$z$$ axes acting on a vector $$\\begin{pmatrix} x, y, z \\end{pmatrix}$$ are represented by the following matrices. $R_{x}(\\theta) = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & \\cos\\theta & -\\sin\\theta \\\\ 0 & \\sin\\theta & \\cos\\theta \\end{pmatrix} \\label{9.6a}$ $R_{y}(\\theta) = \\begin{pmatrix} \\cos\\theta & 0 & -\\sin\\theta \\\\ 0 & 1 & 0 \\\\ \\sin\\theta & 0 & \\cos\\theta \\end{pmatrix} \\label{9.6b}$ $R_{z}(\\theta) = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta & 0 \\\\ \\sin\\theta & \\cos\\theta & 0 \\\\", "### Home > PC3 > Chapter 9 > Lesson 9.2.2 > Problem9-115 9-115. ​Write the matrix associated with reflecting across the $x$-axis and then rotating $90^\\circ$ counterclockwise. Review your work from problem 9-114. Multiply the transformation matrices in the opposite order you did in problem 9-114.", "The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially. To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed. In the end you obtain $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}\\mapsto \\frac1{110}\\begin{pmatrix} -\\sqrt{11} + 99 & -3 \\, \\sqrt{11} - 33 & 10 \\, \\sqrt{11} \\\\ -3 \\, \\sqrt{11} - 33 & -9 \\, \\sqrt{11} + 11 & 30 \\, \\sqrt{11} \\\\ -10 \\, \\sqrt{11} & -30 \\, \\sqrt{11} & -10 \\, \\sqrt{11} \\end{pmatrix}\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}- \\frac1{110}\\begin{pmatrix}", "in the $$y$$ axis maps $$x$$ to $$-x$$. The appropriate matrix is very like the identity matrix but with a change in sign for the appropriate element. Reflection in the $$x$$ axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} x, -y \\end{pmatrix}$$, and the appropriate matrix is $\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\end{pmatrix}$ Reflection in the y axis transforms the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ to $$\\begin{pmatrix} -x, y \\end{pmatrix}$$, and the appropriate matrix is: $\\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -x \\\\ y \\end{pmatrix} \\label{9.3}$ More generally, matrices can be used to represent reflections in any plane (or line in 2D). For example, reflection in the 45° axis shown below maps $$\\begin{pmatrix} x, y \\end{pmatrix}$$ onto $$\\begin{pmatrix} -y, -x \\end{pmatrix}$$. $\\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -y \\\\ -x \\end{pmatrix} \\label{9.4}$ ## The $$n$$-fold Rotation Operation, $$C_n$$ The counterclockwise rotation about $$z$$ is: $C_n\\bf{u} = \\begin{pmatrix} \\cos{\\frac{2\\pi}{n}} & -\\sin{\\frac{2\\pi}{n}} & 0 \\\\ \\sin{\\frac{2\\pi}{n}} & \\cos{\\frac{2\\pi}{n}} & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x' \\\\ y' \\\\ z \\end{pmatrix}$ For clockwise", "=$$\\begin{pmatrix}3\\\\-2\\\\ \\end{pmatrix}$$ The image of (2, -3) is (3, -2) The given matrix represent reflection on the line y = -x. The reflection matrix on y = -x is$$\\begin{pmatrix}0&-1\\\\-1&0\\\\ \\end{pmatrix}$$. The point P(-4, 6) represent in matrix form$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$. The image of point P$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0&-1\\\\-1&0\\\\ \\end{pmatrix}$$$$\\begin{pmatrix}-4\\\\6\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}-6\\\\4\\\\ \\end{pmatrix}$$ ∴ The image of P(-4, 6) is P'(-6, 4) Ans. Here, The matrix$$\\begin{bmatrix}0&-1\\\\1&0\\\\ \\end{bmatrix}$$ represents the rotation about origin through positive quarter turn. The point (a, b) write in matrix form is $$\\begin{pmatrix}a\\\\b\\\\ \\end{pmatrix}$$ The transformation matrix is$$\\begin{pmatrix}0&-1\\\\1&0\\\\ \\end{pmatrix}$$ Image of the object =$$\\begin{pmatrix}0&-1\\\\1&0\\\\ \\end{pmatrix}$$ $$\\begin{pmatrix}a\\\\b\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0 * a - 1 * b\\\\1 * a + 0 * b\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0 - b\\\\a + 0\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}-b\\\\a\\\\ \\end{pmatrix}$$ Hence, the image of (a, b) = (-b, a) Ans. Given vertices areA(0, 0), B(1, 0), C(1, 1) and D(0, 1) and the image of square areA'(0, 0), B'(3, 0), C'(4, 1) and D'(1, 1). Let the transformation matrix (M) =$$\\begin{pmatrix}a&b\\\\c&d\\\\ \\end{pmatrix}$$ Vertices of square write in matrix form $$\\begin{pmatrix}A&B&C&D\\\\0&1&1&0\\\\0&0&1&1\\\\ \\end{pmatrix}$$ and$$\\begin{pmatrix}A'&B'&C'&D'\\\\0&3&4&1\\\\0&0&1&1\\\\ \\end{pmatrix}$$ $$\\begin{pmatrix}a&b\\\\c&d\\\\ \\end{pmatrix}$$$$\\begin{pmatrix}0&1&1&0\\\\0&0&1&1\\\\ \\end{pmatrix}$$ =$$\\begin{pmatrix}0&3&4&1\\\\0&0&1&1\\\\ \\end{pmatrix}$$ Comparing the corresponding elements of equal matrix. a = 3, b = 1, c = 0 and d = 1 ∴ The required matrix =$$\\begin{pmatrix}3&1\\\\0&1\\\\ \\end{pmatrix}$$. or,$$\\begin{pmatrix}0 + 0&a +0&a + b&0 + b\\\\0 + 0&c + 0&c + d&0 +", "# Show by using matrix method that a reflection about the line y=x followed by rotation about origin through 90° +ve is equivalent to reflection about y-axis.? Jul 14, 2018 #### Explanation: The matrix of the reflection in the line $y = x$ is ${A}_{1} = \\left(\\begin{matrix}0 & 1 \\\\ 1 & 0\\end{matrix}\\right)$ The matrix of the rotation anticlockwise by ${90}^{\\circ}$ is ${A}_{2} = \\left(\\begin{matrix}0 & - 1 \\\\ 1 & 0\\end{matrix}\\right)$ The combination of matrices ${A}_{1}$ and ${A}_{2}$ is ${A}_{2} \\cdot {A}_{1} = \\left(\\begin{matrix}0 & - 1 \\\\ 1 & 0\\end{matrix}\\right) \\cdot \\left(\\begin{matrix}0 & 1 \\\\ 1 & 0\\end{matrix}\\right)$ $= \\left(\\begin{matrix}- 1 & 0 \\\\ 0 & 1\\end{matrix}\\right)$ The matrix of the reflection in the y-axis is ${A}_{3} = \\left(\\begin{matrix}- 1 & 0 \\\\ 0 & 1\\end{matrix}\\right)$ Therefore, ${A}_{1} \\cdot {A}_{2} = {A}_{3}$ For verification, you can take a vector $\\left(\\begin{matrix}x \\\\ y\\end{matrix}\\right)$ and apply the matrices.", "2). (-2 , 3) reflection in y = -x ; (-1, 2) reflection in y = -x ; (-2 , 3) reflection in y = y ; (-4 , 2) • ### Find the 2 ( imes) 2 matrix which transforms A(-3 , 1) to A(3 , 1). \begin{pmatrix} 2 & 1 \\ 1 & 1 \\ end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \\ end{pmatrix} \begin{pmatrix} -1 & -2 \\ 9 & 1 \\ end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & -1 \\ end{pmatrix} • ### If a point (x , y) is transformed into (y , x) by a 2 ( imes) 2 transformation matrix , then find the matrix . \begin{pmatrix} 0 & -2 \\ 0 & -2 \\ end{pmatrix} \begin{pmatrix} 0 & -4 \\ 0 & -1 \\ end{pmatrix} \begin{pmatrix} 0 & 0 \\ 5 & 8 \\ end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ end{pmatrix} • ### If a point (a , b) is transformed into ( b ,-a) by a 2 ( imes) 2 transformation matrix , find the matrix . \begin{pmatrix} 1 & 1 \\ 0 & 0 \\ end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ end{pmatrix} \begin{pmatrix} 1 & 1 \\", "# find the matrix for the reflection over a line that goes through the origin and makes the angle pi/17 with the x-axis. I already know how to find the matrix of reflection over the $x$-axis (in the plane) (ON-base). By looking at what happens to the standard basis vectors $\\hat{e}_1 = (1,0)$ and $\\hat{e}_2 = (0,1)$. The matrix for the transformation will be: $$\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}$$ That's an easy example. But I wan't to find the matrix for the reflection over a line that goes through the origin and makes the angle $\\pi/17$ with the x-axis. How can I approach this problem? • Exactly the same way. – saulspatz May 3 '18 at 13:04 The same way. Look at the standard basis and find out the images of the standard basis under your transformation. These should involve sines and cosines of $\\pi/17$... UPDATE OK, let's reflect $(1,0)$ over that line. Notice in a sense you would be just rotating the point by $2\\pi/17$ radians around the origin, can you figure out the result? Similarly, from $(0,1)$ to the line is $\\pi/2 - \\pi/17 = 15\\pi/34$ radians, so the result after reflection is a negative rotation of $2 \\cdot 15\\pi/34$ radians from the $\\pi/2$ angle, resulting in the final angle of$$\\frac\\pi2 -", "If you find any mistakes, please make a comment! Thank you. ## Solution to Linear Algebra Done Wrong Exercise 1.3.2 Let a linear transformation in $\\mathbb R^2$ be the reflection in the line $x_1=x_2$. Find its matrix. Solution: Let us think $x_1=x$ and $x_2=y$. In the $(x,y)$-coordinate plane, the graph of the inverse function of a function $f(x)$ is obtained by reflecting the graph of $f(x)$ in the line $y=x$. Also note the the inverse function interchanges $x$ and $y$. Denote the linear transformation in $\\mathbb R^2$ corresponding to the reflection in the line $x_1=x_2$ by $T$. With the above observation, we see that $$T(x_1,x_2)^T=(x_2,x_1)^T.$$Therefore, we have $$T\\begin{pmatrix}x_1\\\\ x_2\\end{pmatrix}=x_1\\begin{pmatrix}0\\\\ 1\\end{pmatrix}+x_2\\begin{pmatrix}1\\\\ 0\\end{pmatrix}.$$Hence the matrix is given by $\\begin{pmatrix}0 & 1\\\\ 1 & 0\\end{pmatrix}$.", "# Reflection across a line? The linear transformation matrix for a reflection across the line $y = mx$ is: $$\\frac{1}{1 + m^2}\\begin{pmatrix}1-m^2&2m\\\\2m&m^2-1\\end{pmatrix}$$ My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks • When you say \"completely new\", does that mean too new to know about bases and basis changes? – Daniel Fischer Oct 13 '13 at 21:20 • @DanielFischer Yup – dsd Oct 13 '13 at 21:23 You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be: The line $y = mx$ is parametrised by $t \\cdot \\begin{pmatrix}1\\\\m\\end{pmatrix}$. The line orthogonal to it is parametrised by $r \\cdot \\begin{pmatrix}-m\\\\1\\end{pmatrix}$. The line $y = mx$ shall be fixed, the line orthogonal to it shall be reflected, so you want a matrix $R$ with $$R \\begin{pmatrix}1 & -m\\\\ m & 1\\end{pmatrix} = \\begin{pmatrix}1 & m\\\\ m & -1\\end{pmatrix},$$ and that means \\begin{align} R &= \\begin{pmatrix}1 & m\\\\m&-1\\end{pmatrix} \\begin{pmatrix}1&-m\\\\m&1\\end{pmatrix}^{-1}\\\\ & = \\begin{pmatrix}1&m\\\\m&-1\\end{pmatrix}\\cdot \\frac{1}{1+m^2}\\begin{pmatrix}1&m\\\\-m&1\\end{pmatrix}\\\\ &= \\frac{1}{1+m^2} \\begin{pmatrix}1 - m^2 & 2m\\\\2m &m^2-1\\end{pmatrix}. \\end{align} Another way: to reflect along a line", "# Find the standard matrix of a transformation Let $$M:\\mathbb{R}^2\\ \\rightarrow\\mathbb{R}^2$$ be the linear transformation that first reflects it through the line $$x_1=x_{2}$$, and then rotates each point counterclockwise around the origin by $$\\frac{5 \\pi}{4}$$ radians. Find the standard matrix of $$M$$ So I was thinking, since it's the line $$x_1 = x_2$$, looking at it graphically, I would have $$\\begin{pmatrix} cos(x)&-sin(x)\\\\ sin(x)&cos(x)\\\\ \\end{pmatrix}$$ which then becomes $$\\begin{pmatrix} sin(x)&cos(x)\\\\ cos(x)&-sin(x)\\\\ \\end{pmatrix}$$ which then becomes (using $$x = \\frac{5\\pi}{4}$$) $$\\begin{pmatrix} -\\frac{\\sqrt{2}}{2}&-\\frac{\\sqrt{2}}{2}\\\\ -\\frac{\\sqrt{2}}{2}&\\frac{\\sqrt{2}}{2}\\\\ \\end{pmatrix}$$ But that's the wrong answer. Would really appreciate some help with these, I don't fully understand how to algebraically find / convert these reflections yet. Thanks! The reflection through the line $$x_1=x_2$$ maps $$\\left(1,1\\right)$$ into itself and it maps $$\\left(-1,1\\right)$$ into $$\\left(1,-1\\right)$$. So, its matrix with respect to the standard basis is $$\\left[\\begin{smallmatrix}0&1\\\\1&0\\end{smallmatrix}\\right]$$. And the matrix of a counterclockwise rotation around the origin with angle $$\\frac{5\\pi}4$$ is$$\\begin{bmatrix}-\\frac1{\\sqrt2}&\\frac1{\\sqrt2}\\\\-\\frac1{\\sqrt2}&-\\frac1{\\sqrt2}\\end{bmatrix}.$$So, take$$\\begin{bmatrix}-\\frac1{\\sqrt2}&\\frac1{\\sqrt2}\\\\-\\frac1{\\sqrt2}&-\\frac1{\\sqrt2}\\end{bmatrix}.\\begin{bmatrix}0&1\\\\1&0\\end{bmatrix}=\\begin{bmatrix}\\frac1{\\sqrt2}&-\\frac1{\\sqrt2}\\\\-\\frac1{\\sqrt2}&-\\frac1{\\sqrt2}\\end{bmatrix}.$$ • Why is the first row, second column the one that is positive rather than second row second column? We are first supposed to reflect, correct? If so, $-sin(x)$ will be in the second row second column I believe? Jun 6, 2021 at 23:23 • Because it's the matrix$$\\begin{bmatrix}\\cos\\left(\\frac{5\\pi}4\\right)&-\\sin\\left(\\frac{5\\pi}4\\right)\\\\\\sin\\left(\\frac{5\\pi}4\\right)&\\cos\\left(\\frac{5\\pi}4\\right)\\end{bmatrix}.$$ Jun 7, 2021 at 6:15 • No, I am not reflecting first. The matrix that I wrote is the", "the matrix representing a reflection across the yz-plane. a. Write A and B. b. Evaluate AB. What does this matrix represent? AB is a reflection across the yz-plane followed by a dilation of $$\\frac{1}{2}$$. c. Let x=$$\\left[\\begin{array}{l} 5 \\\\ 6 \\\\ 4 \\end{array}\\right]$$, y=$$\\left[\\begin{array}{l} -1 \\\\ 3 \\\\ 2 \\end{array}\\right]$$, and z=$$\\left[\\begin{array}{l} 8 \\\\ -2 \\\\ -4 \\end{array}\\right]$$. Find (AB)x, (AB)y, and (AB)z. Question 2. Let A be the matrix representing a rotation of 30° about the x-axis, and let B be the matrix representing a dilation of 5. a. Write A and B. b. Evaluate AB. What does this matrix represent? AB is a dilation of 5 followed by a rotation of 30° about the x-axis. c. Let x=$$\\left[\\begin{array}{l} 1 \\\\ 0 \\\\ 0 \\end{array}\\right]$$, y=$$\\left[\\begin{array}{l} 0 \\\\ 1 \\\\ 0 \\end{array}\\right]$$, z=$$\\left[\\begin{array}{l} 0 \\\\ 0 \\\\ 1 \\end{array}\\right]$$. Find (AB)x, (AB)y, and (AB)z. Question 3. Let A be the matrix representing a dilation of 3, and let B be the matrix representing a reflection across the plane y=x. a. Write A and B. b. Evaluate AB. What does this matrix represent? AB=$$\\left[\\begin{array}{lll} 0 & 3 & 0 \\\\ 3 & 0 & 0 \\\\ 0 & 0 & 3 \\end{array}\\right]$$ AB is a reflection across the y=x plane followed by a dilation of 3. c. Let", "I) • For any matrix, A, it follows from the definition of I that I × A = A . ## Reflections . … … $\\left[ \\begin{matrix} -1&0 \\\\ 0&1 \\\\ \\end{matrix} \\right]$ … causes a reflection in the x direction (across the y-axis or on the y-axis) . … … $\\left[ \\begin{matrix} 1&0 \\\\ 0&-1 \\\\ \\end{matrix} \\right]$ … causes a reflection in the y direction (across the x-axis or on the x-axis)} . A reflection in the x-direction (across the y-axis or on the y-axis) could be written as: … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} -1&0 \\\\ 0&1 \\\\ \\end{matrix} \\right] \\, \\left[ \\begin{matrix} x \\\\ y \\\\ \\end{matrix} \\right]$ . so {by multiplying matrices} … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} -x \\\\ y \\\\ \\end{matrix} \\right]$ or … … $x' = -x$ … and … … $y' = y$ For example, if we start with the point $(5, \\; 3)$ under this transformation, the image is $(-5, \\; 3)$ . Compare this to starting with the standard cubic graph of $y = x^3$ under this transformation, the image is $y' = (-x')^3$ . ## Dilations . … … $\\left[ \\begin{matrix} a&0 \\\\ 0&1 \\\\ \\end{matrix} \\right]$ … causes a dilation by" ]

[ "+0 # help 0 46 1 What is the area of the parallelogram with vertices $$\\begin{pmatrix} 3 \\\\ 1\\\\ 2 \\end{pmatrix},$$ $$\\begin{pmatrix} 2 \\\\ 1\\\\4\\end{pmatrix}$$$$\\begin{pmatrix} 4\\\\2\\\\4 \\end{pmatrix},$$and $$\\begin{pmatrix} 5\\\\2\\\\2 \\end{pmatrix}?$$ Apr 29, 2020", "of the parallelogram with edges ${v=(3,2)}$ and ${w=(1,4)}$. • Find the area of the triangle with sides ${v,w,}$and ${v+w}$.Draw it. • Find the area of the triangle with sides ${v,w,}$and ${w-v}$.Draw it. Solution: • $\\displaystyle \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=10.$ • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture omitted. • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture Omitted. $\\Box$ Exercise 4.4.28 A box has edges from ${(0,0,0)}$ to ${(3,1,1),(1,3,1),}$and ${(1,1,3)}$.Find its volumn and also find the area of each parallelogram face. Solution: The volumn of the box is $\\displaystyle \\begin{vmatrix} 3&1&1\\\\ 1&3&1\\\\ 1&1&3 \\end{vmatrix}=20.$ The area of each parallelogram face is $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 3&1 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 1&3\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2}.$ In order to compute the area of each parallelogram face,we can also compute $\\displaystyle \\sqrt{\\det A^TA}=\\sqrt{\\det \\left(\\begin{bmatrix} 3&1&1\\\\ 1&3&1 \\end{bmatrix} \\begin{bmatrix} 3&1\\\\ 1&3\\\\ 1&1 \\end{bmatrix}\\right)}= 6 \\sqrt{2}$ etc.$\\Box$ Exercise 4.4.35 An ${n}$– dimensional cube has how many corners?How many edges?How many ${n-1}$-dimensional faces?The ${n}$– cube whose edges are the rows of ${2I}$ has volumn ${\\underline{~~~~~~~~~~}}$.A hypercube computer has parallel processors at the corners with connections along the edeges. Solution: An ${n}$-dimensional cube", "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", ", 6 )$ $u = AB = \\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 6 \\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1 } \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 5 & 6 \\\\ 2 & 4 \\end{bmatrix}= 20 \\: – \\: 12 = 8$ The area of the parallelogram is $8$. ## Numerical Result The area of the parallelogram is $8$. ## Example Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$ Solution A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram. $A = ( 0 , 0 )$ $B = ( 6 , 8 )$ $C = ( 5 , 4 )$ $D = ( 11 , 6 )$ $u = AB = \\begin{bmatrix} 6\\\\ 8 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 5\\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1", "# Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cristiano Sears ###### Not exactly what you’re looking for? content_user Vertices are $$A(-3,0),\\ B(-1,6),\\ C(8,5)$$ and $$D(6,-1)$$ So $$\\vec{AB}=<-1-(-3),6-0>=<2,6>$$ $$\\vec{AD}=<6-(-3),-1-0>=<9,-1>$$ Hence area of parallelogram $$=|\\vec{AB}\\times\\vec{AD}|$$ $$\\vec{AB}\\times\\vec{AD}=\\begin{bmatrix}i&j&k\\\\2&6&0\\\\9&-1&0\\end{bmatrix}$$ $$=i(0-0)+j(0-0)+k(-2-54)$$ $$=-56k$$ Hence Area $$=|\\vec{AB}\\times\\vec{AD}|$$ $$=\\sqrt{(-56)^2}=56$$", "find the area of the parallelogram whose vertices are listed # find the area of the parallelogram whose vertices are listed H 240 points In Exercises 19–22, find the area of the parallelogram whose vertices are listed. (0,-2),(5,-2),(-3,1),(2,1) find the hansel 8.7k points Translate the coordinates so that one of them is at the origin, doing so does not change the area of the parallelogram, but allows us to find the area. In this case I will add the vector (0,2) to each vertex which gives, (0,0),(5,0),(-3,3),(2,3) let Now the area of the parallelogram is the determinant of the matrix formed by the column vectors (5,0) and (-3,3), $$\\begin{vmatrix} 5 & -3 \\\\ 0 & 3 \\end{vmatrix}=15$$ Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.", "# Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4 Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4), (11,6) ## Want to know more about Matrix transformations? • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Aniqa O'Neill Parallelogram can be described with 4 vertices or 2 vectors. Since we are given 4 vertices(ABCD), let's find vectors u, v that describe parallelogram. Graph these points to see how vectors are described A=(0,0) B=(5,2) C=(6,4) D=(11,6) $$u=AB=\\begin{bmatrix}5\\\\2\\end{bmatrix}$$ $$u=AC=\\begin{bmatrix}6\\\\4\\end{bmatrix}$$ Area of parallelogram is absolute value of determinant $$\\begin{bmatrix}u_1 & v_1 \\\\u_2 &v_2 \\end{bmatrix}$$ $$\\begin{bmatrix}u_1 & v_1 \\\\u_2 &v_2 \\end{bmatrix}=det\\begin{bmatrix}5& 6 \\\\2 &4 \\end{bmatrix}=20-12=8$$ Result: 8", "Find the area of the parallelogram with vertices at $(5, -3),$ $(-2, -1),$ $(-3, 4),$ and $(-10, 6).$ Area = .", "as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The formula for the area of a parallelogram is base x height. Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [closed], Soft question on the area of a parallelogram using vectors. Area of the Parallelogram: A parallelogram is a geometric structure having one side having opposite sides parallel to each other. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. #vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj# First of all, to simplify the task, let's move it to a position when its vertex A coincides with the origin of coordinates. So this's a B C on G E. Now the area of", "# Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). I can't figure out what I'm doing wrong.. I let: $a = |DC| = (-2,-1) - (-6,3) = (4,-4)$ $b = |AD| = (-1,1) - (-5,5) = (4,-4)$ so: $\\begin{bmatrix} 4&-4\\\\4&-4\\end{bmatrix} \\implies$ Area $= -16 + 16 = 0$ ? The determinant is the area of the parallelogram is it not? • Not sure how you're labelling vertices, but $a$ appears to be $|CD|$, while $b$ appears to be $|AB|$. You get a result of zero because those lines are parallel. – Joey Zou Jul 6 '16 at 22:08 • You are correct that was the problem. The answer is $|-12|$ – Yusha Jul 6 '16 at 22:11", "# Math Help - Parallelogram... 1. ## Parallelogram... Let f(t) be the area of the parallelogram with vertices v=(1,2t), w = (2t,1), (0,0), and v+w = (2t+1,2t+1). Find a formula for f(t). I'm assuming one can't simply 'stand up' the parallelogram so that it's a rectangle... 2. Let $O(0,0), \\ A(1,2t), \\ B(2t,1), C(1+2t, 1+2t)$ $Area[OACB]=2\\cdot Area[OAB]=2\\cdot\\frac{1}{2}|\\Delta|$ where $\\Delta=\\begin{vmatrix}0 & 0 & 1\\\\1 & 2t & 1\\\\2t & 1 & 1\\end{vmatrix}=1-4t^2$ Therefore $f(t)=|1-4t^2|$ 3. Another way to do it is using the cross product of v and w. That also works with 3D coordinates. 4. Hello, PTL! Yet another approach . . . Did you make a sketch? Let $f(t)$ be the area of the parallelogram with vertices: $\\vec v\\,=\\,(1,2t),\\;\\vec w\\,=\\,(2t,1),\\;(0,0)$, and $\\overrightarrow{v+w} \\:=\\:(2t+1,2t+1).$ Find a formula for $f(t).$ Code: | 2t+1 + - - - - - - - - - - o v+w | * *: | * * : | v * * : 2t + - - o * : | * * : 1+ * o w : | * * : : | * * : : |* * : : - - o - - + - - - - + - - * - - (0,0) 1 2t 2t+1 We have a rhombus. If $d_1,\\:d_2$ are diagonals of a rhombus,", "the area of the paralellogram and the original square. Since the original square has area $1$, the area of the parallelogram is the determinant of $\\bc{\\A}$. Working this out requires only a small amount of basic geometry. Here’s the simplest way to do it. We first name the four elements of our matrix: $\\bc{\\A} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ We can now name the four corners of our parallelogram in terms of these four scalars, by multiplying the corner points of the unit square by $\\bc{\\A}$: $\\begin{matrix} \\times & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}1\\\\ 0\\end{pmatrix} & \\begin{pmatrix}0\\\\ 1\\end{pmatrix} & \\begin{pmatrix}1\\\\ 1\\end{pmatrix} \\\\ \\bc{\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}} & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}a\\\\ c\\end{pmatrix} & \\begin{pmatrix}b\\\\ d\\end{pmatrix} & \\begin{pmatrix}a+b\\\\ c+d\\end{pmatrix} \\\\ \\end{matrix}$ This gives us the following picture. Here we can see the area of the paralellogram clearly: there is one large rectangle of area $\\oc{ad}$. To get from this area to the area of our parallelogram, we should subtract the area of the green triangle in the bottom, which is part of the rectangle but not the parallelogram. But then, there’s a green triangle at the top, with the same size, which is (mostly) part of the parallelogram but not of the rectangle, so these cancel each other out. We", "# Find the area of the parallelogram with vertices A(-3,0), B(-1,7), Find the area of the parallelogram with vertices A(-3,0), B(-1,7), C(9,6), and D(7,-1) • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Phisecome Area of parallelogram with vertices (-3,0), (-1,7), (9,6) and (7,-1) is given by $$\\displaystyle{A}={\\frac{{{1}}}{{{2}}}}{\\left[{x}_{{1}}{y}_{{2}}+{x}_{{2}}{y}_{{3}}+{x}_{{3}}{y}_{{4}}+{x}_{{4}}{y}_{{1}}-{x}_{{2}}{y}_{{1}}-{x}_{{3}}{y}_{{1}}-{x}_{{3}}{y}_{{2}}-{x}_{{4}}{y}_{{3}}-{x}_{{1}}{y}_{{4}}\\right]}$$ Here, $$\\displaystyle{\\left({x}_{{1}},{y}_{{1}}\\right)}={\\left(-{3},{0}\\right)}$$ $$\\displaystyle{\\left({x}_{{2}},{y}_{{2}}\\right)}={\\left(-{1},{7}\\right)}$$ $$\\displaystyle{\\left({x}_{{3}},{y}_{{3}}\\right)}={\\left({9},{6}\\right)}$$ $$\\displaystyle{\\left({x}_{{4}},{y}_{{4}}\\right)}={\\left({7},-{1}\\right)}$$ $$\\displaystyle{A}={\\frac{{{1}}}{{{2}}}}{\\mid}{\\left(-{3}\\right)}{\\left({7}\\right)}+{\\left(-{1}\\right)}{6}+{9}\\times{\\left(-{1}\\right)}+{7}\\times{\\left({0}\\right)}-{9}\\times{7}-{7}\\times{6}-{\\left(-{3}\\right)}{\\left(-{1}\\right)}$$ $$\\displaystyle={\\frac{{{1}}}{{{2}}}}{\\left|-{21}-{6}-{9}+{0}-{0}-{63}-{42}-{3}\\right|}$$ $$\\displaystyle={\\frac{{{1}}}{{{2}}}}{\\left({144}\\right)}={72}$$ ###### Have a similar question? • Questions are typically answered in as fast as 30 minutes", "? Free Version Easy # Area of Parallelogram MVCALC-TSPEOK Let $A=(3,2,1)$ be a vertex of a parallelogram, and let $B=(-1,0,4)$ and $D=(2,3,0)$ be the two neighboring vertices of the parallelogram. What is the area of this parallelogram? A $2$ B $\\sqrt{2}/2$ C $\\sqrt{2}$ D $1$ E $\\sqrt{2}/3$", "back and think about what properties a mathematical object, describing area should have. Suppose we have two (possibly infinitesimal) vectors, forming a two-dimensional parallelogram and we need the mathematical object that tells us the area of this parallelogram: \\mathrm{area}(V,W) = \\text{ area of the parallelogram formed by V and W}. An obious property is that the number $\\mathrm{area}(V,W)$ ought to double if we double the length of one vector: \\mathrm{area}(2V,W) = 2 \\times \\text{ area of the parallelogram formed by V and W} \\mathrm{area}(V,W+Z) = \\mathrm{area}(V,W) + \\mathrm{area}(V,Z) A pictorial proof can be seen in the following figure: Together these properties are what we called linearity above, and we see that this fits perfectly with the defintion of a $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. Another important propery is that $\\mathrm{area}(V,W)$ must vanish if $V$ and $W$ are parallel, i.e. if $W=aV$ for some number $a$, we have \\mathrm{area}(V,W) =\\mathrm{area}(V,aV) \\stackrel{!}{=} 0. .Because we have $\\mathrm{area}(V,aV)=a \\times \\mathrm{area}(V,V) \\stackrel{!}{=} 0$, we start with $\\mathrm{area}(V,V) \\stackrel{!}{=} 0$ in the next step to see that $\\mathrm{area}(V,W)$ must be an antisymmetric $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. The proof is simple: From \\mathrm{area}(V,V) = 0, if we write $V= U+W$, it follows \\mathrm{area}(U+W,U+W) = 0. Using the linearity yields \\mathrm{area}(U,U) + \\mathrm{area}(U,W) + \\mathrm{area}(W,U) + \\mathrm{area}(W,W) = 0. Using now $\\mathrm{area}(V,V)", "} \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 6 & 5 \\\\ 8 & 4 \\end{bmatrix}= 24 \\: – \\: 40 = 16$ The area of the parallelogram is $16$. 5/5 - (5 votes)", "Additionally, $$SN = SR$$. You need to prove that $$NPTS$$ is a parallelogram. Redraw the diagram and mark all given and known information: Study the diagram below; it is not necessarily drawn to scale. Quadrilateral $$XWST$$ is a parallelogram and $$TV$$ and $$XW$$ have lengths $$b$$ and $$2b$$, respectively, as shown. You need to prove that $$\\triangle TVU \\equiv \\triangle SVW$$. Redraw the diagram and fill in all given and known information. $$T \\text{ and } V \\text{ are mid-points}$$ $\\begin{array}{|l | l|} \\hline \\text{Steps} & \\text{Reasons} \\\\ \\hline WV = VU & \\text{definition of mid-point} \\\\ T\\hat{V}U = S\\hat{V}W & \\text{vert opp } \\angle \\text{s} = \\\\ TV + VS = XW & \\text{opp sides parm are equal} \\\\ b + VS = 2b & \\text{substitute given values: } TV = b \\text{ and } XW = 2 b \\\\ VS = b = VT & \\text{solve for } VS; \\text{ note that it is equal to } VT \\\\ \\triangle TVU \\equiv \\triangle SVW & \\text{SAS} \\\\ \\hline \\end{array}$", "\\hat{ i } + \\hat{ j } - 2 \\hat{ k } \\text{ and } \\hat{ i } - 3 \\hat{ j } + 4 \\hat{ k }$ . Ex. 25.1 | Q 8.4 | Page 29 Find the area of the parallelogram determined by the vector $\\hat{ i } - 3 \\hat{ j } + \\hat{ k } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k } .$ Ex. 25.1 | Q 9.1 | Page 30 Find the area of the parallelogram whose diagonals are $4 \\hat{ i } - \\hat{ j } - 3 \\hat{ k } \\text{ and } - 2 \\hat{ j } + \\hat{ j } - 2 \\hat{ k }$ Ex. 25.1 | Q 9.2 | Page 30 Find the area of the parallelogram whose diagonals are $2 \\hat{ i }+ \\hat{ k } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Ex. 25.1 | Q 9.3 | Page 30 Find the area of the parallelogram whose diagonals are $3 \\hat{ i } + 4 \\hat{ j } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Ex. 25.1 | Q 9.4 | Page 30 Find the area of the parallelogram whose diagonals are $2 \\hat{", "diagonal, length 3, of the enclosed square. Now that we have the position vectors of vertices A and B we could find the length of the line AB using Pythagoras. However, you might notice that the area of the square will be given by $AB^2$ and we can write this down directly, $AB^2 = 2 \\times \\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ You could also have noticed that the right-angled triangle joining points A and B is isosceles. This tells us that the square could be divided up into four of these triangles. We can now show that the area of the square is again equal to $4 \\times \\frac{1}{2}\\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ What can we now say about the positions of the other vertices and the diagonals of the square? Find a vector equation for each of the two diagonals. In order to write down a vector equation of a line we need to know the position of a point on that line and a direction vector. For the vertical diagonal of the square we know that point B, with position vector $\\begin{pmatrix}2\\\\4\\end{pmatrix}$ lies on the line and that the direction vector is $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. A vector equation of the vertical diagonal of the square is therefore $\\mathbf{r}=\\begin{pmatrix}2\\\\4\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix}.$ We could also have written $\\mathbf{r}=\\begin{pmatrix}2\\\\1\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix},$ or any vector equation", "one) intersects the other vector. Then find the magnitude of a vector that would be the height of the parallelogram. To determine the area of our parallelogram, we need the length of its base #|AB|# and the altitude #|DH|# from vertex #D# to point #H# on side #AB# (that is, #DH_|_AB#). #S = |AB|*|DH|=|U_B*V_D-V_B*U_D|#, In terms of original coordinates, it looks like this: Protection against an aboleths enslave ability. #C[U_C=x_C-x_A,V_C=y_C-y_A]# #|AD|=sqrt(U_D^2+V_D^2)# A rectangle is 12cm longer than its wide. {\\displaystyle 15in^ {2}}, since. We note that the area of a triangle defined by two vectors $\\vec{u}, \\vec{v} \\in \\mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. 1. The parallelograms are differentiated by the size of the adjacent sides as well as by the angles, however, the opposite sides are equal. #sin^2(/_BAD)=1-cos^2(/_BAD)=# The length #AD# is the length of vector #q#: Is there a bias against mentioning your name on presentation slides? So in your case we have to write the points in $\\mathbb{R}^2$ as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ Thus we have the same" ]

[ "$CDP$. Therefore the area of $CDP$ is $\\frac{1}{2} \\cdot 13 \\cdot 84=\\boxed{546}$.", "For the parallelogram OPQR shown in the sketch, $\\overrightarrow{OP}=a \\hat{i}+b \\hat{j}$ and $\\overrightarrow{OR}=c \\hat{i} +d \\hat{j}$. The area of the parallelogram is 1. $ad-bc$ 2. $ac+bd$ 3. $ad+bc$ 4. $ab-cd$", "\\vec c|$ is the area of the parallelogram. Formula: . $\\cos60^o \\:=\\:\\frac{\\vec d \\cdot \\vec c}{|\\vec d||\\vec c|}$ So we have: . $\\frac{1}{2} \\:=\\:\\frac{50}{|\\vec d||\\vec c|} \\quad\\Rightarrow\\quad |\\vec d||\\vec c| \\:=\\:100$ The area of $\\Delta AC\\!D$ is given by: . $\\tfrac{1}{2}|\\vec d||\\vec c|\\sin60^o$ The area of parallelogram $AC\\!BD$ is: . $|\\vec d||\\vec c|\\sin60^o \\;=\\;(100)\\left(\\tfrac{\\sqrt{3}}{2}\\right) \\;=\\;50\\sqrt{3}$ Therefore: . $|\\vec d \\times \\vec c| \\;=\\;50\\sqrt{3}$", "? Free Version Difficult # Perimeter and Area of a Parallelogram GEOM-LQAEJL Parallelogram $EUCD$ has an area of $180\\sqrt { 3 }$ square inches. What is the perimeter of $EUCD$? A $12\\sqrt { 2 }$ inches B $15\\sqrt { 2 }$ inches C $54\\sqrt { 2 }$ inches D $18\\sqrt { 2 }+6\\sqrt { 6 }$ inches E $54\\sqrt { 2 }+6\\sqrt { 6 }$ inches", "# Area of parallelogram Algebra Level 3 Find the area of the parallelogram with vertices $A=(-2, 1,3), B=(0, 4,3), C=(4, 2,3), D=(2, -1,3).$ ×", "## Elementary Technical Mathematics $180 \\ cm^2$ The area of the rectangle is given as follows $$(6.5 \\ cm ) \\cdot (28.3 \\ cm ) =183.95 \\ cm^2\\approx180 \\ cm^2 .$$", "+0 # Please help! +1 105 3 +59 In the figure below, $ABCD$ is a parallelogram, $AD=18$ and $DC=27$. Segments $\\overline{EG}$ and $\\overline{FH}$ are perpendicular to sides of the parallelogram as shown, and $EG=16$. What is the area of the parallelogram? Apr 11, 2021 ### 3+0 Answers #1 -1 The area of the parallelogram is 540. Apr 11, 2021 #2 +1 Area = DC * EG Guest Apr 11, 2021 #3 +121004 +1 Note that the area = DC * EG = 27 * 16 = 432 Apr 11, 2021", "the area of the parallelogram is $11\\sqrt{5}$ sq. units. edited May 24, 2013", "the parallelogram is equal to $17.5$.", "# Area Search Geometry Level pending Point $$E$$ lies on segment $$AB$$, and point $$C$$ lies on segment $$FG$$. The area of parallelogram $$ABCD$$ is 20 square units. What’s the area of parallelogram $$EFGD$$? ×", "arcs. Each of the rectangles has area $5\\cdot 1=5$. The four arcs together form a circle with radius $1$. Therefore the total area she can see is $16 + 4\\cdot 5 + \\pi\\cdot 1^2 = 36+\\pi \\simeq 39.14$, which rounded to the nearest integer is $39$. $\\boxed{C}$", "The separation is $4 - \\left(- 2\\right) = 6$ linear units. So the area of the polygon ABCD, a parallelogram, is ${S}_{A B C D} = b a s e \\cdot h e i g h t = 5 \\cdot 6 = 30$", "vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$. The area of the triangle is $\\frac{1}{2}\\left| {\\left| {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right|} \\right| = \\frac{1}{2}\\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right|$. Example 2: Calculate the area of the triangle with vertices $(-1, 4)$, $(3,1)$ and $(2, 6)$. Solution: The area is $\\frac{1}{2}\\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} { - 1}&4&1\\\\ 3&1&1\\\\ 2&6&1 \\end{array}} \\right]} \\right)} \\right| = \\frac{1}{2} \\cdot \\left| {17} \\right| = \\frac{1}{2} \\cdot 17 = 8.5$ . Parallelogram: Suppose we have a parallelogram with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ and $(x_4, y_4)$, Then, the area of the parallelogram is $2 \\cdot \\frac{1}{2} \\cdot \\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right| = \\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right|$", "# Math Help - area of parallelogram 1. ## area of parallelogram I got everything except for the area can you please help me find the area of the parallelogram? 2. Originally Posted by jwein4492 I got everything except for the area can you please help me find the area of the parallelogram? By pathagoras you ahve $2x^2=8\\sqrt{2}$ solving we get $x=2^{\\frac{5}{4}}$ so now looking at the sides the area is $A_{rectangle}=2^{\\frac{5}{4}}\\cdot(11+2^{\\frac{5}{ 4}})$", "\\cdot 1.0353\\right) = 8$ Total Surface Area = Lateral surface area + Base area. T S A 77.7944 + 8 = 85.7944#", "Find the area of the parallelogram with vertices at $(5, -3),$ $(-2, -1),$ $(-3, 4),$ and $(-10, 6).$ Area = .", "# Two opposite sides of a parallelogram each have a length of 9 . If one corner of the parallelogram has an angle of (2 pi)/3 and the parallelogram's area is 18 , how long are the other two sides? Jul 8, 2017 $\\text{Missing side} = 2.31$ #### Explanation: ${\\left(\\frac{2 \\pi}{3}\\right)}^{\\text{c}} = {60}^{\\circ}$ If the whole area is made of two equal triangles, and has an area of 18 in total, then the area of one triangle is $\\frac{18}{2} = 9$ Area of a triangle $= \\left(\\frac{a \\cdot b \\cdot \\sin \\left(C\\right)}{2}\\right)$, where $a$ and $b$ are two sides, and $C$ is the angle directly opposite to the unknown side. In this case, $\\text{Area of triangle} = 9 = \\frac{a \\cdot 9 \\cdot \\sin \\left(60\\right)}{2}$, then $a = \\frac{18}{9 \\sin \\left(60\\right)} = \\frac{2}{\\sin} \\left(60\\right) = 2.31 \\left(\\text{rounded to 2 d.p.}\\right)$", "# Deploying laser beams Geometry Level 3 Given a parallelogram $ABCD$ with area of $24\\text{ cm}^2$ with points $E$ and $F$ as the midpoint of the lines $AB$ and $BC$ respectively. What is the area of the quadrilateral $EFGH$? ×", "### Area of parallelogram $ABCD$ is $x \\space cm^2.$ If $E, F, G,$ and $H$ are mid-points of the sides, find the area of $EFGH.$ $\\dfrac{ x }{ 2 } cm^2$ Step by Step Explanation: 1. It is given that$, E, F, G$ and $H$ are respectively the mid-points of the sides of the parallelogram $ABCD.$ 2. Let's join the midpoints $E, F, G$ and $H$ and join $HF, EG.$ 3. Line $HF$ drawn from the midpoints of the parallelogram$ABCD$ devides the parallelogram into two equal parts. $$\\therefore \\text{ Area of the parallelogram } HFCD = \\dfrac { \\text{ Area of the parallelogram } ABCD } { 2 } = \\dfrac { x } { 2 }$$ 4. Lines $HF$ and $EG$ drawn from the midpoints of the parallelogram $ABCD$ bisect each other other at the $O.$ $$\\therefore \\text{ Area of the parallelogram } HOGD = \\dfrac{ \\text { Area of the parallelogram } HFCD } { 2 } = \\dfrac { x }{ 4 }$$ 5. Diagonal $GH$ of the parallelogram $HOGD$ devides the parallelogram into two equal parts. $$\\therefore \\text{ Area of HOG } = \\dfrac { \\text{ Area of the parallelogram HOGD } } { 2 } = \\dfrac { x }{ 8 }$$ 6. Area of $FOG =$ Area of $FOE =$ Area of", "area of $\\dfrac{1}{2} \\cdot 2 + 1 \\cdot 2 + 4 \\cdot 2 = 11 \\text{ ft}^2.$ Thus, E is the correct answer." ]

[ "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", "of the parallelogram with edges ${v=(3,2)}$ and ${w=(1,4)}$. • Find the area of the triangle with sides ${v,w,}$and ${v+w}$.Draw it. • Find the area of the triangle with sides ${v,w,}$and ${w-v}$.Draw it. Solution: • $\\displaystyle \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=10.$ • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture omitted. • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture Omitted. $\\Box$ Exercise 4.4.28 A box has edges from ${(0,0,0)}$ to ${(3,1,1),(1,3,1),}$and ${(1,1,3)}$.Find its volumn and also find the area of each parallelogram face. Solution: The volumn of the box is $\\displaystyle \\begin{vmatrix} 3&1&1\\\\ 1&3&1\\\\ 1&1&3 \\end{vmatrix}=20.$ The area of each parallelogram face is $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 3&1 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 1&3\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2}.$ In order to compute the area of each parallelogram face,we can also compute $\\displaystyle \\sqrt{\\det A^TA}=\\sqrt{\\det \\left(\\begin{bmatrix} 3&1&1\\\\ 1&3&1 \\end{bmatrix} \\begin{bmatrix} 3&1\\\\ 1&3\\\\ 1&1 \\end{bmatrix}\\right)}= 6 \\sqrt{2}$ etc.$\\Box$ Exercise 4.4.35 An ${n}$– dimensional cube has how many corners?How many edges?How many ${n-1}$-dimensional faces?The ${n}$– cube whose edges are the rows of ${2I}$ has volumn ${\\underline{~~~~~~~~~~}}$.A hypercube computer has parallel processors at the corners with connections along the edeges. Solution: An ${n}$-dimensional cube", "# Math Help - Parallelogram... 1. ## Parallelogram... Let f(t) be the area of the parallelogram with vertices v=(1,2t), w = (2t,1), (0,0), and v+w = (2t+1,2t+1). Find a formula for f(t). I'm assuming one can't simply 'stand up' the parallelogram so that it's a rectangle... 2. Let $O(0,0), \\ A(1,2t), \\ B(2t,1), C(1+2t, 1+2t)$ $Area[OACB]=2\\cdot Area[OAB]=2\\cdot\\frac{1}{2}|\\Delta|$ where $\\Delta=\\begin{vmatrix}0 & 0 & 1\\\\1 & 2t & 1\\\\2t & 1 & 1\\end{vmatrix}=1-4t^2$ Therefore $f(t)=|1-4t^2|$ 3. Another way to do it is using the cross product of v and w. That also works with 3D coordinates. 4. Hello, PTL! Yet another approach . . . Did you make a sketch? Let $f(t)$ be the area of the parallelogram with vertices: $\\vec v\\,=\\,(1,2t),\\;\\vec w\\,=\\,(2t,1),\\;(0,0)$, and $\\overrightarrow{v+w} \\:=\\:(2t+1,2t+1).$ Find a formula for $f(t).$ Code: | 2t+1 + - - - - - - - - - - o v+w | * *: | * * : | v * * : 2t + - - o * : | * * : 1+ * o w : | * * : : | * * : : |* * : : - - o - - + - - - - + - - * - - (0,0) 1 2t 2t+1 We have a rhombus. If $d_1,\\:d_2$ are diagonals of a rhombus,", "vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$. The area of the triangle is $\\frac{1}{2}\\left| {\\left| {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right|} \\right| = \\frac{1}{2}\\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right|$. Example 2: Calculate the area of the triangle with vertices $(-1, 4)$, $(3,1)$ and $(2, 6)$. Solution: The area is $\\frac{1}{2}\\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} { - 1}&4&1\\\\ 3&1&1\\\\ 2&6&1 \\end{array}} \\right]} \\right)} \\right| = \\frac{1}{2} \\cdot \\left| {17} \\right| = \\frac{1}{2} \\cdot 17 = 8.5$ . Parallelogram: Suppose we have a parallelogram with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ and $(x_4, y_4)$, Then, the area of the parallelogram is $2 \\cdot \\frac{1}{2} \\cdot \\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right| = \\left| {\\det \\left( {\\left[ {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\\\ {{x_2}}&{{y_2}}&1\\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right]} \\right)} \\right|$", "find the area of the parallelogram whose vertices are listed # find the area of the parallelogram whose vertices are listed H 240 points In Exercises 19–22, find the area of the parallelogram whose vertices are listed. (0,-2),(5,-2),(-3,1),(2,1) find the hansel 8.7k points Translate the coordinates so that one of them is at the origin, doing so does not change the area of the parallelogram, but allows us to find the area. In this case I will add the vector (0,2) to each vertex which gives, (0,0),(5,0),(-3,3),(2,3) let Now the area of the parallelogram is the determinant of the matrix formed by the column vectors (5,0) and (-3,3), $$\\begin{vmatrix} 5 & -3 \\\\ 0 & 3 \\end{vmatrix}=15$$ Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.", ", 6 )$ $u = AB = \\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 6 \\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1 } \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 5 & 6 \\\\ 2 & 4 \\end{bmatrix}= 20 \\: – \\: 12 = 8$ The area of the parallelogram is $8$. ## Numerical Result The area of the parallelogram is $8$. ## Example Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$ Solution A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram. $A = ( 0 , 0 )$ $B = ( 6 , 8 )$ $C = ( 5 , 4 )$ $D = ( 11 , 6 )$ $u = AB = \\begin{bmatrix} 6\\\\ 8 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 5\\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1", "# Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). I can't figure out what I'm doing wrong.. I let: $a = |DC| = (-2,-1) - (-6,3) = (4,-4)$ $b = |AD| = (-1,1) - (-5,5) = (4,-4)$ so: $\\begin{bmatrix} 4&-4\\\\4&-4\\end{bmatrix} \\implies$ Area $= -16 + 16 = 0$ ? The determinant is the area of the parallelogram is it not? • Not sure how you're labelling vertices, but $a$ appears to be $|CD|$, while $b$ appears to be $|AB|$. You get a result of zero because those lines are parallel. – Joey Zou Jul 6 '16 at 22:08 • You are correct that was the problem. The answer is $|-12|$ – Yusha Jul 6 '16 at 22:11", "# Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cristiano Sears ###### Not exactly what you’re looking for? content_user Vertices are $$A(-3,0),\\ B(-1,6),\\ C(8,5)$$ and $$D(6,-1)$$ So $$\\vec{AB}=<-1-(-3),6-0>=<2,6>$$ $$\\vec{AD}=<6-(-3),-1-0>=<9,-1>$$ Hence area of parallelogram $$=|\\vec{AB}\\times\\vec{AD}|$$ $$\\vec{AB}\\times\\vec{AD}=\\begin{bmatrix}i&j&k\\\\2&6&0\\\\9&-1&0\\end{bmatrix}$$ $$=i(0-0)+j(0-0)+k(-2-54)$$ $$=-56k$$ Hence Area $$=|\\vec{AB}\\times\\vec{AD}|$$ $$=\\sqrt{(-56)^2}=56$$", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "C(9,6)$ and $D(13,8)$. We let $x_1=13,y_1=8,x_2=9,y_2=6,x_3=4,y_3=2,x_4=8,y_4=4$ and the area is given by $$A=\\frac 12|13\\cdot 6+9\\cdot 2+4\\cdot 4+8\\cdot 8-9\\cdot 8-4\\cdot 6-8\\cdot 2-13\\cdot 4|\\\\=\\frac 12\\cdot 12=6$$ • You answered with another person's answer? xD – Billy Rubina Oct 13 '16 at 8:15 • This is hitting a nail with a sledge hammer. There aren't many polygons that are as simple to handle as parallelograms. Or to be more specific, it is very probably the computation for parallelograms (and derived from that, triangles) that serves as basis for deriving the shoelace formula. – Marc van Leeuwen Oct 13 '16 at 9:14 The absolute value of the cross product of two vectors $\\vec{a}, \\vec{b} \\in \\mathbb{R}^3$ spanning the parallelogram is its area: $$A_\\text{parallelogram}= \\left|\\vec{a}\\times\\vec{b}\\right|$$ So in your case we have to write the points in $\\mathbb{R}^2$ as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$ $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$ $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$ You might have noticed that this simplifies to $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ $$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$ There are plenty of ways, such as the Shoelace Theorem and Pick's Theorem. If you have a graph, you can also simply draw a rectangle around the shape and subtract the parts you don't want. I think this is", "+0 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) +1 136 1 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) part (a): Let $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$find a*b and the answer was 10 Apr 2, 2022 #1 +30 +2 $\\sin^2 θ = 1 - \\cos^2 θ = 1 - \\frac{4}{13} = \\frac{9}{13}$ every angle in a parallelogram is between 0 and π radians, so θ must be positive, so $\\sin θ = \\sqrt{\\frac{9}{13}} = \\frac{3}{\\sqrt{13}}$ area of parallelogram is $\\|{a}\\| \\|{b}\\| \\sin \\theta = \\sqrt{65} \\cdot \\sqrt{5} \\cdot \\frac{3}{\\sqrt{13}} = \\boxed{15}$ Jun 25, 2022", "as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The formula for the area of a parallelogram is base x height. Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [closed], Soft question on the area of a parallelogram using vectors. Area of the Parallelogram: A parallelogram is a geometric structure having one side having opposite sides parallel to each other. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. #vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj# First of all, to simplify the task, let's move it to a position when its vertex A coincides with the origin of coordinates. So this's a B C on G E. Now the area of", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "Find the area of the parallelogram with vertices at $(5, -3),$ $(-2, -1),$ $(-3, 4),$ and $(-10, 6).$ Area = .", "the area of the paralellogram and the original square. Since the original square has area $1$, the area of the parallelogram is the determinant of $\\bc{\\A}$. Working this out requires only a small amount of basic geometry. Here’s the simplest way to do it. We first name the four elements of our matrix: $\\bc{\\A} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ We can now name the four corners of our parallelogram in terms of these four scalars, by multiplying the corner points of the unit square by $\\bc{\\A}$: $\\begin{matrix} \\times & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}1\\\\ 0\\end{pmatrix} & \\begin{pmatrix}0\\\\ 1\\end{pmatrix} & \\begin{pmatrix}1\\\\ 1\\end{pmatrix} \\\\ \\bc{\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}} & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}a\\\\ c\\end{pmatrix} & \\begin{pmatrix}b\\\\ d\\end{pmatrix} & \\begin{pmatrix}a+b\\\\ c+d\\end{pmatrix} \\\\ \\end{matrix}$ This gives us the following picture. Here we can see the area of the paralellogram clearly: there is one large rectangle of area $\\oc{ad}$. To get from this area to the area of our parallelogram, we should subtract the area of the green triangle in the bottom, which is part of the rectangle but not the parallelogram. But then, there’s a green triangle at the top, with the same size, which is (mostly) part of the parallelogram but not of the rectangle, so these cancel each other out. We", "is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[6] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V = \\begin{bmatrix} a_1 & a_2 & \\dots", "Example Find the area of an triangle with its vertices located at (-2,2), (1,3) and (3,0) (this example is also shown in our video lesson). We plug our coordinates for the vertices into our area formula $A=\\frac{1}{2}\\begin{vmatrix} -2 & 2 & 1\\\\ 1 & 3 & 1\\\\ 3 & 0 & 1 \\end{vmatrix}$ and continues with $=\\frac{1}{2}(-2\\begin{vmatrix} 3 & 1\\\\ 0 & 1 \\end{vmatrix}-2\\begin{vmatrix} 1 & 1\\\\ 3 & 1 \\end{vmatrix}+1\\begin{vmatrix} 1 & 3\\\\ 3 & 0 \\end{vmatrix})=\\\\ \\\\ =\\frac{1}{2}(-2(3\\cdot1-1\\cdot 0 )-2(1\\cdot 1-1\\cdot 3)+1(1\\cdot 0-3\\cdot 3))=\\\\ \\\\ =\\frac{1}{2}(-6+4-9)=\\frac{-11}{2}=-5.5$ We received a negative value for A and an area cannot be negative, therefore we must take the absolute value for A: $\\mid A\\mid =\\mid -5.5\\mid =5.5\\;$ So the triangles area is 5.5 square units. ## Video lesson Find the area from the example above.", "\\hat{ i } + \\hat{ j } - 2 \\hat{ k } \\text{ and } \\hat{ i } - 3 \\hat{ j } + 4 \\hat{ k }$ . Ex. 25.1 | Q 8.4 | Page 29 Find the area of the parallelogram determined by the vector $\\hat{ i } - 3 \\hat{ j } + \\hat{ k } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k } .$ Ex. 25.1 | Q 9.1 | Page 30 Find the area of the parallelogram whose diagonals are $4 \\hat{ i } - \\hat{ j } - 3 \\hat{ k } \\text{ and } - 2 \\hat{ j } + \\hat{ j } - 2 \\hat{ k }$ Ex. 25.1 | Q 9.2 | Page 30 Find the area of the parallelogram whose diagonals are $2 \\hat{ i }+ \\hat{ k } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Ex. 25.1 | Q 9.3 | Page 30 Find the area of the parallelogram whose diagonals are $3 \\hat{ i } + 4 \\hat{ j } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Ex. 25.1 | Q 9.4 | Page 30 Find the area of the parallelogram whose diagonals are $2 \\hat{", "2), Q (3, 3, 3d), R (7, 5, 8), and S (5, 2, 7). The minimum area parallelogram enclosing a convex polygon can be found in linear time, linear in the number of vertices (for your case, a small constant). ABC= bc- ac/2 -(b-a)(c-d)/2 -bd/2 and double it and you should get Change the name (also URL address, possibly the category) of the page. See the answer Give your answer to one decimal place. By using this website, you agree to our Cookie Policy. Favorite Answer. Solution for Find the area of the parallelogram with vertices A(−4, 5), B(−2, 8), C(2, 6), and D(0, 3). The area of this is equal to the absolute value of the determinant of A. Here’s our parallelogram. to #1: Calculate the midpoint of the diagonals (because you don't know which vertices are opposite you have to check in 2 … From the details to the question: > Given points P,Q,R w/position vectors p(1,4,1), q(3,1,2), r(3,8,7). Thus we can give the area of a triangle with the following formula: The Areas of Parallelograms and Triangles in 3-Space, \\begin{align} A = \\| \\vec{u} \\| \\| \\vec{v} \\| \\sin \\theta \\\\ \\blacksquare \\end{align}, \\begin{align} \\| \\vec{u} \\times \\vec{v} \\|^2 = \\|\\vec{u}\\|^2 \\|\\vec{v} \\|^2 - (\\vec{u} \\cdot \\vec{v})^2 \\end{align}, \\begin{align} \\| \\vec{u} \\times \\vec{v} \\|^2 =", "prove the following theorem. Area of a Parallelogram If we locate vectors $$\\vecs u$$ and $$\\vecs v$$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $$‖\\vecs u×\\vecs v‖$$ (Figure $$\\PageIndex{5}$$). Proof We show that the magnitude of the cross product is equal to the base times height of the parallelogram. \\begin{align*} \\text{Area of a parallelogram} &= \\text{base} × \\text{height} \\\\[4pt] &=‖\\vecs u‖(‖\\vecs v‖\\sin θ) \\\\[4pt] &=‖\\vecs u×\\vecs v‖ \\end{align*} Example $$\\PageIndex{9}$$: Finding the Area of a Triangle Let $$P=(1,0,0),Q=(0,1,0),$$ and $$R=(0,0,1)$$ be the vertices of a triangle (Figure $$\\PageIndex{6}$$). Find its area. Solution We have $$\\vecd{PQ}=⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩$$ and $$\\vecd{PR}=⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩$$. The area of the parallelogram with adjacent sides $$\\vecd{PQ}$$ and $$\\vecd{PR}$$ is given by $$∥\\vecd{PQ}×\\vecd{PR}∥$$: \\begin{align*} \\vecd{PQ} \\times \\vecd{PR} &= \\begin{vmatrix}\\mathbf{\\hat i} & \\mathbf{\\hat j} & \\mathbf{\\hat k}\\\\−1 & 1 & 0\\\\−1 & 0 & 1\\end{vmatrix} \\\\[4pt] &=(1−0)\\mathbf{\\hat i}−(−1−0)\\mathbf{\\hat j}+(0−(−1))\\mathbf{\\hat k} \\\\[4pt] &=\\mathbf{\\hat i}+\\mathbf{\\hat j}+\\mathbf{\\hat k} \\\\[10pt] ∥\\vecd{PQ}×\\vecd{PR}∥ &=∥⟨1,1,1⟩∥ \\\\[4pt] &=\\sqrt{1^2+1^2+1^2} \\\\[4pt] &=\\sqrt{3}. \\end{align*} The area of $$ΔPQR$$ is half the area of the parallelogram or $$\\sqrt{3}/2 \\, \\text{units}^2$$. Exercise $$\\PageIndex{9}$$ Find the area of the parallelogram $$PQRS$$ with vertices $$P(1,1,0)$$, $$Q(7,1,0)$$, $$R(9,4,2)$$, and $$S(3,4,2)$$. Hint Sketch the parallelogram and identify two vectors that form adjacent sides of the parallelogram. $$6\\sqrt{13}\\, \\text{units}^2$$ ### The Triple Scalar Product Because the cross" ]

[ "Browse Questions # If the point represented by the complex number $iz$ is rotated about the origin through the angle $\\large\\frac{\\pi}{2}$ in the counter clockwise direction then the complex number representing the new position is $\\begin{array}{1 1}(1)iz&(2)-iz\\\\(3)-z&(4)z\\end{array}$ The complex number iz is rotated about the origin through a angle $\\large\\frac{\\pi}{2}$ i the counter clock wise direction then the complex number representing the new position is $i(iz)=-z$ Hence 3 is the correct answer.", "### Rotation of 180 ° about the origin In the diagram above, triangle $$ABC$$ is rotated 180 ° about the origin: $$A(1, 2) \\rightarrow A^\\prime (−1, −2); B(4, 3) \\rightarrow B^\\prime (−4, −3); C(5, 1) \\rightarrow C^\\prime (−5, −1)$$. In general, $$P(x,y)\\rightarrow P^\\prime (−x, −y)$$. ### Combinations of rotations of 90 ° It is easy to see that two separate rotations of 90 ° in a clockwise direction is the same as rotating by 180 °.", "about the origin? 1. This is a $135^\\circ$ clockwise rotation about point $B$ . You could also say this is a $225^\\circ$ counterclockwise rotation about point $B$ . 2. You can see that there is a straight line $(180^\\circ)$ passing through the origin connecting each pair of corresponding points. $(x, y) \\rightarrow (-x, -y)$ . 3. There would be no difference. A rotation $180^\\circ$ clockwise is the same as a rotation $180^\\circ$ counterclockwise. Practice Describe each of the following rotations in TWO ways. 1. 2. 3. 4. What does a $360^\\circ$ rotation look like? 5. Rotate the rectangle below $90^\\circ$ clockwise about the origin. 6. Use the previous example to help you describe what happens to a point $(x, y)$ that is rotated $90^\\circ$ clockwise about the origin. 7. (3, 2) is rotated $90^\\circ$ clockwise about the origin. Where does it end up? 8. (3, 2) is rotated $180^\\circ$ clockwise about the origin. Where does it end up? Hint: Use Guided Practice #2 for help. 9. (3, 2) is rotated $90^\\circ$ counterclockwise about the origin. Where does it end up? Hint: Use Example C. 10. What do circles have to do with rotations? 11. $\\Delta ABC$ is rotated $130^\\circ$ clockwise about point $O$ to create $\\Delta A^\\prime B^\\prime C^\\prime$ . Describe two connections between the points $B, O$", "is just a $90\\deg$ rotation. Essentially, we will be dealing with complex differential forms! Also, many interesting things involving linear algebra deal with our field being over the complex numbers.", "it has been rotated $90$ degrees clockwise about $(1,1)$?", "a negative value? A rotation of more than $360^\\circ$360° means we have completed more than a full lap around the unit circle. But when we're finished rotating we will land on a point that we had already passed over during the first rotation. A negative rotation happens in the opposite direction, clockwise rather than anti-clockwise, stopping in a position that matches some equivalent positive rotation between $0^\\circ$0° and $360^\\circ$360°. We can rotate as much as we like in either direction, with each ratio changing in a predictable and repeating way - we use the word periodic to capture this property - and land on some point. The smallest amount of positive rotation to reach that point will produce the same values for the trigonometric ratios, and we can then use everything we learned before to find what we're looking for. In practice we add or subtract multiples of $360^\\circ$360° to obtain an angle between $0^\\circ$0° and $360^\\circ$360°. After determining what quadrant the angle lies in, we proceed as before. #### Worked examples ##### Example 1 Express $\\cos117^\\circ$cos117° in terms of a first quadrant angle. Think: The angle $117^\\circ$117° is between $90^\\circ$90° and $180^\\circ$180°, so it is in the second quadrant. The point representing $117^\\circ$117° on the unit circle diagram, where the radius cuts the circle, must have a", "(-4, 3) 2. (5, -4) 3. (-5, -4) 4. (3, 3) 5. (-8, -9) Describe the following rotations: 1. Why is it not necessary to specify the direction when rotating $180^\\circ$ ?", "3. A rotation of $160^\\circ$ counterclockwise is the same as what clockwise rotation? 1. Because a rotation is an isometry that produces congruent figures, we can set up two equations to solve for $x$ and $y$ . $2y &= 80^\\circ && \\ 2x-3 =15\\\\y &= 40^\\circ && \\qquad 2x = 18\\\\&&& \\qquad \\ \\ x = 9$ 2. There are $360^\\circ$ around a point. So, an $80^\\circ$ rotation clockwise is the same as a $360^\\circ-80^\\circ=280^\\circ$ rotation counterclockwise. 3. $360^\\circ-160^\\circ=200^\\circ$ clockwise rotation. ### Practice In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter $p \\ 180^\\circ$ counterclockwise, what letter would you have? 2. If you rotated the letter $p \\ 180^\\circ$ clockwise, what letter would you have? Why do you think that is? 3. A $90^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 4. A $270^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 5. Rotating a figure $360^\\circ$ is the same as what other rotation? Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin. 1. $180^\\circ$ 2. $90^\\circ$ 3. $180^\\circ$ 4. $270^\\circ$ 5. $90^\\circ$ 6. $270^\\circ$ 7. $180^\\circ$ 8. $270^\\circ$ 9. $90^\\circ$ Algebra Connection Find the measure of $x$ in the rotations below. The blue figure is the", "the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. Vocabulary Language: English Rotation Rotation A rotation is", "B 270 degrees counterclockwise about the origin. 2. Rotate the triangle ABC 60 degrees clockwise about the origin. 3. Rotate the triangle $130$ degrees clockwise about the circumcenter. ### Open Problem Solutions Images/mathematical drawings are created with GeoGebra.", "of the class to be the origin. 2. Standing at the origin, I’ll face the positive real axis, spin in an angle of $5\\pi/6 = 150^\\circ$, and take two steps to arrive at the point $-\\sqrt{3} + i$. 3. Returning to the origin, I’ll face the positive real axis, spin the other direction in an angle of $-210^\\circ$, and take two steps to arrive at the same point. 4. Returning to the origin, I’ll face the positive real axis, spin in an angle of $510^\\circ$ (getting more than a little dizzy while doing so), and take two steps to arrive at the same point. 5. Returning to the origin, I’ll face the positive real axis, spin in an angle of only $-30^\\circ$, and take two steps backwards (while doing the moonwalk) to arrive at the same point. We will need to use this concept of writing a complex number in trigonometric form in order to explain the calculator’s results. For completeness, here’s the movie that I used to begin this series of posts. # Calculators and complex numbers (Part 1) What is $\\ln(-5)$? Or $(-8)^{1/3}$? Easy, right? Well, let’s plug into a calculator and find out. (Click anywhere in the image below to start the movie. The important stuff is the screen at the top; you can see", "rotated about the point $A \\ 225^\\circ$ . Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon. Answers: 1. Notice the direction of the angle is clockwise, therefore the angle measure is $60^\\circ$ CW or $-60^\\circ$ . 2. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $155^\\circ$ CCW or $155^\\circ$ . 3. The measure of $\\angle BAB^\\prime = m \\angle BAE^\\prime + m \\angle E^\\prime AB^\\prime$ . Therefore $\\angle BAB^\\prime = 111.80^\\circ + 113.20^\\circ$ or $225^\\circ$ . Notice the direction of the angle is counter-clockwise, therefore the angle measure is $225^\\circ$ CCW or $225^\\circ$ . ### Explore More 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3.", "the points at $(1,2)$ and $(-1,0)$, which are indeed related by a rotation through $180^\\circ$ about $(0,1)$, note a) that these don't correspond to each other in the shapes and b) that no other pairs of points are related by this rotation.", "3 0.00000000000000000000 0.00000000000000000000 Since $(a, b)$ is the origin (the center of rotation), a rotation does not change its coordinates. Sample Input 4 15 5 360 Sample Output 4 15.00000000000000177636 4.99999999999999555911 A $360$-degree rotation does not change the coordinates of a point. Sample Input 5 -505 191 278 Sample Output 5 118.85878514480690171240 526.66743699786547949770", "\\alpha_2=180,\\quad \\alpha_3=330.$ How does this scheme draw a spiral arm? Let’s look at the figure below. We begin at the origin, and then draw a segment at $0^\\circ,$ labeled “1” in the figure. Next, we turn counterclockwise $30^\\circ$ (since $\\alpha_1=30$) and draw the segment labeled “2”. The next turn is $180^\\circ,$ so we turn completely around and begin retracing the arm in the opposite direction. Then turn counterclockwise $330^\\circ$ and move forward. Note that since $330=360-30,$ this has the effect of taking us right back to the origin. What next? It is at this point that we invoke a recursive call — and so the next angle tells us what direction the next arm will be drawn in. Of course, like before, the three angles after that will continue drawing the arm and then bring us back to the origin, awaiting the next recursive call. This behavior can be illustrated in the following chart. Here, the angles turned are listed in rows of four, one after another. The first three angles in each row are the same, and their job is to complete the arm once a direction is chosen. But the direction chosen is determined by the fourth column (separated by the divider), whose behavior is not periodic and highly recursive. So where do we go from", "Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. ### Vocabulary Language: English Rotation Rotation A rotation is a transformation that turns a figure on the coordinate plane a certain number of degrees about a given point without changing the shape or size of the figure. Please wait... Please wait...", "$1$ rotated through $45$ degrees, that is, $\\cos(\\pi/4)+i\\sin(\\pi/4)$. -", "simply be $0^\\circ$, $45^\\circ$, $90^\\circ$, $135^\\circ$, $180^\\circ$, and so on? - First thing to notice: your text speaks of $30^\\circ$ increments, but the pictures seem to be taken from angles differing by $45^\\circ$, since the third picture is already a straight side view of the car. That would indeed change in $45^\\circ$ increments (or $30^\\circ$ if you go by the text not the pictures), starting at $0^\\circ$ since it started out pointing north. The only thing that remains is getting the sign straight. Usually you have azimuth increasing from north to east, with $90^\\circ$ corresponding to due east, so if you move around your car in a clockwise fashion, your viewing azimuth would increase. Your sequence however is counter-clockwise, therefore you'd have azimuths $0^\\circ, -45^\\circ, -90^\\circ$ and so on, or equivalently $0^\\circ, 315^\\circ, 270^\\circ, \\dots$.", "# Going Around In A Discrete Harmonic Spiral Calculus Level 5 A man is standing at the origin. First, he walks $$1$$ unit in the positive $$x$$ direction and reaches $$(1,0)$$. Then, he turns $$60^o$$ counterclockwise and walks $$1/2$$ units. Again, he turns $$60^o$$ counterclockwise and walks $$1/3$$ units. He continues this endlessly. Find the distance between his initial and final position. ×", "$\\alpha_1=30,\\quad \\alpha_2=180,\\quad \\alpha_3=330.$ How does this scheme draw a spiral arm? Let’s look at the figure below. We begin at the origin, and then draw a segment at $0^\\circ,$ labeled “1” in the figure. Next, we turn counterclockwise $30^\\circ$ (since $\\alpha_1=30$) and draw the segment labeled “2”. The next turn is $180^\\circ,$ so we turn completely around and begin retracing the arm in the opposite direction. Then turn counterclockwise $330^\\circ$ and move forward. Note that since $330=360-30,$ this has the effect of taking us right back to the origin. What next? It is at this point that we invoke a recursive call — and so the next angle tells us what direction the next arm will be drawn in. Of course, like before, the three angles after that will continue drawing the arm and then bring us back to the origin, awaiting the next recursive call. This behavior can be illustrated in the following chart. Here, the angles turned are listed in rows of four, one after another. The first three angles in each row are the same, and their job is to complete the arm once a direction is chosen. But the direction chosen is determined by the fourth column (separated by the divider), whose behavior is not periodic and highly recursive. So where do we go" ]

[ "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)", "$270^\\circ$ (or $-90^\\circ$ ) $(x, y)$ $(y, -x)$ #### Example A Line $\\overline{AB}$ drawn from (-4, 2) to (3, 2) has been rotated about the origin at an angle of $90^\\circ$ CW. Draw the preimage and image and properly label each. Solution: #### Example B The diamond $ABCD$ is rotated $145^\\circ$ CCW about the origin to form the image $A^\\prime B^\\prime C^\\prime D^\\prime$ . On the diagram, draw and label the rotated image. Solution: Notice the direction is counter-clockwise. #### Example C The following figure is rotated about the origin $200^\\circ$ CW to make a rotated image. On the diagram, draw and label the image. Solution: Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is $160^\\circ$ . #### Concept Problem Revisited Quadrilateral $WXYZ$ has coordinates $W(-5, -5), X(-2, 0), Y(2, 3)$ and $Z(-1, 3)$ . Draw the quadrilateral on the Cartesian plane. Rotate the image $110^\\circ$ counterclockwise about the point $X$ . Show the resulting image. ### Guided Practice 1. Line $\\overline{ST}$ drawn from (-3, 4) to (-3, 8) has been rotated $60^\\circ$ CW about the point $S$ . Draw the preimage and image and properly label each. 2. The polygon below has been rotated $155^\\circ$ CCW about the origin. Draw the rotated image and properly label each. 3. The purple pentagon is", "3. A rotation of $160^\\circ$ counterclockwise is the same as what clockwise rotation? 1. Because a rotation is an isometry that produces congruent figures, we can set up two equations to solve for $x$ and $y$ . $2y &= 80^\\circ && \\ 2x-3 =15\\\\y &= 40^\\circ && \\qquad 2x = 18\\\\&&& \\qquad \\ \\ x = 9$ 2. There are $360^\\circ$ around a point. So, an $80^\\circ$ rotation clockwise is the same as a $360^\\circ-80^\\circ=280^\\circ$ rotation counterclockwise. 3. $360^\\circ-160^\\circ=200^\\circ$ clockwise rotation. ### Practice In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter $p \\ 180^\\circ$ counterclockwise, what letter would you have? 2. If you rotated the letter $p \\ 180^\\circ$ clockwise, what letter would you have? Why do you think that is? 3. A $90^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 4. A $270^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 5. Rotating a figure $360^\\circ$ is the same as what other rotation? Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin. 1. $180^\\circ$ 2. $90^\\circ$ 3. $180^\\circ$ 4. $270^\\circ$ 5. $90^\\circ$ 6. $270^\\circ$ 7. $180^\\circ$ 8. $270^\\circ$ 9. $90^\\circ$ Algebra Connection Find the measure of $x$ in the rotations below. The blue figure is the", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "0) $270^\\circ$ (or $-90^\\circ$ ) $(x, y)$ $(y, -x)$ Example A Line $\\overline{AB}$ drawn from (-4, 2) to (3, 2) has been rotated about the origin at an angle of $90^\\circ$ CW. Draw the preimage and image and properly label each. Solution: Example B The diamond $ABCD$ is rotated $145^\\circ$ CCW about the origin to form the image $A^\\prime B^\\prime C^\\prime D^\\prime$ . On the diagram, draw and label the rotated image. Solution: Notice the direction is counter-clockwise. Example C The following figure is rotated about the origin $200^\\circ$ CW to make a rotated image. On the diagram, draw and label the image. Solution: Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is $160^\\circ$ . Concept Problem Revisited Quadrilateral $WXYZ$ has coordinates $W(-5, -5), X(-2, 0), Y(2, 3)$ and $Z(-1, 3)$ . Draw the quadrilateral on the Cartesian plane. Rotate the image $110^\\circ$ counterclockwise about the point $X$ . Show the resulting image. Vocabulary Center of rotation A center of rotation is the fixed point that a figure rotates about when undergoing a rotation. Rotation A rotation is a transformation that rotates (turns) an image a certain amount about a certain point. Image In a transformation, the final figure is called the image . Preimage In a transformation, the original figure is called", "$5$ left to insert, we place them such that we don't have the two pairs adjacent. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"1\", A, N); label(\"3\", B, NW); label(\"2\", C, SW); label(\"4\", D, SE); label(\"5\", E, NE); [/asy]$ This is the only solution to make $7$ \"bad.\" We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\\boxed{\\textbf {(B) }2}$.", "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", "# Example Construct a \"multiplication table\" for only the rotational symmetries of a pentagon. The transformations we use must be limited to rotations where our pentagon ultimately ends up in the same area it originally occupied after the rotation. Given that a pentagon has 5 sides, the smallest such rotation (besides rotating by $0^{\\circ}$) is a rotation of $360^{\\circ} / 5=72^{\\circ}$. One could rotate the pentagon by any multiple of $72^{\\circ}$ and occupy its original area, but notice $5 \\cdot 72^{\\circ} = 360^{\\circ}$ is equivalent to no rotation at all. As such, we really only have 5 distinct transformations: rotating by $0^{\\circ}$ (doing nothing), $72^{\\circ}$, $144^{\\circ}$, $216^{\\circ}$, or $288^{\\circ}$. Calling these rotations $R_{0}, R_{72}, R_{144}, R_{216}$, and $R_{288},$ we can flesh out the table for their combinations: $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{72}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{144}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{216}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{288}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$", "4. Recalling that $i^2=-1$, this is quickly done: . $1$ $i$ $-i$ $-1$ $1$ $1$ $i$ $-i$ $-1$ $i$ $i$ $-1$ $1$ $-i$ $-i$ $-i$ $1$ $-1$ $i$ $-1$ $-1$ $-i$ $i$ $1$ 5. The transformations we use must be limited to rotations where our pentagon ultimately ends up in the same area it originally occupied after the rotation. Given that a pentagon has 5 sides, the smallest such rotation (besides rotating by $0^{\\circ}$) is a rotation of $360^{\\circ} / 5=72^{\\circ}$. One could rotate the pentagon by any multiple of $72^{\\circ}$ and occupy its original area, but notice $5 \\cdot 72^{\\circ} = 360^{\\circ}$ is equivalent to no rotation at all. As such, we really only have 5 distinct transformations: rotating by $0^{\\circ}$ (doing nothing), $72^{\\circ}$, $144^{\\circ}$, $216^{\\circ}$, or $288^{\\circ}$. Calling these rotations $R_{0}, R_{72}, R_{144}, R_{216}$, and $R_{288},$ we can flesh out the table for their combinations: $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{72}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{144}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{216}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{288}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ 6. In both cases, we simply do the normal operation (addition or multiplication) and then find the remainder upon division by 5 of the result. This remainder is our answer. So for example: $$4 \\cdot 3", "Common Rotations • Rotation of $180^\\circ$ : If $(x, y)$ is rotated $180^\\circ$ around the origin, then the image will be $(-x, -y)$ . • Rotation of $90^\\circ$ : If $(x, y)$ is rotated $90^\\circ$ around the origin, then the image will be $(-y, x)$ . • Rotation of $270^\\circ$ : If $(x, y)$ is rotated $270^\\circ$ around the origin, then the image will be $(y, -x)$ . While we can rotate any image any amount of degrees, only $90^\\circ, 180^\\circ$ and $270^\\circ$ have special rules. To rotate a figure by an angle measure other than these three, you must use the process from the Investigation. #### Example A Rotate $\\triangle ABC$ , with vertices $A(7, 4), B(6, 1),$ and $C(3, 1) \\ 180^\\circ$ . Find the coordinates of $\\triangle A'B'C'$ . It is very helpful to graph the triangle. If $A$ is $(7, 4)$ , that means it is 7 units to the right of the origin and 4 units up. $A'$ would then be 7 units to the left of the origin and 4 units down. The vertices are: $A(7,4) \\rightarrow A'(-7,-4)\\\\B(6,1) \\rightarrow B'(-6,-1)\\\\C(3,1) \\rightarrow C'(-3,-1)$ #### Example B Rotate $\\overline{ST} \\ 90^\\circ$ counter-clockwise about the origin. Using the $90^\\circ$ rotation rule, $T'$ is (8, 2). #### Example C Find the coordinates of $ABCD$ after a", "- y)$ 7. Rotation of $180^\\circ$ - G. $(x, y)$ 8. Rotation of $90^\\circ$ - H. $(-x, -y)$ 9. Rotation of $270^\\circ$ - I. $(y, -x)$ 10. Rotation of $360^\\circ$ - J. $(y, x)$ ## Texas Instruments Resources In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9697. 8 , 9 , 10 Feb 22, 2012 Dec 11, 2014", "(ac - bd) + (ad + bc + bd)h. These operations have the same geometric interpretation as their complex counterparts: addition is vector addition, and multiplication is scaling and rotation. In particular, to rotate a hexgonal number 60° counter-clockwise, we multiply it by h: (a + bh) · h = -b + (a + b)h, and to rotate the same number 60° clockwise, we divide by h: (a + bh) / h = (a + bh) · (1 - h) = (a + b) - ah. For example, we can take the unit hexagonal number pointing right, 1 = (1, 0), a full circle, going counter-clockwise, by multiplying it by h six times: (1, 0) · h = (0, 1); (0, 1) · h = (-1, 1); (-1, 1) · h = (-1, 0); (-1, 0) · h = (0, -1); (0, -1) · h = (1, -1); (1, -1) · h = (1, 0). The program uses hexagonal numbers in this fashion to represent the current position in the maze and the current direction, to advance in the specified direction, and to rotate the direction to the left and to the right. # Hexagony, 2437 bytes The long-awaited program is here: (.=$>({{{({}}{\\>6'%={}{?4=$/./\\_><./,{}}{<$<\\?{&P'_(\"'\"#<\".>........_..\\></</(\\.|/<>}{0/'$.}_.....><>)<.$)).><$./$\\))'\"<$_.)><.>%'2{/_.>(/)|_>}{{}./..>#/|}.'/$|\\})'%.<>{=$_.\\<))<>(|\\4?<.{.%.|/</{=..../<>/...'..._.>'\"'_/<}....({{>%'))}/.><../{}{\\>\\|(<><?..\\\\<.}_>=<._..$$/.//..\\}\\.)))))<...2/|$${}/{..><>).../_.._>{0#{{((((/|#.}><..>.<_.$$//>))<(/.\\.\\})'\"#4?#\\_=_-..=.>(<...(..>(/\\\")<((.\\=}}}\\>{}{?<,|{>/...(...>(>{)<.>{=P&/(>//(_.)\\}=#=\\4#|)__.>\"'()'\\.'..\".(\\&P'&</'&$_></}{)<\\<0|\\<.}.\\\"\\.(.(.(/(\\..{.>}=P/|><.(...(...\"/<.{\"_{{=..)..>})<|><}}/\\}}&P<\\(/._...>$'/.>}/{}}{)..|/(\\'.<(\\''\"\")/{{}})<..'...}}>3#./$<}|.}|...><{{}/>.}}{{<>(\"\"''/..>){<}\\?=}{\\._=//=_>)\\{_\\._..>)</{\\=._.....>((>}}<.>5#.\\/}>)<>-/(.....{\\<>}}{{/)$>=}}}))<...=...(\\?{{{?<\\<._...}.><..\\}}/..>'P&//(\\......(..\\})\"'/./&P'&P{}}&P'<{}\\{{{({{{(.\\&P=<.><\"&1}(./'\"?&'&\"\\.|>}{?&\"?&'P&/|{/&P''</(\\..>P&{/&/}{}&'&},/\"&P'&?<.|\\}{&?\"&P'&P'<._.>\"&}\\(>))<\\=}{}<.{/}&?\"&\"&/\"&\"?&}\\.|>?&\"?&{{}}?&//x'&{((<._\\(|(}.\\/}{/>=&'P&\"&/\".{3?<.|\\\"&P'&P}{}&P'<.>&{}}?&\"&'P&\\=}}<.}/2?\".?''5?\"/?1{(}\\.\"..../{},<../&//&\"&P'&P'&\"&\"</{}}{{/>\"?1''?.'({/}}{}<..>?&\"?&}}>)|P/<.>\"&'P&'P&\"&\"&{/........._/\"$#1}/._.........|,(<'\"}'?/_P#\"0'{)}))|........(>/\\.#1?<<|.....>?&}>=?&\"?&/1..>I;n;v;a;l;i;d;P0;m;a\\|\\\"(}}({=/.._...\\\"&P=},}}&P'<.|><....................;...>1\"(}}){=/_....>'P&'P&}}_?&/#.>}?4'%\\/<...@;1P;e;z<._><>\"))'?=<.=..\\&P}{&</\\\"><_'|/'&=}<.>{{.<.........|>(/>3\")}}){=/=/_.>}P&\"?/\"<).}_.>?4{=:<.|_...........$\\2'>4\")}}({/.\"\\{&P'&?/><.?|>P....\"/=(>(/./(}{{\\..>(<>(<>?5'\"((..'/...#,</,}{{\\.......;.F>..\\(...}....._.._..._..._........__..'......\\.<R...>))<}{{&P'&?}<.$.\\...................\\.<>L\\.\\(('_\"\\>}P&\"?&{/__/=(.(<.>_)..<...>....\\..._.<.....&?=\\}=&?\"&<..\"'>.\\>))<.|>))\\.|.>&\"?&{{}=P&}?&=}/{\\.>&{{P/{\">)<|\\{<(|\\(_(<>\\_\\?\"&P'&P}{{{&<=_.>\\&\\?\"&?<|'{/(/>{{/_>.{/=/\\\\.>'P&\"?&\"?&\"?/._($$\\\\>?&\"/_|.>/.$/|$..\\\\><..\\&?}{{}&P'&<}.._>{<|\\{}<._$>-\")<.>_)<|{)$|..>}=P&\"?&\"?/...{\"./>'P&/=_\\{?(/>(<>$$|)__.\\&?}}{}&P<}..\\&P'&P'&<\\})))&=<$$<'.'_,><.>\"?&'P&'/.|>?&{{}?&\"?/>&\"?&\"?&}}<.\".(\\\\\\&?={&P<{..\\\"&?\"&P'&<.?....|.$'$/\\\"/.,.>{}{}=/..>&'P&}{{}P/\\{}&P{(&?\"&?\"<'.(\\&?\"&<}..\\?\"&?\"&<.>P&={}}?&}}P&'P&/.'.>&\"?/..>P&}}{{P/\\}&P'&?={&?<}=\\\".\"\\P'<{..\\'&P'&<....>'P&{}P&\"?&{{<\\\\..>&/.>}{?&\"?/|'&.P&P$'&P={(/..\\P\\\\.\\{&?\"&?\\...\\?{{}=<$&P'&P<.,./<?\\...{}=P\\\"&<.>=P&\"\"'?&'P&'/$.#1.>{?1#=$\\&'P/\\}&P'&?={(,}<._?_&\\&?{=&{*=}4<.>P&\"?&\"?&'P&/1_$>}?&}}=?&){?/\\{}&P'&?={&?#<$ Try it online! \"Readable\" version: ( . =$ > ( { { { ( {", "4-6. • Questions 34-37 are a review. • Question 38 is similar to Example 4. In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter $p \\ 180^\\circ$ counterclockwise, what letter would you have? 2. If you rotated the letter $p \\ 180^\\circ$ clockwise, what letter would you have? 3. A $90^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 4. A $270^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 5. A $210^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 6. A $120^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 7. A $340^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 8. Rotating a figure $360^\\circ$ is the same as what other rotation? 9. Does it matter if you rotate a figure $180^\\circ$ clockwise or counterclockwise? Why or why not? 10. When drawing a rotated figure and using your protractor, would it be easier to rotate the figure $300^\\circ$ counterclockwise or $60^\\circ$ clockwise? Explain your reasoning. Using Investigation 12-1, rotate each figure around point $P$ the given angle measure. 1. $50^\\circ$ 2. $120^\\circ$ 3. $200^\\circ$ 4. $330^\\circ$ 5. $75^\\circ$ 6. $170^\\circ$ Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin. 1. $180^\\circ$ 2. $90^\\circ$", "1: Rotate $\\triangle ABC$, with vertices $A(7, 4), B(6, 1),$ and $C(3, 1) \\ 180^\\circ$. Find the coordinates of $\\triangle A'B'C'$. Solution: You can either use Investigation 12-1 or the hint given above to find $\\triangle A'B'C'$. It is very helpful to graph the triangle. Using the hint, if $A$ is $(7, 4)$, that means it is 7 units to the right of the origin and 4 units up. $A'$ would then be 7 units to the left of the origin and 4 units down. The vertices are: $A(7,4) \\rightarrow A'(-7,-4)\\\\B(6,1) \\rightarrow B'(-6,-1)\\\\C(3,1) \\rightarrow C'(-3,-1)$ The image has vertices that are the negative of the preimage. This will happen every time a figure is rotated $180^\\circ$. Rotation of $180^\\circ$: If $(x, y)$ is rotated $180^\\circ$ around the origin, then the image will be $(-x, -y)$. From this example, we can also see that a rotation is an isometry. This means that $\\triangle ABC \\cong \\triangle A'B'C'$. You can use the distance formula to verify that our assertion holds true. $90^\\circ$ Rotation Similar to the $180^\\circ$ rotation, a $90^\\circ$ rotation (counterclockwise) is an isometry. Each image will be the same distance away from the origin as its preimage, but rotated $90^\\circ$. Example 2: Rotate $\\overline{ST} \\ 90^\\circ$. Solution: When we rotate something $90^\\circ$, you can use Investigation 12-1. Draw", "$z^3$ form a equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$. Therefore, we note that either $z^3 = z \\text{cis} 60^\\circ{}$ or $z \\text{cis}(-60^\\circ{}) = z^3$ The first equation in $z$ (meaning $z^3 = z \\text{cis} 60^\\circ{}$) gives us: $z^2 = cis 60^\\circ{}$, which gives 2 solutions in $z$. The second equation in $z$ (which is $z \\text{cis} (-60^\\circ{}) = z^3$) gives us $z^2 = (\\text{cis} -60^\\circ{})$, which must give another 2 solutions in $z$. Therefore, there are $\\boxed{E) 4}$ solutions for $z$. (Professor-Mom) ## Video Solution For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc", "write the following: $(x,y) \\rightarrow (-y,x) \\ \\text{or} \\ R_{90^\\circ}(x,y)=(-y,x)$ In this concept you will write rules for rotation as well as draw graphs from these rules. For this concept you will be using the second of the above notations, namely $R_{90^\\circ}(x,y)=(-y,x)$. Guidance The figure below shows a pattern of two fish. Write the mapping rule for the rotation of Image A to Image B. Notice that the angle measure is $90^\\circ$ and the direction is clockwise. Therefore the Image A has been rotated $-90^\\circ$ to form Image B. To write a rule for this rotation you would write: $R_{270^\\circ}(x,y)=(y,-x)$. Examples Example A Find an image of the point (3, 2) that has undergone a rotation in: a) about the origin at $90^\\circ$, b) about the origin at $180^\\circ$, and c) about the origin at $270^\\circ$. Write the notation to describe the rotation. a) Rotation about the origin at $90^\\circ : R_{90^\\circ}(x,y)=(-y,x)$ b) Rotation about the origin at $180^\\circ : R_{180^\\circ}(x,y)=(-x,-y)$ c) Rotation about the origin at $270^\\circ : R_{270^\\circ}(x,y)=(y,-x)$ Example B Rotate Image A in the diagram below: d) about the origin at $90^\\circ$, and label it $B$. e) about the origin at $180^\\circ$, and label it $O$. f) about the origin at $270^\\circ$, and label it $Z$. Write notation for each to indicate the type of", "$$270^{\\circ}$$ about the origin? What would its new coordinates be? Example $$\\PageIndex{1}$$ A rotation of $$80^{\\circ}$$ clockwise is the same as what counterclockwise rotation? Solution There are $$360^{\\circ}$$ around a point. So, an $$80^{\\circ}$$ rotation clockwise is the same as a $$360^{\\circ}−80^{\\circ}=280^{\\circ}$$ rotation counterclockwise. Example $$\\PageIndex{2}$$ A rotation of $$160^{\\circ} counterclockwise is the same as what clockwise rotation? Solution \\(360^{\\circ}−160^{\\circ}=200^{\\circ}$$ clockwise rotation. Example $$\\PageIndex{3}$$ Rotate $$\\overline{ST} 90^{\\circ}$$. Solution Example $$\\PageIndex{4}$$ The rotation of a quadrilateral is shown below. What is the measure of $$x$$ and $$y$$? Solution Because a rotation produces congruent figures, we can set up two equations to solve for $$x$$ and $$y$$. $$y=4\\begin{array}{rr} 2 y=80^{\\circ} & 2 x-3=15 \\\\ y=40^{\\circ} & 2 x=18 \\\\ & x=9 \\end{array}$$ Example $$\\PageIndex{5}$$ Rotate $$\\Delta ABC$$, with vertices $$A(7,4)$$, $$B(6,1)$$, and $$C(3,1)$$, $$180^{\\circ}$$ about the origin. Find the coordinates of $$\\Delta A′B′C′$$. Solution Use the rule above to find $$\\Delta A′B′C′$$. \\begin{aligned}A(7,4)&\\rightarrow A′(−7,−4) \\\\ B(6,1)&\\rightarrow B′(−6,−1) \\\\ C(3,1)&\\rightarrow C′(−3,−1)\\end{aligned} ## Review In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter p 180^{\\circ}\\) counterclockwise, what letter would you have? 2. If you rotated the letter p $$180^{\\circ}$$ c$$ockwise, what letter would you have? 3. A \\(90^{\\circ}$$ clockwise rotation is the same as what counterclockwise rotation? 4. A $$270^{\\circ}$$ clockwise rotation is the", "Rotation of $180^{\\circ}$ is going into two quadrants. $270^{\\circ}$ will go into three quadrants. And we don’t do $360^{\\circ}$ because it’s right back where we started. So, don’t worry about rotating $360^{\\circ}$ because we’ll end exactly where we started. You might also see rotations for $-90^{\\circ}$, rotations of $-180^{\\circ}$, and rotations of $-270^{\\circ}$. If $90^{\\circ}$ is counterclockwise, then $-90^{\\circ}$ is clockwise direction. $-90^{\\circ}$ is the same as $270^{\\circ}$. $-180^{\\circ}$ is the same as $180^{\\circ}$. And rotating $-270^{\\circ}$ is the same as $90^{\\circ}$. Now that we have an idea of what quadrant we’d end up in, let’s take a look at the specific rules that tells exactly where each coordinate will go. $R_{90} (x, y) \\rightarrow (-y, x)$ $R_{180} (x, y) \\rightarrow (-x, -y)$ $R_{270} (x, y) \\rightarrow (y, -x)$ Let’s try rotating this triangle $90^{\\circ}$. $R_{90} \\triangle {ABC}$ Let’s first write down what the rule is. $R_{90} (x, y) \\rightarrow (-y, x)$ $A (1, 5) \\rightarrow A^\\prime (-5, 1)$ $B (3, 2) \\rightarrow B^\\prime (-2, 3)$ $C (1, 2) \\rightarrow C^\\prime (-2, 1)$ What happened is that the $x$ coordinate of $A$ became the $y$ coordinate of $A^\\prime$ and the $y$ coordinate became the $x$ coordinate after we negated it. Let’s graph this and then draw our triangle. Now, we have the image of triangle $ABC$ when we", "through $90^{\\circ},$ and around $K\\,$ through $180^{\\circ}.$ For example, $M_{90^{\\circ}}(A) = O,$ $N_{90^{\\circ}}(O) = P,$ $K_{180^{\\circ}}(P) = A.$ The composition of the three rotations has a fixed point at $A$ and is a rotation through $360^{\\circ} = 0^{\\circ}\\;(\\text{mod}\\;360^{\\circ}).$ It is therefore a translation with a fixed point, i.e. the identity transformation. Denote the image of $M$ under $N_{90^{\\circ}}$ as $M':$ $M' = N_{90^{\\circ}}(M).$ ($\\Delta MNM'$ is isosceles with angle $N$ being $90^{\\circ}.)\\;$ Then $M = K_{180^{\\circ}}(N_{90^{\\circ}}(M_{90^{\\circ}}(M))) = K_{180^{\\circ}}(N_{90^{\\circ}}(M)) = K_{180^{\\circ}}(M'),$ It follows that $M = K_{180^{\\circ}}(M').$ But $K_{180^{\\circ}}(M')$ is a reflection in $K.$ Therefore $K$ is the midpoint of $MM'.\\;$ Therefore $\\Delta MNK$ is isosceles with angle $K$ being $90^{\\circ}.$ ### Remark 1. Squares $OABC$ and $OPQR$ are constructed on the sides $OA$ and OP of $\\Delta AOP.\\;$ We have just showed that the center of the square built on the line $MN$ of their centers coincides with the midpoint $K$ of $AP.\\;$ This fact is known as Neuberg's Theorem. 2. The Finsler-Hadwiger theorem is also a special case of the Fundamental Theorem of Directly Similar Figures 3. The theorem is directly equivalent to one of Thébault's statements. ### References 1. R. L. Finney, Dynamic Proofs of Euclidean Theorems, Math Magazine, 43, pp. 177-185. 2. R. Honsberger, In Pólya's Footsteps, MAA, 1997" ]

[ "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)", "$270^\\circ$ (or $-90^\\circ$ ) $(x, y)$ $(y, -x)$ #### Example A Line $\\overline{AB}$ drawn from (-4, 2) to (3, 2) has been rotated about the origin at an angle of $90^\\circ$ CW. Draw the preimage and image and properly label each. Solution: #### Example B The diamond $ABCD$ is rotated $145^\\circ$ CCW about the origin to form the image $A^\\prime B^\\prime C^\\prime D^\\prime$ . On the diagram, draw and label the rotated image. Solution: Notice the direction is counter-clockwise. #### Example C The following figure is rotated about the origin $200^\\circ$ CW to make a rotated image. On the diagram, draw and label the image. Solution: Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is $160^\\circ$ . #### Concept Problem Revisited Quadrilateral $WXYZ$ has coordinates $W(-5, -5), X(-2, 0), Y(2, 3)$ and $Z(-1, 3)$ . Draw the quadrilateral on the Cartesian plane. Rotate the image $110^\\circ$ counterclockwise about the point $X$ . Show the resulting image. ### Guided Practice 1. Line $\\overline{ST}$ drawn from (-3, 4) to (-3, 8) has been rotated $60^\\circ$ CW about the point $S$ . Draw the preimage and image and properly label each. 2. The polygon below has been rotated $155^\\circ$ CCW about the origin. Draw the rotated image and properly label each. 3. The purple pentagon is", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "$z^3$ form a equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$. Therefore, we note that either $z^3 = z \\text{cis} 60^\\circ{}$ or $z \\text{cis}(-60^\\circ{}) = z^3$ The first equation in $z$ (meaning $z^3 = z \\text{cis} 60^\\circ{}$) gives us: $z^2 = cis 60^\\circ{}$, which gives 2 solutions in $z$. The second equation in $z$ (which is $z \\text{cis} (-60^\\circ{}) = z^3$) gives us $z^2 = (\\text{cis} -60^\\circ{})$, which must give another 2 solutions in $z$. Therefore, there are $\\boxed{E) 4}$ solutions for $z$. (Professor-Mom) ## Video Solution For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc", "0) $270^\\circ$ (or $-90^\\circ$ ) $(x, y)$ $(y, -x)$ Example A Line $\\overline{AB}$ drawn from (-4, 2) to (3, 2) has been rotated about the origin at an angle of $90^\\circ$ CW. Draw the preimage and image and properly label each. Solution: Example B The diamond $ABCD$ is rotated $145^\\circ$ CCW about the origin to form the image $A^\\prime B^\\prime C^\\prime D^\\prime$ . On the diagram, draw and label the rotated image. Solution: Notice the direction is counter-clockwise. Example C The following figure is rotated about the origin $200^\\circ$ CW to make a rotated image. On the diagram, draw and label the image. Solution: Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is $160^\\circ$ . Concept Problem Revisited Quadrilateral $WXYZ$ has coordinates $W(-5, -5), X(-2, 0), Y(2, 3)$ and $Z(-1, 3)$ . Draw the quadrilateral on the Cartesian plane. Rotate the image $110^\\circ$ counterclockwise about the point $X$ . Show the resulting image. Vocabulary Center of rotation A center of rotation is the fixed point that a figure rotates about when undergoing a rotation. Rotation A rotation is a transformation that rotates (turns) an image a certain amount about a certain point. Image In a transformation, the final figure is called the image . Preimage In a transformation, the original figure is called", "$5$ left to insert, we place them such that we don't have the two pairs adjacent. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"1\", A, N); label(\"3\", B, NW); label(\"2\", C, SW); label(\"4\", D, SE); label(\"5\", E, NE); [/asy]$ This is the only solution to make $7$ \"bad.\" We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\\boxed{\\textbf {(B) }2}$.", "(ac - bd) + (ad + bc + bd)h. These operations have the same geometric interpretation as their complex counterparts: addition is vector addition, and multiplication is scaling and rotation. In particular, to rotate a hexgonal number 60° counter-clockwise, we multiply it by h: (a + bh) · h = -b + (a + b)h, and to rotate the same number 60° clockwise, we divide by h: (a + bh) / h = (a + bh) · (1 - h) = (a + b) - ah. For example, we can take the unit hexagonal number pointing right, 1 = (1, 0), a full circle, going counter-clockwise, by multiplying it by h six times: (1, 0) · h = (0, 1); (0, 1) · h = (-1, 1); (-1, 1) · h = (-1, 0); (-1, 0) · h = (0, -1); (0, -1) · h = (1, -1); (1, -1) · h = (1, 0). 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A rotation of $160^\\circ$ counterclockwise is the same as what clockwise rotation? 1. Because a rotation is an isometry that produces congruent figures, we can set up two equations to solve for $x$ and $y$ . $2y &= 80^\\circ && \\ 2x-3 =15\\\\y &= 40^\\circ && \\qquad 2x = 18\\\\&&& \\qquad \\ \\ x = 9$ 2. There are $360^\\circ$ around a point. So, an $80^\\circ$ rotation clockwise is the same as a $360^\\circ-80^\\circ=280^\\circ$ rotation counterclockwise. 3. $360^\\circ-160^\\circ=200^\\circ$ clockwise rotation. ### Practice In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter $p \\ 180^\\circ$ counterclockwise, what letter would you have? 2. If you rotated the letter $p \\ 180^\\circ$ clockwise, what letter would you have? Why do you think that is? 3. A $90^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 4. A $270^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 5. Rotating a figure $360^\\circ$ is the same as what other rotation? Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin. 1. $180^\\circ$ 2. $90^\\circ$ 3. $180^\\circ$ 4. $270^\\circ$ 5. $90^\\circ$ 6. $270^\\circ$ 7. $180^\\circ$ 8. $270^\\circ$ 9. $90^\\circ$ Algebra Connection Find the measure of $x$ in the rotations below. The blue figure is the", "an angle of $$90^{\\circ}$$ CW. Now, let's draw and label the rotated image for the following rotations: 1. The diamond $$ABCD$$ is rotated $$145^{\\circ}$$ CCW about the origin to form the image $$A′B′C′D′$$. Notice the direction is counter-clockwise. 1. The following figure is rotated about the origin $$200^{\\circ}$$ CW to make a rotated image. Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is $$160^{\\circ}$$. Example $$\\PageIndex{1}$$ Earlier, you were asked about the quadrilateral $$WXYZ$$ has coordinates $$W(−5,−5)$$, $$X(−2,0)$$, Y(2,3)\\) and $$Z(−1,3)$$. Draw the quadrilateral on the Cartesian plane. Rotate the image $$110^{\\circ}$$ counterclockwise about the point X\\). Show the resulting image. Solution Example $$\\PageIndex{2}$$ Line $$\\overline{ST}$$ drawn from $$(-3, 4)$$ to $$(-3, 8)$$ has been rotated $$60^{\\circ}$$ CW about the point $$S$$. Draw the preimage and image and properly label each. Solution Notice the direction of the angle is clockwise, therefore the angle measure is $$60^{\\circ}$$ CW or $$−60^{\\circ}$$. Example $$\\PageIndex{3}$$ The polygon below has been rotated $$155^{\\circ}$$ CCW about the origin. Draw the rotated image and properly label each. Solution Notice the direction of the angle is counter-clockwise, therefore the angle measure is $$155^{\\circ}$$ CCW or $$155^{\\circ}$$. Example $$\\PageIndex{4}$$ The purple pentagon is rotated about the point A $$225^{\\circ}$$. Find the coordinates of the purple pentagon. On the diagram, draw and label", "about the origin? 1. This is a $135^\\circ$ clockwise rotation about point $B$ . You could also say this is a $225^\\circ$ counterclockwise rotation about point $B$ . 2. You can see that there is a straight line $(180^\\circ)$ passing through the origin connecting each pair of corresponding points. $(x, y) \\rightarrow (-x, -y)$ . 3. There would be no difference. A rotation $180^\\circ$ clockwise is the same as a rotation $180^\\circ$ counterclockwise. Practice Describe each of the following rotations in TWO ways. 1. 2. 3. 4. What does a $360^\\circ$ rotation look like? 5. Rotate the rectangle below $90^\\circ$ clockwise about the origin. 6. Use the previous example to help you describe what happens to a point $(x, y)$ that is rotated $90^\\circ$ clockwise about the origin. 7. (3, 2) is rotated $90^\\circ$ clockwise about the origin. Where does it end up? 8. (3, 2) is rotated $180^\\circ$ clockwise about the origin. Where does it end up? Hint: Use Guided Practice #2 for help. 9. (3, 2) is rotated $90^\\circ$ counterclockwise about the origin. Where does it end up? Hint: Use Example C. 10. What do circles have to do with rotations? 11. $\\Delta ABC$ is rotated $130^\\circ$ clockwise about point $O$ to create $\\Delta A^\\prime B^\\prime C^\\prime$ . Describe two connections between the points $B, O$", "$$270^{\\circ}$$ about the origin? What would its new coordinates be? Example $$\\PageIndex{1}$$ A rotation of $$80^{\\circ}$$ clockwise is the same as what counterclockwise rotation? Solution There are $$360^{\\circ}$$ around a point. So, an $$80^{\\circ}$$ rotation clockwise is the same as a $$360^{\\circ}−80^{\\circ}=280^{\\circ}$$ rotation counterclockwise. Example $$\\PageIndex{2}$$ A rotation of $$160^{\\circ} counterclockwise is the same as what clockwise rotation? Solution \\(360^{\\circ}−160^{\\circ}=200^{\\circ}$$ clockwise rotation. Example $$\\PageIndex{3}$$ Rotate $$\\overline{ST} 90^{\\circ}$$. Solution Example $$\\PageIndex{4}$$ The rotation of a quadrilateral is shown below. What is the measure of $$x$$ and $$y$$? Solution Because a rotation produces congruent figures, we can set up two equations to solve for $$x$$ and $$y$$. $$y=4\\begin{array}{rr} 2 y=80^{\\circ} & 2 x-3=15 \\\\ y=40^{\\circ} & 2 x=18 \\\\ & x=9 \\end{array}$$ Example $$\\PageIndex{5}$$ Rotate $$\\Delta ABC$$, with vertices $$A(7,4)$$, $$B(6,1)$$, and $$C(3,1)$$, $$180^{\\circ}$$ about the origin. Find the coordinates of $$\\Delta A′B′C′$$. Solution Use the rule above to find $$\\Delta A′B′C′$$. \\begin{aligned}A(7,4)&\\rightarrow A′(−7,−4) \\\\ B(6,1)&\\rightarrow B′(−6,−1) \\\\ C(3,1)&\\rightarrow C′(−3,−1)\\end{aligned} ## Review In the questions below, every rotation is counterclockwise, unless otherwise stated. 1. If you rotated the letter p 180^{\\circ}\\) counterclockwise, what letter would you have? 2. If you rotated the letter p $$180^{\\circ}$$ c$$ockwise, what letter would you have? 3. A \\(90^{\\circ}$$ clockwise rotation is the same as what counterclockwise rotation? 4. A $$270^{\\circ}$$ clockwise rotation is the", "1: Rotate $\\triangle ABC$, with vertices $A(7, 4), B(6, 1),$ and $C(3, 1) \\ 180^\\circ$. Find the coordinates of $\\triangle A'B'C'$. Solution: You can either use Investigation 12-1 or the hint given above to find $\\triangle A'B'C'$. It is very helpful to graph the triangle. Using the hint, if $A$ is $(7, 4)$, that means it is 7 units to the right of the origin and 4 units up. $A'$ would then be 7 units to the left of the origin and 4 units down. The vertices are: $A(7,4) \\rightarrow A'(-7,-4)\\\\B(6,1) \\rightarrow B'(-6,-1)\\\\C(3,1) \\rightarrow C'(-3,-1)$ The image has vertices that are the negative of the preimage. This will happen every time a figure is rotated $180^\\circ$. Rotation of $180^\\circ$: If $(x, y)$ is rotated $180^\\circ$ around the origin, then the image will be $(-x, -y)$. From this example, we can also see that a rotation is an isometry. This means that $\\triangle ABC \\cong \\triangle A'B'C'$. You can use the distance formula to verify that our assertion holds true. $90^\\circ$ Rotation Similar to the $180^\\circ$ rotation, a $90^\\circ$ rotation (counterclockwise) is an isometry. Each image will be the same distance away from the origin as its preimage, but rotated $90^\\circ$. Example 2: Rotate $\\overline{ST} \\ 90^\\circ$. Solution: When we rotate something $90^\\circ$, you can use Investigation 12-1. Draw", "write the following: $(x,y) \\rightarrow (-y,x) \\ \\text{or} \\ R_{90^\\circ}(x,y)=(-y,x)$ In this concept you will write rules for rotation as well as draw graphs from these rules. For this concept you will be using the second of the above notations, namely $R_{90^\\circ}(x,y)=(-y,x)$. Guidance The figure below shows a pattern of two fish. Write the mapping rule for the rotation of Image A to Image B. Notice that the angle measure is $90^\\circ$ and the direction is clockwise. Therefore the Image A has been rotated $-90^\\circ$ to form Image B. To write a rule for this rotation you would write: $R_{270^\\circ}(x,y)=(y,-x)$. Examples Example A Find an image of the point (3, 2) that has undergone a rotation in: a) about the origin at $90^\\circ$, b) about the origin at $180^\\circ$, and c) about the origin at $270^\\circ$. Write the notation to describe the rotation. a) Rotation about the origin at $90^\\circ : R_{90^\\circ}(x,y)=(-y,x)$ b) Rotation about the origin at $180^\\circ : R_{180^\\circ}(x,y)=(-x,-y)$ c) Rotation about the origin at $270^\\circ : R_{270^\\circ}(x,y)=(y,-x)$ Example B Rotate Image A in the diagram below: d) about the origin at $90^\\circ$, and label it $B$. e) about the origin at $180^\\circ$, and label it $O$. f) about the origin at $270^\\circ$, and label it $Z$. Write notation for each to indicate the type of", "Common Rotations • Rotation of $180^\\circ$ : If $(x, y)$ is rotated $180^\\circ$ around the origin, then the image will be $(-x, -y)$ . • Rotation of $90^\\circ$ : If $(x, y)$ is rotated $90^\\circ$ around the origin, then the image will be $(-y, x)$ . • Rotation of $270^\\circ$ : If $(x, y)$ is rotated $270^\\circ$ around the origin, then the image will be $(y, -x)$ . While we can rotate any image any amount of degrees, only $90^\\circ, 180^\\circ$ and $270^\\circ$ have special rules. To rotate a figure by an angle measure other than these three, you must use the process from the Investigation. #### Example A Rotate $\\triangle ABC$ , with vertices $A(7, 4), B(6, 1),$ and $C(3, 1) \\ 180^\\circ$ . Find the coordinates of $\\triangle A'B'C'$ . It is very helpful to graph the triangle. If $A$ is $(7, 4)$ , that means it is 7 units to the right of the origin and 4 units up. $A'$ would then be 7 units to the left of the origin and 4 units down. The vertices are: $A(7,4) \\rightarrow A'(-7,-4)\\\\B(6,1) \\rightarrow B'(-6,-1)\\\\C(3,1) \\rightarrow C'(-3,-1)$ #### Example B Rotate $\\overline{ST} \\ 90^\\circ$ counter-clockwise about the origin. Using the $90^\\circ$ rotation rule, $T'$ is (8, 2). #### Example C Find the coordinates of $ABCD$ after a", "rotated about the point $A \\ 225^\\circ$ . Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon. Answers: 1. Notice the direction of the angle is clockwise, therefore the angle measure is $60^\\circ$ CW or $-60^\\circ$ . 2. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $155^\\circ$ CCW or $155^\\circ$ . 3. The measure of $\\angle BAB^\\prime = m \\angle BAE^\\prime + m \\angle E^\\prime AB^\\prime$ . Therefore $\\angle BAB^\\prime = 111.80^\\circ + 113.20^\\circ$ or $225^\\circ$ . Notice the direction of the angle is counter-clockwise, therefore the angle measure is $225^\\circ$ CCW or $225^\\circ$ . ### Explore More 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3.", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "origin at $$270^{\\circ}$$ Rotation about the origin at 270^{\\circ}: $$R_270^{\\circ}(x,y)=(y,−x)$$ Now let's perform the following rotations on Image $$A$$ shown below in the diagram below and describe the rotations: 1. about the origin at $$90^{\\circ}$$, and label it $$B$$. Rotation about the origin at $$90^{\\circ}$$: $$R_{90^{\\circ}}A\\rightarrow B=R_{90^{\\circ}}(x,y)\\rightarrow (−y,x)$$ 1. about the origin at $$180^{\\circ}$$, and label it $$O$$. Rotation about the origin at $$180^{\\circ}$$: $$R_{180^{\\circ}}A\\rightarrow O=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$ 1. about the origin at $$270^{\\circ}$$, and label it $$Z$$. Rotation about the origin at $$270^{\\circ}$$: $$R_{270^{\\circ}}A\\rightarrow Z=R270^{\\circ}(x,y)\\rightarrow (y,−x)$$ Finally, let's write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below: First, pick a point in the diagram to use to see how it is rotated. $$E:(−1,2)\\qquad E′:(1,−2)$$ Notice how both the $$x$$- and $$y$$-coordinates are multiplied by -1. This indicates that the preimage $$A$$ is reflected about the origin by $$180^{\\circ}$$CCW to form the rotated image J. Therefore the notation is $$R_{180^{\\circ}}A\\rightarrow J=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$. Example $$\\PageIndex{1}$$ Earlier, you were given the figure below which shows a pattern of two fish. Write the mapping rule for the rotation of Image $$A$$ to Image $$B$$. Solution Notice that the angle measure is $$90^{\\circ}$$ and the direction is clockwise. Therefore the Image $$A$$ has been rotated $$−90^{\\circ}$$ to form Image $$B$$. To write", "$x$ and $y$ ? 1. 2. Using the rule, we have: $(x,y) & \\rightarrow (y,-x)\\\\ A(-4,5) & \\rightarrow A'(5,4)\\\\B(1,2) & \\rightarrow B'(2,-1)\\\\C(-6,-2) & \\rightarrow C'(-2,6)\\\\D(-8,3) & \\rightarrow D'(3,8)$ 3. Because a rotation produces congruent figures, we can set up two equations to solve for $x$ and $y$ . $2y &= 80^\\circ && 2x-3=15\\\\ y &= 40^\\circ && \\quad \\ \\ 2x=18\\\\& && \\qquad \\ x = 9$ ### Practice In the questions below, every rotation is counterclockwise , unless otherwise stated. 1. If you rotated the letter $p \\ 180^\\circ$ counterclockwise, what letter would you have? 2. If you rotated the letter $p \\ 180^\\circ$ clockwise , what letter would you have? 3. A $90^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 4. A $270^\\circ$ clockwise rotation is the same as what counterclockwise rotation? 5. A $210^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 6. A $120^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 7. A $340^\\circ$ counterclockwise rotation is the same as what clockwise rotation? 8. Rotating a figure $360^\\circ$ is the same as what other rotation? 9. Does it matter if you rotate a figure $180^\\circ$ clockwise or counterclockwise? Why or why not? 10. When drawing a rotated figure and using your protractor, would it be easier to rotate the", "Home > INT2 > Chapter 3 > Lesson 3.2.3 > Problem3-97 3-97. On graph paper, plot $ABCD$ if $A(0,3),B(2,5),C(6,3)$, and $D(4,1)$. 1. Rotate $ABCD\\ 90°$ clockwise ($↻$) about the origin to form $A'B'C'D'$. Name the coordinates of $B'$. $(5, −2)$ 2. Translate $A'B'C'D'$ up $8$ units and left $7$ units to form $A''B''C''D''$. Name the coordinates of $C''$. $(−4, 2)$ 3. After rotating $ABCD$ $180^\\circ$ to form $A'''B'''C'''D'''$, Arah noticed that the image lined up exactly with the original. In this case, about what point was $ABCD$ rotated? How did you locate it? Use the eTool below to solve each part. Click the link at right for the full version of the eTool. Int2 3-97 HW eTool", "you were to describe the rotated image using notation, you would write the following: $R_{90^\\circ}(x,y)=(-y,x)$ #### Example A Find an image of the point (3, 2) that has undergone a clockwise rotation: a) about the origin at $90^\\circ$ , b) about the origin at $180^\\circ$ , and c) about the origin at $270^\\circ$ . Write the notation to describe the rotation. Solution: a) Rotation about the origin at $90^\\circ : R_{90^\\circ}(x,y)=(-y,x)$ b) Rotation about the origin at $180^\\circ : R_{180^\\circ}(x,y)=(-x,-y)$ c) Rotation about the origin at $270^\\circ : R_{270^\\circ}(x,y)=(y,-x)$ #### Example B Rotate Image A in the diagram below: a) about the origin at $90^\\circ$ , and label it $B$ . b) about the origin at $180^\\circ$ , and label it $O$ . c) about the origin at $270^\\circ$ , and label it $Z$ . Write notation for each to indicate the type of rotation. Solution: a) Rotation about the origin at $90^\\circ$ : $R_{90^\\circ}A \\rightarrow B=R_{90^\\circ}(x,y) \\rightarrow (-y,x)$ b) Rotation about the origin at $180^\\circ$ : $R_{180^\\circ}A \\rightarrow O=R_{180^\\circ}(x,y) \\rightarrow (-x,-y)$ c) Rotation about the origin at $270^\\circ$ : $R_{270^\\circ}A \\rightarrow Z=R_{270^\\circ}(x,y) \\rightarrow (y,-x)$ #### Example C Write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below. First, pick a point in the diagram to use to" ]

[ "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "+0 +1 48 1 +140 Find the integer $$n$$$$0 \\le n \\le 5,$$ such that $$n \\equiv -3736 \\pmod{6}.$$ Mar 29, 2019 #1 +21977 +2 Find the integer $$n,\\ 0 \\le n \\le 5$$, such that $$n \\equiv -3736 \\pmod{6}$$. Mar 29, 2019", "+0 # help +1 84 1 Let $$z = 3[\\cos(85^{\\circ})+i\\sin(85^{\\circ})]$$. Find the smallest positive value of $$r$$ such that $$r[\\cos(45^{\\circ})+i\\sin(45^{\\circ})] = z^n$$ for some positive integer $$n$$. Dec 20, 2019", "7 '18 at 10:08 All possibilities are wrong! Try $$A=270^{\\circ}$$, $$B=-90^{\\circ}$$ and $$C=0^{\\circ}.$$", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$,", "# 1996 USAMO Problems/Problem 1 ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ ###", "SEARCH HOME Math Central Quandaries & Queries Question from nahla, a student: Hello !! I am not a native english speaker and I am currently at grade 11. I have done all my maths studies in arabic, but there has been this problem that I came across and that really bothered me, I cannot seem to find the answer, although I have some hints of how it should be done, I cannot seem to get it right, here it is, if you do not mind clearing it up ! f: IN --> IN n --> f(n) for every n that belongs the IN : fof(n) = 4n - 3 and for every n that belongs to IN f(2^n) = 2^(n+1) -1 Calculate f(993) Thank you so much ! And greetings ! Hi Nahla, I'll give you some hints also and hopefully they will help. Start with the first condition and see if $993$ can be the result of $f \\circ f(n)$ because if is is than $f(993) = f \\circ f \\circ f(n).$ If $f \\circ (n) = 993$ then $4n - 3 = 993$ and $n = 249.$ Hence $f \\circ f(249) = 993.$ What about $249?$ If $f \\circ (n) = 249$ then $4n - 3 = 249$ and $n = 63.$ Hence $f \\circ f(63)", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "##r## vary you get the ##45^\\circ## line.", "$0 = (X + 1)(2X - 1)$. So $X = -1$ or $X = \\frac{1}{2}$. This means $\\sin{2x} = -1$ or $\\sin{2x} = \\frac{1}{2}$. Case 1: $\\sin{2x} = -1$ $2x = 270^{\\circ} + 360^{\\circ}n$, where $n$ is an integer. $x = 135^{\\circ} + 180^{\\circ}n$, where $n$ is an integer. Case 2: $\\sin{2x} = \\frac{1}{2}$ Since sine is positive in the first and second quadrants: $2x = \\left\\{30^{\\circ}, 180^{\\circ} - 30^{\\circ} \\right\\} + 360^{\\circ} n$ $2x = \\left\\{ 30^{\\circ}, 150^{\\circ}\\right\\} + 360^{\\circ} n$ $x = \\left \\{ 15^{\\circ}, 75^{\\circ} \\right\\} + 180^{\\circ} n$. So putting it together: $x = \\left\\{ 15^{\\circ}, 75^{\\circ}, 135^{\\circ}\\right\\} + 180^{\\circ}n$. Upon substituting $n = 0$ and $n = 1$, we find all the possible solutions in the domain $0^{\\circ} \\leq x \\leq 360^{\\circ}$: $x = \\left\\{ 15^{\\circ}, 75^{\\circ}, 135^{\\circ}, 195^{\\circ}, 255^{\\circ}, 315^{\\circ} \\right\\}$. 3. Thanks a lot, I appreciate it. Very clear. 4. Just to give an alternative way $\\cos{4x} = \\sin{2x}$ $\\cos{4x} = \\cos{(90^{\\circ}-2x)}$ Case 1: $4x = 90^{\\circ}-2x+ 360^{\\circ} n$ $6x = 90^{\\circ} + 360^{\\circ} n$ $x = 15^{\\circ} + 60^{\\circ} n$ $x = \\left\\{ 15^{\\circ}, 75^{\\circ}, 135^{\\circ}, 195^{\\circ}, 255^{\\circ}, 315^{\\circ} \\right\\}$ Case 2: $4x = -(90^{\\circ}-2x) + 360^{\\circ} n$ $4x = -90^{\\circ}+2x + 360^{\\circ} n$ $2x = -90^{\\circ} + 360^{\\circ} n$ $x = -45^{\\circ} + 180^{\\circ} n$ $x = \\left\\{135^{\\circ}, 315^{\\circ}", "Find the least positive integer $$n$$ such that $$\\frac 1{\\sin 45^\\circ\\sin 46^\\circ}+\\frac 1{\\sin 47^\\circ\\sin 48^\\circ}+\\cdots+\\frac 1{\\sin 133^\\circ\\sin 134^\\circ}=\\frac 1{\\sin n^\\circ}.$$ (第十八届AIME2 2000 第15题)", "## Where n(n – 1) and n(n + 1) are both pandigital Find an integer $n$ such that both $n \\,(n - 1)$ and $n \\,(n + 1)$ are pandigital Antonio Roldán found n = 55152", "$$1 - \\sin 10 ^ \\circ - 2 \\sin^2 10^\\circ = \\sin N^\\circ$$, where $$0 \\leq N \\leq 90$$. What is $$N$$?", "- 2)180^\\circ}{n}$. So now substitute some values for $n$ and see which ones that you have listed apply.", "# Find N Let $$N$$ be an integer between $$200$$ and $$250$$ such that $N^3 - N^2 - N + 1$ is a perfect square. Find $$N$$. ×", "we get $$(305)753+(-6954)33=183$$ Thus, all integer solutions are given by $$x=-6954+251k$$ for some $k\\in\\mathbb{Z}$. By adjusting $k$, we can reduce this to $$x=74+251k$$ for some $k\\in\\mathbb{Z}$.", "heard of it before) to find the fractional roots of the expressions above! For $\\cos 240^\\circ + i\\sin 240$ we have $(\\cos 240^\\circ + i\\sin 240^\\circ)^{1/3}$ $\\Rightarrow$ $\\cos 80^\\circ + i\\sin 80^\\circ, \\cos 200^\\circ + i\\sin200^\\circ,$ and $\\cos 320^\\circ + i\\sin320^\\circ.$ Similarly for $(\\cos 120^\\circ + i\\sin 120^\\circ)^{1/3}$, we have $\\cos 40^\\circ + i\\sin 40^\\circ, \\cos 160^\\circ + i\\sin 160^\\circ,$ and $\\cos 280^\\circ + i\\sin 280^\\circ.$ The only argument out of all these roots that fits the description is $\\theta = \\boxed{160}$ Note: We can add $120$ to the angles of the previous solutions to get new solutions because De Moivre's formula says that $(\\cos \\theta + i \\sin \\theta)^n = \\cos n\\theta + i \\sin n\\theta$ and $\\frac{360}{3} = 120$. ~programmeruser ~ blueballoon", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "45.0$$^\\circ$$ 3.00 $$\\vec{A}=k\\vec{u}$$ $$\\vec{u} =$$ 0.71 $$\\hat{i}$$ + 0.71 $$\\hat{j}$$ $$k =$$ number of u's $$=$$ 2.00 $$\\vec{A} =$$ 1.41 $$\\hat{i}$$ + 1.41 $$\\hat{j}$$" ]

[ "Trigonometry 7th Edition $\\theta=\\{60^o+120^ok\\;\\;\\}$ $cos(3\\theta)=-1$ $3\\theta=cos^{-1}(-1)$ The period of the cosine function is $360^o$ $3\\theta=180^o+360^ok\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=60^o+120^ok\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=\\{60^o+120^ok\\;\\;\\},\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;when \\;\\;\\;K=\\{0,1,2,3\\}$", "series: $\\cos 35^\\circ = \\cos \\left({35 \\over 180} \\pi \\right) \\approx \\cos 0.610865$ Then, substitute into the Taylor series of cosine above: $\\cos (0.610865) \\approx 1 - {0.610865^2 \\over 2!} + {0.610865^4 \\over 4!}$ Here we have written the 4th-degree Taylor polynomial, but this should be enough to show us something. The right side of the equation can be reduced to the four simple operations, so we can easily calculate its value: $\\cos (0.523599) \\approx 0.81922$ We can compare this to the value given by the calculator. The calculator's value, actually, is also an approximation obtained by a similar method, but we can expect it to be accurate for all displayed decimal places. $\\cos 35^\\circ = 0.81915 \\cdots$ So our approximating value agrees with the \"actual\" value to three decimal places, which is good accuracy for a basic approximation. As above, better accuracy can be attained by using more terms in the Taylor series. This result can be observed if we zoom in the graph at the intersection point, as shown in Figure 1. In a large graph, the functions look almost identical at the intersection point, but there is indeed a difference between these two functions, as the graph shows. # A More Mathematical Explanation Note: understanding of this explanation requires: *Calculus ## The general form of", "0.866. So, the cosine of $25^\\circ$ will be between the values of 0.866 and 0.940. Since the range for the cosine is greater, than the range for the sine, it can be assumed that the cosine of $25^\\circ$ will be greater than the sine of $25^\\circ$. Notice that as the angle measures approach $90^\\circ$, sine approaches the value of 1. Similarly, as the value of the angles approaches $90^\\circ$, the cosine approaches the value of 0. In other words, as the sine gets greater, the cosine loses value. The tangent, on the other hand, increases rapidly from a small value to a large value approaching $90^\\circ$. ## Real-Life Example Completed The Ramp Dilemma Mr. Watson’s homeroom decided to do a community service repairing the ramp outside the shed. The fresh coat of paint shone in the bright sunlight and Mr. Watson walked across the grass with all of his students to look at the ramp outside the door of the shed. “Has that always been there?” asked Dan. “No, in fact it was just brought out here yesterday,” Mr. Watson explained. “Well if it’s brand new, then why do we need to fix it?” Emily asked. “Because it doesn’t fit neatly under the door.” Sure enough, the students looked and could see that the back of the ramp", "1:23 You're plotting the function $\\cos n\\pi/4$, which has period $8$, and thus $8$ full periods from $0$ to $64$, but you're only substituting values from $0$ to $16\\pi$. Since $16\\pi\\approx50$, you're missing a bit less than two of the periods. From what it seems you're trying to do, you should be plotting the function from $0$ to $64$, i. e. replace 2*pi*8 by 64 in the first line. @Raven: A clue that your linspace() expression was faulty was that your n consists of $(0,16\\pi/63,\\dots,16\\pi)$. Replacing the n in cos(2*pi*8*n/64) with, for example, 16*pi/63 gives you something with a pi^2 within the cosine... – Ｊ. Ｍ. Sep 16 '11 at 4:59", "## Precalculus (10th Edition) $1$ I know that $\\cos$ has a period of $360^\\circ$, hence $\\cos{\\theta}=\\cos{(\\theta-360^\\circ)}.$ Hence: $\\cos{400^\\circ}\\sec{40^\\circ}=\\cos{(400^\\circ-360^\\circ)}\\sec{40^\\circ}=\\cos{40^\\circ}\\sec{40^\\circ}$. We also know that $\\cos{\\theta}\\sec{\\theta}=1$. Thus: $\\cos{40^\\circ}\\sec{40^\\circ}=1$.", "## a1234 one year ago Evaluate the two trigonometric functions using its period as an aid. 1. cos (9pi/2) = ______ Answer - 0 2. cos (-9pi/2) = ______ Answer - 0 1. anonymous $\\cos \\frac{ 9 \\pi }{ 2 }=\\cos \\left( 4 \\pi+\\frac{ \\pi }{ 2 } \\right)=\\cos \\frac{ \\pi }{ 2 }=0$ $\\cos \\left( -\\theta \\right)=\\cos\\theta\\$ 2. a1234 So are both my answers correct? 3. anonymous yes", "cosine functions. #### Graphical For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of$t. t.$ 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. #### Technology For the following exercises, use a graphing calculator to evaluate. 80. $sin 5π 9 sin 5π 9$ 81. $cos 5π 9 cos 5π 9$ 82. $sin π 10 sin π 10$ 83. $cos π 10 cos π 10$ 84. $sin 3π 4 sin 3π 4$ 85. $cos 3π 4 cos 3π 4$ 86. $sin 98° sin 98°$ 87. $cos 98° cos 98°$ 88. $cos 310° cos 310°$ 89. $sin 310° sin 310°$ #### Extensions For the following exercises, evaluate. 90. $sin( 11π 3 ) cos( −5π 6 ) sin( 11π 3 ) cos( −5π 6 )$ 91. $sin( 3π 4 ) cos( 5π 3 ) sin( 3π 4 ) cos( 5π 3 )$ 92. $sin( − 4π 3 ) cos( π 2 ) sin( − 4π 3 ) cos( π 2 )$ 93. $sin( −9π 4 ) cos( −π 6 ) sin( −9π 4 ) cos( −π 6 )$ 94. $sin( π 6 ) cos( −π 3 ) sin( π 6 ) cos( −π 3 )$ 95. $sin( 7π", "cos(-x)= cos(x) so $[-2\\pi/3, 0]$ will also work- together, $[-2\\pi/3, 2\\pi/3]$. But, finally, cosine has period $2\\pi$: $cos(x)\\ge -1/2$ for all x in $[-2\\pi/3+ 2k\\pi, 2\\pi/3+ 2k\\pi]$ for k any integer, exactly what you have!", "## Trigonometry 7th Edition $\\theta=\\{60^o,180^o,300^o\\}$ $cos(3\\theta)=-1$ $3\\theta=cos^{-1}(-1)$ We know $cos(\\theta)$ is negative in quardent $II$ and quardent $III$ The period of the cosine function is $360^o$ $3\\theta=180^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;3\\theta=360^o+180^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;3\\theta=720^o+180^o\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=60^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;\\theta=180\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;\\theta=300^o\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=\\{60^o,180^o,300^o\\}$", "and cosine graph, the tangent graph has a period of 180°, so you can simply add or subtract 180° to the base solution. Solution 2 is: $56.3 + 180 = 236^{\\circ} \\text{ (3s.f.)}$ The two solutions for $2tanx = 3$, where 0° ≤ x ≤ 360° are: $x = 56.3^{\\circ} \\text{ and } 236^{\\circ} \\text{ (3s.f.)}$ ### Example Question 3 Solve $\\boldsymbol{5\\cos{y} + 7 = 10}$ for -360° ≤ y ≤ 360°, giving your answer correct to 1 decimal place. Start by rearranging the equation so it is in a similar for to the previous questions: $5\\cos{y} + 7 = 10 \\\\ 5\\cos{y} = 3 \\\\ \\cos{y} = 0.6$ Next, find the base solution: $y = \\cos{^{-1}}(0.6) = 53.1^{\\circ}$ Then, consider the cosine graph for the given interval to look for other solutions: Use the symmetry of the graph to find the second positive solution. You can see that solution 2 is the same distance from 360° as solution 1 is from 0°, so solution 2 is: $360-53.1=306.9^{\\circ} \\text{ (1d.p.)}$ You can then use the periodic nature of the graph to find solutions 3 and 4. Since the graph repeats every 360°, you can add or subtract 360° from your existing solutions to find others in the given interval. Therefore, solutions 3 and 4 are: $53.1 -360 =-306.9^{\\circ}", "-\\cos^{-1}(x)$ $-x = cos(-\\cos^{-1}(x))$ the cosine is an even function so $\\cos(-x)=\\cos(x)$ thus $-x = \\cos(\\cos^{-1}(x)) = x$ and thus as Country Boy keenly noted your statement isn't true. skeeter April 18th, 2018 11:10 AM evaluate the cosine of each side ... $\\cos[\\cos^{-1}(-x)] = -x$ $\\cos[\\pi - \\cos^{-1}(x)] = \\cos(\\pi)\\cos[\\cos^{-1}(x)] + \\sin(\\pi)\\sin[\\cos^{-1}(x)] = (-1)(x) + 0 = -x$ All times are GMT -8. The time now is 06:51 AM. Copyright © 2019 My Math Forum. All rights reserved.", "## Elementary Technical Mathematics The value of $56.1 \\%$ of $360{}^\\circ$ is $202{}^\\circ$. \\begin{align} & 56.1 \\% \\times 360{}^\\circ =\\frac{561}{1000}\\times 360{}^\\circ \\\\ & =201.96{}^\\circ \\end{align} Now round off the result, which gives $202{}^\\circ$. Thus, the value of $56.1 \\%$ of ${{360}^{\\circ }}$is $202{}^\\circ$.", "Given that the period of cosecant is $$2\\pi,$$ we get $\\csc \\left( { - \\frac{{27\\pi }}{4}} \\right) = \\csc \\left( {\\frac{{5\\pi }}{4} - 2\\pi \\times 4} \\right) = \\csc \\frac{{5\\pi }}{4}.$ The angle $${\\frac{{5\\pi }}{4}}$$ is in the $$3\\text{rd}$$ quadrant, in which cosecant is negative. The reference angle of $${\\frac{{5\\pi }}{4}}$$ is $${\\frac{{\\pi }}{4}}.$$ Then we have $\\csc \\left( { - \\frac{{27\\pi }}{4}} \\right) = \\csc \\frac{{5\\pi }}{4} = - \\csc \\frac{\\pi }{4} = - \\sqrt 2 .$ ### Example 3. Find the value of the expression $\\cos {630^\\circ} - \\sin {1470^\\circ} - \\cot {1125^\\circ}.$ Solution. Cosine and sine are periodic functions, with period $$360^\\circ.$$ Therefore $\\cos {630^\\circ} = \\cos \\left( {{{270}^\\circ} + {{360}^\\circ}} \\right) = \\cos {270^\\circ} = 0,$ $\\sin {1470^\\circ} = \\sin \\left( {{{30}^\\circ} + {{1440}^\\circ}} \\right) = \\sin \\left( {{{30}^\\circ} + {{360}^\\circ} \\times 4} \\right) = \\sin {30^\\circ} = \\frac{1}{2}.$ The period of cotangent is $$180^\\circ.$$ Hence, $\\cot {1125^\\circ} = \\cot \\left( {{{45}^\\circ} + {{1080}^\\circ}} \\right) = \\cot \\left( {{{45}^\\circ} + {{180}^\\circ} \\times 6} \\right) = \\cot {45^\\circ} = 1.$ Substitute these values into the initial expression: $\\cos {630^\\circ} - \\sin {1470^\\circ} - \\cot {1125^\\circ} = 0 - \\frac{1}{2} - 1 = - \\frac{3}{2}.$ ### Example 4. Find the value of the expression $\\tan {1800^\\circ} - \\sin {495^\\circ} + \\cos {945^\\circ}.$ Solution. Calculate the tangent:", "cosine have different signs 4. $\\cos(330^\\circ)$ 5. Possible response: $\\cos(45^\\circ) = \\cos(315^\\circ); \\cos(60^\\circ) = \\cos(300^\\circ)$ 6. $\\tan(225^\\circ)$ 7. Possible response: $\\tan(60^\\circ) = \\tan(240^\\circ), \\tan(30^\\circ) = \\tan(210^\\circ)$ 8. Answers will vary. Sample response: The values in the first quadrant are repeated in the other quadrants, but have different signs. 9. Answers will vary. Sample response: The values of sine and cosine switch within a quadrant, such as $\\sin(30^\\circ) = \\cos(60^\\circ)$, but the signs may be opposite in some quadrants. Extension $\\theta$ $\\sin \\theta$ $\\cos \\theta$ $\\tan \\theta$ $30^\\circ$ 1.155 2 1.732 $45^\\circ$ 1.414 1.414 1 $60^\\circ$ 2 1.155 0.577 $90^\\circ$ Undefined 1 0 $120^\\circ$ -2 1.155 -1.732 $135^\\circ$ -1.414 1.414 -1 $150^\\circ$ -1.155 2 -1.732 $180^\\circ$ -1 Undefined Undefined $210^\\circ$ -1.155 -2 1.732 $225^\\circ$ -1.414 -1.414 1 $240^\\circ$ -2 -1.155 0.577 $270^\\circ$ Undefined -1 0 $300^\\circ$ 2 -1.155 -0.577 $315^\\circ$ 1.414 -1.414 -1 $330^\\circ$ 1.144 -2 -1.732 $360^\\circ$ 1 Undefined Undefined Feb 23, 2012 Nov 04, 2014", "From $180^\\circ$ to $360^\\circ$, you remember, it was always negative, and you can readily see, it has a minimum value of $-1$ at $270^\\circ$. Beyond $360^\\circ$, of course, it just repeats. The function is what we call an oscillatory function, and in your studies in physics, you will find it most important to appreciate this property, particularly in the study of waves and alternating current in electricity and electronics. How does the cosine function behave? Panel 10 again shows you. At $0^\\circ$, $q = p$ and so the value is $1$. At $90^\\circ$, $p = 0$, and so $\\cos(90^\\circ) = 0$. This is shown in Panel 10b. You can work out the rest of it for yourself. You see that the cosine function is exactly the same as the sine function if you slide the sine function graph back $90^\\circ$. This says that $\\sin(90^\\circ + q) = \\cos (\\theta)$. The tangent curve looks quite different. Panel 11 shows that $\\tan(0) = r/p = 0$, but $\\tan(90^\\circ) = r/p = \\infty$ since $r$ is finite but $p$ has gone to $0$. Panel 11 shows this function. The tangent function is a repeating one but not oscillatory. One last point which you will find used many times are the approximate values of these functions when $\\theta$ is very small. Look", "cosine rule to get $a^2 = 2.618$approximately equal to $1 + \\phi$ and from here recall that $1 + \\phi = \\phi^{2}$it is easy to get $a = \\phi$. - 7 years ago Log in to reply Right. This is the way I first discovered it. - 7 years ago Log in to reply this is the way I think - 7 years ago Log in to reply I will attach an image after I know how to attach an image - 7 years ago Log in to reply I attach the image already - 7 years ago Log in to reply it is very interesting to know that the equivalent resistance of thiscircuit where all the resistances are 1 ohm is GOLDEN RATIO. I think this is because of the existence of the ratio in the radical form - 7 years ago Log in to reply Very interesting indeed. - 7 years ago Log in to reply × Problem Loading... Note Loading... Set Loading...", "to $360^{\\circ}$ which sine has value of $\\frac 12$ We also know that the following rule holds in the first quadrant. $$\\sin \\alpha = \\cos (90^{\\circ} - \\alpha)$$ $$\\sin 30^{\\circ} = \\cos 60^{\\circ} = \\cos 300^{\\circ} = \\frac 12$$ These are the only angles from $0^{\\circ}$ to $360^{\\circ}$ which cosine has value of $\\frac 12$ And for angles of $540^{\\circ}$ to $720^{\\circ}$ no angle has sine value of $\\frac 12$, beacuse in this region sine function has negative values. While the only angle which has cosine value of $\\frac 12$ is $\\alpha = 660^{\\circ}$ -", "# Using Differentials to Approximate Cos 44 In this tutorial we shall look at the use of differentials to approximate the value of $\\cos {44^ \\circ }$. The nearest number to 44 whose cosine value can be taken is 45, so let us consider that $x = {45^ \\circ }$ and $\\delta x = dx = - {1^ \\circ }$. Now consider Since $y = \\cos x,\\,\\,\\delta y \\approx dy$, putting these values in equation (ii), we have Taking the differential of equation (i), we have Putting this value in equation (ii), we have", "you should know that the first (positive) zero of the cosine function is at α = 90°. Use the arccos function (or maybe you say the cos^(-1) function or the inverse function of the cosine): $314^\\circ \\cdot t -60^\\circ = 90^\\circ~\\Longrightarrow~ 314^\\circ \\cdot t = 150^\\circ~\\Longrightarrow~t\\approx 0.4777\\ s$ to ii) The next positive zero occurs at 270°. Your equation becomes now: $314^\\circ \\cdot t -60^\\circ = 270^\\circ~\\Longrightarrow~ 314^\\circ \\cdot t = 330^\\circ~\\Longrightarrow~t\\approx 1.0560\\ s$ The first operation took place at 0.4777 s. The operation time is 40 ms = 0.04 s. The next operation takes place at 1.0560 s. The elapsed time is: 1.0560 s - (0.4777 s + 0.04 s) = 0.5383 s 3. ## Thanks Hey, thanks for the reply, unfortunately I sat the paper yesterday before i got the reply and failed it miserably I think it was unfair because he (the lecturer - who is hated by students and staff alike) gave us no study material, no past papers or anything, then to make it worse, he totally changed the structure of this exam which hasn't been done by any of the lecturers in a long time. Not to worry, what's done is done.", "constant, since $\\,d\\,$, $\\,e\\,$ and $\\,f\\,$ are all known. For the moment, suppose that this right-hand side evaluates to (say) $\\,0.5\\,$. Then, we'd need to solve: $$\\cos D = 0.5$$ So the question is: What angle(s) have cosine equal to 0.5? Of course, there are infinitely many: any angle whose terminal point has an $x$-value of $\\,0.5\\,$. (Look right and/or below to see a bunch of them.) \\begin{alignat}{4} \\cos 60^\\circ &= 0.5& \\qquad \\cos(-60^\\circ) &= 0.5& \\qquad \\cos(-300^\\circ) &= 0.5& \\qquad \\cos 300^\\circ &= 0.5\\cr \\cos 420^\\circ &= 0.5& \\qquad \\cos(-420^\\circ) &= 0.5& \\qquad \\cos(660^\\circ) &= 0.5& \\qquad \\cos(-1740^\\circ) &= 0.5 \\end{alignat} and so on!! The arccosine function can be abbreviated as ‘$\\,\\arccos\\,$’, but is still pronounced ‘arc-co-sign’. The arccosine function helps us to ‘undo’ the cosine! Arccosine is a function: each input has exactly one output. When we ask for $\\,\\text{arccos}(0.5)\\,$, we can't be given all the angles with a cosine of $\\,0.5\\,$! We need $\\,\\text{arccos}(0.5)\\,$ to represent exactly one angle. So, which one? Answer: the one between $\\,0^\\circ\\,$ and $\\,180^\\circ\\,$! That is, by definition: $$\\text{arccos}(x) = \\text{the angle between \\,0^\\circ\\, and \\,180^\\circ\\, whose cosine is x}$$ So, what exactly is $\\,\\text{arccos}(x)\\,$? Answer: It is the angle between $\\,0^\\circ\\,$ and $\\,180^\\circ\\,$ that has a cosine of $\\,x\\,$. So: $$\\text{arccos}(0.5) \\ =\\ \\text{the angle between \\,0^\\circ\\, and \\,180^\\circ\\," ]

[ "+0 # law of cosines +5 390 2 +11 whats 400=169+121-286CosC . i keep getting 74.something and the answer is 112degrees37seconds? i can't find my calculator nessapierce9 Jan 30, 2012 Sort: #1 0 112.61986494804282480511 is what I come up with. Not sure how many decimals you need to round to, but that's it. my final equation was: C=acos((169+121-400)/286) Guest Jan 30, 2012 #2 +6925 0 $$400 = 169 + 121 - 286\\cos C\\\\ -286\\cos C = 400 - 169 - 121\\\\ \\cos C = -\\dfrac{110}{286}\\\\ C = 112.6198649^{\\circ} = 112^{\\circ}37'11.51''$$ the arccosine of a negative number is in the range 90° < x $$\\leqslant$$ 180° MaxWong Aug 21, 2016 ### 22 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "series: $\\cos 35^\\circ = \\cos \\left({35 \\over 180} \\pi \\right) \\approx \\cos 0.610865$ Then, substitute into the Taylor series of cosine above: $\\cos (0.610865) \\approx 1 - {0.610865^2 \\over 2!} + {0.610865^4 \\over 4!}$ Here we have written the 4th-degree Taylor polynomial, but this should be enough to show us something. The right side of the equation can be reduced to the four simple operations, so we can easily calculate its value: $\\cos (0.523599) \\approx 0.81922$ We can compare this to the value given by the calculator. The calculator's value, actually, is also an approximation obtained by a similar method, but we can expect it to be accurate for all displayed decimal places. $\\cos 35^\\circ = 0.81915 \\cdots$ So our approximating value agrees with the \"actual\" value to three decimal places, which is good accuracy for a basic approximation. As above, better accuracy can be attained by using more terms in the Taylor series. This result can be observed if we zoom in the graph at the intersection point, as shown in Figure 1. In a large graph, the functions look almost identical at the intersection point, but there is indeed a difference between these two functions, as the graph shows. # A More Mathematical Explanation Note: understanding of this explanation requires: *Calculus ## The general form of", "## Elementary Technical Mathematics The value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$. Step 1: Open Ti-83. Step 2: Press $cos$ button. Step 3: Enter the value like this, $\\cos \\left( 256{}^\\circ \\right)=-0.2419$ The obtained value is $-0.2419$. Hence the value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$.", "cos(-x)= cos(x) so $[-2\\pi/3, 0]$ will also work- together, $[-2\\pi/3, 2\\pi/3]$. But, finally, cosine has period $2\\pi$: $cos(x)\\ge -1/2$ for all x in $[-2\\pi/3+ 2k\\pi, 2\\pi/3+ 2k\\pi]$ for k any integer, exactly what you have!", "# Using Differentials to Approximate Cos 44 In this tutorial we shall look at the use of differentials to approximate the value of $\\cos {44^ \\circ }$. The nearest number to 44 whose cosine value can be taken is 45, so let us consider that $x = {45^ \\circ }$ and $\\delta x = dx = - {1^ \\circ }$. Now consider Since $y = \\cos x,\\,\\,\\delta y \\approx dy$, putting these values in equation (ii), we have Taking the differential of equation (i), we have Putting this value in equation (ii), we have", "+0 # law of cosines +5 813 2 +11 whats 400=169+121-286CosC . i keep getting 74.something and the answer is 112degrees37seconds? i can't find my calculator Jan 30, 2012 #1 0 112.61986494804282480511 is what I come up with. Not sure how many decimals you need to round to, but that's it. my final equation was: C=acos((169+121-400)/286) Jan 30, 2012 #2 +7531 0 $$400 = 169 + 121 - 286\\cos C\\\\ -286\\cos C = 400 - 169 - 121\\\\ \\cos C = -\\dfrac{110}{286}\\\\ C = 112.6198649^{\\circ} = 112^{\\circ}37'11.51''$$ the arccosine of a negative number is in the range 90° < x $$\\leqslant$$ 180° Aug 21, 2016", "$$g(360^{\\circ})$$ from $$f(360^{\\circ})$$. \\begin{align*} f(360^{\\circ}) - g(360^{\\circ}) &= 2 - 0 \\\\ &=2 \\end{align*} Given the following graph. State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = ( 90^{\\circ}; 1),~B = ( 180^{\\circ}; -3),~C = (270 ^{\\circ};-1 ) \\text{ and } D = ( 360^{\\circ}; 3)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$2$$ What is the amplitude of $$g(x)$$. $$3$$ Evaluate: $$f(90^{\\circ}) - g(90^{\\circ})$$ . Read off the value of $$f(90^{\\circ})$$ and $$g(90^{\\circ})$$ from the graph. Then subtract $$g(90^{\\circ})$$ from $$f(90^{\\circ})$$. \\begin{align*} f(90^{\\circ}) - g(90^{\\circ}) &= 1 - 0 \\\\ &=1 \\end{align*} Given the following graph: State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = (90 ^{\\circ};3 ),~B = (90 ^{\\circ};2 ),~C = (180 ^{\\circ}; -4) \\text{ and } D = ( 270^{\\circ};2)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$3$$ What is the amplitude of $$g(x)$$. $$2$$ Evaluate: $$f(270^{\\circ}) - g(270^{\\circ})$$ . Read off the value of $$f(270^{\\circ})$$ and $$g(270^{\\circ})$$ from the graph. Then subtract $$g(270^{\\circ})$$ from $$f(270^{\\circ})$$. \\begin{align*} f(270^{\\circ}) - g(270^{\\circ}) &= -3 - (-2)\\\\ &=-1 \\end{align*}", "Trigonometry 7th Edition $\\theta=\\{60^o+120^ok\\;\\;\\}$ $cos(3\\theta)=-1$ $3\\theta=cos^{-1}(-1)$ The period of the cosine function is $360^o$ $3\\theta=180^o+360^ok\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=60^o+120^ok\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=\\{60^o+120^ok\\;\\;\\},\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;when \\;\\;\\;K=\\{0,1,2,3\\}$", "1:23 You're plotting the function $\\cos n\\pi/4$, which has period $8$, and thus $8$ full periods from $0$ to $64$, but you're only substituting values from $0$ to $16\\pi$. Since $16\\pi\\approx50$, you're missing a bit less than two of the periods. From what it seems you're trying to do, you should be plotting the function from $0$ to $64$, i. e. replace 2*pi*8 by 64 in the first line. @Raven: A clue that your linspace() expression was faulty was that your n consists of $(0,16\\pi/63,\\dots,16\\pi)$. Replacing the n in cos(2*pi*8*n/64) with, for example, 16*pi/63 gives you something with a pi^2 within the cosine... – Ｊ. Ｍ. Sep 16 '11 at 4:59", "## Trigonometry 7th Edition $\\theta=\\{60^o,180^o,300^o\\}$ $cos(3\\theta)=-1$ $3\\theta=cos^{-1}(-1)$ We know $cos(\\theta)$ is negative in quardent $II$ and quardent $III$ The period of the cosine function is $360^o$ $3\\theta=180^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;3\\theta=360^o+180^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;3\\theta=720^o+180^o\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=60^o\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;\\theta=180\\;\\;\\;\\;or\\;\\;\\;\\;\\;\\;\\theta=300^o\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $\\theta=\\{60^o,180^o,300^o\\}$", "## Trigonometry 7th Edition $\\theta=60^{\\circ}+180^{\\circ}k,90^{\\circ}+180^{\\circ}k,120^{\\circ}+180^{\\circ}k$ where k is any integer. $2(\\cos2\\theta)^{2}+3\\cos2\\theta+1=0$ or, $(2cos2\\theta+1)(\\cos2\\theta+1)=0$ either $cos2\\theta=-\\frac{1}{2}$ or $-1$ For, $\\cos2\\theta=-\\frac{1}{2}$, $2\\theta=\\cos^{-1}(-\\frac{1}{2})$, Since, $\\cos2\\theta$ is negative in both quadrants 2 and 3, $2\\theta=180^{\\circ}-60^{\\circ}=120^{\\circ} \\,\\,or \\,\\,180^{\\circ}+60^{\\circ}=240^{\\circ},$ To find all values of $2\\theta$, we can add $360^{\\circ}k$ where k is any integer, since the period of cos function is $360^{\\circ}$. So, $2\\theta=120^{\\circ}+360^{\\circ}k\\,\\,or \\,\\,240^{\\circ}+360^{\\circ}k,$ So, $\\theta=60^{\\circ}+180^{\\circ}k\\,\\,or \\,\\,120^{\\circ}+180^{\\circ}k,$ For, $\\cos2\\theta=-1$, $2\\theta=\\cos^{-1}(-1)$, $2\\theta=180^{\\circ}$ To find all values of $2\\theta$, we can add $360^{\\circ}k$ where k is any integer, since the period of cos function is $360^{\\circ}$. So, $2\\theta=180^{\\circ}+360^{\\circ}k\\,\\,$ So, $\\theta=90^{\\circ}+180^{\\circ}k\\,\\,$ So, all degree solutions are $\\theta=60^{\\circ}+180^{\\circ}k,90^{\\circ}+180^{\\circ}k,120^{\\circ}+180^{\\circ}k$", "Precalculus (10th Edition) $0$ I know that $\\sin$ is an odd function, which means $f(-\\theta)=-f(\\theta).$ Also, I know that $\\cos$ has a period of $360^\\circ$, hence $\\cos{\\theta}=\\cos{(\\theta-360^\\circ)}.$ Therefore $\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{358^\\circ}+\\sin{359^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{(358^\\circ-360^\\circ)}+\\sin{(359^\\circ-360^\\circ)}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{-2^\\circ}+\\sin{-1^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...-\\sin{2^\\circ}-\\sin{1^\\circ}=0$.", "all the other solutions for $0^\\circ \\leq \\theta \\leq 500^\\circ.$ As well as the above solutions, we can find others by adding integer multiples of $360^\\circ$. We find that $119.5^\\circ$, $346.7^\\circ$ and $479.5^\\circ$ are all the solutions within the given range. 1. Hence give the positive values of $\\theta$ less than $500^\\circ$ for which $3\\cos \\theta + 4\\sin \\theta > 2$. A graph of the function $y=3\\cos \\theta + 4\\sin \\theta$ looks like this. Note that the function switches between being above and below $y=2$ at each of the intersections found in part (c). When $\\theta = 0$, the function is greater than $2$. This must mean that $3\\cos \\theta + 4\\sin \\theta > 2$ on the intervals $0 < \\theta < 119.5^\\circ$ and $346.7^\\circ < \\theta < 479.5^\\circ$.", "From $180^\\circ$ to $360^\\circ$, you remember, it was always negative, and you can readily see, it has a minimum value of $-1$ at $270^\\circ$. Beyond $360^\\circ$, of course, it just repeats. The function is what we call an oscillatory function, and in your studies in physics, you will find it most important to appreciate this property, particularly in the study of waves and alternating current in electricity and electronics. How does the cosine function behave? Panel 10 again shows you. At $0^\\circ$, $q = p$ and so the value is $1$. At $90^\\circ$, $p = 0$, and so $\\cos(90^\\circ) = 0$. This is shown in Panel 10b. You can work out the rest of it for yourself. You see that the cosine function is exactly the same as the sine function if you slide the sine function graph back $90^\\circ$. This says that $\\sin(90^\\circ + q) = \\cos (\\theta)$. The tangent curve looks quite different. Panel 11 shows that $\\tan(0) = r/p = 0$, but $\\tan(90^\\circ) = r/p = \\infty$ since $r$ is finite but $p$ has gone to $0$. Panel 11 shows this function. The tangent function is a repeating one but not oscillatory. One last point which you will find used many times are the approximate values of these functions when $\\theta$ is very small. Look", "can see quite clearly that • just like the cosine curve: it oscillates between its minimum value $$-1$$ and maximum value $$1$$. This confirms the fact that the range of $$sin(x)$$ is: $-1 \\leq sin(x) \\leq 1$ • it is periodic, with period $$T = 360^{\\circ}$$. Indeed no matter which point we look at on the curve: its • It reaches its maximum value $$cos(x) = 1$$ when $$x=90^{\\circ}, 450^{\\circ},810^{\\circ}, \\dots$$. • It equals to zero $$sin(x) = 0$$ and cuts the $$x$$-axis when $$x=0^{\\circ}, 180^{\\circ},360^{\\circ}, \\dots$$ (all $$x$$-intercepts are $$180^{\\circ}$$ apart from each other). ## Method: Sketching $$y=cos(x)$$ & $$y=sin(x)$$ To sketch both $$y=cos(x)$$ and $$y=sin(x)$$ the idea is to plot: • points at which the curve reaches the maximum height of $$1$$. For $$y=sin(x)$$: that' • points at which the curve reaches the minimum height of $$-1$$, • points at which the curve cuts the $$x$$-axis. Once this is done, we draw a smooth curve passing through these points. This method is best explained in the tutorial below.", "New Zealand Level 7 - NCEA Level 2 # Period changes for sine and cosine Lesson ## Preliminaries We define the $\\cos$cos and $\\sin$sin functions as the horizontal and vertical coordinates of a point that moves on the unit circle. In the diagrams below, this is shown for an angle $\\alpha$α in the first and second quadrants. If we imagine the point moving anticlockwise on the unit circle so that the radius from the point makes an ever-increasing angle with the positive horizontal axis, eventually the angle exceeds $360^\\circ$360°; but the values of the $\\cos$cos and $\\sin$sin functions repeat the values of the coordinates from the angle $360^\\circ$360° smaller. We say $\\sin$sin and $\\cos$cos are periodic functions with period $360^\\circ$360°. Thus, for any angle $\\alpha$α, there is a sequence of angles with the same value of $\\sin\\alpha$sinα. $...,\\alpha-720^\\circ,\\alpha-360^\\circ,\\alpha,\\alpha+360^\\circ,\\alpha+720^\\circ,\\alpha+1080^\\circ,...$...,α720°,α360°,α,α+360°,α+720°,α+1080°,... ## Transformations involving period Again, consider the angle $\\alpha$α made by the point moving around the unit circle. If a new angle $\\alpha'$α is defined by $\\alpha'=k\\alpha$α=kα, We know that $\\sin\\alpha'$sinα has period $360^\\circ$360°, but we see that $\\alpha'$α reaches $360^\\circ$360° when $\\alpha=\\frac{360^\\circ}{k}$α=360°k. So, $\\sin k\\alpha$sinkα and $\\cos k\\alpha$coskα must have period $\\frac{360^\\circ}{k}$360°k with respect to $\\alpha$α. #### Example 1 The function $\\sin2x$sin2x begins to repeat when $2x=360^\\circ$2x=360°. That is, when $x=180^\\circ$x=180°. So, $\\sin2x$sin2x has period $180^\\circ$180°. The period is", "and in fact is much harder, than solving your original equation $\\cos3\\theta = \\tfrac{1}{2}$. Solving this equations is really very easy. If you want to find all solutions with $0 \\le \\theta \\le 360^{\\circ}$ then you need to find all solutions with $0 \\le 3\\theta \\le 1080^{\\circ}$. You can do this by plotting a graph, or by drawing a CAST quadrant diagram. You will find that: $$3\\theta = \\ldots,60^{\\circ}, 300^{\\circ}, 420^{\\circ}, 660^{\\circ}, \\ldots$$ Dividing both sides by $3$ gives $$\\theta = \\ldots, 20^{\\circ}, 100^{\\circ}, 140^{\\circ}, 220^{\\circ}, \\ldots$$ - Hint: Let $\\phi=3\\theta$. What angles $\\phi$ between $0^\\circ$ and $(3)(360^\\circ)$ satisfy $\\cos\\phi=\\frac{1}{2}$? Remark: What is needed here could be called formulas, though I think of them as facts about the geometry of the cosine function. We switch to radians since you used them in your comment. We use the periodicity of cosine, which can be written as the formula $\\cos(x+2k\\pi)=\\cos x$. The fact that $\\cos(\\pi/3)=\\frac{1}{2}$ is one of those \"special angle\" facts that at one stage one is expected to remember. We also used the fact that $\\cos(2\\pi-x)=\\cos x$, which is the fact that the cosine function is symmetric about $\\pi$. - I have the 5 solutions which are theta= pie/9, 5pie/9, 7pie/9, 11pie/9, and 13pie/9. I just can't remember what formula I am supposed to use. – rick Mar", "recalling the known exact values, we would recognise that $\\alpha=60^\\circ$α=60° and $\\beta=45^\\circ$β=45° are solutions in the first quadrant. There should also be a second quadrant solution to $\\sin\\alpha=\\frac{\\sqrt{3}}{2}$sinα=32. It is found by subtracting the first quadrant solution from $180^\\circ$180°. That is, $\\alpha=180^\\circ-60^\\circ=120^\\circ$α=180°60°=120°. This can be verified by reference to the unit circle diagram: it is always true that $\\sin\\alpha\\equiv\\sin\\left(180^\\circ-\\alpha\\right)$sinαsin(180°α) We now have the two solutions $\\alpha=60^\\circ$α=60° and $\\alpha=120^\\circ$α=120°. Any number of further solutions can be obtained by adding or subtracting multiples of $360^\\circ$360° to these basic solutions. For the equation $\\cos\\beta=\\frac{1}{\\sqrt{2}}$cosβ=12, there should be a fourth quadrant solution as well as the one in the first quadrant. We find it by subtracting the first quadrant $\\beta$β from $360^\\circ$360°. Thus, $\\beta=45^\\circ$β=45° or $\\beta=315^\\circ$β=315° . Again, this can be verified by referring to the unit circle diagram. It is always true that $\\cos\\beta\\equiv\\cos\\left(360^\\circ-\\beta\\right)$cosβcos(360°β) As with the sine function, as many further solutions as are needed are found by adding or subtracting multiples of $360^\\circ$360° to the basic solutions. We say the sine and cosine functions are periodic with a period of $360^\\circ$360° because the function values recur regularly at that interval. #### Example: Find all the solutions that lie between $0^\\circ$0° and $360^\\circ$360° of the equation $\\sin2\\theta=\\frac{\\sqrt{3}}{2}$sin2θ=32 In the first quadrant, we have $2\\theta=60^\\circ$2θ=60°. It must also be true that", "with values and so we plot each of these points and join them with a smooth curve. State the coordinates at $$E$$ and the range of the function. To find the coordinates of $$E$$ we read the value off the graph. To find the range we note that this is a cosine graph and so the maximum value occurs at $$0^\\circ$$ (and at $$360^\\circ$$). The minimum value occurs at $$180^\\circ$$. So we read off the value of $$y$$ at $$0^\\circ$$ and at $$180^\\circ$$. Therefore $$E(180^{\\circ};-2) \\text{ and } -2 \\le y \\le 0$$. To find the coordinates of $$E$$ we read the value off the graph. To find the range we note that this is a sine graph and so the maximum value occurs at $$90^\\circ$$. The minimum value occurs at $$270^\\circ$$. So we read off the value of $$y$$ at $$90^\\circ$$ and at $$270^\\circ$$. Therefore $$E(360^{\\circ};2) \\text{ and } 0 \\le y \\le 4$$ To find the coordinates of $$E$$ we read the value off the graph. To find the range we note that this is a sine graph and so the maximum value occurs at $$90^\\circ$$. The minimum value occurs at $$270^\\circ$$. So we read off the value of $$y$$ at $$90^\\circ$$ and at $$270^\\circ$$. Therefore $$E(90^{\\circ};-\\text{0,5}) \\text{ and } -\\text{2,5} \\le y \\le -\\text{0,5}$$ To", "# If $\\sin32^{\\circ}=k$ and $\\cos x=1-2k^2, \\alpha$ and $\\beta$ are the two values of $x$ between $0^{\\circ}$ and $360^{\\circ}$ with $\\alpha<\\beta$ A)$$\\alpha+\\beta=180^{\\circ}$$ B)$$\\beta-\\alpha=200^{\\circ}$$ C)$$\\beta=4\\alpha+40^{\\circ}$$ D)$$\\beta=5\\alpha-20^{\\circ}$$ I solved it by taking $$\\cos x=1-2\\sin^232^{\\circ}$$ Therefore $$x=64$$ So the two values of $$x$$ can be $$64$$ and $$334$$, so B) should be the right answer, but the answer is actually c. Where am I going wrong? Note that: $$\\cos x=\\cos 64^\\circ \\Rightarrow x=\\pm64^\\circ+ 360^\\circ n \\\\ n=0 \\Rightarrow x=64^\\circ\\\\ n=1 \\Rightarrow x=-64^\\circ +360^\\circ =296^\\circ$$ $$360-64=296=\\beta$$ $$4\\alpha=4\\cdot64=256$$ Since you found that $$x=64$$, you could draw the reference triangle for $$\\cos(64^\\circ)$$ Taking $$360-64=296$$ would produce the only angle for $$0\\leq x \\leq 2\\pi$$ which would have the same cosine value. The reference triangle for $$\\cos(296^\\circ)$$ is If you wanted to compute the values of $$\\cos(64^\\circ)$$ and $$\\cos(296^\\circ)$$ then you would find that they are both $$\\sin\\big(\\frac{13\\pi}{90}\\big)$$." ]

[ "\\frac{AC}5$ or $AB = \\frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \\frac{50}7$. We can plug this back in to find $AB = \\frac{20}7$. 2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \\frac{20\\sqrt{3}}3, CP = \\frac{10\\sqrt{3}}3$. Find the value of $m\\angle BAP - m\\angle CAP$. Solution: First, we notice that $\\frac{AB}{BP}=\\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0. 3. Part (b), 1959 IMO Problems/Problem 5.", "? Free Version Moderate # Trapezoids and Triangles ACTMAT-OTVKEY In trapezoid ABCD below, diagonals${\\overline {AC}}$ and${\\overline {BD}}$ intersect at E. If $AE = 4$, $BE = 6$, $AB = 8$, and $DC = 24$, what is the length of${\\overline {EC}}$? A $6$ B $8$ C $10$ D $12$ E $18$", "# Geometry problem In triangle ABC, $\\measuredangle$ ABC is bisected by $\\overline{BE}$ and $\\measuredangle$ ACB is bisected by $\\overline{CD}$. $\\overline{BE} = \\overline{CD} = x$. Show that $\\overline{BD} = \\frac{ac}{a + b}$ where $\\overline{AB} = c$, $\\overline{BC} = a$ and $\\overline{AC} = b.$ A high school student asked me this question today and I was completely stumped. He goes to a magnet school in the city. The topics he went over in class were Stewart's Theorem and Ptolemy's Theorem. - Do you know of the angle bisector theorem? – Sawarnik Dec 11 '14 at 9:47 Since $$\\dfrac{BD}{DA}=\\frac{\\triangle CBD}{\\triangle CDA}=\\frac{CB\\cdot CD\\sin\\angle BCD}{CD\\cdot CA\\sin\\angle DCA}=\\frac{BC}{CA}=\\frac ab$$ therefore $$BD=c\\cdot\\frac a{a+b}$$ Q.E.D", "+0 # heElp 0 90 1 +4 In triangle $XYZ$, $\\angle X = 60^\\circ$ and $\\angle Y = 45^\\circ$. The bisector of $\\angle X$ intersects $\\overline{YZ}$ at $W.$ If $XW = 24,$ then find the area of triangle $XYZ$. Jun 22, 2020", "}} + {\\text{ 4 }} = \\overline {DG}$$ $${\\text{14}} = \\overline {DG}$$ 3432 x Each figure shows a triangle with one of its angle bisectors. This free worksheet contains 10 assignments each with 24 questions with answers. Example of one question: Watch below how to solve this example: 2744 x Each figure shows a triangle with one of its angle bisectors. This free worksheet contains 10 assignments each with 24 questions with answers. Example of one question: Watch below how to solve this example: 2095 x Each figure shows a triangle with its three angle bisectors intersecting at point P. This free worksheet contains 10 assignments each with 24 questions with answers. Example of one question: Watch below how to solve this example: ### Geometry Circles Congruent Triangles Constructions Parallel Lines and the Coordinate Plane Properties of Triangles ### Algebra and Pre-Algebra Beginning Algebra Beginning Trigonometry Equations Exponents Factoring Linear Equations and Inequalities Percents Polynomials", "# Triangle and Bisector Geometry Level 3 In $$\\triangle ABC$$ above, $$M$$ is the midpoint of side $$BC,$$ line segment $$AN$$ bisects $$\\angle BAC ,$$ and $$\\overline{ AN } \\bot \\overline{ BN }.$$ If $$\\overline{AB} =14$$ and $$\\overline{AC} =19,$$ what is $$\\overline{MN}?$$ ×", "# Extended Bisector Geometry Level 3 Triangle $$ABC$$ is inscribed in a unit circle. The three bisectors of the angle $$A$$, $$B$$ and $$C$$ are extended to intersect the circle at $$A'$$, $$B'$$ and $$C'$$ respectively. Determine the value of $\\frac{AA'\\cos\\frac{A}{2}+BB'\\cos\\frac{B}{2}+CC'\\cos\\frac{C}{2}}{\\sin A+\\sin B+\\sin C}$ × Problem Loading... Note Loading... Set Loading...", "# Angle Bisector of an Inscribed Triangle Geometry Level 3 In the above figure $$\\triangle$$ABC is inscribed in the circle and DF is the angle bisector of $$\\angle$$ACE. If AD = 8 and AC = 6, find BD Note: Figure is not drawn to scale ×", "the angle bisector theorem, $\\frac{AB}2 = \\frac{AC}5$ or $AB = \\frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \\frac{50}7$. We can plug this back in to find $AB = \\frac{20}7$. 2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \\frac{20\\sqrt{3}}3, CP = \\frac{10\\sqrt{3}}3$. Find the value of $m\\angle BAP - m\\angle CAP$. Solution: First, we notice that $\\frac{AB}{BP}=\\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0. 3. Part (b), 1959 IMO Problems/Problem 5.", "# Cool problem Geometry Level 3 Let ABC be a triangle and D be a point on the segment BC such that DC=2BD. Let E be the mid-point of AC. Let AD and BE intersect in P. Find $\\frac{BP}{PE} \\times \\frac{AP}{PD}$ ×", "$$\\frac { 12 }{ 5 }$$ (or) 2.4 cm and y = $$\\frac { 10 }{ 3 }$$ (or) 3.3 cm Question 3. O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA. Solution: In ∆ABC the bisector meets AB at D, BC at E and AC at F. The angle bisector AO, BO and CO intersect at “O”. By Cevas Theorem $$\\frac { AD }{ DB }$$ × $$\\frac { BE }{ EC }$$ × $$\\frac { CF }{ AF }$$ = 1 AD × BE × CF = DB × EC × AF Hence it is proved Question 4. In the figure, ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle. Solution: ∠B = ∠C (Given AB = AC) AD + DB = AE + EC BD = EC (Given AD = AE) DE parallel BC Since AEC is a straight line. ∠AED + ∠CED = 180° ∠CBD + ∠CED = 180°", "# Difference between revisions of \"2011 AIME I Problems/Problem 4\" ## Problem 4 In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\\overline{BK}$ and $\\overline{AL}$, respectively. Find $MN$. ## Solution Extend ${CM}$ and ${CN}$ such that they intersects lines ${AB}$ at points $P$ and $Q$, respectively. Since ${BM}$ is the angle bisector of angle B,and ${CM}$ is perpendicular to ${BM}$ ,so , $BP=BC=120$, M is the midpoint of ${CP}$ .For the same reason,$AQ=AC=117$,N is the midpoint of ${CQ}$. Hence$MN=\\frac{PQ}{2}$.But $PQ=BP+AQ-AB=120+117-125=112$,so$MN=\\boxed{56}$.", "clarification and other similar things (and so this is spoiling the fun for those who read comments, but not answers). Posting this as an answer would be definitely a better idea ;-) – dtldarek May 23 '12 at 23:20 I remember a similar problem but not where I saw it. Can you remember where you saw this one? – plm May 23 '12 at 23:27 $Q$ is the intersection of the two angle bisectors of the triangle $ABP$ at $A$ and $B$. Therefore, $Q$ is the incenter of $ABP$ and $PQ$ is the bisector of $\\angle BPA$ as claimed.", "Internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. In Triangle ABC, if AD is the bisector of ∠A then $\\frac{AB}{AC}=\\frac{BD}{DC}$ SIMILARITY OF TRIANGLES 🙏", "# Math Help - Triangle 1. ## Triangle Given a $\\triangle ABC$ and $D$ a point on the extension of side $BC$ (from $B$) such that $\\overline{BD}=\\overline{BA}$ and let $M$ be the mid point of $\\overline{AC}$. The angle bisector of $\\measuredangle\\,ABC$ intersects to $\\overline{DM}$ in $P$. Prove that $\\measuredangle\\,BAP=\\measuredangle\\,ACB.$ • Let $N$ be the foot of the angle bisector of $B$ on the side $AC$. Since $\\overline{AD}\\parallel\\overline{BN}\\implies\\triang le ADM\\sim\\triangle NPM$, it yields $\\frac{{\\overline {AD} }} {{\\overline {NP} }} = \\frac{{\\overline {AM} }} {{\\overline {MN} }}$ • On the other hand, $\\triangle BCN\\sim\\triangle DCA$, then $\\frac{{\\overline {BN} }} {{\\overline {AD} }} = \\frac{{\\overline {CN} }} {{2 \\cdot \\overline {AM} }} \\implies \\frac{{2 \\cdot \\overline {MN} }} {{\\overline {NP} }} = \\frac{{\\overline {CN} }} {{\\overline {BN} }} \\implies \\frac{{\\overline {AN} }} {{\\overline {CN} }} = \\frac{{\\overline {BP} }} {{\\overline {BN} }}$ • Since $\\frac{{\\overline {AN} }} {{\\overline {CN} }} = \\frac{{\\overline {AB} }} {{\\overline {BC} }} \\implies \\frac{{\\overline {BP} }} {{\\overline {BN} }} = \\frac{{\\overline {AB} }} {{\\overline {BC} }}$ We have $\\measuredangle\\,ABN=\\measuredangle\\,CBN\\implies\\tr iangle{ABP}\\sim\\triangle{CBN}\\,\\therefore\\,\\measur edangle\\,BAP=\\measuredangle\\,ACB$", "intersecting at P. If $\\angle BPC = 4x - 5^\\circ$ and $\\angle APD = 3x + 15^\\circ,$ find the value of $x$. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q5 Subjective Medium In a triangle ABC, the angle bisectors of angle ABC and ACB meet at point O. If $\\angle$BAC $=32^o$, find $\\angle$ BOC. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "angle triangle sin$$\\theta$$ = $$p\\over h$$; cos$$\\theta$$ = $$b\\over h$$; tan$$\\theta$$ = $$p\\over b$$; cosec$$\\theta$$ = $$h\\over p$$; sec$$\\theta$$ = $$h\\over b$$ and cot$$\\theta$$ = $$b\\over p$$ where ‘p’ is perpendicular ; ‘b’ is base and …", "previous equation to get $\\frac{\\sin \\angle ADB}{\\sin \\angle BAD} = \\frac bn$. We conclude that $\\frac{\\sin \\angle ADB}{\\sin \\angle BAD} = \\frac cm = \\frac bn$, which was what we wanted. In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds. ## Examples 1. Let ABC be a triangle with angle bisector AD with D on line segment BC. If $BD = 2, CD = 5,$ and $AB + AC = 10$, find AB and AC. Solution: By the angle bisector theorem, $\\frac{AB}2 = \\frac{AC}5$ or $AB = \\frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \\frac{50}7$. We can plug this back in to find $AB = \\frac{20}7$. 2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \\frac{20\\sqrt{3}}3, CP = \\frac{10\\sqrt{3}}3$. Find the value of $m\\angle BAP - m\\angle CAP$. Solution: First, we notice that $\\frac{AB}{BP}=\\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0. 3. Part (b), 1959 IMO Problems/Problem 5. ACS WASC ACCREDITED SCHOOL Our Team Our History Jobs", "In triangle ABC there is point D on line segment AB, so that $|AD|:|DB|=2:1$, and there is point E on BC, so that $|BE|:|EC|=1:4$. Point F is the point where line segments AE and DC intersect. Express the relationship (ratio) between $|AF|:|FE|$ ( $\\vec{AB}=\\vec{a}$, $\\vec{BC}=\\vec{b}$).", "BC. The line DE cuts AC extended at F. If AD=5, CE=3 and EF=8 find DE. +1 vote If the medians of triangle ABC intersect at P , then evaluate `````` AB^2+AC^2+BC^2 ---------------- PA^2+PB^2+PC^2 `````` +1 vote In the figure AC is a common tangent of two touching circles. If B is touching point of both circles then find angle B." ]

[ "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "as the center and draw segment $ON$ perpendicular to $AM$, $ON\\cap AM=N$, link $OM$. Then we have $OM\\parallel AD$. So $\\angle DAM=\\angle OMA$. Since $AD=2AM=2OM=1$, we have $\\cos\\angle DAM=\\cos\\angle OMP=\\frac{2}{\\sqrt{5}}$. As a result, $NM=OM\\cos\\angle OMP=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{5}}=\\frac{1}{\\sqrt{5}}.$ Thus $PM=2NM=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. Since $AM=\\frac{\\sqrt{5}}{2}$, we have $AP=AM-PM=\\frac{\\sqrt{5}}{10}$. Put $\\boxed{B}$. ~FANYUCHEN20020715 ## Solution 5 (Similar Triangles) Call the midpoint of $\\overline{AB}$ point $N$. Draw in $\\overline{NM}$ and $\\overline{NP}$. Note that $\\angle{NPM}=90^{\\circ}$ due to Thales's Theorem. $[asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw(circle((0.5,0.5),0.5)); draw((0,0)--(0,0.5)--(1,0.5)--cycle); label(\"A\",(0,0),SW); label(\"B\",(0,1),NW); label(\"C\",(1,1),NE); label(\"D\",(1,0),SE); label(\"M\",(1,0.5),E); label(\"P\",(0.2,0.1),S); label(\"N\",(0,0.5),W); draw((0,0.5)--(0.2,0.1)); markscalefactor=0.007; draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); [/asy]$ Using the Pythagorean theorem, $AM=\\frac{\\sqrt{5}}{2}$. Now we just need to find $AP$ using similar triangles. $$\\triangle APN\\sim\\triangle ANM\\Rightarrow\\frac{AP}{AN}=\\frac{AN}{AM}\\Rightarrow\\frac{AP}{\\frac{1}{2}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}}\\Rightarrow AP=\\boxed{\\textbf{(B) }\\frac{\\sqrt{5}}{10}}$$ ~QIDb602 ## Solution 6 (Intersecting Chords) Label the midpoint of $AD$ as $N$, and let $Q$ be the intersection of $ON$ and $AM$ Then $MQ=AQ=\\frac{1}{2}AM = \\frac{\\sqrt{5}}{4}$ and $NQ=\\frac{1}{4}$ Then $\\frac{1}{4}*\\frac{3}{4}=PQ*QM$ and $AP=AQ-PQ$ ~ERMSCoach ## Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem) Let $N$ be the midpoint of $\\overline{AB},$ from which $\\angle ANM=90^\\circ.$ Note that $\\angle NPM=90^\\circ$ by the Inscribed Angle Theorem. We have the following diagram: Since $AN=\\frac12$ and $NM=1,$ we get $AM=\\frac{\\sqrt5}{2}$ by the Pythagorean Theorem. Let $AP=x.$ It follows that $PM=\\frac{\\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\\triangle ANP$ gives $NP^2=\\left(\\frac12\\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\\triangle MNP$ gives $NP^2=1^2-\\left(\\frac{\\sqrt5}{2}-x\\right)^2.$ Equating the expressions for $NP^2$ produces \\begin{align*}", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "so the maximum length of $XY$ is $\\sin(a)\\times\\frac{\\sin(t')}{\\sin(t' + a) + \\cos(a)} = 7 - 4\\sqrt{3}$, and the answer is $7 + 4 + 3 = \\boxed{014}$. ### Solution 3 $[asy] unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label(\"A\",A,N); label(\"B\",B,NNW); label(\"M\",M,W); label(\"C\",C,SSE); label(\"N\",n,E); label(\"D\",D[0],SE); label(\"E\",e[0],SW); label(\"x\",(M+C)/2,SW); label(\"y\",(n+C)/2,SE); [/asy]$ Suppose $\\overline{AC}$ and $\\overline{BC}$ intersect $\\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\\overline{CA}$ bisects $\\angle MCN$, and we get $$\\frac{MC}{MD} = \\frac{NC}{ND}\\Rightarrow MD = \\frac{x}{x + y}.$$ To find $ME$, we note that $\\triangle BNE\\sim\\triangle MCE$ and $\\triangle BME\\sim\\triangle NCE$, so \\begin{align*} \\frac{BN}{NE} &= \\frac{MC}{CE} \\\\ \\frac{ME}{BM} &= \\frac{CE}{NC}. \\end{align*} Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get $$\\frac{4(ME)}{3 - 3(ME)} = \\frac{x}{y}\\Rightarrow ME = \\frac{3x}{3x + 4y}.$$ The value we wish to maximize is \\begin{align*} DE &= MD - ME \\\\ &= \\frac{x}{x + y} - \\frac{3x}{3x + 4y} \\\\ &= \\frac{xy}{3x^2 + 7xy + 4y^2} \\\\ &= \\frac{1}{3(x/y) + 4(y/x) + 7}. \\end{align*} By the AM-GM inequality, $3(x/y) + 4(y/x)\\geq 2\\sqrt{12} = 4\\sqrt{3}$, so $$DE\\leq \\frac{1}{4\\sqrt{3} +", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}" ]

[ "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "Difference between revisions of \"2016 AMC 12A Problems/Problem 12\" Problem 12 In $\\triangle ABC$, $AB = 6$, $BC = 7$, and $CA = 8$. Point $D$ lies on $\\overline{BC}$, and $\\overline{AD}$ bisects $\\angle BAC$. Point $E$ lies on $\\overline{AC}$, and $\\overline{BE}$ bisects $\\angle ABC$. The bisectors intersect at $F$. What is the ratio $AF$ : $FD$? $[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,1.5*N); [/asy]$ $\\textbf{(A)}\\ 3:2\\qquad\\textbf{(B)}\\ 5:3\\qquad\\textbf{(C)}\\ 2:1\\qquad\\textbf{(D)}\\ 7:3\\qquad\\textbf{(E)}\\ 5:2$ Solution 1 By the angle bisector theorem, $\\frac{AB}{AE} = \\frac{CB}{CE}$ $\\frac{6}{AE} = \\frac{7}{8 - AE}$ so $AE = \\frac{48}{13}$ Similarly, $CD = 4$. Now, we use mass points. Assign point $C$ a mass of $1$. $mC \\cdot CD = mB \\cdot DB$ , so $mB = \\frac{4}{3}$ Similarly, $A$ will have a mass of $\\frac{7}{6}$ $mD = mC + mB = 1 + \\frac{4}{3} = \\frac{7}{3}$ So $\\frac{AF}{AD} = \\frac{mD}{mA} = \\boxed{\\textbf{(C)}\\; 2 : 1}$ Solution 2 Denote $[\\triangle{ABC}]$ as the area of triangle ABC and let $r$ be the inradius. Also, as above, use the angle bisector theorem to find that $BD = 3$. There are two ways to continue from here: $1.$ Note that $F$ is the incenter. Then, $\\frac{AF}{FD} = \\frac{[\\triangle{AFB}]}{[\\triangle{BFD}]} = \\frac{AB * \\frac{r}{2}}{BD * \\frac{r}{2}} = \\frac{AB}{BD} = \\boxed{\\textbf{(C)}\\; 2 : 1}$ $2.$ Apply", "AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this", "+0 # Some old geometry problem for you to solve 0 70 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "+0 # Some old geometry problem for you to solve 0 133 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022" ]

[ "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$ $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ Level 4-5 ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "multiply both of its components by $2$: $2\\mathbf{a}=2\\times\\begin{pmatrix}3\\\\8\\end{pmatrix}=\\begin{pmatrix}6\\\\16\\end{pmatrix}$ Then, to add this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$. $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ ## Related Topics #### Coordinates and Ratios Level 4-5Level 6-7KS3 #### Direct and Inverse Proportion Level 4-5Level 6-7KS3 ## Worksheet and Example Questions ### (NEW) Column Vectors Exam Style Questions - MMe Level 4-5 GCSENewOfficial MME ## You May Also Like... ### GCSE Maths Revision Cards Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC. £8.99 ### GCSE Maths Revision Guide The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!", "\\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ heureka Apr 30, 2020", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "= k$, and $\\begin{pmatrix} 4 \\\\ 8 \\\\ -12 \\end{pmatrix} = 4 \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix}$. Since $\\mathbf{b} = 4 \\mathbf{a}$, $\\mathbf{a}$ is parallel to $\\mathbf{b}$. The presentation/phrasing style for \"show/prove\" questions are slightly different from the more common solve questions. Try to mimic the style above and consult with your teachers/me about your presentation when you attempt such questions.", "B) \\times (\\mathbf A + \\mathbf B)$$ Now the cross product $\\times$ is linear, but you also get terms $\\mathbf A \\times \\mathbf B$ and $\\mathbf B \\times \\mathbf A$, which cancel by a property of the cross product. So in the end you get $$\\mathbf C \\times \\mathbf C = \\mathbf A \\times \\mathbf A + \\mathbf B \\times \\mathbf B$$ Which just says that $0=0+0$. Another way to see that your conclusion does not hold, is simply to see that $\\mathbf A \\times \\mathbf B$ is non-zero in general, so by taking the magnitude, we find that $|\\mathbf A \\circ \\mathbf B|$ is non-zero in general, and hence $\\mathbf A \\circ \\mathbf B$ is non-zero in general. 5. Aug 4, 2016 ### Devil Moo Now it is defined as the angle between $\\mathbf A$ and $\\mathbf B$ where $\\theta$ is smaller than or equal to $\\pi$ and it is commutative. How do you find out that it is not distributive? Um... How are these two examples related to the derivation? 6. Aug 4, 2016 ### Staff: Mentor By looking at an example. $$A = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, B = \\begin{pmatrix} -1 \\\\ 0 \\\\ 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ $$(A+B) \\circ C = 0 \\neq A", "of $\\mathbb{R}^{d\\times 1}.$ So, for all $x\\in \\mathbb{R}^{n\\times 1}$ $$(Xx)^T(Xx)=(w_1,\\ldots,w_d) \\begin{pmatrix}{w_1}\\\\{\\vdots}\\\\{w_d}\\end{pmatrix} =w_1^2+\\ldots+w_d^2\\geq 0$$ Ok I get the point, thank you.", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "$[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A} \\nonumber$ ##### Example $$\\PageIndex{3}$$ Given $$\\mathbf{A}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}$$ and $$\\mathbf{B}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}$$ Calculate the commutator $$[\\mathbf{A},\\mathbf{B}]$$ ###### Solution $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}\\nonumber$ $\\mathbf{A}\\mathbf{B}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}=\\begin{pmatrix} 3\\times 1+1\\times (-1)&3\\times 0 +1\\times 2 \\\\ 2\\times 1+0\\times (-1)&2\\times 0+ 0\\times 2 \\end{pmatrix}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}\\nonumber$ $\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}=\\begin{pmatrix} 1\\times 3+0\\times 2&1\\times 1 +0\\times 0 \\\\ -1\\times 3+2\\times 2&-1\\times 1+2\\times 0 \\end{pmatrix}=\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}\\nonumber$ $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}-\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}\\nonumber$ $\\displaystyle{\\color{Maroon}[\\mathbf{A},\\mathbf{B}]=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}}\\nonumber$ ## Multiplication of a vector by a scalar The multiplication of a vector $$\\vec{v_1}$$ by a scalar $$n$$ produces another vector of the same dimensions that lies in the same direction as $$\\vec{v_1}$$; $n\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} nx \\\\ ny \\end{pmatrix} \\nonumber$ The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]). ## Multiplication of a square matrix by a vector The multiplication of a vector $$\\vec{v_1}$$ by a square matrix produces another vector of the same dimensions of $$\\vec{v_1}$$. For example, we can multiply a $$2\\times 2$$ matrix and a 2-dimensional vector: $\\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by \\\\ cx+dy \\end{pmatrix} \\nonumber$ For example, consider the matrix $\\mathbf{A}=\\begin{pmatrix} -2 &0", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "\\times \\mathbf{j} = \\mathbf{k},\\;$$ $$\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i},\\;$$ $$\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}.$$", "in what context can I apply this trick? Thanks If $B \\in \\mathbb R^{(m+n)\\times(m+n)}$ is of the form $$B = \\pmatrix{B_{11}&0\\\\0&B_{22}}$$ with $B_{11}\\in\\mathbb R^{m\\times m}$ and $B_{22}\\in\\mathbb R^{n\\times n}$ both invertible, then $B$ is invertible with $$B^{-1} = \\pmatrix{B_{11}^{-1}&0\\\\0&B_{22}^{-1}}$$ To prove it, note that $$\\pmatrix{A_{11}&0\\\\0&A_{22}}\\pmatrix{B_{11}&0\\\\0&B_{22}}=\\pmatrix{A_{11}B_{11}&0\\\\0&A_{22}B_{22}}$$", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "+0 help 0 94 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 #1 +25565 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times", "11 questions linked to/from calculating $a^b \\!\\mod c$ 18k views ### Modular exponentiation by hand ($a^b\\bmod c$) How do I efficiently compute $a^b\\bmod c$: When $b$ is huge, for instance $5^{844325}\\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for ... 5k views ### Proof of this result related to Fibonacci numbers: $\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}^n=\\begin{pmatrix}F_{n+1}&F_n\\\\F_n&F_{n-1}\\end{pmatrix}$? $$\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}^n=\\begin{pmatrix}F_{n+1}&F_n\\\\F_n&F_{n-1}\\end{pmatrix}$$ Somebody has any idea how to go about proving this result? I know a proof by ... 477 views ### exponential equation $$\\sqrt{(5+2\\sqrt6)^x}+\\sqrt{(5-2\\sqrt6)^x}=10$$ So I have squared both sides and got: $$(5-2\\sqrt6)^x+(5+2\\sqrt6)^x+2\\sqrt{1^x}=100$$ $$(5-2\\sqrt6)^x+(5+2\\sqrt6)^x+2=100$$ I don't know what to do ... ### Calculate $7^{154} \\pmod{341}$ [duplicate] how to calculate remainder of $7^ {154}$ when it is divided by $341$. Could you please state which method or theorem to use.", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "+0 # help 0 52 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} .$$ Find the area of the triangle whose vertices are $$\\mathbf{a}, \\mathbf{b}$$ and $$\\mathbf{c}.$$ Apr 29, 2020 ### Best Answer #1 +24983 +3 Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} = \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix}$$. Find the area of the triangle whose vertices are $$\\mathbf{a}$$$$\\mathbf{b}$$ and $$\\mathbf{c}$$. $$\\text{Let the area of the triangle =A }$$ $$\\begin{array}{|rcll|} \\hline 2A &=& |~ (\\mathbf{a}-\\mathbf{b})\\times (\\mathbf{a}-\\mathbf{c}) ~| \\\\ 2A &=& |~ \\mathbf{a}\\times\\mathbf{a}+\\mathbf{a}\\times(\\mathbf{-c})+(\\mathbf{-b})\\times\\mathbf{a}+(\\mathbf{-b})\\times(\\mathbf{-c}) ~| \\\\ && \\boxed{ \\mathbf{a}\\times\\mathbf{a} = 0 \\\\ \\mathbf{a}\\times(\\mathbf{-c}) = \\mathbf{c}\\times\\mathbf{a} \\\\ (\\mathbf{-b})\\times\\mathbf{a} = \\mathbf{a}\\times\\mathbf{b} \\\\ (\\mathbf{-b})\\times(\\mathbf{-c}) = \\mathbf{b}\\times\\mathbf{c} } \\\\ 2A &=& |~ 0+\\mathbf{c}\\times\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b} +\\mathbf{b}\\times\\mathbf{c} ~| \\\\ 2A &=& |~ \\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a} ~| \\\\ 2A &=& |~ \\begin{pmatrix} -3 \\\\ -12 \\\\ 4 \\end{pmatrix} ~| \\\\ 2A &=& \\sqrt{(-3)^2+(-12)^2+(4)^2} \\\\ 2A &=& \\sqrt{9+144+16} \\\\ 2A &=& \\sqrt{169} \\\\ 2A &=& 13 \\\\ A &=& \\dfrac{13}{2} \\\\ \\mathbf{A} &=& \\mathbf{6.5} \\\\ \\hline \\end{array}$$ Apr 30, 2020 ### 1+0 Answers #1 +24983 +3 Best Answer Let $$\\mathbf{a}$$, $$\\mathbf{b}$$, $$\\mathbf{c}$$ be vectors such that $$\\mathbf{a}", "$a$, $b$ with $\\mathbf w=a\\mathbf u+b\\mathbf v$. Then $\\mathbf u\\times(\\mathbf v\\times\\mathbf w)=a\\mathbf u\\times(\\mathbf v\\times\\mathbf u)$ and $\\mathbf v\\times(\\mathbf w\\times\\mathbf u)=b\\mathbf v\\times(\\mathbf v\\times\\mathbf u)$ and so $\\mathbf u\\times(\\mathbf v\\times\\mathbf w)+\\mathbf v\\times(\\mathbf w\\times\\mathbf u)=(a\\mathbf u+b\\mathbf v)\\times(\\mathbf v\\times\\mathbf u)=\\mathbf w\\times(\\mathbf v\\times\\mathbf u)=-\\mathbf w\\times(\\mathbf u\\times\\mathbf v)$, proving the result in this case. Now assume that $\\mathbf u$, $\\mathbf v$ and $\\mathbf w$ are linearly independent, thus forming a basis for $\\mathbb R^3$. We have $\\mathbf u*[\\mathbf u\\times(\\mathbf v\\times\\mathbf w)]=0$ and $\\mathbf u*[\\mathbf v\\times(\\mathbf w\\times\\mathbf u)]=(\\mathbf u\\times\\mathbf v)*(\\mathbf w\\times\\mathbf u)=-\\mathbf u*[\\mathbf w\\times(\\mathbf u\\times\\mathbf v)]$. Hence $\\mathbf u*[\\mathbf u\\times(\\mathbf v\\times\\mathbf w)+\\mathbf v\\times(\\mathbf w\\times\\mathbf u)+\\mathbf w\\times(\\mathbf u\\times\\mathbf v)]=0$. Similarly $\\mathbf v*[\\mathbf u\\times(\\mathbf v\\times\\mathbf w)+\\mathbf v\\times(\\mathbf w\\times\\mathbf u)+\\mathbf w\\times(\\mathbf u\\times\\mathbf v)]=0$ and $\\mathbf w*[\\mathbf u\\times(\\mathbf v\\times\\mathbf w)+\\mathbf v\\times(\\mathbf w\\times\\mathbf u)+\\mathbf w\\times(\\mathbf u\\times\\mathbf v)]=0$. Since $\\mathbf u$, $\\mathbf v$ and $\\mathbf w$ form a basis and $\\mathbf u\\times(\\mathbf v\\times\\mathbf w)+\\mathbf v\\times(\\mathbf w\\times\\mathbf u)+\\mathbf w\\times(\\mathbf u\\times\\mathbf v)$ is orthogonal to all of them, the result follows. • September 13th 2009, 02:16 AM NonCommAlg Quote: Originally Posted by halbard Warning: I'm relying on the identity $\\mathbf a*(\\mathbf b\\times\\mathbf c)=(\\mathbf a\\times\\mathbf b)*\\mathbf c$. This is proved by considering the identity $(\\mathbf c+\\mathbf a)*[(\\mathbf c+\\mathbf a)\\times \\mathbf b]=0$. First assume that $\\mathbf u$, $\\mathbf v$ and $\\mathbf w$ are linearly dependent. WLOG assume there are scalars $a$, $b$ with $\\mathbf w=a\\mathbf", "+0 # Help please, if you could explain it that would be even better. 0 60 1 Let $$\\mathbf{A} = {\\begin{pmatrix} \\sqrt{3}/2& -1/2\\\\ 1/2 & \\sqrt{3}/2 \\end{pmatrix}}$$ Then what is $$\\mathbf{A}^{2018} \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix}$$? Aug 7, 2020" ]

[ "b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1\\right) \\\\ &=& \\left(16,0,48\\right). \\end{eqnarray}$$ Wikipedia has a decent explanation of the cross product. - By definition, the product of vectors $a= (a_x, a_y, a_z)$ and $b = (b_x, b_y, b_z)$ equals $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ a_x & a_y & a_z\\\\ b_x & b_y & b_z \\end{pmatrix},$$ where $\\mathbf{i}$, $\\mathbf{j}$, $\\mathbf{k}$ is the standard basis. So we have $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 6 & 0 & -2\\\\ 0 & 8 & 0 \\end{pmatrix} = 16\\, \\mathbf{i} + 0\\, \\mathbf{j} + 48\\, \\mathbf{k}.$$ -", "B) \\times (\\mathbf A + \\mathbf B)$$ Now the cross product $\\times$ is linear, but you also get terms $\\mathbf A \\times \\mathbf B$ and $\\mathbf B \\times \\mathbf A$, which cancel by a property of the cross product. So in the end you get $$\\mathbf C \\times \\mathbf C = \\mathbf A \\times \\mathbf A + \\mathbf B \\times \\mathbf B$$ Which just says that $0=0+0$. Another way to see that your conclusion does not hold, is simply to see that $\\mathbf A \\times \\mathbf B$ is non-zero in general, so by taking the magnitude, we find that $|\\mathbf A \\circ \\mathbf B|$ is non-zero in general, and hence $\\mathbf A \\circ \\mathbf B$ is non-zero in general. 5. Aug 4, 2016 ### Devil Moo Now it is defined as the angle between $\\mathbf A$ and $\\mathbf B$ where $\\theta$ is smaller than or equal to $\\pi$ and it is commutative. How do you find out that it is not distributive? Um... How are these two examples related to the derivation? 6. Aug 4, 2016 ### Staff: Mentor By looking at an example. $$A = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, B = \\begin{pmatrix} -1 \\\\ 0 \\\\ 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ $$(A+B) \\circ C = 0 \\neq A", "the definition of the cross product, a × x are represented as follows. ${\\displaystyle \\mathbf {a} \\times \\mathbf {x} ={\\begin{pmatrix}a_{2}x_{3}-a_{3}x_{2}\\\\a_{3}x_{1}-a_{1}x_{3}\\\\a_{1}x_{2}-a_{2}x_{1}\\end{pmatrix}}}$ Therefore, when we apply the operator a × to each of the standard basis, we obtain following, and thus, the matrix representation of a × as shown in (★1). ${\\displaystyle \\mathbf {a} \\times \\mathbf {e} _{1}={\\begin{pmatrix}0\\\\a_{3}\\\\-a_{2}\\end{pmatrix}},\\quad \\mathbf {a} \\times \\mathbf {e} _{2}={\\begin{pmatrix}-a_{3}\\\\0\\\\a_{1}\\end{pmatrix}},\\quad \\mathbf {a} \\times \\mathbf {e} _{3}={\\begin{pmatrix}a_{2}\\\\-a_{1}\\\\0\\end{pmatrix}}}$ ${\\displaystyle \\mathbf {a} \\times \\mathbf {x} ={\\begin{pmatrix}0&-a_{3}&a_{2}\\\\a_{3}&0&-a_{1}\\\\-a_{2}&a_{1}&0\\end{pmatrix}}{\\begin{pmatrix}x_{1}\\\\x_{2}\\\\x_{3}\\end{pmatrix}}}$ (★1) Next, let ${\\displaystyle A=(a_{ij})}$ is a 3 × 3 matrix and using the following substitution, (a1, ... , a3 are components of a), ${\\displaystyle a_{1}={a}_{32}-a_{23}}$ (★2-1) ${\\displaystyle a_{2}=a_{13}-a_{31}}$ (★2-2) ${\\displaystyle a_{3}=a_{21}-a_{12}}$ (★2-3) Then we obtain following (★3) from the (★1). ${\\displaystyle (A-{}^{t}A)\\mathbf {x} =\\mathbf {a} \\times \\mathbf {x} }$ (★3) We substitute the JF to above mentioned A, under the substitution of (★2-1), (★2-2), and (★2-3), we obtain the following (★4) ${\\displaystyle \\mathbf {a} =(\\nabla \\times \\mathbf {F} )}$ (★4) The (★0) is obvious from (★3) and (★4). 5. ^ There are a number of theorems with the same name, however they are not necessarily the same. 6. ^ Typical definition of homotopy and homotope are as follows. 7. ^ In some textbooks such as Conlon, Lawrence (2008). Differentiable Manifolds. Modern Birkhauser Classics. Boston: Birkhaeuser. use the term of homotopy and homotope", "## Algebra 1 Use the distributive property to restate a more complex product as two simpler products that can be solved with mental math. $3.9\\times6=(4-0.1)\\times6=24-0.6=23.4$", "and $s$ and $t$ will be arbitrary real numbers. • $\\mathbf{a}\\times\\mathbf{a}=0$. • $\\mathbf{a}\\times(\\mathbf{b}\\times\\mathbf{c})+\\mathbf{b}\\times(\\mathbf{c}\\times% \\mathbf{a})+\\mathbf{c}\\times(\\mathbf{a}\\times\\mathbf{b})=0$. • The cross product is a bilinear map. This means that $(s\\mathbf{a})\\times(t\\mathbf{b})=(st)(\\mathbf{a}\\times\\mathbf{b})$, and that the cross product is distributive over vector addition, that is, $\\mathbf{a}\\times(\\mathbf{b}+\\mathbf{c})=\\mathbf{a}\\times\\mathbf{b}+\\mathbf{a}% \\times\\mathbf{c}$ and $(\\mathbf{b}+\\mathbf{c})\\times\\mathbf{a}=\\mathbf{b}\\times\\mathbf{a}+\\mathbf{c}% \\times\\mathbf{a}$. • The three properties above mean that the cross product makes $\\mathbb{R}^{3}$ into a Lie algebra. • $\\mathbf{a}\\times\\mathbf{b}$ is orthogonal to both $\\mathbf{a}$ and $\\mathbf{b}$. • $\\mathbf{a}\\times\\mathbf{b}=-\\mathbf{b}\\times\\mathbf{a}$. • The length of $\\mathbf{a}\\times\\mathbf{b}$ is the area of the parallelogram spanned by $\\mathbf{a}$ and $\\mathbf{b}$, so $|\\mathbf{a}\\times\\mathbf{b}|=|\\mathbf{a}||\\mathbf{b}|\\sin\\theta$, where $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{b}$. This gives us an expression for the area of a triangle in $\\mathbb{R}^{3}$: if the vertices are at $\\mathbf{a}$, $\\mathbf{b}$ and $\\mathbf{c}$, then the area is $\\frac{1}{2}|(\\mathbf{a}-\\mathbf{c})\\times(\\mathbf{b}-\\mathbf{c})|$, which can be written more symmetrically as $\\frac{1}{2}|\\mathbf{a}\\times\\mathbf{b}+\\mathbf{b}\\times\\mathbf{c}+\\mathbf{c}% \\times\\mathbf{a}|$. • From the above, you can see that the cross product of any vector with $\\mathbf{0}$ is $\\mathbf{0}$. More generally, the cross product of two parallel vectors is $\\mathbf{0}$, since $\\sin 0=0$. • One can also see that $|\\mathbf{a}\\times\\mathbf{b}|^{2}=|\\mathbf{a}|^{2}|\\mathbf{b}|^{2}-|\\mathbf{a}% \\cdot\\mathbf{b}|^{2}$. • $\\mathbf{a}\\times(\\mathbf{b}\\times\\mathbf{c})=(\\mathbf{a}\\cdot\\mathbf{c})% \\mathbf{b}-(\\mathbf{a}\\cdot\\mathbf{b})\\mathbf{c}$. This is the vector triple product. • The cross product is rotationally invariant (http://planetmath.org/RotationalInvarianceOfCrossProduct). That is, for any $3\\times 3$ rotation matrix $M$ we have $M(\\mathbf{a}\\times\\mathbf{b})=(M\\mathbf{a})\\times(M\\mathbf{b})$. Title cross product Canonical name CrossProduct Date of creation 2013-03-22 11:58:52 Last modified on 2013-03-22 11:58:52", "## Calculus: Early Transcendentals 8th Edition The cross product follows the distributive property of addition, and the cross product of a vector and itself is zero. Thus, $(u+v)\\times v=u\\times v+v \\times v$ $= u \\times v +0$ $(u+v)\\times v=0$ Hence, the statement is true.", "## Re: Cross product problem You certainly can do that. In particular, If you multiply the last equation by 2 and subtract it from the first equation, you get $6b_1+ 3b_3= 3$. From the second equation, $b_3= 1- 2b_1$ so $6b_1+ 3b_3= 6b_1+3- 6b_1= 3$ which is true for all x. Taking $b_1= t$ gives $b2= 1- 3t$ and $b_3= 1- 2t$, exactly what you give above.", "## Elementary Algebra $\\frac{-2}{x-1}$ = $\\frac{-3}{x+4}$ To solve this equation, we equate the cross products. -2 $\\times$ (x + 4) = -3 $\\times$ (x-1) Use the distributive property. -2x -8 = -3x + 3 -2x + 3x = 3 + 8 x = 11", "Study guide and 1 practice problem on: ## Algebra of cross products The cross product of a sum is the sum of cross products: $$\\bfa \\times (\\bfb + \\bfc) = \\bfa \\times \\bfb + \\bfa \\times \\bfc.$$", "Study guide and 1 practice problem on: Algebra of cross products The cross product of a sum is the sum of cross products: $$\\bfa \\times (\\bfb + \\bfc) = \\bfa \\times \\bfb + \\bfa \\times \\bfc.$$", "three whole numbers, then $a\\times \\left( b + c \\right)=a\\times b+a\\times c$. The given product is $728\\times 101$ $=728\\times \\left( 100+1 \\right)$ $=728\\times 100+728\\times 1$, using the distributive property. $=72800+728$, multiplying the numbers. $=73528$. (ii) $\\mathbf{5437}\\times \\mathbf{1001}$ Ans. The given product is $5437\\times 1001$ $=5437\\times \\left( 1000+1 \\right)$ $=5437\\times 1000+5437\\times 1$, using the distributive property. $=5437000+5437$, multiplying the numbers. $=5442437$. (iii) $\\mathbf{824}\\times \\mathbf{25}$ Ans. The given product is $824\\times 25$ $=824\\times \\left( 20+5 \\right)$ $=824\\times 20+824\\times 5$, using the distributive property. $=16480+4120$, multiplying the numbers. $=20600$. (iv) $\\mathbf{4275}\\times \\mathbf{125}$ Ans. The given product is $4275\\times 125$ $=4275\\times \\left( 100+20+5 \\right)$, $=4275\\times 100+4275\\times 20+4275\\times 5$, using the distributive property. $=427500+85500+21375$, multiplying the numbers. $=534375$. (v) $\\mathbf{504}\\times \\mathbf{35}$ Ans. The given product is $504\\times 35$ $=\\left( 500+4 \\right)\\times 35$ $=500\\times 35+4\\times 35$, applying the distributive property. $=17500+140$, multiplying the numbers. $=17640$. 5. Study the pattern : $\\mathbf{1\\times 8+1=9}$; $\\mathbf{12}\\times \\mathbf{8}+\\mathbf{2}=\\mathbf{98}$; $\\mathbf{123}\\times \\mathbf{8}+\\mathbf{3}=\\mathbf{987}$; $\\mathbf{1234}\\times \\mathbf{8}+\\mathbf{4}=\\mathbf{9876}$; $\\mathbf{12345\\times 8+5=98765}$ Write the next two steps. Can you say how the pattern works? (Hint: $\\mathbf{12345=11111+1111+111+11+1}$). Ans. The next two steps will be $123456\\times 8+6=987654$; $1234567\\times 8+7=9876543$. The pattern described in the question will be like the following way: $\\left( 1 \\right)\\times 8+1=9$ $\\left( 12 \\right)\\times 8+2=\\left( 11+1 \\right)\\times 8+2=98$ $\\left( 123 \\right)\\times 8+3=\\left( 111+11+1 \\right)\\times 8+3=987$ $\\left( 1234 \\right)\\times 8+4=\\left( 1111+111+11+1 \\right)\\times 8+4=9876$ $\\left( 12345", "$\\textbf{x}\\times \\textbf{y} = \\textbf{0}$, and $$\\textbf{x}\\times \\textbf{y} = \\pmatrix{\\textbf{e}_1 & \\textbf{e}_2 & \\textbf{e}_3\\\\x_1 & x_2 & x_3\\\\y_1 & y_2 & y_3}$$ (where $\\textbf{e}_n$ is the unit vector in the $n$th dimension) Alternate hint in absence of cross product: What happens if $\\textbf{x} = a\\textbf{y}$ for some constant $a$? • I am not quite sure if we are allowed to use the cross product as we haven't proved it yet. – Samuli Lehtonen Jan 19 '14 at 14:02 • See my edit, it provides an alternate hint. – Glen O Jan 19 '14 at 14:03", "# Cross product proof Three vectors $a$, $b$, and $c$ are given. Prove that if $a \\perp b$, $a \\perp c$, $b \\perp c$, and $|a|=|b|=|c|=1$, then $$a=b \\times c$$ $$b=c \\times a$$ $$c=a \\times b$$ - This does not seem to be true, e.g. taking $e_1,e_2,e_3 \\in \\mathbb{R}^3$, then $e_2 \\times e_3 = -e_1$ instead of $e_1$ – anegligibleperson Nov 28 '12 at 16:51 $$a = \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix} \\quad b=\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\quad c=\\begin{pmatrix}0\\\\0\\\\-1\\end{pmatrix}$$ Then they are orthogonal, but $a \\times b = \\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix} =-c$ also $c \\times a = -b$ and $b \\times c = - a$.", "## Elementary Algebra We know that cross products are equal. So, following this rule, we obtain: ($x^{2}$ + 7x + 12) $\\times$ 1 = (x + 3) $\\times$ (x + 4) ($x^{2}$ + 7x + 12) = (x + 3)(x + 4) So, this statement is true.", "Find the cross product $\\mathbf a \\times \\mathbf b$ where $\\mathbf a = \\left<5,2,1\\right>$ and $\\mathbf b = \\left<3,4,3\\right>$. $\\mathbf a \\times \\mathbf b =$ Entering-Vectors.html Your overall score for this problem is", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "# A Cross Product Formula $\\mathbf{V}_1 \\times \\mathbf{V}_2 = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\frac{\\partial X}{\\partial u} & \\frac{\\partial Y}{\\partial u} & 0 \\\\ \\frac{\\partial X}{\\partial v} & \\frac{\\partial Y}{\\partial v} & 0 \\end{vmatrix}$", "\\times \\mathbf{b} = \\mathbf{0}$$ Scalar multiplication is conserved in the vector cross product $$\\left(c\\mathbf{a}\\right)\\times \\mathbf{b} = c\\left(\\mathbf{a} \\times \\mathbf{b}\\right) = \\mathbf{a} \\times \\left(c\\mathbf{b}\\right)$$ The cross product is also anti-commutative. That means $$\\mathbf{a} \\times \\mathbf{b} = -\\left(\\mathbf{b} \\times \\mathbf{a}\\right)$$ It is also distributive but not associative $$\\left(\\mathbf{a} + \\mathbf{b}\\right)\\times \\mathbf{p} = \\mathbf{a} \\times \\mathbf{p} + \\mathbf{b} \\times \\mathbf{p}$$ $$\\mathbf{p} \\times\\left(\\mathbf{a} + \\mathbf{b}\\right) = \\mathbf{p} \\times \\mathbf{a} + \\mathbf{p} \\times \\mathbf{b}$$ Another formula for calculating the cross product is $$\\|\\mathbf{a}\\|\\|\\mathbf{b}\\|\\sin(\\theta)\\hat{\\mathbf{n}}$$ Where $\\theta$ is the angle between the two vectors, and $\\hat{\\mathbf{n}}$ is the normal vector in the direction of the cross product. Therefore the length of the cross product is always $$\\|\\mathbf{a} \\times \\mathbf{b}\\| = \\|\\mathbf{a}\\|\\|\\mathbf{b}\\|\\sin\\theta$$ This is what the cross product looks like for the standard basis vectors $\\mathbf{i}$ and $\\mathbf{j}$ See that $$\\mathbf{i} \\times \\mathbf{j} = \\left|\\begin{array}{ccc} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0 \\end{array}\\right| = \\mathbf{k}$$ and by the anti-commutativity property of the cross product $$\\mathbf{j} \\times \\mathbf{i} = -\\left( \\mathbf{i} \\times \\mathbf{j} \\right) = -\\mathbf{k}$$ ## The intersection of two planes Since this topic is all about three-dimensional space, we look at planes as well as lines. A plane in three-dimensional $(x,y,z)$ space is a flat surface that can be expressed as $$f(x,y,z) = c$$ where", "... of the cross product (though neither follows easily from the definition given above), are sufficient to determine the cross product of any two vectors a and b. Each vector can be defined as the sum of three orthogonal components parallel to the standard basis vectors: :$\\begin \\mathbf &= a_1\\mathbf &&+ a_2\\mathbf &&+ a_3\\mathbf \\\\ \\mathbf &= b_1\\mathbf &&+ b_2\\mathbf &&+ b_3\\mathbf \\end$ Their cross product can be expanded using distributivity: :$\\begin \\mathbf\\times\\mathbf = &\\left(a_1\\mathbf + a_2\\mathbf + a_3\\mathbf\\right) \\times \\left(b_1\\mathbf + b_2\\mathbf + b_3\\mathbf\\right)\\\\ = &a_1b_1\\left(\\mathbf \\times \\mathbf\\right) + a_1b_2\\left(\\mathbf \\times \\mathbf\\right) + a_1b_3\\left(\\mathbf \\times \\mathbf\\right) + \\\\ &a_2b_1\\left(\\mathbf \\times \\mathbf\\right) + a_2b_2\\left(\\mathbf \\times \\mathbf\\right) + a_2b_3\\left(\\mathbf \\times \\mathbf\\right) + \\\\ &a_3b_1\\left(\\mathbf \\times \\mathbf\\right) + a_3b_2\\left(\\mathbf \\times \\mathbf\\right) + a_3b_3\\left(\\mathbf \\times \\mathbf\\right)\\\\ \\end$ This can be interpreted as the decomposition of into the sum of nine simpler cross products involving vectors aligned with i, j, or k. Each one of these nine cross products operates on two vectors that are easy to handle as they are either parallel or orthogonal to each other. From this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain: :$\\begin \\mathbf\\times\\mathbf = &\\quad\\ a_1b_1\\mathbf + a_1b_2\\mathbf - a_1b_3\\mathbf \\\\ &- a_2b_1\\mathbf + a_2b_2\\mathbf + a_2b_3\\mathbf \\\\ &+ a_3b_1\\mathbf\\ - a_3b_2\\mathbf\\ + a_3b_3\\mathbf \\\\ = &\\left(a_2b_3 - a_3b_2\\right)\\mathbf + \\left(a_3b_1", "(4, 2, 3) Cross product is a * b = $\\begin{bmatrix} i & j & k\\\\ 2&-3 & 5\\\\ 4& 2& 3 \\end{bmatrix}$ = i(-9 - 10) - j (6 - 20) + k (4 + 12) = -19i + 14j + 16k" ]

[ "B) \\times (\\mathbf A + \\mathbf B)$$ Now the cross product $\\times$ is linear, but you also get terms $\\mathbf A \\times \\mathbf B$ and $\\mathbf B \\times \\mathbf A$, which cancel by a property of the cross product. So in the end you get $$\\mathbf C \\times \\mathbf C = \\mathbf A \\times \\mathbf A + \\mathbf B \\times \\mathbf B$$ Which just says that $0=0+0$. Another way to see that your conclusion does not hold, is simply to see that $\\mathbf A \\times \\mathbf B$ is non-zero in general, so by taking the magnitude, we find that $|\\mathbf A \\circ \\mathbf B|$ is non-zero in general, and hence $\\mathbf A \\circ \\mathbf B$ is non-zero in general. 5. Aug 4, 2016 ### Devil Moo Now it is defined as the angle between $\\mathbf A$ and $\\mathbf B$ where $\\theta$ is smaller than or equal to $\\pi$ and it is commutative. How do you find out that it is not distributive? Um... How are these two examples related to the derivation? 6. Aug 4, 2016 ### Staff: Mentor By looking at an example. $$A = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, B = \\begin{pmatrix} -1 \\\\ 0 \\\\ 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ $$(A+B) \\circ C = 0 \\neq A", "b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1\\right) \\\\ &=& \\left(16,0,48\\right). \\end{eqnarray}$$ Wikipedia has a decent explanation of the cross product. - By definition, the product of vectors $a= (a_x, a_y, a_z)$ and $b = (b_x, b_y, b_z)$ equals $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ a_x & a_y & a_z\\\\ b_x & b_y & b_z \\end{pmatrix},$$ where $\\mathbf{i}$, $\\mathbf{j}$, $\\mathbf{k}$ is the standard basis. So we have $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 6 & 0 & -2\\\\ 0 & 8 & 0 \\end{pmatrix} = 16\\, \\mathbf{i} + 0\\, \\mathbf{j} + 48\\, \\mathbf{k}.$$ -", "and $s$ and $t$ will be arbitrary real numbers. • $\\mathbf{a}\\times\\mathbf{a}=0$. • $\\mathbf{a}\\times(\\mathbf{b}\\times\\mathbf{c})+\\mathbf{b}\\times(\\mathbf{c}\\times% \\mathbf{a})+\\mathbf{c}\\times(\\mathbf{a}\\times\\mathbf{b})=0$. • The cross product is a bilinear map. This means that $(s\\mathbf{a})\\times(t\\mathbf{b})=(st)(\\mathbf{a}\\times\\mathbf{b})$, and that the cross product is distributive over vector addition, that is, $\\mathbf{a}\\times(\\mathbf{b}+\\mathbf{c})=\\mathbf{a}\\times\\mathbf{b}+\\mathbf{a}% \\times\\mathbf{c}$ and $(\\mathbf{b}+\\mathbf{c})\\times\\mathbf{a}=\\mathbf{b}\\times\\mathbf{a}+\\mathbf{c}% \\times\\mathbf{a}$. • The three properties above mean that the cross product makes $\\mathbb{R}^{3}$ into a Lie algebra. • $\\mathbf{a}\\times\\mathbf{b}$ is orthogonal to both $\\mathbf{a}$ and $\\mathbf{b}$. • $\\mathbf{a}\\times\\mathbf{b}=-\\mathbf{b}\\times\\mathbf{a}$. • The length of $\\mathbf{a}\\times\\mathbf{b}$ is the area of the parallelogram spanned by $\\mathbf{a}$ and $\\mathbf{b}$, so $|\\mathbf{a}\\times\\mathbf{b}|=|\\mathbf{a}||\\mathbf{b}|\\sin\\theta$, where $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{b}$. This gives us an expression for the area of a triangle in $\\mathbb{R}^{3}$: if the vertices are at $\\mathbf{a}$, $\\mathbf{b}$ and $\\mathbf{c}$, then the area is $\\frac{1}{2}|(\\mathbf{a}-\\mathbf{c})\\times(\\mathbf{b}-\\mathbf{c})|$, which can be written more symmetrically as $\\frac{1}{2}|\\mathbf{a}\\times\\mathbf{b}+\\mathbf{b}\\times\\mathbf{c}+\\mathbf{c}% \\times\\mathbf{a}|$. • From the above, you can see that the cross product of any vector with $\\mathbf{0}$ is $\\mathbf{0}$. More generally, the cross product of two parallel vectors is $\\mathbf{0}$, since $\\sin 0=0$. • One can also see that $|\\mathbf{a}\\times\\mathbf{b}|^{2}=|\\mathbf{a}|^{2}|\\mathbf{b}|^{2}-|\\mathbf{a}% \\cdot\\mathbf{b}|^{2}$. • $\\mathbf{a}\\times(\\mathbf{b}\\times\\mathbf{c})=(\\mathbf{a}\\cdot\\mathbf{c})% \\mathbf{b}-(\\mathbf{a}\\cdot\\mathbf{b})\\mathbf{c}$. This is the vector triple product. • The cross product is rotationally invariant (http://planetmath.org/RotationalInvarianceOfCrossProduct). That is, for any $3\\times 3$ rotation matrix $M$ we have $M(\\mathbf{a}\\times\\mathbf{b})=(M\\mathbf{a})\\times(M\\mathbf{b})$. Title cross product Canonical name CrossProduct Date of creation 2013-03-22 11:58:52 Last modified on 2013-03-22 11:58:52", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "a hard time figuring out what they are supposed to do. Depending on students' tolerance for reading and deciphering problems of this nature, it might be advisable for the teacher to go through an example with different numbers first so students understand what is expected. A slightly different, more visual approach to these ideas can be found here: https://www.illustrativemathematics.org/tasks/1986. Both tasks focus on products of negative numbers and the distributive property: this task focuses on the arithmetic of the distributive property while the other emphasizes geometry. ## Solution 1. The value of $3\\times (-5)$ should be $(-5)+(-5)+(-5) = -15$. 2. Applying the distributive property gives $$3\\times(5+(-5))=3\\times5+3\\times(-5).$$ Since $5+(-5)=0$ and $3\\times5=15$, \\begin{align} 3\\times0=&3\\times (5+(-5))\\\\ =&3\\times5+3\\times(-5)\\\\ =&15+3\\times(-5) \\end{align} Since $3\\times 0 = 0$, these equations show that $0 = 15+ 3\\times(-5)$. From here, we see that $3\\times(-5)=-15$. 3. Extending the commutative property of multiplication to negative numbers, the product $(-5)\\times3$ should have the same value as $3 \\times (-5)$ so it should be -15. Furthermore, the distributive property gives $(-5)\\times(3+(-3)) = (-5)\\times3 + (-5)\\times(-3)$. 4. Using the equality from part (c), $$(-5)\\times(3+(-3)) = (-5)\\times3 + (-5)\\times(-3),$$ we know the left side is 0 and we know $(-5)\\times3 = -15$. So $$0 = -15 + (-5)\\times(-3)$$ and $(-5)\\times(-3) = 15$.", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "# Cross product proof Three vectors $a$, $b$, and $c$ are given. Prove that if $a \\perp b$, $a \\perp c$, $b \\perp c$, and $|a|=|b|=|c|=1$, then $$a=b \\times c$$ $$b=c \\times a$$ $$c=a \\times b$$ - This does not seem to be true, e.g. taking $e_1,e_2,e_3 \\in \\mathbb{R}^3$, then $e_2 \\times e_3 = -e_1$ instead of $e_1$ – anegligibleperson Nov 28 '12 at 16:51 $$a = \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix} \\quad b=\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\quad c=\\begin{pmatrix}0\\\\0\\\\-1\\end{pmatrix}$$ Then they are orthogonal, but $a \\times b = \\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix} =-c$ also $c \\times a = -b$ and $b \\times c = - a$.", "# A Cross Product Formula $\\mathbf{V}_1 \\times \\mathbf{V}_2 = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\frac{\\partial X}{\\partial u} & \\frac{\\partial Y}{\\partial u} & 0 \\\\ \\frac{\\partial X}{\\partial v} & \\frac{\\partial Y}{\\partial v} & 0 \\end{vmatrix}$", "the definition of the cross product, a × x are represented as follows. ${\\displaystyle \\mathbf {a} \\times \\mathbf {x} ={\\begin{pmatrix}a_{2}x_{3}-a_{3}x_{2}\\\\a_{3}x_{1}-a_{1}x_{3}\\\\a_{1}x_{2}-a_{2}x_{1}\\end{pmatrix}}}$ Therefore, when we apply the operator a × to each of the standard basis, we obtain following, and thus, the matrix representation of a × as shown in (★1). ${\\displaystyle \\mathbf {a} \\times \\mathbf {e} _{1}={\\begin{pmatrix}0\\\\a_{3}\\\\-a_{2}\\end{pmatrix}},\\quad \\mathbf {a} \\times \\mathbf {e} _{2}={\\begin{pmatrix}-a_{3}\\\\0\\\\a_{1}\\end{pmatrix}},\\quad \\mathbf {a} \\times \\mathbf {e} _{3}={\\begin{pmatrix}a_{2}\\\\-a_{1}\\\\0\\end{pmatrix}}}$ ${\\displaystyle \\mathbf {a} \\times \\mathbf {x} ={\\begin{pmatrix}0&-a_{3}&a_{2}\\\\a_{3}&0&-a_{1}\\\\-a_{2}&a_{1}&0\\end{pmatrix}}{\\begin{pmatrix}x_{1}\\\\x_{2}\\\\x_{3}\\end{pmatrix}}}$ (★1) Next, let ${\\displaystyle A=(a_{ij})}$ is a 3 × 3 matrix and using the following substitution, (a1, ... , a3 are components of a), ${\\displaystyle a_{1}={a}_{32}-a_{23}}$ (★2-1) ${\\displaystyle a_{2}=a_{13}-a_{31}}$ (★2-2) ${\\displaystyle a_{3}=a_{21}-a_{12}}$ (★2-3) Then we obtain following (★3) from the (★1). ${\\displaystyle (A-{}^{t}A)\\mathbf {x} =\\mathbf {a} \\times \\mathbf {x} }$ (★3) We substitute the JF to above mentioned A, under the substitution of (★2-1), (★2-2), and (★2-3), we obtain the following (★4) ${\\displaystyle \\mathbf {a} =(\\nabla \\times \\mathbf {F} )}$ (★4) The (★0) is obvious from (★3) and (★4). 5. ^ There are a number of theorems with the same name, however they are not necessarily the same. 6. ^ Typical definition of homotopy and homotope are as follows. 7. ^ In some textbooks such as Conlon, Lawrence (2008). Differentiable Manifolds. Modern Birkhauser Classics. Boston: Birkhaeuser. use the term of homotopy and homotope", "this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$ $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ Level 4-5 ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "multiply both of its components by $2$: $2\\mathbf{a}=2\\times\\begin{pmatrix}3\\\\8\\end{pmatrix}=\\begin{pmatrix}6\\\\16\\end{pmatrix}$ Then, to add this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$. $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ ## Related Topics #### Coordinates and Ratios Level 4-5Level 6-7KS3 #### Direct and Inverse Proportion Level 4-5Level 6-7KS3 ## Worksheet and Example Questions ### (NEW) Column Vectors Exam Style Questions - MMe Level 4-5 GCSENewOfficial MME ## You May Also Like... ### GCSE Maths Revision Cards Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC. £8.99 ### GCSE Maths Revision Guide The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!", "... of the cross product (though neither follows easily from the definition given above), are sufficient to determine the cross product of any two vectors a and b. Each vector can be defined as the sum of three orthogonal components parallel to the standard basis vectors: :$\\begin \\mathbf &= a_1\\mathbf &&+ a_2\\mathbf &&+ a_3\\mathbf \\\\ \\mathbf &= b_1\\mathbf &&+ b_2\\mathbf &&+ b_3\\mathbf \\end$ Their cross product can be expanded using distributivity: :$\\begin \\mathbf\\times\\mathbf = &\\left(a_1\\mathbf + a_2\\mathbf + a_3\\mathbf\\right) \\times \\left(b_1\\mathbf + b_2\\mathbf + b_3\\mathbf\\right)\\\\ = &a_1b_1\\left(\\mathbf \\times \\mathbf\\right) + a_1b_2\\left(\\mathbf \\times \\mathbf\\right) + a_1b_3\\left(\\mathbf \\times \\mathbf\\right) + \\\\ &a_2b_1\\left(\\mathbf \\times \\mathbf\\right) + a_2b_2\\left(\\mathbf \\times \\mathbf\\right) + a_2b_3\\left(\\mathbf \\times \\mathbf\\right) + \\\\ &a_3b_1\\left(\\mathbf \\times \\mathbf\\right) + a_3b_2\\left(\\mathbf \\times \\mathbf\\right) + a_3b_3\\left(\\mathbf \\times \\mathbf\\right)\\\\ \\end$ This can be interpreted as the decomposition of into the sum of nine simpler cross products involving vectors aligned with i, j, or k. Each one of these nine cross products operates on two vectors that are easy to handle as they are either parallel or orthogonal to each other. From this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain: :$\\begin \\mathbf\\times\\mathbf = &\\quad\\ a_1b_1\\mathbf + a_1b_2\\mathbf - a_1b_3\\mathbf \\\\ &- a_2b_1\\mathbf + a_2b_2\\mathbf + a_2b_3\\mathbf \\\\ &+ a_3b_1\\mathbf\\ - a_3b_2\\mathbf\\ + a_3b_3\\mathbf \\\\ = &\\left(a_2b_3 - a_3b_2\\right)\\mathbf + \\left(a_3b_1", "Find the cross product $\\mathbf a \\times \\mathbf b$ where $\\mathbf a = \\left<5,2,1\\right>$ and $\\mathbf b = \\left<3,4,3\\right>$. $\\mathbf a \\times \\mathbf b =$ Entering-Vectors.html Your overall score for this problem is", "\\mathbf{e}_\\theta = \\mathbf{e}_\\phi \\\\ \\mathbf{e}_\\theta \\times \\mathbf{e}_\\phi = \\mathbf{e}_r \\\\ \\mathbf{e}_\\phi \\times \\mathbf{e}_r = \\mathbf{e}_\\theta,$$ as are cartesian coordinates (x,y,z):$$\\mathbf{i} \\times \\mathbf{j} = \\mathbf{k} \\\\ \\mathbf{j} \\times \\mathbf{k} = \\mathbf{i} \\\\ \\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}$$ Learning these results and using distributivity of the cross product over vector addition results in better conceptual understanding than using determinants.", "\\mathbf{z} \\cdot (\\mathbf{x} \\times \\mathbf{y})$ 3. Malcev identity:[8] $(\\mathbf{x} \\times \\mathbf{y}) \\times (\\mathbf{x} \\times \\mathbf{z}) = ((\\mathbf{x} \\times \\mathbf{y}) \\times \\mathbf{z}) \\times \\mathbf{x} + ((\\mathbf{y} \\times \\mathbf{z}) \\times \\mathbf{x}) \\times \\mathbf{x} + ((\\mathbf{z} \\times \\mathbf{x}) \\times \\mathbf{x}) \\times \\mathbf{y}$ $\\mathbf{x} \\times (\\mathbf{x} \\times \\mathbf{y}) = -|\\mathbf{x}|^2 \\mathbf{y} + (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{x}.$ Other properties follow only in the three-dimensional case, and are not satisfied by the seven-dimensional cross product, notably, 1. Vector triple product: $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) = (\\mathbf{x} \\cdot \\mathbf{z}) \\mathbf{y} - (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{z}$ 2. Jacobi identity:[8] $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) + \\mathbf{y} \\times (\\mathbf{z} \\times \\mathbf{x}) + \\mathbf{z} \\times (\\mathbf{x} \\times \\mathbf{y}) = 0$ ## Coordinate expressions To define a particular cross product, an orthonormal basis {ej} may be selected and a multiplication table provided that determines all the products {ei × ej}. One possible multiplication table is described in the Example section, but it is not unique.[5] Unlike three dimensions, there are many tables because every pair of unit vectors is perpendicular to five other unit vectors, allowing many choices for each cross product. Once we have established a multiplication table, it is then applied to general vectors x and y by expressing x and y in terms of the basis and expanding x × y through bilinearity. × e1", "cross product of vectors $$\\mathbf{a}$$ and $$\\mathbf{b}:$$ $\\mathbf{a} \\times \\mathbf{b} = \\left({\\mathbf{i} + \\mathbf{j} + \\mathbf{k}}\\right) \\times \\left({\\mathbf{j} - \\mathbf{i}}\\right) = \\left({\\mathbf{i} + \\mathbf{j} + \\mathbf{k}}\\right) \\times \\mathbf{j} - \\left({\\mathbf{i} + \\mathbf{j} + \\mathbf{k}}\\right) \\times \\mathbf{i}.$ Since cross product is anticommutative, we have $\\mathbf{a} \\times \\mathbf{b} = \\mathbf{i} \\times \\left({\\mathbf{i} + \\mathbf{j} + \\mathbf{k}}\\right) - \\mathbf{j} \\times \\left({\\mathbf{i} + \\mathbf{j} + \\mathbf{k}}\\right).$ By the distributive property, $\\mathbf{a} \\times \\mathbf{b} = \\mathbf{i} \\times \\mathbf{i} + \\mathbf{i} \\times \\mathbf{j} + \\mathbf{i} \\times \\mathbf{k} - \\mathbf{j} \\times \\mathbf{i} - \\mathbf{j} \\times \\mathbf{j} - \\mathbf{j} \\times \\mathbf{k}.$ Substitute the results of multiplication of the unit vectors: $\\mathbf{a} \\times \\mathbf{b} = \\mathbf{0} + \\mathbf{k} + \\left({-\\mathbf{j}}\\right) - \\left({-\\mathbf{k}}\\right) - \\mathbf{0} - \\mathbf{i} = -\\mathbf{i} - \\mathbf{j} + 2\\mathbf{k}.$ Now we can calculate the area of the parallelogram: $A = \\Vert{\\mathbf{a} \\times \\mathbf{b}}\\Vert = \\Vert{-\\mathbf{i} - \\mathbf{j} + 2\\mathbf{k}}\\Vert = \\sqrt{\\left({-1}\\right)^2 + \\left({-1}\\right)^2 + 2^2} = \\sqrt{6}.$ ### Example 5. Vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ are not collinear. Find the values of $$\\lambda$$ for which the vectors $$\\lambda\\mathbf{a} + \\mathbf{b}$$ and $$2\\mathbf{a} + \\lambda\\mathbf{b}$$ are collinear. Solution. If two vectors are collinear then their cross product (denote it by c) is equal to the zero vector. Hence $\\mathbf{c} = \\left({\\lambda\\mathbf{a} + \\mathbf{b}}\\right) \\times \\left({2\\mathbf{a} + \\lambda\\mathbf{b}}\\right) = \\mathbf{0}.$ Using the distributive property, we can write:", "# Vector Cross Product Distributes over Addition ## Theorem The vector cross product is distributive over addition. That is, in general: $\\mathbf a \\times \\paren {\\mathbf b + \\mathbf c} = \\paren {\\mathbf a \\times \\mathbf b} + \\paren {\\mathbf a \\times \\mathbf c}$ for $\\mathbf a, \\mathbf b, \\mathbf c \\in \\R^3$. ## Proof Let: $\\mathbf a = \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix}$, $\\mathbf b = \\begin{bmatrix} b_x \\\\ b_y \\\\ b_z \\end{bmatrix}$, $\\mathbf c = \\begin{bmatrix} c_x \\\\ c_y \\\\ c_z \\end{bmatrix}$ Then: $\\displaystyle \\mathbf a \\times \\paren {\\mathbf b + \\mathbf c}$ $=$ $\\displaystyle \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix} \\times \\paren {\\begin{bmatrix} b_x \\\\ b_y \\\\ b_z \\end{bmatrix} + \\begin{bmatrix} c_x \\\\ c_y \\\\ c_z \\end{bmatrix} }$ $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix} \\times {\\begin{bmatrix} b_x + c_x \\\\ b_y + c_y \\\\ b_z + c_z \\end{bmatrix} }$ Definition of Vector Addition $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_y \\paren {b_z + c_z} - a_z \\paren {b_y + c_y} \\\\ a_z \\paren {b_x + c_x} - a_x \\paren {b_z + c_z} \\\\ a_x \\paren {b_y + c_y} - a_y \\paren {b_x + c_x} \\end{bmatrix}$ Definition of Vector Cross Product $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_y b_z + a_y c_z - a_z b_y - a_z c_y \\\\ a_z b_x + a_z c_x - a_x", "Vector Cross Product Distributes over Addition Theorem The vector cross product is distributive over addition. That is, in general: $\\mathbf a \\times \\paren {\\mathbf b + \\mathbf c} = \\paren {\\mathbf a \\times \\mathbf b} + \\paren {\\mathbf a \\times \\mathbf c}$ for $\\mathbf a, \\mathbf b, \\mathbf c \\in \\R^3$. Proof Let: $\\mathbf a = \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix}$, $\\mathbf b = \\begin{bmatrix} b_x \\\\ b_y \\\\ b_z \\end{bmatrix}$, $\\mathbf c = \\begin{bmatrix} c_x \\\\ c_y \\\\ c_z \\end{bmatrix}$ Then: $\\displaystyle \\mathbf a \\times \\paren {\\mathbf b + \\mathbf c}$ $=$ $\\displaystyle \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix} \\times \\paren {\\begin{bmatrix} b_x \\\\ b_y \\\\ b_z \\end{bmatrix} + \\begin{bmatrix} c_x \\\\ c_y \\\\ c_z \\end{bmatrix} }$ $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_x \\\\ a_y \\\\a_z \\end{bmatrix} \\times {\\begin{bmatrix} b_x + c_x \\\\ b_y + c_y \\\\ b_z + c_z \\end{bmatrix} }$ Definition of Vector Addition $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_y \\paren {b_z + c_z} - a_z \\paren {b_y + c_y} \\\\ a_z \\paren {b_x + c_x} - a_x \\paren {b_z + c_z} \\\\ a_x \\paren {b_y + c_y} - a_y \\paren {b_x + c_x} \\end{bmatrix}$ Definition of Vector Cross Product $\\displaystyle$ $=$ $\\displaystyle \\begin{bmatrix} a_y b_z + a_y c_z - a_z b_y - a_z c_y \\\\ a_z b_x + a_z c_x - a_x b_z - a_x", "three whole numbers, then $a\\times \\left( b + c \\right)=a\\times b+a\\times c$. The given product is $728\\times 101$ $=728\\times \\left( 100+1 \\right)$ $=728\\times 100+728\\times 1$, using the distributive property. $=72800+728$, multiplying the numbers. $=73528$. (ii) $\\mathbf{5437}\\times \\mathbf{1001}$ Ans. The given product is $5437\\times 1001$ $=5437\\times \\left( 1000+1 \\right)$ $=5437\\times 1000+5437\\times 1$, using the distributive property. $=5437000+5437$, multiplying the numbers. $=5442437$. (iii) $\\mathbf{824}\\times \\mathbf{25}$ Ans. The given product is $824\\times 25$ $=824\\times \\left( 20+5 \\right)$ $=824\\times 20+824\\times 5$, using the distributive property. $=16480+4120$, multiplying the numbers. $=20600$. (iv) $\\mathbf{4275}\\times \\mathbf{125}$ Ans. The given product is $4275\\times 125$ $=4275\\times \\left( 100+20+5 \\right)$, $=4275\\times 100+4275\\times 20+4275\\times 5$, using the distributive property. $=427500+85500+21375$, multiplying the numbers. $=534375$. (v) $\\mathbf{504}\\times \\mathbf{35}$ Ans. The given product is $504\\times 35$ $=\\left( 500+4 \\right)\\times 35$ $=500\\times 35+4\\times 35$, applying the distributive property. $=17500+140$, multiplying the numbers. $=17640$. 5. Study the pattern : $\\mathbf{1\\times 8+1=9}$; $\\mathbf{12}\\times \\mathbf{8}+\\mathbf{2}=\\mathbf{98}$; $\\mathbf{123}\\times \\mathbf{8}+\\mathbf{3}=\\mathbf{987}$; $\\mathbf{1234}\\times \\mathbf{8}+\\mathbf{4}=\\mathbf{9876}$; $\\mathbf{12345\\times 8+5=98765}$ Write the next two steps. Can you say how the pattern works? (Hint: $\\mathbf{12345=11111+1111+111+11+1}$). Ans. The next two steps will be $123456\\times 8+6=987654$; $1234567\\times 8+7=9876543$. The pattern described in the question will be like the following way: $\\left( 1 \\right)\\times 8+1=9$ $\\left( 12 \\right)\\times 8+2=\\left( 11+1 \\right)\\times 8+2=98$ $\\left( 123 \\right)\\times 8+3=\\left( 111+11+1 \\right)\\times 8+3=987$ $\\left( 1234 \\right)\\times 8+4=\\left( 1111+111+11+1 \\right)\\times 8+4=9876$ $\\left( 12345", "is: $4 \\times (1 + 1 + 1)$. Again, we can solve this in two different ways: • We could distribute the multiplier over the sum inside the parentheses first. This gives us $4 \\times 3 = 12$. • Alternatively, we could multiply each addend separately and then sum. Using this method we get $4 \\times 1 + 4 \\times 1 + 4 \\times 1 = 4 + 4 + 4 = 12$. The result is always the same. As you can see, we can always increase the number of addends: $c \\times (a + b + d) = c \\times a + c \\times b + c \\times d$. The variable $d$ stands for Dave. In contrast, increasing the number of multipliers is not possible and would not make sense at all. If you want to verify this fact, try it out for yourself! • ### Use the Distributive Property to demonstrate what happens when Mrs. Harper bakes fewer pancakes. Hints The Distributive Property works with subtraction, too. We can express that with variables: $c \\times (a - b) = c \\times a - c \\times b$ In general we can say $c \\times \\left( a \\pm b \\right) = c\\times a \\pm c \\times b$. The sign $\\pm$ stands for either $+$ or $-$. Solution The" ]

[ "+0 # help +1 52 1 for a vector v, let r be the reflection of v over the line $$x = t \\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a 2 X 2 matrix R such that r = Rv for all2-dimensional vectors v. Find R. Mar 29, 2020 The matrix is $$\\begin{pmatrix} 1 & 2 \\\\ -2 & 4 \\end{pmatrix}$$", "down the 2 by 2 matrices which 1. reverse the direction for every vector 2. project every vector onto the $x_2$-axis 3. turn every vector counterclowise through $90^\\circ$ 4. reflect every vector through the $45^\\circ$ line $x_1 = x_2$", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "### Home > PC3 > Chapter 9 > Lesson 9.2.2 > Problem9-115 9-115. ​Write the matrix associated with reflecting across the $x$-axis and then rotating $90^\\circ$ counterclockwise. Review your work from problem 9-114. Multiply the transformation matrices in the opposite order you did in problem 9-114.", "Reflection xyz plane I have this matrix: $$A=\\dfrac17\\begin{bmatrix}6 & 3 &-2\\\\3 & -2 & 6 \\\\ -2 & 6 & 3\\end{bmatrix}$$ We have that $$A^TA=I$$. Since $$\\det A=-1$$, its is a reflection matrix. My question is what is the reflecting plane? A key observation is that the normal vector $$n$$ of the reflecting plane gets sent to $$-n$$ by $$A$$. We find $$An = -n = (-I)n$$, where $$I$$ is the identity matrix, hence $$(A + I)n = 0$$. Solving this for $$n$$ we find $$n = (1, -3, 2)$$ as a solution (up to a scalar multiple), so the reflecting plane is given by $$x - 3y + 2z = 0$$.", "Is a reflection matrix a rotation matrix? QuestionsCategory: QuestionsIs a reflection matrix a rotation matrix? I wonder if a reflection matrix is a rotation as well? Nasreenkanwal replied 3 years ago How to sum one matrix into other matrix? [Unrelated link removed by moderator, see posting rules.] Tomas Akenine-Möller Staff answered 3 years ago A reflection matrix is usually not considered to be a rotation matrix, since rotation matrices have $\\det\\mathbf M=1$ and a reflection matrix can have $\\det\\mathbf M=-1$. An example of a reflection matrix is \\mathbf M = \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix}, whose determinant is $-1$. Interestingly, the product of two reflection matrices become a rotation matrix. Assume we have another reflection matrix, i.e., \\mathbf N = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} Their product becomes \\mathbf M\\mathbf N = \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1 \\end{pmatrix}, and as can be seen $\\det(\\mathbf M\\mathbf N)=1$.", "If you find any mistakes, please make a comment! Thank you. ## Solution to Linear Algebra Done Wrong Exercise 1.3.2 Let a linear transformation in $\\mathbb R^2$ be the reflection in the line $x_1=x_2$. Find its matrix. Solution: Let us think $x_1=x$ and $x_2=y$. In the $(x,y)$-coordinate plane, the graph of the inverse function of a function $f(x)$ is obtained by reflecting the graph of $f(x)$ in the line $y=x$. Also note the the inverse function interchanges $x$ and $y$. Denote the linear transformation in $\\mathbb R^2$ corresponding to the reflection in the line $x_1=x_2$ by $T$. With the above observation, we see that $$T(x_1,x_2)^T=(x_2,x_1)^T.$$Therefore, we have $$T\\begin{pmatrix}x_1\\\\ x_2\\end{pmatrix}=x_1\\begin{pmatrix}0\\\\ 1\\end{pmatrix}+x_2\\begin{pmatrix}1\\\\ 0\\end{pmatrix}.$$Hence the matrix is given by $\\begin{pmatrix}0 & 1\\\\ 1 & 0\\end{pmatrix}$.", "# Homework Help: Matrix of reflection 1. Feb 17, 2016 ### DanZ123 1. The problem statement, all variables and given/known data Let u1,u2,u3 be an orthonormal basis for R3 and consider M as the plane with equation x1+2x2-2x3=0. Find the matrix of orthogonal reflection in that plane with respect to the given basis. 2. Relevant equations 3. The attempt at a solution In previous exercises , I had a matrix A given and was asked to find the equation of the plane that the matrix was projected or reflected on. To do that I solved the equation (A-I)x=0 . (The nullspace/kernel minus the identity matrix) .. I was thinking that maybe to solve this current exercise, I could maybe use the method from the previous exercises but use it backwards ... and find the matrix A? But im not sure if im thinking right, or how to attack the problem... Would appreciate help...thanks. 2. Feb 17, 2016 ### geoffrey159 The direction of your reflection is $M^\\perp =\\mathbb{R} (1,2,-2)$. Find an orthonormal basis of $\\mathbb{R}^3$ starting with a unit vector of $M^\\perp$. With respect to that basis, the matrix of reflection is $\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$. What do you need to complete the exercise", "of the reflection matrix. To reflect a vector $v$, one way is to take the line through the endpoint of $v$, with direction equal to a normal vector of the plane, and find the point on the plane where the line intersects the plane. That point is the midpoint of the line segment between the endpoint of $v$ and the endpoint of its reflection. Alternatively, you can project $v$ onto the plane and find the orthogonal complement, $q$ say. Then the reflection of $v$ is just $v-2q$. If you want to construct the matrix using eigenvectors, you need all three. You’ve got two eigenvectors (of $1$) that span the plane, but you also need an eigenvector of $-1$, which will be normal to the plane. You can read a normal to the plane directly from its equation, which can be written in vector form as $\\mathbf n\\cdot\\mathbf x=0$. This gives us $(1,1,1)^T$. In the eigenbasis, the matrix of a reflection is $\\operatorname{diag}(1,1,-1)$, and a change of basis will get us the matrix relative to the standard basis: $$\\begin{bmatrix}-1&_-1&1\\\\1&0&1\\\\0&1&1\\end{bmatrix}\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&-1\\end{bmatrix}\\begin{bmatrix}-1&_-1&1\\\\1&0&1\\\\0&1&1\\end{bmatrix}^{-1}.$$ I’ll leave multiplying this out to you. What you had, however, is almost the way to construct the reflection using a normal to the plane. This requires only a single matrix multiplication (and much simpler one than the previous construction,", "a vector, $x \\text{,}$ and reflect it with respect to the mirror defined by $u \\text{,}$ and you then reflect the result with respect to the same mirror, you should get the original vector $x$ back. Hence, the matrix that represents the reflection should be its own inverse. ###### Homework2.2.5.6. Let $M: \\R^3 \\rightarrow \\R^3$ be defined by $M(x ) = (I - 2 u u^T) x \\text{,}$ where $\\| u \\|_2 = 1 \\text{.}$ Show that the matrix that represents it is unitary (or, rather, orthogonal since it is in $\\R^{3 \\times 3}$). Solution Pushing through the math we find that \\begin{equation*} \\begin{array}{l} (I - 2 u u^T)^T ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt ( A + B )^T = A^T + B^T \\gt \\\\ ( I^T - ( 2 u u^T )^T ) ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt ( \\alpha A B^T )^T = \\alpha B A^T \\gt \\\\ ( I - 2 u u^T ) ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt \\mbox{ distributivity } \\gt \\\\ ( I - 2 u u^T ) - ( I - 2 u u^T ) ( 2 u u^T ) \\\\ ~~~=~~~~ \\lt \\mbox{ distributivity } \\gt \\\\ I - 2 u u^T - 2 u", "10:07 This is the correct answer $$\\begin{bmatrix} 0 & -1 \\\\ -1 & 0 \\end{bmatrix}$$ The first column is the reflection of $(1,0)$ and the second column is the reflection of $(0,1)$.", "y\\end{matrix}\\right) = \\left(\\begin{matrix}- x \\\\ y\\end{matrix}\\right)$ That is a reflection in the y-axis.", "From reflection plane to “reflection” matrix How do I determine the \"reflection\" matrix, relative to the canonical basis, of a reflection on the plane $\\{x+y+z = 0\\}$ in $\\mathbb{R}^3$ ? I figured I should take the two eigenvectors which should be $n =(-1, 1, 0)$ and $v= (-1, 0, 1)$. Take one of the two and compute the Matrix with this: $M = I-2n^Tn$, but it's not working. If $\\bf n$ is a unit normal to the plane (as a column vector), the reflection across that plane is $I - 2 {\\bf n} {\\bf n}^T$. EDIT: In your case $${\\bf n} = \\pmatrix{1/\\sqrt{3}\\cr 1/\\sqrt{3}\\cr 1/\\sqrt{3}\\cr}$$ • So n shouldn't be the eigenvector ?! – Leroy Aug 26 '17 at 0:52 • $\\bf n$ is an eigenvector with eigenvalue $-1$. – Robert Israel Aug 26 '17 at 0:53 • Can you shortly explain how you found \"your\" n ?! – Leroy Aug 26 '17 at 0:54 • @Leroy You can read a normal to the plane directly from its equation, then normalize. Alternatively, take the cross product of your two vectors that lie in the plane and normalize that. – amd Aug 26 '17 at 1:07 A fairly straightforward way is to reflect the standard unit vectors directly. The reflected vectors (in standard basis order) will be the columns", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "find that the image in this case is spanned by the matrices $$\\begin{pmatrix} 1 & 0\\\\2 & 0\\end{pmatrix} \\quad \\text{ and } \\quad \\begin{pmatrix} 0 & 1\\\\0 & 2\\end{pmatrix}.$$", "# Reflection Matrix linear algebra I am practicing some linear algebra question to prepare for my test. I have come across one question that has given me much trouble. It states: If $\\lVert u\\rVert = 1$, then $Q = I - 2uu^T$ is a reflection matrix. I have established this: If you reflect twice of across the same axis twice $Q^2 = I$. First, it says to compute $Qu$ and simplify as much as possible. Does this just mean move the equation around to get $Qu$? Suppose $v$ is orthogonal to $u$. Compute $Qv$. Im not sure how this is done. Then im asked to explain in plain English which subspace $Q$ is reflecting across. The last one states: Compute the reflection matrix $Q_1 = I - 2u_1 u_1^T$ where $u_1 = (0,1)$. Compute $Q_1 x_1$, where $x_1 = (0,1)$ and sketch the vectors $u1$, $x1$ and $Q_1x_1$ in the plane. I'm not sure how to even start that one. If anyone could clear this up for me that would be much help ! • Please review my edits to see if I interpreted your text correctly, and also take note of correct Mathjax usage. – dfeuer Oct 31 '13 at 16:38 • Yes that all makes sense. – user104536 Oct 31 '13 at 16:39 ## 1 Answer", "+0 # Matrices 0 80 1 The matrix for reflecting through a certain plane P which passes through the origin, is given by Find the normal vector of plane P. Aug 8, 2020 edited by Guest Aug 8, 2020", "Finding $\\theta$ and unit vector for a reflection matrix I'm given a transformation matrix, and am told that it is a reflection combined with scaling: \\begin{pmatrix} 4 & 3 \\\\ 3 & -4 \\end{pmatrix} I'm asked to do two things: 1) Find $\\theta$: how do I understand an angle theta in relation to a reflection? This is intuitive to me for a rotation, but not for a reflection. Is it twice the angle between a given vector and the line of reflection, so that the reflection across the line is viewed as a rotation? How might I find this? 2) Find the unit vector for the line of reflection. Since this is a reflection, I set $2u_1^2 - 1 = 4$ to get $u_1 = \\pm\\frac 52$, but then $2u_2^2 - 1 = -4$ nets an imaginary number. What am I doing wrong? I have heard that reflections involve imaginary numbers and we haven't covered that yet, but I'm lost on how else I might find the unit vector. I'm looking for pointers/concept explanations, rather than outright answers. Appreciate any help! • Where did the expression $2u_1^2 - 1 = 4$ come from? – amd Sep 21 '17 at 21:31 The line of reflection bisects the angle between a vector and its image. Pick any convenient unit vector", "0 \\ end{pmatrix} denote ? Using the given matrix , find the image of the point A (5 , -7) reflection in y = x , (7 , 5) reflection in y = -x , (7 , -5) reflection in x = y , (6 , 3) reflection in y = x , (3 , 4) • ### What transformation does the matrix \begin{pmatrix} 0 & -1 \\ -1 & 0 \\ end{pmatrix} denote ? Using the given matrix , find the image of the point . reflection in y = x , (5 , 8) reflection in y = -x , (0 , -1) reflection in y = -x , (-5 , -8) reflection in y = X , (6 , 5) • ### Which transformation isrepresented by the matrix \begin{pmatrix} 0 & -1 \\ -1 & 0 \\ end{pmatrix} Find the image (5 , -7) using this matrix. reflection in y = y ; (-7 , 4) reflection in y = -x ; (-4 , 1) reflection in y = -x ; (7 , -5) reflection in y = x ; (-7 , 5) • ### What transformation does the matrix \begin{pmatrix} 0 & -1 \\ -1 & 0 \\ end{pmatrix} denote ? Using the given matrix , find the image of the point (-3 ,", "2\\theta, \\sin 2\\theta)$ and $(0,1)$ gets reflected to $(\\sin 2\\theta, -\\cos 2\\theta)$. Another way is to observe that we can rotate an arbitrary mirror line onto the x-axis, then reflect across the x-axis, and rotate back. (The matrix product $[T]_{xy}$ can be seen as operating this way.) We took neither of these two approaches, because to justify them rigorously takes a bit of work, that is avoided by the pure linear algebra approach. Note also that $[T]_{uv}$ and $[T]_{xy}$ are orthogonal matrices, with determinant $-1$, as expected. \\section*{Reflection across a line of given direction vector} Suppose instead of being given an angle $\\theta$, we are given the unit direction vector $u$ to reflect the vector $w$. We can derive the matrix for the reflection directly, without involving any trigonometric functions. In the decomposition $\\vw = a \\vu + b\\vv$, we note that $b = \\vw \\cdot \\vv$. Therefore $\\vw' = (a\\vu + b\\vv) - 2b\\vv = \\vw - 2(\\vw \\cdot \\vv) \\vv\\,.$ (In fact, this is the formula used in the \\PMlinkescapetext{source code} to draw the diagram in this entry.) To derive the matrix with respect to $x,y$ coordinates, we resort to a trick: $\\vw' = I\\vw - 2\\vv (\\vw \\cdot \\vv) = I\\vw - 2\\vv (\\transpose{\\vv} \\vw) = I\\vw - 2(\\vv \\transpose{\\vv}) \\vw\\,.$ Therefore the matrix of" ]

[ "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "{e} _{x}\\\\\\mathbf {b} _{y}&=&{\\frac {\\mathbf {e} _{z}}{\\mathbf {d} _{z}}}\\mathbf {d} _{y}-\\mathbf {e} _{y}\\\\\\end{array}}.}$ Or, in matrix form using homogeneous coordinates, the system ${\\displaystyle {\\begin{bmatrix}\\mathbf {f} _{x}\\\\\\mathbf {f} _{y}\\\\\\mathbf {f} _{z}\\\\\\mathbf {f} _{w}\\\\\\end{bmatrix}}={\\begin{bmatrix}1&0&-{\\frac {\\mathbf {e} _{x}}{\\mathbf {e} _{z}}}&0\\\\0&1&-{\\frac {\\mathbf {e} _{y}}{\\mathbf {e} _{z}}}&0\\\\0&0&1&0\\\\0&0&1/\\mathbf {e} _{z}&0\\\\\\end{bmatrix}}{\\begin{bmatrix}\\mathbf {d} _{x}\\\\\\mathbf {d} _{y}\\\\\\mathbf {d} _{z}\\\\1\\\\\\end{bmatrix}}}$ in conjunction with an argument using similar triangles, leads to division by the homogeneous coordinate, giving ${\\displaystyle {\\begin{array}{lcl}\\mathbf {b} _{x}&=&\\mathbf {f} _{x}/\\mathbf {f} _{w}\\\\\\mathbf {b} _{y}&=&\\mathbf {f} _{y}/\\mathbf {f} _{w}\\\\\\end{array}}.}$ The distance of the viewer from the display surface, ${\\displaystyle \\mathbf {e} _{z}}$, directly relates to the field of view, where ${\\displaystyle \\alpha =2\\cdot \\tan ^{-1}(1/\\mathbf {e} _{z})}$ is the viewed angle. (Note: This assumes that you map the points (-1,-1) and (1,1) to the corners of your viewing surface) The above equations can also be rewritten as: ${\\displaystyle {\\begin{array}{lcl}\\mathbf {b} _{x}=(\\mathbf {d} _{x}\\mathbf {s} _{x})/(\\mathbf {d} _{z}\\mathbf {r} _{x})\\mathbf {r} _{z}\\\\\\mathbf {b} _{y}=(\\mathbf {d} _{y}\\mathbf {s} _{y})/(\\mathbf {d} _{z}\\mathbf {r} _{y})\\mathbf {r} _{z}\\\\\\end{array}}.}$ In which ${\\displaystyle \\mathbf {s} _{x,y}}$ is the display size, ${\\displaystyle \\mathbf {r} _{x,y}}$ is the recording surface size (CCD or film), ${\\displaystyle \\mathbf {r} _{z}}$ is the distance from the recording surface to the entrance pupil (camera center), and ${\\displaystyle \\mathbf {d} _{z}}$ is the distance, from the 3D point being projected, to", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection", "# Proof: Every H-Projection is product of H-Reflections I have some problems with the proof of the statement above (sorry for a bad translation I guess). First of all some definitions: Definition 1 (möbius-transformation): Let $A \\in \\mathbb{C}^{2 \\times 2}$ with $det A=ad-bc \\neq 0$ then 1. for $c \\neq 0$: $$\\varphi_A: \\begin{cases} \\mathbb{C}_{\\backslash{\\lbrace -d/c \\rbrace}} &\\to \\mathbb C \\\\ z &\\to \\frac{az+b}{cz+d}=\\frac 1 c \\cdot (a-\\frac{ad-bc}{cz+d})\\end{cases}$$ 2. for $c=0$: $$\\varphi_A: \\begin{cases} \\mathbb{C} &\\to \\mathbb C \\\\ z &\\to \\frac{az+b}{cz+d}=\\frac a d \\cdot z + \\frac b d\\end{cases}$$ is called möbius-tranformation. Definition 2 (H-Projection): Let $\\mathcal{B}$ the set of all H-Projections with $\\beta: \\mathbb H \\to \\mathbb H$ and 1. (typ 1): $\\beta:=\\frac{az+b}{cz+d}$ with $a,b,c,d \\in \\mathbb R$ and $\\left|\\begin{array} aa & b \\\\ c & d \\end{array}\\right|>0$ 2. (typ 2): $\\beta:=\\frac{a\\overline z+b}{c\\overline z+d}$ with $a,b,c,d \\in \\mathbb R$ and $\\left|\\begin{array} aa & b \\\\ c & d \\end{array}\\right|<0$ Definition 3 (H-reflection): There are two kinds of H-reflections: 1. (typ 1): Let $g$ with $Re(z)=a$ be a H-Line. Then we have with $\\sigma_g(z)=-\\overline z+2a$ a H-transformation. 2. (typ 2): Let $g$ with $|z-a|^2=r^2$ be a H-Line. Then we have with $\\sigma_g(z)=\\frac{a \\overline z+(r^2-a^2)}{\\overline z -a}=\\frac{r^2}{\\overline z -a}+a$ a H-transformation. Theorem 1: Every H-Projection is product of H-Reflections Proof: Given $\\beta \\in \\mathcal B$ with $\\beta=\\frac{az+b}{cz+d}$ ($a,b,c,d \\in \\mathbb", "a ball at position $(0,0,1)$, i.e. on the unit vector $\\mathbf e_z$, its shadow appears at the point $(-1,0,0)$. If you place it on the table, of course the shadow will \"coincide\" with the ball, i.e. $\\mathbf e_x$ and $\\mathbf e_y$ will be fixed, as before. So the matrix you are looking for is $$C = \\begin{pmatrix} 1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ and you may check that the null-space of $C$ is spanned by $(x,0,x)$, i.e. it is a diagonal line. This kinds of matrices represent linear operators called projectors, i.e. such that if you apply them twice you get the same result ($PP = P$: this identity is evident in the vertical-shadow example – try multiplying matrix $A$ by itself).", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "\\mathbf {u} =\\mathbf {x} +\\alpha \\mathbf {e} _{1},}$ and ${\\displaystyle \\mathbf {v} ={\\mathbf {u} \\over \\|\\mathbf {u} \\|}.}$ Here, ${\\displaystyle \\alpha =-14}$ and ${\\displaystyle \\mathbf {x} =\\mathbf {a} _{1}=(12,6,-4)^{T}}$ Therefore ${\\displaystyle {\\mathbf {u} }=(-2,6,-4)^{T}=({2})(-1,3,-2)^{T}}$ and ${\\displaystyle \\mathbf {v} ={1 \\over {\\sqrt {14}}}(-1,3,-2)^{T}}$, and then ${\\displaystyle Q_{1}=I-{2 \\over {\\sqrt {14}}{\\sqrt {14}}}{\\begin{pmatrix}-1\\\\3\\\\-2\\end{pmatrix}}{\\begin{pmatrix}-1&3&-2\\end{pmatrix}}}$ ${\\displaystyle =I-{1 \\over 7}{\\begin{pmatrix}1&-3&2\\\\-3&9&-6\\\\2&-6&4\\end{pmatrix}}}$ ${\\displaystyle ={\\begin{pmatrix}6/7&3/7&-2/7\\\\3/7&-2/7&6/7\\\\-2/7&6/7&3/7\\\\\\end{pmatrix}}.}$ Now observe: ${\\displaystyle Q_{1}A={\\begin{pmatrix}14&21&-14\\\\0&-49&-14\\\\0&168&-77\\end{pmatrix}},}$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to ${\\displaystyle A'=M_{11}={\\begin{pmatrix}-49&-14\\\\168&-77\\end{pmatrix}}.}$ By the same method as above, we obtain the matrix of the Householder transformation ${\\displaystyle Q_{2}={\\begin{pmatrix}1&0&0\\\\0&-7/25&24/25\\\\0&24/25&7/25\\end{pmatrix}}}$ after performing a direct sum with 1 to make sure the next step in the process works properly. Now, we find ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}6/7&-69/175&58/175\\\\3/7&158/175&-6/175\\\\-2/7&6/35&33/35\\end{pmatrix}}}$ Then ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}0.8571&-0.3943&0.3314\\\\0.4286&0.9029&-0.0343\\\\-0.2857&0.1714&0.9429\\end{pmatrix}}}$ ${\\displaystyle R=Q_{2}Q_{1}A=Q^{T}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&-35\\end{pmatrix}}.}$ The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition. Using Givens rotations QR decompositions can also be computed with a series of Givens rotations. Each rotation zeros an element in the subdiagonal of the matrix, forming the R matrix. The concatenation of all the Givens rotations forms the orthogonal Q matrix. In practice, Givens rotations are not actually performed by building a whole matrix and doing a matrix multiplication. A Givens rotation procedure is", "fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}. since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W. finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so $P = \\begin{bmatrix}\\frac{1}{2}&0&-\\frac{1}{2}\\\\0&1&0\\\\ -\\frac{1}{2}&0&\\frac{1}{2} \\end{bmatrix}$ one can easily verify that P2 = P (this is the defining property of reflections). now our reflection R is a \"reflection about W\", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w. we can write this in terms of P like so: for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w. that is: u = v - P(v), and w = P(v), so: R(v) = -u + w = -(v - P(v)) + P(v) = -v + P(v) + P(v) = 2P(v) - I(v) = (2P- I)(v), so that R = 2P - I or: $R = \\begin{bmatrix}0&0&-1\\\\0&1&0\\\\-1&0&0 \\end{bmatrix}$ i always like to \"make sure\" my P and R do what the problem says they should do. so", "of the algorithm (in terms of floating point multiplications) is given by ${\\displaystyle {\\frac {2}{3}}n^{3}+n^{2}+{\\frac {1}{3}}n-2=O(n^{3}).}$ #### Example Let us calculate the decomposition of ${\\displaystyle A={\\begin{pmatrix}12&-51&4\\\\6&167&-68\\\\-4&24&-41\\end{pmatrix}}.}$ First, we need to find a reflection that transforms the first column of matrix A, vector ${\\displaystyle \\mathbf {a} _{1}=(12,6,-4)^{T}}$, into ${\\displaystyle \\|\\mathbf {a} _{1}\\|\\;\\mathbf {e} _{1}=(14,0,0)^{T}.}$ Now, ${\\displaystyle \\mathbf {u} =\\mathbf {x} -\\alpha \\mathbf {e} _{1},}$ and ${\\displaystyle \\mathbf {v} ={\\mathbf {u} \\over \\|\\mathbf {u} \\|}.}$ Here, ${\\displaystyle \\alpha =14}$ and ${\\displaystyle \\mathbf {x} =\\mathbf {a} _{1}=(12,6,-4)^{T}}$ Therefore ${\\displaystyle \\mathbf {u} =(-2,6,-4)^{T}=({2})(-1,3,-2)^{T}}$ and ${\\displaystyle \\mathbf {v} ={1 \\over {\\sqrt {14}}}(-1,3,-2)^{T}}$, and then ${\\displaystyle Q_{1}=I-{2 \\over {\\sqrt {14}}{\\sqrt {14}}}{\\begin{pmatrix}-1\\\\3\\\\-2\\end{pmatrix}}{\\begin{pmatrix}-1&3&-2\\end{pmatrix}}}$ ${\\displaystyle =I-{1 \\over 7}{\\begin{pmatrix}1&-3&2\\\\-3&9&-6\\\\2&-6&4\\end{pmatrix}}}$ ${\\displaystyle ={\\begin{pmatrix}6/7&3/7&-2/7\\\\3/7&-2/7&6/7\\\\-2/7&6/7&3/7\\\\\\end{pmatrix}}.}$ Now observe: ${\\displaystyle Q_{1}A={\\begin{pmatrix}14&21&-14\\\\0&-49&-14\\\\0&168&-77\\end{pmatrix}},}$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to ${\\displaystyle A'=M_{11}={\\begin{pmatrix}-49&-14\\\\168&-77\\end{pmatrix}}.}$ By the same method as above, we obtain the matrix of the Householder transformation ${\\displaystyle Q_{2}={\\begin{pmatrix}1&0&0\\\\0&-7/25&24/25\\\\0&24/25&7/25\\end{pmatrix}}}$ after performing a direct sum with 1 to make sure the next step in the process works properly. Now, we find ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}6/7&-69/175&58/175\\\\3/7&158/175&-6/175\\\\-2/7&6/35&33/35\\end{pmatrix}}}$ Or, to four decimal digits, ${\\displaystyle Q=Q_{1}^{T}Q_{2}^{T}={\\begin{pmatrix}0.8571&-0.3943&0.3314\\\\0.4286&0.9029&-0.0343\\\\-0.2857&0.1714&0.9429\\end{pmatrix}}}$ ${\\displaystyle R=Q_{2}Q_{1}A=Q^{T}A={\\begin{pmatrix}14&21&-14\\\\0&175&-70\\\\0&0&-35\\end{pmatrix}}.}$ The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition. The use of Householder transformations is", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "B. $\\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ E. $\\begin{pmatrix} 0 &-1 \\\\-1 & 0 \\end{pmatrix}$ C. $\\begin{pmatrix} 1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ Solution The reflection across the line $y = mx$ can be represented by the matrix in form of $\\begin{pmatrix} m & 0 \\\\ 0 & m \\end{pmatrix}$. For the reflection across the line $y=\\dfrac13x$, the representation matrix is given by $T_1 = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix}$ For the reflection across the line $y=-3x$, the representation matrix is given by $T_1 = \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix}$ Hence, the representation matrix for $T$ can be expressed as follows. \\begin{aligned} T & = T_2 \\cdot T_1 \\\\ & = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix} \\cdot \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix} \\end{aligned} [collapse] Problem Number 21 A photocopier can make copies of images/writing of different sizes. A rectangular image is copied with certain settings. If this setting can be compared to the transformation process of the matrix $\\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix}$, then dilated in the origin by scale factor of $3$, then the area of ​​the rectangular image will be $\\cdots$ times the original area. A. $12$ C.", "_{i=1}^{m}b_{ji}a_{ij}=\\sum _{j=1}^{n}\\left(\\mathbf {B} \\mathbf {A} \\right)_{jj}=\\operatorname {tr} (\\mathbf {B} \\mathbf {A} ).}$ 2. ^ For example, if ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix}},\\quad \\mathbf {B} ={\\begin{pmatrix}0&0\\\\1&0\\end{pmatrix}},}$ then the product is ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}1&0\\\\0&0\\end{pmatrix}},}$ and the traces are tr(AB) = 1 ≠ 0 ⋅ 0 = tr(A)tr(B). 3. ^ Proof: Let ${\\displaystyle e_{ij}}$ the standard basis and note that ${\\displaystyle f\\left(e_{ij}\\right)=0}$ if and only if ${\\displaystyle i\\neq j}$ and ${\\displaystyle f\\left(e_{jj}\\right)=f\\left(e_{11}\\right)}$ ${\\displaystyle f(\\mathbf {A} )=\\sum _{i,j}[\\mathbf {A} ]_{ij}f\\left(e_{ij}\\right)=\\sum _{i}[\\mathbf {A} ]_{ii}f\\left(e_{11}\\right)=f\\left(e_{11}\\right)\\operatorname {tr} (\\mathbf {A} ).}$ More abstractly, this corresponds to the decomposition ${\\displaystyle {\\mathfrak {gl}}_{n}={\\mathfrak {sl}}_{n}\\oplus k,}$ as ${\\displaystyle \\operatorname {tr} (AB)=\\operatorname {tr} (BA)}$ (equivalently, ${\\displaystyle \\operatorname {tr} ([A,B])=0}$) defines the trace on ${\\displaystyle {\\mathfrak {sl}}_{n},}$ which has complement the scalar matrices, and leaves one degree of freedom: any such map is determined by its value on scalars, which is one scalar parameter and hence all are multiple of the trace, a nonzero such map. 4. ^ Proof: ${\\displaystyle {\\mathfrak {sl}}_{n}}$ is a semisimple Lie algebra and thus every element in it is a linear combination of commutators of some pairs of elements, otherwise the derived algebra would be a proper ideal. 5. ^ This follows from the fact that tr(A*A) = 0 if and only if A = 0. ## References 1. \"Rank, trace, determinant, transpose, and inverse of", "\\begin{alignat}{2} 10&x+&3&y = 2 \\\\\\\\ 3&x+&13&y = 4 \\end{alignat} \\begin{gather} a=b \\\\\\\\ e=b+c \\end{gather} \\begin{CD} A @>a\\>> B \\\\\\\\ @VbVV @AAcA \\\\\\\\ C @= D \\end{CD} The rendered output looks like this: $$c = \\pm\\sqrt{a^2 + b^2}$$ $f(x)=\\int_{-\\infty}^{\\infty} \\hat{f}(\\xi) e^{2 \\pi i \\xi x} d \\xi$ \\begin{equation*} \\rho \\frac{\\mathrm{D} \\mathbf{v}}{\\mathrm{D} t}=\\nabla \\cdot \\mathbb{P}+\\rho \\mathbf{f} \\end{equation*} $$\\mathbf{E}=\\sum_{i} \\mathbf{E}_{i}=\\mathbf{E}_{1}+\\mathbf{E}_{2}+\\mathbf{E}_{3}+\\cdots$$ \\begin{align} a&=b+c \\\\ d+e&=f \\end{align} \\begin{alignat}{2} 10&x+&3&y = 2 \\\\ 3&x+&13&y = 4 \\end{alignat} \\begin{gather} a=b \\\\ e=b+c \\end{gather} \\begin{CD} A @>a>> B \\\\ @VbVV @AAcA \\\\ C @= D \\end{CD} Tip You can add more inline and block delimiters in your site configuration. #### Copy-tex Copy-tex is an extension for $\\KaTeX$. By the extension, when selecting and copying $\\KaTeX$ rendered elements, copies their $\\LaTeX$ source to the clipboard. Set the property copyTex = true under [params.math] in your site configuration to enable Copy-tex. Select and copy the formula rendered in the previous section, and you can find that the copied content is the $\\LaTeX$ source code. #### mhchem mhchem is an extension for $\\KaTeX$. By the extension, you can write beautiful chemical equations easily in the article. Set the property mhchem = true under [params.math] in your site configuration to enable mhchem. $$\\ce{CO2 + C -> 2 CO}$$ $$\\ce{Hg^2+ ->[I-] HgI2 ->[I-] [Hg^{II}I4]^2-}$$ The rendered output looks like this: $$\\ce{CO2", "+0 # I will really appreciate any help. 0 28 1 I will really appreciate any help. Let $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ be the following matrices: \\begin{align*} &\\mathbf{A} = \\begin{pmatrix} 0 & 0 \\\\ 0& 0 \\end{pmatrix},\\\\ &\\mathbf{B} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1\\end{pmatrix}, \\\\ &\\mathbf{C} = \\begin{pmatrix} 1/2 & 1/2 \\\\ 1/2 & 1/2\\end{pmatrix}, \\\\ &\\mathbf{D} = \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix}. \\end{align*} Also, consider the following functions: \\begin{align*} &\\text{Reflection across a line with direction vector \\begin{pmatrix}1\\\\-1 \\end{pmatrix}},\\\\ &\\text{The function that sends every vector to the zero vector}, \\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{The function that sends every vector to its negative},\\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{Projection onto \\begin{pmatrix}1\\\\1 \\end{pmatrix}}. \\end{align*} How do we order the list $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ to match the order of the functions above? I am having a hard time understanding problem such as these. Thank you in advance for any help Jul 20, 2022", "$f$ is an isometry of $\\mathbb{R}^n$ hence $\\forall x,y\\in\\mathbb{R}^n:d(x,y)=d(f(x),f(y))$, in particular this holds for very $x,y\\in V$ . Let $E$ be the projection $E:\\mathbb{R}^n\\to\\mathbb{R}^{n-1}$. $E(V)=\\mathbb{R}^{n-1}$and $E\\circ f|_{V}$ is an isometry of $\\mathbb{R}^{n-1}$ hence be the induction hypothesis it can be represented as $R=R_{1}\\circ...\\circ R_{n-1}$. We now expand $R$ to $\\mathbb{R}^n$ in this manner: denote the matrix representin $R$ as $R_{M}$ ($R_{M}\\in M_{n-1}(\\mathbb{R})$) then $R$ expanded to $\\mathbb{R}^n$ is represented by $\\begin{pmatrix}R & 0\\\\ 0 & 1 \\end{pmatrix}$that is : on the first $n-1$coordinates we act the same as $R$ did and we keep the last coordinate (note that $|R_{M}|=1$ and that it is straightforward to check that $R_{M}$ is an orthogonal matrix, $R_{1},...,R_{n}$are expanded from $\\mathbb{R}^{n-1}$to $R^{n}$ in the same manner). Denote $\\varphi=R^{-1}\\circ f$ (here $R$ is the expended isometry to $\\mathbb{R}^n$). if $\\varphi(e_{n})=e_{n}$ then we are done, otherwise there exist a reflection $\\alpha$ s.t $\\alpha\\circ\\varphi(e_{n})=e_{n}$and s.t $\\alpha|_{V}\\equiv f|_{V}$. (explicitly: $\\alpha$ is a rotation around the plane consistent of all points $v\\in\\mathbb{R}^n$ s.t $\\langle v,e_{n}-f(e_{n})\\rangle=0$). $\\varphi$ is at most $n-1$ reflections hence $\\alpha\\circ\\varphi$ is at most $n$ reflections. In te first case $(f(0)\\neq0)$ we are doing the process with $\\tau\\circ f$ hence in this case $f$ is represented by at most $n+1$ reflections. Any comment about the style of the proof is also welcomed, this is one of", "tells us that ker(P) = <(1,0,1)>, and this is a normal vector to the plane x + z = 0 (we can divide by 2 in the plane equation). let's call ker(P), U, and our plane x + z = 0, W = im(P). the fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}. since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W. finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so $P = \\begin{bmatrix}\\frac{1}{2}&0&-\\frac{1}{2}\\\\0&1&0\\\\ -\\frac{1}{2}&0&\\frac{1}{2} \\end{bmatrix}$ one can easily verify that P2 = P (this is the defining property of reflections). now our reflection R is a \"reflection about W\", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w. we can write this in terms of P like so: for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w. that is: u = v - P(v), and w = P(v), so: R(v) = -u + w = -(v - P(v)) +", "ray is closer. Remark 3. One might be tempted to apply the following method of 3D drawing. Take all segments/rays/lines to be drawn, sort them using some sort method utilizing the above (or anyone other's) algorithm as a subroutine (callback), and draw the lines starting from the \"outermost\" and finishing with the \"closest\" one. I must warn against doing this because the visual precedence discussed here is nontransitive and therefore it cannot be used in any kind of sorting. Take, for example, 3 lines: $$A=(2,p,1), \\; B=(1,2,q), \\; C=(r,1,2), \\quad p,q,r \\in \\mathbb R.$$ Let the viewpoint be at the origin and the eyeline be $(1,1,1)$. Here $$\\left[ \\begin{array}{c|c|c|c|c|c} \\mathring{\\mathbf a} & \\mathbf{\\tilde a} & \\mathring{\\mathbf b} & \\mathbf{\\tilde b} & \\mathring{\\mathbf c} & \\mathbf{\\tilde c} \\end{array} \\right]= \\begin{bmatrix} 2 & 0 & 1 & 0 & 0 & 1 \\\\ 0 & 1 & 2 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 1 & 2 & 0 \\\\ \\end{bmatrix}.$$ Comparing $A$ and $B$, we get $\\Delta_{\\;a}^{AB}=2, \\Delta_{\\;b}^{AB}=1$, so $B \\prec A$, where the sign $\\prec$ means \"in front of\". Similarly, $\\Delta_{\\;b}^{BC}=2, \\Delta_{\\;c}^{BC}=1$ and $\\Delta_{\\;a}^{AC}=-1, \\Delta_{\\;c}^{AC}=-2$. We see that $C \\prec B$, but $A \\prec C$. This creates nontransitive cycle $B \\prec A \\prec C \\prec B$. Any optimized sorting algorithm would", "\\tfrac{1}{4}\\left[(a+d)^2 - \\left((a-d)^2+4b^2\\right)\\right] - a(a+d) + \\left(a^2+b^2\\right)\\\\ &= \\tfrac{1}{4}\\left[a^2+2ad+d^2 - a^2+2ad-d^2-4b^2\\right] - a^2-ad + a^2+b^2\\\\ &= \\tfrac{1}{4}\\left[4ad-4b^2\\right] -ad+b^2\\\\ &=0 \\end{align*} from which the result follows. Now if we construct a matrix $$\\mathbf{V}$$ whose columns are the eigenvectors $$\\mathbf{v}_0$$ and $$\\mathbf{v}_1$$ Derivation 2: The Product Of VT And V \\begin{align*} \\mathbf{V}^\\mathrm{T} \\times \\mathbf{V} &= \\begin{pmatrix} v_{00} & v_{01}\\\\ v_{10} & v_{11} \\end{pmatrix} \\times \\begin{pmatrix} v_{00} & v_{10}\\\\ v_{01} & v_{11} \\end{pmatrix} \\\\&= \\begin{pmatrix} v_{00} \\times v_{00} + v_{01} \\times v_{01} & v_{00} \\times v_{10} + v_{01} \\times v_{11}\\\\ v_{10} \\times v_{00} + v_{11} \\times v_{01} & v_{10} \\times v_{10} + v_{11} \\times v_{11} \\end{pmatrix} \\end{align*} Now the elements of this matrix are simply the products of the eigenvectors, so \\begin{align*} \\mathbf{V}^\\mathrm{T} \\times \\mathbf{V} &= \\begin{pmatrix} \\mathbf{v}_0\\times\\mathbf{v}_0 & \\mathbf{v}_0\\times\\mathbf{v}_1\\\\ \\mathbf{v}_1\\times\\mathbf{v}_0 & \\mathbf{v}_1\\times\\mathbf{v}_1 \\end{pmatrix} \\\\&= \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\end{align*} by the properties of the eigenvectors proven above. \\begin{align*} \\mathbf{v}_0 &= \\begin{pmatrix} v_{00}\\\\ v_{01} \\end{pmatrix} \\\\ \\mathbf{v}_1 &= \\begin{pmatrix} v_{10}\\\\ v_{11} \\end{pmatrix} \\\\ \\mathbf{V} &= \\begin{pmatrix} v_{00} & v_{10}\\\\ v_{01} & v_{11} \\end{pmatrix} \\end{align*} we have $\\mathbf{V}^\\mathrm{T} \\times \\mathbf{V} = \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix}$ as shown in derivation 2, and hence the transpose of the matrix is equal to its inverse. If we also construct a diagonal matrix $$\\mathbf{\\Lambda}$$ whose diagonal elements are", "${\\displaystyle ({\\begin{smallmatrix}{\\frac {\\phi }{2}},{\\frac {1}{2}},{\\frac {\\phi -1}{2}}\\end{smallmatrix}})}$n (0,1,0)n (φ,1,0)axis (1,1,1)axis (1,0,0)axis ### Affine rank 3 A simple example affine group is [4,4] () (p4m), can be given by three reflection matrices, constructed as a reflection across the x axis (y=0), a diagonal (x=y), and the affine reflection across the line (x=1). [4,4]+ () (p4) is generated by S0,1 S1,2, and S0,2. [4+,4+] () (pgg) is generated by 2-fold rotation S0,2 and transreflection V0,1,2. [4+,4] () (p4g) is generated by S0,1 and R3. The group [(4,4,2+)] () (cmm), is generated by 2-fold rotation S1,3 and reflection R2. [4,4], Reflections Rotations Rotoreflection Name R0 R1 R2 S0,1 S1,2 S0,2 V0,1,2 Group Order 2 2 2 4 2 Matrix ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&-1&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\1&0&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}-1&0&2\\\\0&1&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\-1&0&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\-1&0&2\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}-1&0&2\\\\0&-1&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\1&0&-2\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ### Rank 4 A irreducible 4-dimensional finite reflective group is hyperoctahedral group (or hexadecachoric group (for 16-cell), B4=[4,3,3], order 384, . The reflection generators matrices are R0, R1, R2, R3. R02=R12=R22=R32=(R0×R1)4=(R1×R2)3=(R2×R3)3=(R0×R2)2=(R1×R3)2=(R0×R3)2=Identity. Chiral hyperoctahedral symmetry, [4,3,3]+, () is generated by 3 of 6 rotations: S0,1, S1,2, S2,3, S0,2, S1,3, and S0,3. Hyperpyritohedral symmetry [4,(3,3)+], () is generated by reflection R0 and rotations S1,2 and S2,3. An 8-fold double rotation is generated by W0,1,2,3, the product of all 4 reflections. [4,3,3], Reflections Rotations Rotoreflection Double rotation Name R0 R1 R2", "$$\\operatorname{det}(A) = \\pm 1 \\in { \\mathbf{F} }_p^{\\times}$$, so the vectors $$\\mathbf{e}_1 - \\mathbf{e}_i$$ are still linearly independent and thus $$\\dim E_{-1} = p-1$$ • So there are $$p-1$$ Jordan blocks for $$\\lambda = 0$$. Summary: \\begin{align*} \\min_{A, { \\mathbf{F} }_p}(x) &= (x+1)^2 \\\\ \\chi_{A, { \\mathbf{F} }_p}(x) &\\equiv (x+1)^p \\\\ \\dim E_{-1} &= p-1 .\\end{align*} Thus \\begin{align*} JCF_{{ \\mathbf{F} }_p}(A) = J_{-1}^{2} \\oplus (p-2)J_{-1}^1 = \\left[\\begin{array}{rr|r|r|r|r} -1 & 1 & 0 & \\cdots & 0 & 0 \\\\ 0& -1 & 0 & 0 & 0 & 0 \\\\ \\hline 0& 0 & -1 & 0 & 0 & 0 \\\\ \\hline 0& 0 & 0 & \\ddots & \\ddots & 0 \\\\ \\hline 0& 0 & 0 & \\cdots & -1 & 0 \\\\ \\hline 0& 0 & 0 & \\cdots & 0 & -1 \\\\ \\end{array}\\right] .\\end{align*} To obtain a basis for $$E_{\\lambda = 0}$$, first note that the matrix $$P = [\\mathbf{v}_1, \\mathbf{p}_1, \\cdots , \\mathbf{p}_{p-1}]$$ from part (a) is singular over $${ \\mathbf{F} }_p$$, since \\begin{align*} \\mathbf{v}_1 + \\mathbf{p}_1 + \\mathbf{p}_2 + \\cdots + \\mathbf{p}_{p-2} &= [p-1, 0, 0, \\cdots, 0, 1] \\\\ &= [-1, 0,0,\\cdots, 0, 1] \\\\ &= - \\mathbf{p}_{p-1} .\\end{align*} We still have a linearly independent set given by the first $$p-1$$ columns of $$P$$, so we can extend this" ]

[ "that it equals $|\\mathbf{H}| = -1$ ### Tridiagonalisation The advantage of reflections over the rotations that were used in the Jacobi algorithm is that we can use them to transform an entire row or column of a matrix in a single step. In particular, if we have a reflection that leaves the first element of every vector as it is and transforms a specific vector $$\\mathbf{x}$$ so that its second element equals $$s_0$$ and all of the remaining elements equal zero $\\mathbf{H}_0 \\times \\begin{pmatrix}x_0\\\\x_1\\\\x_2\\\\x_3\\\\\\vdots\\\\x_{n-1}\\end{pmatrix} = \\begin{pmatrix}x_0\\\\s_0\\\\0\\\\0\\\\\\vdots\\\\0\\end{pmatrix}$ then, if $$\\mathbf{x}$$ equals the first row and column of a symmetric matrix $$\\mathbf{M}$$, we have $\\mathbf{H}_0 \\times \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ m_{0,1} & m_{1,1} & m_{1,2} & m_{1,3} & \\dots & m_{1,n-1}\\\\ m_{0,2} & m_{1,2} & m_{2,2} & m_{2,3} & \\dots & m_{2,n-1}\\\\ m_{0,3} & m_{1,3} & m_{2,3} & m_{3,3} & \\dots & m_{3,n-1}\\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ m_{0,n-1} & m_{1,n-1} & m_{2,n-1} & m_{3,n-1} & \\dots & m_{n-1,n-1} \\end{pmatrix} = \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ s_0 & m^\\prime_{1,1} & m^\\prime_{1,2} & m^\\prime_{1,3} & \\dots & m^\\prime_{1,n-1}\\\\ 0 & m^\\prime_{1,2} & m^\\prime_{2,2} & m^\\prime_{2,3} & \\dots & m^\\prime_{2,n-1}\\\\ 0 & m^\\prime_{1,3} & m^\\prime_{2,3} & m^\\prime_{3,3} & \\dots", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "# arrow_back Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. 7 views Is it possible to have $\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$ Explain. If $\\mathbf{n}=(a, b, c)$ is a unit vector, use this formula to show that (perhaps surprisingly) the orthogonal projection of $\\mathbf{x}$ into the plane perpendicular to $\\mathbf{n}$ is given bya) Let $\\mathbf{v}:=(a, b, c)$ and $\\mathbf{x}:=(x, y, z)$ be any vectors in $\\mathbb{R}^{3}$. Viewed as column vectors, find a $3 \\times 3$ m ... Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\mathbf{a}=(-1,0,2,1)$ and $\\mathbf{w}_{2}$ is orthogonal to a. Express the vector $\\mathbf{u}=(2,3,1,2)$ in the form $\\mathbf{u}=\\mathbf{w}_{1}+\\mathbf{w}_{2}$, where $\\mathbf{w}_{1}$ is a scalar multiple of $\\m ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Prove the orthogonal projection theorem 1 answer 7 views Prove the orthogonal projection theoremProve the orthogonal projection theorem ... close 1 answer 8 views Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the positive$x$-axis in$R^{2}$.Find the orthogonal projections of the vectors$\\mathbf{e}_{1}=(1,0)$and$\\mathbf{e}_{2}=(0,1)$on the line$L$that makes an angle$\\theta$with the ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss symmetry. 0 answers 7 views Find the intercepts on the$x$- and$y$-axes (if any) for the function$f(x)=\\frac{x^{2}-3}{x+2}$and discuss", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "a ball at position $(0,0,1)$, i.e. on the unit vector $\\mathbf e_z$, its shadow appears at the point $(-1,0,0)$. If you place it on the table, of course the shadow will \"coincide\" with the ball, i.e. $\\mathbf e_x$ and $\\mathbf e_y$ will be fixed, as before. So the matrix you are looking for is $$C = \\begin{pmatrix} 1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ and you may check that the null-space of $C$ is spanned by $(x,0,x)$, i.e. it is a diagonal line. This kinds of matrices represent linear operators called projectors, i.e. such that if you apply them twice you get the same result ($PP = P$: this identity is evident in the vertical-shadow example – try multiplying matrix $A$ by itself).", "+0 # I will really appreciate any help. 0 28 1 I will really appreciate any help. Let $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ be the following matrices: \\begin{align*} &\\mathbf{A} = \\begin{pmatrix} 0 & 0 \\\\ 0& 0 \\end{pmatrix},\\\\ &\\mathbf{B} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1\\end{pmatrix}, \\\\ &\\mathbf{C} = \\begin{pmatrix} 1/2 & 1/2 \\\\ 1/2 & 1/2\\end{pmatrix}, \\\\ &\\mathbf{D} = \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix}. \\end{align*} Also, consider the following functions: \\begin{align*} &\\text{Reflection across a line with direction vector \\begin{pmatrix}1\\\\-1 \\end{pmatrix}},\\\\ &\\text{The function that sends every vector to the zero vector}, \\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{The function that sends every vector to its negative},\\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{Projection onto \\begin{pmatrix}1\\\\1 \\end{pmatrix}}. \\end{align*} How do we order the list $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ to match the order of the functions above? I am having a hard time understanding problem such as these. Thank you in advance for any help Jul 20, 2022", "that there are other ways of “deriving” this result. One is by the use of a diagram, which would show that $(1,0)$ gets reflected to $(\\cos 2\\theta,\\sin 2\\theta)$ and $(0,1)$ gets reflected to $(\\sin 2\\theta,-\\cos 2\\theta)$. Another way is to observe that we can rotate an arbitrary mirror line onto the x-axis, then reflect across the x-axis, and rotate back. (The matrix product $[T]_{xy}$ can be seen as operating this way.) We took neither of these two approaches, because to justify them rigorously takes a bit of work, that is avoided by the pure linear algebra approach. Note also that $[T]_{uv}$ and $[T]_{xy}$ are orthogonal matrices, with determinant $-1$, as expected. ## Reflection across a line of given direction vector Suppose instead of being given an angle $\\theta$, we are given the unit direction vector $u$ to reflect the vector $w$. We can derive the matrix for the reflection directly, without involving any trigonometric functions. In the decomposition $\\mathbf{w}=a\\mathbf{u}+b\\mathbf{v}$, we note that $b=\\mathbf{w}\\cdot\\mathbf{v}$. Therefore $\\mathbf{w}^{\\prime}=(a\\mathbf{u}+b\\mathbf{v})-2b\\mathbf{v}=\\mathbf{w}-2(% \\mathbf{w}\\cdot\\mathbf{v})\\mathbf{v}\\,.$ (In fact, this is the formula used in the to draw the diagram in this entry.) To derive the matrix with respect to $x,y$ coordinates, we resort to a trick: $\\mathbf{w}^{\\prime}=I\\mathbf{w}-2\\mathbf{v}(\\mathbf{w}\\cdot\\mathbf{v})=I% \\mathbf{w}-2\\mathbf{v}(\\mathbf{v}^{\\textrm{tr}}\\mathbf{w})=I\\mathbf{w}-2(% \\mathbf{v}\\mathbf{v}^{\\textrm{tr}})\\mathbf{w}\\,.$ Therefore the matrix of the transformation is $I-2\\mathbf{v}\\mathbf{v}^{\\textrm{tr}}=\\begin{bmatrix}u_{x}^{2}-u_{y}^{2}&2u_{x% }u_{y}\\\\ 2u_{x}u_{y}&u_{y}^{2}-u_{x}^{2}\\end{bmatrix}\\,,\\quad\\mathbf{u}=(u_{x},u_{y})^{% \\textrm{tr}}\\,,\\quad\\mathbf{v}=(-u_{y},u_{x})^{\\textrm{tr}}\\,.$ If $u$ was not a unit vector to", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "$$\\operatorname{det}(A) = \\pm 1 \\in { \\mathbf{F} }_p^{\\times}$$, so the vectors $$\\mathbf{e}_1 - \\mathbf{e}_i$$ are still linearly independent and thus $$\\dim E_{-1} = p-1$$ • So there are $$p-1$$ Jordan blocks for $$\\lambda = 0$$. Summary: \\begin{align*} \\min_{A, { \\mathbf{F} }_p}(x) &= (x+1)^2 \\\\ \\chi_{A, { \\mathbf{F} }_p}(x) &\\equiv (x+1)^p \\\\ \\dim E_{-1} &= p-1 .\\end{align*} Thus \\begin{align*} JCF_{{ \\mathbf{F} }_p}(A) = J_{-1}^{2} \\oplus (p-2)J_{-1}^1 = \\left[\\begin{array}{rr|r|r|r|r} -1 & 1 & 0 & \\cdots & 0 & 0 \\\\ 0& -1 & 0 & 0 & 0 & 0 \\\\ \\hline 0& 0 & -1 & 0 & 0 & 0 \\\\ \\hline 0& 0 & 0 & \\ddots & \\ddots & 0 \\\\ \\hline 0& 0 & 0 & \\cdots & -1 & 0 \\\\ \\hline 0& 0 & 0 & \\cdots & 0 & -1 \\\\ \\end{array}\\right] .\\end{align*} To obtain a basis for $$E_{\\lambda = 0}$$, first note that the matrix $$P = [\\mathbf{v}_1, \\mathbf{p}_1, \\cdots , \\mathbf{p}_{p-1}]$$ from part (a) is singular over $${ \\mathbf{F} }_p$$, since \\begin{align*} \\mathbf{v}_1 + \\mathbf{p}_1 + \\mathbf{p}_2 + \\cdots + \\mathbf{p}_{p-2} &= [p-1, 0, 0, \\cdots, 0, 1] \\\\ &= [-1, 0,0,\\cdots, 0, 1] \\\\ &= - \\mathbf{p}_{p-1} .\\end{align*} We still have a linearly independent set given by the first $$p-1$$ columns of $$P$$, so we can extend this", "need to find a reflection that transforms the first column of matrix A, vector $\\mathbf{a}_1 = (12, 6, -4)^T$, to $\\|\\mathbf{a}_1\\| \\;\\mathrm{e}_1 = (14, 0, 0)^T.$ Now, $\\mathbf{u} = \\mathbf{x} + \\alpha\\mathbf{e}_1,$ and $\\mathbf{v} = {\\mathbf{u}\\over\\|\\mathbf{u}\\|}.$ Here, $\\alpha = -14$ and $\\mathbf{x} = \\mathbf{a}_1 = (12, 6, -4)^T$ Therefore $\\mathbf{u} = (-2, 6, -4)^T=({2})(-1, 3, -2)^T$ and $\\mathbf{v} = {1 \\over \\sqrt{14}}(-1, 3, -2)^T$, and then $Q_1 = I - {2 \\over \\sqrt{14} \\sqrt{14}} \\begin{pmatrix} -1 \\\\ 3 \\\\ -2 \\end{pmatrix}\\begin{pmatrix} -1 & 3 & -2 \\end{pmatrix}$ $= I - {1 \\over 7}\\begin{pmatrix} 1 & -3 & 2 \\\\ -3 & 9 & -6 \\\\ 2 & -6 & 4 \\end{pmatrix}$ $= \\begin{pmatrix} 6/7 & 3/7 & -2/7 \\\\ 3/7 &-2/7 & 6/7 \\\\ -2/7 & 6/7 & 3/7 \\\\ \\end{pmatrix}.$ Now observe: $Q_1A = \\begin{pmatrix} 14 & 21 & -14 \\\\ 0 & -49 & -14 \\\\ 0 & 168 & -77 \\end{pmatrix},$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to $A' = M_{11} = \\begin{pmatrix} -49 & -14 \\\\ 168 & -77 \\end{pmatrix}.$ By the same method as above, we obtain the matrix of the Householder transformation $Q_2 = \\begin{pmatrix} 1 & 0 & 0", "the dot product with \\cdot \\vec u \\cdot \\vec v $\\vec u \\cdot \\vec v$ Angle brackets can be used \\vec u = \\left< u_x, u_y, u_z \\right> $\\vec u = \\left< u_x, u_y, u_z \\right>$ The nabla operator \\nabla f = \\operatorname{grad} f $\\nabla f = \\operatorname{grad} f$ Divergence of a vector field \\nabla \\cdot \\mathbf F = \\operatorname{div} \\mathbf F $\\nabla \\cdot \\mathbf F = \\operatorname{div} \\mathbf F$ For repeated use, add this once … \\DeclareMathOperator{\\div}{div} \\div \\mathbf F $\\DeclareMathOperator{\\div}{div} \\div \\mathbf F$ … and use it as many times as you want. \\div \\mathbf A = \\div \\mathbf B $\\div \\mathbf A = \\div \\mathbf B$ Curl of a vector field \\nabla \\times \\mathbf F = \\operatorname{curl} \\mathbf F $\\nabla \\times \\mathbf F = \\operatorname{curl} \\mathbf F$ # Matrices Make a matrix with parentheses \\begin{pmatrix} 0 & 1\\\\ 1 & 1 \\end{pmatrix} $\\begin{pmatrix} 0 & 1\\\\ 1 & 1 \\end{pmatrix}$ Make a matrix with brackets \\begin{bmatrix} 0 & 1\\\\ 1 & 1 \\end{bmatrix} $\\begin{bmatrix} 0 & 1\\\\ 1 & 1 \\end{bmatrix}$ \\operatorname{adj} A $\\operatorname{adj} A$ Transposing a matrix A^\\top \\text{ or } A^\\intercal $A^\\top \\text{ or } A^\\intercal$ The determinant \\det A = \\begin{vmatrix} 0 & 1\\\\ 1 & 1 \\end{vmatrix} = -1 $\\det A = \\begin{vmatrix} 0 & 1\\\\ 1 & 1 \\end{vmatrix} =", "{\\color{red}2/3}&{\\color{red}2/3}&{\\color{red}1/3}& \\\\ {\\color{blue}1/3}&{\\color{blue}-2/3}&{\\color{blue}2/3} \\\\ -2/3&1/3&2/3 \\end{pmatrix} \\nonumber$ $\\mathbf{M}^T\\mathbf{M}=\\begin{pmatrix} 1&0&0 \\\\ 0&1&0 \\\\ 0&0&1 \\end{pmatrix}=\\mathbf{I} \\nonumber$ Because $$\\mathbf{M}^T\\mathbf{M}=\\mathbf{I}$$, the matrix is orthogonal. We now how to prove that the columns for a set of orthonormal vectors. The vectors are: $\\mathbf{a}=2/3\\mathbf{i}+2/3\\mathbf{j}+1/3\\mathbf{i}= \\begin{pmatrix}2/3\\\\2/3\\\\1/3\\end{pmatrix}\\nonumber$ $\\mathbf{b}=1/3\\mathbf{i}-2/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}1/3\\\\-2/3\\\\2/3\\end{pmatrix}\\nonumber$ $\\mathbf{c}=-2/3\\mathbf{i}+1/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}-2/3\\\\1/3\\\\2/3\\end{pmatrix}\\nonumber$ The modulii of these vectors are: $|\\mathbf{a}|^2=(2/3)^2+(2/3)^2+(1/3)^2=1 \\nonumber$ $|\\mathbf{b}|^2=(1/3)^2+(-2/3)^2+(2/3)^2=1 \\nonumber$ $|\\mathbf{c}|^2=(-2/3)^2+(1/3)^2+(2/3)^2=1 \\nonumber$ which proves that the vectors are normalized. The dot products of the three pairs of vectors are: $\\mathbf{a}\\cdot\\mathbf{b}=(2/3)(1/3)+(2/3)(-2/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{a}\\cdot\\mathbf{c}=(2/3)(-2/3)+(2/3)(1/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{c}\\cdot\\mathbf{b}=(-2/3)(1/3)+(1/3)(-2/3)+(2/3)(2/3)=0 \\nonumber$ which proves they are mutually orthogonal. Because the vectors are normalized and mutually orthogonal, they form an orthonormal set. Orthogonal matrices, when thought as operators that act on vectors, are important because they produce transformations that preserve the lengths of the vectors and the relative angles between them. For example, in two dimensions, the matrix $\\mathbf{M}_1=1/\\sqrt{2}\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix} \\nonumber$ is an operator that rotates a vector by $$\\pi/4$$ in the counterclockwise direction (Figure $$\\PageIndex{1}$$) preserving the lengths of the vectors and their relative orientation. In other words, an orthogonal matrix rotates a shape without distorting it. If the columns are orthogonal vectors that are not normalized, as in $\\mathbf{M}_2=\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix}, \\nonumber$ the object changes in size but the shape is not distorted. If, however, the two columns are non-orthogonal vectors, the transformation will distort", "B. $\\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ E. $\\begin{pmatrix} 0 &-1 \\\\-1 & 0 \\end{pmatrix}$ C. $\\begin{pmatrix} 1 & 0 \\\\ 0 &-1 \\end{pmatrix}$ Solution The reflection across the line $y = mx$ can be represented by the matrix in form of $\\begin{pmatrix} m & 0 \\\\ 0 & m \\end{pmatrix}$. For the reflection across the line $y=\\dfrac13x$, the representation matrix is given by $T_1 = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix}$ For the reflection across the line $y=-3x$, the representation matrix is given by $T_1 = \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix}$ Hence, the representation matrix for $T$ can be expressed as follows. \\begin{aligned} T & = T_2 \\cdot T_1 \\\\ & = \\begin{pmatrix} \\dfrac13 & 0 \\\\ 0 & \\dfrac13 \\end{pmatrix} \\cdot \\begin{pmatrix}-3 & 0 \\\\ 0 &-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix}-1 & 0 \\\\ 0 &-1 \\end{pmatrix} \\end{aligned} [collapse] Problem Number 21 A photocopier can make copies of images/writing of different sizes. A rectangular image is copied with certain settings. If this setting can be compared to the transformation process of the matrix $\\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix}$, then dilated in the origin by scale factor of $3$, then the area of ​​the rectangular image will be $\\cdots$ times the original area. A. $12$ C.", "of some vector $\\mathbf{x}$ onto the line $L$ spanned by some non-zero vector $\\mathbf{v}$. But what if you did want to compute the reflection of $\\mathbf{x}$ in the line $L$? What would this mean? Well, in general, suppose that $S$ is a subspace of $\\mathbb{R}^3$ (e.g., a line through the origin or a plane through the origin), so that for any vector $\\mathbf{x} \\in \\mathbb{R}^3$, we have $$\\mathbf{x} = \\operatorname{Proj}_S(\\mathbf{x}) + \\operatorname{Proj}_{S^\\perp}(\\mathbf{x}),$$ where $\\operatorname{Proj}_S(\\mathbf{x})$ is the orthogonal projection of $\\mathbf{x}$ onto $S$ and $\\operatorname{Proj}_{S^\\perp}(\\mathbf{x})$ is the orthogonal projection of $\\mathbf{x}$ onto the orthogonal complement $$S^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\text{for every \\mathbf{z} \\in S}, \\; \\mathbf{y} \\cdot \\mathbf{z} = 0\\}$$ of $S$; in particular, observe that $$\\operatorname{Proj}_{S^\\perp}(\\mathbf{x}) = \\mathbf{x} - \\operatorname{Proj}_S(\\mathbf{x}).$$ Then, geometrically, the reflection $\\operatorname{Refl}_S(\\mathbf{x})$ of $\\mathbf{x}$ in $S$ is given by fixing the component of $\\mathbf{x}$ in $S$ and flipping the direction of the component of $\\mathbf{x}$ in $S^\\perp$, i.e., $$\\operatorname{Refl}_S(\\mathbf{x}) = \\operatorname{Proj}_S(\\mathbf{x}) - \\operatorname{Proj}_{S^\\perp}(\\mathbf{x}) = \\operatorname{Proj}_S(\\mathbf{x}) - \\left( \\mathbf{x} - \\operatorname{Proj}_S(\\mathbf{x})\\right) = 2\\operatorname{Proj}_S(\\mathbf{x}) - \\mathbf{x}.$$ So, suppose that $L$ is the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$. On the one hand, the reflection of $\\mathbf{x}$ in $L$ is given by $$\\operatorname{Refl}_L(\\mathbf{x}) = 2\\operatorname{Proj}_L(\\mathbf{x}) - \\mathbf{x} = 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v} - \\mathbf{x},$$ which in your case yields $$\\operatorname{Refl}_L(\\mathbf{x}) = 2 \\cdot", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "2\\end{pmatrix}\\end{align*}, while A.^2 will return \\begin{align*}\\begin{pmatrix}1^2 & 1^2\\\\ 1^2 & 1^2\\end{pmatrix} = \\begin{pmatrix}1 & 1\\\\ 1 & 1\\end{pmatrix}\\end{align*}. Example 5 Given a vector x , compute a vector y having elements \\begin{align*}y (n) = \\tfrac{1}{sin(x(n))}\\end{align*}. This can be easily be done with the command y=1./sin(x) Note that using / in place of ./ would result in the (common) error \"Matrix dimensions must agree\". ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:", "tells us that ker(P) = <(1,0,1)>, and this is a normal vector to the plane x + z = 0 (we can divide by 2 in the plane equation). let's call ker(P), U, and our plane x + z = 0, W = im(P). the fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}. since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W. finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so $P = \\begin{bmatrix}\\frac{1}{2}&0&-\\frac{1}{2}\\\\0&1&0\\\\ -\\frac{1}{2}&0&\\frac{1}{2} \\end{bmatrix}$ one can easily verify that P2 = P (this is the defining property of reflections). now our reflection R is a \"reflection about W\", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w. we can write this in terms of P like so: for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w. that is: u = v - P(v), and w = P(v), so: R(v) = -u + w = -(v - P(v)) +", "x_i + b \\,, \\end{align} with $$w_i$$ denoting the $$i$$-th so-called weight, i.e. parameter, of this block and $$b$$ denoting the so-called bias (also a parameter), i.e., an additive term not depending on the input. When one input is simultaneously processed by $$K$$ linear blocks, i.e., \\begin{align} y_k = \\sum\\limits^{N}_{i=1} w^k_i \\cdot x_i + b^k = \\underbrace{\\left( w^k_1, w^k_2, \\ldots, w^k_N, b^k \\right)}_{\\mathbf{w}\\transp_k} \\cdot \\underbrace{\\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_N \\\\ 1 \\end{pmatrix}}_{\\mathbf{x}} \\,, \\end{align} for block $$k$$, this can be expressed compactly via matrix-vector multiplications: \\begin{align} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_K \\end{pmatrix} &= \\begin{pmatrix} \\qquad \\mathbf{w}\\transp_1 \\qquad \\\\ \\mathbf{w}\\transp_2 \\\\ \\vdots \\\\ \\mathbf{w}\\transp_K \\end{pmatrix} \\cdot \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_N \\\\ 1 \\end{pmatrix} \\\\ &= \\qquad \\quad \\mathbf{W} \\quad \\qquad \\cdot \\quad \\mathbf{x} \\,, \\end{align} with $$\\mathbf{W} \\in \\mathbb{R}^{K \\times (N+1)}$$ and $$\\mathbf{x} \\in \\mathbb{R}^{(N+1) \\times 1}$$. This expression can be further extended for the case when multiple input vectors $$\\mathbf{x}_1, \\mathbf{x}_2, \\ldots, \\mathbf{x}_B$$, i.e., a so-called batch of size $$B$$, have to be processed simultaneously: \\begin{align} \\begin{pmatrix} \\qquad \\mathbf{w}\\transp_1 \\qquad \\\\ \\mathbf{w}\\transp_2 \\\\ \\vdots \\\\ \\mathbf{w}\\transp_K \\end{pmatrix} \\cdot \\begin{pmatrix} \\\\ \\\\ \\mathbf{x}_1 & \\mathbf{x}_2 & \\cdots & \\mathbf{x}_B \\\\ \\\\ \\\\ \\end{pmatrix} &= \\begin{pmatrix} \\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_1 & \\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_2 &\\cdots &\\mathbf{w}\\transp_1 \\cdot \\mathbf{x}_B \\\\ \\mathbf{w}\\transp_2 \\cdot" ]

[ "($\\pi/4$, the left side is 2 while the right side is $\\sqrt{2}$. You can prove that $tan^2 x+ 1= sec^2(x)$ for all x.", "$\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\left(1-n\\right)\\mathrm{tan}\\alpha$ What are you looking for?", "+ \\tan x) \\right]_0^{\\pi/4}$$", "\\frac{\\tan 2 \\theta}{2 \\theta} \\times 2 = 2 \\left(1\\right) \\times 2 = 4$", "\\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}$", "\\right]' = 2 \\\\\\\\ \\boxed{ \\left[\\ {\\tan^{-1}(2x+1)} \\ \\right]' = \\frac{2} {1+(2x+1)^2} } }$$", "\\left(x\\right)$ $\\frac{2 \\tan \\left(x\\right) + 2 {\\sin}^{3} \\frac{x}{\\cos} \\left(x\\right)}{\\cos} ^ 4 \\left(x\\right) = \\frac{2 \\tan \\left(x\\right) + 2 {\\sin}^{2} \\left(x\\right) \\tan \\left(x\\right)}{\\cos} ^ 4 \\left(x\\right)$", "y}}{a}} \\right) = x + y - x + c\\;$$ $${\\tan ^{ - 1}}\\left( {\\frac{{x + y}}{a}} \\right) = a(y+c)$$ $$\\left( {\\frac{{x + y}}{a}} \\right) = \\tan \\left( {a(y+c)} \\right)$$ ∴ $$x + y = a\\tan \\left( {a(y+c)} \\right)$$", "# How do you differentiate f(x) = x^3/(xtanx-2) using the quotient rule? $f ' \\left(x\\right) = \\frac{- {x}^{2} \\left(6 + {x}^{2} \\sec {\\left(x\\right)}^{2} - 2 x \\tan \\left(x\\right)\\right)}{- 2 + x \\cdot \\tan \\left(x\\right)} ^ 2$ $f ' \\left(x\\right) = \\frac{3 {x}^{2} \\left(x \\tan \\left(x\\right) - 2\\right) - {x}^{3} \\left(\\tan \\left(x\\right) + x {\\sec}^{2} \\left(x\\right)\\right)}{x \\tan \\left(x\\right) - 2} ^ 2$", "# If f(x)=sec(x), how do I find f''(π/4)? Mar 23, 2015 First find the expression for $f ' ' \\left(x\\right)$, then evaluate it at $\\frac{\\pi}{4}$. $f \\left(x\\right) = \\sec x$. $f ' \\left(x\\right) = \\sec x \\tan x$. Use the product rule to find $f ' ' \\left(x\\right)$. $f ' ' \\left(x\\right) = \\left(\\sec x \\tan x\\right) \\tan x + \\sec x \\left({\\sec}^{2} x\\right)$. So, $f ' ' \\left(x\\right) = \\sec x {\\tan}^{2} x + {\\sec}^{3} x$. Evaluate: $f ' ' \\left(\\frac{\\pi}{4}\\right) = \\sec \\left(\\frac{\\pi}{4}\\right) {\\tan}^{2} \\left(\\frac{\\pi}{4}\\right) + {\\sec}^{3} \\left(\\frac{\\pi}{4}\\right)$ $= \\left(\\sqrt{2}\\right) {\\left(1\\right)}^{2} + {\\left(\\sqrt{2}\\right)}^{3} = \\sqrt{2} + 2 \\sqrt{2} = 3 \\sqrt{2}$.", "# Be careful to the learnt formulas... Geometry Level 2 Find the value of $\\left( \\tan^{-1}1+\\tan^{-1}2+\\tan^{-1}3 \\right)$. ×", "# Prove that $\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$ [duplicate] Prove that $$\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$$", "{2 - \\frac{1}{2}} \\right) = m\\left( {1 - 0} \\right)$$ ⇒ m = 3 / 2 ⇒ tan θ = 3 / 2 $$\\theta = {\\tan ^{ - 1}}\\left( {\\frac{3}{2}} \\right)$$", "{\\rm{x}}} \\right)}}} \\right) = {\\tan ^{ - 1}}\\left( {\\frac{1}{0}} \\right)$$ $$\\Rightarrow \\frac{2}{{1 - 1 + {{\\rm{x}}^2}}} = \\frac{1}{0}$$ ⇒ x2 = 0 ⇒ x = 0", "$x=\\frac{{\\pi}}{12},\\ \\frac{5{\\pi}}{12},\\ \\frac{9{\\pi}}{12}$ Then t=tanx", "13 } +cos\\frac { 5\\pi }{ 13 }=0$ . 1. If cosec A + sec A = cosec B + sec B, prove that cot$\\left( \\frac { A+B }{ 2 } \\right)$=tanA tanB 2. If y = 500 e7x + 600 e-7x then show that y2 - 49y = 0.", "Thus, we consider $f \\left(x\\right) = {\\tan}^{2} \\left(x\\right) = \\tan \\left(x\\right) \\tan \\left(x\\right)$ so that we can deal with $\\tan \\left(x\\right)$, for which we know the derivative. Utilizing the derivative of $\\tan \\left(x\\right)$, namely $\\frac{d}{\\mathrm{dx}} \\tan \\left(x\\right) = {\\sec}^{2} \\left(x\\right)$, and the Chain Rule $\\frac{\\mathrm{df}}{\\mathrm{dx}} = g ' \\left(x\\right) h \\left(x\\right) + g \\left(x\\right) h ' \\left(x\\right)$, we obtain: $f ' \\left(x\\right) = \\left[\\frac{d}{\\mathrm{dx}} \\left(\\tan \\left(x\\right)\\right)\\right] \\tan \\left(x\\right) + \\tan \\left(x\\right) \\left[\\frac{d}{\\mathrm{dx}} \\left(\\tan \\left(x\\right)\\right)\\right]$ $\\frac{d}{\\mathrm{dx}} \\tan \\left(x\\right) = {\\sec}^{2} \\left(x\\right)$, so... $f ' \\left(x\\right) = {\\sec}^{2} \\left(x\\right) \\tan \\left(x\\right) + \\tan \\left(x\\right) {\\sec}^{2} \\left(x\\right) = 2 \\tan \\left(x\\right) {\\sec}^{2} \\left(x\\right)$", "# Prove That: Sec 4 a ( 1 − Sin 4 a ) − 2 Tan 2 a = 1 Proove that : $$\\sec^4 A\\left( 1 - \\sin^4 A \\right) - 2 \\tan^2 A = 1$$", "How do you differentiate f(x)=(5e^x+tanx)(x^2-2x) using the product rule? Apr 25, 2018 $f ' \\left(x\\right) = \\left(5 {e}^{x} + {\\sec}^{2} x\\right) \\left({x}^{2} - 2 x\\right) + \\left(5 {e}^{x} + \\tan x\\right) \\left(2 x - 2\\right)$ For $f \\left(x\\right) = \\left(5 {e}^{x} + \\tan x\\right) \\left({x}^{2} - 2 x\\right)$, we find $f ' \\left(x\\right)$ by doing: $f ' \\left(x\\right) = \\frac{d}{\\mathrm{dx}} \\left[5 {e}^{x} + \\tan x\\right] \\left({x}^{2} - 2 x\\right) + \\left(5 {e}^{x} + \\tan x\\right) \\frac{d}{\\mathrm{dx}} \\left[{x}^{2} - 2 x\\right]$ $f ' \\left(x\\right) = \\left(5 {e}^{x} + {\\sec}^{2} x\\right) \\left({x}^{2} - 2 x\\right) + \\left(5 {e}^{x} + \\tan x\\right) \\left(2 x - 2\\right)$", "pierdoodsu 2021-12-31 Here are all the results I got $\\mathrm{tan}\\left(A+B+C\\right)=a-b$ And $\\left(1+{\\mathrm{tan}}^{2}A\\right)\\left(1+{\\mathrm{tan}}^{2}B\\right)\\left(1+{\\mathrm{tan}}^{2}C\\right)={\\left(\\frac{1}{\\mathrm{cos}A\\mathrm{cos}B\\mathrm{cos}C}\\right)}^{2}$ And $\\mathrm{cot}A+\\mathrm{cot}B+\\mathrm{cot}C=0$ How should I use these results? veiga34 Expert Write $x=\\mathrm{tan}A,y=\\mathrm{tan}B,z=\\mathrm{tan}C$ By Vieta’s formula, we have Now you want to find the value of $\\left(1+{x}^{2}\\right)\\left(1+{y}^{2}\\right)\\left(1+{z}^{2}\\right)=1+{x}^{2}+{y}^{2}+{z}^{2}+{x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}+{x}^{2}{y}^{2}{z}^{2}$ Note that ${a}^{2}={\\left(x+y+z\\right)}^{2}={x}^{2}+{y}^{2}+{z}^{2}+2\\left(xy+yz+zx\\right)$ so, ${x}^{2}+{y}^{2}+{z}^{2}={a}^{2}$. Similarly, $0={\\left(xy+yz+zx\\right)}^{2}={x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}+2xyz\\left(x+y+z\\right)$ implies ${x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}=2ab$. Hence $\\left(1+{x}^{2}\\right)\\left(1+{y}^{2}\\right)\\left(1+{z}^{2}\\right)=1+{a}^{2}+2ab+{b}^{2}=1+{\\left(a+b\\right)}^{2}$ lovagwb Expert Let $p\\left(x\\right)={x}^{3}-a{x}^{2}+b$, then $p\\left(x\\right)=\\left(x-\\mathrm{tan}A\\right)\\left(x-\\mathrm{tan}B\\right)\\left(x-\\mathrm{tan}C\\right)$ Now you need $p\\left(i\\right)\\cdot p\\left(-i\\right)$ Vasquez Expert Let $y={x}^{2}+1$ so that ${x}^{2}=y-1$ and substitute as much as possible in the original cubic: $x\\left(y-1\\right)-a\\left(y-1\\right)+b=0$ Rearrange to isolate x and obtain: $x\\left(y-1\\right)=a\\left(y-1\\right)+b=ay-\\left(a+b\\right)$ Now square this: ${x}^{2}\\left(y-1{\\right)}^{2}=\\left(y-1{\\right)}^{3}=\\left(ay-\\left(a+b\\right){\\right)}^{2}$ so that ${y}^{3}-3{y}^{2}+3y+1-{a}^{2}{y}^{2}+2a\\left(a+b\\right)y-\\left(a+b{\\right)}^{2}=0$ Do you have a similar question?" ]

[ "the formula \\begin{align*} \\tan{2t} &=\\frac{2 \\cdot \\tan{t}}{1 - \\tan^{2}{t}} \\\\ &= \\frac{2 \\sqrt{x}}{1-x} \\qquad \\Bigl[ \\text{since} \\ x=\\tan^{2}{t} \\ \\Bigr] \\end{align*}", "Double and Half Angle Identity Proofs $$\\displaystyle \\frac{{\\sin \\left( {2x} \\right)}}{2}=\\frac{{\\tan x}}{{{{{\\tan }}^{2}}x+1}}$$ \\require{cancel} \\displaystyle \\begin{align}&=\\frac{{\\tan x}}{{{{{\\sec }}^{2}}x}}\\\\&=\\frac{{\\sin x}}{{\\cancel{{\\cos x}}}}\\cdot \\frac{{{{{\\cancel{{{{{\\cos }}^{2}}x}}}}^{{\\cos x}}}}}{1}\\\\&=\\sin x\\cos x\\\\&=\\sin x\\cos x\\cdot \\frac{2}{2}\\\\\\frac{{\\sin \\left( {2x} \\right)}}{2}&=\\frac{{2\\sin x\\cos x}}{2}\\,\\,\\,\\,\\,\\surd \\end{align} Note: To get $$\\displaystyle \\sin x\\cos x$$ into the right form for the identity $$\\displaystyle (2\\sin x\\cos x)$$, we had to multiply by $$\\displaystyle \\frac{2}{2}$$. $$\\displaystyle \\frac{{1-\\cos \\left( {2A} \\right)}}{{\\sin \\left( {2A} \\right)}}=\\tan A$$ \\displaystyle \\begin{align}\\frac{{1-\\left( {1-2{{{\\sin }}^{2}}A} \\right)}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{2{{{\\sin }}^{2}}A}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{{}^{{\\sin A}}\\cancel{{2{{{\\sin }}^{2}}A}}}}{{\\cancel{{2\\sin A}}\\cos A}}&=\\tan A\\,\\,\\,\\,\\,\\surd \\end{align} Note: We knew to use the $$\\displaystyle 1-2{{\\sin }^{2}}A$$ (instead of $$\\displaystyle 2{{\\cos }^{2}}A-1$$) for $$\\cos 2A$$ to get rid of the 1’s, and also because we have sin in the denominator instead of cos. $$\\displaystyle \\tan \\left( {\\frac{1}{2}\\theta } \\right)=\\csc \\theta -\\cot \\theta$$ \\displaystyle \\begin{align}&=\\frac{1}{{\\sin \\theta }}-\\frac{{\\cos \\theta }}{{\\sin \\theta }}\\\\\\tan \\left( {\\frac{1}{2}\\theta } \\right)&=\\frac{{1-\\cos \\theta }}{{\\sin \\theta }}\\,\\,\\,\\,\\,\\surd \\end{align} Note: Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos. $$\\displaystyle \\frac{{\\cos \\left( {2\\theta } \\right)}}{{\\cos \\theta -\\sin \\theta }}=\\cos \\theta +\\sin \\theta$$ \\displaystyle \\begin{align}\\frac{{{{{\\cos }}^{2}}\\theta -{{{\\sin }}^{2}}\\theta }}{{\\cos \\theta -\\sin \\theta }}&=\\\\\\frac{{\\cancel{{\\left( {\\cos \\theta -\\sin \\theta } \\right)}}\\left( {\\cos \\theta +\\sin \\theta } \\right)}}{{\\cancel{{\\cos \\theta -\\sin \\theta }}}}&=\\cos \\theta +\\sin \\theta \\,\\,\\,\\,\\,\\surd \\end{align}", "=\\frac{1}{f'(\\pi/4)}\\left[1+\\tan^{2}(\\pi/4)\\right]\\\\ & =\\frac{1}{5}\\left[1+1^{2}\\right]\\\\ & =\\frac{2}{5}.\\end{align}", "x$$ $$\\displaystyle f'(x) = 2\\sin x\\cos x$$ Step 1 Rewrite the function so it is easier to see the need for the chain rule. $$f(x) = \\sec^2(3x) = (\\sec 3x)^2$$ Step 2 Differentiate using the chain rule. \\begin{align*} f'(x) & = 2(\\sec 3x)^1\\cdot 3\\sec 3x\\tan 3x\\\\[6pt] & = 6\\sec 3x\\sec 3x\\tan 3x\\\\[6pt] & = 6\\sec^2 3x \\tan 3x \\end{align*} Step 3 Evaluate $$f'\\left(\\frac\\pi 4\\right)$$ \\begin{align*} f'\\left(\\frac \\pi 4\\right) & = 6\\sec^2\\left(3\\cdot \\frac \\pi 4\\right)\\tan\\left(3\\cdot \\frac \\pi 4\\right)\\\\[6pt] & = 6\\left[\\sec\\left(\\frac{3\\pi}4\\right)\\right]^2\\tan\\left(\\frac{3\\pi} 4\\right)\\\\[6pt] & = 6\\left[-\\sqrt 2\\right]^2(-1)\\\\[6pt] & = 6(2)(-1)\\\\[6pt] & = -12 \\end{align*} $$\\displaystyle f'\\left(\\frac \\pi 4\\right) = -12$$ Step 1 Differentiate using the chain rule. $$f'(x) = e^{\\sin(x^4)}\\cdot \\cos(x^4)\\cdot 4x^3$$ $$\\displaystyle f'(x) = 4x^3\\cos(x^4)\\,e^{\\sin(x^4)}$$ Step 1 Rewrite the function to emphasize that the cosine is being raised to the 4th power. $$f(x) = \\left[\\cos\\left(e^{1-x^2}\\right)\\right]^4$$ Step 2 Differentiate using the chain rule. \\begin{align*} f'(x) & = 4\\left[\\cos\\left(e^{1-x^2}\\right)\\right]^3\\cdot \\frac d {dx}\\left[\\cos\\left(e^{1-x^2}\\right)\\right]\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot \\frac d {dx}\\left(e^{1-x^2}\\right)\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}\\cdot \\frac d {dx}\\left(1-x^2\\right)\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}\\cdot \\left(-2x\\right)\\\\[6pt] & = -8x\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2} \\end{align*} $$\\displaystyle f'(x) = -8x\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}$$", "# Thread: Double angle formula 1. ## Double angle formula Is it possible to input $\\displaystyle \\frac{2\\;tan\\;\\frac{\\pi}{5}}{1-tan^2\\;\\frac{\\pi}{5}}$ on a calculator? The $\\displaystyle tan^2$ in the denominator is throwing me off. 2. Originally Posted by TI-84 Is it possible to input $\\displaystyle \\frac{2\\;tan\\;\\frac{\\pi}{5}}{1-tan^2\\;\\frac{\\pi}{5}}$ on a calculator? The $\\displaystyle tan^2$ in the denominator is throwing me off. just enter $\\displaystyle \\left( \\tan \\frac {\\pi}5 \\right)^2$", "QXP=|\\pi/2-\\angle C|$ $BP=\\frac {BD}{2}\\cdot \\sec \\angle B$ $XP=\\frac {BD}{2}\\cdot (\\tan \\angle B+ \\tan \\angle C)$ $XQ=\\frac {BC}{2}\\cdot \\sec \\angle C$ Therefore, \\begin{align*}\\frac {BP}{XP}&=\\frac {\\sec \\angle B}{\\tan \\angle B+\\tan \\angle C}\\\\ &=\\frac {\\cos \\angle C}{\\sin \\angle A}\\\\ &=\\frac {1}{\\sin \\angle A\\cdot \\sec \\angle C}\\\\ &=\\frac {2R}{BC\\cdot \\sec \\angle C}\\\\ &=\\frac {R}{XQ}\\\\ &=\\frac {BO}{XQ}\\end{align*} Hence $XPQ\\sim BPO \\Longrightarrow \\angle OPQ=\\angle BPX=\\angle OAQ\\Longrightarrow A,P,O,Q$ concyclic. Welcome to BdMO Online Forum. Check out Forum Guides & Rules", "\\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get $$\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}$$ after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff \\boxed{129}.$ ## Solution 2 $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]$ Call $O_1$ and $O_2$ the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\\triangle{O_1BN} \\sim", "we have $$\\tan {\\angle PAR} = \\tan (\\angle RAD - \\angle PAD) = \\frac{\\frac{PR}{h}}{1 + \\frac{DP \\cdot DR}{h^2}} = \\frac{PR \\cdot h}{h^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{AQ^2 - DQ^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{\\frac{a^2}{4} - PQ^2} = \\frac{\\frac{a}{n} \\cdot h}{\\frac{a^2}{4} - \\frac{a^2}{4n^2}} = \\frac{4nh}{(n^2-1)a}.$$", "addition formulas). • Using that $$\\cos$$ is even and that $$\\sin$$ is odd, the addition formulas give the subtraction formulas: \\begin{align*} \\cos(x-y)=&\\cos(x)\\cos(y)+\\sin(x)\\sin(y) \\\\ \\sin(x-y)=&\\sin(x)\\cos(y)-\\cos(x)\\sin(y). \\end{align*} • Dividing the formulas we get: $\\tan(x\\pm y)=\\frac{\\sin(x\\pm y)}{\\cos(x\\pm y)}=\\frac{\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y)}=\\frac{\\tan(x)\\pm\\tan(y)}{1\\mp\\tan(x)\\tan(y)}.$ • Putting $$x=y$$ in the addition formulas, we get the double-angle formulas: \\begin{align*} \\cos(2x)=&(\\cos x)^2-(\\sin x)^2 \\\\ \\sin(2x)=&2\\sin(x)\\cos(x) \\end{align*} • Using the identity $$(\\cos x)^2+(\\sin x)^2=1$$ in the double-angle formula for $$\\cos$$, we get the following alternative forms. \\begin{align*} \\cos(2x)=&(\\cos x)^2-(1-(\\cos x)^2)=2(\\cos x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2 \\end{align*} # Trigonometric identities -- from the addition formulas 2 • Once again, the addition and subtraction formulas are \\begin{align*} \\cos(x\\pm y)=&\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y) \\\\ \\sin(x\\pm y)=&\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y), \\end{align*} and the alternative double-angle formulas for $$\\cos$$ are \\begin{align*} \\cos(2x)=&\\cos(x)^2-(1-\\cos(x)^2)=2\\cos(x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2. \\end{align*} • Rearranging the latter, we get the following half-angle formulas. \\begin{align*} (\\cos x)^2=&\\frac{\\cos(2x)+1}{2} \\\\ (\\sin x)^2=&\\frac{1-\\cos(2x)}{2} \\end{align*} • Finally, rearranging the addition formulas, we get the following product formulas. \\begin{align*} \\cos(x)\\cos(y)=&\\frac{\\cos(x+y)+\\cos(x-y)}{2} \\\\ \\sin(x)\\cos(y)=&\\frac{\\sin(x+y)+\\sin(x-y)}{2} \\\\ \\sin(x)\\sin(y)=&\\frac{\\cos(x-y)-\\cos(x+y)}{2}. \\end{align*} • The half-angle and product formulas will be used frequently in integral calculus, since they are linearization formulas. • Exercises: D.45, 47, 67 # Exponentiation by rational numbers • An exponential function is a function of the form $$f(x)=b^x$$ for some fixed positive number $$b>0$$. Let's recall what the expression $$b^x$$ means. • If $$n$$ is a", "# 3.3: Double-Angle and Half-Angle Formulas A special case of the addition formulas is when the two angles being added are equal, resulting in the double-angle formulas: \\begin{align} \\sin\\;2\\theta ~&=~ 2\\;\\sin\\;\\theta ~ \\cos\\;\\theta\\label{eqn:doublesin}\\\\ \\cos\\;2\\theta ~&=~ \\cos^2 \\;\\theta ~-~ \\sin^2 \\;\\theta\\label{eqn:doublecos}\\\\ \\tan\\;2\\theta ~&=~ \\frac{2\\;\\tan\\;\\theta}{1 ~-~ \\tan^2 \\;\\theta}\\label{eqn:doubletan} \\end{align} To derive the sine double-angle formula, we see that $\\sin\\;2\\theta ~=~ \\sin\\;(\\theta+\\theta) ~=~ \\sin\\;\\theta ~ \\cos\\;\\theta ~+~ \\cos\\;\\theta ~ \\sin\\;\\theta ~=~ 2\\;\\sin\\;\\theta ~ \\cos\\;\\theta~. \\nonumber$ Likewise, for the cosine double-angle formula, we have $\\cos\\;2\\theta ~=~ \\cos\\;(\\theta+\\theta) ~=~ \\cos\\;\\theta~\\cos\\;\\theta ~-~ \\sin\\;\\theta~\\sin\\;\\theta ~=~ \\cos^2 \\;\\theta ~-~ \\sin^2 \\;\\theta~,\\nonumber$ and for the tangent we get $\\tan\\;2\\theta ~=~ \\tan\\;(\\theta+\\theta) ~=~ \\frac{\\tan\\;\\theta ~+~ \\tan\\;\\theta}{1 ~-~ \\tan\\;\\theta ~ \\tan\\;\\theta} ~=~ \\frac{2\\;\\tan\\;\\theta}{1 ~-~ \\tan^2 \\;\\theta} \\nonumber$ Using the identities $$\\;\\sin^2 \\;\\theta = 1 - \\cos^2 \\;\\theta$$ and $$\\;\\cos^2 \\;\\theta = 1 - \\sin^2 \\;\\theta$$, we get the following useful alternate forms for the cosine double-angle formula: \\begin{align} \\cos\\;2\\theta ~&=~ 2\\;\\cos^2 \\;\\theta ~-~ 1\\label{eqn:doublecosalt1}\\\\ &=~ 1 ~-~ 2\\;\\sin^2 \\;\\theta\\label{eqn:doublecosalt2} \\end{align} \\nonumber ## Example 3.13 Prove that $$\\;\\sin\\;3\\theta ~=~ 3\\;\\sin\\;\\theta ~-~ 4\\;\\sin^3 \\;\\theta\\;$$. ###### Solution Using $$3\\theta = 2\\theta + \\theta$$, the addition Equation for sine, and the double-angle Equations \\ref{eqn:doublesin} and \\ref{eqn:doublecosalt2}, we get: \\begin{align*} \\sin\\;3\\theta ~&=~ \\sin\\;(2\\theta+\\theta)\\\\ \\nonumber &=~ \\sin\\;2\\theta~\\cos\\;\\theta ~+~ \\cos\\;2\\theta~\\sin\\;\\theta\\\\ \\nonumber &=~ (2\\;\\sin\\;\\theta~\\cos\\;\\theta)\\;\\cos\\;\\theta ~+~ (1 - 2\\;\\sin^2 \\;\\theta)\\;\\sin\\;\\theta\\\\ \\nonumber &=~ 2\\;\\sin\\;\\theta~\\cos^2 \\;\\theta ~+~ \\sin\\;\\theta ~-~ 2\\;\\sin^3", "Thinking Mathematically (6th Edition) (a) According to the distributive property of addition, $a\\left( b+c \\right)=ab+ac$ Use the above property in the given formula, \\begin{align} & \\frac{D\\left( A+1 \\right)}{24}=\\frac{D\\cdot A+D\\cdot 1}{24} \\\\ & =\\frac{DA+D}{24} \\end{align} Both formulae are same as shown above. (b) Substitute $200$ for $D$ and $12$ for $A$ in the first formula. \\begin{align} & \\frac{D\\left( A+1 \\right)}{24}=\\frac{200\\left( 12+1 \\right)}{24} \\\\ & =\\frac{200\\left( 13 \\right)}{24} \\\\ & =108.33 \\\\ & \\approx 108 \\end{align} Substitute $200$ for $D$ and $12$ for $A$ in the second formula. \\begin{align} & \\frac{DA+D}{24}=\\frac{200\\cdot 12+200}{24} \\\\ & =\\frac{2400+200}{24} \\\\ & =108.33 \\\\ & \\approx 108 \\end{align} With both expressions, the proper dose is approximately$108\\,\\text{mg}$. It is easier to use the first formula as an additional step of addition is reduced.", "Angle Identities \\begin{array}{c} \\sin\\frac{\\theta}{2}=\\pm\\sqrt{\\frac{1-\\cos\\theta}{2}}\\\\ \\cos\\frac{\\theta}{2}=\\pm\\sqrt{\\frac{1+\\cos\\theta}{2}}\\\\ \\tan\\frac{\\theta}{2}=\\frac{\\sin\\theta}{1+\\cos\\theta}=\\frac{1-\\cos\\theta}{\\sin\\theta}\\\\ \\end{array} #### Sum/Difference Identities \\begin{array}{c} \\sin(\\theta\\pm\\phi) = \\sin\\theta\\cdot\\cos\\phi \\pm \\cos\\theta\\cdot\\sin\\phi\\\\ \\cos(\\theta\\pm\\phi) = \\cos\\theta\\cdot\\cos\\phi \\mp \\sin\\theta\\cdot\\sin\\phi\\\\ \\tan(\\theta\\pm\\phi) = \\frac{\\tan\\theta\\pm\\tan\\phi}{1\\mp\\theta\\theta\\cdot\\tan\\phi}\\\\ \\end{array}", "\\left(\\frac{1}{3}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2} + \\frac{1}{7}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}+\\quad \\cdots\\right) ### Transformation to current notation Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get • \\theta = \\tan \\theta - \\frac{\\tan^3 \\theta}{3} + \\frac{\\tan^5\\theta}{5} - \\frac{\\tan^7 \\theta}{7} + \\quad \\cdots . Letting tan θ = q we finally have • \\tan^{-1} q = q - \\frac{q^3}{3} + \\frac{q^5}{5} - \\frac{q^7}{7} + \\quad \\cdots ### Another formula for the circumference of a circle The second part of the quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows. c= \\sqrt{12 d^2} - \\frac{\\sqrt{12 d^2}}{3\\cdot 3} + \\frac{\\sqrt{12 d^2}}{3^2 \\cdot 5} - \\frac{\\sqrt{12 d^2}}{3^3 \\cdot 7}+ \\quad \\cdots Since c = π d this can be reformulated as a formula to compute π as follows. \\pi = \\sqrt{12}\\left( 1 - \\frac{1}{3\\cdot3}+\\frac{1}{3^2\\cdot 5} -\\frac{1}{3^3\\cdot 7} +\\quad \\cdots\\right) This is obtained by substituting q = 1/\\sqrt{3} (therefore θ = π / 6) in the power series expansion for tan−1 q above. Comparison of the convergence of two Madhava series", "the Inscribed Angle Theorem, as the next step: \\displaystyle\\begin{align} \\frac{XM}{YM}&=\\frac{AM}{CM}\\cdot\\frac{CM}{YM}\\cdot\\frac{XM}{AM}\\\\ &=\\frac{AM}{CM}\\cdot\\frac{\\sin(CYM)}{\\sin(AXM)}\\\\ &=\\frac{AM}{CM}\\cdot\\frac{BM^-CM^2}{CB\\cdot OM}\\cdot\\frac{AD\\cdot OM}{DM^2-AM^2}\\\\ &=\\frac{AM}{CM}\\cdot\\frac{AD}{CB}\\cdot\\frac{BM^2-CM^2}{DM^2-AM2}\\\\ &=\\frac{1}{\\alpha}\\cdot\\frac{1}{\\alpha}\\cdot\\alpha^2\\\\ &=1. \\end{align}", "cdot {\\ sqrt {\\ left (25 + 11 \\ cdot {\\ sqrt {5}} \\ right) \\ cdot 10}} \\ right) \\\\ & = \\ arccos \\ left (- {\\ sqrt {{\\ frac {1} {10}} \\ cdot \\ left (5 - {\\ sqrt {5}} \\ right)} } \\ right) \\; \\ approx 121 ^ {\\ circ} \\; 43 ^ {\\ prime} \\; 3 ^ {\\ prime \\ prime}. \\ end {aligned}}} #### Solid angles in the corners The following formula, described in Platonic Solids, shows a solution for the solid angle${\\ displaystyle \\ Omega}$ ${\\ displaystyle \\ Omega = 2 \\ pi -2n \\ cdot \\ arcsin \\ left (\\ cos \\ left ({\\ frac {\\ pi} {n}} \\ right) \\ cdot {\\ sqrt {\\ tan ^ {2} \\ left ({\\ frac {\\ pi} {n}} \\ right) - \\ tan ^ {2} \\ left ({\\ frac {\\ alpha} {2}} \\ right)}} \\ right)}$ With the number of edges / faces at a corner and the interior angle of the regular pentagon applies ${\\ displaystyle {n = 3}}$${\\ displaystyle {a = 108 ^ {\\ circ}}}$ {\\ displaystyle {\\ begin {aligned} {\\ mathsf {(1)}} \\; \\ Omega & = 2 \\ pi -2 \\ cdot 3 \\ cdot \\ arcsin \\ left (\\ cos \\ left ({\\ frac {\\ pi }", "on $\\triangle CAO$ with respect to $\\angle CAO$ yields \\begin{align*}\\left(r+\\frac{1}{18}\\right)^2 &= r^2 + \\left(r\\sqrt{11}\\right)^2 - 2 \\cdot r\\sqrt{11} \\cdot r \\cos \\angle BAO \\\\ \\frac{r}{9} + \\frac{1}{324} &= 11r^2 - r\\sqrt{11} \\cdot \\left(\\frac{\\sqrt{11}}{2}\\right) \\\\ 0 &= 1782 r^2 - 36r - 1 \\\\ r &= \\frac{36 \\pm \\sqrt{36^2 + 4 \\cdot 1782 \\cdot 1}}{2 \\cdot 1782} = \\frac{2 + \\sqrt{26}}{198}\\end{align*} The answer is thus $a+b+c+d = 2 + 1 + 26 + 198 = \\boxed{227}$.", "\\cos (x) + i \\sin(x)$. We have that \\begin{align*} \\cos (x+y) + i \\sin (x+y) &= e^{i(x+y)} \\\\ &= e^{ix} \\cdot e^{iy} \\\\ &= (\\cos (x) + i \\sin (x))(\\cos (y) + i \\sin (y)) \\\\ &= (\\cos (x) \\cos (y) - \\sin (x) \\sin(y)) + i(\\sin (x) \\cos(y) + \\cos(x) \\sin(y)) \\end{align*} By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by $\\cos (x) \\cos (y)$, and simplify. \\begin{align*} \\tan (x+y) &= \\frac{\\sin (x+y)}{\\cos (x+y)} \\\\ &= \\frac{\\sin (x) \\cos(y) + \\cos(x) \\sin(y)}{\\cos (x) \\cos (y) - \\sin (x) \\sin(y)} \\\\ &= \\frac{\\frac{\\sin(x)}{\\cos(x)} + \\frac{\\sin(y)}{\\cos(x)}}{1 - \\frac{\\sin (x) \\sin(y)}{\\cos (x) \\cos(y)}} \\\\ &= \\frac{\\tan (x) + \\tan (y)}{1 - \\tan (x) \\tan(y)} \\end{align*} as desired. ## Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$. Doing so yields: • $\\sin (2x) = 2\\sin (x) \\cos (x)$ • $\\cos (2x) = \\cos^2 (x) - \\sin^2 (x)$ • $\\tan (2x) = \\frac{2\\tan (x)}{1-\\tan^2 (x)}$ ### Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle", "\\right)}$ ### Transformation to current notation Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get • ${\\displaystyle \\theta =\\tan \\theta -{\\frac {\\tan ^{3}\\theta }{3}}+{\\frac {\\tan ^{5}\\theta }{5}}-{\\frac {\\tan ^{7}\\theta }{7}}+\\quad \\cdots }$. Letting tan θ = q we finally have • ${\\displaystyle \\tan ^{-1}q=q-{\\frac {q^{3}}{3}}+{\\frac {q^{5}}{5}}-{\\frac {q^{7}}{7}}+\\quad \\cdots }$ ### Another formula for the circumference of a circle The second part of the quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows. ${\\displaystyle c={\\sqrt {12d^{2}}}-{\\frac {\\sqrt {12d^{2}}}{3\\cdot 3}}+{\\frac {\\sqrt {12d^{2}}}{3^{2}\\cdot 5}}-{\\frac {\\sqrt {12d^{2}}}{3^{3}\\cdot 7}}+\\quad \\cdots }$ Since c = π d this can be reformulated as a formula to compute π as follows. ${\\displaystyle \\pi ={\\sqrt {12}}\\left(1-{\\frac {1}{3\\cdot 3}}+{\\frac {1}{3^{2}\\cdot 5}}-{\\frac {1}{3^{3}\\cdot 7}}+\\quad \\cdots \\right)}$ This is obtained by substituting q = ${\\displaystyle 1/{\\sqrt {3}}}$ (therefore θ = π / 6) in the power series expansion for tan−1 q above. ## Comparison of convergence of various infinite series for π Comparison of the convergence of two Madhava series (the one with 12", "Compound angle formula A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of $(A+B)$ or $(A-B)$ in terms of trigonometric functions of $A$ and $B$. The three basic formulae are: \\begin{align*} \\cos(A\\pm B)&=\\cos A\\cos B \\mp \\sin A\\sin B\\\\ \\sin(A\\pm B)&=\\sin A\\cos B \\pm \\cos A\\sin B\\\\ \\tan(A\\pm B)&=\\dfrac{\\tan A\\pm\\tan B}{1\\mp\\tan A\\tan B} \\end{align*} There are corresponding results for hyperbolic functions which can be obtained by applying Osborn’s Rule: \\begin{align*} \\cosh(A\\pm B)&=\\cosh A\\cosh B \\pm \\sinh A\\sinh B\\\\ \\sinh(A\\pm B)&=\\sinh A\\cosh B \\pm \\cosh A\\sinh B\\\\ \\tanh(A\\pm B)&=\\dfrac{\\tanh A\\pm\\tanh B}{1\\pm\\tanh A\\tanh B} \\end{align*}", "double angle formula is being exploited. Therefore, with an eye toward a telescoping sum, I started with \\begin{align} \\... 12 Let f(x)=\\prod_{n=0}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}\\tag{1} $$and$$ g(x)=\\prod_{n=0}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}\\tag{2} $$Then$$ \\begin{align} f(x)\\,g(x) &=\\prod_{n=0}^\\infty\\left(1-x^{2^{n+1}}\\right)^{1/2^n}\\\\ &=\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{2/2^n}\\\\ &=\\left(\\frac{f(x)}{1-x}\\right)^2\\tag{3} \\end{align} $$from ... 12 Hint: use the fact that \\cot{x}-2\\cot{2 x}=\\tan{x} 12$$\\tan rx=\\frac{\\binom r1\\tan x-\\binom r3\\tan^3x+\\cdots}{1-\\binom r2\\tan^2x+\\cdots}\\text{ (Proof below) }$$Now, if \\tan rx=\\tan r\\theta\\implies rx=n\\pi+r\\theta where n is any integer \\implies x=n\\frac{\\pi}r+\\theta If r is even, =2m(say)$$\\tan2m\\theta=\\tan2mx=\\frac{\\binom {2m}1\\tan x-\\binom {2m}3\\tan^3x+\\cdots+\\binom {2m}{2m-1}(-1)^{m-1}\\tan^{... 12 Since: $$\\arctan\\frac{3n^2}{2n^4-1}=\\arctan\\frac{1}{n^2}+\\arctan\\frac{1}{2n^2}$$ then: $$\\sum_{n=1}^{+\\infty}\\arctan\\frac{3n^2}{2n^4-1} = \\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{n^2}\\right)+\\arg\\prod_{n=1}^{+\\infty}\\left(1+\\frac{i}{2n^2}\\right).\\tag{1}$$ Since, by the Weierstrass product for the $\\sinh$ function, $$\\frac{\\sinh(\\pi z)}{\\pi z}=\\prod_{n=1}^{+... 11 Viète's formula gives:$$\\prod_{k\\geq 1} \\cos\\left(\\frac{x}{2^k}\\right)=\\frac{\\sin x}{x}\\Rightarrow \\prod_{k\\geq 1} \\sec\\left(\\frac{x}{2^k}\\right)=\\frac{x}{\\sin x}\\Rightarrow\\sum_{k\\geq 1} \\ln \\sec\\left(\\frac{x}{2^k}\\right)=\\ln x-\\ln \\sin x$$Differentiate twice and you have your result. 11 It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function \\operatorname{dn}(z,k) has Fourier series$$\\operatorname{dn}(z,k)=\\frac{\\pi}{2K}\\left[1+2\\sum_{n=1}^{\\infty}\\frac{\\cos n\\pi\\frac{z}{K}}{\\cosh n \\pi \\... 10 Actually, this sum converges for every $\\alpha>0$. Step I. For every $x\\in\\mathbb R$, the sequence $s_n=\\sum_{k=1}^n\\sin kx$ is bounded. Indeed, if $x=m\\pi$, then $s_n=0$. If $x\\ne m\\pi$, then $\\sin(x/2)\\ne 0$, and $$s_n=\\sum_{k=1}^n\\sin kx=\\mathrm{Im}\\left(\\mathrm{e}^{xi}+\\mathrm{e}^{2xi}+\\cdots\\mathrm{e}^{nxi}\\right)= \\mathrm{Im}\\left(\\mathrm{e}^{... 9 Use:$$ \\sin \\frac{\\theta}{2} \\cdot \\cos(k \\theta) = \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k+\\frac{1}{2}\\right)\\theta\\right)}_{f_{k+1}} - \\underbrace{\\frac{1}{2} \\sin\\left( \\left(k-\\frac{1}{2}\\right)\\theta\\right)}_{f_{k}} $$Thus$$ \\begin{eqnarray} \\sin \\frac{\\theta}{2} \\cdot \\sum_{k=1}^n \\cos(k \\theta) &=& \\sum_{k=1}^n \\left(f_{k+1}" ]

[ "Double and Half Angle Identity Proofs $$\\displaystyle \\frac{{\\sin \\left( {2x} \\right)}}{2}=\\frac{{\\tan x}}{{{{{\\tan }}^{2}}x+1}}$$ \\require{cancel} \\displaystyle \\begin{align}&=\\frac{{\\tan x}}{{{{{\\sec }}^{2}}x}}\\\\&=\\frac{{\\sin x}}{{\\cancel{{\\cos x}}}}\\cdot \\frac{{{{{\\cancel{{{{{\\cos }}^{2}}x}}}}^{{\\cos x}}}}}{1}\\\\&=\\sin x\\cos x\\\\&=\\sin x\\cos x\\cdot \\frac{2}{2}\\\\\\frac{{\\sin \\left( {2x} \\right)}}{2}&=\\frac{{2\\sin x\\cos x}}{2}\\,\\,\\,\\,\\,\\surd \\end{align} Note: To get $$\\displaystyle \\sin x\\cos x$$ into the right form for the identity $$\\displaystyle (2\\sin x\\cos x)$$, we had to multiply by $$\\displaystyle \\frac{2}{2}$$. $$\\displaystyle \\frac{{1-\\cos \\left( {2A} \\right)}}{{\\sin \\left( {2A} \\right)}}=\\tan A$$ \\displaystyle \\begin{align}\\frac{{1-\\left( {1-2{{{\\sin }}^{2}}A} \\right)}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{2{{{\\sin }}^{2}}A}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{{}^{{\\sin A}}\\cancel{{2{{{\\sin }}^{2}}A}}}}{{\\cancel{{2\\sin A}}\\cos A}}&=\\tan A\\,\\,\\,\\,\\,\\surd \\end{align} Note: We knew to use the $$\\displaystyle 1-2{{\\sin }^{2}}A$$ (instead of $$\\displaystyle 2{{\\cos }^{2}}A-1$$) for $$\\cos 2A$$ to get rid of the 1’s, and also because we have sin in the denominator instead of cos. $$\\displaystyle \\tan \\left( {\\frac{1}{2}\\theta } \\right)=\\csc \\theta -\\cot \\theta$$ \\displaystyle \\begin{align}&=\\frac{1}{{\\sin \\theta }}-\\frac{{\\cos \\theta }}{{\\sin \\theta }}\\\\\\tan \\left( {\\frac{1}{2}\\theta } \\right)&=\\frac{{1-\\cos \\theta }}{{\\sin \\theta }}\\,\\,\\,\\,\\,\\surd \\end{align} Note: Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos. $$\\displaystyle \\frac{{\\cos \\left( {2\\theta } \\right)}}{{\\cos \\theta -\\sin \\theta }}=\\cos \\theta +\\sin \\theta$$ \\displaystyle \\begin{align}\\frac{{{{{\\cos }}^{2}}\\theta -{{{\\sin }}^{2}}\\theta }}{{\\cos \\theta -\\sin \\theta }}&=\\\\\\frac{{\\cancel{{\\left( {\\cos \\theta -\\sin \\theta } \\right)}}\\left( {\\cos \\theta +\\sin \\theta } \\right)}}{{\\cancel{{\\cos \\theta -\\sin \\theta }}}}&=\\cos \\theta +\\sin \\theta \\,\\,\\,\\,\\,\\surd \\end{align}", "x \\\\ \\frac d {dx}\\left(\\csc x\\right) = -\\csc x\\cot x$$ ### Examples ##### Example 1 Find $$\\displaystyle \\frac d {dx}\\left(\\cos(x^2+9)\\right)$$. Step 1 Use the chain rule to differentiate. $$\\frac d {dx}\\left(\\cos(x^2+9)\\right) = -\\sin(x^2+9)\\cdot \\frac d {dx}\\left(x^2 + 9\\right) = -\\sin(x^2+9)\\cdot 2x = -2x\\sin(x^2+9)$$ $$\\displaystyle \\frac d {dx}\\left(\\cos(x^2+9)\\right) = -2x\\sin(x^2+9)$$ ##### Example 2 Suppose $$f(x) = x^4\\tan 3x$$. Find $$f'(x)$$. Step 1 Identify the factors in the function. $$f(x) = \\blue{x^4}\\red{\\tan 3x}$$ Step 2 Differentiate using the product rule. \\begin{align*}% f'(x) & = \\blue{4x^3}\\tan 3x + x^4\\cdot \\red{3\\sec^2 3x}\\\\ & = 4x^3\\tan 3x + 3x^4\\sec^2 3x \\end{align*} Step 3 Simplify by factoring. \\begin{align*} f'(x) & = 4\\blue{x^3}\\tan 3x + 3\\blue{x^4}\\sec^2 3x\\\\ & = \\blue{x^3}\\left(4\\tan 3x + 3x\\sec^2 3x\\right) \\end{align*} $$\\displaystyle f'(x) = x^3\\left(4\\tan 3x + 3x\\sec^2 3x\\right)$$. ##### Example 3 Find $$\\displaystyle \\frac d {dx}\\left(\\frac{\\sin 8x}{1 + \\sec 8x}\\right)$$. Step 1 Differentiate using the quotient rule. The parts in $$\\blue{blue}$$ are related to the numerator. \\begin{align*} \\frac d {dx}\\left(\\frac{\\sin 8x}{1 + \\sec 8x}\\right) & = \\frac{(1 + \\sec 8x)\\cdot\\blue{8\\cos 8x} - \\blue{\\sin 8x}\\cdot 8\\sec 8x\\tan 8x}{(1 + \\sec 8x)^2} \\end{align*} Step 2 Simplify the numerator. \\begin{align*} \\frac d {dx}\\left(\\frac{\\sin 8x}{1 + \\sec 8x}\\right) & = \\frac{(1 + \\sec 8x)\\cdot \\blue 8\\cos 8x - \\sin 8x\\cdot \\blue 8\\sec 8x\\tan 8x}{(1 + \\sec 8x)^2}\\\\[6pt] & = \\frac{\\blue 8(1 + \\red{\\sec 8x})\\cos", "## Precalculus (6th Edition) Blitzer Let us consider the left side: $\\tan x+\\tan y$ We know that, $\\tan x=\\frac{\\sin x}{\\cos x}$. It implies: \\begin{align} & \\tan x+\\tan y=\\frac{\\sin x}{\\cos x}+\\frac{\\sin y}{\\cos y} \\\\ & =\\frac{\\sin x\\cos y+\\sin y\\cos x}{\\cos x\\cos y} \\end{align} By using the identity $\\sin \\left( x+y \\right)=\\sin x\\cos y+\\sin y\\cos x$, we obtain: $\\tan x+\\tan y=\\frac{\\sin \\left( x+y \\right)}{\\cos x\\cos y}$ Thus, $\\tan x+\\tan y=\\frac{\\sin \\left( x+y \\right)}{\\cos x\\cos y}$", "x$$ $$\\displaystyle f'(x) = 2\\sin x\\cos x$$ Step 1 Rewrite the function so it is easier to see the need for the chain rule. $$f(x) = \\sec^2(3x) = (\\sec 3x)^2$$ Step 2 Differentiate using the chain rule. \\begin{align*} f'(x) & = 2(\\sec 3x)^1\\cdot 3\\sec 3x\\tan 3x\\\\[6pt] & = 6\\sec 3x\\sec 3x\\tan 3x\\\\[6pt] & = 6\\sec^2 3x \\tan 3x \\end{align*} Step 3 Evaluate $$f'\\left(\\frac\\pi 4\\right)$$ \\begin{align*} f'\\left(\\frac \\pi 4\\right) & = 6\\sec^2\\left(3\\cdot \\frac \\pi 4\\right)\\tan\\left(3\\cdot \\frac \\pi 4\\right)\\\\[6pt] & = 6\\left[\\sec\\left(\\frac{3\\pi}4\\right)\\right]^2\\tan\\left(\\frac{3\\pi} 4\\right)\\\\[6pt] & = 6\\left[-\\sqrt 2\\right]^2(-1)\\\\[6pt] & = 6(2)(-1)\\\\[6pt] & = -12 \\end{align*} $$\\displaystyle f'\\left(\\frac \\pi 4\\right) = -12$$ Step 1 Differentiate using the chain rule. $$f'(x) = e^{\\sin(x^4)}\\cdot \\cos(x^4)\\cdot 4x^3$$ $$\\displaystyle f'(x) = 4x^3\\cos(x^4)\\,e^{\\sin(x^4)}$$ Step 1 Rewrite the function to emphasize that the cosine is being raised to the 4th power. $$f(x) = \\left[\\cos\\left(e^{1-x^2}\\right)\\right]^4$$ Step 2 Differentiate using the chain rule. \\begin{align*} f'(x) & = 4\\left[\\cos\\left(e^{1-x^2}\\right)\\right]^3\\cdot \\frac d {dx}\\left[\\cos\\left(e^{1-x^2}\\right)\\right]\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot \\frac d {dx}\\left(e^{1-x^2}\\right)\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}\\cdot \\frac d {dx}\\left(1-x^2\\right)\\\\[6pt] & = 4\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}\\cdot \\left(-2x\\right)\\\\[6pt] & = -8x\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2} \\end{align*} $$\\displaystyle f'(x) = -8x\\cos^3\\left(e^{1-x^2}\\right)\\cdot\\sin\\left(e^{1-x^2}\\right)\\cdot e^{1-x^2}$$", "\\begin{align} \\sum_{k=0}^{89}\\tan^2\\left(\\frac{k\\pi}{180}\\right) &=\\frac12\\left(\\frac{32402}{3}-180\\right)\\\\ &=\\frac{15931}{3} \\end{align} This same method also gives $$\\sum_{k=0}^{89}\\tan^4\\left(\\frac{k\\pi}{180}\\right)=\\frac{524560037}{45}$$ and $$\\sum_{k=0}^{89}\\tan^6\\left(\\frac{k\\pi}{180}\\right)=\\frac{4855740968543}{135}$$", "# Thread: Double angle formula 1. ## Double angle formula Is it possible to input $\\displaystyle \\frac{2\\;tan\\;\\frac{\\pi}{5}}{1-tan^2\\;\\frac{\\pi}{5}}$ on a calculator? The $\\displaystyle tan^2$ in the denominator is throwing me off. 2. Originally Posted by TI-84 Is it possible to input $\\displaystyle \\frac{2\\;tan\\;\\frac{\\pi}{5}}{1-tan^2\\;\\frac{\\pi}{5}}$ on a calculator? The $\\displaystyle tan^2$ in the denominator is throwing me off. just enter $\\displaystyle \\left( \\tan \\frac {\\pi}5 \\right)^2$", "\\left( x \\ln\\frac{\\tan x}{x}+ x\\ln x \\right) \\end{align} where our goal was to make more manageable limits \"appear.\" In particular: \\begin{align} \\frac{\\sin x}{x} &= \\frac{\\sin x-\\sin 0}{x-0} \\xrightarrow[x\\to 0^+]{} \\sin' 0 = \\cos 0 =1 \\tag{1}\\\\ \\frac{\\tan x}{x} &= \\frac{\\tan x-\\tan 0}{x-0} \\xrightarrow[x\\to 0^+]{} \\tan' 0 = \\frac{1}{\\cos^2 0} =1 \\tag{2}\\\\ x\\ln x &\\xrightarrow[x\\to 0^+]{} 0 \\tag{3}\\\\ \\end{align} and therefore we have no indeterminate form anymore, as \\begin{align} \\sin x \\cdot \\ln(\\tan x) &= \\frac{\\sin x}{x}\\cdot \\left( x \\ln\\frac{\\tan x}{x}+ x\\ln x \\right) \\xrightarrow[x\\to 0^+]{(1), (2), (3)} 1\\cdot \\left(0\\cdot \\ln 1+0\\right) = 0 \\tag{\\dagger} \\end{align} and, finally, $$5(\\tan x)^{\\sin x} = 5e^{\\sin x \\cdot \\ln(\\tan x)} \\xrightarrow[x\\to 0^+]{(\\dagger)} 5e^0 = \\boxed{5}.$$ If you want to use L'Hopital's Rule, here's one way you can find a familiar indeterminant form. First note that $u = e^{\\ln u}$ since $\\ln$ is the inverse of natural exponentiation. So we can write \\begin{align} 5\\left(\\tan x \\right)^{\\sin x} &= 5e^{\\ln\\left[(\\tan x)^{\\sin x} \\right]} \\\\ &= 5e^{\\sin x \\ln(\\tan x)} \\end{align} Now, \\begin{align} \\sin x \\ln(\\tan x) &= \\frac{\\ln(\\tan x)}{\\csc x} \\sim \\frac{-\\infty}{\\infty} \\ \\leftarrow \\textrm{Indeterminant form} \\end{align} as $x \\to 0^+$. By L'Hopital's Rule: \\begin{align} \\lim_{x \\to 0^+} \\frac{\\ln(\\tan x)}{\\csc x} &= \\lim_{x \\to 0^+} \\frac{\\frac{1}{\\tan x} \\cdot \\sec^2 x}{-\\csc x \\cot x} \\ \\ \\ \\ \\textrm{(Derivative of top and bottom)} \\\\ &=", "you do this: $$\\displaystyle \\sec x=\\cot x+\\tan x$$ \\require{cancel} \\displaystyle \\begin{align}{{\\sec }^{2}}x&={{\\left( {\\cot x+\\tan x} \\right)}^{2}}\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2\\cancel{{\\cot x\\tan x}}+{{\\tan }^{2}}x\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2+{{\\tan }^{2}}x\\\\1+\\cancel{{{{{\\tan }}^{2}}x}}&={{\\cot }^{2}}x+2+\\cancel{{{{{\\tan }}^{2}}x}}\\\\{{\\cot }^{2}}x&=-1\\end{align} No solution, since we can’t square something and have it be a negative number. $$\\displaystyle \\frac{{{{{\\cos }}^{2}}\\left( x \\right)}}{{{{{\\sin }}^{2}}\\left( x \\right)}}=3$$ $$\\displaystyle \\begin{array}{c}{{\\cot }^{2}}\\left( x \\right)=3\\\\\\cot \\left( x \\right)=\\pm \\sqrt{3}\\end{array}$$ $$\\displaystyle x=\\left\\{ {\\frac{\\pi }{3},\\,\\,\\frac{{2\\pi }}{3},\\,\\,\\frac{{4\\pi }}{3},\\,\\frac{{5\\pi }}{3}} \\right\\}$$ Use Quotient Identity to convert $$\\displaystyle \\frac{{\\cos }}{{\\sin }}$$ into cot Take square roots of both sides, and don’t forget the $$\\pm$$ Here’s another one where we have to check for extraneous solutions. Solve over the reals: Solving with Trig Identities Hints $$\\displaystyle \\frac{{1-\\cos x}}{{\\sin x}}=\\frac{{\\sin x}}{{1+\\cos x}}$$ \\displaystyle \\begin{align}\\frac{{1-\\cos x}}{{\\sin x}}-\\frac{{\\sin x}}{{1+\\cos x}}&=0\\\\\\frac{{\\left( {1-\\cos x} \\right)\\left( {1+\\cos x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{\\left( {\\sin x} \\right)\\left( {\\sin x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{1-{{{\\cos }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\0&=0\\end{align} This appears to be all real numbers, but since we have denominators that can’t be 0, we have to check for those cases. Get all terms on one side Find common denominator to add terms Turn $$\\left( {1-{{{\\cos }}^{2}}x} \\right)$$ into $${{\\sin }^{2}}x$$ We get", "To show that this is the case, use the angle sum formula for tangent: \\begin{align*}tan(A + B) = \\frac{tan(A) + tan(B)}{1 - tan(A)tan(B)} \\end{align*} If you use angle \\begin{align*}A = tan ^ {-1} \\left ( \\frac{120}{119} \\right )\\end{align*} and \\begin{align*}B = tan ^ {-1} \\left ( \\frac{-1}{239} \\right )\\end{align*} the assertion above is proven. All that is required is a little bit of arithmetic as all of the tangent and tangent inverses cancel each other. Much the same way, we now need to show: \\begin{align*}4tan ^ {-1} \\left ( \\frac{1}{5} \\right ) = tan ^ {-1} \\left ( \\frac{120}{119} \\right ) \\end{align*} This is easiest to show in two steps. First show: \\begin{align*}2tan ^ {-1} \\left ( \\frac{1}{5} \\right ) = tan ^ {-1} \\left ( \\frac{5}{12} \\right ) \\end{align*} Again by using the angle addition rule with\\begin{align*}A =B = tan ^ {-1} \\left ( \\frac{1}{5} \\right ) \\end{align*} Then show that: \\begin{align*}2tan ^ {-1} \\left ( \\frac{5}{12} \\right ) = tan ^ {-1} \\left ( \\frac{120}{119} \\right )\\end{align*} With the same arithmetic techniques for a third time with \\begin{align*} A =B = tan ^ {-1} \\left ( \\frac{5}{12} \\right ) \\end{align*} It should all come together now substituting back to the top. It is also useful to remember that negatives inside of a tangent become negatives outside", "on the final step, but your evaluation is correct.$$-\\tan\\left(-\\frac\\pi3\\right)=\\sqrt 3$$", "+ x^2} \\: dx \\end{align} Now note that $\\frac{x^2}{1 + x^2} = \\frac{x^2 + 1 - 1}{1 + x^2} = \\frac{(x^2 + 1) - 1}{(x^2 + 1)} = 1 - \\frac{1}{1 + x^2}$. Therefore: (8) \\begin{align} \\int 2x \\tan^{-1} x \\: dx = x^2\\tan^{-1} x - \\int 1 - \\frac{1}{1 + x^2} \\: dx \\\\ \\int 2x \\tan^{-1} x \\: dx = x^2\\tan^{-1} x - \\left ( \\int 1 \\: dx - \\int \\frac{1}{1 + x^2} \\: dx \\right )\\\\ \\int 2x \\tan^{-1} x \\: dx = x^2\\tan^{-1} x - x + \\tan ^{-1} x + C \\end{align}", "=\\frac{1}{f'(\\pi/4)}\\left[1+\\tan^{2}(\\pi/4)\\right]\\\\ & =\\frac{1}{5}\\left[1+1^{2}\\right]\\\\ & =\\frac{2}{5}.\\end{align}", "## Precalculus (6th Edition) Blitzer We have $\\tan x+\\frac{1}{\\tan x}=\\frac{1}{\\sin x\\cos x}$ By evaluating the left side, \\begin{align} & \\text{Left side}=\\tan x+\\frac{1}{\\tan x} \\\\ & =\\frac{{{\\tan }^{2}}x+1}{\\tan x} \\\\ & =\\frac{{{\\sec }^{2}}x}{\\tan x} \\\\ & =\\frac{{}^{1}/{}_{{{\\cos }^{2}}x}}{{\\sin x}/{\\cos x}\\;} \\end{align} $=\\frac{1}{\\sin x\\cos x}$ Thus, $\\text{left side = right side}$. So, we can say that the identity is true. Hence, verified.", "2. Start by using the double angle identity for $\\displaystyle \\cos{2x}$ and the triple angle identity for $\\displaystyle \\sin{3x}$. 3. Start by expanding $\\displaystyle (\\sin{\\theta} + \\cos{\\theta})^2$. 10. ## Re: Algebra review Qs pt 2: divsion with several var./exp. & abs. val. inequal. Thanks, got it. Really appreciate all the help. Btw just so you guys know you are not doing all my homework haha, I do about 10-15 problems in between these. Here's the other two I don't know how to do. I completely forgot rules of trig so I don't know how to cancel stuff out or the trig formulas. 11. ## Re: Algebra review Qs pt 2: divsion with several var./exp. & abs. val. inequal. For the first one consider $\\displaystyle \\tan x = \\frac{\\sin x}{\\cos x}$ $\\frac{\\sin x +\\tan x \\cos x}{\\tan x} =\\frac{\\sin x }{\\tan x}+\\frac{\\tan x \\cos x}{\\tan x} = \\cdots$ For the second make $\\sin 3x = \\sin (2x+x)$ and apply the compound angle formula For the third one, expand the LHS, what do you get? Finally $\\displaystyle \\cot 2x = \\frac{-1}{\\sqrt{3}}\\implies \\frac{1}{\\tan x} = \\frac{-1}{\\sqrt{3}}\\implies \\tan x = -\\sqrt{3}$ Maybe time you started a new thread?", "} \\right) = \\left( { - \\frac{\\pi }{4}} \\right)$$. Chapter 2 Ex.2.1 Question 11 Find the value of $${\\tan ^{ - 1}}\\left( 1 \\right) + {\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) + {\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right)$$. Solution Let, $${\\tan ^{ - 1}}\\left( 1 \\right) = x$$ Hence, \\begin{align}\\tan x &= 1\\\\&= \\tan \\left( {\\frac{\\pi }{4}} \\right)\\end{align} Therefore, $${\\tan ^{ - 1}}\\left( 1 \\right) = \\left( {\\frac{\\pi }{4}} \\right)$$ Now, let $${\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = y$$ Hence, \\begin{align}\\cos y &= - \\frac{1}{2}\\\\&= - \\cos \\left( {\\frac{\\pi }{3}} \\right)\\\\&= \\cos \\left( {\\pi - \\frac{\\pi }{3}} \\right)\\\\&= \\cos \\left( {\\frac{{2\\pi }}{3}} \\right)\\end{align} Therefore, $${\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = \\frac{{2\\pi }}{3}$$ Again, let $${\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = z$$ Hence, \\begin{align}\\sin z &= - \\frac{1}{2}\\\\&= - \\sin \\left( {\\frac{\\pi }{6}} \\right)\\\\&= \\sin \\left( { - \\frac{\\pi }{6}} \\right)\\end{align} Therefore, $${\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = - \\frac{\\pi }{6}$$ Thus, \\begin{align}{\\tan ^{ - 1}}\\left( 1 \\right) + {\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) + {\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) &= \\frac{\\pi }{4} + \\frac{{2\\pi }}{3} - \\frac{\\pi }{6}\\\\&= \\frac{{3\\pi + 8\\pi - 2\\pi }}{{12}}\\\\&= \\frac{{9\\pi }}{{12}}\\\\&= \\frac{{3\\pi }}{4}\\end{align} Chapter 2 Ex.2.1 Question 12 Find the value", "That is to say $$\\tan x \\cdot \\left(\\tan \\frac x2\\right)^2 + 2\\tan \\frac x2 - \\tan x = 0$$ and $$\\tan x \\cdot \\left(-\\cot \\frac x2\\right)^2 - 2\\cot \\frac x2 - \\tan x = 0$$ • Wow this is great thank you! – Saran Wrap Oct 15 '19 at 12:47", "nx}{n\\tan mx}\\\\\\\\ &=\\lim_{x\\to 0^+}\\left(\\frac{m}{\\tan mx}\\cdot\\frac{\\tan nx}n\\right)\\\\\\\\ &=1\\cdot 1\\\\\\\\ &=1\\;, \\end{align*} since $\\displaystyle\\lim_{y\\to 0}\\frac{\\tan y}y=1$. -", "sin3x/1 1/sin5x cos5x1/cos3x. He then somehow found a way to get a 3x to the denominator of the first term after factoring out a 3/5, and add 5x to the numerator of the second term. I'm not sure how he managed that, could someone help me to understand? Thanks! Kevin First, you need to know that \\displaystyle \\begin{align*} \\lim_{x \\to 0}\\frac{\\tan{(x)}}{x} = 1 \\end{align*}. This should be obvious if you already know the standard limit \\displaystyle \\begin{align*} \\lim_{x \\to 0}\\frac{\\sin{(x)}}{x} = 1 \\end{align*}. Then \\displaystyle \\begin{align*} \\lim_{x \\to 0} \\frac{\\tan{(3x)}}{\\tan{(5x)}} &= \\lim_{x \\to 0} \\left[ \\frac{15x}{15x} \\cdot \\frac{\\tan{(3x)}}{\\tan{(5x)}} \\right] \\\\ &= \\frac{3}{5} \\lim_{x \\to 0} \\left[ \\frac{\\tan{(3x)}}{3x} \\cdot \\frac{5x}{\\tan{(5x)}} \\right] \\end{align*} I'm sure you can go from here.", "$v = \\tan(z)\\text{.}$ So \\begin{align*} \\int z \\sec^2(z) dz \\amp = z \\tan(z) - \\int \\tan(z) \\,dz\\\\ \\int z \\sec^2(z) dz \\amp = z \\tan(z) - \\int \\frac{\\sin(z)}{\\cos(z)} \\,dz\\\\ \\amp = z \\tan(z) + \\ln |\\cos(z)| + C \\end{align*} We check using differentiation as follows: \\begin{align*} \\frac{d}{dz} \\left( z \\tan(z) + \\ln |\\cos(z)| + C \\right) \\amp = z\\sec^2(z) + \\tan(z) - \\frac{\\sin(z)}{\\cos(z)}\\\\ \\amp = z\\sec^2(z) + \\tan(z) - \\tan(z)\\\\ \\amp = z \\sec^2(z) \\end{align*} 4. Using $u = \\ln(x)$ and $dv = x dx\\text{,}$ we obtain $du = \\dfrac{1}{x} dx$ and $v = \\dfrac{1}{2}x^2\\text{.}$ So \\begin{align*} \\int x\\ln(x) dx \\amp = \\frac{1}{2}x^2 \\ln(x) - \\frac{1}{2} \\int x dx\\\\ \\amp = \\frac{1}{2}x^2 \\ln(x) - \\frac{1}{4}x^2 + C \\end{align*} We check using differentiation as follows: \\begin{align*} \\frac{d}{dx} \\left( \\frac{1}{2}x^2 \\ln(x) - \\frac{1}{4}x^2 + C \\right) \\amp = \\frac{1}{2}x + x \\ln(x) - \\frac{1}{2} x\\\\ \\amp = x \\ln(x) \\end{align*} ### SubsectionSome Subtleties with Integration by Parts Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating \\begin{equation*} \\int \\arctan(x) \\, dx\\text{.} \\end{equation*} Initially, this problem seems ill-suited to integration by parts, since there does not appear to be", "using the fact that $\\tan\\left(\\frac{\\pi}{2}-x\\right)=\\frac{1}{\\tan(x)}$. \\begin{align} \\sum_{k=1}^\\infty\\tan^{-1}\\left(\\frac{1}{2k^2}\\right) &=\\sum_{k=1}^\\infty\\tan^{-1}\\left(\\frac{1}{2k-1}\\right)-\\tan^{-1}\\left(\\frac{1}{2k+1}\\right)\\\\ &=\\tan^{-1}(1)\\\\ &=\\frac{\\pi}{4} \\end{align} Of course, then $\\tan\\left(\\frac{\\pi}{4}\\right)=1$ -" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "6 & 0 & 5 \\end{pmatrix}$", ": $w=\\begin{pmatrix} 6 \\\\ -6 \\\\ -3 \\end{pmatrix}$", "\\begin {pmatrix} 12 & 6 & -2 \\\\ -6 & -3 & 1 \\\\ 6 & 3 & -1 \\end {pmatrix}$", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "& 4 \\end{pmatrix} \\begin{pmatrix} 3 & 6 & 9 \\end{pmatrix} \\begin{pmatrix} 5 & 8 \\end{pmatrix}$.", "= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "1 + 1 \\cdot 3 + 3 \\cdot 5 = 17$$ The $2,2$ entry of $\\bfA\\bfB$ is the product of the second row of $\\bfA$ with the second column of $\\bfB$: $$(\\bfA \\bfB)_{22} = \\begin{pmatrix} -1 & 1 & 3 \\end{pmatrix} \\begin{pmatrix}2\\\\4\\\\6\\end{pmatrix} = -1\\cdot 2 + 1 \\cdot 4 + 3 \\cdot 6 = 20$$ Hence, $$\\begin{pmatrix}1 & 0 & 2 \\\\ -1 & 1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \\end{pmatrix} = \\begin{pmatrix}11 & 14 \\\\ 17 & 20 \\end{pmatrix}.$$", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "where $$A= \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\mbox{ and } I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}.$$", "1 & -2 & -5 \\end{pmatrix},$$ $$(A-6I)(A - 12 I) = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix},$$", "0 & 1 & 0 & 1 & 1 \\end{pmatrix}.$$ -", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "0 & 0 & 0\\\\ 0 & 3 + i\\sqrt3 & 1\\\\ 0 & 0 & 3 + i\\sqrt3 \\end{pmatrix}.$$", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03" ]

[ "-4 & 1 & -3 & 0 \\\\ 3 & -3 & -2 & -3 & 3 & 5 & 0 & -3 & 0 & -1 & -2 & 0 & 3 & 2 & 6 \\end{pmatrix} \\\\ T &=", "4 \\\\ 3 \\end{pmatrix}.$ ×", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "5 & -8 -4 & 12 + 3 \\\\ 3 \\times 2 + 5 & -6 - 4 & 9 +3\\\\7 \\times 2 + 5 & -14 - 4 & 21 + 3\\end{pmatrix} \\\\ & = \\begin{pmatrix}13 & -12 & 15 \\\\ 11 & -10 & 12 \\\\ 19 & -18 & 24\\end{pmatrix}\\end{align*} Similarly: CB = \\begin{pmatrix}23 & -2 & 3 \\\\ 0 & 0 & 0 \\\\ 29 & -7 & 4\\end{pmatrix} 1 Like", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", ". & . & . & . \\\\ . & . & . & . & . & . & 2 & . & . & . \\\\ . & . & . & . & . & . & . & 3 & . & . \\\\ . & . & . & . & . & . & . & . & 4 & . \\end{pmatrix} = \\begin{pmatrix} 1 & . & . & . & . & . & . & . & . & . \\\\ -4 & 1 & . & . & . & . & . & . & . & . \\\\ 6 & -3 & 1 & . & . & . & . & . & . & . \\\\ -4 & 3 & -2 & 1 & . & . & . & . & . & . \\\\ 1 & -1 & 1 & -1 & 1 & . & . & . & . & . \\\\ . & . & . & . & . & 1 & . & . & . & . \\\\ . & . & . & . & . & 1 & 1 & . & . & . \\\\ . & . & . & . &", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "\\end{pmatrix} + } + & + \\color{red} { + MS_{6} = + \\begin{pmatrix} ​ + 3 & 21 & 19 & 12 & 10 \\\\ + 22 & 20 & 13 & 6 & 4 \\\\ + 16 & 14 & 7 & 5 & 23 \\\\ + 15 & 18 & 1 & 24 & 17 \\\\ + 9 & 2 & 25 & 18 & 11 + \\end{pmatrix} + } + \\end{align} + + + + \\begin{align} + \\color{green} { + MS_{7} = + \\begin{pmatrix} ​ + 11 & 10 & 4 & 23 & 17 \\\\ + 18 & 12 & 6 & 5 & 24 \\\\ + 25 & 19 & 13 & 7 & 1 \\\\ + 2 & 21 & 20 & 14 & 8 \\\\ + 9 & 3 & 22 & 16 & 15 + \\end{pmatrix} + } + & + \\color{red} { + MS_{8} = + \\begin{pmatrix} ​ + 10 & 4 & 23 & 17 & 11 \\\\ + 12 & 6 & 5 & 24 & 18 \\\\ + 19 & 13 & 7 & 1 & 25 \\\\ + 21 & 20 & 14 & 8 & 2 \\\\ + 3 & 22 & 16 & 15 & 9 + \\end{pmatrix} + } + \\end{align} +", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "\\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} 1^3 - 3\\times 1^2\\end{pmatrix} - \\begin{pmatrix} \\begin{pmatrix}\\frac{1}{2} \\end{pmatrix}^3 - 3\\begin{pmatrix}\\frac{1}{2} \\end{pmatrix}^2\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} 1 - 3\\end{pmatrix} - \\begin{pmatrix} \\frac{1}{8} - 3\\times \\frac{1}{4}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix} -2\\end{pmatrix} - \\begin{pmatrix} \\frac{1}{8} - \\frac{3}{4}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 - \\begin{pmatrix} \\frac{1}{8} - \\frac{6}{8}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 - \\begin{pmatrix} - \\frac{5}{8}\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} -2 + \\frac{5}{8} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} - \\frac{11}{8} \\end{bmatrix} \\\\ & = \\frac{11}{32} \\\\ P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = 0.344 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, between $$x=0.5$$ and $$x=1$$ is equal to $$0.344$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} = 0.344$$ 3. To find $$P\\begin{pmatrix}X \\geq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}X \\geq 1\\end{pmatrix} & = \\int_1^{+\\infty}f(x)dx \\\\ & = \\int_1^2 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4}\\int_1^2 x(x-2)dx \\\\ & = -\\frac{3}{4}\\int_1^2 \\begin{pmatrix} x^2 - 2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix}\\frac{x^3}{3}-x^2 \\end{bmatrix}_1^2 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix}x^3-3x^2 \\end{bmatrix}_1^2 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}2^3-3\\times 2^2\\end{pmatrix} - \\begin{pmatrix}1^3-3\\times 1^2\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}8-12\\end{pmatrix} - \\begin{pmatrix}1-3\\end{pmatrix} \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\begin{pmatrix}-4\\end{pmatrix} - \\begin{pmatrix}-2\\end{pmatrix} \\end{bmatrix} \\\\ & =", "to choose 4 from the 8 with cracks and 1 from the remaining 17. 5. Jan 19, 2010 ### exitwound b.)$$\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix} = \\frac{8!}{4!4!} * \\frac{17!}{1!16!}= 1190$$ c.) $$\\frac{\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} = \\frac{\\frac{8!}{4!4!} * \\frac{17!}{1!16!}}{\\frac{25!}{5!20!}} = .0224$$ d.) $$\\frac{\\begin{pmatrix} 8\\\\ 5 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 0 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} + AnswerToPartC = \\frac{\\frac{8!}{5!3!} * \\frac{17!}{0!17!}}{\\frac{25!}{5!20!}} + .0224 = .0235$$ Is this how it's done? 6. Jan 19, 2010 ### vela Staff Emeritus Yup, perfect.", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", ": $w=\\begin{pmatrix} 6 \\\\ -6 \\\\ -3 \\end{pmatrix}$", "4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} , \\quad (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We see that the numbers in the parentheses between both cycles are difference. Hence $(23)$ and $(165)$ are disjoint cycles. Also notice that: (2) \\begin{align} \\quad (23) \\circ (165) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 3 & 2 & 4 & 1 & 5 \\end{pmatrix} \\end{align} We also have that: (3) \\begin{align} \\quad (165) \\circ (23) = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 6 & 2 & 3 & 4 & 1 & 5 \\end{pmatrix} \\circ \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\\\ 1 & 3 & 2 & 4 & 5 & 6 \\end{pmatrix} \\circ = \\begin{pmatrix} 1 & 2 &", "\\;=\\;\\begin{pmatrix}(\\text{-}4499-\\frac{11}{2}\\Delta) + (4522+7\\Delta) \\\\ (3272 + 4\\Delta) + (\\text{-}3230 - 5\\Delta) \\end{pmatrix} \\;=\\;\\begin{pmatrix}23 + \\frac{3}{2}\\Delta \\\\ 42 - \\Delta \\end{pmatrix}$ If $x = 35$, we have: . $23+\\tfrac{3}{2}\\Delta \\:=\\:35 \\quad\\Rightarrow\\quad \\boxed{\\Delta \\,=\\,8}$ Then: . $y \\:=\\:42 - 8 \\quad\\Rightarrow\\quad\\boxed{ y \\,=\\,34}$", "\\times 7 = \\boxed{70} \\end{gather*}$$" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "\\\\ 0 & 0 & -1 & 0 \\\\ 1 & 1 & 1-\\mathrm{i} & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix},$$ $$A^{(4)}_6= \\begin{pmatrix} 1 & 0 & 1 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_7= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_8= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix}.$$ $A_0^{(4)}$ $A_1^{(4)}$ $A_2^{(4)}$ $A_3^{(4)}$ $A_4^{(4)}$ $A_5^{(4)}$ $A_6^{(4)}$ $A_7^{(4)}$ $A_8^{(4)}$", "so $\\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ -1 \\end{pmatrix}s =\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix}(2-(1/3)s).$ 2. These two are equal. To show that $S_1\\subseteq S_2$, we check that for any t,s we can find an appropriate m,n so that these hold. $\\begin{array}{*{2}{rc}r} 4m &- &4n &= &1t+2s \\ 7m &- &2n &= &3t+1s \\ 7m &- &10n &= &1t+5s \\end{array}$ Use Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 7 &-2 &3t+1s \\ 7 &-10 &1t+5s \\end{array}\\right) &\\xrightarrow[(-7/4)\\rho_1+\\rho_3]{(-7/4)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &-3 &-(3/4)t+(6/4)s \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to conclude that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 1 \\\\ 5 \\end{pmatrix}s =\\begin{pmatrix} 4 \\\\ 7 \\\\ 7 \\end{pmatrix}((1/2)t) +\\begin{pmatrix} -4 \\\\ -2 \\\\ -10 \\end{pmatrix}((1/4)t-(1/2)s)$ and so $S_1\\subseteq S_2$. For $S_2\\subseteq S_1$, solve $\\begin{array}{*{2}{rc}r} 1t &+ &2s &= &4m-4n \\ 3t &+ &1s &= &7m-2n \\ 1t &+ &5s &= &7m-10n \\end{array}$ with Gaussian reduction $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 3 &1 &7m-2n \\ 1 &5 &7m-10n \\end{array}\\right) &\\xrightarrow[-\\rho_1+\\rho_3]{-3\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &3 &3m-6n \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "&=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&-7\\\\ 0&0&0&7\\end{pmatrix}\\tag{R_3=R_3-R_1}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&7\\tag{R_2=-\\frac17 R_2}\\end{pmatrix}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&0\\tag{R_3=R_3 -7R_2}\\end{pmatrix} \\end{align}", "\\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "Example Using the method we just saw and without using a calculator, calculate each of the following: 1. $$101^2$$ 2. $$27^2$$ 3. $$23^2$$ 4. $$99^2$$ 5. $$104^2$$ 6. $$48^2$$ ## Solution Without Working 1. $$101^2 = 10201$$ 2. $$27^2 = 711$$ 3. $$23^2 = 529$$ 4. $$99^2 = 9801$$ 5. $$104^2 = 10816$$ 6. $$48^2 = 2304$$ ## Solution With Working 1. We calculate $$101^2$$ as follows: \\begin{aligned} 101^2 & = \\begin{pmatrix} 100+1\\end{pmatrix}^2 \\\\ & = 100^2 + 2 \\times 100 \\times 1 + 1^2 \\\\ & = 10000 + 200 + 1\\\\ 101^2 &= 10201 \\end{aligned} 2. We calculate $$27^2$$ as follows: \\begin{aligned} 27^2 & = \\begin{pmatrix} 30 - 3 \\end{pmatrix}^2 \\\\ & = 30^2 - 2 \\times 30 \\times 3 - 3^2 \\\\ & = 900 - 180 - 9 \\\\ & = 720 - 9 \\\\ 27^2 & = 711 \\end{aligned} 3. We calculate $$23^2$$ as follows: \\begin{aligned} 23^2 & = \\begin{pmatrix} 20 + 3 \\end{pmatrix}^2 \\\\ & = 20^2 + 2 \\times 20 \\times 3 + 3^2 \\\\ & = 400 + 120 + 9 \\\\ 23^2 & = 529 \\end{aligned} 4. We calculate $$99^2$$ as follows: \\begin{aligned} 99^2 & = \\begin{pmatrix} 100 - 1 \\end{pmatrix} \\\\ & = 100^2 - 2\\times 100 \\times 1 + 1^2 \\\\ & = 10000 - 200 +", "8 & 12 & 0 & 7 & 3 \\\\\\ 1 & 4 & 7 & 0 & 2 \\\\\\ 4 & 9 & 3 & 2 & 0 \\end{pmatrix}$$", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03", "\\stackrel{E_7}{\\sim} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} \\stackrel{E_6E_5E_4E_3E_2E_1}{\\sim} L \\end{eqnarray*} \\begin{eqnarray*} E_7= \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} =E_7^{-1} \\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}" ]

[ "# For a right triangle ABC, how do you find the sine, cosine and tangent of angle A? $\\sin A = \\text{oppo. side \" / \"hypotenuse} = \\frac{a}{h}$ $\\cos A = \\text{adj. side \" / \"hypotenuse} = \\frac{b}{h}$ $\\tan A = \\text{oppo. side \" / \"adj. side} = \\frac{a}{b}$", "# Math Help - sums and differences of angles 1. ## sums and differences of angles If the angles A, B of triangle ABC are such that $cos A = {-4/5}$ and $cos B = {12/13}$, find without using tables the value of Cos C' Thanks. 2. $C = \\pi - A - B$ $\\cos C = \\cos(\\pi - A - B)$ Using identities: $\\cos C = \\cos(\\pi - A)\\cos B + \\sin(\\pi - A)\\sin B$ $\\cos C = -\\cos A \\cos B + \\sin A \\sin B$ You should be able to calculate $\\sin A$ and $\\sin B$.", "to choose two adjacent coins $A$ and $B$, and then flip all coins on the line $AB$ (and its extension). Characterize all starting configuration such that it's always possible to get all tails. ## Thursday, February 3, 2011 ### Find angle C From AMC12 1999: In triangle $ABC$, $3 \\sin A + 4 \\cos B = 6$ $4 \\sin B + 3 \\cos A = 1$ Find $\\angle C$", "# Thread: triangle give 2sides and a angle -cosine? 1. ## triangle give 2sides and a angle -cosine? Craka_photo/math - Photobucket - Video and Image Hosting give triangle ABC with a=7 b=x c=5 and angle A = 60 degrees. Find x 2. Apply the cosine rule: $a^2=b^2+c^2-2bc\\cos A$ Then, $49=x^2+25-5x\\Rightarrow x^2-5x-24=0$ Find the roots and choose the positive one.", "Krishna 0 Step 1: Understand the question. NOTE: Right angle triangle ABC, right angle at C. So, A and B are complimentary angles. Step 2: Determine the unknown value NOTE: If \\angle A \\text{ and } \\angle B are the acute angles of a right triangle, sin A = cos B Plugging the known values x + 0.1 = 2x - 0.4 2x - x = 0.1 + 0.4 x = 0.5", "## Algebra 1 Recall, for a right triangle, cos B equals the length of the side adjacent to angle B divided by the length of the hypotenuse of the triangle. When we do this for the triangle in the problem, we find that cos B equals 6/10, which reduces to $3/5$.", "# sums and differences of angles • February 20th 2010, 03:47 PM deltaxray sums and differences of angles If the angles A, B of triangle ABC are such that $cos A = {-4/5}$ and $cos B = {12/13}$, find without using tables the value of Cos C' Thanks. • February 20th 2010, 03:53 PM icemanfan $C = \\pi - A - B$ $\\cos C = \\cos(\\pi - A - B)$ Using identities: $\\cos C = \\cos(\\pi - A)\\cos B + \\sin(\\pi - A)\\sin B$ $\\cos C = -\\cos A \\cos B + \\sin A \\sin B$ You should be able to calculate $\\sin A$ and $\\sin B$.", "Alex has drawn a triangle with sides of length $a$, $b$, and $c$ such that $ab+bc+ac=247$. Find the range of possible values for the triangle’s perimeter.", "### বর্গের ভিতরে কোণ ##### Score: 4 Points In square $ABCD$ we select a point $P$ such that $AP=a$, $BP=2a$, $CP=3a$. What is the value of $\\angle APB$ in degrees? Geometry #### Statistics Tried 212 Solved 124 First Solve @Gourab", "Home > Triangles > Triangle problem – finding height with given area and angles. ## Triangle problem – finding height with given area and angles. If the area of triangle ABC is ${\\cal S}$ and angles $\\angle A= \\alpha$ and $\\angle B= \\beta$ then find a altitude of triangle which drawn from C to AB. HINT: Let altitude be h. $AC = \\frac{h}{ \\sin \\beta } \\; and \\; BC = \\frac{h}{ \\sin \\alpha } \\; \\angle ACB = \\pi - \\alpha - \\beta$", "c) Sin 0 = cos (0 – pie/2) for all values of 0. d) If triangle ABC has a right angle … 5. ### maths Choose three options which are true: a) an angle of 150 degrees is equivalent to 2pie/3 radians. b) Cos 0 = cos (0 – pie/2) for al values of 0. c) Sin 0 = cos (0 – pie/2) for all values of 0. d) If triangle ABC has a right angle … 6. ### MATH 1.)Find the exact solution algebriacally, if possible: (PLEASE SHOW ALL STEPS) sin 2x - sin x = 0 2.) GIVEN: sin u = 3/5, 0 < u < ï/2 Find the exact values of: sin 2u, cos 2u and tan 2u using the double-angle formulas. 3.)Use … 7. ### Algebra In triangle GHI, angle H is a right angle, GH=40, and cos G=40/41. Find each value in fraction and in decimal form. a. sin G b. sin I c. cot G d. csc G e. cos I f. sec H 8. ### Algebra 2 math In triangle GHI, angle H is a right angle, GH = 40, and cos G= 40/41. Draw a diagram and find each value in fraction and in decimal form. a) Sin G b. Sin I c. Cot G d. csc G e. cos", "+13 triangle ABC, a=3, angleB=24, angleC=24 1/ You need to find the height(h) of this triangle: $$h = tan(B)(a/2)$$ 2/ Now you can calculate the area of the triangle using this formula: $$Area(A) = h(a/2)$$ civonamzuk May 11, 2015 #4 +27044 +8 civonamzuk has spotted that this is an isosceles triangle, which simplifies the calculation significantly! . Alan May 11, 2015", "= a' \". B B A C A C Note that the angles (3 and 7 determine the positions of O and P, and so determine a and a'. Indeed, we can get an equation relating a, (3 and 7 and another relating a', j3 and 7 as follows: By a harmless ambiguity, let A, B, C denote both vertices and vertex angles of the triangle ABC. Then from the first diagram, applying the sine rule to the three inner triangles, s t t u us sin/3 sin(A —a)' shi7 sin(B — j3)' sina sin(C - 7) leading to /s\\ ( t\\ /u\\ sin a sin (3 sin 7 ~ \\~t) \\u) \\s) ~ s'm(A - a) sin(B - f3) sin(C - 7)' 1987 97 Similarly, from the second diagram, 1=1^ (i x/ \\y sin a' sin f3 sin 7 sin(A - a') sin(S - (3) sin(C - 7) (Since O, P are interior points, none of the relevant angles are zero.) Hence sin a sin a sin(^4 — a) sin(A — a') Cross-multiplying, expanding and cancelling gives sin a cos a' = sin a' cos a, whence sin(a — a') = 0 and so a — a', (a — a' = ±tt is not feasible.) Solution 5, Alternative 1 Setting x\\ = xi = ... = xn", "triangle,state this and indicate how you know A=40 °, a=8, b=10 2. Find all missing sides and angles of the triangle. If there are two solutions, give both. If there is no triangle, state this and indicate how you know. $$A = 40 ^ { \\circ } , a = 8 , b = 10$$ (5 points)", "# Math Help - can someone help please 1. ## can someone help please My calulator is broken and I need this tomorrow. In triangle abc, a=7, b=9 and c=11. Find the three angle measures of the triangle. Using sin 2. Originally Posted by jwein4492 My calulator is broken and I need this tomorrow. In triangle abc, a=7, b=9 and c=11. Find the three angle measures of the triangle. Using sin using sin what? you cannot use the law of sines here. and this is not a right triangle, so Pythagoras or trig ratios won't help us. use the law of cosines. $a^2 = b^2 + c^2 - 2bc \\cos A$ you can use that to find angle A. Similar formulas can get you angles B and C. Or, once you have angle A, you can use the law of Sines to get the other angles. Your choice", "Now continue..... (review the responses above with paper&pencil) Johnny Blaze New member In triangle ABC right angle at A. If Sin B Sin C = 0.2 and Sin (B - C) = 5a/2. Find the value of 5a - 1. Please help me. Thanks. Here's my working Since angle A = 90 Angle (B + C) = 90 Sin (B + C) = Sin 90 = 1 Cos (B + C) = Cos 90 = 0 Sin (B + C) = Sin B Cos C + Cos B Sin C Cos (B + C) = Cos B Cos C - Sin B Sin C 0 = Cos B Cos C - 0.2 Cos B Cos C = 0.2 2 Cos B Cos C = 0.4 Cos (B + C) + Cos (B - C) = 0.4 0 + Cos (B - C) = 0.4 Cos (B - C) = 0.4 = 2/5 Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2 Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5 Since angle A = 90 Angle (B + C) = 90 Sin (B + C) = Sin 90 = 1 Cos (B + C) = Cos 90 = 0 Sin (B + C) =", "triangle $$ABC$$ was multiplied by 1.5 to get the corresponding side length on triangle $$DEF$$.", "triangles is shown here. An isosceles triangle $ABC$ $($with $\\angle B=\\angle C)$ is cut into two smaller isosceles triangles (which don't have to be identical). How many different possible angles $\\angle A$ are there? In each of the cases below, there exists only one solution whereby the sum of $n$ positive integers equals the product. • For $n=2$, we have $2+2=2\\times 2$. • For $n=3$, we have $1+2+3=1\\times2\\times3$. • For $n=4$, we have $1+1+2+4=1\\times 1\\times2\\times4$. If solution(s) exist for $n=5,$ find the sum of all possible sums (or equivalently, products). If such a solution does not exist, submit your answer as 0. ×", "(4x-7)^{\\circ} \\ and\\ (5x-11)^{\\circ}$. Find the value of $x$ and the measures of all the angles. $2x + (4x-7) + (5x-11) = 180^o$ Or $x = 18^o$ Hence the angles are $36^{\\circ}, 65^{\\circ}, \\ and\\ 79^{\\circ}$ $\\\\$ Question 6: In $\\Delta ABC, \\angle A=x^{\\circ}, \\angle B=(2x-15)^{\\circ}, \\ and\\ \\angle C=(3x+21)^{\\circ}$. Find the value of $x$ and the measure of all the angles of the triangle. $x+(2x-15)+(3x+21)=180^o$ or $x=29^o$ Hence $\\angle A=29^{\\circ}, \\angle B=43^{\\circ}, \\ and\\ \\angle C=108^{\\circ}$ $\\\\$ Question 7: In $\\Delta ABC, \\angle A=(2x+15)^{\\circ}, \\angle B=x^{\\circ}, \\ and\\ \\angle C=(3x-15)^{\\circ}$. Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle. $(2x+15)+x+(3x-15)=180^o \\ or\\ x=30^o$ Hence $\\angle A=75^{\\circ}, \\angle B=30^{\\circ}, \\ and\\ \\angle C=75^{\\circ}.$ $\\\\$ Question 8: In $\\Delta ABC, \\angle A=3\\ \\angle B=6 \\ \\angle C$ . Find the angles of the triangle. Let $\\angle A=3\\angle B=6 \\angle C=6x$ $\\Rightarrow \\angle A=6x, \\angle B=2x, and \\angle C=x$ $\\Rightarrow 6x+2x+x=180^o \\ or\\ x=20^o$ $\\Rightarrow \\angle A=120^{\\circ}, \\angle B=40^{\\circ} \\ and\\ \\angle C=20^{\\circ}$ $\\\\$ Question 9: In the adjoining figure, it is given that $\\angle ABC=40^{\\circ}, \\angle ACD=70^{\\circ}, \\angle ACB=x^{\\circ} \\ and\\ \\angle EAC=y$. Find the value of \\$latex $x and y$. $x+70=180^o \\ hence\\ x=110^{\\circ}$ Note: Sum of angles of adjacent angles on a", "rewrite the expression as a product. Simplify if possible. cos 1/2+ cos 1/7 cos 1/2+ cos 1/7= ​(Simplify your answer.Use integers or fractions for any number… Add the given vectors by using the trigonometric functions and the Pythagorean theorem. 19) A = 23.6, Ό = 144.3 ° B = 21.7, Ό = 192.3 ° 1. The population of a certain city increase according to the following formula P = 200,000 e 0.01 t where P is the population and t the number of years, with t = 0 corresponding to the year … need help!!! Example 9. The perimeter of triangle ABC is 24. M is the midpoint of AB such that MC = MA = 5. How many square inches are in the area of A the triangle? (A) 48 (B) 24 (C) 12 (D) 14 (E) 12… Please solve 45 and 51! Show steps! Thank you so much 45. cos 0.8 46. cos 0.6 48. sin 5.4 49. sin 3.8 51. a positive angle whose cosine is —0.65 52. 53. 54. a positive angle whose sine is —O.95 a … Two gears are adjusted so that the smaller gear drives the larger​ one, as shown in the figure. If the smaller gear rotates through an angle of 270​º, through how many degrees will the larger ge…" ]

[ "formula under the assumption that $$\\theta$$ was between $$0^\\circ$$ and $$90^\\circ$$, but the formula in fact holds for all values of $$\\theta$$. We shall see this in the next subsection, where we prove the general expansion formula for $$\\cos(A + B)$$. #### Example Find $$\\cos 22\\frac{1}{2}^\\circ$$ in surd form. #### Solution Putting $$\\theta = 22\\frac{1}{2}^\\circ$$ into the double angle formula $$\\cos 2\\theta = 2\\cos^2\\theta -1$$ and writing $$x$$ for $$\\cos 22\\frac{1}{2}^\\circ$$, we obtain \\begin{align*} \\cos 45^\\circ &= 2x^2-1\\\\ \\dfrac{1}{\\sqrt{2}} &= 2x^2-1 \\\\ 2x^2& =\\dfrac{1+\\sqrt{2}}{\\sqrt{2}}\\\\ x &= \\sqrt{\\dfrac{1+\\sqrt{2}}{2\\sqrt{2}}}, \\end{align*} which simplifies to $$\\dfrac{1}{2}\\sqrt{2 + \\sqrt{2}}$$. We can also find a double angle formula for sine using the same diagram. In this case, we write down formulas for the area of $$\\triangle ABC$$ in two ways: • On the one hand, the area is given by $$\\dfrac{1}{2} AB \\cdot AC \\, \\sin (\\angle BAC) = \\dfrac{1}{2}\\sin 2\\theta$$. • Since $$AD=\\cos\\theta$$, we can alternatively split the triangle into two right-angled triangles and write the area as $$2\\times \\dfrac{1}{2} \\sin\\theta\\, \\cos\\theta = \\sin\\theta\\, \\cos\\theta$$. Equating these two expressions for the area, we obtain $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta.$ As before, we assumed that $$\\theta$$ lies between $$0^\\circ$$ and $$90^\\circ$$, but the formula is valid for all values of $$\\theta$$. Exercise 8 Find $$\\sin 15^\\circ$$ in surd form. From the Pythagorean identity and the", "equivalent to $2-\\sin^2B$, which is $2-\\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\\dfrac{111-4\\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \\boxed{222}$. Note that the problem has a flaw because $\\cos B < 0$ which contradicts with the statement that it's an acute triangle. Would be more accurate to state that $A$ and $C$ are smaller than 90. -Mathdummy ### Solution 2 Let us use the identity $\\cos^2A+\\cos^2B+\\cos^2C+2\\cos A \\cos B \\cos C=1$ . Add \\begin{align*} \\cos^2 C+2(\\cos A\\cos B-\\sin A \\sin B)\\cos C \\end{align*} to both sides of the first given equation. Thus, as \\begin{align*} \\cos A\\cos B-\\sin A\\sin B=\\cos (A+B)=-\\cos C ,\\end{align*} we have \\begin{align*} \\dfrac{15}{8}-2\\cos^2 C +\\cos^2 C=1, \\end{align*} so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$. Similarily, we have $\\sin A =\\dfrac{2}{3}$ and $\\cos A=\\sqrt{\\dfrac{14}{9}-1}=\\sqrt{\\dfrac{5}{9}}$ and the rest of the solution proceeds as above. ### Solution 3 Let \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ ------ (1)}\\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\text{ ------ (2)}\\\\ \\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B &= x \\text{ ------ (3)}\\\\ \\end{align*} Adding (1) and (3)", "then \\begin{align*} 4(\\cos^2 A \\cos^2 B + &\\sin^2 A \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &= 4\\left(\\frac{1}{4} \\times \\frac{3}{4} + \\frac{3}{4} \\times \\frac{1}{4}\\right) - 2 \\times \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2}\\\\ &= \\frac{3}{2} - \\frac{3}{2} \\\\ &= 0. \\end{align*} So we have found a triangle for which the equation does not hold, and we know that this is an equation that we should try to solve, rather than an identity that we need to prove. We have little choice but to think about the left-hand side. We have \\begin{align*} &4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 2\\sin(2A) \\sin(2B) \\\\ &\\ = 4(\\cos^2 A \\cos^2 B + \\sin^2 A + \\sin^2 B) - 8 \\sin A \\cos A \\sin B \\cos B. \\end{align*} We might think of reordering the terms in the final product: we could think of it as $\\sin A \\sin B \\cos A \\cos B$, and it all starts to look a bit as though it involves things like $(\\cos A \\cos B - \\sin A \\sin B)^2 = \\cos^2(A + B)$. We have \\begin{align*} &\\cos^2 (A + B) \\\\ &\\qquad = (\\cos A \\cos B - \\sin A \\sin B)^2 \\\\ &\\qquad = \\cos^2 A \\cos^2 B + \\sin^2 A \\sin^2 B - 2\\sin A \\sin B \\cos A \\cos B, \\end{align*} and", "equivalent to $2-\\sin^2B$, which is $2-\\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\\dfrac{111-4\\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \\boxed{222}$. ### Solution 2 Let us use the identity $\\cos^2A+\\cos^2B+\\cos^2C+2\\cos A \\cos B \\cos C=1$ . Add \\begin{align*} \\cos^2 C+2(\\cos A\\cos B-\\sin A \\sin B)\\cos C \\end{align*} to both sides of the first given equation. Thus, as \\begin{align*} \\cos A\\cos B-\\sin A\\sin B=\\cos (A+B)=-\\cos C ,\\end{align*} we have \\begin{align*} \\dfrac{15}{8}-2\\cos^2 C +\\cos^2 C=1, \\end{align*} so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$. Similarily, we have $\\sin A =\\dfrac{2}{3}$ and $\\cos A=\\sqrt{\\dfrac{14}{9}-1}=\\sqrt{\\dfrac{5}{9}}$ and the rest of the solution proceeds as above. ### Solution 3 Let \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ ------ (1)}\\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\text{ ------ (2)}\\\\ \\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B &= x \\text{ ------ (3)}\\\\ \\end{align*} Adding (1) and (3) we get: $$2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin A( \\sin B \\cos C + \\sin C \\cos B) = \\frac{15}{8} + x$$ or $$2 \\cos^2 A + \\cos^2 B + \\cos^2", "BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041}.$ Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply $$\\frac{\\frac{9\\sqrt3+16\\sqrt3+25\\sqrt3}{4}+3(\\frac{3*4}{2})}{2}=25\\frac{\\sqrt{3}}{4}+9$$ ### Solution 2 Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties: We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let $\\theta$ be the measure of angle $XAC$ so that angle $BAX$ has measure $60-\\theta$. Let $AB=BC=AC=x$. The law of cosines on triangles $BAX$ and $XAC$ yields $\\cos(60-\\theta)=\\frac{x^2+9}{10x}$ and $\\cos\\theta=\\frac{x^2+16}{10x}$. Solving this system will yield the value of $x$. Since $\\cos\\theta=\\frac{x^2+16}{10x}$ we have that $\\sin\\theta=\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$. Substituting these into the equation $\\frac{x^2+9}{10x}=\\cos(60-\\theta)=\\frac{1}{2}\\cos\\theta+\\frac{\\sqrt{3}}{2}\\sin\\theta$ we obtain $\\frac{x^2+9}{10x}=\\frac{1}{2}\\frac{x^2+16}{10x}+\\frac{\\sqrt{3}}{2}\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$.", "but I wouldn't have the term in $x$ less than the coefficient. – Sebastiano Jun 25 '20 at 22:53 $$I=\\int\\frac{2\\cos x-\\sin x}{3\\sin x+5\\cos x }\\,dx$$ $$2\\cos x-\\sin x=\\sqrt5 \\cos(x+a)$$, where $$\\tan a=1/2$$ and $$3\\sin x+5\\cos x=\\sqrt{34}\\cos(x-b)$$, where $$\\tan b=3/5$$ $$I=\\int \\frac{\\sqrt 5 \\cos(x+a) } {\\sqrt{34}\\cos(x-b)} \\, dx$$ Substitute $$t=x-b$$ so that $$I=\\sqrt{5/34}\\int \\frac{\\cos(t+a+b) } {\\cos t}\\, dt$$ The integrand is now: $$\\cos(a+b) - \\tan t\\sin(a+b)$$ Can you take it from here? • My $\\infty$ thank you. I think that this way is more complicated. – Sebastiano Jun 27 '20 at 9:01 • @Sebastiano: You are welcome. This way isn't complicated. It only involves calculating $\\cos (a+b)$. Given $\\tan a, \\tan b$, we can find $\\sin, \\cos$ and after finding $\\sin (a+b)$, we can find $\\cos (a+b)$. Why do you think that it's complicated? – Koro Jun 27 '20 at 9:05 • @Sebastiano: What I would do is imagine a rt. angled triangle (for angle $a$), then its perp. is $1$,base is $2$ and hence hypot. is $\\sqrt 5$. Hence we can find $cos a$, $sin a$ etc. Similarly for angle $b$. You don't have to do $\\sin(arctan (1/2)+arc tan (3/5))$ – Koro Jun 27 '20 at 9:10 • Why there are roots square $\\cos \\theta=\\pm 1/\\sqrt{1+\\tan^2\\theta}$, $\\sin \\theta=\\pm \\tan \\theta/\\sqrt{1+\\tan^2\\theta}$ and the choice of angle depends on the", "$108^\\circ = 3\\pi/5$ into 3 angles of $36^\\circ = \\pi/5$: Again this is easy and fun to show. Combining these facts, we can prove that $\\displaystyle{ \\cos(2\\pi/5) = \\frac{1}{2\\Phi} }$ and $\\displaystyle{ \\cos(\\pi/5) = \\frac{\\Phi}{2} }$ To prove the first equation, chop one of those golden triangles into two right triangles and do things you learned in high school. To prove the second, do the same things to one of the short squat isosceles triangles: Starting from these equations and using $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we can show $\\displaystyle{ \\sin(2\\pi/5) = \\frac{1}{2}\\sqrt{2 + \\Phi}}$ and, just for completeness (we don’t need it here): $\\displaystyle{ \\sin(\\pi/5) = \\frac{1}{2}\\sqrt{3 - \\Phi}}$ These require some mildly annoying calculations, where it helps to use the identity $\\displaystyle{\\frac{1}{\\Phi^2} = 2 - \\Phi }$ Okay, that’s all for now! But if you want more fun, try a couple of puzzles: Puzzle 1. We’ve gotten formulas for pi starting from a square or a regular pentagon. What formula do you get starting from an equilateral triangle? Puzzle 2. Using the generalized Viète formula, prove Euler’s formula $\\displaystyle{ \\frac{\\sin x}{x} = \\cos\\frac{x}{2} \\cdot \\cos\\frac{x}{4} \\cdot \\cos\\frac{x}{8} \\cdots }$ Conversely, use Euler’s formula to prove the generalized Viète formula. So, one might say that the real point of Viète’s formula, and its generalized version,", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "# If $A+B+C=180^{\\circ}$, then show that: $\\cos B=\\sin A\\sin C-\\cos A\\cos C$ Here is the question : If $A+B+C = 180^{\\circ}$, then show that: $\\cos B = \\sin A \\sin C - \\cos A \\cos C$. EDIT : Here is my reviewed working : $$\\cos B=-\\cos (A+C)$$ Since $$\\space A+B+C = 180^\\circ, \\space B =180^\\circ-(A+C)$$ And \\begin{align} -\\cos B &=\\cos (180+B) \\\\ -\\cos B &=\\cos(180+(180-(A+C)) \\\\ -\\cos B &=\\cos(360-(A+C)) \\\\ -\\cos B &=\\cos(A+C) \\\\ -\\cos B &=\\cos A \\cos C - \\sin A \\sin C \\\\ \\cos B &= \\sin A \\sin C - \\cos A \\cos C \\\\ \\cos B &= -\\cos (A+C) \\end{align} Can someone confirm that my working is correct? Thanks! • $\\cos(B)$ does not equal $\\cos(A+C)$. You know $B=180^{\\circ}-A-C$ and you can verify $\\cos(180^{\\circ}-\\theta)=-\\cos(\\theta)$ by looking at a circle, so you're missing a sign at the beginning. – anon Mar 4 '18 at 0:16 • @anon right, but he never used the first \"identity\" – Paolo Leonetti Mar 4 '18 at 0:22 • @PaoloLeonetti OP began with $\\cos(B)=\\cos(A+C)$ in the original version of the question. – anon Mar 4 '18 at 5:22 Your first step $\\cos(B) = \\cos(A+C)$ is incorrect. It should be $\\cos(B) = -\\cos(A+C)$. Your first line should be deleted, ok. But the main mistake is at the last line. You write", "important fact is that the pentagram trisects the interior angle of the regular pentagon, breaking the interior angle of $108^\\circ = 3\\pi/5$ into 3 angles of $36^\\circ = \\pi/5$: Again this is easy and fun to show. Combining these facts, we can prove that $\\displaystyle{ \\cos(2\\pi/5) = \\frac{1}{2\\Phi} }$ and $\\displaystyle{ \\cos(\\pi/5) = \\frac{\\Phi}{2} }$ To prove the first equation, chop one of those golden triangles into two right triangles and do things you learned in high school. To prove the second, do the same things to one of the short squat isosceles triangles: Starting from these equations and using $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we can show $\\displaystyle{ \\sin(2\\pi/5) = \\frac{1}{2}\\sqrt{2 + \\Phi}}$ and, just for completeness (we don’t need it here): $\\displaystyle{ \\sin(\\pi/5) = \\frac{1}{2}\\sqrt{3 - \\Phi}}$ These require some mildly annoying calculations, where it helps to use the identity $\\displaystyle{\\frac{1}{\\Phi^2} = 2 - \\Phi }$ Okay, that’s all for now! But if you want more fun, try a couple of puzzles: Puzzle 1. We’ve gotten formulas for pi starting from a square or a regular pentagon. What formula do you get starting from an equilateral triangle? Puzzle 2. Using the generalized Viète formula, prove Euler’s formula $\\displaystyle{ \\frac{\\sin x}{x} = \\cos\\frac{x}{2} \\cdot \\cos\\frac{x}{4} \\cdot \\cos\\frac{x}{8} \\cdots }$ Conversely, use Euler’s formula to prove", "(\\text{after some simplification}) = \\sin^2x + {\\cos^2x \\over \\cos^2x} = {1\\over \\cos^2x} = \\sec^2x$. The same issue happens with $1+\\cot^2x=\\csc^2x$. To my understanding, $\\sin^2+\\cos^2=1$ is the same as $a^2+b^2=c^2$. Am I right or wrong? I think there is a huge concept that I am missing between the UNIT CIRCLE and just TRIANGLES. • cos(x)=3/5 and sin(x)=4/5; I don't think you are using these definitions correctly. cos=adj/hyp and sin=opp/hyp – randomgirl Feb 26 '15 at 1:31 • Try reviewing your statement \" let's say a triangle where a=3, b=4, and c=5, or a=cos b=sin and c=hypotenuse.\". – John Joy Feb 27 '15 at 21:37 • @gebruiker: FWIW, OP can accept the best answer, but at $6$ points, does not yet have enough reputation to upvote the others. – Brian Tung Mar 15 '16 at 17:58 Your issue is your understanding of the relationship between your basic trigonometry values and the sides of a triangle. You must keep in mind that the values of trigonometric functions are utilising the sides of your triangles in the following manner: $\\sin(x) = \\frac{opp}{hyp}$, $\\cos(x) = \\frac{adj}{hyp}$, and $\\tan(x) = \\frac{\\sin(x)}{\\cos(x)}$. Let us use your example of a $3$, $4$, $5$ triangle to prove our identities. Let us go ahead and figure out our values of $\\cos(x)$, $\\sin(x)$, and $\\tan(x)$: $$\\sin(x) = \\frac{4}{5}\\\\ \\cos(x)", "you would need to know to solve the quadrilateral? Explain your answer. ## 2.4 Exercises For Exercises 1-6, find the area of the triangle $$\\triangle\\,ABC$$. 2.4.1 $$A = 70^\\circ$$, $$b = 4$$, $$c = 12$$ 2.4.2 $$a = 10$$, $$B = 95^\\circ$$, $$c = 35$$ 2.4.3 $$A = 10^\\circ$$, $$B = 48^\\circ$$, $$C = 122^\\circ$$, $$c = 11$$ 2.4.4 $$A = 171^\\circ$$, $$B = 1^\\circ$$, $$C = 8^\\circ$$, $$b = 2$$ 2.4.5 $$a = 2$$, $$b = 3$$, $$c = 4$$ 2.4.6 $$a = 5$$, $$b=6$$, $$c = 5$$ 2.4.7 Find the area of the quadrilateral in Figure 2.4.3 below. Figure 2.4.3 Exercise 7 Figure 2.4.4 Exercise 8 2.4.8 Let $$ABCD$$ be a quadrilateral which completely contains its two diagonals, as in Figure 2.4.4 above. Show that the area $$K$$ of $$ABCD$$ is equal to half the product of its diagonals and the sine of the angle they form, i.e. $$K = \\frac{1}{2}\\,AC\\,\\cdot\\,BD\\;\\sin\\;\\theta\\;$$. 2.4.9 From Equation 2.26 derive the following formula for the area of a triangle $$\\triangle\\,ABC$$: $\\text{Area} ~=~ K ~=~ \\frac{a^2 \\;\\sin\\;B \\;\\sin\\;C}{2\\;\\sin\\;(B+C)}$ 2.4.10 Show that the triangle area formula $\\text{Area} ~=~ K ~=~ \\tfrac{1}{4}\\,\\sqrt{(a + (b+c))\\,(c - (a-b))\\,(c + (a-b))\\,(a + (b-c))}$ is equivalent to Heron's formula. (Hint: In Heron's formula replace $$s$$ by $$\\frac{1}{2}(a+b+c)$$.) 2.4.11 Show that the triangle area Equation 2.34 is equivalent", "$AD$ and $BC$ so that they meet at $E$, then $\\angle E=\\angle ECD=15^\\circ$, $CD=DE=\\sqrt2$ we have now constructed our $15^\\circ$-$75^\\circ$-$90^\\circ$ triangle. By law of cosines we could work out $AD=\\sqrt6$, now, $\\cos{75^\\circ}=\\sin{15^\\circ}=\\frac{BA}{AE}=\\frac{1}{\\sqrt6+\\sqrt2}=\\frac{\\sqrt6-\\sqrt2}{4}$ $\\sin{75^\\circ}=\\cos{15^\\circ}=\\sqrt{1-\\cos^2{75^\\circ}}=\\frac{\\sqrt6+\\sqrt2}{4}$ $\\tan{75^\\circ}=\\cot{15^\\circ}=\\frac{\\sin{75^\\circ}}{\\cos{75^\\circ}}=2+\\sqrt3$ $\\cot{75^\\circ}=\\tan{15^\\circ}=\\frac{1}{\\tan{75^\\circ}}=2-\\sqrt3$ ## $18^\\circ$ or $72^\\circ$ Consider a regular pentagon $ABCDE$, connect $BE$ and $AC$ so that they intersect at $F$, let the sides of this pentagon be $x$ and $AF$ be $y$, its obvious that $AE=AB=EF=x$, $AF=FB=y$. In $\\triangle AEF$, by law of cosines, $x^2=x^2+y^2-2xy\\cos{72^\\circ}$ $\\therefore y=2x\\cos{72^\\circ}$ In $\\triangle ABE$, by law of cosines, $(x+y)^2=x^2+x^2-2x^2\\cos{108^\\circ}$ $2xy+y^2=x^2+2x^2\\cos{72^\\circ}$ $4x^2\\cos{72^\\circ}+4x^2\\cos^2{72^\\circ}=x^2(1+2\\cos{72^\\circ})$ $4\\cos{72^\\circ}+4\\cos^2{72^\\circ}=1+2\\cos{72^\\circ}$ $4\\cos^2{72^\\circ}+2\\cos{72^\\circ}-1=0$ Solving this we will have $\\cos{72^\\circ}=\\sin{18^\\circ}=\\frac{\\sqrt5-1}{4}$ Hence, $\\sin{72^\\circ}=\\cos{18^\\circ}=\\sqrt{1-\\cos^2{72^\\circ}}=\\frac{\\sqrt{10+2\\sqrt5}}{4}$ $\\tan{72^\\circ}=\\cot{18^\\circ}=\\frac{\\sin{72^\\circ}}{\\cos{72^\\circ}}=\\frac{\\sqrt5-1}{\\sqrt{10+2\\sqrt5}}$ $\\cot{72^\\circ}=\\tan{18^\\circ}=\\frac{1}{\\tan{72^\\circ}}=\\frac{\\sqrt{10+2\\sqrt5}}{\\sqrt5-1}$ That's all for now, I hope you find this helpful. ##### This is one part of 1+1 is not = to 3. Note by Kenneth Tan 3 years, 8 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print \"hello world\" #", "ACD is a 60^\\circ-30^\\circ-90^\\circ triangle.$$AD = 12, CD = 12\\sqrt 3$$\\triangle BCD is a 45^\\circ-45^\\circ-90^\\circ triangle.$$BD = CD = 12\\sqrt 3$$The area is ... 1 Since$$\\measuredangle CTA=45^{\\circ}+30^{\\circ}=75^{\\circ}=\\measuredangle C,$$we obtain:$$AC=AT=24.$$Now, let TK and TN be altitudes of \\Delta ABT and \\Delta ACT respectively. Thus,$$TK=TN=\\frac{1}{2}AT=12$$and$$S_{\\Delta ATC}=\\frac{12\\cdot 24}{2}=144.$$Also,$$S_{\\Delta ABT}=\\frac{12\\cdot(12+\\sqrt{24^2-12^2})}{2}.$$Can you end it now? I ... 1 When you are applying the quadratic equation, you are solving for x^2, not x since this is a fourth degree polynomial equation which is quadratic in form. The first solution is not possible because x_1^2 cannot be negative. So that leaves the second solution x_2^2=a^2. So you get two solutions for x. You get x=\\pm a. So to get the area ... 1 It's$$\\frac{24\\cdot\\sqrt{15^2+8^2}}{2}.$$1 Consider a half of the lens shaped lune cut along a central vertical line. Sector angle at center of either circle is 2 \\theta$$ \\frac{2 \\theta}{2 \\pi} .\\pi r^2 -\\frac12 r \\sin \\theta. r \\cos\\theta = \\pi r^2/4$$Simplifying$$ \\theta - \\sin \\theta \\cos \\theta =\\frac{\\pi}{4} $$A transcendental equation this is.. can be solved by numerical methods e.g., ... 1 Colorado is not perfectly rectangular, and its northern border (assuming that it runs along a parallel) is shorter than the southern one. To see this it is sufficient to consider the Earth as an ideal sphere, in which case:$$ L_{E/W}=R|\\theta_N-\\theta_S|,\\quad L_{S/N}=R|\\phi_W-\\phi_E|\\cos\\theta_{S/N},\\quad A=R^2|(\\phi_W-\\phi_E)(\\sin\\theta_N-\\sin\\theta_S)|,", "s + z &= v + w \\\\ \\end{align} To get a third equation, note that the total of the interior angles of any quadrilateral add up to $2\\pi$: $$s + t + u + v + w + x + y + z = 2\\pi$$ Solving this system of three equations gives us: \\begin{align} w &= \\pi - t - u - v \\\\ y &= t + u - x \\\\ z &= \\pi - s - t - u \\end{align} Now, we can use the law of sines to write a second set of relations: \\begin{align} \\frac{\\overline{HS}}{\\sin s} &= \\frac{\\overline{CH}}{\\sin x} & \\frac{\\overline{CH}}{\\sin u} &= \\frac{\\overline{BC}}{\\sin z} & \\frac{\\overline{BC}}{\\sin w} &= \\frac{\\overline{SB}}{\\sin t} & \\frac{\\overline{SB}}{\\sin y} &= \\frac{\\overline{HS}}{\\sin v} \\end{align} Multiplying them together, the lengths cancel: $$\\sin s \\sin u \\sin w \\sin y = \\sin t \\sin v \\sin x \\sin z$$ Adding in our substitutions from before (and remembering $\\sin(\\pi-x)=\\sin x$): $$\\sin s \\sin u \\sin (t+u+v) \\sin (t+u-x) = \\sin t \\sin v \\sin x \\sin (s+t+u)$$ (This approach is adapted from the webpage Angular Angst that Fimpellizieri posted in this comment on the question.) Now proving that the answer is $x=34^{\\circ}$ (oops, did I say that out loud?) reduces to proving the following trigonometric identity: $$\\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 110^{\\circ}", "as follows: Let $AB=1$. From symmetry and some algebra, we get $AC=BC=\\dfrac{\\sqrt{5}+1}{4}=\\phi$ and $AD=BD=1$. Then apply the Pythagorean Theorem on the triangle. Calculations should be tedious but we can arrive at the result. Or we can do this: \\begin{aligned} \\because \\cos 18^\\circ \\gt 0, \\\\ \\therefore \\cos 18^\\circ &=\\sqrt{1-\\sin ^2 18^\\circ}\\\\ &=\\sqrt{1-\\frac{(5-2\\sqrt{5} +1)}{16}}~~~\\text{the complete square formula helps} \\\\ &=\\sqrt{\\dfrac{10+2\\sqrt{5}}{16}} \\\\ &=\\dfrac{\\sqrt{10+2\\sqrt{5}}}{4}.\\end{aligned} It is still a tedious job to calculate $\\tan 18^\\circ$ though. $20^\\circ$ We must use triple angle identities to find the value of $\\sin 20^\\circ$ and $\\cos 20^\\circ$. This shall be separated into two sections. $\\huge \\text{COSINE}$Perhaps we can try $\\cos 20^\\circ$ first as $\\cos (3\\times 20^\\circ )=\\dfrac{1}{2}$. Let $\\cos 20^\\circ =x$, then $4x^3-3x=\\dfrac{1}{2}.$ Therefore $x^3-\\frac{3}{4}x-\\dfrac{1}{8}=0.$ So we can use the cubic formula to solve this equation. If the function has real coefficients $x^3+ax^2+bx+c=0$, then one real root would be: $\\sqrt[3]{-q+\\sqrt{r}}+\\sqrt[3]{-q-\\sqrt{r}}-\\frac{a}{3},$ Where $q=\\dfrac{1}{27}a^3-\\dfrac{1}{6}ab+\\dfrac{1}{2}c$, $r=q^2+\\dfrac{1}{27}(b-\\dfrac{1}{3}a^2)^3$. Plug in $\\color{#D61F06} a=0, ~~b=\\dfrac{3}{4}, ~~c=\\dfrac{1}{8}$ to get the result: $q=\\dfrac{1}{16},~~r=\\frac{5}{256}.$ $\\sqrt{r}=\\dfrac{\\sqrt5}{16}.$ $\\sqrt[3]{-q+\\sqrt{r}}+\\sqrt[3]{-q-\\sqrt{r}}-\\frac{a}{3}=\\sqrt[3]{\\dfrac{\\sqrt5-1}{16}}+\\sqrt[3]{\\dfrac{-\\sqrt5-1}{16}}=\\dfrac{\\sqrt[3]{\\sqrt5-1}+\\sqrt[3]{-\\sqrt5-1}}{2\\sqrt[3]{2}}=-0.161092677...$ Check if this is the expected root. $\\cos 20^\\circ =0.93969262...$, therefore we must try another root but things get complicated. $\\huge \\text{SINE}$ Let $y=\\sin 20^\\circ$ and do the same thing. This time $3y-4y^3=\\dfrac{\\sqrt3}{2}.$ Therefore $y^3-\\dfrac{3}{4}y+\\dfrac{\\sqrt3}{2}=0.$ Again plug in the equation. $30^\\circ$ $\\dfrac{1}{2}$ $\\dfrac{\\sqrt{3}}{2}$ $\\dfrac{\\sqrt{3}}{3}$ N/A $36^\\circ$ $\\dfrac{\\sqrt{10-2\\sqrt{5}}}{4}$ $\\dfrac{\\sqrt{5}+1}{4}$ $\\sqrt{5-2\\sqrt5}$ Use formulae $(4),(5),(6)$ on the values of $18^\\circ$ to get the", "# 1999 AHSME Problems/Problem 27 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem In triangle $ABC$, $3 \\sin A + 4 \\cos B = 6$ and $4 \\sin B + 3 \\cos A = 1$. Then $\\angle C$ in degrees is $\\mathrm{(A) \\ }30 \\qquad \\mathrm{(B) \\ }60 \\qquad \\mathrm{(C) \\ }90 \\qquad \\mathrm{(D) \\ }120 \\qquad \\mathrm{(E) \\ }150$ ## Solution Square the given equations and add (simplifying with the Pythagorean identity $\\sin^2 x + \\cos^2 x = 1$): \\begin{align*} 9\\sin^2 A + 16\\cos^2 B + 24 \\sin A \\cos B & = 36 \\\\ + 9\\cos^2 A + 16\\sin^2 B + 24 \\sin B \\cos A & = 1 \\\\ \\Longrightarrow 25 + 24(\\sin A \\cos B + \\sin B \\cos A ) & = 37 \\end{align*} Thus $\\frac 12 = \\sin A \\cos B + \\sin B \\cos A$. This is the sine addition identity, so $\\frac 12 = \\sin (A + B) = \\sin (180 - C) = \\sin C$. Thus either $C = 30^{\\circ}, 150^{\\circ}$. If $C = 150$, then $A + B = 30 \\Longrightarrow A,B < 30$, and $\\sin A < \\frac 12, \\cos A < 1$. The first equation implies $6 = 3 \\sin A + 4\\cos B < 3\\left(\\frac 12\\right) + 4(1)", "= \\overline{ZF}\\cdot k$$ $$\\sin(a-30) + \\sin(60-a) = k\\cdot 41(\\sqrt{3}-1)$$ And now expanding using our trig formulas, we get: $$(\\sin a + \\cos a)(\\frac{\\sqrt{3}-1}{2} = k\\cdot 41(\\sqrt{3}-1)$$ $$\\sin a + \\cos a = 82k$$ $$\\sin a = 42k$$ And so now we have a triangle where $\\cos a = 40k$ and $\\sin a = 42k$. Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get: $$\\sqrt{(40k)^2 + (42k)^2} = 1$$ $$3364k^2 = 1$$ $$k = \\frac{1}{58}$$ And since $k = \\frac{\\sin(120^{\\circ})}{x}$, then: $$x = \\frac{\\sqrt{3}}{2}\\cdot58$$ $$x = \\boxed{29\\sqrt{3}}$$ Solution by IronicNinja ~dolphin7", "value of $$\\sin B$$ instead to specify one of them. Probably, when you computing the value of angle $$B$$ you will get $$165^\\circ$$ and $$15^\\circ$$, but as the sum of angles is equal to $$180^\\circ$$, $$15^\\circ$$ will be accepted. By law of cosines we obtain: $$\\cos\\alpha=\\frac{(\\sqrt3+1)^2+(\\sqrt6)^2-(\\sqrt3-1)^2}{2(\\sqrt3+1)\\sqrt6}=\\frac{\\sqrt3+1}{2\\sqrt2},$$ which gives $$\\alpha=15^{\\circ}$$ and from here $$\\beta=105^{\\circ}.$$ Playing around with the angle sum and the Law of Sines ($$|\\angle A|>|\\angle B|$$): $$|\\angle A|+|\\angle B|=120°$$ $$(\\sin A)/(\\sin B)=(\\sqrt{3}+1)/(\\sqrt{3}-1)=2+\\sqrt{3}$$ Put $$|\\angle A|=120°-|\\angle B|$$ and apply the formula for the wine of a difference: $$((\\sqrt{3}/2)(\\cos B)-(1/2)(\\sin B))/(\\sin B)=2+\\sqrt{3}$$ $$(\\cos B)/(\\sin B)=\\cot B = 2+\\sqrt{3}$$ $$\\tan B=2-\\sqrt{3}$$ Then $$\\tan A = \\tan (120°-B)=\\dfrac{-\\sqrt{3}-(2-\\sqrt{3})}{1+(-\\sqrt{3})((2-\\sqrt{3}))}=-(2+\\sqrt{3})$$ Then $$\\tan A \\tan B=-1$$ so the angle measures must differ by a right angle. So $$|\\angle A|+|\\angle B|=120°$$ $$|\\angle A|-|\\angle B|=90°$$ $$|\\angle A|=105°, |\\angle B|=15°$$", "\\sin^2 B + \\sin^2 C) &=6 ,\\\\ \\sin^2 A + \\sin^2 B + \\sin^2 C &=2 ,\\\\ 4R^2\\sin^2 A + 4R^2\\sin^2 B + 4R^2\\sin^2 C &=2\\cdot4R^2 ,\\\\ a^2+b^2+c^2&=8R^2 ,\\\\ 2\\rho^2-8r\\,R-2r^2&=8R^2 ,\\\\ \\rho^2-(r+2R)^2&=0 ,\\\\ \\frac{\\rho^2-(r+2R)^2}{4R^2} =0 . \\end{align} And since \\begin{align} \\cos A \\cos B \\cos C &=\\frac{\\rho^2-(r+2R)^2}{4R^2} , \\end{align} one of $\\cos A,\\cos B,\\cos C$ must be 0, hence, one of the angles must be $\\tfrac\\pi2$. COMMENT.- First note that if $C=90^{\\circ}$ then you have $\\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\\circ}$ you can consider that $\\sin C=\\sin (180^{\\circ}-(A+B))=\\sin(A+B)$ and $\\cos C= \\cos (180^{\\circ}-(A+B))=-\\cos(A+B)$ so you get $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B)=2(\\cos^2A+\\cos ^2 B+\\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish. EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B) = 2\\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\\dfrac{\\pi}{2}$ (the \"if\" is obvious and we need the \"only if\"). So put $A + B = \\dfrac{\\pi}{2}\\pm h$ with $h\\ne 0$ and look at the functions $$\\sin^2(x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h-x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h),\\quad0\\lt x\\lt\\dfrac{\\pi}{2}$$ which becomes for $h$ positive and $h$" ]

[ "formula under the assumption that $$\\theta$$ was between $$0^\\circ$$ and $$90^\\circ$$, but the formula in fact holds for all values of $$\\theta$$. We shall see this in the next subsection, where we prove the general expansion formula for $$\\cos(A + B)$$. #### Example Find $$\\cos 22\\frac{1}{2}^\\circ$$ in surd form. #### Solution Putting $$\\theta = 22\\frac{1}{2}^\\circ$$ into the double angle formula $$\\cos 2\\theta = 2\\cos^2\\theta -1$$ and writing $$x$$ for $$\\cos 22\\frac{1}{2}^\\circ$$, we obtain \\begin{align*} \\cos 45^\\circ &= 2x^2-1\\\\ \\dfrac{1}{\\sqrt{2}} &= 2x^2-1 \\\\ 2x^2& =\\dfrac{1+\\sqrt{2}}{\\sqrt{2}}\\\\ x &= \\sqrt{\\dfrac{1+\\sqrt{2}}{2\\sqrt{2}}}, \\end{align*} which simplifies to $$\\dfrac{1}{2}\\sqrt{2 + \\sqrt{2}}$$. We can also find a double angle formula for sine using the same diagram. In this case, we write down formulas for the area of $$\\triangle ABC$$ in two ways: • On the one hand, the area is given by $$\\dfrac{1}{2} AB \\cdot AC \\, \\sin (\\angle BAC) = \\dfrac{1}{2}\\sin 2\\theta$$. • Since $$AD=\\cos\\theta$$, we can alternatively split the triangle into two right-angled triangles and write the area as $$2\\times \\dfrac{1}{2} \\sin\\theta\\, \\cos\\theta = \\sin\\theta\\, \\cos\\theta$$. Equating these two expressions for the area, we obtain $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta.$ As before, we assumed that $$\\theta$$ lies between $$0^\\circ$$ and $$90^\\circ$$, but the formula is valid for all values of $$\\theta$$. Exercise 8 Find $$\\sin 15^\\circ$$ in surd form. From the Pythagorean identity and the", "BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041}.$ Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply $$\\frac{\\frac{9\\sqrt3+16\\sqrt3+25\\sqrt3}{4}+3(\\frac{3*4}{2})}{2}=25\\frac{\\sqrt{3}}{4}+9$$ ### Solution 2 Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties: We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let $\\theta$ be the measure of angle $XAC$ so that angle $BAX$ has measure $60-\\theta$. Let $AB=BC=AC=x$. The law of cosines on triangles $BAX$ and $XAC$ yields $\\cos(60-\\theta)=\\frac{x^2+9}{10x}$ and $\\cos\\theta=\\frac{x^2+16}{10x}$. Solving this system will yield the value of $x$. Since $\\cos\\theta=\\frac{x^2+16}{10x}$ we have that $\\sin\\theta=\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$. Substituting these into the equation $\\frac{x^2+9}{10x}=\\cos(60-\\theta)=\\frac{1}{2}\\cos\\theta+\\frac{\\sqrt{3}}{2}\\sin\\theta$ we obtain $\\frac{x^2+9}{10x}=\\frac{1}{2}\\frac{x^2+16}{10x}+\\frac{\\sqrt{3}}{2}\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$.", "but I wouldn't have the term in $x$ less than the coefficient. – Sebastiano Jun 25 '20 at 22:53 $$I=\\int\\frac{2\\cos x-\\sin x}{3\\sin x+5\\cos x }\\,dx$$ $$2\\cos x-\\sin x=\\sqrt5 \\cos(x+a)$$, where $$\\tan a=1/2$$ and $$3\\sin x+5\\cos x=\\sqrt{34}\\cos(x-b)$$, where $$\\tan b=3/5$$ $$I=\\int \\frac{\\sqrt 5 \\cos(x+a) } {\\sqrt{34}\\cos(x-b)} \\, dx$$ Substitute $$t=x-b$$ so that $$I=\\sqrt{5/34}\\int \\frac{\\cos(t+a+b) } {\\cos t}\\, dt$$ The integrand is now: $$\\cos(a+b) - \\tan t\\sin(a+b)$$ Can you take it from here? • My $\\infty$ thank you. I think that this way is more complicated. – Sebastiano Jun 27 '20 at 9:01 • @Sebastiano: You are welcome. This way isn't complicated. It only involves calculating $\\cos (a+b)$. Given $\\tan a, \\tan b$, we can find $\\sin, \\cos$ and after finding $\\sin (a+b)$, we can find $\\cos (a+b)$. Why do you think that it's complicated? – Koro Jun 27 '20 at 9:05 • @Sebastiano: What I would do is imagine a rt. angled triangle (for angle $a$), then its perp. is $1$,base is $2$ and hence hypot. is $\\sqrt 5$. Hence we can find $cos a$, $sin a$ etc. Similarly for angle $b$. You don't have to do $\\sin(arctan (1/2)+arc tan (3/5))$ – Koro Jun 27 '20 at 9:10 • Why there are roots square $\\cos \\theta=\\pm 1/\\sqrt{1+\\tan^2\\theta}$, $\\sin \\theta=\\pm \\tan \\theta/\\sqrt{1+\\tan^2\\theta}$ and the choice of angle depends on the", "equivalent to $2-\\sin^2B$, which is $2-\\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\\dfrac{111-4\\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \\boxed{222}$. Note that the problem has a flaw because $\\cos B < 0$ which contradicts with the statement that it's an acute triangle. Would be more accurate to state that $A$ and $C$ are smaller than 90. -Mathdummy ### Solution 2 Let us use the identity $\\cos^2A+\\cos^2B+\\cos^2C+2\\cos A \\cos B \\cos C=1$ . Add \\begin{align*} \\cos^2 C+2(\\cos A\\cos B-\\sin A \\sin B)\\cos C \\end{align*} to both sides of the first given equation. Thus, as \\begin{align*} \\cos A\\cos B-\\sin A\\sin B=\\cos (A+B)=-\\cos C ,\\end{align*} we have \\begin{align*} \\dfrac{15}{8}-2\\cos^2 C +\\cos^2 C=1, \\end{align*} so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$. Similarily, we have $\\sin A =\\dfrac{2}{3}$ and $\\cos A=\\sqrt{\\dfrac{14}{9}-1}=\\sqrt{\\dfrac{5}{9}}$ and the rest of the solution proceeds as above. ### Solution 3 Let \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ ------ (1)}\\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\text{ ------ (2)}\\\\ \\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B &= x \\text{ ------ (3)}\\\\ \\end{align*} Adding (1) and (3)", "ACD is a 60^\\circ-30^\\circ-90^\\circ triangle.$$AD = 12, CD = 12\\sqrt 3$$\\triangle BCD is a 45^\\circ-45^\\circ-90^\\circ triangle.$$BD = CD = 12\\sqrt 3$$The area is ... 1 Since$$\\measuredangle CTA=45^{\\circ}+30^{\\circ}=75^{\\circ}=\\measuredangle C,$$we obtain:$$AC=AT=24.$$Now, let TK and TN be altitudes of \\Delta ABT and \\Delta ACT respectively. Thus,$$TK=TN=\\frac{1}{2}AT=12$$and$$S_{\\Delta ATC}=\\frac{12\\cdot 24}{2}=144.$$Also,$$S_{\\Delta ABT}=\\frac{12\\cdot(12+\\sqrt{24^2-12^2})}{2}.$$Can you end it now? I ... 1 When you are applying the quadratic equation, you are solving for x^2, not x since this is a fourth degree polynomial equation which is quadratic in form. The first solution is not possible because x_1^2 cannot be negative. So that leaves the second solution x_2^2=a^2. So you get two solutions for x. You get x=\\pm a. So to get the area ... 1 It's$$\\frac{24\\cdot\\sqrt{15^2+8^2}}{2}.$$1 Consider a half of the lens shaped lune cut along a central vertical line. Sector angle at center of either circle is 2 \\theta$$ \\frac{2 \\theta}{2 \\pi} .\\pi r^2 -\\frac12 r \\sin \\theta. r \\cos\\theta = \\pi r^2/4$$Simplifying$$ \\theta - \\sin \\theta \\cos \\theta =\\frac{\\pi}{4} $$A transcendental equation this is.. can be solved by numerical methods e.g., ... 1 Colorado is not perfectly rectangular, and its northern border (assuming that it runs along a parallel) is shorter than the southern one. To see this it is sufficient to consider the Earth as an ideal sphere, in which case:$$ L_{E/W}=R|\\theta_N-\\theta_S|,\\quad L_{S/N}=R|\\phi_W-\\phi_E|\\cos\\theta_{S/N},\\quad A=R^2|(\\phi_W-\\phi_E)(\\sin\\theta_N-\\sin\\theta_S)|,", "In triangle ABC, prove the following: Question: In triangle ABC, prove the following: $\\frac{\\cos ^{2} B-\\cos ^{2} C}{b+c}+\\frac{\\cos ^{2} C-\\cos ^{2} A}{c+a}+\\frac{\\cos ^{2} A-\\cos ^{2} B}{a+b}=0$ Solution: Let $\\frac{a}{\\sin A}=\\frac{b}{\\sin B}=\\frac{c}{\\sin C}=k$ Then, Consider the LHS of the equation $\\frac{\\cos ^{2} B-\\cos ^{2} C}{b+c}+\\frac{\\cos ^{2} C-\\cos ^{2} A}{c+a}+\\frac{\\cos ^{2} A-\\cos ^{2} B}{a+b}=0$. $\\mathrm{LHS}=\\frac{\\cos ^{2} B-\\cos ^{2} C}{b+c}+\\frac{\\cos ^{2} C-\\cos ^{2} A}{c+a}+\\frac{\\cos ^{2} A-\\cos ^{2} B}{a+b}$ Now, $\\frac{\\cos ^{2} B-\\cos ^{2} C}{b+c}=\\frac{\\cos ^{2} B-\\cos ^{2} C}{k(\\sin B+\\sin C)}$ $=\\frac{(\\cos B+\\cos C)(\\cos B-\\cos C)}{k(\\sin B+\\sin C)} \\quad\\left(\\because \\cos ^{2} B-\\cos ^{2} C=(\\cos B+\\cos C)(\\cos B-\\cos C)\\right)$ $=\\frac{\\left[2 \\cos \\left(\\frac{B+C}{2}\\right) \\cos \\left(\\frac{B-C}{2}\\right)\\right]\\left[-2 \\sin \\left(\\frac{B+C}{2}\\right) \\sin \\left(\\frac{B-C}{2}\\right)\\right]}{2 k \\sin \\left(\\frac{B+C}{2}\\right) \\sin \\left(\\frac{B-C}{2}\\right)}$ $=\\frac{-2 \\cos \\left(\\frac{B+C}{2}\\right) \\sin \\left(\\frac{B-C}{2}\\right)}{k}=\\frac{-(\\sin B-\\sin C)}{k}=\\frac{\\sin C-\\sin B}{k}$ Also, $\\frac{\\cos ^{2} C-\\cos ^{2} A}{c+a}=\\frac{\\cos ^{2} C-\\cos ^{2} A}{k(\\sin C+\\sin A)}$ $=\\frac{(\\cos C+\\cos A)(\\cos C-\\cos A)}{k(\\sin C+\\sin A)}$ $=\\frac{\\left[2 \\cos \\left(\\frac{C+A}{2}\\right) \\cos \\left(\\frac{C-A}{2}\\right)\\right]\\left[-2 \\sin \\left(\\frac{C+A}{2}\\right) \\sin \\left(\\frac{C-A}{2}\\right)\\right]}{2 k(\\sin C+\\sin A) k}$ $=\\frac{-2 \\cos \\left(\\frac{C+A}{2}\\right) \\cos \\left(\\frac{C-A}{2}\\right)}{k}=\\frac{-(\\sin C-\\sin A)}{k}=\\frac{\\sin A-\\sin C}{k}$ Similarly, $\\frac{\\cos ^{2} A-\\cos ^{2} B}{a+b}=\\frac{\\sin B-\\sin A}{k}$ Thus, LHS $=\\frac{\\sin A-\\sin C}{k}+\\frac{\\sin B-\\sin A}{k}+\\frac{\\sin C-\\sin B}{k}$ $=0=\\mathrm{RHS}$ Hence, in any triangle $A B C, \\frac{\\cos ^{2} B-\\cos ^{2} C}{b+c}+\\frac{\\cos ^{2} C-\\cos ^{2} A}{c+a}+\\frac{\\cos ^{2} A-\\cos ^{2} B}{a+b}=0$.", "you would need to know to solve the quadrilateral? Explain your answer. ## 2.4 Exercises For Exercises 1-6, find the area of the triangle $$\\triangle\\,ABC$$. 2.4.1 $$A = 70^\\circ$$, $$b = 4$$, $$c = 12$$ 2.4.2 $$a = 10$$, $$B = 95^\\circ$$, $$c = 35$$ 2.4.3 $$A = 10^\\circ$$, $$B = 48^\\circ$$, $$C = 122^\\circ$$, $$c = 11$$ 2.4.4 $$A = 171^\\circ$$, $$B = 1^\\circ$$, $$C = 8^\\circ$$, $$b = 2$$ 2.4.5 $$a = 2$$, $$b = 3$$, $$c = 4$$ 2.4.6 $$a = 5$$, $$b=6$$, $$c = 5$$ 2.4.7 Find the area of the quadrilateral in Figure 2.4.3 below. Figure 2.4.3 Exercise 7 Figure 2.4.4 Exercise 8 2.4.8 Let $$ABCD$$ be a quadrilateral which completely contains its two diagonals, as in Figure 2.4.4 above. Show that the area $$K$$ of $$ABCD$$ is equal to half the product of its diagonals and the sine of the angle they form, i.e. $$K = \\frac{1}{2}\\,AC\\,\\cdot\\,BD\\;\\sin\\;\\theta\\;$$. 2.4.9 From Equation 2.26 derive the following formula for the area of a triangle $$\\triangle\\,ABC$$: $\\text{Area} ~=~ K ~=~ \\frac{a^2 \\;\\sin\\;B \\;\\sin\\;C}{2\\;\\sin\\;(B+C)}$ 2.4.10 Show that the triangle area formula $\\text{Area} ~=~ K ~=~ \\tfrac{1}{4}\\,\\sqrt{(a + (b+c))\\,(c - (a-b))\\,(c + (a-b))\\,(a + (b-c))}$ is equivalent to Heron's formula. (Hint: In Heron's formula replace $$s$$ by $$\\frac{1}{2}(a+b+c)$$.) 2.4.11 Show that the triangle area Equation 2.34 is equivalent", "(\\text{after some simplification}) = \\sin^2x + {\\cos^2x \\over \\cos^2x} = {1\\over \\cos^2x} = \\sec^2x$. The same issue happens with $1+\\cot^2x=\\csc^2x$. To my understanding, $\\sin^2+\\cos^2=1$ is the same as $a^2+b^2=c^2$. Am I right or wrong? I think there is a huge concept that I am missing between the UNIT CIRCLE and just TRIANGLES. • cos(x)=3/5 and sin(x)=4/5; I don't think you are using these definitions correctly. cos=adj/hyp and sin=opp/hyp – randomgirl Feb 26 '15 at 1:31 • Try reviewing your statement \" let's say a triangle where a=3, b=4, and c=5, or a=cos b=sin and c=hypotenuse.\". – John Joy Feb 27 '15 at 21:37 • @gebruiker: FWIW, OP can accept the best answer, but at $6$ points, does not yet have enough reputation to upvote the others. – Brian Tung Mar 15 '16 at 17:58 Your issue is your understanding of the relationship between your basic trigonometry values and the sides of a triangle. You must keep in mind that the values of trigonometric functions are utilising the sides of your triangles in the following manner: $\\sin(x) = \\frac{opp}{hyp}$, $\\cos(x) = \\frac{adj}{hyp}$, and $\\tan(x) = \\frac{\\sin(x)}{\\cos(x)}$. Let us use your example of a $3$, $4$, $5$ triangle to prove our identities. Let us go ahead and figure out our values of $\\cos(x)$, $\\sin(x)$, and $\\tan(x)$: $$\\sin(x) = \\frac{4}{5}\\\\ \\cos(x)", "triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$, we can represent these areas using $\\sin \\theta$ as follows: \\begin{align*} 30 &=[ABCD] \\\\ &=[AXB] + [BXC] + [CXD] + [DXA] \\\\ &=\\frac{1}{2} ab \\sin (\\angle AXB) + \\frac{1}{2} bc \\sin (\\angle BXC) + \\frac{1}{2} cd \\sin (\\angle CXD) + \\frac{1}{2} da \\sin (\\angle AXD) \\\\ &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta). \\end{align*} We know that $\\sin (180^\\circ - \\theta) = \\sin \\theta$. Therefore it follows that: \\begin{align*} 30 &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2} ab \\sin (\\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (\\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2}\\sin\\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\sin \\theta = \\frac{60}{ab + bc + cd + da}$. Now we need to find $\\cos \\theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations: \\begin{align*} 5^2 &= a^2 + b^2 - 2ab\\cos(180^\\circ-\\theta), \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, \\\\ 9^2 &= c^2 + d^2 - 2cd\\cos(180^\\circ-\\theta),", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "$$\\triangle ABC$$, we get $$180^\\circ-\\theta$$ + $$\\frac{\\theta}{3}$$ + $$\\angle ABD$$ = $$180^\\circ$$ Than calucating the value, I got $$\\angle ABD$$ = $$\\frac{2\\theta}{3}$$ and so, from isosceles $$\\triangle ABD$$, $$\\angle BAD$$ = $$180^\\circ$$ - $$\\frac{4\\theta}{3}$$ and from here we can write that $$\\angle DAI$$ = $$\\frac{\\theta}{3}$$. In $$\\triangle ADI$$, $$AD$$ = $$r$$ and $$AI$$ = $$r$$. So, $$\\triangle ADI$$ is an isosceles and $$\\angle DIA$$ = $$\\frac{540-\\theta}{6}$$. And now from right angled $$\\triangle DIG$$, we get $$\\frac{DG}{IG}$$ = $$\\tan(\\frac{540-\\theta}{6}$$) $$\\implies$$ $$IG$$ = $$\\frac{r\\sin\\phi}{\\tan(\\frac{540-\\theta}{6})}$$ Subtracting |$$IG$$| from |$$CG$$|, $$|IC|$$ = $$r\\cos\\phi$$ - $$\\frac{r\\sin\\phi}{\\tan(\\frac{540-\\theta}{6})}$$ $$\\implies$$ $$|IC|$$ = $$r\\cos(\\frac{\\theta}{3})$$ - $$\\frac{r\\sin(\\frac{\\theta}{3})}{\\tan(\\frac{540-\\theta}{6})}$$ Here we get the value of $$IC$$, and so the point $$C$$ has an exact and individual location depending on the radius of the circle and the value of the $$\\theta$$. Is my formula\\answer correct? If it so, then I would like to see another method except trigonometry. And my query is that how can we find the length of $$IC$$ without applying trigonometry or in how many ways the angle $$\\theta$$ can be trisected equally? If I have a mistake, then please forgive me and help me correct the error. Thanks in advance. I'm not sure what you mean by \"reverse proof\". Assuming $$\\phi=\\theta/3$$ and then reaching conclusions from there can't convince me that $$\\phi$$ must be $$\\theta/3$$, since I", "In triangle ABC, prove the following: Question: In triangle ABC, prove the following: $a \\cos A+b \\cos B+c \\cos C=2 b \\sin A \\sin C=2 c \\sin A \\sin B$ Solution: Let $\\frac{a}{\\sin A}=\\frac{b}{\\sin B}=\\frac{c}{\\sin C}=k \\quad \\ldots$ (1) Consider the LHS of the equation $a \\cos A+b \\cos B+c \\cos C$. $a \\cos A+b \\cos B+c \\cos C=k(\\sin A \\cos A+\\sin B \\cos B+\\sin C \\cos C)$ $=\\frac{k}{2}(2 \\sin A \\cos A+2 \\sin A \\cos A+2 \\sin C \\cos C)$ $=\\frac{k}{2}(\\sin 2 A+\\sin 2 B+\\sin 2 C)$ $=\\frac{k}{2}[2 \\sin (A+B) \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{k}{2}[2 \\sin (\\pi-C) \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{k}{2}[2 \\sin C \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{2 k \\sin C}{2}[\\cos (A-B)+\\cos C]$ $=k \\sin C[\\cos (A-B)+\\cos \\{\\pi-(\\mathrm{A}+\\mathrm{B})\\}]$ $=k \\sin C[\\cos (A-B)-\\cos (\\mathrm{A}+\\mathrm{B})]$ $=k \\sin C[2 \\sin A \\sin B]$ $=2 k \\sin A \\sin B \\sin C \\quad \\ldots(1)$", "# If $A+B+C=180^{\\circ}$, then show that: $\\cos B=\\sin A\\sin C-\\cos A\\cos C$ Here is the question : If $A+B+C = 180^{\\circ}$, then show that: $\\cos B = \\sin A \\sin C - \\cos A \\cos C$. EDIT : Here is my reviewed working : $$\\cos B=-\\cos (A+C)$$ Since $$\\space A+B+C = 180^\\circ, \\space B =180^\\circ-(A+C)$$ And \\begin{align} -\\cos B &=\\cos (180+B) \\\\ -\\cos B &=\\cos(180+(180-(A+C)) \\\\ -\\cos B &=\\cos(360-(A+C)) \\\\ -\\cos B &=\\cos(A+C) \\\\ -\\cos B &=\\cos A \\cos C - \\sin A \\sin C \\\\ \\cos B &= \\sin A \\sin C - \\cos A \\cos C \\\\ \\cos B &= -\\cos (A+C) \\end{align} Can someone confirm that my working is correct? Thanks! • $\\cos(B)$ does not equal $\\cos(A+C)$. You know $B=180^{\\circ}-A-C$ and you can verify $\\cos(180^{\\circ}-\\theta)=-\\cos(\\theta)$ by looking at a circle, so you're missing a sign at the beginning. – anon Mar 4 '18 at 0:16 • @anon right, but he never used the first \"identity\" – Paolo Leonetti Mar 4 '18 at 0:22 • @PaoloLeonetti OP began with $\\cos(B)=\\cos(A+C)$ in the original version of the question. – anon Mar 4 '18 at 5:22 Your first step $\\cos(B) = \\cos(A+C)$ is incorrect. It should be $\\cos(B) = -\\cos(A+C)$. Your first line should be deleted, ok. But the main mistake is at the last line. You write", "$108^\\circ = 3\\pi/5$ into 3 angles of $36^\\circ = \\pi/5$: Again this is easy and fun to show. Combining these facts, we can prove that $\\displaystyle{ \\cos(2\\pi/5) = \\frac{1}{2\\Phi} }$ and $\\displaystyle{ \\cos(\\pi/5) = \\frac{\\Phi}{2} }$ To prove the first equation, chop one of those golden triangles into two right triangles and do things you learned in high school. To prove the second, do the same things to one of the short squat isosceles triangles: Starting from these equations and using $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we can show $\\displaystyle{ \\sin(2\\pi/5) = \\frac{1}{2}\\sqrt{2 + \\Phi}}$ and, just for completeness (we don’t need it here): $\\displaystyle{ \\sin(\\pi/5) = \\frac{1}{2}\\sqrt{3 - \\Phi}}$ These require some mildly annoying calculations, where it helps to use the identity $\\displaystyle{\\frac{1}{\\Phi^2} = 2 - \\Phi }$ Okay, that’s all for now! But if you want more fun, try a couple of puzzles: Puzzle 1. We’ve gotten formulas for pi starting from a square or a regular pentagon. What formula do you get starting from an equilateral triangle? Puzzle 2. Using the generalized Viète formula, prove Euler’s formula $\\displaystyle{ \\frac{\\sin x}{x} = \\cos\\frac{x}{2} \\cdot \\cos\\frac{x}{4} \\cdot \\cos\\frac{x}{8} \\cdots }$ Conversely, use Euler’s formula to prove the generalized Viète formula. So, one might say that the real point of Viète’s formula, and its generalized version,", "## 15 Degree Triangle #### Problem In the diagram below, $AB$ represents the diameter, $C$ lies on the circumference of the circle, and you are given that (Area of Circle) / (Area of Triangle) = $2\\pi$. Prove that the two smaller angles in the triangle are exactly $15^o$ and $75^o$ respectively. #### Solution We shall consider three quite different solutions to this problem. Method 1 As triangle $ABC$ is in a semi-circle, angle $ACB$ is a right angle. Let angle $CBA = \\theta, AB = 2r, AC = 2r sin(\\theta),$ and $BC = 2r cos(\\theta)$. $\\therefore A\\Delta = \\frac{1}{2}4r^2 sin(\\theta) cos(\\theta) = 2r^2 sin(\\theta) cos(\\theta),$ and using the double angle identity: $sin(2\\theta) = 2 sin(\\theta) cos(\\theta),$ we get, $A\\Delta = r^2 sin(2\\theta)$. So, (Area of Circle) / (Area of Triangle) = $\\dfrac{\\pi r^2}{r^2 sin(2\\theta)} = \\dfrac{\\pi}{sin(2\\theta)} = 2\\pi$ $\\therefore sin(2\\theta) = \\frac{1}{2} \\Rightarrow 2\\theta = 30^o \\Rightarrow \\theta = 15^o$ Hence the complementary angle must be 75 degrees. Method 2 In the diagram, $CD$ is perpendicular to $AB$. Let the diameter, $AB = 2r, OC = r, OD = x,$ and $CD = y$. $\\therefore A\\Delta = \\frac{1}{2}2ry = ry$ So, (Area of Circle) / (Area of Triangle) $= \\dfrac{\\pi r^2}{ry} = \\dfrac{\\pi r}{y} = 2\\pi \\Rightarrow r = 2y$ Using the Pythagorean Theorem, $x^2 + y^2 =", "of the triangle: the horizontal side has length $x$ and the vertical side has length $y$. If we make the circle have radius $1$, then the hypotenuse of the triangle is $1$, and we have \\begin{align*} \\cos \\theta &= \\frac{x}{1} = x \\\\ \\sin \\theta &= \\frac{y}{1} = y \\end{align*} and our point $(x,y)$ can be written as $(\\cos \\theta, \\sin \\theta)$. Immediately, we notice two important things. First, using the Pythagorean theorem on the triangle, we see that \\begin{align*} \\sin^2 \\theta + \\cos^2 \\theta = 1. \\end{align*} This is a handy equation that can be useful in simplifying trigonometric expressions. For example, the expression $(\\sin \\theta + \\cos \\theta)^2 + (\\sin \\theta - \\cos \\theta)^2$ is actually just equivalent to $2$. \\begin{align*} &(\\sin \\theta + \\cos \\theta)^2 + (\\sin \\theta - \\cos \\theta)^2 \\\\ &= (\\sin^2 \\theta + 2\\sin \\theta \\cos \\theta + \\cos^2 \\theta) \\\\ &\\hspace{1cm} + (\\sin^2 \\theta - 2 \\sin \\theta \\cos \\theta + \\cos^2 \\theta) \\\\ &= 2 \\sin^2 \\theta + 2 \\cos^2 \\theta \\\\ &= 2 \\end{align*} Second, angles repeat every $360^\\circ$, since going $360^\\circ$ around the circle brings us back to the starting point of $0^\\circ$. That means, for example, that $\\sin 390^\\circ$ and $\\sin -330^\\circ$ are both equivalent to $\\sin 30^\\circ$. \\begin{align*} 390^\\circ &= 30^\\circ + 360^\\circ \\\\ -330^\\circ &=", "$3-2k = \\cos 2A + \\cos 2B + \\cos (2A+2B)$ i.e $3-2k = 2\\cos(A+B)\\cos(A-B) + 2\\cos^2(A+B) -1$ i.e $4-2k = 4\\cos(A+B)\\cos A\\cos B$ So if $k > 2$, then $\\cos(A+B) \\lt 0$ hence acute. $k = 2$, then $\\cos(A+B) = 0$ hence right triangled. $k < 2$, then $\\cos(A+B) \\gt 0$, hence obtuse. In fact, we can go further and show that the maximum possible value of $k$ is $k = \\frac{9}{4}$ which corresponds to $\\triangle{ABC}$ being equilateral, as follows: $4\\sin^2 A + 4\\sin^2 B + 4\\sin^2 C = 9 + \\delta$ i.e. $(2 - 2\\cos2A) + (2-2\\cos 2B) + 4(1- \\cos^2 (A+B)) = 9 + \\delta$ i.e. $1 + 2\\cos2A + 2\\cos 2B + 4\\cos^2(A+B) = -\\delta$ i.e. $1 + 4\\cos(A+B)\\cos(A-B) + 4\\cos^2(A+B) = -\\delta$ i.e. $\\sin^2(A-B) + \\cos^2(A-B) + 4\\cos(A+B)\\cos(A-B) + 4\\cos^2(A+B) = -\\delta$ i.e. $\\sin^2(A-B) + (\\cos (A-B) + 2\\cos(A+B))^2 = -\\delta$. Hence $\\delta \\le 0$ and so $\\sin^2 A + \\sin^2 B + \\sin^2 C \\le \\frac{9}{4}$ The case $\\delta = 0$ gives us $\\sin(A-B) = 0$ and $\\cos(A+B) = \\frac{-1}{2}$. Hence $A=B=C$. Thus the max value of $\\sin^2 A + \\sin^2 B + \\sin^2 C$ is $\\frac{9}{4}$ and is achieved when $A=B=C$. - $$\\sin^2A+\\sin^2B+\\sin^2C$$ $$=1-(\\cos^2A-\\sin^2B)+1-\\cos^2C$$ $$=2-\\cos(A+B)\\cos(A-B)-\\cos C\\cdot\\cos C$$ $$=2-\\cos(\\pi-C)\\cos(A-B)-\\cos\\{\\pi-(A+B)\\}\\cdot\\cos C$$ $$=2+\\cos C\\cos(A-B)+\\cos(A+B)\\cdot\\cos C\\text{ as }\\cos(\\pi-x)=-\\cos x$$ $$=2+\\cos C\\{\\cos(A-B)+\\cos(A+B)\\}$$ $$=2+2\\cos A\\cos B\\cos C$$ $(1)$", "# In triangle ABC, prove the following: Question: In triangle ABC, prove the following: $a \\cos A+b \\cos B+c \\cos C=2 b \\sin A \\sin C=2 c \\sin A \\sin B$ Solution: Let $\\frac{a}{\\sin A}=\\frac{b}{\\sin B}=\\frac{c}{\\sin C}=k \\quad \\ldots$ (1) Consider the LHS of the equation $a \\cos A+b \\cos B+c \\cos C$. $a \\cos A+b \\cos B+c \\cos C=k(\\sin A \\cos A+\\sin B \\cos B+\\sin C \\cos C)$ $=\\frac{k}{2}(2 \\sin A \\cos A+2 \\sin A \\cos A+2 \\sin C \\cos C)$ $=\\frac{k}{2}(\\sin 2 A+\\sin 2 B+\\sin 2 C)$ $=\\frac{k}{2}[2 \\sin (A+B) \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{k}{2}[2 \\sin (\\pi-C) \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{k}{2}[2 \\sin C \\cos (A-B)+2 \\sin C \\cos C]$ $=\\frac{2 k \\sin C}{2}[\\cos (A-B)+\\cos C]$ $=k \\sin C[\\cos (A-B)+\\cos \\{\\pi-(\\mathrm{A}+\\mathrm{B})\\}]$ $=k \\sin C[\\cos (A-B)-\\cos (\\mathrm{A}+\\mathrm{B})]$ $=k \\sin C[2 \\sin A \\sin B]$ $=2 k \\sin A \\sin B \\sin C \\quad \\ldots(1)$ Now, on putting $k \\sin C=C$ in equation (1), we get: $2 c \\sin A \\sin B$ and on putting $k \\sin B=b$ in equation (1), we get: $2 b \\sin A \\sin C$ So, from (1), we have $a \\cos A+b \\cos B+c \\cos C=2 b \\sin A \\sin C=2 c \\sin A \\sin B$ Hence proved. Leave a comment", "# Circumradius of triangle ABC where H is orthocentre and $(AH)(BH)(CH) = 3$ and $(AH)^2 + (BH)^2 + (CH)^2 = 7$ Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC. Given $(AH)(BH)(CH) = 3$ and $(AH)^2 + (BH)^2 + (CH)^2 = 7$ Then what is the circumradius of triangle? I know that AH = 2RcosA and so but I get stuck after two or three steps. As you wrote, we have $$AH=2R\\cos A,\\quad BH=2R\\cos B,\\quad CH=2R\\cos C$$ Now using that $$\\cos^2A+\\cos^2B+\\cos^2C+2\\cos A\\cos B\\cos C=1\\tag1$$ (the proof for $(1)$ is written at the end of this answer) we get $$\\frac{7}{4R^2}+2\\cdot\\frac{3}{8R^3}=1,$$ i.e. $$(R+1)(2R-3)(2R+1)=0$$ to have $$\\color{red}{R=\\frac 32}$$ Let us prove $(1)$. We have \\begin{align}\\cos 2A+\\cos 2B+\\cos 2C&=2\\cos(A+B)\\cos(A-B)+\\cos 2C\\\\\\\\&=-2\\cos C\\cos(A-B)+2\\cos^2C-1\\\\\\\\&=-2\\cos C(\\cos(A-B)-\\cos C)-1\\\\\\\\&=-2\\cos C\\ (\\cos(A-B)+\\cos(A+B))-1\\\\\\\\&=-4\\cos A\\cos B\\cos C-1\\end{align} from which we have $$(2\\cos^2A-1)+(2\\cos^2B-1)+(2\\cos^2C-1)=-4\\cos A\\cos B\\cos C-1$$", "# Thread: Find Value of Sin C 1. ## Find Value of Sin C In triangle ABC, sin (A + B) = 3/5. What is the value of sin C? 2. Originally Posted by magentarita In triangle ABC, sin (A + B) = 3/5. What is the value of sin C? In triangle $ABC, C = 180^\\circ - (A + B)^\\circ$ So use the formula $\\sin{C} = \\sin{[180^\\circ - (A + B)^\\circ]} = \\sin{180^\\circ}\\cos{(A + B)^\\circ} - \\cos{180^\\circ}\\sin{(A + B)^\\circ}$ and solve for $\\sin{C}$. 3. ## Values for... Originally Posted by Prove It In triangle $ABC, C = 180^\\circ - (A + B)^\\circ$ So use the formula $\\sin{C} = \\sin{[180^\\circ - (A + B)^\\circ]} = \\sin{180^\\circ}\\cos{(A + B)^\\circ} - \\cos{180^\\circ}\\sin{(A + B)^\\circ}$ and solve for $\\sin{C}$. I can plug and chug but what are the values of A and B? How do I find the values of A and B. Do I need the values of A and B? 4. You don't need them. What does $\\sin{180^\\circ}$ equal? Once you've found that, what does $\\sin{180^\\circ}\\cos{(A + B)^\\circ}$ equal? What does $\\cos{180^\\circ}$ equal? Since you're GIVEN the value of $\\sin{(A + B)^\\circ}$, what is $\\cos{180^\\circ}\\sin{(A + B)^\\circ}$? 5. ## ok Originally Posted by Prove It You don't need them. What does $\\sin{180^\\circ}$ equal? Once you've found that, what does" ]

[ "$$\\mu = 5.5$$. 2. To calculate the variance we use the formula: $Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2 \\end{pmatrix} - \\mu^2$ Where: $E\\begin{pmatrix} X^2 \\end{pmatrix} = \\sum x^2.P\\begin{pmatrix} X = x \\end{pmatrix}$ We start by calculating $$E\\begin{pmatrix}X^2\\end{pmatrix}$$. Using the values $$x$$ that the discrete random variable can take the formula leads to: \\begin{aligned} E\\begin{pmatrix} X^2 \\end{pmatrix} & = 3^2\\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 4^2 \\times P\\begin{pmatrix}X = 3 \\end{pmatrix} + 6^2 \\times P\\begin{pmatrix}X = 6 \\end{pmatrix} + 7^2 \\times P\\begin{pmatrix}X = 7 \\end{pmatrix} \\\\ & = 3^2 \\times \\frac{3}{20} + 4^2 \\times \\frac{4}{20} + 6^2\\times \\frac{6}{20} + 7^2 \\times \\frac{7}{20} \\\\ & = 9 \\times \\frac{3}{20} + 16 \\times \\frac{4}{20} + 36 \\times \\frac{6}{20} + 49 \\times \\frac{7}{20} \\\\ & = \\frac{27}{20} + \\frac{64}{20} + \\frac{216}{20} + \\frac{343}{20} \\\\ & = \\frac{650}{20} \\\\ E\\begin{pmatrix} X^2 \\end{pmatrix} & = 32.5 \\end{aligned} Now that we know the value of $$E\\begin{pmatrix} X^2 \\end{pmatrix} = 32.5$$, we can calculate the variance: \\begin{aligned} Var\\begin{pmatrix} X \\end{pmatrix} = E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 32.5 - 5.5^2 \\\\ & = 32.5 - 30.25 \\\\ Var\\begin{pmatrix} X \\end{pmatrix} & = 2.25 \\end{aligned} The variance is $$2.25$$. Now that we know the variance, we can calculate this discrete random variable's standard deviation: \\begin{aligned} \\sigma & = \\sqrt{Var\\begin{pmatrix} X \\end{pmatrix}} \\\\ & = \\sqrt{2.25}\\\\ \\sigma & =", "at 19:02 • The vectors to multiply by $s$ and $t$ should solve the equation $8x+3y-7z=0$. Again, not even close. Wouldn't you like to know what wrong step you took so you can avoid taking it again on some other problem? That would be a useful question rather than just another instance of \"what's the answer to my homework?\" – David K Feb 15 '16 at 19:09 • Dear @Lovsovs, if you really want to edit, then do it properly: you should have replaced the image with equivalent MathJax code. – Alex M. Feb 15 '16 at 21:30 • @AlexM. You do have a point, but since my edits must go through peer-review, I see my edit as being accepted by the community. I respect that you'd want to use the extra time to code up (i.e. other than the basics) a downvoted answer regarding homework, but I must admit that I probably wouldn't. – Bobson Dugnutt Feb 15 '16 at 21:39 Your equation is $8x+3y-7z=12$, hence $\\displaystyle x=\\frac{12-3y+7z}{8}=\\frac{3}{2}-\\frac{3}{8}y+\\frac{7}{8}z$, thus the set of solutions is $\\displaystyle S=\\left\\{\\begin{pmatrix}\\frac{3}{2}-\\frac{3}{8}t+\\frac{7}{8}s \\\\t\\\\s\\end{pmatrix}\\right\\}$ or equivalently $\\displaystyle \\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}=\\begin{pmatrix}\\frac{3}{2}\\\\0\\\\0\\end{pmatrix}+t\\begin{pmatrix}-\\frac{3}{8}\\\\1\\\\0\\end{pmatrix}+s\\begin{pmatrix}\\frac{7}{8}\\\\0\\\\1\\end{pmatrix}$.", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "= \\frac{1}{\\textbf{h}^{3T}\\textbf{x}}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\textbf{h}^{1T}\\textbf{x}}{(\\textbf{h}^{3T}\\textbf{x})^2} \\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Substituting $x’ = \\textbf{h}^{1T}\\textbf{x}$ and $w’ = \\textbf{h}^{3T}\\textbf{x}$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{x’}{(w’)^2}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Finally, substituting $\\tilde{x’} = x’/w’$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\tilde{x’} }{w’}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ $$= \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’} \\textbf{x}^T \\end{pmatrix}$$ Performing a similar process for $\\tilde{y’}$, we get $$\\partial\\tilde{y}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} 0 & \\textbf{x}^T & -\\tilde{y’} \\textbf{x}^T \\end{pmatrix}$$ Hence, we have established that $$\\partial{\\tilde{\\textbf{x}}}’ / \\partial{\\textbf{h}} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’}\\textbf{x}^T \\\\ 0 & \\textbf{x}^T & -\\tilde{y’}\\textbf{x}^T\\end{pmatrix}$$ #### VI. Sampson error with non-isotropic error distributions. The derivation of Sampson error in section 4.2.6(p98) assumed that points were measured with circular error distributions. In the case where the point $\\textbf{X} = (x, y, x’, y’)$ is measured with covariance matrix $\\Sigma_\\textbf{x}$ it is appropriate instead to minimize the Mahalanobis norm $\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\boldsymbol{\\delta}^T_\\textbf{x}\\Sigma^{-1}_\\textbf{x}\\boldsymbol{\\delta}_\\textbf{x}$. Show that in this case the formulae corresponding to (4.11-p99) and (4.12-p99) are $$\\boldsymbol{\\delta}_\\textbf{x} = -\\Sigma_\\textbf{x}J^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ and $$\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\epsilon^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ Note that if the measurements in the two images are independent, then the covariance matrix $\\Sigma_\\textbf{x}$ will be block-diagonal with two $2\\times2$ diagonal blocks", "All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $cos{\\theta}=\\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}|}{\\sqrt{9} \\sqrt{49}}=\\frac{16}{21}$ $\\theta = {40.4}^{\\circ}$ (ii) Using Graphing Calculator, equation of line, $r = \\begin{pmatrix}{-\\frac{1}{6}}\\\\{-\\frac{2}{3}}\\\\0\\end{pmatrix} + \\lambda \\begin{pmatrix}7\\\\10\\\\6\\end{pmatrix}, \\lambda \\in \\mathbb{R}$ (iii) Distance A to $P_1 = \\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} - 1|}{\\sqrt{9}} = \\frac{|1+c|}{3}$ Distance A to $P_1 = \\frac{|\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} + 1|}{\\sqrt{49}} = \\frac{|2c-14|}{7}$ $\\frac{|1+c|}{3} = \\frac{|2c-14|}{7}$ $49(1+c)^{2} = 9(2c-14)^{2}$ $c = -49 \\mathrm{~or~} \\frac{35}{13}$", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "\\\\ 5, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 16, & 3 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{3--7}{16-5} && \\text{Fill in the values}\\\\& m =\\frac{3+7}{16-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{10}{11} && \\text{Calculate the value of the numerator and the denominator}$ ii) (–6, –7) and (–1, –4) $& \\begin{pmatrix} x_1, & y_1 \\\\ -6, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -1, & -4 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-4--7}{-1--6} && \\text{Fill in the values}\\\\& m =\\frac{-4+7}{-1+6} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{3}{5} && \\text{Calculate the value of the numerator and the denominator}$ iii) (5, –12) and (0, –6) $& \\begin{pmatrix} x_1, & y_1 \\\\ 5, & -12 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 0, & -6 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-6--12}{0-5} && \\text{Fill in the values}\\\\& m =\\frac{-6+12}{0-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\& m =\\frac{6}{-5} && \\text{Calculate the value of the numerator and the denominator}\\\\& m =\\frac{-6}{5} && \\text{Do not leave the negative sign in the denominator}$ 3. i) You are not given the", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "to choose 4 from the 8 with cracks and 1 from the remaining 17. 5. Jan 19, 2010 ### exitwound b.)$$\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix} = \\frac{8!}{4!4!} * \\frac{17!}{1!16!}= 1190$$ c.) $$\\frac{\\begin{pmatrix} 8\\\\ 4 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 1 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} = \\frac{\\frac{8!}{4!4!} * \\frac{17!}{1!16!}}{\\frac{25!}{5!20!}} = .0224$$ d.) $$\\frac{\\begin{pmatrix} 8\\\\ 5 \\end{pmatrix} \\begin{pmatrix} 17\\\\ 0 \\end{pmatrix}}{\\begin{pmatrix} 25\\\\ 5 \\end{pmatrix}} + AnswerToPartC = \\frac{\\frac{8!}{5!3!} * \\frac{17!}{0!17!}}{\\frac{25!}{5!20!}} + .0224 = .0235$$ Is this how it's done? 6. Jan 19, 2010 ### vela Staff Emeritus Yup, perfect.", "we saw above we can write: \\begin{aligned} \\frac{dy}{dx} & = 3 \\times \\frac{u'(x)}{u(x)} \\\\ & = 3 \\times \\frac{3x^2 -6x}{x^3 - 3x^2 + 6} \\\\ & = \\frac{3\\begin{pmatrix} 3x^2 - 6x \\end{pmatrix}}{x^3 - 3x^2 + 6} \\\\ \\frac{dy}{dx} & = \\frac{ 9x^2 - 18x}{x^3 - 3x^2 + 6} \\end{aligned} 3. The function $$f(x) = -2 ln \\begin{pmatrix} 4x+5 \\end{pmatrix}$$ can be written: $f(x) = -2ln \\begin{pmatrix} u(x) \\end{pmatrix}$ where $u(x) = 4x+5$ and $u'(x) = 4$ So \\begin{aligned} f'(x) & = -2 \\times \\frac{u'(x)}{u(x)} \\\\ & = -2 \\times \\frac{4}{4x+5} \\\\ & = - \\frac{2\\times 4}{4x+5} \\\\ f'(x) & = - \\frac{-8}{4x+5} \\end{aligned} 4. The function $$y = 4.ln \\begin{pmatrix} \\sqrt{x} \\end{pmatrix}$$ can be written: $y =4.ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = \\sqrt{x}$ and $u'(x) = \\frac{1}{2 \\sqrt{x}}$ So \\begin{aligned} \\frac{dy}{dx} & = 4 \\times \\frac{u'(x)}{u(x)} \\\\ & = 4 \\times \\frac{ \\frac{1}{2\\sqrt{x}}}{\\sqrt{x}} \\\\ & = 4\\times \\frac{1}{2\\sqrt{x}} \\times \\frac{1}{\\sqrt{x}} \\\\ & = \\frac{4}{2 \\sqrt{x} \\times \\sqrt{x}} \\\\ & = \\frac{4}{2x} \\\\ \\frac{dy}{dx} &= \\frac{2}{x} \\end{aligned} 5. The function $$f(x) = ln \\begin{pmatrix} \\frac{1}{x} \\end{pmatrix}$$ can be written: $f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = \\frac{1}{x}$ and $u'(x) = - \\frac{1}{x^2}$ So \\begin{aligned} f'(x) & = \\frac{u'(x)}{u(x)} \\\\ & = \\frac{ - \\frac{1}{x^2}}{\\frac{1}{x}} \\\\ & = - \\frac{1}{x^2}\\times x \\\\ & = - \\frac{x}{x^2} \\\\", "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "how to prove the formula. – làntèrn Aug 6 '15 at 16:38 • What if $a=0$?. – Michael Galuza Aug 6 '15 at 16:43 The fast approach, as I see it is (in case $a \\neq 0$): you have $ax+by = c$ , just plug in $y=0+ta$, and you get $$ax + bta = c$$ $$x= \\frac ca - bt$$. alternatively, in case you don't have the solution in advance, use same approach and you will get: $$x = \\frac ca - \\frac ba y$$, so the general solution will be $$(x,y) = (\\frac ca - \\frac bat , t)$$ $$(x,y) = (\\frac ca, 0) + (-\\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result. • So, using $t$ here is redundant? – làntèrn Aug 6 '15 at 16:58 • I've edited my answer, I should have you $t$ ,, $y$ is somewhat confusing. – d_e Aug 6 '15 at 17:03 • So $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} -b \\\\ a \\\\ \\end{pmatrix}$ would be same with $\\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} \\frac{c}{a} \\\\ 0 \\\\ \\end{pmatrix}+t \\begin{pmatrix} \\frac{-b}{a} \\\\ 1 \\\\ \\end{pmatrix}$ – làntèrn Aug 6 '15 at 17:27", "to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded", "\\frac{1}{x^2} \\end{pmatrix}'= -2x^{-2-1} = -2x^{-3}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' &= 3f'(x) +4g'(x) \\\\ & = 3\\times 3x^{3-1} + 4\\times (-2)x^{-2-1} \\\\ & = 9x^2 - 8 x^{-3} \\\\ \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' & = 9x^2-\\frac{8}{x^3} \\end{aligned} 2. Remembering that $$\\begin{pmatrix} \\sqrt{x} \\end{pmatrix}' = \\frac{1}{2\\sqrt{x}}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= g'(x) - 6h'(x) \\\\ & = \\frac{1}{2\\sqrt{x}} - 6\\times (-2)x^{-2-1} \\\\ & = \\frac{1}{2\\sqrt{x}} + 12x^{-3} \\\\ \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= \\frac{1}{2\\sqrt{x}} + \\frac{12}{x^3} \\end{aligned} 3. To differentiate $$\\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x)$$, we use all that we have seen here to write: \\begin{aligned} \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' &= \\frac{5}{6}f'(x)+\\frac{1}{8}h'(x)-\\frac{3}{4}g'(x) \\\\ &= 3\\times \\frac{5}{6} x^{3-1} -2\\times \\frac{1}{8}x^{-2-1} - \\frac{3}{4} \\times \\frac{1}{2}\\sqrt{x} \\\\ & = \\frac{15}{6}x^2-\\frac{2}{8}x^{-3} - \\frac{3}{8\\sqrt{x}} \\\\ & = \\frac{15}{6}x^2-\\frac{1}{4}\\frac{1}{x^3} - \\frac{3}{8\\sqrt{x}} \\\\ \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' & = \\frac{5}{2}x^2-\\frac{1}{4x^3} - \\frac{3}{8\\sqrt{x}} \\end{aligned}", "+ \\frac{1}{8}\\cdot\\frac{3}{4} + 0\\cdot0 \\end{pmatrix}\\\\ = \\begin{pmatrix} \\frac{9}{16} & \\frac{1}{16} & \\frac{6}{16} \\\\ \\frac{21}{32} & \\frac{7}{32} & \\frac{4}{32} \\\\ \\frac{1}{32} & \\frac{14}{32} & \\frac{17}{32} \\end{pmatrix}$$ ### Regular Stochastic Matrices All entries after some power Pm are positive. (In the above example, m is 2). They have the following three properties: 1. A unique fixed probability vector t. 2. P, P2, P3, … approaches T whose rows are all t 3. For any probability vector p, pP, pP2, pP3, … approaches t For the above example, t = (0.3866666667, 0.24, 0.3733333333) $$t = tP\\\\ \\begin{pmatrix} t_1 & t_2 & t_3 \\end{pmatrix} = \\begin{pmatrix} t_1 & t_2 & t_3 \\end{pmatrix} \\begin{pmatrix} 0 & \\frac{1}{2} & \\frac{1}{2}\\\\ \\frac{1}{4} & 0 & \\frac{3}{4}\\\\ \\frac{7}{8} & \\frac{1}{8} & 0 \\end{pmatrix}\\\\ t_1 = \\frac{t_2}{4} + \\frac{7t_3}{8}\\\\ t_2 = \\frac{t_1}{2} + \\frac{t_3}{8}\\\\ t_3 = \\frac{t_1}{2} + \\frac{3t_2}{4}\\\\ t_1 + t_2 + t_3 = 1$$ Substituting t1 from the first equation into the second, then t2 into the third $$t_2 = \\frac{9t_3}{14}\\\\ t_3 = \\frac{28t_1}{29}\\\\ t_1 = 1 - t_2 - t_3\\\\ t_1 = 1 - \\frac{9t_3}{14} - \\frac{28t_1}{29} \\\\ t_1 = 1 - \\frac{9}{14}\\cdot\\frac{28t_1}{29} - \\frac{28t_1}{29}\\\\ t_1 = \\frac{29}{75}\\\\ t_2 = \\frac{18}{75}\\\\ t_3 = \\frac{28}{75}$$ Last updated on 27 Jul 2021 Published on 27 Jul 2021 Content Footer", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$." ]

[ "that a direction vector is $$(3/2,1)$$. Of course, there's nothing wrong with using our method of chapter 1: if $$(a,b)$$ is a direction vector, then $$\\pm(-b,a)$$ are normal vectors, and vice versa. Example. Consider the line with point-parallel form equation $\\mathbf r(t)=\\begin{pmatrix}2\\\\-1\\end{pmatrix}+t\\begin{pmatrix}-3\\\\1\\end{pmatrix}.$ We can see that the line contains the point $$P(2,-1)$$, and that a direction vector is $$\\mathbf u=(-3,1)$$. Therefore, a normal vector is $$\\mathbf n=(1,3)$$, and from the point-normal form equation $\\left(\\begin{pmatrix}x & y\\end{pmatrix}-\\begin{pmatrix}2 & -1\\end{pmatrix}\\right)\\cdot\\begin{pmatrix}1\\\\3\\end{pmatrix}=0$ we get the standard form equation $x+3y=1.$ Note that a standard equation of a hyperplane in $$\\mathbf R^n$$ gives a $$1\\times n$$ matrix with one nonzero row. Therefore, the rank is 1, and you get an ($$n-1$$)-parameter family of solutions. This is why in $$\\mathbf R^2$$, from one normal vector you get one directional vector (unique up multiplication by a nonzero scalar). Example. Consider the plane in $$\\mathbf R^3$$ with standard equation $$2x-y+z=2$$. Solving the SLE, you get the 2-parameter family of solutions \\begin{align*} x&=s/2-t/2+2\\\\ y&=s\\\\ z&=t \\end{align*} That is, we have the parametrization $\\mathbf r(s,t)=\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}=\\begin{pmatrix} s/2-t/2+2 \\\\ s \\\\ t \\end{pmatrix}=\\begin{pmatrix}2\\\\0\\\\0\\end{pmatrix}+\\begin{pmatrix}1/2\\\\1\\\\0\\end{pmatrix}+\\begin{pmatrix}-1/2\\\\0\\\\1\\end{pmatrix}.$ You can see that a plane is determined by one point and two nonparallel direction vectors. To get the normal vector back from the two direction vectors, you need to find a nonzero vector that is", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "found using the following formula[1]: ${\\displaystyle proj_{\\overrightarrow {B}}{\\overrightarrow {A}}={\\frac {{\\overrightarrow {A}}\\cdot {\\overrightarrow {B}}}{|{\\overrightarrow {B}}|^{2}}}{\\overrightarrow {B}}}$ ## Vector Equations A basic vector equation consists of a vector and a variable coefficient of another vector. The standalone vector is known as the position vector and the vector with the variable coefficient is known as the direction vector. The position vector indicates where in the 3D space the vector is located, and the direction vector shows which way the vector pointing. The vector for the position vector can be the position vector of any point on the line. A typical equations might look like this: ${\\displaystyle {\\textbf {r}}={\\textbf {a}}+x{\\textbf {b}}}$ or ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}2\\\\2\\\\2\\end{pmatrix}}+x{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ ### Checking if lines are parallel Lines are parallel when their direction vectors are scalar multiples of each other. Example: Line 1: ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+\\lambda {\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ Line 2: ${\\displaystyle {\\textbf {s}}={\\begin{pmatrix}2\\\\5\\\\7\\end{pmatrix}}+\\mu {\\begin{pmatrix}-2\\\\4\\\\-6\\end{pmatrix}}}$ They are parallel as ${\\displaystyle -2{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2\\\\4\\\\-6\\end{pmatrix}}}$ . ### Checking if a point lies on a vector line equation So if the point has the position vector ${\\displaystyle {\\textbf {P}}={\\begin{pmatrix}2\\\\4\\\\4\\end{pmatrix}}}$ and the line equation is ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+t{\\begin{pmatrix}2\\\\4\\\\3\\end{pmatrix}}}$ and you want to see if P lies on the line, then you'll need to do the following: Set the point equal to the equation: ${\\displaystyle {\\begin{pmatrix}2\\\\4\\\\4\\end{pmatrix}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+t{\\begin{pmatrix}2\\\\4\\\\3\\end{pmatrix}}}$ From this you can then make 3 separate equations from each", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "lines have the same $x$-intercept. • These lines both go through the point $(1, 5)$. • These lines … Can you sort them into the correct pairs and complete the final description? While sketching the lines could be helpful, 12 might be too many on one graph. Are there any pairs we can match up by just looking at the equations? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\ 6 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$ Equations 7 and 10 have the same position vector which also happens to be their $y$-interept. Do you think there are particular features of lines that are easier to spot when the equation is written in vector form compared to Cartesian form? $\\mathbf{r}= \\begin{pmatrix} 2 \\\\ 9.5 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -2 \\\\ -8 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ 4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ Equations 3 and 6 are parallel lines, but it looks like they have different direction vectors. How do you know these lines are parallel? What is the same about the two direction vectors? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -6 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -1 \\\\ 4 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 1 \\\\ 1 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\", "+ 1.5 \\times \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix}$ $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix} + \\begin{pmatrix} 1.5 \\\\ 4.5 \\end{pmatrix}$ $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 1.5 \\\\ 6.5 \\end{pmatrix}$ However, there’s no such value of $\\text{scalar}$ for the point (2, 9), because it’s not on the line. Two slightly unexpected features of this representation are that it doesn’t matter which point on the line you pick for the expression, and you can multiply the direction vector by any constant without it changing the line. So the above line is equivalent to either of the following: $\\begin{pmatrix} 1 \\\\ 5 \\end{pmatrix} + \\text{scalarB} \\times \\begin{pmatrix} -1 \\\\ -3 \\end{pmatrix}$ $\\begin{pmatrix} -10 \\\\ -28 \\end{pmatrix} + \\text{scalarC} \\times \\begin{pmatrix} 6 \\\\ 18 \\end{pmatrix}$ #### Finite line: two endpoints The simplest way to represent a finite line is to simply write down its two endpoints. So you might have a line from (1, 3) to (6, 7). That’s a good, simple, unambiguous definition of the line. However, to do any calculations involving the line, you usually need to convert it to another representation, such as the last on our list: #### Finite line: point and direction You can take the two endpoints of a line and turn it into a form that looks identical", "with the parameter $t$ has its whole body in the line— it is a direction vector for the line. Note that points on the line to the left of $x=1$ are described using negative values of $t$. In $\\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set of (endpoints of) vectors of this form and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$ consists of (endpoints of) the vectors in $\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} +t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} +s\\cdot\\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R} \\}$ (the column vectors associated with the parameters $\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ -3 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} \\qquad \\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ 4 \\\\ 0.5 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix}$ are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative $t$'s or negative $s$'s or", "for example in $\\mathbb{R}^2,$ $$\\begin{pmatrix}a\\\\b\\end{pmatrix} = a\\begin{pmatrix}1\\\\0\\end{pmatrix}+b\\begin{pmatrix}0\\\\1\\end{pmatrix}$$ Other bases are possible, such as $$\\mathbb{R}^2 = \\text{span}\\left\\{\\begin{pmatrix}3\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\2\\end{pmatrix}\\right\\} = \\text{span}\\left\\{\\begin{pmatrix}1\\\\1\\end{pmatrix},\\begin{pmatrix}1\\\\-1\\end{pmatrix}\\right\\}$$ Although 'direction giving' can be a little more complicated here, its still possible. As long as the set can describe (using linear combinations) any coordinate in the space and there are no redundant vectors included, the set forms a basis. Although $$\\mathbb{R}^2 =\\text{span}\\left\\{\\begin{pmatrix}1\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\1\\end{pmatrix}, \\begin{pmatrix}1\\\\1\\end{pmatrix}\\right\\}$$ is true, we have more vectors than necessary. The third is redundant, since the directions to get to (1,1) are possible with just the first two. This is not a basis. For vector spaces, bases have a unique size, and that is the definition of dimension.", "to our earlier point and direction form. So using our (1, 3) to (6, 7) example, you get: $\\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + \\text{scalar} \\times \\begin{pmatrix} 5 \\\\ 4 \\end{pmatrix}$ The difference now is that the line extends from $\\text{scalar}=0$ (the start-point) to $\\text{scalar}=1$ (the end-point). Any values for the scalar variable outside the 0–1 range would be on the line if it’s infinite, but are not when it’s finite. ## End of the Line Today’s post was a lot of theory without an actual application, but we’ll be using it in our next post when we’ll look at finding out if two finite lines intersect.", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "start at the origin and end at a point in space. For example, below is the vector (we’ll name it $$\\mathbf{v}$$) from above in 2-D space. Also pictured is the vector $\\mathbf{u} = \\begin{pmatrix} -2 \\\\ -2 \\end{pmatrix}$ on the same 2-D coordinate system. Often, we might not specify the actual values of a vector because we’re working generically with any vector in 2-dimensional space. In that case, we would use a variable such as $$\\mathbf{x}$$ to indicate this vector and we would write $$\\mathbf{x} \\in \\mathbb{R}^{2}$$ to indicate that $$\\mathbf{x}$$ is a 2-dimensional vector whose entries are real numbers. This means that they can take on any value between $$-\\infty$$ and $$\\infty$$. In our posts, we will represent vectors with bold face, lower case letters. If we write $$\\mathbf{a} \\in \\mathbb{R}^{5}$$, this means that the vector $$\\mathbf{a}$$ is a $$5$$-dimensional vector whose entries are real numbers. If we don’t specify the entries in $$\\mathbf{a}$$, then we indicate its entries with non-bold letters followed by the entry position in subscript $\\mathbf{a} = \\begin{pmatrix} a_{1} \\\\ a_{2} \\\\ a_{3} \\\\ a_{4} \\\\ a_{5} \\end{pmatrix}.$ Throughout our posts, we will always assume that vectors such as $$\\mathbf{x}$$ and $$\\mathbf{a}$$ are column vectors as we presented them above. If we want to indicate a row vector, we will specify it with", "$t$ and the variable given in the answer $t'$. Then let's see when the two vectors are equal. $\\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-\\tfrac{1}{2} \\\\ \\tfrac{1}{2} \\\\ 1\\end{pmatrix}t = \\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}t'$ when $t=2t'$. Since the map $t \\mapsto 2t$ is a bijection, you wind up with the same line either way. Same thing for line 2. so if i i do my ans in this way. wouldnt my ans be wrong? 4. ## Re: vector (simplified b1 and b2 ) Originally Posted by delso so if i i do my ans in this way. wouldnt my ans be wrong? I just showed that you get the same line either way. Why would you get the wrong answer that way? If two line intersect and you have two different formulas for each line, it is still the same two lines. It doesn't matter that the formulas are not identical. They still represent the same lines, so they will still intersect in the same place. 5. ## Re: vector (simplified b1 and b2 ) do u mean t=2t\" just different in terms of length ? but it's still the same line with same gradient?", "plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 • How did you parameterize? How did you find the normal vectors? What is the upward caret operator? – rocksNwaves May 1 '18 at 14:02", "vector: $$\\boldsymbol{v}=\\begin{pmatrix} 1\\\\2\\\\ \\end{pmatrix}$$ Show that this vector does not form a basis for $\\mathbb{R}^2$. Solution. Since we only have one vector, this vector must be linearly independent. However, $\\boldsymbol{v}$ clearly does not span $\\mathbb{R}^2$ because some vectors in $\\mathbb{R}^2$ cannot be expressed as a scalar multiple of $\\boldsymbol{v}$. Therefore, $\\boldsymbol{v}$ does not form a basis for $\\mathbb{R}^2$. We have already covered the standard basis of $\\mathbb{R}^2$ in the earlier sectionlink. We can also generalize the idea to $\\mathbb{R}^n$. Definition. # Standard basis of Rn The standard basis vectors of $\\mathbb{R}^n$ is defined as a set containing the following $n$ vectors: $$\\left\\{ \\begin{pmatrix} 1\\\\0\\\\\\vdots\\\\0 \\end{pmatrix}, \\begin{pmatrix} 0\\\\1\\\\\\vdots\\\\0 \\end{pmatrix},\\;\\cdots\\;, \\begin{pmatrix} 0\\\\0\\\\\\vdots\\\\1 \\end{pmatrix} \\right\\}$$ Edited by 0 others thumb_up thumb_down Comment Citation Ask a question or leave a feedback... thumb_up 0 thumb_down 0 chat_bubble_outline 0 settings Enjoy our search Hit / to insta-search docs and recipes!", "# Vectors, Linear Transformations, and Matrix Forms A two-dimensional vector, $\\bar{v} = \\begin{pmatrix}x\\\\y\\end{pmatrix}$, can be thought of representing the displacement that, when starting at the origin, would move something to the point $(x,y)$. Consequently, a natural interpretation for the sum of two vectors would be the total displacement that results from their individual applications. This can be seen in the picture below, where the red vector represents the total displacement resulting from the application of the blue and green vectors. Fortunately, as seen below, whether we follow the the displacement arising from the blue vector with that from the green vector (the solid segments), or vice-versa (the dashed segments), the overall displacement is the same. (Hence, the addition of vectors is commutative.) As indicated in the picture, this suggests how we might find the sum of two vectors algebraically as well: $$\\begin{pmatrix}x_1\\\\y_1\\end{pmatrix} + \\begin{pmatrix}x_2\\\\y_2\\end{pmatrix} = \\begin{pmatrix}x_1+x_2\\\\y_1+y_2\\end{pmatrix}$$ Further, understanding how vectors get added together suggests a definition for a constant (or scalar) multiple of a vector. For example, it would be natural to assume that $3\\begin{pmatrix}x\\\\y\\end{pmatrix}$ is a displacement three times as large as $\\begin{pmatrix}x\\\\y\\end{pmatrix}$. Consequently, $$3\\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}x\\\\y\\end{pmatrix} + \\begin{pmatrix}x\\\\y\\end{pmatrix} + \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}x+x+x\\\\y+y+y\\end{pmatrix} = \\begin{pmatrix}3x\\\\3y\\end{pmatrix}$$ In keeping with this example, we define the scalar multiple of a two-dimensional vector, for any real value $c$ (which is called", "the latter is parallel to the $y$-axis ($y$ can change independently of $z$ and $x$). The line itself is the intersection of these two planes, and it can be projected along the planes indicated by the individual equations in (2) onto the $yz$-plane and the $zx$-plane, respectively. In a narrower meaning, the equation $y=mz\\!+\\!p$ represents the projection line of $l$ onto the $yz$-plane and the equation $x=nz\\!+\\!q$ the projection line of $l$ onto the $zx$-plane. A line in space can be represented also without using planes. We can derive another way by starting with two vectors: one vector $\\vec{r}_{0}$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called direction vector) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as $\\displaystyle\\vec{r}\\;=\\;\\vec{r}_{0}\\!+\\!t\\vec{u}.$ (3) This is the vector-formed equation of the line. Since every vector can be given by three coordinates of the end-point, we may write $\\vec{r}\\;=\\;\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix},\\quad\\vec{r}_{0}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix},\\quad\\vec{u}\\;=\\;\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!.$ $\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix}+t\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!,$ and we obtain three scalar equations $\\displaystyle\\begin{cases}x\\;=\\;x_{0}\\!+\\!ta\\\\ y\\;=\\;y_{0}\\!+\\!tb\\\\ z\\;=\\;z_{0}\\!+\\!tc.\\end{cases}$ (4) These are the parametric equations of the line (3), with the parameter $t$ taking", "+0 # shortest distance between two lines via vector equation 0 255 4 +844 For a i got r= (-2,3,4) + λ(3,2,-7) for b so far i wanted to identify a point A on l1 and a point b on l2 i assumed the two lines are parallel as they have the same direction vector $$A=\\begin{pmatrix} -2+3λ\\\\ 3+2λ\\\\ 4-7λ \\end{pmatrix}$$ $$B= \\begin{pmatrix} 3\\\\ 1+2λ\\\\ -2-7λ \\end{pmatrix}$$ used this to get $$AB= \\begin{pmatrix} 5-3λ\\\\ -2\\\\ -6 \\end{pmatrix}$$ used this and my direction vector to find λ=53/9 via dot product substituted and calculated the magnitude which gave me 14.26 but my answer should be 3.57 please identify what went wrong, thanks Feb 10, 2019 #1 +2 The equation of the first line can be written as r1 = (-2, 3, 4) + s(3, 2, -7), (as you calculated), and the equation of the second as r2 = (0, 1, -2) + t(3, 2, -7), s and t arbitrary parameters. A vector d connecting two points, one on each line, would be given by d = r2 - r1 = (2, -2, -6) + (t - s)(3, 2 ,-7) = (2, -2, -6) + w(3, 2, -7), w for convenience. The distance between the two points will be the magnitude of this vector, so what you are looking for is the", "a set of vectors is the set of all linear combinations of these vectors. So the span of $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ would be the set of all linear combinations of them, which is $\\mathbb{R}^2$. The span of $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is also $\\mathbb{R}^2$, although we don't need $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ to be so. So both these two sets are said to be the spanning sets of $\\mathbb{R}^2$. However, only the first set $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is a basis of $\\mathbb{R}^2$, because the $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ makes the second set linearly dependent. Also, the set $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ can also be a basis for $\\mathbb{R}^2$. Because its span is also $\\mathbb{R}^2$ and it is linearly independent. For another example, the span of the set $\\{\\begin{pmatrix}1\\\\1\\end{pmatrix}\\}$ is the set of all vectors in the form of $\\begin{pmatrix}a\\\\a\\end{pmatrix}$. • So a basis of $\\mathbb R^n$ is a smallest spanning set of $\\mathbb R^n$, i.e any spanning set of $\\mathbb R^n$ with cardinality $n$? – Zaz Oct 26 '18 at 23:17 • @Zaz: That is correct. Oct 30 '18 at 10:28 The basis is a combination of vectors which are linearly independent and which spans the whole vector V. Suppose we take a system of $R^2$. Now as you said,$(1,0)$ and $(0,1)$ are the basis in this system and we want to find any $(x,y)$ in this system. $$(x,y)=x(1,0)+y(0,1)$$", "0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •" ]

[ "& 2\\\\1 & -2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 \\, .$ This gives us the equalities $$-x +2y = 0$$ and $$x -2y = 0$$ which both give the line $$y = \\frac{1}{2}x$$. Any point on this line, so for example $$\\begin{pmatrix}2\\\\1\\end{pmatrix}$$, is an eigenvector with eigenvalue $$\\lambda = 5$$. Now lets find the eigenvector for $$\\lambda = 2$$ $\\begin{pmatrix}4-2 & 2\\\\1 & 3-2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}2 & 2\\\\1 & 1 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 ,$ which gives the equalities $$2x+2y = 0$$ and $$x+y = 0$$. (Notice that these equations are not independent of one another, so our eigenvalue must be correct.) This means any vector $$v = \\begin{pmatrix}x\\\\y\\end{pmatrix} where \\(y = -x$$ , such as $$\\begin{pmatrix}1\\\\-1\\end{pmatrix}, or any scalar multiple of this vector, \\(\\textit{i.e.}$$ any vector on the line $$y = -x$$ is an eigenvector with eigenvalue 2. This solution could be written neatly as $\\lambda_{1} = 5, \\, v_{1}= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\, \\text{ and } \\lambda_{2} = 2, \\, v_{2}=\\begin{pmatrix}1\\\\-1\\end{pmatrix}.$ Jordan Block Example Consider the matrix $J_{2} = \\begin{pmatrix}\\lambda & 1\\\\ 0 & \\lambda\\end{pmatrix},$ and we note that we can just read off the eigenvector $$e_{1}$$ with eigenvalue $$\\lambda$$. However the characteristic polynomial of $$J_{2}$$ is $$P_{J_{2}}(\\mu) = (\\mu - \\lambda)^{2}$$ so the only possible eigenvalue is $$\\lambda$$, but we claim it does not", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "$$(A + iB)(x + iy) = \\lambda x + i \\lambda y, \\tag 9$$ or $$(Ax - By) + i(Ay + Bx) = \\lambda x + i \\lambda y; \\tag{10}$$ equating the real and imaginary parts of either side yields $$Ax - By = \\lambda x, \\tag{11}$$ $$Ay + Bx = \\lambda y. \\tag{12}$$ We can now in fact find two real orthogonal eigenvectors for $$M = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix}, \\tag{13}$$ each corresponding to the eigenvalue $$\\lambda$$; set $$w = \\begin{pmatrix} x \\\\ y \\end{pmatrix}; \\tag{14}$$ $$Mw = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} Ax - By \\\\ Bx + Ay \\end{pmatrix} = \\begin{pmatrix} \\lambda x \\\\ \\lambda y \\end{pmatrix} = \\lambda \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\lambda w; \\tag{15}$$ now set $$v = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}; \\tag{16}$$ $$Mv = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\begin{pmatrix} -Ay - Bx \\\\ -By + Ax \\end{pmatrix} = \\begin{pmatrix} -\\lambda y \\\\ \\lambda x \\end{pmatrix} = \\lambda \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\lambda v; \\tag{17}$$ furthermore, $$\\langle v, w \\rangle = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}^T \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -y^T & x^T \\end{pmatrix} \\begin{pmatrix} x \\\\ y", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "\\right\\rbrace }$ Example (plane through the origin as a span) Let ${\\displaystyle (5,0,0)^{T}}$ and ${\\displaystyle (0,3,0)^{T}}$ be two vectors from ${\\displaystyle \\mathbb {R} ^{3}}$. The span of these two vectors is the ${\\displaystyle xy}$-plane. The following transformation shows {\\displaystyle {\\begin{aligned}\\operatorname {span} \\left(\\left\\lbrace {\\begin{pmatrix}5\\\\0\\\\0\\end{pmatrix}},{\\begin{pmatrix}0\\\\3\\\\0\\end{pmatrix}}\\right\\rbrace \\right)&=\\left\\lbrace \\lambda \\cdot {\\begin{pmatrix}5\\\\0\\\\0\\end{pmatrix}}+\\mu \\cdot {\\begin{pmatrix}0\\\\3\\\\0\\end{pmatrix}}{\\Bigg |}\\ \\lambda ,\\mu \\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&=\\,\\left\\lbrace 5\\cdot \\lambda \\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+3\\cdot \\mu \\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}{\\Bigg |}\\ \\lambda ,\\mu \\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&\\ {\\color {OliveGreen}\\left\\downarrow {\\text{set }}{{\\tilde {\\lambda }}=5\\cdot \\lambda ,\\ {\\tilde {\\mu }}=3\\cdot \\mu }\\ {\\text{ bzw. }}\\ \\lambda ={\\frac {1}{5}}\\cdot {\\tilde {\\lambda }},\\ \\mu ={\\frac {1}{3}}\\cdot {\\tilde {\\mu }}\\right.}\\\\[0.3em]&=\\,\\left\\lbrace {\\tilde {\\lambda }}\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+\\cdot {\\tilde {\\mu }}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}{\\Bigg |}\\ {\\tilde {\\lambda }},{\\tilde {\\mu }}\\in \\mathbb {R} \\right\\rbrace \\\\[0.3em]&=\\,\\left\\{{\\begin{pmatrix}{\\tilde {\\lambda }}\\\\{\\tilde {\\mu }}\\\\0\\end{pmatrix}}{\\Bigg |}\\ {\\tilde {\\lambda }},{\\tilde {\\mu }}\\in \\mathbb {R} \\right\\}\\end{aligned}}} ## Overview: Properties of the span Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space, ${\\displaystyle M}$, ${\\displaystyle N\\subseteq V}$ subsets of ${\\displaystyle V}$ and ${\\displaystyle W\\subseteq V}$ a subspace of ${\\displaystyle V}$. Then, we have • For a vector ${\\displaystyle v\\in V}$ we have ${\\displaystyle \\operatorname {span} (\\{v\\})=\\{\\lambda \\cdot v|\\lambda \\in K\\}}$ • If ${\\displaystyle N\\subseteq M}$, then ${\\displaystyle \\operatorname {span} (N)\\subseteq \\operatorname {span} (M)}$ • From ${\\displaystyle \\operatorname {span} (N)=\\operatorname {span} (M)}$ one can usually not conclude ${\\displaystyle N=M}$ • ${\\displaystyle M\\subseteq \\operatorname {span} (M)}$ •", "\\\\ 9, & 5 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -1, & 6 \\end{pmatrix}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{6-5}{-1-9}\\\\ & m =-\\frac{1}{10}\\end{align*} 1. \\begin{align*}(-2, 7)\\end{align*} and \\begin{align*}(-3, -1)\\end{align*}. \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ -2, & 7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -3, & -1 \\end{pmatrix}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{-1-7}{-3--2}\\\\ & m =\\frac{-1-7}{-3+2}\\\\ & m =\\frac{-8}{-1}\\\\ & m =8\\end{align*} ### Examples #### Example 1 Earlier, you were asked to find the slope of a line with an \\begin{align*}x\\end{align*}-intercept of 4 and \\begin{align*}y\\end{align*}-intercept of –3. \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ 4, & 0 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 0, & -3 \\end{pmatrix} && \\text{Express the} \\ x- \\text{and} \\ y \\text{-intercepts as coordinates of a point.}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{-3-0}{0-4}\\\\ & m =\\frac{-3}{-4}\\\\ & m =\\frac{3}{4}\\end{align*} #### Example 2 Calculate the slope of the line that passes through the points: (5, –7) and (16, 3) \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ 5, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 16, & 3 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{3--7}{16-5} && \\text{Fill in the values}\\\\ & m =\\frac{3+7}{16-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\ & m =\\frac{10}{11} && \\text{Calculate the value of", "\\\\ Y\\end{bmatrix} \\qquad (x,y) = [X,Y]P^T$ $q(\\mathbf{v}) = (x,y)A\\begin{pmatrix}x \\\\ y\\end{pmatrix} = [X,Y]P^T AP\\begin{bmatrix}X \\\\ Y\\end{bmatrix}$ If we can find an orthogonal change of basis $(P^T = P^{-1})$ then $P$ ... Eigenvalues of $A = \\begin{pmatrix}2 & 1 \\\\ 1 & 2\\end{pmatrix}$ \\begin{aligned} 0 &= \\begin{vmatrix}\\lambda I - A \\end{vmatrix} = \\begin{vmatrix} \\lambda -2 & -1 \\\\ -1 & \\lambda -2\\end{vmatrix}= (\\lambda -2)^2 - 1 \\\\ 0 &= (\\lambda -2)^2 - 1^2 = ((\\lambda -2)-1)((\\lambda -2)+1) \\\\ 0 &= (\\lambda -3)(\\lambda -1)\\end{aligned} $\\begin{cases}\\lambda_1 = 3 \\lambda_2 = 1\\end{cases}$ eigenvalues $3X^2 + Y^2$ is $q$ in the new system, Eigenvalue $\\lambda_1 = 3$ $\\begin{pmatrix} 3-2 & -1 \\\\ -1 & 3-2 \\end{pmatrix} \\begin{pmatrix}\\alpha \\\\ \\beta\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0\\end{pmatrix} \\begin{cases} \\alpha - \\beta = 0 \\\\ -\\alpha + \\beta = 0 \\end{cases} so \\alpha = \\beta \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$ will do Note $<\\begin{pmatrix}1 \\\\ 1\\end{pmatrix},\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}> = 0$ orthogonal basis of eigenvectors $\\|\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}\\|^2 = 2 =\\|\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}\\|^2$ Corresponding orthonormal basis: $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}-\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ But another can be $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ and two more candidates... We may take $P = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} = (P^{-1})^T$ Note $det P = \\pm 1$ (rotation) Claim $\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} \\begin{pmatrix} 2 & 1", "to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "\\frac{1}{x^2} \\end{pmatrix}'= -2x^{-2-1} = -2x^{-3}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' &= 3f'(x) +4g'(x) \\\\ & = 3\\times 3x^{3-1} + 4\\times (-2)x^{-2-1} \\\\ & = 9x^2 - 8 x^{-3} \\\\ \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' & = 9x^2-\\frac{8}{x^3} \\end{aligned} 2. Remembering that $$\\begin{pmatrix} \\sqrt{x} \\end{pmatrix}' = \\frac{1}{2\\sqrt{x}}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= g'(x) - 6h'(x) \\\\ & = \\frac{1}{2\\sqrt{x}} - 6\\times (-2)x^{-2-1} \\\\ & = \\frac{1}{2\\sqrt{x}} + 12x^{-3} \\\\ \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= \\frac{1}{2\\sqrt{x}} + \\frac{12}{x^3} \\end{aligned} 3. To differentiate $$\\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x)$$, we use all that we have seen here to write: \\begin{aligned} \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' &= \\frac{5}{6}f'(x)+\\frac{1}{8}h'(x)-\\frac{3}{4}g'(x) \\\\ &= 3\\times \\frac{5}{6} x^{3-1} -2\\times \\frac{1}{8}x^{-2-1} - \\frac{3}{4} \\times \\frac{1}{2}\\sqrt{x} \\\\ & = \\frac{15}{6}x^2-\\frac{2}{8}x^{-3} - \\frac{3}{8\\sqrt{x}} \\\\ & = \\frac{15}{6}x^2-\\frac{1}{4}\\frac{1}{x^3} - \\frac{3}{8\\sqrt{x}} \\\\ \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' & = \\frac{5}{2}x^2-\\frac{1}{4x^3} - \\frac{3}{8\\sqrt{x}} \\end{aligned}", "2: The entries with index $1$ are not part of the two-dimensional subspace. To better identify the four subspaces see the blue colored components $\\color{blue}{00},\\color{blue}{01},\\color{blue}{10}$ and $\\color{blue}{11}$. \\begin{align*} W_4=\\left\\{ \\begin{pmatrix}\\color{blue}{0}\\\\1\\\\0\\end{pmatrix} \\begin{pmatrix}\\color{blue}{0}\\\\0\\\\1\\end{pmatrix} \\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix} \\right\\} &\\qquad W_5=\\left\\{ \\begin{pmatrix}\\color{blue}{0}\\\\1\\\\0\\end{pmatrix} \\begin{pmatrix}\\color{blue}{1}\\\\0\\\\1\\end{pmatrix} \\begin{pmatrix}1\\\\1\\\\1\\end{pmatrix} \\right\\}\\\\ W_6=\\left\\{ \\begin{pmatrix}\\color{blue}{1}\\\\1\\\\0\\end{pmatrix} \\begin{pmatrix}\\color{blue}{0}\\\\0\\\\1\\end{pmatrix} \\begin{pmatrix}1\\\\1\\\\1\\end{pmatrix} \\right\\} &\\qquad W_7=\\left\\{ \\begin{pmatrix}\\color{blue}{1}\\\\1\\\\0\\end{pmatrix} \\begin{pmatrix}\\color{blue}{1}\\\\0\\\\1\\end{pmatrix} \\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix} \\right\\} \\end{align*} • I have to digest this. – zhoraster Feb 27 '17 at 12:59 • What do you mean by the \"first dimension\" being part of a subspace? – darij grinberg Mar 3 '17 at 2:11 • @darijgrinberg: The first coordinate. In fact the intention was the same as in the other answer, but in my example a projection with the first coordinate discarded. – Markus Scheuer Mar 3 '17 at 7:24 As proposed, I define $\\binom nk_q$ as the number of $k$-dim subspaces of $n$-dim space over the field $\\mathbb Z_q$. Take a $k$-dim subspace S of a given $(n+1)$-dim space $\\mathbb Z_q^{n+1}$ and take its projection to $\\mathbb Z_q^n$ (last coordinate discarded). Denote the projected $S$ as $S'$. There are two cases: 1. $\\dim(S') = \\dim(S)-1$. In this case, $S'$ uniquely determines $S$: $S$ is the preimage of $S'$ in the projection. So it corresponds to the second summand $\\binom n{k-1}_q$ since it counts the number of all possible $(k-1)$-dim subspaces $S'$. 2. $\\dim(S') = \\dim(S)$.", "move upward six spaces in a positive direction. \\begin{align*}m&=\\frac{\\text{rise}}{\\text{run}}\\\\ m&=\\frac{6}{{\\color{blue}10}}\\\\ m&=\\frac{3}{{\\color{blue}5}}\\end{align*} 2. i) (5, –7) and (16, 3) \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ 5, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 16, & 3 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{3--7}{16-5} && \\text{Fill in the values}\\\\ & m =\\frac{3+7}{16-5} && \\text{Simplify the numerator and denominator (if possible)}\\\\ & m =\\frac{10}{11} && \\text{Calculate the value of the numerator and the denominator}\\end{align*} ii) (–6, –7) and (–1, –4) \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ -6, & -7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -1, & -4 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{-4--7}{-1--6} && \\text{Fill in the values}\\\\ & m =\\frac{-4+7}{-1+6} && \\text{Simplify the numerator and denominator (if possible)}\\\\ & m =\\frac{3}{5} && \\text{Calculate the value of the numerator and the denominator}\\end{align*} iii) (5, –12) and (0, –6) \\begin{align*}& \\begin{pmatrix} x_1, & y_1 \\\\ 5, & -12 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 0, & -6 \\end{pmatrix} && \\text{Designate the points as to the first point and the second point.}\\\\ & m =\\frac{y_2-y_1}{x_2-x_1}\\\\ & m =\\frac{-6--12}{0-5} && \\text{Fill in the values}\\\\ & m =\\frac{-6+12}{0-5} && \\text{Simplify the numerator and", "3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix}$ and so $A = \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix}^{-1}$ #### Example 2 a) Diagonalize A, i.e find P (invertible) and B (diagonal) such that AP = PB b) Compute A5 $A = \\begin{pmatrix} 1&2\\\\ -1&4 \\end{pmatrix}$ ' Solution a) We are solving Ax = λx, where λ is a constant and x a column vector. Firstly $Ax - \\lambda Ix = 0$ $(A - \\lambda I)x = 0$ since x ≠ 0 we have $det(A - \\lambda I) = 0$ i.e. $\\det\\begin{pmatrix} 1-\\lambda&2\\\\ -1&4-\\lambda \\end{pmatrix} = 0$ $\\begin{matrix} (1-\\lambda)(4-\\lambda) + 2 &=& 0\\\\ \\lambda^2-5\\lambda + 6 &=& 0 \\end{matrix}$ $\\lambda = 3,\\ 2$ For λ = 3, $(A-3I)x = 0$ $\\begin{pmatrix} -2&2\\\\ -1&1 \\end{pmatrix} \\begin{pmatrix} x\\\\ y \\end{pmatrix} = 0$ Clearly $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}$ is a solution. Note that we do not accept x = 0 as an solution, because we assume x ≠ 0. Note also that $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} t\\\\ t \\end{pmatrix}$ for some constant t is also a solution. Indeed we could use x = y = 2, 3 or 4 as a solution, but for convenience we choose the simplest i.e. x = y = 1. For λ = 2, $(A-2I)x", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "is, this holds. $\\begin{array}{*{2}{rc}r} 4m &- &n &= &3 \\\\ 1m &- &3n &= &-3 \\\\ & &0 &= &0 \\end{array}$ In the general case, to show that any vector from $S_1$ is a member of $S_2$ we must show that for any $c$ and $d$ there are appropriate $m$ and $n$. We follow the pattern of the examples; fix $\\begin{pmatrix} c+d \\\\ -c+d \\\\ 0 \\end{pmatrix}\\in S_1$ and look for $m$ and $n$ such that $\\begin{pmatrix} 4 \\\\ 1 \\\\ 0 \\end{pmatrix}m +\\begin{pmatrix} -1 \\\\ -3 \\\\ 0 \\end{pmatrix}n =\\begin{pmatrix} c+d \\\\ -c+d \\\\ 0 \\end{pmatrix}$ that is, this is true. $\\begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\\\ m &- &3n &= &-c+d \\\\ & &0 &= &0 \\end{array}$ Applying Gauss' method $\\begin{array}{rcl} \\begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\\\ m &- &3n &= &-c+d \\end{array} &\\xrightarrow[]{-(1/4)\\rho_1+\\rho_2} &\\begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\\\ & &-(11/4)n &= &-(5/4)c+(3/4)d \\end{array} \\end{array}$ gives $n=(5/11)c-(3/11)d$ and $m=(4/11)c+(2/11)d$. This shows that for any choice of $c$ and $d$ there are appropriate $m$ and $n$. We conclude any member of $S_1$ is a member of $S_2$ because it can be rewritten in this way: $\\begin{pmatrix} c+d \\\\ -c+d \\\\ 0 \\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 1 \\\\ 0 \\end{pmatrix}((4/11)c+(2/11)d)+ \\begin{pmatrix} -1 \\\\ -3 \\\\ 0 \\end{pmatrix}((5/11)c-(3/11)d).$ For the other inclusion, $S_2\\subseteq S_1$, we", "# 27.1 Initial value problem #### bigwave did a c/p just avoid typos ok the example did this $$A=\\begin{pmatrix}1&2\\\\3&2 \\end{pmatrix}$$ and $$y'=Ay$$ with initial value $$y(0)=\\begin{pmatrix}y_1(0)\\\\y_2(0) \\end{pmatrix}$$ ok I'm ???? Last edited: #### romsek MHF Helper $y = c_1 e^{\\lambda_1 t}v_1 + c_2 e^{\\lambda_2 t} v_2$ where $\\lambda_k, v_k$ are the $kth$ eigenvalues/vectors of $A$ Then use the initial conditions to solve for $c_1, c_2$ 2 people #### romsek MHF Helper $A=\\begin{pmatrix}1 &2\\\\3 &2\\end{pmatrix}$ $y(0)=\\begin{pmatrix}1\\\\3\\end{pmatrix}$ Find the eigensystem. $\\left|\\begin{pmatrix}1-\\lambda &2\\\\3 &2-\\lambda\\end{pmatrix}\\right| = \\lambda^2-3\\lambda -4$ $\\lambda^2-3\\lambda-4 = 0\\\\ (\\lambda-4)(\\lambda+1) = 0\\\\ \\lambda = 4,~-1$ $\\lambda_1=4 \\Rightarrow v_1 = \\begin{pmatrix}2\\\\3\\end{pmatrix}$ $\\lambda_2=-1 \\Rightarrow v_2 = \\begin{pmatrix}1\\\\-1\\end{pmatrix}$ $y(t) = c_1 e^{4t}\\begin{pmatrix}2\\\\3\\end{pmatrix} + c_2 e^{-t}\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ $y(0) = c_1 \\begin{pmatrix}2\\\\3\\end{pmatrix} + c_2 \\begin{pmatrix}1\\\\-1\\end{pmatrix} = \\begin{pmatrix}1\\\\3\\end{pmatrix}$ $(c_1,c_2) = \\left(\\dfrac 4 5, -\\dfrac 3 5\\right)$ $y(t) = \\dfrac 4 5 e^{4t}\\begin{pmatrix}2\\\\3\\end{pmatrix} - \\dfrac 3 5 e^{-t}\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ $y(t) = \\dfrac 1 5 \\begin{pmatrix}8 e^{4t}-3 e^{-t}\\\\12 e^{4t} +3 e^{-t}\\end{pmatrix}$ 2 people #### bigwave Appreciate much Have deal more with this tomorro To hard with just cell phone. #### bigwave how did you get $$\\displaystyle v_1$$ and $$\\displaystyle v_2$$ Last edited: #### romsek MHF Helper how did you get $$\\displaystyle v_1$$ and $$\\displaystyle v_2$$ If you've advanced to the point of solving systems of linear diff eqs via eigensystems then you should know how to find eigenvectors given eigenvalues." ]

[ "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "## Matrix inverses Show that $\\mathbf{R_1} = \\begin{pmatrix}-1 & 1 \\\\ 6& -4 \\\\ 4& -3\\end{pmatrix}$ and $\\mathbf{R_2} = \\begin{pmatrix}1 & -1 \\\\ -4& 6 \\\\ -4& 5\\end{pmatrix}$ are both right-inverses of the matrix $\\mathbf{A} = \\begin{pmatrix}1 &1 &-1 \\\\ 4&0 &1 \\end{pmatrix}$. Use the right-inverses $\\mathbf{R_1}$ and $\\mathbf{R_2}$ to find two solutions $\\mathbf{x_1}$ and $\\mathbf{x_2}$ of the equation $\\mathbf{Ax = b}$, where $\\mathbf{b} =\\begin{pmatrix}0\\\\ 8\\end{pmatrix}$. By what I understand, the only way to solve Ax = b is with an inverse: $\\mathbf{A^{-1}Ax = A^{-1}b}$ $\\mathbf{x = A^{-1}b}$ and matrix $\\mathbf{A}$ doesnt have an inverse but the question asks to use the right-inverse and this is what I dont understand", "We partition A, B and C into equally sized block matrices $\\mathbf{A} = \\begin{bmatrix} \\mathbf{A}_{1,1} & \\mathbf{A}_{1,2} \\\\ \\mathbf{A}_{2,1} & \\mathbf{A}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{B} = \\begin{bmatrix} \\mathbf{B}_{1,1} & \\mathbf{B}_{1,2} \\\\ \\mathbf{B}_{2,1} & \\mathbf{B}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{C} = \\begin{bmatrix} \\mathbf{C}_{1,1} & \\mathbf{C}_{1,2} \\\\ \\mathbf{C}_{2,1} & \\mathbf{C}_{2,2} \\end{bmatrix}$ with $\\mathbf{A}_{i,j}, \\mathbf{B}_{i,j}, \\mathbf{C}_{i,j} \\in R^{2^{n-1} \\times 2^{n-1}}$ then $\\mathbf{C}_{1,1} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{1,2} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,2}$ $\\mathbf{C}_{2,1} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{2,2} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,2}$ With this construction we have not reduced the number of multiplications. We still need 8 multiplications to calculate the Ci,j matrices, the same number of multiplications we need when using standard matrix multiplication. Now comes the important part. We define new matrices $\\mathbf{M}_{1} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{2,2}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{2,2})$ $\\mathbf{M}_{2} := (\\mathbf{A}_{2,1} + \\mathbf{A}_{2,2}) \\mathbf{B}_{1,1}$ $\\mathbf{M}_{3} := \\mathbf{A}_{1,1} (\\mathbf{B}_{1,2} - \\mathbf{B}_{2,2})$ $\\mathbf{M}_{4} := \\mathbf{A}_{2,2} (\\mathbf{B}_{2,1} - \\mathbf{B}_{1,1})$ $\\mathbf{M}_{5} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{1,2}) \\mathbf{B}_{2,2}$ $\\mathbf{M}_{6} := (\\mathbf{A}_{2,1} - \\mathbf{A}_{1,1}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{1,2})$ $\\mathbf{M}_{7} := (\\mathbf{A}_{1,2} - \\mathbf{A}_{2,2}) (\\mathbf{B}_{2,1} + \\mathbf{B}_{2,2})$ only using 7 multiplications (one for each Mk) instead of 8. We may now express the Ci,j in terms of Mk, like this: $\\mathbf{C}_{1,1} = \\mathbf{M}_{1} + \\mathbf{M}_{4} - \\mathbf{M}_{5} + \\mathbf{M}_{7}$ $\\mathbf{C}_{1,2} = \\mathbf{M}_{3} + \\mathbf{M}_{5}$ $\\mathbf{C}_{2,1} =", "### Solving Matrix Equations #### Definition Given matrices $\\mathbf{A}$ and $\\mathbf{B}$, we can solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$ for the unknown matrix $\\mathbf{X}$ when the matrix inverse of $\\mathbf{A}$ exists. We do this by multiplying both sides of the equation on the left by $\\mathbf{A}^{-1}$, providing it exists, to give \\begin{align} \\mathbf{A} \\mathbf{X} &= \\mathbf{B} \\\\ \\mathbf{A}^{-1} \\mathbf{A} \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\\\ \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\end{align} as $\\mathbf{A}^{-1} \\mathbf{A} = \\mathbf{I}$ and multiplying any matrix by the identity matrix leaves it unaltered. It is important that both sides of the equation are multiplied on the left by $\\mathbf{A}^{-1}$ because matrix multiplication does not commute - multiplying on the right could give a different result. #### Worked Example ###### Example 1 Let \\begin{align} \\mathbf{A} &= \\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}, \\\\ \\mathbf{B} &= \\begin{pmatrix}4\\\\1\\end{pmatrix}. \\end{align} Solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$. ###### Solution We have the equation $\\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}\\mathbf{X} = \\begin{pmatrix}4\\\\1\\end{pmatrix}$ To get $\\mathbf{X}$ on its own, multiply both sides by the inverse of $\\mathbf{A}$. Note that $\\lvert \\mathbf{A} \\rvert = -5-6 = -11$. Therefore, $\\mathbf{A}^{-1} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}$ So $\\mathbf{X} = \\mathbf{A}^{-1} \\mathbf{B} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}$ Now, use matrix multiplication to find $\\mathbf{X}$. \\begin{align} \\mathbf{X} &= -\\frac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}(-20)+(-2)\\\\(-12)+1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}-22\\\\-11\\end{pmatrix}\\\\ &= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\end{align} #### Simultaneous Equations We can apply this method to solving a system of", "# Matrix Entrywise Addition/Examples/Real 2 x 2 Let $\\mathbf A = \\begin {pmatrix} p & q \\\\ r & s \\end {pmatrix}$ and $\\mathbf B = \\begin {pmatrix} w & x \\\\ y & z \\end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix sum of $\\mathbf A$ and $\\mathbf B$ is given by: $\\mathbf A + \\mathbf B = \\begin {pmatrix} p + w & q + x \\\\ r + y & s + z \\end {pmatrix}$", "+0 # Long solution matrix part 2 0 237 1 Let A be a $$2 \\times 2$$ matrix. Suppose that for every two-dimensional vector v, there exists a two-dimensional vector w such that $$\\mathbf{A} \\mathbf{w} = \\mathbf{v}$$ Show that we can find a matrix B such that $$\\mathbf{A} \\mathbf{B} = \\mathbf{I}$$. Mar 13, 2019 $$\\exists w_1 \\ni A w_1 = \\begin{pmatrix}1\\\\0\\end{pmatrix}\\\\ \\exists w_2 \\ni A w_2 = \\begin{pmatrix}0\\\\1\\end{pmatrix}\\\\ B = \\begin{pmatrix}w_1 &w_2\\end{pmatrix}$$", "Question: Find the group G generated under matrix multiplication by the matrices $A=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, B=\\begin{pmatrix} 0 & i \\\\ i & 0 \\end{pmatrix}.$ Determine its proper subgroups, and verify for each of them that its cosets exhaust G. Step-by-step Before we can draw up a group multiplication table to search for subgroups, we must determine the multiple products of A and B with themselves and with each other: ${A}^{2}=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}=\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}=I$. Since B = iA, it follows that ${B}^{2} = −I$, that AB = iI = BA, and that ${B}^{3} = −B$. In brief, A is of order 2, B is of order 4, and A and B commute. The eight distinct elements of the group are therefore: $I,A,B,{B}^{2},{B}^{3}, AB,A{B}^{2}$ and $A{B}^{3}$. The group multiplication table is By inspection, the subgroups and their cosets are as follows: {I,A} : {I,A}, {B,AB}, {${B}^{2},A{B}^{2}$}, {${B}^{3},A{B}^{3}$}; {$I,{B}^{2}$} : {$I,{B}^{2}$}, {$A,A{B}^{2}$}, {$B,{B}^{3}$}, {$AB,A{B}^{3}$}; {$I,A{B}^{2}$} : {$I,A{B}^{2}$}, {$A,{B}^{2}$}, {$B,A{B}^{3}$}, {${B}^{3},AB$}; {$I,B,{B}^{2},{B}^{3}$} : {$I,B,{B}^{2},{B}^{3}$}, {$A,AB,A{B}^{2},A{B}^{3}$}; {$I,AB,{B}^{2},A{B}^{3}$} : {$I,AB,{B}^{2},A{B}^{3}$}, {$A,B,A{B}^{2},{B}^{3}$}. As expected, in each case the cosets exhaust the group, with each element in one and only one coset.", "\\map {\\MM_R} n: \\mathbf {A B} \\ne \\mathbf {B A}$ and by definition (conventional) matrix multiplication over $\\map {\\MM_R} n$ is not commutative. $\\blacksquare$ ## Examples ### Arbitrary $2 \\times 2$ Matrices Consider the matrices: $\\ds \\mathbf A$ $=$ $\\ds \\begin {pmatrix} 1 & 2 \\\\ -1 & 0 \\end {pmatrix}$ $\\ds \\mathbf B$ $=$ $\\ds \\begin {pmatrix} 1 & -1 \\\\ 0 & 1 \\end {pmatrix}$ We have: $\\ds \\mathbf A \\mathbf B$ $=$ $\\ds \\begin {pmatrix} 1 & 1 \\\\ -1 & 1 \\end {pmatrix}$ $\\ds \\mathbf B \\mathbf A$ $=$ $\\ds \\begin {pmatrix} 2 & 2 \\\\ -1 & 0 \\end {pmatrix}$ and it is seen that $\\mathbf A \\mathbf B \\ne \\mathbf B \\mathbf A$.", "# Thread: Left inverse of matrix 1. ## Left inverse of matrix Consider the matrix $\\mathbf{C} = \\begin{pmatrix}1 & -1\\\\ 1&1 \\\\ 0&1 \\end{pmatrix}$ Find a left-inverse of C. Attempt: $\\begin{pmatrix}x1 & y1 & z1\\\\ x2&y2 &z2 \\end{pmatrix}\\begin{pmatrix}1 & -1\\\\ 1&1\\\\ 0&1 \\end{pmatrix} = \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ $\\begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\\\ x2 + y2& -x2 + y2 + z2 \\end{pmatrix}= \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ I know that you suppose to somehow use gauss reduction, and thought about gauss reducing: $x1 + y1 = 1$ $-x1 + y1 + z1 = 0$ and $x2 + y2 = 0$ $-x2 + y2 + z2 = 1$ but didnt get any luck 2. Only square matrices have inverses. To solve a matrix equation of the form $\\mathbf{Y} = \\mathbf{AX}$ for $\\mathbf{X}$ where $\\mathbf{A}$ is not square, you need to premultiply both sides by $\\mathbf{A}^T$ first. $\\mathbf{Y} = \\mathbf{AX}$ $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$. Now $\\mathbf{A}^T\\mathbf{A}$ is a square matrix, so you can premultiply both sides by $(\\mathbf{A}^T\\mathbf{A})^{-1}$. $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = (\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{IX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{X}$. 3. Originally Posted by Prove It Only square matrices have inverses. Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both. 4. You have", "a b= \\pi \\sqrt{1/\\lambda_1}\\sqrt{1/\\lambda_2} = \\frac{\\pi}{\\sqrt{\\det A}}$$. I will give one final example where the matrices can be ordered: The two matrices in this case were: $$A =\\begin{pmatrix}2/3 & 1/5 \\\\ 1/5 & 3/4\\end{pmatrix}, \\quad B=\\begin{pmatrix} 1& 1/7 \\\\ 1/7& 1 \\end{pmatrix}$$", "It doesn't immediately seem clear why this should be the way to multiply matrices. In this blog post, we look at why this is the definition of matrix multiplication. ### Simultaneous equations Matrices can be thought of as representing a system of simultaneous equations. For example, solving the matrix problem $$\\begin{bmatrix}2&5&2\\\\1&0&-2\\\\3&1&1\\end{bmatrix} \\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix}14\\\\-16\\\\-4\\end{pmatrix}$$ is equivalent to solving the following simultaneous equations. \\begin{align*} 2x+5y+2z&=14\\\\ 1x+0y-2z&=-16\\\\ 3x+1y+1z&=-4 \\end{align*} ### Two matrices Now, let $$\\mathbf{A}$$ and $$\\mathbf{C}$$ be two 3×3 matrices, let $$\\mathbf{b}$$ by a vector with three elements, and let $$\\mathbf{x}=(x,y,z)$$. We consider the equation $$\\mathbf{A}\\mathbf{C}\\mathbf{x}=\\mathbf{b}.$$ In order to understand what this equation means, we let $$\\mathbf{y}=\\mathbf{C}\\mathbf{x}$$ and think about solving the two simuntaneous matrix equations, \\begin{align*} \\mathbf{A}\\mathbf{y}&=\\mathbf{b}\\\\ \\mathbf{C}\\mathbf{x}&=\\mathbf{y}. \\end{align*} We can write the entries of $$\\mathbf{A}$$, $$\\mathbf{C}$$, $$\\mathbf{x}$$, $$\\mathbf{y}$$ and $$\\mathbf{b}$$ as \\begin{align*} \\mathbf{A}&=\\begin{bmatrix} a_{11}&a_{12}&a_{13}\\\\ a_{21}&a_{22}&a_{23}\\\\ a_{31}&a_{32}&a_{23} \\end{bmatrix} & \\mathbf{C}&=\\begin{bmatrix} c_{11}&c_{12}&c_{13}\\\\ c_{21}&c_{22}&c_{23}\\\\ c_{31}&c_{32}&c_{23} \\end{bmatrix} \\end{align*} \\begin{align*} \\mathbf{x}&=\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\end{pmatrix} & \\mathbf{y}&=\\begin{pmatrix}y_1\\\\y_2\\\\y_3\\end{pmatrix} & \\mathbf{b}&=\\begin{pmatrix}b_1\\\\b_2\\\\b_3\\end{pmatrix} \\end{align*} We can then write out the simultaneous equations that $$\\mathbf{A}\\mathbf{y}=\\mathbf{b}$$ and $$\\mathbf{C}\\mathbf{x}=\\mathbf{y}$$ represent: \\begin{align} a_{11}y_1+a_{12}y_2+a_{13}y_3&=b_1& c_{11}x_1+c_{12}x_2+c_{13}x_3&=y_1\\\\ a_{21}y_1+a_{22}y_2+a_{23}y_3&=b_2& c_{21}x_1+c_{22}x_2+c_{23}x_3&=y_2\\\\ a_{31}y_1+a_{32}y_2+a_{33}y_3&=b_3& c_{31}x_1+c_{32}x_2+c_{33}x_3&=y_3\\\\ \\end{align} Substituting the equations on the right into those on the left gives: \\begin{align} a_{11}(c_{11}x_1+c_{12}x_2+c_{13}x_3)+a_{12}(c_{21}x_1+c_{22}x_2+c_{23}x_3)+a_{13}(c_{31}x_1+c_{32}x_2+c_{33}x_3)&=b_1\\\\ a_{21}(c_{11}x_1+c_{12}x_2+c_{13}x_3)+a_{22}(c_{21}x_1+c_{22}x_2+c_{23}x_3)+a_{23}(c_{31}x_1+c_{32}x_2+c_{33}x_3)&=b_2\\\\ a_{31}(c_{11}x_1+c_{12}x_2+c_{13}x_3)+a_{32}(c_{21}x_1+c_{22}x_2+c_{23}x_3)+a_{33}(c_{31}x_1+c_{32}x_2+c_{33}x_3)&=b_3\\\\ \\end{align} Gathering the terms containing $$x_1$$, $$x_2$$ and $$x_3$$ leads to: \\begin{align} (a_{11}c_{11}+a_{12}c_{21}+a_{13}c_{31})x_1 +(a_{11}c_{12}+a_{12}c_{22}+a_{13}c_{32})x_2 +(a_{11}c_{13}+a_{12}c_{23}+a_{13}c_{33})x_3&=b_1\\\\ (a_{21}c_{11}+a_{22}c_{21}+a_{23}c_{31})x_1 +(a_{21}c_{12}+a_{22}c_{22}+a_{23}c_{32})x_2 +(a_{21}c_{13}+a_{22}c_{23}+a_{23}c_{33})x_3&=b_2\\\\ (a_{31}c_{11}+a_{32}c_{21}+a_{33}c_{31})x_1 +(a_{31}c_{12}+a_{32}c_{22}+a_{33}c_{32})x_2 +(a_{31}c_{13}+a_{32}c_{23}+a_{33}c_{33})x_3&=b_3 \\end{align} We can write this as a matrix:", "8 \\end{pmatrix}$, $\\mathbf{x}= \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix}$ and $\\mathbf{b}= \\begin{pmatrix} 15 \\\\ 50 \\end{pmatrix}$ . To find $\\mathbf{x}$, we need to find $$\\mathbf{x} = \\mathbf{A}^{-1}\\mathbf{b},$$ where $\\mathbf{A}^{-1}$ is the inverse of the matrix $\\mathbf{A}$. In general, finding the inverse of matrices is a computationally expensive task. In the remaining part of this article we will discuss how to efficiently solve this matrix inverse problem; however, before that let's look at some common types of matrices that arise frequently in practice. ### Types of Matrices Diagonal Matrix: All the off-diagonal elements in a diagonal matrix are zero, i.e., $$A_{i,j} = 0\\; \\forall\\; i\\neq j\\,.$$ $$A = \\begin{pmatrix} a_{1,1} & 0 & 0 \\\\ 0 & a_{2,2} & 0 \\\\ 0 & 0 & a_{3,3} \\end{pmatrix}$$ The inverse of a diagonal matrix is obtained by simply taking the reciprocal of the diagonal elements, i.e., $A^{-1}_{i,j}\\;=\\;1/A_{i,j}\\; \\forall\\; i=j$. $$A^{-1} = \\begin{pmatrix} \\frac{1}{a_{1,1}} & 0 & 0 \\\\ 0 & \\frac{1}{a_{2,2}} & 0 \\\\ 0 & 0 & \\frac{1}{a_{3,3}} \\end{pmatrix}$$ Identity Matrix: An identity matrix is a diagonal matrix with the diagonal values equal to 1, i.e., $$I_{i,j} = 1\\; \\forall\\; i = j \\,.$$ The identity matrix is the inverse of itself. $$I = I^{-1} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0", "\\mathbf{M}{O} = \\mathbf{O}$ and $\\mathbf{O}+\\mathbf{M}=\\mathbf{M}$ ## Inverse Matrices If you have two matrices such that $\\mathbf{AB} = \\mathbf{I}$ then you can think to yourself that A is equal to $\\frac{I}{B}$, which is analogous to $A=\\frac{1}{B}$. However, since we do not specifically define division for matrices, we must always use the notation B-1 instead of 1/B. So we have: $\\mathbf{A} = \\mathbf{B^{-1}I} = \\mathbf{B^{-1}}$ A matrix X-1 is called the inverse matrix of X - if you multiply the two together you get the identity matrix. ## 2x2 Matrices ### Inverse For two by two matrices the process of finding an inverse is very simple: $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}^{-1} = \\frac{1}{ad-bc}\\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix}$ This implies that not all square matrices do have an inverse, since if $ad-bc$ is zero then the inverse will be undefined. A matrix with no inverse is known as singular. The formula above is derived as follows: $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}\\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ $\\Rightarrow \\begin{cases} ae + bg = 1\\\\ af + bh = 0\\\\ ce + dg = 0\\\\ cf + dh = 1 \\end{cases}$ $e = \\tfrac{d}{ad - bc}$ $f =", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "$$\\alpha=\\frac{\\mathbf u\\cdot\\mathbf v}{\\|\\mathbf u\\|\\cdot\\|\\mathbf v\\|}$$. If $$\\cos\\alpha=1$$, then we say $$\\mathbf u$$ and $$\\mathbf v$$ have the same direction. If $$\\cos\\alpha=-1$$, then we say $$\\mathbf u$$ and $$\\mathbf v$$ have the opposite direction. If $$\\cos\\alpha=\\pm1$$, then we say that $$\\mathbf u$$ and $$\\mathbf v$$ are parallel. Is $$\\cos\\alpha=0$$, then we say that $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal. Method. We have 3 ways of deciding whether $$\\mathbf u$$ and $$\\mathbf v$$ are parallel. 1. Check whether $$\\mathbf u\\cdot\\mathbf v=\\pm\\|\\mathbf u\\|\\cdot\\|\\mathbf v\\|$$. 2. Suppose that $$u_j\\ne0$$. Then check whether $$\\frac{v_j}{u_j}\\mathbf u=\\mathbf v$$. 3. Check whether the matrix with rows $$\\mathbf u$$ and $$\\mathbf v$$ has rank 1. Example. Let $$\\mathbf u=(2,-1,3)$$ and $$\\mathbf v=(4,-2,-6)$$. Then by the reduction step $\\begin{pmatrix}2 & -1 & 3 \\\\ 4 & -2 & -6\\end{pmatrix}\\xrightarrow{R_2-2R_1}\\begin{pmatrix}2 & -1 & 3 \\\\ 0 & 0 & -12\\end{pmatrix}$ we get an echelon for matrix and thus see that the matrix with rows $$\\mathbf u$$ and $$\\mathbf v$$ has rank 2. Thus, $$\\mathbf u$$ and $$\\mathbf v$$ are not parallel. Example. Let $$\\mathbf u=(2,-1,3)$$ and $$\\mathbf v=(4,k,6)$$. Then $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal, precisely when $0=\\mathbf u\\cdot\\mathbf v=26-k.$ Therefore, $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal, if $$k=26$$. The cross product of $$\\mathbf u$$ and $$\\mathbf v$$ is $$\\left|\\begin{smallmatrix}\\mathbf i & u_1 & v_1 \\\\", "two different Möbius transformations of this type? ## Group structure 7. We can associate a matrix to a Möbius transformation. In problem 6 we would have the matrix $\\mathbf{M} = \\begin{pmatrix} a & 2b \\\\ \\bar{b} & \\bar{a} \\end{pmatrix}$. We would like to show that the collection of matrices $\\{ \\mathbf{M} = \\begin{pmatrix} a & 2b \\\\ \\bar{b} & \\bar{a}\\end{pmatrix} | |a|^2-2|b|^2=1 \\}$ forms a group. We would need to show that the collection is closed under multiplication, has an identity element, has an inverse and is associative. a. Closed under multiplication: Show that given two arbitrary elements, the product belongs to the collection. I.e. show that $\\begin{pmatrix} a_1 & 2b_1 \\\\ \\bar{b}_1 & \\bar{a}_1 \\end{pmatrix} * \\begin{pmatrix} a_2 & 2b_2 \\\\ \\bar{b}_2 & \\bar{a}_2 \\end{pmatrix}$ is a matrix that belongs to the collection. b. Identity element: Show that $\\mathbf{I} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ belongs to our collection. c. Inverses: Given $\\mathbf{M} = \\begin{pmatrix} a & 2b \\\\ \\bar{b} & \\bar{a} \\end{pmatrix}$, show that the inverse also belongs to the collection. d. Associativity: Show that $(\\mathbf{M_1} \\mathbf{M_2})\\mathbf{M_3} = \\mathbf{M_1} (\\mathbf{M_2} \\mathbf{M_3})$", "matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & 0 & 1 & 1 \\\\ 0 & 0 & 3 & 2 \\\\ 0 & 1 & 6 & 10 \\end {pmatrix}$ ### Arbitrary Matrix $3$ Let $\\mathbf A$ be the matrix: $\\mathbf A = \\begin {pmatrix} 0 & 4 & 7 & 10 \\\\ 2 & 5 & 8 & 11 \\\\ 3 & 6 & 9 & 12 \\end {pmatrix}$ The matrix $\\mathbf R$ corresponding to the row operation to clear the first column is: $\\mathbf R = \\begin {pmatrix} 0 & \\dfrac 1 2 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & -\\dfrac 3 2 & 1 \\end {pmatrix}$ and the matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & \\dfrac 5 2 & 4 & \\dfrac {11} 2 \\\\ 0 & 4 & 7 & 10 \\\\ 0 & -\\dfrac 3 2 & -3 & -\\dfrac 9 2 \\end {pmatrix}$ ## Also presented as Some sources are not concerned about making $b_{1 1}$ equal to $1$. Hence they do not use step $3$ of the algorithm.", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "immediately say $$\\frac{b_1}{f_1}=\\frac{b_2}{f_2}\\\\ \\frac{g_1^2-f_1^2}{b_1^2}=\\frac{g_2^2-f_2^2}{b_2^2}$$ The second of these can be rearranged, using the first of them ($b_2=b_1f_2/f_1$), to give $$\\frac{g_1^2}{f_1^2}=\\frac{g_2^2}{f_2^2}$$ So $$\\frac{g_1}{f_1} = \\pm \\frac{g_2}{f_2}$$ We can convert these into regular a,b,c,d form by noting that $a=H-f$ and $d=H+f$, so $f=(d-a)/2$. Similarly, $bc=g^2-f^2=g^2-(d-a)^2/4$, so $g^2=bc+(d-a)^2/4$. And so, we know that $$\\frac{d_1-a_1}{b_1}=\\frac{d_2-a_2}{b_2}\\\\ \\frac{b_1}{c_1}=\\frac{b_2}{c_2}$$ These two conditions immediately tell us that $M_1$ and $M_2$ must commute. Proof: $$\\left(\\begin{matrix}a_1&b_1\\\\c_1&d_1\\end{matrix}\\right)\\left(\\begin{matrix}a_2&b_2\\\\c_1\\!\\!\\frac{b_2}{b_1}&a_2+\\frac{b_2}{b_1}(d_1-a_1)\\end{matrix}\\right) = \\left(\\begin{matrix}a_1a_2+c_1b_2&a_2b_1+b_2d_1\\\\a_2c_1+c_1d_1\\frac{b_2}{b_1}&c_1b_2+d_1a_2+d_1(d_1-a_1)\\frac{b_2}{b_2}\\end{matrix}\\right)=\\left(\\begin{matrix}a_2&b_2\\\\c_1\\!\\!\\frac{b_2}{b_1}&a_2+\\frac{b_2}{b_1}(d_1-a_1)\\end{matrix}\\right)\\left(\\begin{matrix}a_1&b_1\\\\c_1&d_1\\end{matrix}\\right)$$ Since our matrices commute, we can immediately know that $e^{M_1-M_2}=I$. So all we have to do is find a non-zero solution to $e^M=I$. If we let $g=n\\pi i$, with $n$ being an integer other than zero (for $n=0$, the following doesn't apply), then we can see that for $$M = \\left(\\begin{matrix}H-f&b\\\\-\\frac{f^2+n^2\\pi^2}{b}&H+f\\end{matrix}\\right)$$ we can write $$e^M = (-1)^ne^H\\left(\\begin{matrix}1&0\\\\0&1\\end{matrix}\\right)$$ Therefore, as long as $H=i\\pi(2m-n)$ for some integer $m$, and integer $n\\neq 0$, then we have a matrix that satisfies $e^M=I$. For the case of $n=2k$ (integer $k\\neq 0$) and $m=0$, we have $$M = \\left(\\begin{matrix}-f&b\\\\-\\frac{4k^2\\pi^2+f^2}{b}&f\\end{matrix}\\right)$$ As an example that is more fun to test, consider $f=b=\\pi$ and $k=1$. Now we have $$M = \\left(\\begin{matrix}-\\pi&\\pi\\\\-5\\pi&\\pi\\end{matrix}\\right)$$ Which satisfies $e^M=I$.", "\\frac{y\\,dy+z\\,dz+w\\,dw}{\\sqrt{1-y^2-z^2-w^2}}\\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}$$ from which you can read off the Pauli matrices at the point $(x,y,z,w)=(1,0,0,0)$." ]

[ "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "a) (c – b)} \\end{pmatrix} = I_{3}$$ $$\\begin{pmatrix} 1&1&1&1 \\\\ a&b&c&d \\\\ a^2&b^2&c^2&d^2 \\\\ a^3&b^3&c^3&d^3 \\end{pmatrix} \\begin{pmatrix} -\\frac{bcd}{(a – b) (a – c)(a-d)} & \\frac{bc + cd + bd}{(a – b) (a – c)(a-d)} &-\\frac{b+c+d}{(a – b) (a – c)(a-d)} & \\frac{1}{(a – b) (a – c)(a-d)}\\\\ -\\frac{a cd}{(b – a) (b – c)(b-d)} & \\frac{ac + ad + cd}{(b – a) (b – c)(b-d)} & -\\frac{a + c + d}{(b – a) (b – c)(b-d)}& \\frac{1}{(b – a) (b – c)(b-d)}\\\\ -\\frac{a bd}{(c – a) (c – b)(c-d)} & \\frac{ab + ad + bd}{(c – a) (c – b)(c-d)} & -\\frac{a + b + d}{(c – a) (c – b)(c-d)}&\\frac{1}{(c – a) (c – b)(c-d)}\\\\ -\\frac{a bc}{(d – a) (d – b)(d-c)} & \\frac{ab + ac + bc}{(d – a) (d – b)(d-c)} & -\\frac{a + b + c}{(d – a) (d – b)(d-c)}&\\frac{1}{(d – a) (d – b)(d-c)} \\end{pmatrix} = I_{4}$$ The identities $f(0),f(1), f(2)$ are the last column of the inverse equation for 3-dim matrices. However, it seems that such matrix argument is not sufficient to prove the case for $n$ general, and that many similar identities (the other places in the matrices) should exist. • How to evaluate the sum of $\\sum_{n=0}^{\\infty}\\frac{1}{3n^{2}+4n+1}$ by Bruh on May 18, 2022 at 5:09 pm I hava an", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "& b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ so I'd guess applying that $2c$ would give simply $$= 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ However, $$= \\begin{pmatrix} c_3 & c_2 & c_1 \\\\ b_3 & b_2 & b_1 \\\\ a_3 & a_2 & a_1 \\end{pmatrix} \\neq 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ So I must have gotten something terribly wrong. Can anyone help? edit: In accordance to Quang's comment, I guess it should be $$= \\begin{pmatrix} c_3 + a_1 & c_2 + a_2 & c_1 + a_3 \\\\ b_3 + b_1 & b_2 + b_2 & b_1 + b_3 \\\\ a_3 + c_1 & a_2 + c_2 & a_1 + c_3 \\end{pmatrix} \\neq 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ • You forgot $\\mathrm{id}_V$: it is $\\mathrm{id}_V+r^2=2c$, not $r^2=2c$. – Quang Hoang Oct 2 '15 at 1:58 • True, but even then I still don't see how the sides equal each other. – linalgglanil Oct 2 '15 at 1:59 • Remember that you are working on the magic squares. Try to show that for such a", "$& x = -14-4y\\\\& x = -14-4 \\left(-\\frac{32}{9}\\right)\\\\& x = -14+\\frac{128}{9}\\\\& x = -\\frac{126}{9}+\\frac{128}{9}\\\\& \\boxed{x = \\frac{2}{9}}\\\\& \\boxed{l_1 \\cap l_2 @ \\left(\\frac{2}{9},-\\frac{32}{9}\\right)}$ Solve each of the following... $& \\begin{Bmatrix}x=-4+y\\\\x=3y-6\\end{Bmatrix}$ $& -4+y = 3y-6\\\\& -4+4+y = 3y-6+4\\\\& y = 3y-2\\\\& y-3y = 3y-3y-2\\\\& -2y = -2\\\\& \\frac{-2y}{-2} = \\frac{-2}{-2}\\\\& \\boxed{y = 1}$ $& x = -4+y\\\\& x = -4+1\\\\& \\boxed{x = -3}\\\\& \\boxed{l_1 \\cap l_2 @ (-3,1)}$ $& \\begin{Bmatrix}2x=5y-12\\\\3x+5y=7\\end{Bmatrix}$ $& 2x = 5y-12\\\\& 2x+12 = 5y-12+12\\\\& 2x+12 = 5y\\\\& \\boxed{5y = 2x+12}$ $& 3x+5y = 7\\\\& 3x-3x+5y = 7-3x\\\\& \\boxed{5y = 7-3x}$ $& \\begin{Bmatrix}5y=2x+12\\\\5y=7-3x\\end{Bmatrix}$ $& 2x+12 = 7-3x\\\\& 2x+12+3x = 7-3x+3x\\\\& 5x+12 = 7\\\\& 5x+12-12 = 7-12\\\\& 5x = -5\\\\& \\frac{5x}{5} = \\frac{-5}{5}\\\\& \\frac{\\cancel{5}x}{\\cancel{5}} = \\frac{\\overset{-1}{\\cancel{-5}}}{\\cancel{5}}\\\\& \\boxed{x = -1}$ $& 5y = 7-3x\\\\& 5y = 7-3(-1)\\\\& 5y = 7+3\\\\& 5y = 10\\\\& \\frac{5y}{5} = \\frac{10}{5}\\\\& \\frac{\\cancel{5}y}{\\cancel{5}} = \\frac{\\overset{2}{\\cancel{10}}}{\\cancel{5}}\\\\& \\boxed{y = 2}\\\\& \\boxed{l_1 \\cap l_2 @ (-1,2)}$ $& \\begin{Bmatrix}\\frac{x+y}{3}+\\frac{x-y}{2} =\\frac{25}{6}\\\\\\frac{x+y-9}{2} =\\frac{y-x-6}{3}\\end{Bmatrix}$ $& \\frac{x+y}{3}+\\frac{x-y}{2} = \\frac{25}{6}\\\\& 6 \\left(\\frac{x+y}{3}\\right)+6 \\left(\\frac{x-y}{2}\\right) = 6 \\left(\\frac{25}{6}\\right)\\\\& \\overset{2}{\\cancel{6}} \\left(\\frac{x+y}{\\cancel{3}}\\right)+\\overset{3}{\\cancel{6}} \\left(\\frac{x-y}{\\cancel{2}}\\right) = \\cancel{6} \\left(\\frac{25}{\\cancel{6}}\\right)\\\\& 2(x+y)+3(x-y) = 25\\\\& 2x+2y+3x-3y = 25\\\\& \\boxed{5x-y = 25}$ $& \\frac{x+y-9}{2} = \\frac{y-x-6}{3}\\\\& 6 \\left(\\frac{x+y-9}{2}\\right) = 6 \\left(\\frac{y-x-6}{3}\\right)\\\\& \\overset{3}{\\cancel{6}} \\left(\\frac{x+y-9}{\\cancel{2}}\\right) = \\overset{2}{\\cancel{6}} \\left(\\frac{y-x-6}{\\cancel{3}}\\right)\\\\& 3(x+y-9) = 2(y-x-6)\\\\& 3x+3y-27 = 2y-2x-12\\\\& 3x+3y-27+27 = 2y-2x-12+27\\\\& 3x+3y = 2y-2x+15\\\\& 3x+3y+2x = 2y-2x+2x+15\\\\& 5x+3y = 2y+15\\\\& 5x+3y-2y = 2y-2y+15\\\\& \\boxed{5x+y = 15}$ $& \\begin{Bmatrix}5x-y=25\\\\5x+y=15\\end{Bmatrix}$ $& 5x-y = 25", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "\\begin{pmatrix} ax + b \\end{pmatrix} dx = \\frac{1}{a} F \\begin{pmatrix} ax+b \\end{pmatrix} + c$ Where: \\begin{aligned} F(x) & = \\int f(x) dx \\\\ & = \\int \\sqrt[3]{x} dx \\\\ & = \\int x^{\\frac{1}{3}}dx \\\\ & = \\frac{1}{1+\\frac{1}{3}}.x^{\\frac{1}{3}+1}+c \\\\ & = \\frac{1}{\\frac{4}{3}}.x^{\\frac{4}{3}}+c \\\\ F(x) & = \\frac{3}{4}.x^{\\frac{4}{3}}+c \\end{aligned} So we can state the answer: \\begin{aligned} \\int \\sqrt[3]{4x - 3} dx & = \\frac{1}{4} \\times \\frac{3}{4}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\\\ & = \\frac{1\\times 3}{4\\times 4}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\\\ \\int \\sqrt[3]{4x - 3} dx & = \\frac{3}{16}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\end{aligned} We could also write the answer in raidcal form: $\\int \\sqrt[3]{4x - 3} dx = \\frac{3}{16}.\\sqrt[3]{\\begin{pmatrix} 4x - 3 \\end{pmatrix}^4}+c$ 4. We find the integral $$\\int 10.\\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$$ as follows: We start by taking the factor of $$10$$ out of the integral: $\\int 10.\\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx = 10 \\int \\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$ We can see that the integral $$\\int \\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$$ is of the type $$\\int f \\begin{pmatrix} ax+b \\end{pmatrix}dx$$, where: • $$ax+b = 3x - 2$$ • $$f(x) = x^4$$ Using the special case 1 formula we can state: $\\int f \\begin{pmatrix} ax + b \\end{pmatrix}dx = \\frac{1}{a}F\\begin{pmatrix} ax + b \\end{pmatrix} + c$ Where: \\begin{aligned} F(x) & = \\int x^4 dx", "\\end {pmatrix}$$ - $$\\frac 13$$$$\\begin {pmatrix} 18 & -6\\\\ 12 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 9 & 15\\\\ -18 & 0\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} 6 & -2\\\\ 4 & 1\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} 9-6 & 15+2\\\\ -18-4 & 0-1\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3 & 17\\\\ -22 & -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} 3A - (\\frac 13) B\\\\ \\end {vmatrix}$$ =$$\\begin {pmatrix} 3 & 17\\\\ -22 & -1\\\\ \\end {pmatrix}$$ = -3 +374 = 371 Ans Let: M =$$\\begin{pmatrix} a & c\\\\ b & d\\\\ \\end {pmatrix}$$ $$\\begin{pmatrix} a & c\\\\ b & d\\\\ \\end {pmatrix}$$$$\\begin{pmatrix} 1 & 1\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin{pmatrix} 4 & 5\\\\ 6 & 2\\\\ \\end {pmatrix}$$ or,$$\\begin{pmatrix} a×1+c×3 & a×1+c×4\\\\ b×1+d×3 & b×1+d×4\\\\ \\end {pmatrix}$$ =$$\\begin{pmatrix} 4 & 5\\\\ 6 & 2\\\\ \\end {pmatrix}$$ or,$$\\begin{pmatrix} a+3c & a+4c\\\\ b+3d & b+4d\\\\ \\end {pmatrix}$$ =$$\\begin{pmatrix} 4 & 5\\\\ 6 & 2\\\\ \\end {pmatrix}$$ Equating both sides, a + 3c = 4 or, a = 4 - 3c .........................(1) a + 4c = 5................................(2) Putting the value of a in equation (2) 4 - 3c + 4c = 5 or, 4 + c = 5 or, c = 5 - 4 ∴ c = 1 Putting the value of c in equation (1) ∴ a = 4 - 3c", "+ e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... \\\\ \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{X^2} = [\\mathbf{1}, \\mathbf{A^1}] = \\begin{bmatrix} 1 & \\frac{1}{1 + e^{-z_{11}}} & \\frac{1}{1 + e^{-z_{12}}} & \\frac{1}{1 + e^{-z_{13}}} \\\\ 1 & \\frac{1}{1 + e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... & ... \\\\ 1 & \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{Z^2} = \\mathbf{X^2}\\mathbf{W^2} = \\begin{bmatrix} z_{11} & z_{12} \\\\ z_{21} & z_{22} \\\\ ... & ... \\\\ z_{N1} & z_{N2} \\end{bmatrix}$ $\\mathbf{\\hat{Y}} = \\mathbf{A^2} = softmax(\\mathbf{Z^2}) = \\begin{bmatrix} \\hat{y}_{11} & \\hat{y}_{12} \\\\ \\hat{y}_{21} & \\hat{y}_{22} \\\\ ... & ... \\\\ \\hat{y}_{N1} & \\hat{y}_{N2} \\end{bmatrix}$ Here, $sigmoid$ is the sigmoid function and similarly $softmax$ is the row-wise softmax function. (Also, note that I've used superscripts to denote the layer of the network for each element.) My goal is to use gradient descent to find the weight matrices $\\mathbf{W^1}$ and $\\mathbf{W^2}$ that minimize categorical cross entropy. Can anyone walk me through this process? I have seen it done here, but I can't understand how the solution generalizes to matrix equations.", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "a \\quad c \\, a \\, b \\quad c \\, b \\, a$ $\\dbinom {1 \\ 2 \\ 3} { 1 \\ 2 \\ 3}, \\dbinom {1 \\ 2 \\ 3} { 1 \\ 3 \\ 2}, \\dbinom {1 \\ 2 \\ 3} { 2 \\ 1 \\ 3}, \\dbinom {1 \\ 2 \\ 3} { 2 \\ 3 \\ 1}, \\dbinom {1 \\ 2 \\ 3} { 3 \\ 1 \\ 2}, \\dbinom {1 \\ 2 \\ 3} { 3 \\ 2 \\ 1}$ ### $4$ from $4$ There are $24$ permutations of $4$ objects taken $4$ at a time: $\\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_2 & a_3 & a_4 \\end{pmatrix}, \\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_2 & a_4 & a_3 \\end{pmatrix}, \\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_3 & a_2 & a_4 \\end{pmatrix}, \\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_3 & a_4 & a_2 \\end{pmatrix}, \\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_4 & a_2 & a_4 \\end{pmatrix}, \\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_1 & a_4 & a_3 & a_2 \\end{pmatrix},$ $\\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\\\ a_2 & a_1 & a_3 &" ]

[ "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "& b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ so I'd guess applying that $2c$ would give simply $$= 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ However, $$= \\begin{pmatrix} c_3 & c_2 & c_1 \\\\ b_3 & b_2 & b_1 \\\\ a_3 & a_2 & a_1 \\end{pmatrix} \\neq 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ So I must have gotten something terribly wrong. Can anyone help? edit: In accordance to Quang's comment, I guess it should be $$= \\begin{pmatrix} c_3 + a_1 & c_2 + a_2 & c_1 + a_3 \\\\ b_3 + b_1 & b_2 + b_2 & b_1 + b_3 \\\\ a_3 + c_1 & a_2 + c_2 & a_1 + c_3 \\end{pmatrix} \\neq 2 \\begin{pmatrix} b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\\\ b_2 & b_2 & b_2 \\end{pmatrix}$$ • You forgot $\\mathrm{id}_V$: it is $\\mathrm{id}_V+r^2=2c$, not $r^2=2c$. – Quang Hoang Oct 2 '15 at 1:58 • True, but even then I still don't see how the sides equal each other. – linalgglanil Oct 2 '15 at 1:59 • Remember that you are working on the magic squares. Try to show that for such a", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "\\begin{pmatrix} ax + b \\end{pmatrix} dx = \\frac{1}{a} F \\begin{pmatrix} ax+b \\end{pmatrix} + c$ Where: \\begin{aligned} F(x) & = \\int f(x) dx \\\\ & = \\int \\sqrt[3]{x} dx \\\\ & = \\int x^{\\frac{1}{3}}dx \\\\ & = \\frac{1}{1+\\frac{1}{3}}.x^{\\frac{1}{3}+1}+c \\\\ & = \\frac{1}{\\frac{4}{3}}.x^{\\frac{4}{3}}+c \\\\ F(x) & = \\frac{3}{4}.x^{\\frac{4}{3}}+c \\end{aligned} So we can state the answer: \\begin{aligned} \\int \\sqrt[3]{4x - 3} dx & = \\frac{1}{4} \\times \\frac{3}{4}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\\\ & = \\frac{1\\times 3}{4\\times 4}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\\\ \\int \\sqrt[3]{4x - 3} dx & = \\frac{3}{16}. \\begin{pmatrix} 4x - 3 \\end{pmatrix}^{\\frac{4}{3}}+c \\end{aligned} We could also write the answer in raidcal form: $\\int \\sqrt[3]{4x - 3} dx = \\frac{3}{16}.\\sqrt[3]{\\begin{pmatrix} 4x - 3 \\end{pmatrix}^4}+c$ 4. We find the integral $$\\int 10.\\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$$ as follows: We start by taking the factor of $$10$$ out of the integral: $\\int 10.\\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx = 10 \\int \\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$ We can see that the integral $$\\int \\begin{pmatrix} 3x - 2 \\end{pmatrix}^4dx$$ is of the type $$\\int f \\begin{pmatrix} ax+b \\end{pmatrix}dx$$, where: • $$ax+b = 3x - 2$$ • $$f(x) = x^4$$ Using the special case 1 formula we can state: $\\int f \\begin{pmatrix} ax + b \\end{pmatrix}dx = \\frac{1}{a}F\\begin{pmatrix} ax + b \\end{pmatrix} + c$ Where: \\begin{aligned} F(x) & = \\int x^4 dx", "$$\\alpha=\\frac{\\mathbf u\\cdot\\mathbf v}{\\|\\mathbf u\\|\\cdot\\|\\mathbf v\\|}$$. If $$\\cos\\alpha=1$$, then we say $$\\mathbf u$$ and $$\\mathbf v$$ have the same direction. If $$\\cos\\alpha=-1$$, then we say $$\\mathbf u$$ and $$\\mathbf v$$ have the opposite direction. If $$\\cos\\alpha=\\pm1$$, then we say that $$\\mathbf u$$ and $$\\mathbf v$$ are parallel. Is $$\\cos\\alpha=0$$, then we say that $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal. Method. We have 3 ways of deciding whether $$\\mathbf u$$ and $$\\mathbf v$$ are parallel. 1. Check whether $$\\mathbf u\\cdot\\mathbf v=\\pm\\|\\mathbf u\\|\\cdot\\|\\mathbf v\\|$$. 2. Suppose that $$u_j\\ne0$$. Then check whether $$\\frac{v_j}{u_j}\\mathbf u=\\mathbf v$$. 3. Check whether the matrix with rows $$\\mathbf u$$ and $$\\mathbf v$$ has rank 1. Example. Let $$\\mathbf u=(2,-1,3)$$ and $$\\mathbf v=(4,-2,-6)$$. Then by the reduction step $\\begin{pmatrix}2 & -1 & 3 \\\\ 4 & -2 & -6\\end{pmatrix}\\xrightarrow{R_2-2R_1}\\begin{pmatrix}2 & -1 & 3 \\\\ 0 & 0 & -12\\end{pmatrix}$ we get an echelon for matrix and thus see that the matrix with rows $$\\mathbf u$$ and $$\\mathbf v$$ has rank 2. Thus, $$\\mathbf u$$ and $$\\mathbf v$$ are not parallel. Example. Let $$\\mathbf u=(2,-1,3)$$ and $$\\mathbf v=(4,k,6)$$. Then $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal, precisely when $0=\\mathbf u\\cdot\\mathbf v=26-k.$ Therefore, $$\\mathbf u$$ and $$\\mathbf v$$ are orthogonal, if $$k=26$$. The cross product of $$\\mathbf u$$ and $$\\mathbf v$$ is $$\\left|\\begin{smallmatrix}\\mathbf i & u_1 & v_1 \\\\", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "\\begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \\end{pmatrix}$ $\\textbf u = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3\\\\...\\\\ b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = \\begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 \\\\ b_2 \\\\b_3 \\\\ ... \\\\b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$ Notice that it only works if $m = n$ and it always results in a matrix of order $1 \\times 1$. The outer product of these two vectors is $\\textbf v \\cdot \\textbf u^T$. $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\\\ ... \\\\ a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \\end{pmatrix}$ $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\\\...& ... & ... & ... & ... \\\\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \\end{pmatrix}$ If you have a column vector of order $m \\times 1$ and a row vector of order $1 \\times n$ respectively, their outer product will be a matrix of order", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V.", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "# Computing a particular finite set of quaternion matrices. Let $B = \\left(\\frac{-1,-11}{\\mathbb{Q}}\\right)$ be a choice of quaternion algebra ramifying at $11$ and consider the maximal order $\\mathfrak{O}=\\mathbb{Z}\\oplus\\mathbb{Z}i\\oplus\\mathbb{Z}\\frac{1+j}{2}\\oplus\\mathbb{Z}\\frac{i+k}{2}$. Let $g = \\begin{bmatrix} 1 & (1+3i)j\\\\ 0 & j\\end{bmatrix}$. I need to get my hands on all $2\\times2$ matrices $A = \\begin{bmatrix} \\alpha & \\beta\\\\ \\gamma & \\delta\\end{bmatrix}$ with entries in $B$ such that: 1) $N(\\alpha)+N(\\beta) = 1$. 2) $N(\\alpha) = N(\\delta)$ and $N(\\beta) = N(\\gamma)$. 3) $\\alpha\\bar{\\gamma}+\\beta\\bar{\\delta} = 0$ 4) $gAg^{-1} \\in$ GL$_2(\\mathfrak{O})$ A priori I know that there are exactly $48$ such matrices but have only managed to find $4$ of them by guessing a huge amount of matrices of the form: 1. $\\begin{bmatrix} \\alpha & 0 \\\\ 0 & \\delta\\end{bmatrix}$ where the entries are units in $\\mathfrak{O} = \\{\\pm1, \\pm i\\}$. 2. $\\begin{bmatrix} 0 & \\beta \\\\ \\gamma & 0\\end{bmatrix}$ wth similar entries. 3. $\\begin{bmatrix} j^{-1}a & j^{-1}b\\\\ j^{-1}c & j^{-1}d\\end{bmatrix}$ where $a,b,c,d\\in\\mathfrak{O}$ have norms between $1$ and $10$, $N(a)+N(b)=11$ and $N(a) = N(d), N(b) = N(c)$. Between case 1 and case 3 I find $4$ matrices. Looking at a calculation done by Ibukiyama for ramified prime $2$ (in the longer version of \"On Symplectic Euler factors of genus two\") I expected the majority of the matrices to be of these forms and are stuck at", "\\\\ 0\\end{pmatrix} + \\beta e^{\\lambda_2 t} T \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}\\\\ & = \\alpha e^{\\lambda_2 t} \\mathbf v_1 + \\beta e^{\\lambda_2 t} \\mathbf v_2. \\end{align*} Suppose $$d{\\mathbf x}/dt = A {\\mathbf x}\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}. \\end{equation*} The eigenvalues of $$A$$ are $$\\lambda_1 = 5$$ and $$\\lambda_2 = -1$$ and the associated eigenvectors are $$(1, 2)$$ and $$(1, -1)\\text{,}$$ respectively. In this case, our matrix $$T$$ is \\begin{equation*} \\begin{pmatrix} 1 & 1 \\\\ 2 & -1 \\end{pmatrix}. \\end{equation*} If $${\\mathbf e}_1 = (1, 0)$$ and $${\\mathbf e}_2 = (0, 1)\\text{,}$$ then $$T {\\mathbf e}_i = {\\mathbf v}_i$$ for $$i = 1, 2\\text{.}$$ Consequently, $$T^{-1} {\\mathbf v}_i = {\\mathbf e}_i$$ for $$i = 1, 2\\text{,}$$ where \\begin{equation*} T^{-1} = \\begin{pmatrix} 1/3 & 1/3 \\\\ 2/3 & -1/3 \\end{pmatrix}. \\end{equation*} Thus, \\begin{equation*} T^{-1} A T = \\begin{pmatrix} 1/3 & 1/3 \\\\ 2/3 & -1/3 \\end{pmatrix} \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix} \\begin{pmatrix} 1 & 1 \\\\ 2 & -1 \\end{pmatrix} = \\begin{pmatrix} 5 & 0 \\\\ 0 & -1 \\end{pmatrix}. \\end{equation*} The eigenvalues of the matrix \\begin{equation*} \\begin{pmatrix} 5 & 0 \\\\ 0 & -1 \\end{pmatrix} \\end{equation*} are $$\\lambda_1 = 5$$ and $$\\lambda_2 = -1$$ with eigenvectors $$(1, 0)$$ and $$(0, 1)\\text{,}$$ respectively. Thus, the general solution", "1\\\\ 3 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 3 × 1 = 4 - 3 = 1 ≠ 0 It has a unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac 11$$ $$\\begin {pmatrix} 2 & -1\\\\ -3 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2 & -1\\\\ -3 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\begin {pmatrix} 2 & -1\\\\ -3 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 3\\\\ 2\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} 2 × 3 -1 × 2\\\\ -3 × 3 + 2 × 2\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} 6 - 2\\\\ -9 + 4\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} 4\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = 4 and y = -5 Ans Here, A = $$\\begin {bmatrix} a & b\\\\ 0 & c\\\\ \\end {bmatrix}$$ and B =$$\\begin {bmatrix} 2 & 0\\\\ 0 & 3\\\\ \\end {bmatrix}$$ AB =$$\\begin {bmatrix} a & b\\\\ 0 & c\\\\ \\end {bmatrix}$$$$\\begin {bmatrix} 2 & 0\\\\ 0 & 3\\\\ \\end {bmatrix}$$ =$$\\begin {bmatrix} a × 2 + b × 0 & a × 0 + b × 3\\\\ 0 × 2 + c × 0& 0 × 0 + c × 3\\\\ \\end {bmatrix}$$ =$$\\begin {bmatrix} 2a & 3b\\\\ 0 & 3c\\\\ \\end {bmatrix}$$ A + B", "+ \\sqrt{2} & 2 - \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} 2 + \\sqrt{2} & 0 \\\\ 0 & 2 - \\sqrt{2} \\end{pmatrix}^n \\begin{pmatrix} \\frac{1}{2}-\\frac{1}{\\sqrt{2}} & \\frac{1}{2\\sqrt{2}} \\\\ \\frac{1}{2}+\\frac{1}{\\sqrt{2}} & -\\frac{1}{2\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} 1/4 \\\\ 1 \\end{pmatrix} \\nonumber\\\\ &= \\begin{pmatrix} \\frac{1}{8}\\left((2+\\sqrt{2})^n(1+\\sqrt{2}) + (2-\\sqrt{2})^n(1-\\sqrt{2})\\right) \\\\ \\frac{1}{8}\\left((2+\\sqrt{2})^{n+1}(1+\\sqrt{2}) + (2-\\sqrt{2})^{n+1}(1-\\sqrt{2})\\right) \\end{pmatrix} \\label{eqn:3b_c_matrix} \\end{align} by diagonalization. Using Equation \\ref{eqn:3b_c_matrix}, Equation \\ref{eqn:3b_p_lim_2} becomes $$\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M) = \\frac{1}{2}(1 - 2a_{n})\\left((1+\\sqrt{2}) + \\left(\\frac{2-\\sqrt{2}}{2+\\sqrt{2}}\\right)^n(1-\\sqrt{2})\\right), \\label{eqn:3b_p_lim_3}$$ where we recall that $a = \\frac{1}{2 + \\sqrt{2}}.$ $|2-\\sqrt{2}| < 1$ and so, taking the limit as $n \\rightarrow \\infty$ of Equation \\ref{eqn:3b_p_lim_3}, the second term goes to $0$: $$\\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M) = \\frac{1-2a}{2}(1+\\sqrt{2}) = \\frac{\\sqrt{2} - 1}{2}(1+\\sqrt{2}) = \\frac{1}{2}. \\label{eqn:3b_p_lim_lim}$$ And finally, by symmetry, \\begin{align} \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 1 \\mid \\tau > M) &= \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 3 \\mid \\tau > M) \\nonumber\\\\ &= \\frac{1 - \\lim_{n\\rightarrow 0}\\lim_{M \\rightarrow \\infty} \\mathbb{P}(X_n = 2 \\mid \\tau > M)}{2} \\nonumber\\\\ &= \\frac{1}{4}. \\label{eqn:3b_p_1_3} \\end{align} Equations \\ref{eqn:3b_p_lim_lim} and \\ref{eqn:3b_p_1_3} combine to give us the limiting distribution. ## Graphs in $\\LaTeX$ Here's the code for the graph. I used a package called tikz. \\documentclass[10pt]{article} \\usepackage[utf8]{inputenc} \\usepackage[T1]{fontenc} \\usepackage{tikz} \\usetikzlibrary{arrows} \\usepackage[active, tightpage]{preview} \\PreviewEnvironment{tikzpicture} \\renewcommand\\PreviewBbAdjust {-0bp -25bp 0bp 25bp} \\begin{document} \\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.75cm, thick,main node/.style={circle,draw}] \\node[main node]", "? & ? & (r-a-b)\\\\ \\end{pmatrix}$$ Using symmetry we obtain these two values: $$\\begin{pmatrix} a & ? & \\frac{r-b}{2}\\\\ ? & b & ? \\\\ \\frac{r-b}{2} & ? & r-a-b\\\\ \\end{pmatrix}$$ Using the sum of the first and last row we deduce: $$\\begin{pmatrix} a & \\frac{r-b-2a}{2} & \\frac{r-b}{2}\\\\ ? & b & ? \\\\ \\frac{r-b}{2} & \\frac{2a+b-r}{2} & r-a-b\\\\ \\end{pmatrix}$$ Using the sum of the first and last column we deduce: $$\\begin{pmatrix} a & \\frac{r+b-2a}{2} & \\frac{r-b}{2}\\\\ \\frac{r+b-2a}{2} & b & \\frac{3b+2a-r}{2} \\\\ \\frac{r-b}{2} & \\frac{2a+3b-r}{2} & r-a-b\\\\ \\end{pmatrix}$$ It is not hard to see that the sum of the middle column is $3b$ so in fact $b=\\frac{r}{3}$ Now notice $a=\\frac{2a+3b-r}{2}$, so $P_3(r)=0$ for all $r$. • So since $P_{3}(r) = 0$ for all $r$ then it is $\\leq (r+1)^{3}$? – user482578 Oct 5 '17 at 21:52", "\\\\ 1 && -2\\end{bmatrix}$$ I'll write the eigenvalue equation as$$\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix} = c \\begin{pmatrix} a \\\\ b\\end{pmatrix}$$. The bottom line of the matrix equation gives $a-2b=cb$. Thus $a=(c+2)b$. So we have shown that the vector $\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix}$ should be an eigenvector. So we can plug this back into the eigenvalue equation: $$c\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix}=\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}(c+2)b \\\\ b\\end{pmatrix} =\\begin{pmatrix}-2cb-3b\\\\ cb \\end{pmatrix}$$ Reading across the top, we see the equation $$c(c+2)b=-2cb-3b$$Or$$0=c^2+4c+3=(c+3)(c+1)$$ so the two eigenvalues are $c=-3$ and $c=-1$. Now we plug these two values back into the eigenvalue equation to find the eigenvectors corresponding to each eigenvalue. For $c=-3$ we have: $$\\begin{bmatrix}-2 && 1 \\\\ 1 && -2\\end{bmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix} = -3 \\begin{pmatrix} a \\\\ b\\end{pmatrix}$$ which implies $b=-a$. This means any vector of the form $\\begin{pmatrix}a \\\\ -a \\end{pmatrix}$ is an eigenvector with eigenvalue -3. We want our states to be normalized though, so we pick $a=\\frac{1}{\\sqrt{2}}$. See if you can find the other eigenvector (the one corresponding to $c=-1$) for yourself, and then see if you can use your answers to solve the original problem with the k's and m's. Last edited: Sep 10, 2013", "\\end{bmatrix} = I$ If we choose $a = 0$ then $A = \\begin{bmatrix} 0&\\frac{1}{2} \\\\ 2&0 \\end{bmatrix}$ and if we choose $a = \\frac{1}{2}$ then $A = \\begin{bmatrix} \\frac{1}{2}&\\frac{1}{2} \\\\ \\frac{3}{2}&-\\frac{1}{2} \\end{bmatrix}$ There are an infinite number of other matrices for which $a_{12} = \\frac{1}{2}$ and $A^2 = I$, corresponding to other values of $a = a_{11}$. (b) We have $a_{12} = \\frac{1}{2}$ and $A^{-1} = A^T$ so that $AA^T = \\begin{bmatrix} a&\\frac{1}{2} \\\\ c&d \\end{bmatrix} \\begin{bmatrix} a&c \\\\ \\frac{1}{2}&d \\end{bmatrix}$ $= \\begin{bmatrix} a^2 + \\frac{1}{4}&ac + \\frac{1}{2}d \\\\ ac + \\frac{1}{2}d&c^2 + d^2 \\end{bmatrix} = \\begin{bmatrix} 1&0 \\\\ 0&1 \\end{bmatrix}$ Since $a^2 + \\frac{1}{4} = 1$ we have $a = \\pm \\frac{\\sqrt{3}}{2}$. We choose $a = \\frac{\\sqrt{3}}{2}$. Since $ac + \\frac{1}{2}d = \\frac{\\sqrt{3}}{2}c + \\frac{1}{2}d = 0$ we have $d = -\\sqrt{3}c$. Finally, since $c^2 + d^2 = c^2 + (-\\sqrt{3}c)^2 = c^2 + 3c^2 = 1$ we have $c^2 = \\frac{1}{4}$ or $c = \\pm \\frac{1}{2}$. We choose $c = \\frac{1}{2}$. Since $a = \\frac{\\sqrt{3}}{2}$ and $c = \\frac{1}{2}$ we thus have $A = \\begin{bmatrix} \\frac{\\sqrt{3}}{2}&\\frac{1}{2} \\\\ \\frac{1}{2}&-\\frac{\\sqrt{3}}{2} \\end{bmatrix}$ and $AA^T = \\begin{bmatrix} \\frac{\\sqrt{3}}{2}&\\frac{1}{2} \\\\ \\frac{1}{2}&-\\frac{\\sqrt{3}}{2} \\end{bmatrix} \\begin{bmatrix} \\frac{\\sqrt{3}}{2}&\\frac{1}{2} \\\\ \\frac{1}{2}&-\\frac{\\sqrt{3}}{2} \\end{bmatrix} = \\begin{bmatrix} 1&0 \\\\ 0&1 \\end{bmatrix} = I$ Note that $A = A^T$ in this case so that we also have $A^2 = I$.", "3$$ or $$2x = -3$$. Which, in turn, leads to: $$x = \\frac{3}{2}$$ or $$x = -\\frac{3}{2}$$. Finally we can state that this quadratic equation has two solutions: $$x = \\frac{3}{2}$$ and $$x = -\\frac{3}{2}$$. 5. We factor and solve $$16x^2 - 81=0$$ in two steps: • Step 1: we notice that $$16x^2 - 81$$ is the difference of two squares. Indeed $$16x^2 - 81 = \\begin{pmatrix}4x\\end{pmatrix}^2 - 9^2$$. So we can write it in factored form: $16x^2 - 81 = \\begin{pmatrix} 4x - 9 \\end{pmatrix}\\begin{pmatrix} 4x + 9 \\end{pmatrix}$ • Step 2: we use the null product law to solve the factored equation $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 3x + 3 \\end{pmatrix} = 0$$. That's: $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 2x + 3 \\end{pmatrix} = 0$$ if and only if: $$2x - 3 = 0$$ or $$2x+3 = 0$$, which leads to: $$2x = 3$$ or $$2x = -3$$. Which, in turn, leads to: $$x = \\frac{3}{2}$$ or $$x = -\\frac{3}{2}$$. Finally we can state that this quadratic equation has two solutions: $$x = \\frac{3}{2}$$ and $$x = -\\frac{3}{2}$$. 6. We factor and solve $$x^2 -5 = 0$$ in two steps: • Step 1: Using the fact that $$5 = \\begin{pmatrix} \\sqrt{5} \\end{pmatrix}^2$$, we can write $$x^2 - 5$$ as the difference of two squares: Indeed $$x^2", "1 \\end{pmatrix}$$. ### Solution 1. To find $$P\\begin{pmatrix}X \\leq 1.5\\end{pmatrix}$$, we use write: \\begin{aligned} P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = \\int_{-\\infty}^{1.5} f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}-\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}\\begin{pmatrix} x^2 - 2x \\end{pmatrix}dx \\\\ & = -\\frac{3}{4} \\begin{bmatrix}\\frac{x^3}{3} - x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}x^3 - 3x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}\\begin{pmatrix} \\frac{3}{2}\\end{pmatrix}^3 - 3 \\times \\begin{pmatrix} \\frac{3}{2} \\end{pmatrix}^2 - 0 \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\frac{27}{8} - \\frac{27}{4} \\end{bmatrix} \\\\ & = - \\frac{1}{4} \\begin{bmatrix} -\\frac{27}{8} \\end{bmatrix} \\\\ & = \\frac{27}{32} \\\\ P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = 0.844 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, from $$-\\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix} X \\leq 1.5\\end{pmatrix} = 0.844$$ 2. To find $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = \\int_{0.5}^1f(x)dx \\\\ & = \\int_{0.5}^1 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{0.5}^1 x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{\\frac{1}{2}}^1 \\begin{pmatrix} x^2-2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix} \\frac{x^3}{3} - x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1" ]

[ "= (4\\alpha-1)\\begin{pmatrix}0\\\\3\\\\1\\end{pmatrix}$, which is constant in direction whatever the value of $\\alpha$.", "+0 0 103 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19818 0 Sep 15, 2019 #1 +19818 0", "+0 0 67 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19832 +1 Sep 15, 2019 #1 +19832 +1", "and $B=(1,3,-2)$. The line that joints them is $$C = \\begin{pmatrix} 1 \\\\ 7 \\\\ 2 \\end{pmatrix} \\times \\begin{pmatrix} 1 \\\\ 3 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -20 \\\\ 4 \\\\ -4 \\end{pmatrix}$$ with the equation $$(-20)+(4)x+(-4)y=0 \\\\ x-y-5 = 0$$ -", "a constant depending only on $$A$$. I don't see an immediate way to write $$\\begin{pmatrix} 0 & a \\\\\\ -a & 0 \\end{pmatrix} \\to \\begin{pmatrix} 0 & a \\\\\\ a & 0 \\end{pmatrix}$$ in \"algebraic way\", i.e. involving only matrix addition, multiplication, and square roots.", "1 & -2 & -5 \\end{pmatrix},$$ $$(A-6I)(A - 12 I) = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix},$$", "of $d.$ If $y = 3$ when $m = 2$ and $d = 4,$ find the constant $k,$ then use $k$ to find $y$ when $m = 25$ and $d = 5.$ Midterm 1: Version C Answer Key", "$$x(t) = \\mathrm e^{2t}\\left(\\int \\mathrm e^{-2t}(f'(t,c_1,c_2)-f(t,c_1,c_2))dt+c_3\\right)$$ and that is where the third constant appears. - So, I put my solution for y $y=C_1e^tcos(t)+C_2e^tsin(t)$ into $z = x + 2y - y'$? Doing that doesn't produce a third constant. – aarnes Feb 18 '12 at 21:44 @aarnes: it does since you have this constant in $x$ – Ilya Feb 19 '12 at 10:11 I think you are saying that solving for $y$ only requires $2$ constants, but to solve for all $3$ unknowns, requires the $3$ constants expected. Is that so? – robjohn Feb 19 '12 at 10:19 @robjohn: yes, that is what I meant – Ilya Feb 19 '12 at 10:33 First of, let me say that I'm sorry for responding so late but I'm probably in a different time zone and last night was late and I slept for quite a while. What I still don't understand is how do I get a solution for x then in order to get that third constant? Putting solution for y (with two constants) inside $z = x + 2y - y'$ gives $z = C_1e^tcos(t) + C_1e^tsin(t) + C_2e^tsin(t) - C_2e^tcos(t) + x(t)$ – aarnes Feb 19 '12 at 18:39 Note that since $$D\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}=\\begin{bmatrix}3&1&-1\\\\1&2&-1\\\\3&3&-1\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}\\tag{1}$$ we have the general solution $$\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}=\\exp\\left(\\begin{bmatrix}3&1&-1\\\\1&2&-1\\\\3&3&-1\\end{bmatrix}t\\right)\\begin{bmatrix}x_0\\\\y_0\\\\z_0\\end{bmatrix}\\tag{2}$$ Computing the Jordan Normal Form, we get \\begin{align} &", "+0 0 47 1 Find constants A and B such that: $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x≠-1 and x≠2. Give your answer as the ordered pair (A,B). Sep 14, 2019 #1 +1 I think I saw this earlier. Check out https://web2.0calc.com/questions/help-please_7890 . Sep 14, 2019", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "+0 +7 259 3 +730 Find constants A and B such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that $$x\\neq -1$$ and $$x\\neq 2$$ . Give your answer as the ordered pair (A,B). MIRB16 Aug 28, 2017 #1 +87301 +1 CPhill Aug 28, 2017 #2 +730 +6 Thanks! MIRB16 Aug 28, 2017 #3 +730 +6 CPhill MIRB16 Aug 28, 2017", "& 24 x & 126 x \\\\ 306 & 42 & 24 & 126 \\end{vmatrix}$ $$829440$$ In this case, it is just a constant, but can be a function of $x$ depending on the functions.", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "+0 # Help me please, tough question 0 82 1 Find constants and such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x is not equal to -1or 2. Give your answer as the ordered pair . May 27, 2020 #1 +25541 +1 Find constants and such that $$\\dfrac{x + 7}{x^2 - x - 2} = \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1}$$", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "& wz-1+z(z+1) \\end{pmatrix}$ $=-\\begin{pmatrix} (w\\times 0)-1 & 0 \\\\ 0 & (z\\times 0)-1 \\end{pmatrix}$ $=\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ Thus A has order 3.", "= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$", "= x\\cdot \\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix} + y\\cdot \\begin{pmatrix} 6 & 5 & 4 \\end{pmatrix} + z\\cdot \\begin{pmatrix} 7 & 8 & 9 \\end{pmatrix}.$$", "have that $$\\theta_{t*}(X_{(x,y)}) = \\begin{pmatrix} e^{2t} & 0 \\\\ 0 & e^{-3t} \\end{pmatrix} \\begin{pmatrix} 2x \\\\ -3y \\end{pmatrix} = \\begin{pmatrix} 2xe^{2t} \\\\ -3ye^{-3y} \\end{pmatrix} .$$ Here again the basis is understood to be $$(\\frac{\\partial}{\\partial x}, \\frac{\\partial}{\\partial y})$$. So we are done." ]

[ "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "or, Q3= 3($$\\frac{N + 1}{4}$$)th item or, Q3= 3($$\\frac{7 + 1}{4}$$)th item or, Q3= 3($$\\frac{8}{4}$$)th item or, Q3= 3 $$\\times$$ 2th item or, Q3= 6th item i.e.Q3= x + 25 Then, or,Q3= x + 25 or, 60 = x + 25 or, x + 60 - 25 $$\\therefore$$ x = 35 Solution: Now, Arranging in ascending order 50, 52, 56, 61, 64, 68, 72 numbers = 7 We know that, or, Q1 = $$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$thitem or, Q1 = $$\\begin{pmatrix}\\frac{7 + 1}{4} \\end{pmatrix}$$thitem or, Q1 = $$\\begin{pmatrix}\\frac{8}{4} \\end{pmatrix}$$thitem or, Q1 = 2thitem or,Q1 = 52 Again, or, Q2= $$\\begin{pmatrix}\\frac{n + 1}{2} \\end{pmatrix}$$thitem or, Q2= $$\\begin{pmatrix}\\frac{7 + 1}{2} \\end{pmatrix}$$thitem or, Q2= $$\\begin{pmatrix}\\frac{8}{2} \\end{pmatrix}$$thitem or, Q2= 4thitem or, Q2= 61 Similarly, or, Q3= 3$$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$thitem or, Q3= 3$$\\begin{pmatrix}\\frac{7 + 1}{4} \\end{pmatrix}$$thitem or, Q3= 3$$\\begin{pmatrix}\\frac{8}{4} \\end{pmatrix}$$thitem or, Q3= 3 $$\\times$$2thitem or, Q3= 6thitem or, Q3 = 68 $$\\therefore$$ The value ofQ1, Q2 and Q3is 52, 61 and 68 respectively. Solution: Marks No. of students(f) cumulative frequency (c.f.) 10 2 2 20 4 6 30 6 12 40 12 24 50 8 32 60 6 38 70 1 39 N = 39 We have, Q1= value of $$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$th term = value of $$\\begin{pmatrix}\\frac{39 + 1}{4} \\end{pmatrix}$$th term = value of 10th term In c.f. column just", "4 \\\\ 3 \\end{pmatrix}.$ ×", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "1 & -2 & -5 \\end{pmatrix},$$ $$(A-6I)(A - 12 I) = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix},$$", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "Example Using the method we just saw and without using a calculator, calculate each of the following: 1. $$101^2$$ 2. $$27^2$$ 3. $$23^2$$ 4. $$99^2$$ 5. $$104^2$$ 6. $$48^2$$ ## Solution Without Working 1. $$101^2 = 10201$$ 2. $$27^2 = 711$$ 3. $$23^2 = 529$$ 4. $$99^2 = 9801$$ 5. $$104^2 = 10816$$ 6. $$48^2 = 2304$$ ## Solution With Working 1. We calculate $$101^2$$ as follows: \\begin{aligned} 101^2 & = \\begin{pmatrix} 100+1\\end{pmatrix}^2 \\\\ & = 100^2 + 2 \\times 100 \\times 1 + 1^2 \\\\ & = 10000 + 200 + 1\\\\ 101^2 &= 10201 \\end{aligned} 2. We calculate $$27^2$$ as follows: \\begin{aligned} 27^2 & = \\begin{pmatrix} 30 - 3 \\end{pmatrix}^2 \\\\ & = 30^2 - 2 \\times 30 \\times 3 - 3^2 \\\\ & = 900 - 180 - 9 \\\\ & = 720 - 9 \\\\ 27^2 & = 711 \\end{aligned} 3. We calculate $$23^2$$ as follows: \\begin{aligned} 23^2 & = \\begin{pmatrix} 20 + 3 \\end{pmatrix}^2 \\\\ & = 20^2 + 2 \\times 20 \\times 3 + 3^2 \\\\ & = 400 + 120 + 9 \\\\ 23^2 & = 529 \\end{aligned} 4. We calculate $$99^2$$ as follows: \\begin{aligned} 99^2 & = \\begin{pmatrix} 100 - 1 \\end{pmatrix} \\\\ & = 100^2 - 2\\times 100 \\times 1 + 1^2 \\\\ & = 10000 - 200 +", "+4\\end{pmatrix}$ we find: $-4x^9 - 10x^8 + 8x^7 - 3x^6 - 26x^5 + 20x^4 - 20x^3 + 20x^2 - 7x + 12$ 4. Given $$p(x) = \\frac{x^4}{2} - 3x^3 + 2x^2 + 5x - 1$$ and $$q(x) = 6x^4 + 4x^2 - 8x + 2$$, we find: $p(x)\\times q(x) = 3x^8 - 18x^7 + 14x^6 + 14x^5 + 27x^4 - 2x^3 - 40x^2 + 18x - 2$. 5. For all the terms in the expansion of the product: $\\begin{pmatrix} 6x^4 - 3x^2 + x - 5 \\end{pmatrix}.\\begin{pmatrix} 2x^3+ 4x^2 - 7x - 2 \\end{pmatrix}$ we find: $12x^7 + 24x^6 - 48x^5 - 24x^4 + 15x^3 - 21x^2 + 33x + 10$ ### Subscribe to Our Channel Subscribe Now and view all of our playlists & tutorials.", "- 5× 2 = 15 - 10 = 5 Ans Here, A = $$\\begin {pmatrix} 4 & 6\\\\ x & 3\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} -4 & 5\\\\ 7 & 8\\\\ \\end {pmatrix}$$ and I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Now, A - B - 5I =$$\\begin {pmatrix} 4 & 6\\\\ x & 3\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} -4 & 5\\\\ 7 & 8\\\\ \\end {pmatrix}$$ - 5$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 4-(-4) & 6-5\\\\ x-7 & 3-8\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} 5 & 0\\\\ 0 & 5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8 & 1\\\\ x-7 & -5\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} 5 & 0\\\\ 0 & 5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8-5 & 1-0\\\\ x-7-0 & -5-5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3 & 1\\\\ x - 7 & -10\\\\ \\end {pmatrix}$$ We have, $$\\begin {vmatrix} A - B - 2I\\\\ \\end {vmatrix}$$ = 14 or,$$\\begin {vmatrix} 3 & 1\\\\ x - 7 & -10\\\\ \\end {vmatrix}$$ = 14 or, 3× -10 - 1× (x-7) = 14 or, -30 - x + 7 = 14 or, -x = 14 - 7 + 30 or, -x = 37 ∴ x = -37 Ans Here, A = $$\\begin {pmatrix} 1 & -2\\\\ 0 & x\\\\", "\\begin {pmatrix} 12 & 6 & -2 \\\\ -6 & -3 & 1 \\\\ 6 & 3 & -1 \\end {pmatrix}$", "& 4 \\end{pmatrix} \\begin{pmatrix} 3 & 6 & 9 \\end{pmatrix} \\begin{pmatrix} 5 & 8 \\end{pmatrix}$.", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "6 & 0 & 5 \\end{pmatrix}$", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$" ]

[ "and $B=(1,3,-2)$. The line that joints them is $$C = \\begin{pmatrix} 1 \\\\ 7 \\\\ 2 \\end{pmatrix} \\times \\begin{pmatrix} 1 \\\\ 3 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -20 \\\\ 4 \\\\ -4 \\end{pmatrix}$$ with the equation $$(-20)+(4)x+(-4)y=0 \\\\ x-y-5 = 0$$ -", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "the same length: • For$m = 1$: the roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$are • (all separate)$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$, • (two pairs)$\\begin{pmatrix}2 & 5\\end{pmatrix} \\begin{pmatrix}8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8\\end{pmatrix} \\begin{pmatrix}5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 14\\end{pmatrix} \\begin{pmatrix}5 & 8\\end{pmatrix}$, • (all together)$\\begin{pmatrix}2 & 5 & 8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 5 & 14 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 14 & 5\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 5 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 8 & 5\\end{pmatrix}$. So there are$1 + 3 + 6 = 10$roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$. • For$m = 2$, the roots of$\\begin{pmatrix}1 & 7\\end{pmatrix} \\begin{pmatrix}4 & 10\\end{pmatrix} \\begin{pmatrix}6 & 18\\end{pmatrix} \\begin{pmatrix}12 & 17\\end{pmatrix}$are$6 \\times 8 = 48$in number (pick one of the$3!$orderings of the last three, then one of the$2^3$orderings of each of the three). • For$m = 3$, the roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$are • (both separate)$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$• (both together)$\\begin{pmatrix}3 & 11 & 9 & 13 & 16 & 15\\end{pmatrix}$,$\\begin{pmatrix}3 & 15 & 9 & 11 & 16 & 13\\end{pmatrix}$,$\\begin{pmatrix}3 & 13 & 9 & 15 & 16 & 11\\end{pmatrix}$. So there are$1 + 3 = 4$roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$. So totally$N(\\sigma) = 10 \\times 48 \\times 4 = 1920$. - In", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "- 5× 2 = 15 - 10 = 5 Ans Here, A = $$\\begin {pmatrix} 4 & 6\\\\ x & 3\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} -4 & 5\\\\ 7 & 8\\\\ \\end {pmatrix}$$ and I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Now, A - B - 5I =$$\\begin {pmatrix} 4 & 6\\\\ x & 3\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} -4 & 5\\\\ 7 & 8\\\\ \\end {pmatrix}$$ - 5$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 4-(-4) & 6-5\\\\ x-7 & 3-8\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} 5 & 0\\\\ 0 & 5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8 & 1\\\\ x-7 & -5\\\\ \\end {pmatrix}$$ -$$\\begin {pmatrix} 5 & 0\\\\ 0 & 5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8-5 & 1-0\\\\ x-7-0 & -5-5\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3 & 1\\\\ x - 7 & -10\\\\ \\end {pmatrix}$$ We have, $$\\begin {vmatrix} A - B - 2I\\\\ \\end {vmatrix}$$ = 14 or,$$\\begin {vmatrix} 3 & 1\\\\ x - 7 & -10\\\\ \\end {vmatrix}$$ = 14 or, 3× -10 - 1× (x-7) = 14 or, -30 - x + 7 = 14 or, -x = 14 - 7 + 30 or, -x = 37 ∴ x = -37 Ans Here, A = $$\\begin {pmatrix} 1 & -2\\\\ 0 & x\\\\", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "1 & -2 & -5 \\end{pmatrix},$$ $$(A-6I)(A - 12 I) = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix},$$", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "+4\\end{pmatrix}$ we find: $-4x^9 - 10x^8 + 8x^7 - 3x^6 - 26x^5 + 20x^4 - 20x^3 + 20x^2 - 7x + 12$ 4. Given $$p(x) = \\frac{x^4}{2} - 3x^3 + 2x^2 + 5x - 1$$ and $$q(x) = 6x^4 + 4x^2 - 8x + 2$$, we find: $p(x)\\times q(x) = 3x^8 - 18x^7 + 14x^6 + 14x^5 + 27x^4 - 2x^3 - 40x^2 + 18x - 2$. 5. For all the terms in the expansion of the product: $\\begin{pmatrix} 6x^4 - 3x^2 + x - 5 \\end{pmatrix}.\\begin{pmatrix} 2x^3+ 4x^2 - 7x - 2 \\end{pmatrix}$ we find: $12x^7 + 24x^6 - 48x^5 - 24x^4 + 15x^3 - 21x^2 + 33x + 10$ ### Subscribe to Our Channel Subscribe Now and view all of our playlists & tutorials.", "write $$A=\\pmatrix{a&b\\\\c&d}$$ such that $ad+bc=1$. Then, $$A^2=\\pmatrix{a^2+bc&b(a+d)\\\\c(a+d)&bc+d^2}$$ $$A^{-1}=\\pmatrix{d&-b\\\\-c&a}$$ Then you see we need things like $a+d=-1$, etc. You could play with these and see if something comes out of it 1 You can even demand that $A$ have integer entries. For example, $$A=\\begin{bmatrix} 0&1\\\\ -1&-1 \\end{bmatrix}\\,.$$ 1 $A$ and $A^T$ are necessarily similar. To show this, it suffices to note that $$\\dim \\ker [(A - \\lambda I)^k] = \\dim \\ker [(A^T - \\lambda I)^k]$$ for all $\\lambda$ taken from the algebraic closure of $K$ and all $k \\in \\Bbb N$. Only top voted, non community-wiki answers of a minimum length are eligible", "x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 16\\\\ 20\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 3 & 5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 16\\\\ 20\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 3 × 3 - 7 × 5 = 9 - 35 = -26 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-26}$$$$\\begin {pmatrix} 3 & -5\\\\ -7 & 3\\\\ \\end {pmatrix}$$ We know that: AX = B X = A-1B or, X =$$\\frac {1}{-26}$$$$\\begin {pmatrix} 3 & -5\\\\ -7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 16\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-26}$$$$\\begin {pmatrix} 48 - 100\\\\ -112 + 60\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-26}$$$$\\begin {pmatrix} -52\\\\ -52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-52}{-26}\\\\ \\frac {-52}{-26}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ 2\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = 2 Ans Here, Given equations are: x = $$\\frac 23$$y i.e. 3x - 2y = 0................(1) 4x - 3y = 1............................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -2\\\\ 4 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 0\\\\ 1\\\\ \\end {pmatrix}$$, AX =", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "1 + 1 \\cdot 3 + 3 \\cdot 5 = 17$$ The $2,2$ entry of $\\bfA\\bfB$ is the product of the second row of $\\bfA$ with the second column of $\\bfB$: $$(\\bfA \\bfB)_{22} = \\begin{pmatrix} -1 & 1 & 3 \\end{pmatrix} \\begin{pmatrix}2\\\\4\\\\6\\end{pmatrix} = -1\\cdot 2 + 1 \\cdot 4 + 3 \\cdot 6 = 20$$ Hence, $$\\begin{pmatrix}1 & 0 & 2 \\\\ -1 & 1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \\end{pmatrix} = \\begin{pmatrix}11 & 14 \\\\ 17 & 20 \\end{pmatrix}.$$", "$1212$ twice, $121212$ thrice and so on. By inclusion-exclusion principle, No. of numbers w/o $12$ = Total no. of numbers - No. of numbers with $1$2 at least once + No. of numbers with $12$ twice - No. of numbers with $12$ thrice + No. of numbers with $12$ four times and so on. Total no: $10,000,000$ • $12$ at least once: $6*10^5$ • $12$ at least twice: $(4+3+2+1)*10^3$ • $12$ at least thrice: $((2+1)+(1))*10^1$ • $12$ four times and so on $=0$ • so ans= $10000000-600000+10000-40=9409960$ - Welcome to MSE! It really helps to format questions/answers using MathJax (see FAQ). Regards – Amzoti May 11 '13 at 16:36 Keep track of the amount of $n$-digit numbers ending in $1$ and not ending in $1$, starting with $n=1$ $$v_1 = \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix}$$ so $1+8 = 9$ numbers in total. The vector for $(n+1)$-digit numbers ending in $1$ and not ending in $1$ that do not contain the sequence \"$12$\" can be obtained by a recursion: $$v_{n+1} = \\begin{pmatrix} 1 & 1 \\\\ 8 & 9 \\end{pmatrix} v_n.$$ If $M$ is the matrix in the recursion above then there are $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}(M^{n-1} + M^{n-2} + \\dotsc + 1) \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix}$$ numbers of at most $n$ digits that do not contain", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "$$x(t) = \\mathrm e^{2t}\\left(\\int \\mathrm e^{-2t}(f'(t,c_1,c_2)-f(t,c_1,c_2))dt+c_3\\right)$$ and that is where the third constant appears. - So, I put my solution for y $y=C_1e^tcos(t)+C_2e^tsin(t)$ into $z = x + 2y - y'$? Doing that doesn't produce a third constant. – aarnes Feb 18 '12 at 21:44 @aarnes: it does since you have this constant in $x$ – Ilya Feb 19 '12 at 10:11 I think you are saying that solving for $y$ only requires $2$ constants, but to solve for all $3$ unknowns, requires the $3$ constants expected. Is that so? – robjohn Feb 19 '12 at 10:19 @robjohn: yes, that is what I meant – Ilya Feb 19 '12 at 10:33 First of, let me say that I'm sorry for responding so late but I'm probably in a different time zone and last night was late and I slept for quite a while. What I still don't understand is how do I get a solution for x then in order to get that third constant? Putting solution for y (with two constants) inside $z = x + 2y - y'$ gives $z = C_1e^tcos(t) + C_1e^tsin(t) + C_2e^tsin(t) - C_2e^tcos(t) + x(t)$ – aarnes Feb 19 '12 at 18:39 Note that since $$D\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}=\\begin{bmatrix}3&1&-1\\\\1&2&-1\\\\3&3&-1\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}\\tag{1}$$ we have the general solution $$\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}=\\exp\\left(\\begin{bmatrix}3&1&-1\\\\1&2&-1\\\\3&3&-1\\end{bmatrix}t\\right)\\begin{bmatrix}x_0\\\\y_0\\\\z_0\\end{bmatrix}\\tag{2}$$ Computing the Jordan Normal Form, we get \\begin{align} &" ]

[ "Browse Questions # For what value of $k$ , the matrix $\\begin{bmatrix} 2-k & 3 \\\\ -5 & 1 \\end{bmatrix}$ is not invertible? Can you answer this question?", "# Show that the $5\\times 5$ matrix is not invertable Find $a$ such that the matrix $$\\begin{pmatrix} 0 & a & 0 & 0 & 0 \\\\ b & 0 & c & 0 & 0 \\\\ 0 & d & 0 & e & 0 \\\\ 0 & 0 & f & 0 & g \\\\ 0 & 0 & 0 & h & 0 \\end{pmatrix}$$ is not invertable, for all values of $a,b,x,d,e,f,g,h$.", "Free Version Moderate # 2x2 Matrix Invertible with Parameter 1 LINALG-SLNB7V For which of the following values of $b$ is the matrix: $$\\begin{pmatrix}\\sqrt{3}&b\\\\\\ b+2&\\sqrt{3}\\end{pmatrix}$$ ...NOT invertible? Select ALL that apply. A $-3$ B $-2$ C $0$ D $1$", "Free Version Moderate # 2x2 Matrix Invertible with Parameter 2 LINALG-CDVRSM Pick the real number value of $b$, if it exists, such that the matrix: $$\\begin{pmatrix}b&1\\\\\\ -1&b+1\\end{pmatrix}$$ ...is not invertible. A $-1$ B $0$ C $1$ D $2$ E This matrix is invertible for any real value of $b$.", "# Find $x$ so that $\\begin{bmatrix} 2 & 0 & 10 \\\\ 0 & x+7 & -3 \\\\ 0 & 4 & x \\end{bmatrix}$ is invertible Find the value of $x$ so that the matrix $$\\begin{bmatrix} 2 & 0 & 10 \\\\ 0 & x+7 & -3 \\\\ 0 & 4 & x \\end{bmatrix}$$ is invertible.", "$$\\pmatrix{2 & 1 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 3}$$ Note that in all these cases the matrix $A^2 -7A+16I_3$ is invertible because the polynomial $x^2-7x+16$ has complex zeros, so $\\frac1{12}A=[A^2-7A+16I_3]^{-1}$ is indeed equivalent to $A^3-7A^2+16A-12I_3=0$.", "# Find the values of α such that the following matrix is invertible I have the following matrix $$\\begin{pmatrix} 1 &-1 &α-1\\\\1 &1 &1\\\\ 2α & 2 & 4 \\end{pmatrix}$$ I only have to set the determinant of this matrix to $$0$$. Then find values of $$α$$ when the determinant equals to $$0$$, and the values of $$\\alpha$$ are correct? • The matrix is invertible when the determinant is not zero. So yes, find the values of $a$ for which the determinant is zero and that will give you the values for which it is not zero. – Mark Oct 11 at 12:14 • There are several questions of this type already here on this site, e,g, here. Please look for duplicates before posting. – Dietrich Burde Oct 11 at 12:40", "careful: the matrices $$\\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix}$$ are not invertible, hence are not elements of $GL_2(\\mathbb R)$. However, the set of matrices of this type $$\\left\\{ \\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix} : \\ a \\in \\mathbb R, \\ a\\ne 0 \\right\\}$$ is an Abelian group under matrix multiplication. Its identity is not the usual identity $I_2$...", "$a\\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if... Close", "also $X=XA^{-1}$. These facts immediately imply that $G$ is a subgroup of the group of invertible $2\\times2$-matrices. In a later course you may learn about group actions and stabilizers. Then you can reinterpret this as stating that $G$ is the stabilizer of the vector $X$, when the group of invertible matrices acts on the (row) vectors by matrix muliplication. The \"easier\" way of doing this is to note the condition on $A$ is equivalent to requiring $A$ to be invertible and satisfy: $$\\begin{pmatrix}1&1\\end{pmatrix}A = \\begin{pmatrix}1&1\\end{pmatrix}$$ Then if $A,B$ satisfy this constraint, show $AB$ satisfies this constraint and that $A^{-1}$ satisfies this constraint. For example, if $A$ is invertible, and $\\begin{pmatrix}1&1\\end{pmatrix}A = \\begin{pmatrix}1&1\\end{pmatrix}$ then $$\\begin{pmatrix}1&1\\end{pmatrix}A^{-1} = \\left(\\begin{pmatrix}1&1\\end{pmatrix}A\\right)A^{-1} = \\begin{pmatrix}1&1\\end{pmatrix} I = \\begin{pmatrix}1&1\\end{pmatrix}$$ So $A^{-1}$ is in our set.", "For Which Choices of $x$ is the Given Matrix Invertible? Determine the values of $x$ so that the matrix \\[A=\\begin{bmatrix} 1 & 1 & x \\\\ 1 &x &x \\\\...", "matrix is $0$ and so the matrix is not invertible.", "4 \\\\ 3 \\end{pmatrix}.$ ×", "0 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 2\\end{pmatrix}\\begin{pmatrix} \\varphi_1 \\\\ \\varphi_2 \\\\ \\varphi_3\\end{pmatrix} = \\begin{pmatrix} 4 & 3 & 1\\end{pmatrix}^T.$$ How do I proceed? I’m used to invert matrices in «front of» the variable, not in the middle of the equation. A: The given matrix $A$ is invertible. Therefore, there exists a unique matrix $P$ such that $A P=I$. We can easily show this by computing the null", "lennox Feb 21 at 17:25 • Yes, it is, at least for $k=\\pm 1$. – Dietrich Burde Feb 21 at 17:26 Take the matrix $$M =\\begin{pmatrix} 1 & 1 & 1 \\\\ -1 & 0 & 1 \\\\ 0 & -1 & 1 \\end{pmatrix}$$ It is clearly of full rank over $$\\mathbb{Z}$$ but the transformed is not. We also have that $$\\det(M) = 3$$ hence it is not invertible $$\\pmod{3}$$.", "pivot point in your augmented matrix, since the $(2,2)$ entry is nonzero. For any $h$ the system will be \\begin{split} x&-y&=4\\\\ -2x&+3y&=h \\end{split} or equivalently $$Ax=b$$ with $$A=\\begin{pmatrix}1 &-1 \\\\-2 & 3\\end{pmatrix}, \\ \\ b=\\begin{pmatrix}4\\\\h\\end{pmatrix}.$$ Since $A$ is invertible the system has a solution for any $h$ (that is $A^{-1}b$).", "matrix $\\begin{bmatrix} 1 & -1 & 0 \\\\ 0 &1 &-1 \\\\ 0 & 0 & 1 \\end{bmatrix}.$ Is the matrix $A$ invertible? If not, then explain why it isn’t invertible. If so, then find the inverse. (The Ohio State University Linear Algebra […]", "'18 at 17:03 • These matrices are called HANKEL matrices. The yours are invertible, and if $A_{2×2}$ is Hankel then $A^{-1}$ is Hankel. Unfortunately, as already stated, this set is not closed under multiplication. – user376343 Sep 17 '18 at 17:23 ## 1 Answer Let $A=\\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\end{pmatrix}$ and $B=\\begin{pmatrix} 1 & 1 \\\\ 1 & 0\\end{pmatrix}$ Then $AB = \\begin{pmatrix} 2 & 2 \\\\ 1 & 0\\end{pmatrix}$. Clearly, $ab_{12}\\neq ab_{21}$. So the set is not closed under multiplication", "# Invertible Matrix ## Invertible Matrix A square matrix A of order n is invertible if there exists a square matrix B of order n such that AB = BA = I, an identity matrix. ### Example Consider the matrices $$A = \\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1 \\end{pmatrix}$$ and $$B = \\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$. The matrix A is invertible, since : $$\\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1\\end{pmatrix}$$ × $$\\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$ = $$\\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$ × $$\\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1\\end{pmatrix} = \\begin{pmatrix}1 & 0 & 0\\\\0 & 1 & 0\\\\ 0 & 0 & 1\\end{pmatrix}$$", "$i$ and $j$. Then we can take $A=I_m$ and $B=I_n$ (and r=0). • $a_{i_0 \\ j_0}\\neq 0$ for some $i_0,j_0$. Remember that doing elementary changes on our matrix $C$ is the same as multiplying $C$ by an elementary matrix (if you are not familiar with this you can check this article on wikipedia). For exemple, if $C'$ is the $3\\times 3$ matrix you obtain after switching rows 1 and 2 of the matrix $C$ then $$C'= \\begin{pmatrix} 0 & 1 & 0\\\\ 1 & 0 &0\\\\ 0&0&1 \\end{pmatrix} C$$ Using this property we can move $a_{i_0 \\ j_0}$ to $a_{11}$ by multiplying $C$ with invertible matrices, let's call them $A_0$ and $B_0$. You can also obtain a matrix of the form (multiplying $C$ by invertible matrices $A_1$ and $B_1$): $$\\begin{pmatrix} 1 & 0 & \\cdots & 0\\\\ 0 & & \\\\ \\vdots & & C^* \\\\ 0 & \\\\ \\end{pmatrix}$$ And we have $$\\begin{pmatrix} 1 & 0 & \\cdots & 0\\\\ 0 & & \\\\ \\vdots & & C^* \\\\ 0 & \\\\ \\end{pmatrix}=A_1\\cdot A_0 \\cdot C\\cdot B_0\\cdot B_1.$$ $A_1\\cdot A_0$ is invertible (because both $A_1$ and $A_0$ are invertible) and $B_0\\cdot B_1$ is also invertible. $C^*$ is a $(m-1)\\times(n-1)$ matrix and due to our induction hypothesis there are invertibles matrices $A^*$ and $B^*$ such that $$A^* C^*" ]

[ "matrix B) $A$is orthogonal matrix C) Determinant$A\\equiv 1$ D) A is not invertible • question_answer170) $R$is a relation from {11,12,13} to {8,10,12} defined by$y=x-3$. Then${{R}^{-1}}$is A) {(8,11), (10,13)} B) {(11,18), (13,10)} C) {(10,13), (8,11)} D) None of the above", "matrix is $0$ and so the matrix is not invertible.", "calculator, and check that the above is true. 3. A matrix is invertible if and only if its determinant is NOT equal to zero. Use the calculator to determine which of the following matrices are invertible. a) $\\begin{bmatrix} 2 & 1 & -1 \\\\ 3 & 1 & 0 \\\\ -1 & 0 & 2 \\end{bmatrix}$ b) $\\begin{bmatrix} -5 & 0 & -1 & 7\\\\ 2 & 2 & 0 & 9 \\\\ -6 & 4 & -2 & 32 \\\\ -8 & 0 & 2 & 1 \\end{bmatrix}$ c) $\\begin{bmatrix} -5 & 1 & -1 & 7 & 5\\\\ -4 & 1 & 0 & 9 &-2 \\\\ -1 & 2 & -2 & 7 & 0\\\\ -8 & 0 & 2 & 1 & -1\\\\ -9 & 2 & -1 & 16 & 3 \\end{bmatrix}$ d) $\\begin{bmatrix} -15 & 12 & -1 & 7 & 5 & 8\\\\ -4 & 11 & 0 & 9 &-2 & 5\\\\ 2 & 2 & -2 & 7 & 0 & 2\\\\ -8 & -2 & 12 & 1 & -1 & -3\\\\ -7 & 2 & -1 & 0 & 3 & 2 \\\\ 0 & 4 & -4 & -1 & 3 & -3 \\end{bmatrix}$", "A matrix is invertible if and only if its determinant is non-zero. If A had determinant 0, then $det(A^2)= det(A)det(A)= 0$. but $A^2= I$ which has determinant 1, not 0. Therefore det(A) is not 0. Same thing for B. Then det(AB)= det(A)det(B) is non-zero. I chose those four matrices because they are the \"standard basis\" for the vector space of 2 by 2 matrices. Every such matrix can be written as a linear combination of them: $\\begin{barray}a & b \\\\ c & d\\end{bmatrix}= a\\begin{bmatrix}1 & 0 \\\\ 0 & 0\\end{bmatrix}+ b\\begin{bmatrix}0 & 1 \\\\ 0 & 0\\end{bmatrix}+ c\\begin{bmatrix}0 & 0 \\\\ 1 & 0\\end{bmatrix}+ d\\begin{bmatrix}0 & 0 \\\\ 0 & 1\\end{bmatrix}$ so they \"represent\" all 2 by 2 matrices.", "## Linear Algebra: A Modern Introduction We know that if the determinant of a matrix $\\begin{bmatrix} a &b \\\\ c & d \\end{bmatrix}$ $, D=ad-bc$ is $0$, then the matrix is not invertible. Here $D=3.55\\cdot0.6-0.25\\cdot8.52=2.13-2.13=0$, thus the matrix is not invertible.", "Browse Questions # For what value of $k$ , the matrix $\\begin{bmatrix} 2-k & 3 \\\\ -5 & 1 \\end{bmatrix}$ is not invertible? Can you answer this question?", "in a matrix, its determinant will always be 0 and it will be invertible. The same goes for columns, if any column is a constant multiple of another column, the determinant is going to be 0 and the matrix is not invertible.", "we have: k = |A| = $$\\rm \\frac{1}{|A^{-1}|}$$ ----(3) We know, A.A-1 = I ⇒ |A.A-1| = |I| = 1 ⇒ |A| |A-1| = 1 ⇒ |A| = 1/ |A-1| ....(4) Now, |A-1| = 1(24 - 3) + 2(9 - 12) + 3(2 - 12) = 21 - 6 - 30 = - 15. |A-1| = -15 Therefore, from equation (3) k = $$\\rm - \\frac1{15}$$. Mistake PointsNote that, we have A-1 matrix, not an A matrix. So to find the value of k, don't you have to use relation |A| = 1/|A-1| #### Adjoint and Inverse of a Square Matrix Question 9 If $$\\begin{bmatrix} 1 & -3 & 2 \\\\\\ 2 & -8 & 5 \\\\\\ 4 & 2 & λ \\end{bmatrix}$$ is not an invertible matrix, then what is the value of λ ? 1. -1 2. 0 3. 1 4. 2 Option 3 : 1 #### Adjoint and Inverse of a Square Matrix Question 9 Detailed Solution Concept: If the matrix A is not an invertible matrix then | A | = 0 If the matrix A is the non-singular matrix then | A | 0 Calculations: Given, A = $$\\begin{bmatrix} 1 & -3 & 2 \\\\\\ 2 & -8 & 5 \\\\\\ 4 & 2 & λ \\end{bmatrix}$$is not an invertible matrix As we", "Finding a Matrix Inverse with its Determinant and Adjoint Theorem: If $A$ is an invertible matrix, that is $\\det (A) ≠ 0$, then it follows that $A^{-1} = \\frac{1}{\\det(A)} \\mathrm{adj}(A)$.", "# [Math] Is not full rank matrix invertible determinantlinear algebralinear-transformationsmatricesmatrix-rank ## Problem $$A$$ is a $$4 \\times 4$$ matrix. It is known that $$\\text{rank}(A)=3$$. Is matrix A invertible ? ## Attempt to solve $$\\text{rank(A)}=3 \\implies \\det(A)=0$$ which implies matrix is $$\\textbf{not}$$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $$0$$ and that some information is lost in this linear transformation. Is my intuition behind this correct ? The following properties are equivalent for a square matrix $$A$$: • $$A$$ has full rank • $$A$$ is invertible • the determinant of $$A$$ is non-zero", "this up? We get 20 plus 16 minus 1. It is equal to 35. We're done. We found the determinant of our 3 by 3 matrix. Not too bad. Right there, so that is equal to the determinant of C. So the fact that this isn't 0 tells you that C is invertible. In the next video, we'll try to extend this to n by n square matrices.", "& 0 & 0\\\\ -1 & 0 & 0 \\\\ 12 & -2 & 3 \\end{bmatrix}$. Which of the following matrices $A , \\; B ,\\; C ,\\; A B ,\\; A C,\\; A + C$ is invertible? Solution A matrix with a determinant not equal to zero in invertible. $Det(A) = (3)(-2)(4) = -24$ , hence matrix $A$ is invertible. $Det(B) = (-2)(-2)(-1) = -4$ , hence matrix $B$ is invertible. $Det(C) = (7)(0)(3) = 0$ , hence matrix $C$ is NOT invertible. The determinant of the product of two matrices is equal to the product of their determinants $Det(A B) = Det(A) Det(B) = (-24)(-4) = 96$ , hence matrix $A B$ is invertible. $Det(A C) = Det(A) Det(C) = (-24)(0) = 0$ , hence matrix $A C$ is NOT invertible. $Det(A+C) = Det \\left(\\begin{bmatrix} 3 & 0 & 0\\\\ 5 & -2 & 0 \\\\ 9 & -2 & 4 \\end{bmatrix} + \\begin{bmatrix} 7 & 0 & 0\\\\ -1 & 0 & 0 \\\\ 12 & -2 & 3 \\end{bmatrix} \\right)$ $\\quad \\quad = Det \\begin{bmatrix} 10 & 0 & 0\\\\ 4 & -2 & 0 \\\\ 21 & -4 & 7 \\end{bmatrix} = (7) (-2) (7) = -98$ , hence matrix $A + C$ is invertible. ## Questions (with solutions given below) • Part", "the set of 2 X 2 matrices with the entries in Z/6Z 2x2 (a) Can you find a matrix M(Z/6Z) whose determinant is non-zero and yet is 2x2 not invertible? This entry was posted in Uncategorized. Bookmark the permalink.", "# Recent questions tagged invertible-matrices ### The number of $3\\times 3$ matrices A whose entries are either 0 or 1 and for which the system $A=\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}=\\begin{bmatrix}1\\\\0\\\\0\\end{bmatrix}$ has exactly two distinct solution is To see more, click for the full list of questions or popular tags.", "# Solution May 23, 2012 The determinant of such a matrix is of the form u + m where $u \\in A - \\mathfrak{m}$ and $m \\in \\mathfrak{m}$, which is invertible.", "the following matrix invertible? \\[A=\\begin{bmatrix} 1 & 1 & 1 \\\\ 1...", "+0 # Matrix 0 61 2 +55 When is a Matrix Invertible?? May 10, 2021 #1 +556 +2 Since $\\mathbb{A}^{-1}=\\frac{1}{det\\mathbb{A}}adj(\\mathbb{A})$, so its invertible whenever the determinant is nonzero. May 10, 2021 #1 +556 +2 Since $\\mathbb{A}^{-1}=\\frac{1}{det\\mathbb{A}}adj(\\mathbb{A})$, so its invertible whenever the determinant is nonzero.", "\"=0\" is not possible since V*V is invertible. Therefor wrt the basis e_i, the matrix of V*V is diagonal with all nonpositive diagonal entries, so it's positive definite.", "# What are the implications of matrix invertibility? Feb 6, 2017 See below for rough outline. #### Explanation: If an nxn matrix is invertible, then the big-picture consequence is that its column and row vectors are linearly independent. It is also (always) true to say that if an nxn matrix is invertible: • (1) its determinant is non-zero, • (2) $m a t h b f x = m a t h b f 0$ is the only solution to $A m a t h b f x = m a t h b f 0$, • (3) $m a t h b f x = {A}^{- 1} m a t h b f b$ is the only solution to $A m a t h b f x = m a t h b f b$, and • (4) it's eigenvalues are non-zero. A singular (non-invertible) matrix has at last one zero eigenvalue. But there is no guarantee that an invertible matrix can be diagonalised or vice versa. Diagonalisation will only happen when a matrix delivers up a full set of eigenvectors (which can occur where an eigenvalue is zero).", "obviously has determinant 0, it is not invertible, which means it has either no solution or an infinite number of solutions. Because the system is homogeneous, Ax= 0, it has the obvious solution, x= 0, which means \"no solution\" cannot apply. We are left with an infinite number of solutions." ]

[ "$$\\pmatrix{2 & 1 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 3}$$ Note that in all these cases the matrix $A^2 -7A+16I_3$ is invertible because the polynomial $x^2-7x+16$ has complex zeros, so $\\frac1{12}A=[A^2-7A+16I_3]^{-1}$ is indeed equivalent to $A^3-7A^2+16A-12I_3=0$.", "Browse Questions # For what value of $k$ , the matrix $\\begin{bmatrix} 2-k & 3 \\\\ -5 & 1 \\end{bmatrix}$ is not invertible? Can you answer this question?", "# 8.1: The Determinant Formula The determinant extracts a single number from a matrix that determines whether its invertibility. Lets see how this works for small matrices first. ## 8.1.1 Simple Examples For small cases, we already know when a matrix is invertible. If $$M$$ is a $$1\\times 1$$ matrix, then $$M=(m) \\Rightarrow M^{-1}=(1/m)$$. Then $$M$$ is invertible if and only if $$m\\neq 0$$. For $$M$$ a $$2\\times 2$$ matrix, chapter 7 section 7.5 shows that if $$M=\\begin{pmatrix} m^{1}_{1} & m^{1}_{2} \\\\ m^{2}_{1} & m^{2}_{2} \\\\ \\end{pmatrix}\\, ,$$ then $$M^{-1}=\\frac{1}{m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}}\\begin{pmatrix} m^{2}_{2} & -m^{1}_{2} \\\\ -m^{2}_{1} & m^{1}_{1} \\\\ \\end{pmatrix}\\, .$$ Thus $$M$$ is invertible if and only if $m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}\\neq 0\\, . For $$2\\times 2$$ matrices, this quantity is called the $$\\textit{determinant of \\(M$$}\\). \\[ \\det M = \\det \\begin{pmatrix} m^{1}_{1} & m^{1}_{2} \\\\ m^{2}_{1} & m^{2}_{2} \\\\ \\end{pmatrix} = m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}\\, .$ Example 92 For a $$3\\times 3$$ matrix, $$M=\\begin{pmatrix} m^{1}_{1} & m^{1}_{2} & m^{1}_{3}\\\\ m^{2}_{1} & m^{2}_{2} & m^{2}_{3}\\\\ m^{3}_{1} & m^{3}_{2} & m^{3}_{3}\\\\ \\end{pmatrix}\\, ,$$ then---see review question 1---$$M$$ is non-singular if and only if $\\det M = m^{1}_{1}m^{2}_{2}m^{3}_{3} - m^{1}_{1}m^{2}_{3}m^{3}_{2} + m^{1}_{2}m^{2}_{3}m^{3}_{1} - m^{1}_{2}m^{2}_{1}m^{3}_{3} + m^{1}_{3}m^{2}_{1}m^{3}_{2} - m^{1}_{3}m^{2}_{2}m^{3}_{1} \\neq 0. Notice that in the subscripts, each ordering of the numbers $$1$$, $$2$$, and $$3$$ occurs exactly once. Each of these is a $$\\textit{permutation}$$ of the set", "# Show that the $5\\times 5$ matrix is not invertable Find $a$ such that the matrix $$\\begin{pmatrix} 0 & a & 0 & 0 & 0 \\\\ b & 0 & c & 0 & 0 \\\\ 0 & d & 0 & e & 0 \\\\ 0 & 0 & f & 0 & g \\\\ 0 & 0 & 0 & h & 0 \\end{pmatrix}$$ is not invertable, for all values of $a,b,x,d,e,f,g,h$.", "# Find the values of α such that the following matrix is invertible I have the following matrix $$\\begin{pmatrix} 1 &-1 &α-1\\\\1 &1 &1\\\\ 2α & 2 & 4 \\end{pmatrix}$$ I only have to set the determinant of this matrix to $$0$$. Then find values of $$α$$ when the determinant equals to $$0$$, and the values of $$\\alpha$$ are correct? • The matrix is invertible when the determinant is not zero. So yes, find the values of $a$ for which the determinant is zero and that will give you the values for which it is not zero. – Mark Oct 11 at 12:14 • There are several questions of this type already here on this site, e,g, here. Please look for duplicates before posting. – Dietrich Burde Oct 11 at 12:40", "# Find $x$ so that $\\begin{bmatrix} 2 & 0 & 10 \\\\ 0 & x+7 & -3 \\\\ 0 & 4 & x \\end{bmatrix}$ is invertible Find the value of $x$ so that the matrix $$\\begin{bmatrix} 2 & 0 & 10 \\\\ 0 & x+7 & -3 \\\\ 0 & 4 & x \\end{bmatrix}$$ is invertible.", "Free Version Moderate # 2x2 Matrix Invertible with Parameter 2 LINALG-CDVRSM Pick the real number value of $b$, if it exists, such that the matrix: $$\\begin{pmatrix}b&1\\\\\\ -1&b+1\\end{pmatrix}$$ ...is not invertible. A $-1$ B $0$ C $1$ D $2$ E This matrix is invertible for any real value of $b$.", "careful: the matrices $$\\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix}$$ are not invertible, hence are not elements of $GL_2(\\mathbb R)$. However, the set of matrices of this type $$\\left\\{ \\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix} : \\ a \\in \\mathbb R, \\ a\\ne 0 \\right\\}$$ is an Abelian group under matrix multiplication. Its identity is not the usual identity $I_2$...", "Free Version Moderate # 2x2 Matrix Invertible with Parameter 1 LINALG-SLNB7V For which of the following values of $b$ is the matrix: $$\\begin{pmatrix}\\sqrt{3}&b\\\\\\ b+2&\\sqrt{3}\\end{pmatrix}$$ ...NOT invertible? Select ALL that apply. A $-3$ B $-2$ C $0$ D $1$", "0 \\\\ 1 & 2\\end{pmatrix}$ (written under the standard choices of bases). This matrix has determinant $4,$ which is non-zero, so we see that $T_{2}$ must therefore be invertible, and hence injective. I'll leave it to you to check that its kernel is trivial, as the theorem says it should be.", "matrix is $0$ and so the matrix is not invertible.", "# 8.1: The Determinant Formula The determinant extracts a single number from a matrix that determines whether its invertibility. Lets see how this works for small matrices first. ## 8.1.1 Simple Examples For small cases, we already know when a matrix is invertible. If $$M$$ is a $$1\\times 1$$ matrix, then $$M=(m) \\Rightarrow M^{-1}=(1/m)$$. Then $$M$$ is invertible if and only if $$m\\neq 0$$. For $$M$$ a $$2\\times 2$$ matrix, chapter 7 section 7.5 shows that if $$M=\\begin{pmatrix} m^{1}_{1} & m^{1}_{2} \\\\ m^{2}_{1} & m^{2}_{2} \\\\ \\end{pmatrix}\\, ,$$ then $$M^{-1}=\\frac{1}{m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}}\\begin{pmatrix} m^{2}_{2} & -m^{1}_{2} \\\\ -m^{2}_{1} & m^{1}_{1} \\\\ \\end{pmatrix}\\, .$$ Thus $$M$$ is invertible if and only if $$m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}\\neq 0\\, .$$ For $$2\\times 2$$ matrices, this quantity is called the $$\\textit{determinant of \\(M$$}\\). $\\det M = \\det \\begin{pmatrix} m^{1}_{1} & m^{1}_{2} \\\\ m^{2}_{1} & m^{2}_{2} \\\\ \\end{pmatrix} = m^{1}_{1}m^{2}_{2}-m^{1}_{2}m^{2}_{1}\\, .$ Example 92 For a $$3\\times 3$$ matrix, $$M=\\begin{pmatrix} m^{1}_{1} & m^{1}_{2} & m^{1}_{3}\\\\ m^{2}_{1} & m^{2}_{2} & m^{2}_{3}\\\\ m^{3}_{1} & m^{3}_{2} & m^{3}_{3}\\\\ \\end{pmatrix}\\, ,$$ then---see review question 1---$$M$$ is non-singular if and only if $$\\det M = m^{1}_{1}m^{2}_{2}m^{3}_{3} - m^{1}_{1}m^{2}_{3}m^{3}_{2} + m^{1}_{2}m^{2}_{3}m^{3}_{1} - m^{1}_{2}m^{2}_{1}m^{3}_{3} + m^{1}_{3}m^{2}_{1}m^{3}_{2} - m^{1}_{3}m^{2}_{2}m^{3}_{1} \\neq 0.$$ Notice that in the subscripts, each ordering of the numbers $$1$$, $$2$$, and $$3$$ occurs exactly once. Each of these is a $$\\textit{permutation}$$ of the set $$\\{1,2,3\\}$$.", "matrix M is invertible if and only if the homogeneous matrix equation Mx=0 does not have any non-trivial solutions. There is a special name for this column, which is basic solution. Watch the recordings here on Youtube! Then, there is a pivot position in every column of the coefficient matrix of $$A$$. Theorem. A linear equation is said to be homogeneous when its constant part is zero. { ( 0 4 0 0 0 ) ⏟ particular solution + w ( 1 − 1 3 1 0 ) + u ( 1 / 2 − 1 1 / 2 0 1 ) ⏟ unrestricted combination | w , u ∈ R } {\\displaystyle \\left\\{\\underbrace {\\begin{pmatrix}0\\\\4\\\\0\\\\0\\\\0\\end{pmatrix}} _{\\begin{array}{c}\\$-19pt]\\scriptstyle {\\text{particular}}\\\\[-5pt]\\s… Therefore by our previous discussion, we expect this system to have infinitely many solutions. © 2021 SOPHIA Learning, LLC. Suppose the system is consistent, whether it is homogeneous or not. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. \\[\\begin{array}{c} x + 4y + 3z = 0 \\\\ 3x + 12y + 9z = 0 \\end{array}$ Find the basic solutions to this system. Example $$\\PageIndex{1}$$: Finding the Rank of a Matrix. If the system has a solution in which not all of the $$x_1, \\cdots, x_n$$ are equal to zero, then we call this solution", "of B is \\[\\Adj(B)=C^{\\trans}=\\begin{bmatrix} 1 & 0 & -2 \\\\ 12 &-5 &-4 \\\\ -3 & 0 & 1 \\end{bmatrix}. It follows from the formula that the inverse matrix of $B$ is $B^{-1}=\\frac{1}{\\det(B)}\\Adj(B)=\\frac{1}{5}\\begin{bmatrix} -1 & 0 & 2 \\\\ -12 &5 &4 \\\\ 3 & 0 & -1 \\end{bmatrix}.$ ### More from my site • For Which Choices of $x$ is the Given Matrix Invertible? Determine the values of $x$ so that the matrix $A=\\begin{bmatrix} 1 & 1 & x \\\\ 1 &x &x \\\\ x & x & x \\end{bmatrix}$ is invertible. For those values of $x$, find the inverse matrix $A^{-1}$. Solution. We use the fact that a matrix is invertible […] • Inverse Matrix Contains Only Integers if and only if the Determinant is $\\pm 1$ Let $A$ be an $n\\times n$ nonsingular matrix with integer entries. Prove that the inverse matrix $A^{-1}$ contains only integer entries if and only if $\\det(A)=\\pm 1$. Hint. If $B$ is a square matrix whose entries are integers, then the […] • Find All Values of $x$ such that the Matrix is Invertible Given any constants $a,b,c$ where $a\\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if $A= \\begin{bmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -1/a &", "calculator, and check that the above is true. 3. A matrix is invertible if and only if its determinant is NOT equal to zero. Use the calculator to determine which of the following matrices are invertible. a) $\\begin{bmatrix} 2 & 1 & -1 \\\\ 3 & 1 & 0 \\\\ -1 & 0 & 2 \\end{bmatrix}$ b) $\\begin{bmatrix} -5 & 0 & -1 & 7\\\\ 2 & 2 & 0 & 9 \\\\ -6 & 4 & -2 & 32 \\\\ -8 & 0 & 2 & 1 \\end{bmatrix}$ c) $\\begin{bmatrix} -5 & 1 & -1 & 7 & 5\\\\ -4 & 1 & 0 & 9 &-2 \\\\ -1 & 2 & -2 & 7 & 0\\\\ -8 & 0 & 2 & 1 & -1\\\\ -9 & 2 & -1 & 16 & 3 \\end{bmatrix}$ d) $\\begin{bmatrix} -15 & 12 & -1 & 7 & 5 & 8\\\\ -4 & 11 & 0 & 9 &-2 & 5\\\\ 2 & 2 & -2 & 7 & 0 & 2\\\\ -8 & -2 & 12 & 1 & -1 & -3\\\\ -7 & 2 & -1 & 0 & 3 & 2 \\\\ 0 & 4 & -4 & -1 & 3 & -3 \\end{bmatrix}$", "we have: k = |A| = $$\\rm \\frac{1}{|A^{-1}|}$$ ----(3) We know, A.A-1 = I ⇒ |A.A-1| = |I| = 1 ⇒ |A| |A-1| = 1 ⇒ |A| = 1/ |A-1| ....(4) Now, |A-1| = 1(24 - 3) + 2(9 - 12) + 3(2 - 12) = 21 - 6 - 30 = - 15. |A-1| = -15 Therefore, from equation (3) k = $$\\rm - \\frac1{15}$$. Mistake PointsNote that, we have A-1 matrix, not an A matrix. So to find the value of k, don't you have to use relation |A| = 1/|A-1| #### Adjoint and Inverse of a Square Matrix Question 9 If $$\\begin{bmatrix} 1 & -3 & 2 \\\\\\ 2 & -8 & 5 \\\\\\ 4 & 2 & λ \\end{bmatrix}$$ is not an invertible matrix, then what is the value of λ ? 1. -1 2. 0 3. 1 4. 2 Option 3 : 1 #### Adjoint and Inverse of a Square Matrix Question 9 Detailed Solution Concept: If the matrix A is not an invertible matrix then | A | = 0 If the matrix A is the non-singular matrix then | A | 0 Calculations: Given, A = $$\\begin{bmatrix} 1 & -3 & 2 \\\\\\ 2 & -8 & 5 \\\\\\ 4 & 2 & λ \\end{bmatrix}$$is not an invertible matrix As we", "## Linear Algebra: A Modern Introduction We know that if the determinant of a matrix $\\begin{bmatrix} a &b \\\\ c & d \\end{bmatrix}$ $, D=ad-bc$ is $0$, then the matrix is not invertible. Here $D=3.55\\cdot0.6-0.25\\cdot8.52=2.13-2.13=0$, thus the matrix is not invertible.", "Bert's blog ## Numerically inverting a matrix Suppose you have a square matrix, like the one below: $$A = \\begin{pmatrix} 0 & 5 & 5 \\\\ 2 & 9 & 0 \\\\ 6 & 8 & 8 \\end{pmatrix},$$ and that you want to compute the inverse of this matrix (numerically). The result should be (thanks WolframAlpha!): $$A^{-1} = \\begin{pmatrix} -\\frac{4}{15} & 0 & \\frac{1}{6} \\\\ \\frac{8}{135} & \\frac{1}{9} & -\\frac{1}{27} \\\\ \\frac{19}{135} & -\\frac{1}{9} & \\frac{1}{27} \\end{pmatrix}$$ A first option to invert this matrix is to realise that this is a $3\\times{}3$ matrix, for which Wikipedia lists a special formula. The problem of this approach is that it is not very general. More generally, Wikipedia lists straightforward inversion algorithms that can be readily executed for matrices of any size (they obviously need to be square to be invertible). These involve a lot of operations, and require a lot of intermediate variables to be stored. Furthermore, these algorithms are not very stable against round off error; if the matrix contains elements with significantly different sizes, then accumulated round off error can lead to a significant loss of accuracy, or to numerical issues that cause the inversion algorithm to fail altogether, even if the matrix is strictly speaking invertible. Fortunately, there is an elegant matrix inversion algorithm that does not", "Browse Questions # Solve system of linear equations, using matrix method: $4x-3y=3$ $3x-5y=7$ $\\begin{array}{1 1} X=-6/11,Y=-19/11 \\\\X=12/11,Y=19/11 \\\\ X=-6/11,Y=19/11 \\\\ X=-12/11,Y=-19/11 \\end{array}$ Toolbox: • A matrix is said to be singular if |A|= 0. • A matrix is said to be invertible if |A|$\\neq 0$. • If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$ • Using this we can solve the system of equation which has unique solution. This can be written in the form AX=B. $\\begin{bmatrix}4 & -3\\\\3 & -5\\end{bmatrix}\\begin{bmatrix}x\\\\y\\end{bmatrix}=\\begin{bmatrix}3\\\\7\\end{bmatrix}$ Where $A=\\begin{bmatrix}4 & -3\\\\3 & -5\\end{bmatrix}\\;X=\\begin{bmatrix}x\\\\y\\end{bmatrix}\\;B=\\begin{bmatrix}3\\\\7\\end{bmatrix}$ Let us see if matrix A is singular and non-singular.Let us find the value of the determinant |A| $|A|=4\\times -5-3\\times -3=-20+9=-11\\neq 0$ Hence it is non -singular. So it is invertible. We know $A^{-1}=\\frac{1}{|A|}(adj \\;A)$ Now let us find out adj A. This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging the symbols of elements of $a_{21}$ and $a_{12}$ Therefore $adj A=\\begin{bmatrix}-5 & 3\\\\-3 & 4\\end{bmatrix}$ and |A|=-11. $A^{-1}=\\frac{1}{-11}\\begin{bmatrix}-5 & 3\\\\-3 & 4\\end{bmatrix}$ Matrix multiplication can be obtained by multiplying the rows of A with the column of B. Now AX=B$\\Rightarrow X=A^{-1}B.$ $X=\\frac{1}{11}\\begin{bmatrix}-5 & 3\\\\-3 & 4\\end{bmatrix}\\begin{bmatrix}3\\\\7\\end{bmatrix}$ $\\begin{bmatrix}x\\\\y\\end{bmatrix}=\\frac{1}{11}\\begin{bmatrix}15-21\\\\9-28\\end{bmatrix}=\\frac{1}{11}\\begin{bmatrix}-6\\\\-19\\end{bmatrix}$ $\\Rightarrow \\begin{bmatrix}x\\\\y\\end{bmatrix}= \\begin{bmatrix}-6/11\\\\-19/11\\end{bmatrix}$ Hence x=-6/11 and y=-19/11. edited Feb 27, 2013", "also $X=XA^{-1}$. These facts immediately imply that $G$ is a subgroup of the group of invertible $2\\times2$-matrices. In a later course you may learn about group actions and stabilizers. Then you can reinterpret this as stating that $G$ is the stabilizer of the vector $X$, when the group of invertible matrices acts on the (row) vectors by matrix muliplication. The \"easier\" way of doing this is to note the condition on $A$ is equivalent to requiring $A$ to be invertible and satisfy: $$\\begin{pmatrix}1&1\\end{pmatrix}A = \\begin{pmatrix}1&1\\end{pmatrix}$$ Then if $A,B$ satisfy this constraint, show $AB$ satisfies this constraint and that $A^{-1}$ satisfies this constraint. For example, if $A$ is invertible, and $\\begin{pmatrix}1&1\\end{pmatrix}A = \\begin{pmatrix}1&1\\end{pmatrix}$ then $$\\begin{pmatrix}1&1\\end{pmatrix}A^{-1} = \\left(\\begin{pmatrix}1&1\\end{pmatrix}A\\right)A^{-1} = \\begin{pmatrix}1&1\\end{pmatrix} I = \\begin{pmatrix}1&1\\end{pmatrix}$$ So $A^{-1}$ is in our set." ]

[ "While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $$x$$ for which the sine of $$x$$ degrees is the same as the sine of $$x$$ radians are $$\\frac{m\\pi}{n-\\pi}$$ and $$\\frac{p\\pi}{q+\\pi}$$, where $$m$$, $$n$$, $$p$$, and $$q$$ are positive integers. Find $$m+n+p+q$$. (第二十届AIME2 2002 第10题)", "Don't have an account Signup Now # Introduction $sin^{–1} x,\\ cos^{–1} x,\\ tan^{–1} x$ etc. represents angles or numbers whose values of sine, cosine and tangent is ‘$x$’, provided that the value in numerical form is smallest. These can be written as $\\arcsin x,\\ arccos x\\$ etc. If two angles whose modulus is equal, in which one is positive and other is negative then we take positive sign.", "+0 # trig question 0 113 1 Find the measure, in degrees, of the smallest positive angle $\\theta$ for which $\\sin 3 \\theta = \\cos 17 \\theta$. Dec 15, 2021 #1 +117437 +1 $$sin\\;3\\theta = cos\\; 17\\theta\\\\ sin\\;3\\theta = sin\\; (90-17\\theta)\\\\$$ you can take it from there. Dec 15, 2021", "# Triggered Trigonometry #2 Algebra Level 2 Find the smallest positive integer value of $$\\theta$$ in degrees such that $81^{sin^2{\\theta}} + 81^{cos^2{\\theta}} = 30.$ ×", "= 11.5 + 17d$ 5. you would have to use degrees if you are using 11.5 as the smallest angle.", "# A geometry problem by Rudraksh Sisodia Geometry Level 3 $\\sin (3a)=4\\sin (a) \\sin (x+a)\\sin (x-a)$ Find the smallest positive angle $$x$$ (in degrees) such that the above is an identity for all values of $$a$$. ×", "Algebra and Trigonometry 10th Edition The $arcsin(0)$ is the angle whose sine is 0. Thus, we see that the expression evaluates to: $$0\\ degrees$$", "## Trigonometry (10th Edition) 1. Since the angle is larger than 360°, to find the smallest positive coterminal angle, we will subtract 360° until we will get positive angle smaller the 360 degrees 541° -360° = 181$^{\\circ}$", "x degrees. If told to find the least positive angle coterminal with 785 degrees you can use the following calculation process shown below.", "# A geometry problem by Aly Ahmed Geometry Level 3 $\\large 2^{\\tan x} + 3^{\\tan x} + 6^{\\tan x} = 1$ Find the smallest positive integer $x$ satisfying the equation above. Clarification: Angles are measured in degrees. ×", "# Think De Moivre Geometry Level 2 Let $a$ and $b$ be positive integers such that $\\dfrac1{\\cos(80^\\circ) + i \\sin(80^\\circ) }$ can be written in the form of $\\cos(a) - i \\sin(b)$, where $a$ and $b$ are minimized, calculate $a - b$. Note: Angles are measured in degrees. ×", "# Yet another curious identity? Geometry Level 5 Find the smallest integer $$n>3$$ such that $$\\displaystyle \\prod_{k=3}^{n}2\\cos(2^k)=1$$, where angles are measured in degrees. If no such $$n$$ exists, enter 666 as your answer. Bonus question: What is the solution for $$\\displaystyle \\prod_{k=2}^{n}2\\cos(2^k)=1$$? Inspiration ×", "degrees.​ You could also subtract 360 degrees to get -325 degrees and subtract 720 degrees to get ​-685 degrees​. 2. Find the smallest positive angle, in degrees, coterminal with -15 radians. Add multiples of 2π until you get a positive angle. Since 2π = 6.28, we need to multiply by 3 to end up with a positive angle:", "and the smallest angles (in degrees)is: A) 20∘ B) 25∘ C) 30∘ D) 35∘", "calculator is in \"degree\" mode, clicking on \" $sin^{-1}$\" gives 22.02431... telling us that the angle is about 22 degrees.", "# The smallest positive value Geometry Level 3 Find the smallest positive value of $$x$$ in radians such that the smallest positive value of $\\sin(x)+\\cos(x)+\\tan(x)+\\cot(x)+\\csc(x)+\\sec(x)$ is attained. Choose the value of $$\\frac{\\pi}{\\text{hi}}$$, where $$\\text{hi}$$ is the required value of $$x$$. Inspired by this. ×", "$f(x)$ is sometimes roughly (but imperfectly) summed up in the statement that a small change in $x$ shall produce a small change in $f(x)$. We shall not consider functions having an infinite number of oscillations in a limited region. 5. If we refer to the table on p. 4 [§ 4], it will be seen that for all angles less than 10° the angle in radians and the sine of the angle are equal to three decimal places. If larger tables are consulted, five-place, say, it will be seen that for all angles less than 2.2° the sine of the angle and the angle itself are equal to four decimal places. From this we may well suspect that $\\lim_{x \\to 0} \\frac{\\sin\\ x}{x} = 1$.", "while multiplying $$sin^2x$$ both sides, we have, $$1=sin^2x$$ and by square rooting both sides, we now have, $$sinx= \\pm1$$ and so, since sin x is both positive and negative, it must lie in all quadrants with alpha 90 degrees. Last edited: Oct 11, 2004 9. Oct 12, 2004 ### omicron Oh so now i know. I didn't know u could $$\\sqrt{sin^2x}$$. Thanks to everyone that helped.", "# smallest 1. ### Smallest interval for existence of unique solution of differential equation The smallest interval on which a unique solution exist for the IVP y′=e^(2y), y(0)=0, is what? A)|x|<=1/2e B)|x|<=2e C)|x|<=2/e D)|x|<=1/e I am confused which one is correct. Please help.:confused: 2. ### [Numerical methods] Smallest positive root With 15 or 16 decimals of precision, compute the smallest positive root of the equation: (x − 2)(x − 3)(x − 4)(x − 5)(x − 6)(x − 7)(x − 8)(x − 9)(x − 10) − ln x = 0 What was the accuracy obtained? 3. ### Find the smallest expression value About a convex hexagon ABCDEF it is known that AB = BC, CD = DE, EF = FA. Find the smallest value of the expression: BC / BE + DE / DA + FA / FC 4. ### How to find the smallest length in a triangle when the angles aren't known? I am confused on how to find the answer for this problem. So far what I believe would apply is the triangle inequality but I'm not sure on how to use it. The figure $ABC$ is a triangle so as $BDC$. It is known that the length of $AB = 8$ inches and $\\angle BAD = 2\\,\\angle BCD$ and... 17. ### Smallest value of expression", "## anonymous one year ago use the unit circle to find the inverse function value in degrees cos -1 (sqrt 3 /2) 1. xapproachesinfinity use the cheat sheet you can find the angle whose cos is root3/2 2. anonymous cos 30 degrees = $\\sqrt{3/2}$ 3. xapproachesinfinity 4. xapproachesinfinity well there are better cheat sheets that give you this just look online :) 5. anonymous No hang on @xapproachesinfinty how would you solve this? 6. anonymous im confused :/" ]

[ "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "x = \\frac {\\sin 2x} 2.$$ There will be either only cos's or only sin's left. Convert them into double-angles with \\begin{align} \\cos^2 x &= \\frac {1 + \\cos 2x} 2 \\\\ \\sin^2 x &= \\frac {1 - \\cos 2x} 2. \\end{align}", "w\\sin 12^\\circ \\\\ W - x &= h\\sin 12^\\circ \\\\ H - y &= h\\cos 12^\\circ \\\\ \\end{align*} Indeed, for the special case where $w = 400$ and $h = 300$, we obtain: \\begin{align*} \\frac{W}{w} &= \\frac{400\\cos 12^\\circ + 300\\sin 12^\\circ}{400} \\approx 113.41\\% \\\\ \\frac{H}{h} &= \\frac{400\\sin 12^\\circ + 300\\cos 12^\\circ}{300} \\approx 125.54\\% \\\\ \\end{align*}", "that $r\\sin(-a)$ is positive. This happens when $r=5$ and $a=-36.87^{\\circ}$; or when $r=-5$ and $a=143.13$. The two choices work, the other two possibilities do not. \\begin{align*} f(x) &= 3 \\sin(x) + 4 \\cos(x) \\\\ f(x) &= 5\\left(\\frac{3}{5} \\sin(x) + \\frac{4}{5} \\cos(x) \\right) \\end{align*} Let $\\sin(y) = 4/5$ and $\\cos(y) =3/5$, for some real number $y$. \\begin{align*} f(x) &= 5\\left(\\cos(y)\\sin(x) + \\sin(y)\\cos(x)\\right)\\\\ f(x) &= 5\\sin(x+y)&\\text{(where }y = \\arcsin(4/5)\\text{)} \\end{align*} Using trigonometric tables, $y = 36.87$. $$f(x) = 5\\sin(x + 36.87^{\\circ})$$ which can also be written as $$f(x) = 5 \\sin (180 - x - 36.87) = -5\\sin(x - 144.13).$$ This justifies the 2 values of $a$ and $r$ in $r\\sin(x−a)$.", "sin(2x) with a double angle formula but I'm not getting the answers. The range is 0 - 360. since $0 < x < 360$ $0 < 2x < 720$ $3\\sin(2x) = 2$ $\\sin(2x) = \\frac{2}{3}$ $2x = \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 180 - \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 360 + \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 540 - \\arcsin\\left(\\frac{2}{3}\\right)$ solve for $x$ in all four equations", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "given below: \\begin{align} &\\sin (A) \\cos (B)=\\dfrac{1}{2}(\\sin (A+B)+\\sin (A-B)) \\\\ &\\cos (A) \\sin (B)=\\dfrac{1}{2}(\\sin (A+B)-\\sin (A-B)) \\\\ &\\sin (A) \\sin (B)=\\dfrac{1}{2}(\\cos (A-B)+\\cos (A+B)) \\\\ &\\cos (A) \\cos (B)=\\dfrac{1}{2}(\\cos (A+B)-\\cos (A-B)) \\end{align} Product To Sum Examples 1. Find the value of sin 45o sin 15o using the product to sum formulas. Solution: Using the third product to sum formulas above we have, $\\sin (A) \\sin (B)=\\dfrac{1}{2}(\\cos (A-B)+\\cos (A+B))$ Substituting the values we get, $\\sin 45^{\\circ} \\sin 15^{\\circ}=\\dfrac{1}{2}\\left(\\cos 30^{\\circ}-\\cos 60^{\\circ}\\right)$ 0.866+0.5=1.366 2. Write the product of the $2 \\cos \\left(\\dfrac{7 x}{2}\\right) \\cos \\left(\\dfrac{3 x}{2}\\right)$ as a sum Solution: Using the 4th identity of the product to sum from above, we get \\begin{align} &\\cos (A) \\cos (B)=\\dfrac{1}{2}(\\cos (A+B)-\\cos (A-B)) \\\\ &=\\frac{1}{2} \\times 2(\\cos 5 x-\\cos 2 x) \\\\ &=(\\cos 5 x-\\cos 2 x) \\end{align} State The Product To Sum Formula and their Derivation To derive the product to sum derivations we need to know the sum and difference trigonometric identities. These are as listed below: \\begin{align} &\\sin (A+B)=\\sin (A) \\cos (B)+\\cos (A) \\sin (B) \\ldots \\ldots(a) \\\\ &\\cos (A+B)=\\cos (A) \\cos (B)-\\sin (A) \\sin (B) \\ldots \\ldots(b) \\\\ &\\sin (A-B)=\\sin (A) \\cos (B)-\\cos (A) \\sin (B) \\ldots \\ldots(c) \\\\ &\\cos (A-B)=\\cos (A) \\cos (B)+\\sin (A) \\sin (B) \\ldots \\ldots(d) \\end{align} (a) Derivation for $\\sin (A) \\cos (B)=\\dfrac{1}{2} \\sin (A+B)+\\sin (A-B)$ Adding", "In fact, the definition $$y=3\\,\\sin\\;x + 4\\,\\cos\\;x$$ may tempt you to think that the amplitude is $$7$$, since the largest that $$3\\,\\sin\\;x$$ could be is $$3$$ and the largest that $$4\\,\\cos\\;x$$ could be is $$4$$, so that the largest their sum could be is $$3+4=7$$. However, $$3\\,\\sin\\;x$$ can never equal $$3$$ for the same $$x$$ that makes $$4\\,\\cos\\;x$$ equal to $$4$$ (why?). There is a useful technique (which we will discuss further in Chapter 6) for showing that the amplitude of $$y=3\\,\\sin\\;x + 4\\,\\cos\\;x$$ is $$5$$. Let $$\\theta$$ be the angle shown in the right triangle in Figure 5.2.8. Then $$\\cos\\;\\theta = \\frac{3}{5}$$ and $$\\sin\\;\\theta = \\frac{4}{5}$$. We can use this as follows: \\nonumber \\begin{align*} y ~&=~ 3\\,\\sin\\;x ~+~ 4\\,\\cos\\;x\\\\ \\nonumber &=~ 5\\,\\left( \\tfrac{3}{5}\\,\\sin\\;x ~+~ \\tfrac{4}{5}\\,\\cos\\;x \\right)\\\\ \\nonumber &=~ 5\\,( \\cos\\;\\theta\\;\\sin\\;x ~+~ \\sin\\;\\theta\\;\\cos\\;x )\\\\ \\nonumber &=~ 5\\,\\sin\\;(x+\\theta)\\quad\\text{(by the sine addition formula)} \\end{align*} Figure 5.2.8 Thus, $$|y| = |5\\,\\sin\\;(x+\\theta)| = |5|\\,\\cdot\\,|\\sin\\;(x+\\theta)| \\le (5)(1) = 5$$, so the amplitude of $$y=3\\,\\sin\\;x + 4\\,\\cos\\;x$$ is $$5$$. In general, a combination of sines and cosines will have a period equal to the lowest common multiple of the periods of the sines and cosines being added. In Example 5.9, $$\\sin\\;x$$ and $$\\cos\\;x$$ each have period $$2\\pi$$, so the lowest common multiple (which is always an integer multiple) is $$1 \\,\\cdot\\, 2\\pi = 2\\pi$$.", "part. 23. Mar 20, 2009 ### Cyosis Barring any mistakes from my side you can do it without using any trigonometric identities: \\begin{align*} \\int \\cos^2 x\\,dx &= \\int 1*\\cos^2 x\\,dx \\\\ &= x \\x cos^2 x + \\int 2x \\cos x \\sin x\\,dx \\\\ &= x \\cos^2 x+(x \\sin^2 x- \\int \\sin^2 x\\,dx) \\\\ &= x \\cos^2x + x \\sin^2 x - (-\\sin x \\cos x- \\int -\\cos^2 x\\,dx) \\\\ &= x (\\cos^2x + \\sin^2 x) +\\sin x \\cos x- \\int \\cos^2 x\\,dx \\end{align*} It follows that: $$2\\int \\cos^2 x\\,dx=\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x \\Rightarrow$$ \\begin{align*} \\int \\cos^2 x\\,dx &= \\frac{1}{2} (\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x) \\\\ &= \\frac{1}{2} (\\x x +\\frac{1}{2} \\sin {2x}) \\\\ &= \\frac{x}{2}+\\frac{1}{4} \\sin {2x} \\end{align} This is the same result you get by using the double angle identity from the start, as it should be. Last edited: Mar 20, 2009 24. Mar 20, 2009 ### itchy8me brilliant! using the coefficient 1. damn thats a sneaky one. thanks guys and girls ;) !! :-D 25. Jan 5, 2010 ### ntc2415 Just use substitution. This makes it all so much easier. First you set u to be cos x$$\\int(cos x)$$2 dx then substitute cos x with u x$$\\int(u)$$2 dx Then you take the derivative of both sides", "4i(e^{-3\\pi i k/4} - e^{3\\pi i k/4})\\right)\\\\ &= \\frac{1}{8} \\left( 8\\cos(\\pi k /4) + 8\\sin(3\\pi k/4) \\right)\\\\ &= \\cos(\\pi k /4) + \\sin(3\\pi k/4). \\end{align}", "## Precalculus (6th Edition) Blitzer The given expression on the left side $\\cos 4t$ can be written as $\\cos 2.2t$. As per the double angle formula, $\\cos 2t=1-2{{\\sin }^{2}}\\theta$. Therefore, using the double angle formula: \\begin{align} & \\cos 4t=\\cos 2.2t \\\\ & =1-2{{\\sin }^{2}}2t \\\\ & =1-2{{\\left( 2\\sin t\\cos t \\right)}^{2}} \\\\ & =1-2.4{{\\sin }^{2}}t{{\\cos }^{2}}t \\end{align} Now, the expression can be further classified as: $1-2.4{{\\sin }^{2}}t{{\\cos }^{2}}t=1-8{{\\sin }^{2}}t{{\\cos }^{2}}t$ Hence, the expression on the left-side is equal to the expression on the right-side.", "angle formulae for sin and cos : $$\\cos 2x = 1 - 2\\sin^2 x = 2\\cos^2 x - 1$$ $$\\sin 2x = 2\\sin x \\cos x$$ $$\\frac{1 - \\cos 2x + \\sin 2x}{1 + \\cos 2x + \\sin 2x} = \\frac{1 - (1 - 2\\sin^2 x) + 2\\sin x \\cos x}{1 + 2 \\cos^2 x - 1 + 2\\sin x \\cos x} = \\frac{2 \\sin x (\\sin x + \\cos x)}{2 \\cos x (\\sin x + \\cos x)} = \\tan x$$ For the second one, as you said, just factorise and apply simple identities : $$\\frac{\\sin^4 x - \\cos^ 4 x}{\\sin^2 x - \\cos^2 x} = \\frac{(\\sin^2 x + \\cos^2 x)(\\sin^2 x - \\cos^2 x)}{\\sin^2 x - \\cos^2 x} = 1$$ Last edited: Mar 19, 2005 11. Mar 19, 2005 ### cepheid Staff Emeritus Sure...just give him the answer why don't you? :grumpy: 12. Mar 19, 2005 ### Curious3141 Well, it looks like he's been trying for long enough, so why not ? 13. Mar 19, 2005 ### Data NEARLY?!?!?!?!?!?! Edit: Now all I need to do is figure out how to get smilies in here! Oh, wait, what are all of those colorful symbols next to this white box I'm writing in? :rofl: 14. Mar 20, 2005 ### scorpa Ok guys thank you so much for the", "possible if the first version is closed since I find this one is helping me better? – Ash Apr 5 '15 at 1:01 $\\sin(x)\\sin(\\frac{\\pi}{2}-x) + \\cos^2(x)\\cot(x) = \\sin(x)\\cos(x) + \\cos^2(x)\\frac{\\cos(x)}{\\sin(x)}$ and now we want to find a common denominator to add the fractions so we get $\\frac{\\sin^2(x)\\cos(x)}{\\sin(x)} + \\frac{\\cos^2(x)\\cos(x)}{\\sin(x)} = \\frac{(\\sin^2(x)+\\cos^2(x))\\cos(x)}{\\sin(x)} = \\frac{\\cos(x)}{\\sin(x)} = \\cot(x)$ • So this is all influencing the LHS (left-hand side of the equation? – Ash Apr 5 '15 at 1:01 • all of this is manipulating the RHS and then the final step is showing that it is equal to the LHS – nosyarg Apr 5 '15 at 1:40 First, note that using the angle addition formula of the sine function reveals \\begin{align} \\sin (\\pi /2 -x)&=\\sin(\\pi /2)\\cos(x)-\\sin(x)\\cos(\\pi /2)\\\\ &=\\cos(x) \\end{align} Then, $\\sin x \\sin (\\pi /2 -x) =\\sin x \\cos x$. Next, $\\cos^2x \\cot x =\\frac{\\cos^3x}{\\sin x}$. Putting it all together \\begin{align} \\sin x \\sin (\\pi /2 -x) +\\sin x \\cos^2x \\cot x&=\\cos(x)+\\frac{\\cos^3x}{\\sin x}\\\\ &=\\cot(x) \\left(\\sin^2 x +\\cos^2 x \\right)\\\\ &=\\cot x \\end{align} as was to be shown! • wait, so where did the sinxcosx come from. Because aren't sin and cos treated as different variables? – Ash Apr 5 '15 at 1:19 • They are multiplied in the first term $\\sin x \\sin (\\pi /2 -x) = \\sin x \\cos x$.", "# Missing and parasite roots in the trigonometric equation. I have this equation: $$\\boxed{\\cos(2x) - \\cos(8x) + \\cos(6x) = 1}$$ RIGHT And its right solution from the textbook is: \\begin{align} \\cos(2x)+\\cos(6x)&=1+\\cos(8x)\\\\\\\\ 2\\cos(4x)*\\cos(2x)&=2\\cos^2(4x)\\\\\\\\ \\cos(4x)*(\\cos(2x)-\\cos(4x))&=0\\\\\\\\ \\cos(4x)*2\\sin(3x)*\\sin(x)=&0 \\end{align} 1. $\\cos(4x) = 0 \\implies 4x = \\frac{\\pi}{2} + \\pi k \\implies \\boxed{x=\\frac{\\pi}{8} + \\frac{\\pi k}{4}}$ 2. $\\sin(3x) = 0 \\implies 3x = \\pi k \\implies \\boxed{x=\\frac{\\pi k}{3}}$ 3. $\\sin(x) = 0 \\implies \\boxed{x=\\pi k}$ WRONG And this is the way how I have been tried to solve it (with different approach) and here I need a help to determine where I'm wrong: \\begin{align} \\cos(2x) - \\cos(8x) + \\cos(6x) &= 1\\\\ \\cos(2x) + \\cos(6x) &= 1 + \\cos(8x) \\end{align} Then using this formula: $$\\bbox[Beige]{\\boxed{\\cos\\alpha + \\cos\\beta = 2\\cos(\\frac{\\alpha + \\beta}{2}) \\cos(\\frac{\\alpha - \\beta}{2})}} \\tag{1}$$ $$2\\cos(4x)\\cos(2x) = 1 + \\cos(8x)$$ And then similarly: $$\\bbox[Beige]{\\boxed{\\cos2\\alpha=2\\cos^2\\alpha-1}} \\tag{2}$$ $$2\\cos(4x)\\cos(2x) = 1 + 2\\cos^2(4x) - 1$$ After that I subtracted $1$ from $1$ in the right part and divided both right and left parts by $2$: $$\\cos(4x)\\cos(2x) = \\cos^2(4x)$$ Then once again I divided both parts. This time by $\\cos(4x)$ and with the aim not to lose a root I checked whether $\\cos(4x)$ can be equal to zero. $$\\cos(4x) = 0 \\implies 4x = \\frac{\\pi}{2} + \\pi k \\implies \\boxed{x=\\frac{\\pi}{8} + \\frac{\\pi k}{4}}$$ so here I found the", "## Precalculus (6th Edition) Blitzer The value of $b$ to the nearest tenth is $8.0$. As the angle $B$ is provided, so, by using the Law of Sines to find $b$, we get: \\begin{align} & \\frac{b}{\\sin B}=\\frac{a}{\\sin A} \\\\ & \\frac{b}{\\sin 68{}^\\circ }=\\frac{4.8}{\\sin 34{}^\\circ } \\\\ & b=\\frac{4.8\\sin 68{}^\\circ }{\\sin 34{}^\\circ } \\\\ & \\approx 8.0 \\end{align} Hence, the value of $b$ is 8.0", "220x^3 − 11x,$$ and $$T_{12}(x)=2048x^{12} − 6144x^{10} + 6912x^8 − 3584x^6 + 840x^4 − 72x^2 + 1.$$ - The formula in the other direction (the powers of $\\cos(x)$ as a linear combination of $\\cos(kx)$) is explained here. – Did Mar 29 '12 at 8:14 add comment You can always repeatedly use \\begin{align*} \\cos(a\\pm b) &= \\cos a\\cos b \\mp \\sin a\\sin b\\\\ \\sin(a\\pm b) &= \\sin a\\cos b \\pm \\cos a\\sin b\\\\ \\sin^2(r) &= 1-\\cos^2(r). \\end{align*} For example, \\begin{align*} \\cos(4x) &= \\cos(2x+2x)\\\\ &= \\cos(2x)^2 - \\sin^2(2x)\\\\ &= \\cos(2x)^2 - (1-\\cos^2(2x))\\\\ &= 2\\cos(2x)^2 - 1\\\\ &= 2(\\cos(x+x))^2 - 1\\\\ &= 2(\\cos x\\cos x - \\sin x\\sin x)^2 - 1\\\\ &= 2(\\cos^2 x - \\sin^2 x)^2 - 1\\\\ &= 2(\\cos^2x - (1-\\cos^2 x))^2 - 1\\\\ &= 2(2\\cos^2x - 1)^2 - 1\\\\ &= 2(4\\cos^4 x - 4\\cos^2 x + 1) - 1\\\\ &= 8\\cos^4 x - 8\\cos^2 x + 1. \\end{align*} - Can you give a general formula – tianzhidaosunyouyu Mar 29 '12 at 6:11 @pxc417: Can you do some of the work? – Arturo Magidin Mar 29 '12 at 14:25 add comment Here is a neat way to derive what the answer will be, using Euler's formula $e^{ix}=\\cos x + i \\sin x$. We have that $\\cos nx$ is the real part of $e^{i(nx)}=\\left(e^{ix}\\right)^n=(\\cos x + i \\sin x)^n$ By", "\\sin^2 B + \\sin^2 C) &=6 ,\\\\ \\sin^2 A + \\sin^2 B + \\sin^2 C &=2 ,\\\\ 4R^2\\sin^2 A + 4R^2\\sin^2 B + 4R^2\\sin^2 C &=2\\cdot4R^2 ,\\\\ a^2+b^2+c^2&=8R^2 ,\\\\ 2\\rho^2-8r\\,R-2r^2&=8R^2 ,\\\\ \\rho^2-(r+2R)^2&=0 ,\\\\ \\frac{\\rho^2-(r+2R)^2}{4R^2} =0 . \\end{align} And since \\begin{align} \\cos A \\cos B \\cos C &=\\frac{\\rho^2-(r+2R)^2}{4R^2} , \\end{align} one of $\\cos A,\\cos B,\\cos C$ must be 0, hence, one of the angles must be $\\tfrac\\pi2$. COMMENT.- First note that if $C=90^{\\circ}$ then you have $\\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\\circ}$ you can consider that $\\sin C=\\sin (180^{\\circ}-(A+B))=\\sin(A+B)$ and $\\cos C= \\cos (180^{\\circ}-(A+B))=-\\cos(A+B)$ so you get $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B)=2(\\cos^2A+\\cos ^2 B+\\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish. EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B) = 2\\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\\dfrac{\\pi}{2}$ (the \"if\" is obvious and we need the \"only if\"). So put $A + B = \\dfrac{\\pi}{2}\\pm h$ with $h\\ne 0$ and look at the functions $$\\sin^2(x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h-x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h),\\quad0\\lt x\\lt\\dfrac{\\pi}{2}$$ which becomes for $h$ positive and $h$", "\\big{(} 1 – cos(2x) \\big{)}$$ to turn all of the sines into cosines of a double angle. Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain. \\begin{align*} \\int cos^{n}(x) \\ dx \\end{align*} If $$n$$ is an odd integer, factor out then use the identity $$cos^{2}(x) \\ = \\ 1 – sin^{2}(x)$$ for the term left after one power has been factored out. The integral you get will be in the form of the ones in the reverse chain rule section. If $$n$$ is an even integer, use $$cos^{2}(x) \\ = \\ \\frac{1}{2} \\big{(} 1 + cos(2x) \\big{)}$$ to turn all of the cosines into sines of a double angle. Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain. ### Example 2 : $$\\int cos^{4} (x) \\ dx$$ \\begin{align*} \\int cos^{4} \\ dx \\ &= \\ \\int \\bigg{(} \\frac{1}{2} \\big{(} 1 + cos(2x) \\big{)} \\bigg{)} ^{2} \\ dx \\\\ &= \\frac{1}{4} \\int \\big{(} 1 + 2 cos(2x) \\ + \\ cos^{2} (2x) \\big{)} \\ dx \\end{align*} We notice here that we need to apply the rule again on $$cos^{2} (2x)$$ \\begin{align*} &= \\frac{1}{4} \\int \\big{(} 1 +", "we have that: (7) \\begin{align} \\quad |f(x, y)| &= \\sqrt{[\\sin (xy + 1)]^2 + [\\cos(x + 2y^2 + 1)]^2} \\\\ & \\leq |\\sin(xy + 1) | + |\\cos(x + 2y^2 + 1)| \\\\ & \\leq 2 \\end{align} Therefore $f (2D^2) \\subseteq 2D^2$. Furthermore, $f$ is continuous. Hence by Brouwer's fixed point theorem there exists a point $(a, b) \\in 2D^2$ such that $f(a, b) = (a,b)$, that is: (8) \\begin{align} \\quad \\sin (ab + 1) = a \\\\ \\quad \\cos (a + 2b^2 + 1) = b \\end{align} Therefore: (9) \\begin{align} \\quad \\sin (ab + 1) - a &= 0 \\\\ \\quad \\cos (a + 2b^2 + 1) - b &= 0 \\end{align} So $(a, b)$ is a solution to the original system.", "# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 37 The value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$ is $\\frac{1}{8}-\\frac{1}{8}\\cos 4x$. #### Work Step by Step We have to find the value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$; various cos and sin formulas are used which represent the answer in the form of cos. \\begin{align} & {{\\sin }^{2}}x{{\\cos }^{2}}x=\\left( \\frac{1-\\cos 2x}{2} \\right)\\left( \\frac{1+\\cos 2x}{2} \\right) \\\\ & =\\frac{1-{{\\cos }^{2}}2x}{4} \\\\ & =\\frac{1}{4}-\\frac{1}{4}{{\\cos }^{2}}2x \\\\ & =\\frac{1}{4}-\\frac{1}{4}\\left( \\frac{1+\\cos \\left( 2\\times 2x \\right)}{2} \\right) \\end{align} So, in step 2 the identity of $\\left( a+b \\right)\\left( a-b \\right)={{a}^{2}}+{{b}^{2}}$ is used. \\begin{align} & {{\\sin }^{2}}x{{\\cos }^{2}}x=\\frac{1}{4}-\\frac{1}{8}\\left( 1+\\cos 4x \\right) \\\\ & =\\frac{1}{4}-\\frac{1}{8}-\\frac{1}{8}\\left( \\cos 4x \\right) \\\\ & =\\frac{1}{8}-\\frac{1}{8}\\left( \\cos 4x \\right) \\end{align} In the last step $\\frac{1}{8}$ is distributed and subtracted from $\\frac{1}{4}$. Thus, the value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$ is $\\frac{1}{8}-\\frac{1}{8}\\cos 4x$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]

[ "x = \\frac {\\sin 2x} 2.$$ There will be either only cos's or only sin's left. Convert them into double-angles with \\begin{align} \\cos^2 x &= \\frac {1 + \\cos 2x} 2 \\\\ \\sin^2 x &= \\frac {1 - \\cos 2x} 2. \\end{align}", "that $r\\sin(-a)$ is positive. This happens when $r=5$ and $a=-36.87^{\\circ}$; or when $r=-5$ and $a=143.13$. The two choices work, the other two possibilities do not. \\begin{align*} f(x) &= 3 \\sin(x) + 4 \\cos(x) \\\\ f(x) &= 5\\left(\\frac{3}{5} \\sin(x) + \\frac{4}{5} \\cos(x) \\right) \\end{align*} Let $\\sin(y) = 4/5$ and $\\cos(y) =3/5$, for some real number $y$. \\begin{align*} f(x) &= 5\\left(\\cos(y)\\sin(x) + \\sin(y)\\cos(x)\\right)\\\\ f(x) &= 5\\sin(x+y)&\\text{(where }y = \\arcsin(4/5)\\text{)} \\end{align*} Using trigonometric tables, $y = 36.87$. $$f(x) = 5\\sin(x + 36.87^{\\circ})$$ which can also be written as $$f(x) = 5 \\sin (180 - x - 36.87) = -5\\sin(x - 144.13).$$ This justifies the 2 values of $a$ and $r$ in $r\\sin(x−a)$.", "angle formulae for sin and cos : $$\\cos 2x = 1 - 2\\sin^2 x = 2\\cos^2 x - 1$$ $$\\sin 2x = 2\\sin x \\cos x$$ $$\\frac{1 - \\cos 2x + \\sin 2x}{1 + \\cos 2x + \\sin 2x} = \\frac{1 - (1 - 2\\sin^2 x) + 2\\sin x \\cos x}{1 + 2 \\cos^2 x - 1 + 2\\sin x \\cos x} = \\frac{2 \\sin x (\\sin x + \\cos x)}{2 \\cos x (\\sin x + \\cos x)} = \\tan x$$ For the second one, as you said, just factorise and apply simple identities : $$\\frac{\\sin^4 x - \\cos^ 4 x}{\\sin^2 x - \\cos^2 x} = \\frac{(\\sin^2 x + \\cos^2 x)(\\sin^2 x - \\cos^2 x)}{\\sin^2 x - \\cos^2 x} = 1$$ Last edited: Mar 19, 2005 11. Mar 19, 2005 ### cepheid Staff Emeritus Sure...just give him the answer why don't you? :grumpy: 12. Mar 19, 2005 ### Curious3141 Well, it looks like he's been trying for long enough, so why not ? 13. Mar 19, 2005 ### Data NEARLY?!?!?!?!?!?! Edit: Now all I need to do is figure out how to get smilies in here! Oh, wait, what are all of those colorful symbols next to this white box I'm writing in? :rofl: 14. Mar 20, 2005 ### scorpa Ok guys thank you so much for the", "$$x(n)=x(n+N)$$, otherwise, the signal is nonperiodic. Simply start with \\begin{align} x(n+N) &= \\cos( \\frac{n}{6} + \\frac{N}{6}) \\\\ &= \\cos(\\frac{n}{6})\\cos(\\frac{N}{6}) - \\sin(\\frac{n}{6})\\sin(\\frac{N}{6}) \\end{align} In order for $$x(n)=x(n+N)$$ to hold, $$\\cos(\\frac{N}{6})=1$$ and $$\\sin(\\frac{N}{6})=0$$. We search for the smallest value of $$N$$. This is true if $$\\frac{N}{6}=2\\pi \\implies N = 12\\pi$$. Indeed, if $$N=12\\pi$$, we get \\begin{align} x(n+N) &= \\cos( \\frac{n}{6} + \\frac{N}{6}) \\\\ &= \\cos(\\frac{n}{6})(1) - \\sin(\\frac{n}{6})(0) \\\\ &= \\cos(\\frac{n}{6}) \\\\ &= x(n) \\end{align} But $$N$$ must be a positive integer, therefore, the signal is nonperiodic.", "\\frac{4\\pi}{5}$ Related angle is $\\pi-\\frac{4\\pi}{5}=\\frac{\\pi}{5}$ $\\sin \\frac{4\\pi}{5}$ is a QII angle. Sin is positive in the first and second quadrants. Therefore, $\\sin -\\frac{4\\pi}{5}=-\\sin \\frac{\\pi}{5}$", "This gives us a minimum point at (π/4, 6.242640687). 2π/9 (40°) negative π/4 (45°) 0 5π/18 (50°) positive x y' This gives us a minimum point at (5π/4, 6.242640687) 11π/9 (220°) negative 5π/4 (225°) 0 23π/18 (230°) positive x y' This gives us a maximum point at (0.844611π, -1.828427125) 5π/6 (150°) positive 0.844611π (152.03°) 0 31π/36 (155°) negative x y' This gives us a maximum point at (1.655389π, -1.828427125) 29π/18 (290°) positive 1.655389π (297.97°) 0 5π/3 (300°) negative Therefore, I can conclude that the minimum value of $$\\displaystyle y=|\\sin x+\\cos x+\\tan x+\\sec x+\\csc x+\\cot x|$$ for all real x is 1.828427125. #### anemone ##### MHB POTW Director Staff member Find the minimum value of $$\\displaystyle |\\sin x+\\cos x+\\tan x+\\cot x+\\sec x+\\csc x|$$ for all real numbers $$\\displaystyle x$$. I just saw another really brilliant way to tackle this problem and since I have a habit to share all that I know with the members of the forum, (ahem, I am quite generous myself), I will post the solution here without further ado. If we let $x=a-135^{\\circ}$, we see that we have: \\begin{align*}\\sin x&=\\sin (a-135^{\\circ})\\\\&=\\dfrac{-\\sin a-\\cos a}{\\sqrt{2}}\\end{align*} \\begin{align*}\\cos x&=\\cos (a-135^{\\circ})\\\\&=\\dfrac{-\\cos a+\\sin a}{\\sqrt{2}}\\end{align*} \\begin{align*}\\sin x+\\cos x&=\\dfrac{-\\sin a-\\cos a}{\\sqrt{2}}+\\dfrac{-\\cos a+\\sin a}{\\sqrt{2}}\\\\&=-\\sqrt{2}\\cos a\\end{align*} \\begin{align*}\\tan x&=\\tan(a-135^{\\circ})\\\\&=\\dfrac{\\sin a+\\cos a}{\\cos a-\\sin a}\\end{align*} \\begin{align*}\\cot x&=\\cot(a-135^{\\circ})\\\\&=\\dfrac{\\cos a-\\sin a}{\\sin a+\\cos a}\\end{align*} \\begin{align*}\\tan x+\\cot x&=\\dfrac{\\sin a+\\cos a}{\\cos a-\\sin a}+\\dfrac{\\cos", "# Thread: Half angle or double angle Identities 1. ## Half angle or double angle Identities I don't understand them at all. The problem is to evaluate tan(22.5) using exact values. How should I do this? 2. $\\tan \\frac{x}{2}=\\frac{\\sin \\frac{x}{2}}{\\cos \\frac{x}{2}}=\\frac{\\sqrt{\\frac{1-\\cos x}{2}}}{\\sqrt{\\frac{1+\\cos x}{2}}}=\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}.$ Actually is $\\sin \\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos x}{2}},$ but the angle is positive, hence our answer is positive, and that's preserves the positive sign. 3. Originally Posted by Krizalid $\\tan \\frac{x}{2}=\\frac{\\sin \\frac{x}{2}}{\\cos \\frac{x}{2}}=\\frac{\\sqrt{\\frac{1-\\cos x}{2}}}{\\sqrt{\\frac{1+\\cos x}{2}}}=\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}.$ Actually is $\\sin \\frac{x}{2}=\\pm \\sqrt{\\frac{1-\\cos x}{2}},$ but the angle is positive, hence our answer is positive, and that's preserves the positive sign. hmm can I assume tan (22.5) is equal to tan 1/2 x? 4. Krizalid is suggesting $x = \\frac{\\pi}{4}$ 5. Well, that was easier than I expected. Thanks.", "the values should be positive or negative, whether they should be $0$ or $1$, or otherwise whether they are \"small\" ($\\frac{1}{2}$), \"medium\" ($\\frac{\\sqrt{2}}{2}$) or \"large\" ($\\frac{\\sqrt{3}}{2}$). • +1! It was only after I was a grad student that I saw this. – pjs36 Dec 1 '15 at 2:12 • This is so AMAZING! – James Smith Dec 1 '15 at 2:26 • In so far as there's a pattern here, it does extend - if you want a bunch of angles with sines $\\frac{\\sqrt{k}}{(2n)^2}$, the sequence of angles is always going to have that the reverse of the sequence is the complement of each angle, the first angle is $0$. The middle angle will be $45$ by the symmetry. The other angles just need to be filled in. (Unfortunately, this extension kind of highlights that the angles on the left don't follow much of a pattern beyond what's necessary - $30$ is just there as the angle whose sine is $\\frac{1}2$) – Milo Brandt Dec 1 '15 at 3:47 • @MiloBrandt: Regarding patterns, this recent answer of mine provides a common form for the sines (and cosines) of multiples of $3^\\circ$. There's no particularly-obvious pattern (certainly not a \"counting\" one), but it seems like the list is trying to tell me something. – Blue Dec 1 '15 at", "them up. for double angles you must know that: $\\cos 2x = \\cos^2 x - \\sin^2 x = 2 \\cos^2 x - 1 = 1 - 2 \\sin^2 x$ $\\sin 2x = 2 \\sin x \\cos x$ ok, continue. tell me if you have any problems Yeah I don't think I can do it. I'm just going in circles. I was never good with functions 15. Originally Posted by FalconPUNCH! Yeah I don't think I can do it. I'm just going in circles. I was never good with functions are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)* $2 \\cos^2 x - \\cos x - 1 = 0$ ....this is a quadratic, see if you can factorize $\\Rightarrow (2 \\cos x + 1)(\\cos x - 1) = 0$ $\\Rightarrow 2 \\cos x + 1 = 0 \\mbox{ or } \\cos x - 1 = 0$ $\\Rightarrow \\cos x = - \\frac 12 \\mbox{ or } \\cos x = 1$ what angles does this happen for? Page 1 of 2 12 Last", "= 1, x = 45^\\circ \\end{align} Remember, there is only one unique solution, however, there are infinite solutions because the unit circle can revolve around the origin infinite times in both a clockwise and counterclockwise direction. For that reason, x = 45deg + 360deg is also a solution, and x = 45 -360 deg is also a solution. It is thus, necessary to write a general solution such that: (65) \\begin{align} x = 45^\\circ \\pm n360^\\circ, n \\in I \\end{align} The notation may look funky, but all the above expression states is that x can equal 45 degrees plus or minus any integer value multiplied by 360 degrees. Note that integers are …-2, -1, 0, 1, 2… ## Solution to Question 9 Remember the unit circle identity such that: (66) $$cos^2(x) + sin^2(x) = 1$$ It can be arranged such that (67) $$cos^2(x) = 1 - sin^2(x)$$ But nothing that this is present within the identity, so we can substitute it in for simplification to get: (68) \\begin{align} tan(x) \\frac{cos^2(x)}{(sin(x))^2} = cot{x} \\end{align} Now, remember the following rule: (69) $$sin^2(x) = (sin(x))^2$$ Not that the following rule is not true but is commonly mistaken as true: (70) $$sin^2(x) =/= sin(x^2)$$ We can now adjust the equation to get: (71) \\begin{align} tan(x) \\frac{cos^2(x)}{sin^2(x)} = cot{x} \\end{align} We know that", "\\big{(} 1 – cos(2x) \\big{)}$$ to turn all of the sines into cosines of a double angle. Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain. \\begin{align*} \\int cos^{n}(x) \\ dx \\end{align*} If $$n$$ is an odd integer, factor out then use the identity $$cos^{2}(x) \\ = \\ 1 – sin^{2}(x)$$ for the term left after one power has been factored out. The integral you get will be in the form of the ones in the reverse chain rule section. If $$n$$ is an even integer, use $$cos^{2}(x) \\ = \\ \\frac{1}{2} \\big{(} 1 + cos(2x) \\big{)}$$ to turn all of the cosines into sines of a double angle. Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain. ### Example 2 : $$\\int cos^{4} (x) \\ dx$$ \\begin{align*} \\int cos^{4} \\ dx \\ &= \\ \\int \\bigg{(} \\frac{1}{2} \\big{(} 1 + cos(2x) \\big{)} \\bigg{)} ^{2} \\ dx \\\\ &= \\frac{1}{4} \\int \\big{(} 1 + 2 cos(2x) \\ + \\ cos^{2} (2x) \\big{)} \\ dx \\end{align*} We notice here that we need to apply the rule again on $$cos^{2} (2x)$$ \\begin{align*} &= \\frac{1}{4} \\int \\big{(} 1 +", "# Missing and parasite roots in the trigonometric equation. I have this equation: $$\\boxed{\\cos(2x) - \\cos(8x) + \\cos(6x) = 1}$$ RIGHT And its right solution from the textbook is: \\begin{align} \\cos(2x)+\\cos(6x)&=1+\\cos(8x)\\\\\\\\ 2\\cos(4x)*\\cos(2x)&=2\\cos^2(4x)\\\\\\\\ \\cos(4x)*(\\cos(2x)-\\cos(4x))&=0\\\\\\\\ \\cos(4x)*2\\sin(3x)*\\sin(x)=&0 \\end{align} 1. $\\cos(4x) = 0 \\implies 4x = \\frac{\\pi}{2} + \\pi k \\implies \\boxed{x=\\frac{\\pi}{8} + \\frac{\\pi k}{4}}$ 2. $\\sin(3x) = 0 \\implies 3x = \\pi k \\implies \\boxed{x=\\frac{\\pi k}{3}}$ 3. $\\sin(x) = 0 \\implies \\boxed{x=\\pi k}$ WRONG And this is the way how I have been tried to solve it (with different approach) and here I need a help to determine where I'm wrong: \\begin{align} \\cos(2x) - \\cos(8x) + \\cos(6x) &= 1\\\\ \\cos(2x) + \\cos(6x) &= 1 + \\cos(8x) \\end{align} Then using this formula: $$\\bbox[Beige]{\\boxed{\\cos\\alpha + \\cos\\beta = 2\\cos(\\frac{\\alpha + \\beta}{2}) \\cos(\\frac{\\alpha - \\beta}{2})}} \\tag{1}$$ $$2\\cos(4x)\\cos(2x) = 1 + \\cos(8x)$$ And then similarly: $$\\bbox[Beige]{\\boxed{\\cos2\\alpha=2\\cos^2\\alpha-1}} \\tag{2}$$ $$2\\cos(4x)\\cos(2x) = 1 + 2\\cos^2(4x) - 1$$ After that I subtracted $1$ from $1$ in the right part and divided both right and left parts by $2$: $$\\cos(4x)\\cos(2x) = \\cos^2(4x)$$ Then once again I divided both parts. This time by $\\cos(4x)$ and with the aim not to lose a root I checked whether $\\cos(4x)$ can be equal to zero. $$\\cos(4x) = 0 \\implies 4x = \\frac{\\pi}{2} + \\pi k \\implies \\boxed{x=\\frac{\\pi}{8} + \\frac{\\pi k}{4}}$$ so here I found the", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "given below: \\begin{align} &\\sin (A) \\cos (B)=\\dfrac{1}{2}(\\sin (A+B)+\\sin (A-B)) \\\\ &\\cos (A) \\sin (B)=\\dfrac{1}{2}(\\sin (A+B)-\\sin (A-B)) \\\\ &\\sin (A) \\sin (B)=\\dfrac{1}{2}(\\cos (A-B)+\\cos (A+B)) \\\\ &\\cos (A) \\cos (B)=\\dfrac{1}{2}(\\cos (A+B)-\\cos (A-B)) \\end{align} Product To Sum Examples 1. Find the value of sin 45o sin 15o using the product to sum formulas. Solution: Using the third product to sum formulas above we have, $\\sin (A) \\sin (B)=\\dfrac{1}{2}(\\cos (A-B)+\\cos (A+B))$ Substituting the values we get, $\\sin 45^{\\circ} \\sin 15^{\\circ}=\\dfrac{1}{2}\\left(\\cos 30^{\\circ}-\\cos 60^{\\circ}\\right)$ 0.866+0.5=1.366 2. Write the product of the $2 \\cos \\left(\\dfrac{7 x}{2}\\right) \\cos \\left(\\dfrac{3 x}{2}\\right)$ as a sum Solution: Using the 4th identity of the product to sum from above, we get \\begin{align} &\\cos (A) \\cos (B)=\\dfrac{1}{2}(\\cos (A+B)-\\cos (A-B)) \\\\ &=\\frac{1}{2} \\times 2(\\cos 5 x-\\cos 2 x) \\\\ &=(\\cos 5 x-\\cos 2 x) \\end{align} State The Product To Sum Formula and their Derivation To derive the product to sum derivations we need to know the sum and difference trigonometric identities. These are as listed below: \\begin{align} &\\sin (A+B)=\\sin (A) \\cos (B)+\\cos (A) \\sin (B) \\ldots \\ldots(a) \\\\ &\\cos (A+B)=\\cos (A) \\cos (B)-\\sin (A) \\sin (B) \\ldots \\ldots(b) \\\\ &\\sin (A-B)=\\sin (A) \\cos (B)-\\cos (A) \\sin (B) \\ldots \\ldots(c) \\\\ &\\cos (A-B)=\\cos (A) \\cos (B)+\\sin (A) \\sin (B) \\ldots \\ldots(d) \\end{align} (a) Derivation for $\\sin (A) \\cos (B)=\\dfrac{1}{2} \\sin (A+B)+\\sin (A-B)$ Adding", "# Find minimum value of $8 \\cos x + 4 \\sin x$ and corresponding value of $x$ Find minimum value of $8 \\cos x + 4 \\sin x$ and corresponding value of $x$ I used the R method to simplify it - $\\sqrt{80} \\cos (x-26.565)$ Minimum value of that = $\\sqrt{80} \\cos (x-26.565) = - \\sqrt{80}$ $\\cos (x-26.565) = -1$ This cosine value lies in the 2nd and 3rd quadrant letting $x-26.565 = y$ y reference angle = $\\cos^-1 (-1) = 180$ 2nd quadrant - $180 - y (ref) = 0$ 3rd quadrant - $180 + y(ref) = 360$ Therefore , $x = 26.565, 386.565$ Why am I wrong ? The minimum value is $206.6$ Your method is perfectly fine, you just made a mistake at the end. $\\cos(u)=-1\\iff u\\equiv 180°\\pmod{360°}$ Here $u=x-x_0$ so you should get $x\\equiv 180°+x_0\\pmod{360°}$ Applying to $x_0=26.565°$ you get $x = 206.565°$ Since $$8^2+4^2=80$$ you know that $8\\cos x+4\\sin x=\\sqrt{80}\\cos(x-\\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\\pi/2$: \\begin{align} 8&=\\sqrt{80}\\cos\\alpha\\\\ 4&=\\sqrt{80}\\sin\\alpha \\end{align} Thus the angle $\\alpha$ is in the first quadrant and so $$\\alpha=\\arcsin\\frac{4}{\\sqrt{80}}=\\arcsin\\frac{1}{\\sqrt{5}}$$ In degrees this is $26.565$. In radians it is $0.464$ (rounding to three decimal digits). The point where the minimum value $-\\sqrt{80}$ is reached is when $x-\\alpha$ is the straight angle. In degrees the value", "sin(2x) with a double angle formula but I'm not getting the answers. The range is 0 - 360. since $0 < x < 360$ $0 < 2x < 720$ $3\\sin(2x) = 2$ $\\sin(2x) = \\frac{2}{3}$ $2x = \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 180 - \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 360 + \\arcsin\\left(\\frac{2}{3}\\right)$ $2x = 540 - \\arcsin\\left(\\frac{2}{3}\\right)$ solve for $x$ in all four equations", "sides is clearly 7 - 3 = 4. The length of the other side is 2 - (-6) = 8. Thus, 4*8 = 32. You may count the number of miniature 1x1 squares to verify, but it is not practical for larger questions. ## Solution to Question 7 This question is a basic recall question. You should know the unit circle by heart. You will only ever be expected to know the basic unit circle angles without the use of a calculator. (58) \\begin{align} sin(60^\\circ) = \\frac{\\sqrt{3}}{2} \\end{align} (59) \\begin{align} tan(\\frac{5 \\pi}{6}) = \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3} \\end{align} (60) \\begin{align} csc(30^\\circ) = \\frac{1}{tan(30^\\circ)} = \\frac{1}{ \\frac{1}{\\sqrt{3}} } = \\sqrt{3} \\end{align} ## Solution to Question 8 There are two ways to do this question. One way is with the unit circle and with the CAST rule. We know that sinx is positive in quadrants 1 and quadrants 2. We also know that cosx is positive in quadrants 1 and quadrants 4. Hence, sinx = cosx has only one unique solution which is in quadrant 1. This only occurs when x = 45 degrees according to the unit circle. (61) \\begin{align} sin(45^\\circ) = cos(45^\\circ) \\end{align} Another way of doing this is algebraically, though some answers may be missed. (62) $$sin(x) = cos(x)$$ (63) \\begin{align} \\frac{sin(x)}{cos(x)} = 1 \\end{align} (64) \\begin{align} tan(x)", "2 d.p.)}$ Now we can write the complex number in polar form. $(-1+4i) = \\sqrt{17}(\\cos 1.82 + i \\sin 1.82)$ So \\begin{align} (-1+4i)^{4} &= \\bigl(\\sqrt{17}(\\cos 1.82+ i\\sin 1.82)\\bigr)^{4}\\\\ &= (\\sqrt{17})^{4} \\times \\bigl(\\cos 1.82 + i\\sin 1.82)\\bigr)^{4}\\\\ &= 17^2 \\times \\bigl( \\cos (4 \\times 1.82)+i\\sin (4 \\times 1.82)\\bigr) & \\text{ (by De Moivre's theorem)}\\\\ &=289\\bigl(\\cos 7.28 + i \\sin 7.28\\bigr) \\text{ in polar form, or}\\\\ &=156.9 +242.7i \\text{ (to 1 d.p.)} \\end{align} Workbooks These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.", "part. 23. Mar 20, 2009 ### Cyosis Barring any mistakes from my side you can do it without using any trigonometric identities: \\begin{align*} \\int \\cos^2 x\\,dx &= \\int 1*\\cos^2 x\\,dx \\\\ &= x \\x cos^2 x + \\int 2x \\cos x \\sin x\\,dx \\\\ &= x \\cos^2 x+(x \\sin^2 x- \\int \\sin^2 x\\,dx) \\\\ &= x \\cos^2x + x \\sin^2 x - (-\\sin x \\cos x- \\int -\\cos^2 x\\,dx) \\\\ &= x (\\cos^2x + \\sin^2 x) +\\sin x \\cos x- \\int \\cos^2 x\\,dx \\end{align*} It follows that: $$2\\int \\cos^2 x\\,dx=\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x \\Rightarrow$$ \\begin{align*} \\int \\cos^2 x\\,dx &= \\frac{1}{2} (\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x) \\\\ &= \\frac{1}{2} (\\x x +\\frac{1}{2} \\sin {2x}) \\\\ &= \\frac{x}{2}+\\frac{1}{4} \\sin {2x} \\end{align} This is the same result you get by using the double angle identity from the start, as it should be. Last edited: Mar 20, 2009 24. Mar 20, 2009 ### itchy8me brilliant! using the coefficient 1. damn thats a sneaky one. thanks guys and girls ;) !! :-D 25. Jan 5, 2010 ### ntc2415 Just use substitution. This makes it all so much easier. First you set u to be cos x$$\\int(cos x)$$2 dx then substitute cos x with u x$$\\int(u)$$2 dx Then you take the derivative of both sides", "# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 37 The value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$ is $\\frac{1}{8}-\\frac{1}{8}\\cos 4x$. #### Work Step by Step We have to find the value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$; various cos and sin formulas are used which represent the answer in the form of cos. \\begin{align} & {{\\sin }^{2}}x{{\\cos }^{2}}x=\\left( \\frac{1-\\cos 2x}{2} \\right)\\left( \\frac{1+\\cos 2x}{2} \\right) \\\\ & =\\frac{1-{{\\cos }^{2}}2x}{4} \\\\ & =\\frac{1}{4}-\\frac{1}{4}{{\\cos }^{2}}2x \\\\ & =\\frac{1}{4}-\\frac{1}{4}\\left( \\frac{1+\\cos \\left( 2\\times 2x \\right)}{2} \\right) \\end{align} So, in step 2 the identity of $\\left( a+b \\right)\\left( a-b \\right)={{a}^{2}}+{{b}^{2}}$ is used. \\begin{align} & {{\\sin }^{2}}x{{\\cos }^{2}}x=\\frac{1}{4}-\\frac{1}{8}\\left( 1+\\cos 4x \\right) \\\\ & =\\frac{1}{4}-\\frac{1}{8}-\\frac{1}{8}\\left( \\cos 4x \\right) \\\\ & =\\frac{1}{8}-\\frac{1}{8}\\left( \\cos 4x \\right) \\end{align} In the last step $\\frac{1}{8}$ is distributed and subtracted from $\\frac{1}{4}$. Thus, the value of ${{\\sin }^{2}}x{{\\cos }^{2}}x$ is $\\frac{1}{8}-\\frac{1}{8}\\cos 4x$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]

[ "# Trigonometry/The sine of 18 degrees The trigonometric functions of angles that are multiples of 3º can all be expressed as surds. This page will only deal with angles that are multiples of 18º. Consider the equation sin(5x) - 1 = 0. Clearly, the only solutions are that 5x = 90º, 90º + 360º, 90º + 2x360º, 90º + 3x360º ... Thus x = 18º, 90º, 162º, 234º or 306º. (There is no need to go further, since increasing x by 360º does not affect its sine.) Sin(x) is, after using various simplifications, sin(18º) twice, 1, -sin(54º) twice. This equation can be solved in another way. Using the formula for sin(a+b), we can expand sin(5x) to get $\\displaystyle 16\\sin^5(x) - 20 \\sin^3(x) + 5 \\sin(x) - 1 = 0$. If we write s = sin(x), this becomes a quintic equation in s, $\\displaystyle 16s^5 - 20s^3 + 5s - 1 = 0$ i.e. $\\displaystyle (s-1)(4s^2+2s-1)^2 = 0$ so either s = 1 (corresponding to x = 90º) or $\\displaystyle s = \\frac{1}{4} (\\pm \\sqrt{5} - 1)$ (corresponding to x = 18º or 54º). Thus $\\displaystyle \\sin(18^\\circ) = \\frac{1}{4} (\\sqrt{5} - 1) \\approx 0.30902$. $\\displaystyle \\sin(54^\\circ) = \\frac{1}{4} (\\sqrt{5} + 1) \\approx 0.80902$. Exercise: More angles By considering the equation sin(5x) = 0 show that $\\displaystyle \\sin(36^\\circ) = \\pm$$\\sqrt", "reasoning is sound) The sine of an angle (specified in radians) can be computed by making use of the approximation $\\sin x \\approx x$ if x is sufficiently small, and the trigonometric identity $\\sin x = 3\\sin{\\frac{x}{3}} - 4\\sin^3{\\frac{x}{3}}$ to reduce the size of the argument of sin. (For purposes of this exercise an angle is considered “sufficiently small” if its magnitude is not greater than 0.1 radians.) These ideas are incorporated in the following procedures: (define (cube x) (* x x x)) (define (p x) (- (* 3 x) (* 4 (cube x)))) (define (sine angle) (if (not (> (abs angle) 0.1)) angle (p (sine (/ angle 3.0))))) 1. How many times is the procedure p applied when (sine 12.15) is evaluated? The way sine works is that when the absolute value of angle is not above 0.1, it returns angle. Otherwise, it invokes p with a recursive call to sine but with the angle having been divided by 3. So I think what will happen is that the procedure will keep building up a deferred chain of p's until the angle is less than 0.1. Starting from 12.15 and dividing by 3 takes 5 steps to get to a value below 0.1, so my initial guess is that the procedure p will be applied 5 times.", "first one should do the trick, the second is being extra careful. -Dan 7. Originally Posted by qbkr21 Here are the directions the computer system is still not taking it: Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2); Thanks! He did not say that: $\\displaystyle \\cos(105^{\\circ})=\\sin(15^{\\circ})$ what he did say was: $\\displaystyle \\cos(105^{\\circ})=-\\sin(15^{\\circ})$ and that: $\\displaystyle \\sin(15^{\\circ})=\\frac{\\sqrt{2-\\sqrt{3}}}{2}$ so: $\\displaystyle \\cos(105^{\\circ})=-\\frac{\\sqrt{2-\\sqrt{3}}}{2}$ RonL 8. Re: Ok thanks Captain Black", "the function does not take angles to numbers; it takes numbers to numbers. The domain of the other sine function is angle measurements; an angle measurement consists of a number and the units of measurement of the angle. Just as a single interval of times or a single linear distances can be written in multiple different ways using various numbers with various units, a single angle can be written in multiple different ways with different units. Hence it is correct to write $$180^\\circ = \\pi\\mbox{ rad},$$ where the symbol $^\\circ$ indicates units of degrees and rad is the symbol for radians, or (if you are in a context where it is appropriate to use the \"other\" function named $\\sin(\\cdot)$) to write $$\\sin(30^\\circ) = \\sin\\left(\\frac\\pi6 \\mbox{ rad}\\right).$$ On the other hand, in a more \"pure\" mathematical context using the function $\\sin: \\mathbb R \\to \\mathbb R,$ strictly speaking we should write $$\\sin\\left(\\frac\\pi6\\right) = \\frac 12 \\neq \\sin(30) \\approx -0.988.$$ In practice, the tendency to interpret the notation $\\sin(30)$ as $\\sin(30^\\circ)$ is so strong that if you type sin(30) as input to Wolfram Alpha (for example) it will return $0.5$ as the answer. On the other hand if you put =sin(30) in a cell in some widely-used spreadsheet programs you may be in for a surprise. One just has to", "### Π 8 {\\Displaystyle {\\Frac {\\Pi } {8}}} Is Half Of. {\\displaystyle {\\frac {\\pi } {4}}.} sin ⁡ π 8 = sin ⁡ ( 1 2. Whenever solving a trig equation set equal to a negative value, we first find our reference angle by setting the equation equal to the same value, only positive. (do not worry about typing the little o degree symbol.) sin(195°) = now evaluate the expression above to give an exact value. ### Without Using The Unit Circle Or Table! To find the exact value of f(x), we suggest the following steps: Remember the ordered pair xysino tt 4. A diagram of the unit circle is shown below:", "basic trigonometry ratios. Then simply use $sin (x-y) = sin\\,x\\,cos\\,y - cos\\,x\\,sin\\,y$ Do the same for question 2. Because we are not considering any obtuse angles (no angle in a right angle triangle can be greater than $\\frac{\\pi}{2}$) , all our angles are within your limits.", "# Exercise 1.15 of SICP Exercise 1.15: The sine of an angle (specified in radians) can be computed by making use of the approximation sin(x) = x if x is sufficiently small, and the trigonometric identity $$\\sin(x) = 3\\sin(\\frac{x}{3})-4\\sin^3(\\frac{x}{3})$$ to reduce the size of the argument of sin. (For purposes of this exercise an angle is considered “sufficiently small” if its magnitude is not greater than 0.1 radians.) These ideas are incorporated in the following procedures: (define (cube x) (* x x x)) (define (p x) (- (* 3 x) (* 4 (cube x)))) (define (sine angle) (if (not (> (abs angle) 0.1)) angle (p (sine (/ angle 3.0))))) a. How many times is the procedure p applied when (sine 12.15) is evaluated? 5 times. (p (sine 4.05)) (p (p (sine 1.35))) (p (p (p (sine 0.45)))) (p (p (p (p (sine 0.15))))) (p (p (p (p (p (sine 0.05)))))) (p (p (p (p (p 0.05))))) (p (p (p (p 0.14950000000000002)))) (p (p (p 0.43513455050000005))) (p (p 0.9758465331678772)) (p -0.7895631144708228) -.39980345741334 b. What is the order of growth in space and number of steps (as a function of a) used by the process generated by the sine procedure when (sine a) is evaluated? Order of in number of steps is $$\\Theta(\\log_3(n))$$ because log3(n) counts the number of factors", "Ask question # Here's what students ask on Trigonometry Trigonometric Functions asked 2021-03-12 ### If the measure of an angle is 23°17′18′′, what is the measure of its supplement? Trigonometric Functions asked 2021-03-06 ### $$\\displaystyle{4}{{\\cos}^{{2}}{x}}−{3}={0}$$ Trigonometric Functions asked 2021-03-06 ### Given the values for sin t and cos t, use reciprocal and quotient identities to find the values of the other trigonometric functions of t. $$\\displaystyle{\\sin{{t}}}=\\frac{{3}}{{4}}{\\quad\\text{and}\\quad}{\\cos{{t}}}=\\frac{\\sqrt{{7}}}{{4}}$$ Trigonometric Functions asked 2021-03-06 ...", "Evaluate $\\dfrac{1}{\\sin{10^°}}-\\dfrac{\\sqrt{3}}{\\cos{10^°}}$ The sine of angle ten degrees and cosine of angle ten degrees are involved in forming a trigonometric expression and we have to evaluate this trigonometric expression by simplification. $\\dfrac{1}{\\sin{(10^°)}}-\\dfrac{\\sqrt{3}}{\\cos{(10^°)}}$ Simplify the Trigonometric expression The two terms that contain trigonometric functions are connected by a minus sign and each term in the expression is in fraction form. Hence, the trigonometric expression can be simplified by the subtraction of fractions. $=\\,\\,\\,$ $\\dfrac{1 \\times \\cos{(10^°)}-\\sqrt{3} \\times \\sin{(10^°)}}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ Simplify the Expression in numerator The rational expression contains trigonometric expressions in both numerator and denominator. So, we have to concentrate on simplifying each trigonometric expression. Now, let’s focus on simplifying the trigonometric expression in the numerator of the function. $\\dfrac{1 \\times \\cos{(10^°)}-\\sqrt{3} \\times \\sin{(10^°)}}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ The trigonometric expression is formed by the subtraction and each term is in product form. It is a good idea to express it in sine angle difference form. Let’s check the possibility. $=\\,\\,\\,$ $\\dfrac{1 \\times \\Big(1 \\times \\cos{(10^°)}-\\sqrt{3} \\times \\sin{(10^°)}\\Big)}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ $=\\,\\,\\,$ $\\dfrac{\\dfrac{2}{2} \\times \\Big(1 \\times \\cos{(10^°)}-\\sqrt{3} \\times \\sin{(10^°)}\\Big)}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ $=\\,\\,\\,$ $\\dfrac{2 \\times \\Bigg(\\dfrac{1 \\times \\cos{(10^°)}-\\sqrt{3} \\times \\sin{(10^°)}}{2}\\Bigg)}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ $=\\,\\,\\,$ $\\dfrac{2 \\times \\Bigg(\\dfrac{1 \\times \\cos{(10^°)}}{2}-\\dfrac{\\sqrt{3} \\times \\sin{(10^°)}}{2}\\Bigg)}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ $=\\,\\,\\,$ $\\dfrac{2 \\times \\Bigg(\\dfrac{1}{2} \\times \\cos{(10^°)}-\\dfrac{\\sqrt{3}}{2} \\times \\sin{(10^°)}\\Bigg)}{\\sin{(10^°)} \\times \\cos{(10^°)}}$ According to trigonometry, sin of 30 degrees is quotient of", "# Sin 1 ## Concept of Sin 1 There are three major ratios in trigonometry upon which the trigonometric functions and formulas are based. The sine function is one of them. The sine function (sin) of an angle gives the ratio of the perpendicular (opposite side of the angle) to the hypotenuse. Similarly, the inverse sine function (sin-1) gives the ratio of the hypotenuse to the perpendicular of an angle. Sin 1 in radians has a value of 0.8414709848. The basic angles which are required very frequently are 0,30,45,60,90 degrees. They can also be expressed in radians as π/2, π/3, π/4, π/6, π, etc. In radians, the value of sine 1 is 0.8414709848. We know, π/3 = 1.047198≈1 $Sin (\\frac{\\pi}{3}) = \\frac{\\sqrt{3}}{2}$ Sin π = 0 Now using these data, we can write; $sin 1 = sin \\frac{\\pi}{3} − (\\frac{\\pi}{3} − 1)$ ⟹ sin1 = $sin \\frac{\\pi}{3} cos(\\frac{\\pi}{3} − 1)−cos \\frac{\\pi}{3} sin(\\frac{\\pi}{3} − 1)$ The angle π/3−1=0.047198 is a very small angle. We know that, for small angles θ, Sinθ ≈ θ and cos⁡θ ≈ 1 hence, Sin 1 ≈ $(\\frac{\\sqrt{3}}{2} \\times 1) − \\frac{1}{2} \\times (\\frac{\\pi}{3} − 1)$ Therefore, sin1 ≈ 0.842427 ### How to find the Value of Sine 1? The sine of an angle can take its argument as either radian or degrees. The rule", "$f(x)$ is sometimes roughly (but imperfectly) summed up in the statement that a small change in $x$ shall produce a small change in $f(x)$. We shall not consider functions having an infinite number of oscillations in a limited region. 5. If we refer to the table on p. 4 [§ 4], it will be seen that for all angles less than 10° the angle in radians and the sine of the angle are equal to three decimal places. If larger tables are consulted, five-place, say, it will be seen that for all angles less than 2.2° the sine of the angle and the angle itself are equal to four decimal places. From this we may well suspect that $\\lim_{x \\to 0} \\frac{\\sin\\ x}{x} = 1$.", "following trigonometry values: Function = sin Angle = $3 \\frac{\\pi}{4}$ Using the Trig Exact Value Calculator, find the trigonometric function given above. ### Solution We may rapidly get the precise value of the trigonometric function using the Trig Exact Value Calculator. First, we use the Trig Exact Value Calculator to select the function we want to use, sin. The function and the angle are entered into the Trig Exact Value Calculator; the angle is $3 \\frac{\\pi}{4}$. After inputting all required information into the Trig Exact Value Calculator, we click the “Submit” button. The calculator computes the results quickly and is presented below the calculator. The following results are taken from the Trig Exact Value Calculator: Input: $3 \\times (\\frac{\\pi}{4})$ Exact Result: $\\frac{1}{\\sqrt{2}}$", "functions. In each case, the trigonometric functions are cofunctions of one another, and the angles are complementary angles. Round your answers to four places past the decimal point. $$\\displaystyle{{\\sec{{6.7}}}^{\\circ},}{\\cos{{e}}}{c}{83.3}^{\\circ}$$ Find the exact value of the trigonometric function $$\\displaystyle\\frac{{\\cos{{\\left({9}\\pi\\right)}}}}{{4}}$$.", "# Question of inverse trigonometry. How do I evaluate $\\sin \\left(\\sec^{-1}\\left(\\frac{-7}{5}\\right)\\right)$? The answer given is $\\frac{2\\sqrt{6}}{7}$, but I have no idea how to get it. Hope someone can explain it for me. Thanks in advance. where the last line is due to Pythagoras Theorem. Also think of which quadrant is $\\cos^{-1}\\left( -\\frac57 \\right)$ located.", "## Trigonometry (11th Edition) Clone $1$ Function Values of Special Angles$:$ $\\displaystyle \\cos 60^{o}=\\frac{1}{2}, \\quad \\sin 30^{o}=\\frac{1}{2}$ $\\displaystyle \\cos 60^{\\mathrm{o}}+2\\sin^{2}30^{\\mathrm{o}}=\\frac{1}{2}+2(\\frac{1}{2})^{2}$ $=\\displaystyle \\frac{1}{2}+2(\\frac{1}{4})$ $=\\displaystyle \\frac{1}{2}+\\frac{1}{2}=1$", "OP = \\frac{1}{{\\sin \\alpha }} \\times 1 = \\csc \\alpha .$ ## Trigonometric Functions of Special Angles There are certain common angles that are more frequently used in trigonometry. The following table contains the values of trigonometric functions for such angles. See solved problems on Page 2.", "- 2\\frac{1}{\\cos{(1)}} = -\\sin{(1)} - 0$ which finally gives me $-0.841470984radians = -48.21273601degrees$ or I'm missing something from basic trigonometry? I'm really frustrated... 4. ## Re: Where the hell is my mistake You are trying to get a numerical value, not an angle. Just plug this into your calculator AS IS. Do NOT use inverse trigonometric functions. I have no idea why you got that 2/cos(1) is 0... 5. ## Re: Where the hell is my mistake Originally Posted by Prove It You are trying to get a numerical value, not an angle. Just plug this into your calculator AS IS. Do NOT use inverse trigonometric functions. I have no idea why you got that 2/cos(1) is 0... that simple $-7.692508626$ , oh my god", "= ½. You know that, I think. Now you are left with :$\\sin^{-1}(\\frac{ 1 }{ 2 })$Doesn't that sound familiar? You are looking for an angle of which the sine is ½. That is also a well known angle...", "p if f(x) = f(x + np), n = 1, 2, 3..... A function with period P will repeat on intervals of length P, and these intervals are referred to as periods. The following table summarizes the names given to periodic functions based on the number of independent periods they posses: Number of Periods Nature of Functions 1 Single periodic function 2 Doubly Periodic function 3 Triply Periodic function ## Trigonometric Functions of Any Angle Trigonometric functions are used to calculate the internal angles of triangles. Trigonometric functions are sine function, cosine function, tangent function, cotangent function, cosecant function and secant function. These functions give a definite value of angles in certain degrees. We can find the trigonometric functions of any angle. Let us go through certain examples in which we will calculate trigonometric functions of some specific angles. Let us take first angle as 00. Sin 0$^{\\circ}$ = 0 Cos 0$^{\\circ}$ = 1 Tan 0$^{\\circ}$ = 0 Cot 0 $^{\\circ}$ = not defined Sec 0$^{\\circ}$ = 1 Csc 0$^{\\circ}$ = not defined So, we can see that values of cot and csc functions are not defined. Similarly, let us calculate values of all above six functions for angle 30$^{\\circ}$. Here, we will find each of the values as below: Sin 30$^{\\circ}$ = $\\frac{1}{2}$ Cos 30$^{\\circ}$ = $\\frac{\\sqrt3}{2}$", "suspect there's more to the original question than was posted. 4. There is nothing more to this question. It is simply to find the exact value of sin 12 degrees. Only thing I could think of was sin(60/5), but these odd numbers really mess with my head. 5. Originally Posted by ISeriouslyNeedHelp What is the exact value of sin 12 degrees? Only hint I was given was to use compound angle formula. Had this on the test but I left it blank, now I just want to know the solution. To do this you need: $\\displaystyle \\sin(12^{\\circ})=\\sin(30^{\\circ}-18^{\\circ})$ We know the values of the trig functions for $\\displaystyle 30^{\\circ}$ and so if we also know the trig function values for $\\displaystyle 18^{\\circ}$ we are home and dry. We can get the values we need for the $\\displaystyle \\sin$ and $\\displaystyle \\cos$ of $\\displaystyle 18^{\\circ }$ from the geometry of a regular pentagon (which we could have guessed as we are dealing with integer fractions of $\\displaystyle 72^{\\circ}=360^{\\circ}/5)$, where we find: $\\displaystyle \\sin(18^{\\circ})=\\frac{1}{2\\Phi}$ where $\\displaystyle \\Phi=\\frac{1+\\sqrt{5}}{2}$ is the Golden number. CB , , , , , , , , , , , , , , # Sin 12= Click on a term to search for related topics." ]

[ "TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-1) Prove the following identities : $\\,1(i).\\,\\,\\,\\sin10^{\\circ}\\sin 50^{\\circ}+\\sin50^{\\circ}\\sin250^{\\circ}\\\\+\\sin250^{\\circ}\\sin10^{\\circ}=-\\frac34$ Sol. $\\,\\,\\,\\sin10^{\\circ}\\sin 50^{\\circ}+\\sin50^{\\circ}\\sin250^{\\circ}\\\\+\\sin250^{\\circ}\\sin10^{\\circ}\\\\=\\frac 12(2\\sin10^{\\circ}\\sin50^{\\circ})+\\frac 12(2\\sin50^{\\circ}\\sin 250^{\\circ})\\\\+\\frac 12(2\\sin250^{\\circ}\\sin10^{\\circ})\\\\=\\frac 12\\left[\\cos(50^{\\circ}-10^{\\circ})-\\cos(50^{\\circ}+10^{\\circ})\\\\+\\cos(250^{\\circ}-50^{\\circ})-\\cos(250^{\\circ}+50^{\\circ})\\\\+\\cos(250^{\\circ}-10^{\\circ})-\\cos(250^{\\circ}+10^{\\circ})\\right]\\\\=\\frac 12[\\cos40^{\\circ}-\\cos60^{\\circ}+\\cos200^{\\circ}-\\cos300^{\\circ}\\\\+\\cos240^{\\circ}-\\cos260^{\\circ}]\\\\=\\frac 12\\left[\\cos40^{\\circ}-\\frac 12+\\cos(2\\times90^{\\circ}+20^{\\circ})\\\\-\\cos(4\\times 90^{\\circ}-60^{\\circ})+\\cos(3 \\times 90^{\\circ}-30^{\\circ})\\\\-\\cos(3\\times 90^{\\circ}-10^{\\circ})\\right]\\\\=\\frac12 \\left[\\cos40^{\\circ}-\\frac 12-\\cos20^{\\circ}-\\cos60^{\\circ} \\\\-\\sin30^{\\circ}+\\sin10^{\\circ}\\right]\\\\=\\frac12\\left(\\cos40^{\\circ}-\\frac 12-\\cos20^{\\circ}-\\frac 12-\\frac 12\\\\+\\cos(90^{\\circ}-10^{\\circ})\\right)\\\\=\\frac 12\\left(\\cos40^{\\circ}+\\cos80^{\\circ}-\\cos20^{\\circ}-\\frac 32\\right)\\\\=\\frac 12\\left(\\sin(90^{\\circ}-40^{\\circ})+2\\sin\\frac{80^{\\circ}+20^{\\circ}}{2}\\sin\\frac{20^{\\circ}-80^{\\circ}}{2}\\\\-\\frac 32\\right)\\\\=\\frac 12\\left(\\sin50^{\\circ}+2\\sin50^{\\circ}\\sin(-30^{\\circ})-\\frac 32\\right)\\\\=\\frac 12\\left(\\sin 50^{\\circ}+2\\sin50^{\\circ}(-\\frac 12)-\\frac 32\\right)\\\\=\\frac 12\\left(\\sin50^{\\circ}-\\sin50^{\\circ}-\\frac 32\\right)\\\\=-\\frac 34\\,\\,\\text{(proved)}$ $\\,1(ii)\\,\\,\\frac{\\sin\\alpha\\sin11\\alpha+\\sin3\\alpha\\sin7\\alpha}{\\sin\\alpha\\cos11\\alpha+\\sin3\\alpha\\cos7\\alpha}=\\tan8\\alpha$ Sol. $\\,\\,\\frac{\\sin\\alpha\\sin11\\alpha+\\sin3\\alpha\\sin7\\alpha}{\\sin\\alpha\\cos11\\alpha+\\sin3\\alpha\\cos7\\alpha}\\\\=\\frac{(2\\sin\\alpha\\sin11\\alpha)+(2\\sin3\\alpha\\sin7\\alpha)}{(2\\sin\\alpha\\cos11\\alpha)+(2\\sin3\\alpha\\cos7\\alpha)}\\\\=\\frac{\\cos(11\\alpha-\\alpha)-\\cos(11\\alpha+\\alpha)+\\cos(7\\alpha-3\\alpha)-\\cos(7\\alpha+3\\alpha)}{\\sin(11\\alpha+\\alpha)-\\sin(11\\alpha-\\alpha)+\\sin(7\\alpha+3\\alpha)-\\sin(7\\alpha-3\\alpha)}\\\\=\\frac{\\cos10\\alpha-\\cos12\\alpha+\\cos4\\alpha-\\cos10\\alpha}{\\sin12\\alpha-\\sin10\\alpha+\\sin10\\alpha-\\sin4\\alpha}\\\\=\\frac{\\cos4\\alpha-\\cos12\\alpha}{\\sin12\\alpha-\\sin4\\alpha}\\\\=\\frac{2\\sin\\frac{12\\alpha+4\\alpha}{2}\\sin\\frac{12\\alpha-4\\alpha}{2}}{2\\sin\\frac{12\\alpha-4\\alpha}{2}\\cos\\frac{12\\alpha+4\\alpha}{2}}\\\\=\\tan8\\alpha$ $\\,1(iii)\\,\\,\\,\\tan\\theta\\tan(\\frac{\\pi}{3}+\\theta)\\tan(\\frac{\\pi}{3}-\\theta)\\\\=\\tan\\theta \\times \\frac{\\sin(\\frac{\\pi}{3}+\\theta)\\sin(\\frac{\\pi}{3}-\\theta)}{\\cos(\\frac{\\pi}{3}+\\theta)\\cos(\\frac{\\pi}{3}-\\theta)}\\\\=\\tan\\theta \\times \\frac{2\\sin(\\frac{\\pi}{3}+\\theta)\\sin(\\frac{\\pi}{3}-\\theta)}{2\\cos(\\frac{\\pi}{3}+\\theta)\\cos(\\frac{\\pi}{3}-\\theta)}\\\\=\\tan\\theta \\times \\frac{\\cos2\\theta-\\cos120^{\\circ}}{\\cos120^{\\circ}+\\cos2\\theta}\\\\=\\tan\\theta\\times \\frac{\\cos2\\theta+\\frac 12}{\\cos2\\theta-\\frac 12}\\\\=\\tan\\theta \\times \\frac{2\\cos2\\theta+1}{2\\cos2\\theta-1}\\\\=\\frac{\\sin\\theta}{\\cos\\theta} \\times \\frac{2\\cos2\\theta+1}{2\\cos2\\theta-1}\\\\=\\frac{2\\sin\\theta\\cos2\\theta+\\sin\\theta}{2\\cos\\theta\\cos2\\theta-\\cos\\theta}\\\\=\\frac{\\sin3\\theta-\\sin\\theta+\\sin\\theta}{\\cos3\\theta+\\cos\\theta-\\cos\\theta}\\\\=\\tan3\\theta$ $\\,2(i)\\,$ If $\\,\\,a\\cos\\phi=b\\cos\\theta,\\,\\,$ show that , $\\,\\,a\\tan\\theta+b\\tan\\phi=(a+b)\\tan\\frac{\\theta+\\phi}{2}.$ Sol. $\\,\\,a\\tan\\theta+b\\tan\\phi\\\\=\\frac{\\left(\\frac{b\\cos\\theta}{\\cos\\phi}\\right)\\frac{\\sin\\theta}{\\cos\\theta}+b\\frac{\\sin\\phi}{\\cos\\phi}}{\\left(\\frac{b\\cos\\theta}{\\cos\\phi}\\right)+b}\\\\=\\frac{b\\sin\\theta+b\\sin\\phi}{b\\cos\\theta+b\\cos\\phi}\\\\=\\frac{2\\sin\\frac{\\theta+\\phi}{2}\\cos\\frac{\\theta-\\phi}{2}}{2\\cos\\frac{\\theta+\\phi}{2}\\cos\\frac{\\theta-\\phi}{2}}\\\\=\\tan\\frac{\\theta+\\phi}{2} \\\\ \\Rightarrow a\\tan\\theta+b\\tan\\phi=(a+b)\\tan\\frac{\\theta+\\phi}{2}$ $\\,2(ii)\\,$ If $\\,\\,p\\sin\\alpha=q\\sin(120^{\\circ}+\\alpha)=r\\sin(240^{\\circ}+\\alpha),\\,\\,$ prove that , $\\,\\,pq+qr+rp=0.$ Sol. Let$\\,\\,p\\sin\\alpha=q\\sin(120^{\\circ}+\\alpha)\\\\=r\\sin(240^{\\circ}+\\alpha)=k(\\neq 0) \\\\ \\therefore \\frac kp+\\frac kq+\\frac kr\\\\=\\sin\\alpha+\\sin(120^{\\circ}+\\alpha)+\\sin(240^{\\circ}+\\alpha)\\\\=\\sin\\alpha+2\\sin(180^{\\circ}+\\alpha)\\cos60^{\\circ}\\,\\,[\\text{By (*)}]\\\\=\\sin\\alpha-2\\sin\\alpha.\\frac 12\\\\=\\sin\\alpha-\\sin\\alpha\\\\=0$ Note[*]: $\\,\\,\\sin(120^{\\circ}+\\alpha)+\\sin(240^{\\circ}+\\alpha)\\\\=2\\sin\\frac{(120^{\\circ}+\\alpha)+(240^{\\circ}+\\alpha)}{2}\\cos\\frac{(240^{\\circ}+\\alpha)-(120^{\\circ}+\\alpha)}{2}\\\\=2\\sin(180^{\\circ}+\\alpha)\\cos60^{\\circ}$ $\\,2(iii)\\,$ If $\\,\\,\\cos\\theta=n\\cos(\\theta+2\\phi),\\,$ show that , $\\,(n+1)\\tan(\\theta+\\phi)=(n-1)\\cot\\phi$ Sol. We have ,$\\,\\,\\cos\\theta=n\\cos(\\theta+2\\phi) \\cdots(1)$ Now, $\\,\\,\\frac{n+1}{n-1}\\\\=\\frac{\\frac{\\cos\\theta}{\\cos(\\theta+2\\phi)}+1}{\\frac{\\cos\\theta}{\\cos(\\theta+2\\phi)}-1},\\,\\,[\\text{By (1)}]\\\\=\\frac{\\cos\\theta+\\cos(\\theta+2\\phi)}{\\cos\\theta-\\cos(\\theta+2\\phi)}\\\\=\\frac{2\\cos(\\theta+\\phi)\\cos\\phi}{2\\sin(\\theta+\\phi)\\sin\\phi}\\\\=\\cot(\\theta+\\phi)\\cot\\phi \\\\ \\therefore (n+1)\\tan(\\theta+\\phi)=(n-1)\\cot\\phi.$ $\\,3.\\,$ Prove that, $\\,\\,\\sin\\alpha+\\sin\\beta+\\sin\\gamma-\\sin(\\alpha+\\beta+\\gamma)\\\\=4\\sin\\frac{\\alpha+\\beta}{2}\\sin\\frac{\\beta+\\gamma}{2}\\sin\\frac{\\gamma+\\alpha}{2}$ Sol. $\\,\\,\\sin\\alpha+\\sin\\beta+\\sin\\gamma-\\sin(\\alpha+\\beta+\\gamma)\\\\=2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2}+2\\cos\\frac{\\alpha+\\beta+2\\gamma}{2}\\sin\\left(-\\frac{\\alpha+\\beta}{2}\\right)\\\\=2\\sin\\frac{\\alpha=\\beta}{2} \\left(\\cos\\frac{\\alpha-\\beta}{2}-\\cos\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)\\\\=4\\sin\\frac{\\alpha+\\beta}{2}\\sin\\frac{\\beta+\\gamma}{2}\\sin\\frac{\\gamma+\\alpha}{2}\\,\\, [**]$ Note[**] : $\\,\\,\\cos C-\\cos D=2\\sin\\frac{C+D}{2}\\sin\\frac{D-C}{2}$ $\\,4.\\,$ Express $\\,\\,4\\sin A \\cos B\\cos C\\,\\,$ as the sum of four sines. Sol. $\\,\\,4\\sin A \\cos B\\cos C \\\\=2(2\\sin A \\cos B)\\cos C\\\\=2\\left[\\sin(A+B)+\\sin(A-B)\\right]\\cos C\\\\=2\\sin(A+B)\\cos C+2\\sin(A-B)\\cos C\\\\=\\sin(A+B+C)+\\sin(A+B-C)\\\\+\\sin(A-B+C)+\\sin(A-B-C)$ $\\,5.\\,$ Express $\\,\\,\\cos\\alpha+\\cos\\beta+\\cos\\gamma+\\cos(\\alpha+\\beta+\\gamma)\\,\\,$ as the product of three cosines. Sol. $\\,\\,\\cos\\alpha+\\cos\\beta+\\cos\\gamma+\\cos(\\alpha+\\beta+\\gamma)\\\\=(\\cos\\alpha+\\cos\\beta)+\\left[\\cos\\gamma+\\cos(\\alpha+\\beta+\\gamma)\\right]\\\\=2\\cos\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2}+2\\cos\\frac{\\alpha+\\beta+2\\gamma}{2}\\cos\\frac{\\alpha+\\beta}{2}\\\\=2\\cos\\frac{\\alpha+\\beta}{2}\\left[\\cos\\frac{\\alpha-\\beta}{2}+\\cos\\frac{\\alpha+\\beta+2\\gamma}{2}\\right]\\\\=2\\cos\\frac{\\alpha+\\beta}{2}.2\\cos\\frac{\\alpha+\\gamma}{2}\\cos\\frac{\\beta+\\gamma}{2}\\\\=4\\cos\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\beta+\\gamma}{2}\\cos\\frac{\\gamma+\\alpha}{2}$ If you want to download Full solution PDF of Chhaya math solution of Transformation of Sums and Products for class XI, click here.", "= \\frac{\\cos 1^\\circ}{\\sin 1^\\circ}. (6)$$ Now we notice that we can multiply both sides by $\\sin 1^\\circ$. Doing so gives us $$\\sin 1^\\circ(2 (\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ) + 1) = \\cos 1^\\circ. (7)$$ Now, simplifying $\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas. Doing so, gives us $$(\\sin 1^\\circ)(2 \\sin 2^\\circ)+(\\sin 1^\\circ)(2\\sin 4^\\circ) +(\\sin 1^\\circ)(2\\sin 6^\\circ) + (\\sin 1^\\circ)(2\\sin 8^\\circ) +(\\sin 1^\\circ)(2 \\sin 10^\\circ) ... + (\\sin 1^\\circ)(2\\sin 88^\\circ) + \\sin 1^\\circ = \\cos 1^\\circ. (8)$$ Now we can use our product of sines to sum formula and see if we can find a pattern. We will start with expanding $(\\sin 1^\\circ)(2 \\sin 2^\\circ)$. Because the format of our formula for the product of sines is $\\sin \\alpha + \\sin \\beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\\sin 1^\\circ)(\\sin 2^\\circ)$. Our formula for the product of sines is $\\sin \\alpha + \\sin \\beta = \\frac{1}{2}[\\cos(\\alpha - \\beta) - \\cos(\\alpha + \\beta)]$. Plugging in our values into the formula and simplifying gives us $2\\left(\\frac{1}{2}(\\cos 1^\\circ", "Sum To Product Identities Reference > Algebra: Trigonometric Identities Description$$\\sin{x}+\\sin{y}=2\\sin{(\\frac{x+y}{2})}\\cos{(\\frac{x-y}{2})}$$ $$\\sin{x}-\\sin{y}=2\\cos{(\\frac{x+y}{2})}\\sin{(\\frac{x-y}{2})}$$ $$\\cos{x}+\\cos{y}=2\\cos{(\\frac{x+y}{2})}\\cos{(\\frac{x-y}{2})}$$ $$\\cos{x}-\\cos{y}=2\\sin{(\\frac{x+y}{2})}\\sin{(\\frac{x-y}{2})}$$", "# Sum To Product Identities ## Reference > Algebra: Trigonometric Identities Description$$\\sin{x}+\\sin{y}=2\\sin{(\\frac{x+y}{2})}\\cos{(\\frac{x-y}{2})}$$ $$\\sin{x}-\\sin{y}=2\\cos{(\\frac{x+y}{2})}\\sin{(\\frac{x-y}{2})}$$ $$\\cos{x}+\\cos{y}=2\\cos{(\\frac{x+y}{2})}\\cos{(\\frac{x-y}{2})}$$ $$\\cos{x}-\\cos{y}=2\\sin{(\\frac{x+y}{2})}\\sin{(\\frac{x-y}{2})}$$", "we have $\\cos(A)\\cos(B) = (\\dfrac{1}{2})[\\cos(A + B) + \\cos(A - B)]$ $\\sin(A)\\sin(B) = (\\dfrac{1}{2})[\\cos(A - B) - \\cos(A + B)]$ $\\sin(A)\\cos(B) = (\\dfrac{1}{2})[\\sin(A + B) + \\sin(A - B)]$ Exercise $$\\PageIndex{1}$$ Find the exact value of $$\\sin(52.5^\\circ)\\sin(7.5^\\circ)$$. To determine the exact value of $$\\sin(52.5^\\circ)\\sin(7.5^\\circ)$$, we will use the Product-to- Sum identity $\\sin(A)\\sin(B) = (\\dfrac{1}{2})[\\cos(A - B) - \\cos(A + B)]$ So we see that $\\sin(52.5^\\circ)\\sin(7.5^\\circ) = (\\dfrac{1}{2})[\\cos(45^\\circ) - \\cos(60^\\circ)] = (\\dfrac{1}{2})[\\dfrac{\\sqrt{2}}{2} - \\dfrac{1}{2}] = \\dfrac{\\sqrt{2} - 1}{4}$ ### Sum-to-Product Identities As our final identities, we derive the reverse of the Product-to-Sum identities. These identities are called the Sum-to-Product identities. For example, to verify the identity $\\cos(A) + \\cos(B) = 2\\cos(\\dfrac{A + B}{2})\\cos(\\dfrac{A - B}{2})$ we first note that $$A = \\dfrac{A + B }{2}$$ and $$B = \\dfrac{A + B}{2} - \\dfrac{A - B}{2}$$. So $\\cos(A) = \\cos(\\dfrac{A + B}{2} + \\dfrac{A - B}{2}) = \\cos(\\dfrac{A + B}{2})\\cos(\\dfrac{A - B}{2}) - \\sin(\\dfrac{A + B}{2})\\sin(\\dfrac{A - B}{2})$ and $\\cos(B) = \\cos(\\dfrac{A + B}{2} + \\dfrac{A - B}{2}) = \\cos(\\dfrac{A + B}{2})\\cos(\\dfrac{A - B}{2}) + \\sin(\\dfrac{A + B}{2})\\sin(\\dfrac{A - B}{2})$ Adding the left and right sides of (8) and (9) results in $\\cos(A) + \\cos(B) = 2\\cos(\\dfrac{A + B}{2})\\cos(\\dfrac{A - B}{2})$ Also, if we subtract the left and right hands sides of (9) from (8) we obtain $\\cos(A) - \\cos(B)", "Math Help - sum-product/product-sum 1. sum-product/product-sum solve the following using sum-product/product-sum identities $cos20+cos100+cos140=0$ $sin19cos11+sin71sin11=\\frac{1}{2}$ $cos20cos40cos60cos180=\\frac{1}{16}$ $cos20cos40cos80=\\frac{1}{8}$ 2. Re: sum-product/product-sum In this case, Q1, $cos20+cos100+cos140=0$ L.H.S = $2cos60cos40+cos140$ = $cos40+cos140$ = $2cos90cos50$ = $0$ (Since cos90=0) =R.H.S Q2, $sin19cos11+sin71sin11=\\frac{1}{2}$ L.H.S = $\\frac{2sin19cos11+2sin71sin11}{2}$ = $\\frac{sin30+sin8+cos60-cos82}{2}$ = $\\frac{\\frac{1}{2}+\\frac{1}{2}+sin(90-82)-cos82}{2}$ = $\\frac{1+cos82-cos82}{2}$ = $\\frac{1}{2}$ =R.H.S Q3, I think you mention cos180 or cos 80 I can't get any answer. Q4, $cos20cos40cos80=\\frac{1}{8}$ L.H.S = $\\frac{2cos20cos40cos80}{2}$ = $\\frac{(cos60+cos20)cos80}{2}$ = $\\frac{(\\frac{1+2cos20}{2})cos80}{2}$ = $\\frac{\\frac{cos80+cos100+cos60}{2}}{2}$ = $\\frac{\\frac{cos(180-100)+cos100+cos60}{2}}{2}$ = $\\frac{\\frac{-cos100+cos100+\\frac{1}{2}}{2}}{2}$ = $\\frac{1}{8}$ =R.H.S 3. Re: sum-product/product-sum $\\frac{2cos20cos40cos80}{2}$ Where did the 2 in the numerator and denominator come from??? 4. Re: sum-product/product-sum you want to multiply by 2 to numerator and denominator,This is a way to get your answer quickly.", "# How do you use the product to sum formulas to write 10cos75^circcos15^circ as a sum or difference? Jun 12, 2017 = 5/2 #### Explanation: Use trig identity: $\\cos a . \\cos b = \\left(\\frac{1}{2}\\right) \\left(\\cos \\left(a + b\\right) + \\cos \\left(a - b\\right)\\right)$. In this case: $10 \\cos 75. \\cos 15 = \\left(\\frac{10}{2}\\right) \\left(\\cos 90 + \\cos 60\\right)$ Since cos 90 = 0, and $\\cos 60 = \\frac{1}{2}$, therefore 10cos 75.cos 15 = 5/2", "(A+B)-cos (A-B)] cos A cos B=(1/2)[cos (A+B)+cos (A-B)] 4. thanks so much. in looking at my handout, i know see the identities you mention under \"product to sum identities.\" thanks. 5. find the exact value in radical form for: cos(45)cos(15) $cos(a-b) = cosacosb+sinasinb$ = cos(45)*(cos(45-30)) = cos(45)*(cos45cos30+sin45sin30) rest is numerical", "the identity. This is the opposite of sine of a sum, where a positive sign is used on the other side of the identity. ### Examples Example 1 Earlier, you were asked to evaluate $$\\sin 15^{\\circ}$$ and $$\\sin 75^{\\circ}$$ exactly without a calculator. To do this you need to use the sine of a difference and sine of a sum. $$\\begin{array}{l} \\sin \\left(45^{\\circ}-30^{\\circ}\\right)=\\sin 45^{\\circ} \\cos 30^{\\circ}-\\cos 45^{\\circ} \\sin 30^{\\circ}=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\\\ \\sin \\left(45^{\\circ}+30^{\\circ}\\right)=\\sin 45^{\\circ} \\cos 30^{\\circ}+\\cos 45^{\\circ} \\sin 30^{\\circ}=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}+\\sqrt{2}}{4} \\end{array}$$ Example 2 Find the exact value of $$\\tan 15^{\\circ}$$ without using a calculator. $$\\tan 15^{\\circ}=\\tan \\left(45^{\\circ}-30^{\\circ}\\right)=\\frac{\\tan 45^{\\circ}-\\tan 30^{\\circ}}{1+\\tan 45^{\\circ} \\tan 30^{\\circ}}=\\frac{1-\\frac{\\sqrt{3}}{3}}{1+1 \\cdot \\frac{\\sqrt{3}}{3}}=\\frac{3-\\sqrt{3}}{3+\\sqrt{3}}$$ A final solution will not have a radical in the denominator. In this case multiplying through by the conjugate of the denominator will eliminate the radical. $$=\\frac{(3-\\sqrt{3}) \\cdot(3-\\sqrt{3})}{(3+\\sqrt{3}) \\cdot(3-\\sqrt{3})}$$ $$=\\frac{(3-\\sqrt{3})^{2}}{9-3}$$ $$=\\frac{(3-\\sqrt{3})^{2}}{6}$$ Example 3 Evaluate the expression exactly without using a calculator. $$\\cos 50^{\\circ} \\cos 5^{\\circ}+\\sin 50^{\\circ} \\sin 5^{\\circ}$$ Once you know the general form of the sum and difference identities then you will recognize this as cosine of a difference. $$\\cos 50^{\\circ} \\cos 5^{\\circ}+\\sin 50^{\\circ} \\sin 5^{\\circ}=\\cos \\left(50^{\\circ}-5^{\\circ}\\right)=\\cos 45^{\\circ}=\\frac{\\sqrt{2}}{2}$$ Example 4 Use a sum or difference identity to find an exact value of $$\\cot \\left(\\frac{5 \\pi}{12}\\right)$$. Start with the definition of cotangent as the inverse of tangent. \\begin{aligned}", "Product form of an equation 1. Oct 12, 2005 ktoz I came up with this compound sum formula $$\\frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}$$ and am attempting to represent it in product notation. Is the form below correct? $$\\frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = (a(n + 1) + cm) \\prod_{j=1}^n \\frac{m + j}{j + 1}$$ Thanks 2. Oct 13, 2005 AKG yes it is correct 3. Oct 13, 2005 Thanks AKG", "$$\\sin (u) + \\sin (-v)$$ Use sum-to-product identity for sine $$2\\text{sin} (\\dfrac{u + (-v)}{2}) \\text{cos} (\\dfrac{u - (-v)}{2})$$ Eliminate the parenthesis $$2\\text{sin} (\\dfrac{u - v}{2}) \\text{cos} (\\dfrac{u + v}{2})$$ Establishing the identity EXAMPLE 9 Evaluate $$\\cos (15{}^\\circ )-\\cos (75{}^\\circ )$$. Solution Using the sum-to-product identity for the difference of cosines, $$\\cos (15{}^\\circ )-\\cos (75{}^\\circ )$$ $$=-2\\sin \\left(\\frac{15{}^\\circ +75{}^\\circ }{2} \\right)\\sin \\left(\\frac{15{}^\\circ -75{}^\\circ }{2} \\right)$$ Simplify $$=-2\\sin \\left(45{}^\\circ \\right)\\sin \\left(-30{}^\\circ \\right)$$ Evaluate $$=-2\\cdot \\frac{\\sqrt{2} }{2} \\cdot \\frac{-1}{2} =\\frac{\\sqrt{2} }{2}$$ EXAMPLE 10 Prove the identity $$\\frac{\\cos (4t)-\\cos (2t)}{\\sin (4t)+\\sin (2t)} =-\\tan (t)$$. Solution Since the left side seems more complicated, we can start there and simplify. $$\\frac{\\cos (4t)-\\cos (2t)}{\\sin (4t)+\\sin (2t)}$$ Use the sum-to-product identities $$=\\frac{-2\\sin \\left(\\frac{4t+2t}{2} \\right)\\sin \\left(\\frac{4t-2t}{2} \\right)}{2\\sin \\left(\\frac{4t+2t}{2} \\right)\\cos \\left(\\frac{4t-2t}{2} \\right)}$$ Simplify $$=\\frac{-2\\sin \\left(3t\\right)\\sin \\left(t\\right)}{2\\sin \\left(3t\\right)\\cos \\left(t\\right)}$$ Simplify further $$=\\frac{-\\sin \\left(t\\right)}{\\cos \\left(t\\right)}$$ Rewrite as a tangent $$=-\\tan (t)$$ Establishing the identity EXAMPLE 11 Solve $$\\sin \\left(\\pi {\\kern 1pt} t\\right)+\\sin \\left(3\\pi {\\kern 1pt} t\\right)=\\cos (\\pi {\\kern 1pt} t)$$ for all solutions with $$0\\le t<2$$. Solution In an equation like th is, it is not immediately obvious how to proceed. One option would be to combine the two sine functions on the left side of the equation. Another would be to move the cosine to the left side of the equation, and combine it with one of the sines.", "e^{-ix}}{2} [32] \\sin(x) = \\frac{e^{ix} - e^{-ix}}{2i} [33] and hence the corollary: \\tan(x) = \\frac{\\sin(x)}{\\cos(x)}= \\frac{e^{ix} - e^{-ix}}{i({e^{ix} + e^{-ix}})} where i^2 = -1. ## Infinite product formulae For applications to special functions, the following infinite product formulae for trigonometric functions are useful:[34][35] ## Identities without variables \\cos 20^\\circ\\cdot\\cos 40^\\circ\\cdot\\cos 80^\\circ=\\frac{1}{8} is a special case of an identity that contains one variable: \\prod_{j=0}^{k-1}\\cos(2^j x)=\\frac{\\sin(2^k x)}{2^k\\sin(x)}. Similarly: \\sin 20^\\circ\\cdot\\sin 40^\\circ\\cdot\\sin 80^\\circ=\\frac{\\sqrt{3}}{8}. The same cosine identity in radians is \\cos\\frac{\\pi}{9}\\cos\\frac{2\\pi}{9}\\cos\\frac{4\\pi}{9} = \\frac{1}{8}, Similarly: \\tan 50^\\circ\\cdot\\tan 60^\\circ\\cdot\\tan 70^\\circ=\\tan 80^\\circ. \\tan 40^\\circ\\cdot\\tan 30^\\circ\\cdot\\tan 20^\\circ=\\tan 10^\\circ. The following is perhaps not as readily generalized to an identity containing variables (but see explanation below): \\cos 24^\\circ+\\cos 48^\\circ+\\cos 96^\\circ+\\cos 168^\\circ=\\frac{1}{2}. Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators: \\begin{align} & \\cos\\left( \\frac{2\\pi}{21}\\right) + \\cos\\left(2\\cdot\\frac{2\\pi}{21}\\right) + \\cos\\left(4\\cdot\\frac{2\\pi}{21}\\right) \\\\[10pt] & {} \\qquad {} + \\cos\\left( 5\\cdot\\frac{2\\pi}{21}\\right) + \\cos\\left( 8\\cdot\\frac{2\\pi}{21}\\right) + \\cos\\left(10\\cdot\\frac{2\\pi}{21}\\right)=\\frac{1}{2}. \\end{align} The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials;", "## Precalculus (6th Edition) Blitzer Some of the identities that are verified using sum-to product formulas are: Verified identities are: \\begin{align} & \\tan x=\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x} \\\\ & -\\tan x=\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x} \\\\ & -\\cot 2x=\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x} \\\\ & \\tan 3x=\\frac{\\sin 2x+\\sin 4x}{\\cos 2x+\\cos 4x} \\end{align} Other identities are: \\begin{align} & \\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2} \\\\ & \\frac{\\sin x+\\sin y}{\\sin x-\\sin y}=\\tan \\frac{x+y}{2}\\cot \\frac{x-y}{2} \\\\ & \\frac{\\sin x-\\sin y}{\\cos x-\\cos y}=-\\cot \\frac{x+y}{2} \\end{align} And, \\begin{align} & \\frac{\\sin x+\\sin 3x}{\\cos x+\\cos 3x}=\\tan 2x \\\\ & \\frac{\\cos 4x-\\cos 2x}{\\sin 2x-\\sin 4x}=\\tan 3x \\end{align} Sum-to product formulas are given below: \\begin{align} & \\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2} \\\\ & \\cos \\alpha +\\cos \\beta =2\\cos \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2} \\end{align} Let us consider the identity: $\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x}=\\tan x$ By using the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin", "divide them by a = 10. Find 2 numbers knowing the sum (b = - 19y) and product (60y^2). They are: (-4y) and (-15y). (sum: -19y and product: 60y^2) Back to f(x)--> $p = \\frac{p '}{a} = - \\frac{4 y}{10} = - \\frac{2 y}{5}$, and $q = - \\frac{15 y}{10} = - \\frac{3 y}{2}$. Factored form: $f \\left(x\\right) = 10 \\left(x - \\frac{2 y}{5}\\right) \\left(x - \\frac{3 y}{2}\\right) = \\left(5 x - 2 y\\right) \\left(2 x - 3 y\\right)$", "8. Product to sum formulas $sin(A)cos(B)=\\frac {1}{2}[cos(A-B)-cos[A+B]$ $cos(A)cos(B)=\\frac {1}{2}[cos(A-B)+cos[A+B]$ $sin(A)cos(B)=\\frac {1}{2}[sin(A+B)+sin[A-B]$ $cos(A)sin(B)=\\frac {1}{2}[sin(A+B)-sin[A-B]$ 9. Power reducing formulas $sin^2 A=\\frac{1-cos(2A)}{2}$ $cos^2 A=\\frac{1+cos(2A)}{2}$ $tan^2 A=\\frac{1-cos(2A)}{1+cos(2A)}$ 10. reciprocal identities $Sin(A)=\\frac{1}{cosec(A)}$ $cos(A)=\\frac{1}{sec(A)}$ $tan(A)=\\frac{1}{cot(A)}$ $cosec(A)=\\frac{1}{sin(A)}$ $Sec(A)=\\frac{1}{cos(A)}$ $cot(A)=\\frac{1}{tan(A)}$ Binomial Theorem $(a+b)^n=C_{0}^{n}a^{n}+C_{1}^{n}a^{n-1}b+C_{2}^{n}a^{n-2}b^2+………+C_{n}^{n}b^{n}$ From the binomial formula, if we let a = 1 and b = x, we can also obtain the binomial series which is valid for any real number n if |x| < 1. (1+x)^n=1+nx+\\frac{n(n-1)}{2!}x^2+\\frac{n(n-1)(n-2)}{3!}x^3+……….. Geometric Series $a,aq,aq^2,aq^3,aq^4………..aq^{n-1}$ where q is not equal to 0, q is the common ratio and a is a scale factor.Formula for the sum of the first n numbers of geometric progression $S_{n}=\\frac{a(1-q^n)}{(1-q)}$ Infinite geometric series where |q| < 1 If |q| < 1 then $a_{n} \\to 0$, when n -> infinity So the sum S of such a infinite geometric progression is: $S=\\frac{1}{(1-x)}$ which is valid only for |x| Arithmetic Progression $a,a+d,a+2d,a+3d……….a+(n-1)d$ The sum S of the first n values of a finite sequence is given by the formula: $S=\\frac{n}{2}[(2a + d(n-1)]$ $ax^2+bx+c=0$ $x=-\\frac{b\\pm?(b2-4ac)}{2a}$", "\\frac{\\pi}{2} + k\\pi$ ## Trigonometric identities Double – angle identities Example 1. Prove that $sin (90^{\\circ}) = 1$ using double angle identities. $sin(90^{\\circ}) = sin(2 * 45^{\\circ}) = 2 sin(⁡45^{\\circ})cos(⁡45^{\\circ}) = 2 * \\frac{\\sqrt{2}}{2} * \\frac{\\sqrt{2}}{2} = 2 * \\frac{1}{2} = 1$ Example 2. Calculate $cos(120^{\\circ})$. $cos(120^{\\circ}) = cos(2 * 60^{\\circ}) = cos^2(⁡60^{\\circ}) – sin^2(60^{\\circ})= \\frac{1}{4} – \\frac{3}{4} = – \\frac{1}{2}$ Half – angle identities Example 1. Calculate $cos(15^{\\circ})$ $cos(⁡15^{\\circ}) = \\frac{cos(30^{\\circ})}{2} = \\pm \\sqrt{\\frac{1 – cos(30^{\\circ})}{2}} = \\pm 0.259$ Example 2. Calculate $tan(22,5^{\\circ})$ $tan(22,5^{\\circ}) = tan(\\frac{45^{\\circ}}{2}) = \\pm \\sqrt{\\frac{(1 – cos(45^{\\circ})}{(1 + cos(45^{\\circ})}} = \\pm ± 0.414$ ## Product identities There are more formulas that can help you to simplify complex trigonometric terms. We have our well known addition formulas: $sin(x + y) = sin(x)cos(y) + cos(x)sin(y)$ $sin⁡(x – y) = sin(x)cos(y) – cos(x)sin(y)$ If we add those two equations and divide it by two, we’ll get $sin(x)cos(y) = \\frac{1}{2}(sin⁡(x + y) + sin⁡(x – y))$ $cos(⁡x)sin(⁡y) = \\frac{1}{2}(sin⁡(x + y) – sin(x – y))$ Similarly we get $cos⁡(x)cos(⁡y) = \\frac{1}{2}(cos⁡(x – y) + cos(x + y))$ $sin(x)sin(y) = \\frac{1}{2}(cos⁡(x – y) – cos⁡(x + y))$ Example 1. Calculate $sin(⁡60^{\\circ})cos(⁡30^{\\circ})$ using the product identities. $sin(⁡60^{\\circ})cos(30^{\\circ}) = \\frac{1}{2}(sin(90^{\\circ}) + sin(⁡30^{\\circ})) = \\frac{1}{2} * \\frac{3}{2} = \\frac{3}{4}$ Example 2. Calculate $sin(60^{\\circ})sin(30^{\\circ})$ using the product identities. $sin(60^{\\circ})sin(30^{\\circ}) = \\frac{1}{2}(cos(30^{\\circ}) –", "# sum-product/product-sum • August 2nd 2012, 12:36 AM hacker804 sum-product/product-sum solve the following using sum-product/product-sum identities $cos20+cos100+cos140=0$ $sin19cos11+sin71sin11=\\frac{1}{2}$ $cos20cos40cos60cos180=\\frac{1}{16}$ $cos20cos40cos80=\\frac{1}{8}$ • August 2nd 2012, 09:04 AM srirahulan Re: sum-product/product-sum In this case, Q1, $cos20+cos100+cos140=0$ L.H.S = $2cos60cos40+cos140$ = $cos40+cos140$ = $2cos90cos50$ = $0$ (Since cos90=0) =R.H.S Q2, $sin19cos11+sin71sin11=\\frac{1}{2}$ L.H.S = $\\frac{2sin19cos11+2sin71sin11}{2}$ = $\\frac{sin30+sin8+cos60-cos82}{2}$ = $\\frac{\\frac{1}{2}+\\frac{1}{2}+sin(90-82)-cos82}{2}$ = $\\frac{1+cos82-cos82}{2}$ = $\\frac{1}{2}$ =R.H.S Q3, I think you mention cos180 or cos 80 I can't get any answer. Q4, $cos20cos40cos80=\\frac{1}{8}$ L.H.S = $\\frac{2cos20cos40cos80}{2}$ = $\\frac{(cos60+cos20)cos80}{2}$ = $\\frac{(\\frac{1+2cos20}{2})cos80}{2}$ = $\\frac{\\frac{cos80+cos100+cos60}{2}}{2}$ = $\\frac{\\frac{cos(180-100)+cos100+cos60}{2}}{2}$ = $\\frac{\\frac{-cos100+cos100+\\frac{1}{2}}{2}}{2}$ = $\\frac{1}{8}$ =R.H.S • August 2nd 2012, 09:55 PM hacker804 Re: sum-product/product-sum Quote: $\\frac{2cos20cos40cos80}{2}$ Where did the 2 in the numerator and denominator come from??? • August 2nd 2012, 11:01 PM srirahulan Re: sum-product/product-sum you want to multiply by 2 to numerator and denominator,This is a way to get your answer quickly.", "Posted by juliak ^ how did you get from 40(-1/2)^9 to (-1)^9(40/2^9) ? The power of a product is the product of the powers: $(ab)^n=a^nb^n$. And the same for ratios: $\\left(\\frac{a}{b}\\right)^n=\\frac{a^n}{b^n}$. Here, we have $40\\left(-\\frac{1}{2}\\right)^9=40(-1)^9\\left(\\frac{1}{2}\\right)^9=40(-1)^9\\frac{1}{2^9}$. Ok?", "\\sin(4x) - \\cos(6x)}{\\cos(2x) + \\sin(4x) - \\cos(6x)} &=& \\frac{ \\sin\\big( x - 30^o \\big) \\cos\\big( x + 30^o \\big) }{ \\sin\\big( x - 30^o \\big) \\cos\\big( x - 30^o \\big) }\\\\ \\end{eqnarray}$ So we obtain $\\begin{eqnarray} \\frac{\\cos(2x) - \\sin(4x) - \\cos(6x)}{\\cos(2x) + \\sin(4x) - \\cos(6x)} &=& \\tan\\big( x - 30^o \\big) \\cot\\big( x + 30^o\\big) \\end{eqnarray}$ - I'd start by using $2x=4x-2x$ and $6x=4x+2x$ to get to $$\\cos(2x)-\\cos(6x)=2\\sin(2x)\\sin(4x)$$ - This is an exercise in the sum-to-product and product-to-sum identities. On the LHS, use $$\\cos a - \\cos b = -2\\sin(\\tfrac12(a+b))\\sin(\\tfrac12(a-b))$$ on the top and bottom (as suggested by Lutzl). For the RHS, write $$\\tan(x-15^\\circ)\\cot(x+15^\\circ) = \\frac{\\sin(x-15^\\circ)\\cos(x+15^\\circ)}{\\sin(x+15^\\circ)\\cos(x-15^\\circ)}$$ and then use $$\\sin a \\cos b = \\tfrac12(\\sin(a+b) + \\sin(a-b))$$ on the top and bottom. - Using Prosthaphaeresis Formula, $$\\frac{\\cos2x-\\cos6x}{\\sin4x}=\\frac{2\\sin4x\\sin2x}{2\\sin4x}=\\frac{\\sin2x}{\\dfrac12}$$ $$\\implies \\frac{\\cos2x-\\cos6x}{\\sin4x}=\\frac{\\sin2x}{\\sin30^\\circ}$$ Now apply Componendo and dividendo and again apply Prosthaphaeresis formulae $$\\sin C\\pm\\sin D$$ -", "Product and Sum. Example: What two numbers have a Product of 24 and a sum of 44? Solution: Given Product of two numbers = 24 Sum of two numbers = 44 Let’s assume the numbers you need to find as x and y. Product(x.y) = 24 Sum(x+y) = 44 y = 44 – x Substitute y in x.y = 24 Substitute the value of y in the equation x.y=24. It’s not important y if you want you to interchange with the value of x too as x and y are substitutable. x. (44-x)=24 On solving the equation, the result of two numbers as 2 and 22 Hence the two numbers whose product is 24 and the sum is 44 are 2, 22. Standard form: 1x^2 -44x + 24 = 0 a = 1, b = -44, and c = 24 $$x = \\frac{ -b \\pm \\sqrt{b^2 – 4ac}}{ 2a }$$ $$x = \\frac{ – (-44) \\pm \\sqrt{(-44)^2 – 4(1)(24)}}{ 2(1) }$$ $$x = \\frac{ 44 \\pm \\sqrt{ 1936 – 96}}{ 2 }$$ $$x = \\frac{44 \\pm \\sqrt{1840}}{2}$$ As $$b^{2}−4ac > 0$$, there are two real roots. $$x₁ = \\frac{ 44 + \\sqrt{1840}}{ 2 },x₁ = { 43.4476}$$ $$x₂ = \\frac{ 44 – \\sqrt{1840}}{ 2 } \\hspace{0.25in} and \\hspace{0.25in} x₂ = { 0.5524}$$ You can verify this answer with" ]

[ "b \\\\\\cos (a \\pm b) &=\\cos a \\cos b \\mp \\sin a \\sin b \\\\\\tan (a \\pm b) &=\\frac{\\tan a \\pm \\tan b}{1 \\mp \\tan a \\tan b}$ Problem Set Find the exact value of the following trig functions. 1. $\\sin 15^\\circ$ 2. $\\cos \\frac{5\\pi}{12}$ 3. $\\tan 345^\\circ$ 4. $\\cos (-255^\\circ)$ 5. $\\sin \\frac{13 \\pi}{12}$ 6. $\\sin \\frac{17\\pi}{12}$ 7. $\\cos 15^\\circ$ 8. $\\tan (-15^\\circ)$ 9. $\\sin 345^\\circ$ 10. Now, use $\\sin 15^\\circ$ from #1, and find $\\sin 345^\\circ$. Do you arrive at the same answer? Why or why not? 11. Using $\\cos 15^\\circ$ from #7, find $\\cos 165^\\circ$. What is another way you could find $\\cos 165^\\circ$? 12. Describe any patterns you see between the sine, cosine, and tangent of these “new” angles. 13. Using your calculator, find the $\\sin 142^\\circ$. Now, use the sum formula and your calculator to find the $\\sin 142^\\circ$ using $83^\\circ$ and $59^\\circ$. 14. Use the sine difference formula to find $\\sin 142^\\circ$ with any two angles you choose. Do you arrive at the same answer? Why or why not? 15. Challenge Using $\\sin (a+b)=\\sin a \\cos b +\\cos a \\sin b$ and $\\cos (a+b)=\\cos a \\cos b - \\sin a \\sin b$, show that $\\tan (a+b)=\\frac{\\tan a + \\tan b}{1-\\tan a \\tan b}$. Simplifying Trig Expressions using Sum and Difference Formulas Objective", "# Trigonometry/The sine of 15 degrees We have $\\displaystyle \\sin(\\theta) = \\sqrt {\\frac{1-\\cos(2\\theta)}{2}}$ $\\displaystyle \\cos(\\theta) = \\sqrt {\\frac{1+\\cos(2\\theta)}{2}}$ If θ = 15º then $\\displaystyle \\cos(2\\theta) = \\cos(30^\\circ) = \\frac{\\sqrt{3}}{2}$ so after some manipulation (left as an exercise), $\\displaystyle \\sin(15^\\circ) = \\frac{\\sqrt{6}-\\sqrt{2}}{4} = \\cos(75^\\circ)$ $\\displaystyle \\cos(15^\\circ) = \\frac{\\sqrt{6}+\\sqrt{2}}{4} = \\sin(75^\\circ)$ These results may be combined with those from the previous section to find the sines and cosines of 3º and its multiples. Last modified on 21 August 2013, at 21:40", "with the positive x-axis using a protractor. • The sin of 15 degrees equals the y-coordinate of the point of intersection of unit circle and r. • If we measure it, we will get, the coordinates of the end point of our arm on the circle as (0.9659, 0.2588). Hence the value of sin 15° = y = 0.2588. • Now, we can convert the radians value to degrees value as we have already seen earlier. The reason why we take y-coordinate of our arm in this case lies in the concept of resolution of angle. Whenever we plot an angle, its cosine value lies on the horizontal x-axis whereas as its sine value lies on the y-axis. Thus, x = 0.9659 is actually the value of cos 15. ## Sin 15° in Terms of Trigonometric Functions All the remaining trigonometric functions can be used for writing sin 15. Let’s see how we can do that. Function Formula $$cos15$$ $$\\pm\\sqrt{(1 – cos^215^{\\circ})}$$ $$tan15$$ $$\\pm\\frac{tan 15^{\\circ}}{\\sqrt{(1 + tan^215^{\\circ})}}$$ $$cot15$$ $$\\pm\\frac{1}{\\sqrt{(1 + cot^215^{\\circ})}}$$ $$sec15$$ $$\\frac{1}{\\cos15}$$ $$cosec15$$ $$\\frac{1}{\\cosec15}$$ ## Alternative Method to Determine Sin 15 Values There is one more method to determine the value of sin 15. It’s simpler but it requires more data. Let’s see it step by step. By using the values of sin 30, sin 45, cos", "$\\sin 90^\\circ= 1$, $c \\left( \\frac{1}{2} \\right) = (10) (1)$ $c = 20$ If the above problem asked to find the radius of the circumcircle of $\\vartriangle ABC$, the law of sines could help to find the diameter. $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2r$ $\\frac{b}{\\sin B} = 2r$ Substituting the values for $b, B$, $\\frac{10}{\\sin 30^\\circ} = 2r$ Since $\\sin 30^\\circ= \\frac{1}{2}$, $\\cfrac{10}{\\left( \\frac{1}{2} \\right)} = 2r$ $20 = 2r$ $10 = r$ # References All images were made by the page's author using Adobe Photoshop and Cinderella2.", "$\\sum_{n=0}^{14}\\tan(12n+1^\\circ)$ I often fail to find trigonometric sums such as the one in the question shown in the following. When I tried the question, I first led $z=e^{i\\pi/180}$. After simple calculations, I obtained $\\sum_{n=0}^{14}\\tan(12n+1)=\\sum_{n=0}^{14}\\dfrac{z^{24n+2}-1}{z^{24n}+z^2}$ How can I proceed calculation further? - This is a special case of the more general identity: $$\\sum_{k=0}^{n-1} \\tan\\left(\\theta + \\frac{k\\pi}{n}\\right) = -n\\cot \\left(\\frac{n\\pi}{2} + n\\theta \\right)$$ In your case of course $\\theta = 1^\\circ$ and $n=15$. A proof of this identity may be found as an answer to a previous question on this site, here. - How beautiful the formula is! Though I knew that a certain sum of sin's whose angle increases linearly like that is proportional or inversely proportional to cot, I didn't know this formula. If I read your page, I may find the closed form of sum of not only tan's but also other trigonometric functions. Thanks so much. – MathOverStackExchange Nov 1 '13 at 1:50", "= \\frac{\\cos 1^\\circ}{\\sin 1^\\circ}. (6)$$ Now we notice that we can multiply both sides by $\\sin 1^\\circ$. Doing so gives us $$\\sin 1^\\circ(2 (\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ) + 1) = \\cos 1^\\circ. (7)$$ Now, simplifying $\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas. Doing so, gives us $$(\\sin 1^\\circ)(2 \\sin 2^\\circ)+(\\sin 1^\\circ)(2\\sin 4^\\circ) +(\\sin 1^\\circ)(2\\sin 6^\\circ) + (\\sin 1^\\circ)(2\\sin 8^\\circ) +(\\sin 1^\\circ)(2 \\sin 10^\\circ) ... + (\\sin 1^\\circ)(2\\sin 88^\\circ) + \\sin 1^\\circ = \\cos 1^\\circ. (8)$$ Now we can use our product of sines to sum formula and see if we can find a pattern. We will start with expanding $(\\sin 1^\\circ)(2 \\sin 2^\\circ)$. Because the format of our formula for the product of sines is $\\sin \\alpha + \\sin \\beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\\sin 1^\\circ)(\\sin 2^\\circ)$. Our formula for the product of sines is $\\sin \\alpha + \\sin \\beta = \\frac{1}{2}[\\cos(\\alpha - \\beta) - \\cos(\\alpha + \\beta)]$. Plugging in our values into the formula and simplifying gives us $2\\left(\\frac{1}{2}(\\cos 1^\\circ", "# Question #19c60 Nov 8, 2014 $\\sin \\left({330}^{\\circ}\\right) + \\cos \\left({315}^{\\circ}\\right) + \\tan \\left({45}^{\\circ}\\right)$ by subtracting ${360}^{\\circ}$ from the first two angles, $= \\sin \\left(- {30}^{\\circ}\\right) + \\cos \\left(- {45}^{\\circ}\\right) + \\tan \\left({45}^{\\circ}\\right)$ since sine is an odd function, and cosine is an even function, $= - \\sin \\left({30}^{\\circ}\\right) + \\cos \\left({45}^{\\circ}\\right) + \\tan \\left({45}^{\\circ}\\right)$ by using special triangles, $= - \\frac{1}{2} + \\frac{\\sqrt{2}}{2} + 1 = \\frac{1 + \\sqrt{2}}{2}$ I hope that this was helpful.", "× Back to all chapters # Sum and Difference Trigonometric Formulas These formulas explain how to add and subtract trigonometric functions (and their arguments). If you've got sum time, see what a difference these formulas will make for your trig toolkit. # Product to Sum Trigonometric Formulas If $\\sin 14^{\\circ} \\cos 56^{\\circ} - \\sin42^{\\circ} \\cos 28^{\\circ} = \\frac{1}{2}(K - \\sin 14^{\\circ}),$ what is the value of $$K?$$ If $$\\cos \\alpha=\\frac{1}{11}$$ and $$\\cos \\beta=\\frac{10}{11}$$, the value of $\\sin (\\alpha+\\beta) \\sin (\\alpha-\\beta)$ can be expressed as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a+b$$? $$N$$ is an integer in the range $$[0, 180]$$ such that $2 \\cos 43 ^\\circ \\cos 26 ^\\circ = \\cos 69 ^ \\circ + \\cos N ^ \\circ.$ What is the value of $$N$$? If $a \\neq 0, \\sin(x+y) = \\frac{12}{a}, \\sin(x-y) = \\frac{5}{a},$ what is $$\\displaystyle{\\frac{\\tan x}{\\tan y}}?$$ The minimum positive value of $$x$$ that satisfies the equation $\\cos5x\\cos2x=\\sin2x\\sin5x$ is $$\\frac{\\pi}{m},$$ where $$m$$ is a positive integer. What is $$m?$$ ×", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "# Sin 1 ## Concept of Sin 1 There are three major ratios in trigonometry upon which the trigonometric functions and formulas are based. The sine function is one of them. The sine function (sin) of an angle gives the ratio of the perpendicular (opposite side of the angle) to the hypotenuse. Similarly, the inverse sine function (sin-1) gives the ratio of the hypotenuse to the perpendicular of an angle. Sin 1 in radians has a value of 0.8414709848. The basic angles which are required very frequently are 0,30,45,60,90 degrees. They can also be expressed in radians as π/2, π/3, π/4, π/6, π, etc. In radians, the value of sine 1 is 0.8414709848. We know, π/3 = 1.047198≈1 $Sin (\\frac{\\pi}{3}) = \\frac{\\sqrt{3}}{2}$ Sin π = 0 Now using these data, we can write; $sin 1 = sin \\frac{\\pi}{3} − (\\frac{\\pi}{3} − 1)$ ⟹ sin1 = $sin \\frac{\\pi}{3} cos(\\frac{\\pi}{3} − 1)−cos \\frac{\\pi}{3} sin(\\frac{\\pi}{3} − 1)$ The angle π/3−1=0.047198 is a very small angle. We know that, for small angles θ, Sinθ ≈ θ and cos⁡θ ≈ 1 hence, Sin 1 ≈ $(\\frac{\\sqrt{3}}{2} \\times 1) − \\frac{1}{2} \\times (\\frac{\\pi}{3} − 1)$ Therefore, sin1 ≈ 0.842427 ### How to find the Value of Sine 1? The sine of an angle can take its argument as either radian or degrees. The rule", "0.016^{\\circ} 0.93^{\\circ} \\begin{array}{l} \\frac{\\sin(B)}{b}=\\frac{\\sin(C)}{c}\\\\\\\\ \\frac{\\sin (\\theta)}{7.4}=\\frac{\\sin (68)}{7.3}\\\\\\\\ \\sin (\\theta)=\\frac{\\sin (68)}{7.3}\\times7.4\\\\\\\\ \\sin (\\theta)=0.9398850033…\\\\\\\\ \\theta=\\sin^{-1}(0.9398850033)\\\\\\\\ \\theta=70^{\\circ}\\quad(2 s.f.) \\end{array} 5. Use trigonometry to calculate the size of angle \\theta . Write your answer to two decimal places. 27.58^{\\circ} 51.90^{\\circ} 28.16^{\\circ} 32.51^{\\circ} \\begin{array}{l} \\frac{\\sin(A)}{a}=\\frac{\\sin(C)}{c}\\\\\\\\ \\frac{\\sin (62)}{8.2}=\\frac{\\sin (\\theta)}{4.3}\\\\\\\\ \\sin (\\theta)=\\frac{\\sin (62)}{8.2}\\times4.3\\\\\\\\ \\sin (\\theta)=0.4630091036…\\\\\\\\ \\theta=\\sin^{-1}(0.4630091036)\\\\\\\\ \\theta=27.58^{\\circ}\\quad(2 d.p.) \\end{array} 6. Calculate the size of angle \\theta . Write your answer to three significant figures. 72^{\\circ} 22.5^{\\circ} 64.7^{\\circ} 55.4^{\\circ} \\begin{array}{l} \\frac{\\sin(A)}{a}=\\frac{\\sin(B)}{b}\\\\\\\\ \\frac{\\sin (\\theta)}{10}=\\frac{\\sin (36)}{6.5}\\\\\\\\ \\sin (\\theta)=\\frac{\\sin (36)}{6.5}\\times10\\\\\\\\ \\sin (\\theta)=0.9042850035…\\\\\\\\ \\theta=\\sin ^{-1}(0.9042850035)\\\\\\\\ \\theta=64.72^{\\circ}\\quad(3s.f.) \\end{array} ### Sine rule GCSE exam questions 1. In triangle ABC, AC = 8.5cm, BC = 23cm and angle BAC = 80^{\\circ}. Calculate the size of the angle marked x. (3 marks) \\frac{\\sin(x)}{8.5}=\\frac{\\sin(80)}{23} (1) \\begin{aligned} \\sin(x)&=\\frac{\\sin(80)}{23} \\times 8.5\\\\\\\\ \\sin(x)&=0.364… \\end{aligned} (1) \\begin{aligned} x&=\\sin^{-1}(0.364…)\\\\\\\\ x&=21.3^{\\circ} \\end{aligned} (1) 2. Work out the size of angle BCD. (5 marks) \\frac{BD}{\\sin(77)}=\\frac{14}{\\sin(51)} (1) \\begin{aligned} BD&=\\frac{14}{\\sin(51)} \\times \\sin(77)\\\\\\\\ BD&= 17.553\\mathrm{cm} \\end{aligned} (1) \\frac{sin(\\theta)}{17.55}=\\frac{sin(82)}{30} (1) \\begin{aligned} \\sin(\\theta)&=\\frac{sin(82)}{30} \\times 17.55\\\\\\\\ sin(\\theta)&=0.579… \\end{aligned} (1) \\begin{aligned} \\theta&=sin^{-1}(0.579…)\\\\\\\\ \\theta&=35.4^{\\circ} \\end{aligned} (1) 3. Calculate the length AB. (4 marks) \\begin{array}{l} \\text{Angle ABE}=42^{\\circ} \\text{ (Alternate angles are equal)}\\\\\\\\ \\text{Angle AEB: }180-46-42=92^{\\circ} \\end{array} (1) \\frac{AB}{\\sin(92)}=\\frac{17}{\\sin(46)} (1) AB=\\frac{17}{\\sin(46)} \\times \\sin(92) (1) AB=23.6cm (1) ## Learning checklist You have now learned how to: • know and apply the sine rule to find unknown lengths and angles ## Still stuck? Prepare", "Categories # Finding smallest positive Integer | AIME I, 1996 Problem 10 Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer. Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer. ## Finding smallest positive Integer – AIME I, 1996 Find the smallest positive integer solution to $tan19x=\\frac{cos96+sin96}{cos96-sin96}$. • is 107 • is 159 • is 840 • cannot be determined from the given information Functions Trigonometry Integers ## Check the Answer AIME I, 1996, Question 10 Plane Trigonometry by Loney ## Try with Hints $\\frac{cos96+sin96}{cos96-sin96}$ =$\\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$ =$\\frac{sin186+sin96}{sin186-sin96}$ =$\\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$ =$\\frac{2sin141cos45}{2cos141sin45}$ =tan141 here $tan(180+\\theta)$=$tan\\theta$ $\\Rightarrow 19x=141+180n$ for some integer n is first equation multiplying equation with 19 gives $x \\equiv 141\\times 19 \\equiv 2679 \\equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159] $\\Rightarrow x=159$.", "# Trigonometry/The sine of 15 degrees We have ${\\displaystyle \\sin(\\theta )={\\sqrt {\\frac {1-\\cos(2\\theta )}{2}}}}$ ${\\displaystyle \\cos(\\theta )={\\sqrt {\\frac {1+\\cos(2\\theta )}{2}}}}$ If ${\\displaystyle \\theta =15^{\\circ }}$ then ${\\displaystyle \\cos(2\\theta )=\\cos(30^{\\circ })={\\frac {\\sqrt {3}}{2}}}$ so after some manipulation (left as an exercise), ${\\displaystyle \\sin(15^{\\circ })={\\frac {{\\sqrt {6}}-{\\sqrt {2}}}{4}}=\\cos(75^{\\circ })}$ ${\\displaystyle \\cos(15^{\\circ })={\\frac {{\\sqrt {6}}+{\\sqrt {2}}}{4}}=\\sin(75^{\\circ })}$ These results may be combined with those from the previous section to find the sines and cosines of ${\\displaystyle =3^{\\circ }}$ and its multiples.", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> Sum and Difference Formulas | CK-12 Foundation 14.4: Sum and Difference Formulas Created by: CK-12 Objective To use and derive the sum and difference formulas. Review Queue Using your calculator, find the value of each trig function below. Round your answer to 4 decimal places. 1. $\\sin 75^\\circ$ 2. $\\cos 15^\\circ$ 3. $\\tan 105^\\circ$ 4. $\\cos 255^\\circ$ Finding Exact Trig Values using Sum and Difference Formulas Objective Here you'll use the sum and difference formulas to find exact values of angles other than the critical angles. Guidance You know that $\\sin 30^\\circ=\\frac{1}{2}, \\cos 135^\\circ=-\\frac{\\sqrt{2}}{2}, \\tan 300 ^\\circ = -\\sqrt{3},$ etc... from the special right triangles. In this concept, we will learn how to find the exact values of the trig functions for angles other than these multiples of $30^\\circ, 45^\\circ,$ and $60^\\circ$. Using the Sum and Difference Formulas, we can find these exact trig values. Sum and Difference Formulas $\\sin(a\\pm b) &=\\sin a \\cos b \\pm \\cos a \\sin b \\\\\\cos(a\\pm b) &=\\cos a \\cos b \\mp \\sin a \\sin b \\\\\\tan(a \\pm b) &=\\frac{\\tan a \\pm \\tan b}{1 \\mp \\tan a \\tan b}$ Example A Find the exact value of $\\sin 75^\\circ$. Solution: This is an example of where we can use the sine sum formula from above, $\\sin(a+b)=\\sin a \\cos b+\\cos a", "field labelled °, but if you want to calculate sin 25 in radians, then you have to press the swap unit button first. ### Calculate sin [degrees] Besides sin25°, similar trigonometric calculations on our site include, but are not limited, to: The identities of sine 25° are as follows: sin25° = cos (90°-25°) = cos 65° = sin (180°-25°) = sin 155° -sin25° = cos (90°+25°) = cos 115° = sin (180°+25°) = sin 205° Note that sin25° is periodic: sin (25° + n × 360°) = sin 25 degrees, n$\\hspace{5px} \\in \\hspace{5px} \\mathbb{Z}$. There are more formulas for the double angle (2 × 25°), half angle ((25/2)°) as well as the sum, difference and products of two angles such as 25° and β. You can locate all of them in the respective article found in the header menu. To find everything about sin -25° click the link. And here is all about cos 25°, including, for instance, a converter. In terms of the other five trigonometric functions, sin of 25° = • $\\pm \\sqrt{1-\\cos^{2} 25 ^\\circ}$ • $\\pm\\frac{\\tan 25^\\circ}{\\sqrt{1 + \\tan^{2} 25^\\circ}}$ • $\\pm\\frac{1}{\\sqrt{1 + \\cot^{2} 25^\\circ}}$ • $\\pm\\frac{\\sqrt{\\sec^{2} 25^\\circ – 1} }{\\sec 25^\\circ}$ • $\\frac{1}{\\csc 25^\\circ}$ As the cosecant function is the reciprocal of the sine function, 1 / csc 25° = sin25°. In the next", "first one should do the trick, the second is being extra careful. -Dan 7. Originally Posted by qbkr21 Here are the directions the computer system is still not taking it: Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2); Thanks! He did not say that: $\\displaystyle \\cos(105^{\\circ})=\\sin(15^{\\circ})$ what he did say was: $\\displaystyle \\cos(105^{\\circ})=-\\sin(15^{\\circ})$ and that: $\\displaystyle \\sin(15^{\\circ})=\\frac{\\sqrt{2-\\sqrt{3}}}{2}$ so: $\\displaystyle \\cos(105^{\\circ})=-\\frac{\\sqrt{2-\\sqrt{3}}}{2}$ RonL 8. Re: Ok thanks Captain Black", "other trigonometric ratios of angle A (using trigonometric identities). Solution: sin A = $$\\\\ \\frac { 3 }{ 5 }$$ In ∆ABC, ∠B = 90° AC = 5 and BC = 3 Question 2. If A is an acute angle and sec A = $$\\\\ \\frac { 17 }{ 8 }$$, find all other trigonometric ratios of angle A (using trigonometric identities). Solution: sec A = $$\\\\ \\frac { 17 }{ 8 }$$ (A is an acute angle) In right ∆ABC sec A = $$\\\\ \\frac { AC }{ AB }$$ = $$\\\\ \\frac { 17 }{ 8 }$$ AC = 17, AB = 8 Question 3. Express the ratios cos A, tan A and sec A in terms of sin A. Solution: cos A = $$\\sqrt { { 1-sin }^{ 2 }A }$$ tan A = $$\\frac { SinA }{ CosA } =\\frac { sinA }{ \\sqrt { { 1-sin }^{ 2 }A } }$$ sec A = $$\\frac { 1 }{ cosA } =\\frac { 1 }{ \\sqrt { { 1-sin }^{ 2 }A } }$$ Question 4. If tan A = $$\\frac { 1 }{ \\sqrt { 3 } }$$, find all other trigonometric ratios of angle A. Solution: tan A = $$\\frac { 1 }{ \\sqrt { 3 } }$$ In right ∆ABC, tan", "\\sin(4x) - \\cos(6x)}{\\cos(2x) + \\sin(4x) - \\cos(6x)} &=& \\frac{ \\sin\\big( x - 30^o \\big) \\cos\\big( x + 30^o \\big) }{ \\sin\\big( x - 30^o \\big) \\cos\\big( x - 30^o \\big) }\\\\ \\end{eqnarray}$ So we obtain $\\begin{eqnarray} \\frac{\\cos(2x) - \\sin(4x) - \\cos(6x)}{\\cos(2x) + \\sin(4x) - \\cos(6x)} &=& \\tan\\big( x - 30^o \\big) \\cot\\big( x + 30^o\\big) \\end{eqnarray}$ - I'd start by using $2x=4x-2x$ and $6x=4x+2x$ to get to $$\\cos(2x)-\\cos(6x)=2\\sin(2x)\\sin(4x)$$ - This is an exercise in the sum-to-product and product-to-sum identities. On the LHS, use $$\\cos a - \\cos b = -2\\sin(\\tfrac12(a+b))\\sin(\\tfrac12(a-b))$$ on the top and bottom (as suggested by Lutzl). For the RHS, write $$\\tan(x-15^\\circ)\\cot(x+15^\\circ) = \\frac{\\sin(x-15^\\circ)\\cos(x+15^\\circ)}{\\sin(x+15^\\circ)\\cos(x-15^\\circ)}$$ and then use $$\\sin a \\cos b = \\tfrac12(\\sin(a+b) + \\sin(a-b))$$ on the top and bottom. - Using Prosthaphaeresis Formula, $$\\frac{\\cos2x-\\cos6x}{\\sin4x}=\\frac{2\\sin4x\\sin2x}{2\\sin4x}=\\frac{\\sin2x}{\\dfrac12}$$ $$\\implies \\frac{\\cos2x-\\cos6x}{\\sin4x}=\\frac{\\sin2x}{\\sin30^\\circ}$$ Now apply Componendo and dividendo and again apply Prosthaphaeresis formulae $$\\sin C\\pm\\sin D$$ -", "your answers to the nearest tenth: . 3. Find the length of the missing sides and round your answers to the nearest tenth: . 1. Using your calculator, you should find that $\\tan 45^\\circ=1$ ? 2. Use tangent for $x$ and cosine for $y$ . $\\tan 28^\\circ &= \\frac{x}{11} && \\quad \\ \\ \\cos 28^\\circ = \\frac{11}{y}\\\\11 \\cdot \\tan 28^\\circ &= x && \\frac{11}{\\cos 28^\\circ} = y\\\\x & \\approx 5.8 && \\qquad \\quad \\ \\ \\ y \\approx 12.5$ 3. Use tangent for $y$ and cosine for $x$ . $\\tan 40^\\circ &= \\frac{y}{16} && \\quad \\ \\ \\cos 40^\\circ = \\frac{16}{x}\\\\16 \\cdot \\tan 40^\\circ &= y && \\frac{16}{\\cos 40^\\circ} = x\\\\y & \\approx 13.4 && \\qquad \\quad \\ \\ \\ x \\approx 20.9$ Practice Use your calculator to find the value of each trig function below. Round to four decimal places. 1. $\\sin 24^\\circ$ 2. $\\cos 45^\\circ$ 3. $\\tan 88^\\circ$ 4. $\\sin 43^\\circ$ 5. $\\tan 12^\\circ$ 6. $\\cos 79^\\circ$ 7. $\\sin 82^\\circ$ Find the length of the missing sides. Round your answers to the nearest tenth.", "is often suitable, then, to take the first-order term of the Taylor series for sin(''x'') as its '''small-angle approximation''': + + :$\\sin (x) \\approx x$ for small ''x'' + + By a similar token, we may obtain small-angle approximations for the other trigonometric functions. In the case of cosine, we go out to the second-order term, since that is the first term that includes ''x'': + + :$\\cos (x) \\approx 1 - {x ^2 \\over 2!}$ for small ''x'' + + Then + + :$\\tan (x) = {\\sin (x) \\over \\cos (x)} \\approx {x\\over 1-{x^2 \\over 2}} \\approx {x\\over 1} = x$ for small ''x'', + + because for small ''x'' we only need the first-order approximations and the first-order approximation of cos(''x'') is 1. + + {{{!}}border=\"0\" cellpadding=5 cellspacing=5 + {{!}}{{Anchor|Reference=Figure9a|Link=[[Image:Small sinx.gif|center|thumb|325px|Figure 9a Comparison of sin(''x'') with its small-angle approximation.]]}}{{!}}{{!}}{{Anchor|Reference=Figure9b|Link=[[Image:Small cosx.gif|center|thumb|325px|Figure 9b Comparison of cos(''x'') with its small-angle approximation.]]}}{{!}}{{!}}{{Anchor|Reference=Figure9c|Link=[[Image:Small tanx.gif|center|thumb|325px|Figure 9c Comparison of tan(''x'') with its small-angle approximation.]]}} + {{!}}} + + What is the small-angle approximation good for? In the simplest respect, the small-angle approximation is \"close enough,\" and it's quicker than evaluating more terms of a Taylor series. However, the small-angle approximation has an additional utility in that it can allow us to solve certain differential equations in closed form. The prime example of this" ]

[ "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "$GH$ is $\\sqrt{3}$. Think of the picture as one large equilateral triangle, $\\triangle{JKL}$ with the sides of $2\\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\\triangle{JKL}$ $\\dfrac{\\sqrt{3}}{4}(2\\sqrt{3}+1)^2=\\dfrac{12+13\\sqrt{3}}{4}$. $[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,W); label(\"E\",E,W); label(\"F\",F,E); label(\"G\",G,E); label(\"H\",H,SE); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,N); label(\"L\",L,SE); [/asy]$ Triangles $\\triangle{DIJ}$, $\\triangle{EFK}$, and $\\triangle{GHL}$ have sides of $\\sqrt{3}$, so their total area is $3(\\dfrac{\\sqrt{3}}{4}(\\sqrt{3})^2)=\\dfrac{9\\sqrt{3}}{4}$. Now, you subtract their total area from the area of $\\triangle{JKL}$: $\\dfrac{12+13\\sqrt{3}}{4}-\\dfrac{9\\sqrt{3}}{4}=\\dfrac{12+13\\sqrt{3}-9\\sqrt{3}}{4}=\\dfrac{12+4\\sqrt{3}}{4}=3+\\sqrt{3}\\implies\\boxed{\\textbf{(C)}\\ 3+\\sqrt3}$ ## Solution 3 We will use, $\\frac{1}{2}ab\\sin x$ to find the area of the following triangles. Since $\\angle A=360$, $\\angle EAF=360-90-90-60=120$. The Area of $\\triangle AEF$ is $\\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin(120)$. Noting, $\\sin (2x) = 2\\sin (x)\\cos (x)$, Area of $\\triangle AEF = \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot 2 \\cdot \\sin(60) \\cdot \\cos(60) = \\dfrac{\\sqrt{3}}{4}$, Area of $\\triangle ABC = \\frac{1}{2} \\cdot 1 \\cdot 1", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "real b, real c){ real p=(a+b+c)/2; return sqrt(p*(p-a)*(p-b)*(p-c)); } currentprojection=orthographic((1,1.6,1),center=true,zoom=.95); // Step 1: construct the base A,B,C on the plane z=0 real a=6, b=5, c=4; triple B=(0,0,0), C=(a,0,0); // Abc is the projection of A on the segment BC triple Abc=barycentric(B,C,O,1/(a^2+c^2-b^2),1/(a^2+b^2-c^2),0); real bt=abs(Abc-B); real hbc=sqrt(c*c-bt*bt); triple A=Abc+hbc*dir(90,90); draw(A--Abc,red); draw(A--B--C--cycle); label(\"$A$\",A,plain.S); label(\"$B$\",B,plain.E); label(\"$C$\",C,plain.W); label(\"$A_{bc}$\",Abc,plain.N,red); // Step 2: get the top point D real at=6, bt=7, ct=4; // H is the projection of D on the base ABC real Sdab=Heron(at,bt,c); real Sdbc=Heron(bt,ct,a); real Sdca=Heron(ct,at,b); triple H=barycentric(A,B,C,1/(Sdab^2+Sdca^2-Sdbc^2),1/(Sdab^2+Sdbc^2-Sdca^2),1/(Sdca^2+Sdbc^2-Sdab^2)); real ha=abs(H-A); real hd=sqrt(at*at-ha*ha); triple D=H+hd*Z; draw(D--A^^D--B^^D--C); draw(D--H,blue); label(\"$D$\",D,plain.N); label(\"$H$\",H,plain.W,blue); dot(H); PS1: Why Asymptote and why not TikZ? That drawing way of using barycentric coordinate can be coded in several drawing languages. TikZ does has barycentric coordinate; but its computation is quite weak, Dimension too large error may happen, even in 2D when I had tried drawing the Euler line of a triangle)! Asymptote has better accuracy, and available for 3D. PS2: There is another way based on origami, described in a book of Polya. I will do it in free time later. • You wrote \"Full code (still some mistake ^^)\" and \" I will do it in free time later.\" When you post correct answer? Oct 6, 2021 at 0:38 • Peter: As I said: \"in free time later\". There is a", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). ## Solution 3 $[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot(\"C\", C, dir(0)); dot(\"A\", A, dir(90));dot(\"B\", B, dir(180)); dot(\"D\", D, NE); dot(\"E\", E, dir(90));dot(\"F\", F, dir(270)); dot(\"M\", M, NE); dot(\"N\", N, dir(270));dot(\"L\", L, NW); dot(\"X\", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]$ Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\\sin(60)$. Similarly, $AB=3\\sin(60)$, and $BC=5\\sin(60)$. So $\\triangle ABC$ is a 3-4-5 right triangle", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "# Carnot's Theorem \"\" Carnot's Theorem\"\" states that in a triangle $ABC$, the signed sum of perpendicular distances from the circumcenter $O$ to the sides (i.e., signed lengths of the pedal lines from $O$) is: $OO_A+OO_B+OO_C=R+r$ $[asy] pair a,b,c,O,i,d,f,g; a=(0,0); b=(4,0); c=(1,3); O=circumcenter(a,b,c); i=incenter(a,b,c); draw(a--b--c--cycle); draw(circumcircle(a,b,c)); draw(incircle(a,b,c)); dot(i); dot(O); label(\"A\",a,W); label(\"B\",b,E); label(\"C\",c,N); label(\"I\",i,N); label(\"O\",O,N); d=foot(O,b,c); dot(d); draw(O--d); label(\"O_A\",d,N); draw(rightanglemark(O,d,b)); f=foot(O,a,b); dot(f); draw(O--f); draw(rightanglemark(O,f,a)); label(\"O_C\",f,S); g=foot(O,c,a); dot(g); draw(O--g); draw(rightanglemark(O,g,a)); label(\"O_B\",g,W); [/asy]$ where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. Explicitly, $OO_A+OO_B+OO_C=\\frac{abc(|\\cos{A}|+|\\cos{B}|+|\\cos{C}|)}{4|\\Delta|}$ where $\\Delta$ is the area of triangle $\\Delta ABC$. Weisstein, Eric W. \"Carnot's Theorem.\" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html # Below is not Carnot's Theorem Carnot's Theorem states that in a triangle $ABC$ with $A_1\\in BC$, $B_1\\in AC$, and $C_1\\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$. #### Proof Only if: Assume that the given perpendiculars are concurrent at $M$. Then, from the Pythagorean Theorem, $A_1B^2=BM^2-MA_1^2$, $C_1A^2=AM^2-MC_1^2$, $B_1C^2=CM^2-MB_1^2$, $A_1C^2=MC^2-MA_1^2$, $C_1B^2=MB^2-MC_1^2$, and $B_1A^2=AM^2-MB_1^2$. Substituting each and every one of these in and simplifying gives the desired result. If: Consider the intersection of the perpendiculars from $A_1$ and $B_1$. Call this intersection point $N$,", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "$\\triangle ABC$ is$$s = \\frac{AB + BC + CA}2 = \\frac{15}2,$$and Heron's Formula gives the area of $\\triangle ABC$ as$$\\sqrt{s(s-AB)(s-BC)(s-CA)} = \\frac{15\\sqrt7}4.$$This area is also $rs$ implying that $r = \\frac{\\sqrt7}2$. Stewart's Theorem gives $AD =3\\sqrt2$. Because the ratio of the areas of $\\triangle IBC$ and $\\triangle ABC$ is $\\frac{ID}{AD},$ it follows that$$ID = AD\\cdot\\frac{\\frac{r\\cdot BC}2}{\\frac{15\\sqrt7}4} = \\sqrt2.$$Thus $IM = MD - ID = \\frac{\\sqrt2}2$. $[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); dot(\"I\", I, dir(120)); dot(\"M\", M, W); [/asy]$ Note that $\\angle EFI = 90^{\\circ} - \\angle FIA = 90^{\\circ} - \\angle CID = 90^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle C}{2} = \\frac{\\angle B}{2} = \\angle IBC$. Thus $\\triangle IBC \\sim \\triangle IFE$. The height of $\\triangle IBC$ to $I$ is $r=\\frac{\\sqrt7}2$, and the height of $\\triangle IFE$ to $I$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "D, E, F, G, H; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); G = rotate(90,(A + C)/2)*(E); H = rotate(90,(A + C)/2)*(F); draw(A--B--C--D--cycle); draw(E--G--F--H--cycle); dot(\"A\", A, N); dot(\"B\", B, dir(0)); dot(\"C\", C, S); dot(\"D\", D, W); dot(\"E\", E, N); dot(\"F\", F, S); dot(\"G\", G, W); dot(\"H\", H, dir(0));[/asy]$ $\\overline{EF}$ is the diagonal of the square, with length $17\\sqrt{2}$; the answer is $EF^2 = (17\\sqrt{2})^2 = 578$. Solution 3 A slightly more analytic/brute-force approach: Drop perpendiculars from $E$ and $F$ to $I$ and $J$, respectively; construct right triangle $EKF$ with right angle at K and $EK || BC$. Since $2[CDF]=DF*CF=CD*JF$, we have $JF=5\\times12/13 = \\frac{60}{13}$. Similarly, $EI=\\frac{60}{13}$. Since $\\triangle DJF \\sim \\triangle DFC$, we have $DJ=\\frac{5JF}{12}=\\frac{25}{13}$. Now, we see that $FK=DC-(DJ+IB)=DC-2DJ=13-\\frac{50}{13}=\\frac{119}{13}$. Also, $EK=BC+(JF+IE)=BC+2JF=13+\\frac{120}{13}=\\frac{289}{13}$. By the Pythagorean Theorem, we have $EF=\\sqrt{\\left(\\frac{289}{13}\\right)^2+\\left(\\frac{119}{13} \\right)^2}=\\frac{\\sqrt{(17^2)(17^2+7^2)}}{13}$$=\\frac{17\\sqrt{338}}{13}=\\frac{17(13\\sqrt{2})}{13}=17\\sqrt{2}$. Therefore, $EF^2=(17\\sqrt{2})^2=578$. Solution 4 Based on the symmetry, we know that $F$ is a reflection of $E$ across the center of the square, which we will denote as $O$. Since $\\angle BEA$ and $\\angle AOB$ are right, we can conclude that figure $AOBE$ is a cyclic quadrilateral. Pythagorean Theorem yields that $BO=AO=\\frac{13\\sqrt{2}}{2}$. Now, using Ptolemy's Theorem, we get that $$AO\\cdot BE + BO\\cdot AE = AB\\cdot AO$$ $$\\frac{13\\sqrt{2}}{2}\\cdot 5+\\frac{13\\sqrt{2}}{2}\\cdot 12 = 13\\cdot OE$$" ]

[ "## anonymous one year ago ΔABC with vertices A(-3, 0), B(-2, 3), C(-1, 1) is rotated 180° clockwise about the origin. It is then reflected across the line y = -x. What are the coordinates of the vertices of the image? A. A'(0, 3), B'(2, 3), C'(1, 1) B. A'(0, -3), B'(3, -2), C'(1, -1) C. A'(-3, 0), B'(-3, 2), C'(-1, 1) D. A'(0, -3), B'(-2, -3), C'(-1, -1) 1. mathmate Hints: Rotation $$R_{180}: (x,y)\\rightarrow (-x,-y)$$ Reflection $$S_{y=-x}:(x,y)\\rightarrow (-y,-x)$$ Transform separately, or combine them: $$S_{y=-x}\\circ R_{180}: (x,y)\\rightarrow (y,x)\\equiv S_{y=x}$$ For example, Point (7,1) after rotation of 180 becomes (-7,-1) followed by a reflection about y=-x gives (1,7) as the final position. Alternatively, using a single reflection about y=x also transforms from (7,1) to (1,7), as explained above. 2. anonymous So Its B? 3. mathmate I don't usually work with letters. I will be glad to confirm if you post the values represented by option \"B\".", "is $C(x,y)=(x-c,y)$. Next, reflect about the line $x=0$. The map of this reflection is $M(x,y)=(-x,y)$. Finally, translate the point $(0,0)$ back to $(c,0)$. The map of this translation is $C^{-1}(x,y)=(x+c,y)$. The fact that $E=C^{-1}\\circ M\\circ C$ can be used to obtain a formula for $E$: $\\begin{array}[]{rl}E(x,y)&=(C^{-1}\\circ M\\circ C)(x,y)\\\\ &=(C^{-1}\\circ M)(x-c,y)\\\\ &=C^{-1}(c-x,y)\\\\ &=(2c-x,y).\\end{array}$ Title determining rotations and reflections in $\\mathbb{R}^{2}$ DeterminingRotationsAndReflectionsInmathbbR2 2013-03-22 17:14:34 2013-03-22 17:14:34 Wkbj79 (1863) Wkbj79 (1863) 4 Wkbj79 (1863) Derivation msc 51A15 msc 51A10 msc 15A04 Symmetry2", "# Fermat's Hexagon ### Problem Let $AA',\\;$ $BB',\\;$ and $CC'\\;$ be three equal line segments lying on concurrent lines at $120^{\\circ}$ to each other. Define centroids $G_A,G_B,G_C,K_A,K_B,K_C\\;$ of triangles $BCA',\\;$ $CAB',\\;$ $ABC',\\;$ $AB'C',\\;$ $A'BC',\\;$ $A'B'C,\\;$ respectively. The hexagon $G_AK_CG_BK_AG_CK_B$ is regular. ### Hint A look at a related page may prove useful. But so are complex numbers. ### Solution I am going to associate complex numbers with the six given points. Assuming $g$ is the the rotation around the origin through $120^{\\circ}$ degrees in the positive direction: $g^3=1,$ and $u,v,w,d$ are fixed real numbers, let's say, \\begin{align} A &= u, \\\\ A' &= u+d, \\\\ B &= gv, \\\\ B' &= g(v+d), \\\\ C &= -gw, \\\\ C' &= -g(w+d). \\end{align} Calculating the centroids gives (without the fator $1/3$): \\begin{align} G_A &= u+d+gv-gw, \\\\ G_B &= u+g(v+d)-gw, \\\\ G_C &= u+gv-g(w+d), \\\\ K_A &= u+g(v+d)-g(w+d), \\\\ K_B &= u+d+gv-g(w+d), \\\\ K_C &= u+d+g(v+d)-gw. \\end{align} Now it is easy to verify that, say, \\begin{align} G_AK_B &= -gd = G_BK_A, \\\\ G_AK_C &= gd = G_CK_A, \\\\ G_BK_C &= d = G_CK_B, \\end{align} which is actually exactly what is needed to establish the assertion. ### Acknowledgment The problem is a generalized reformulation of a post by Dao Thanh Oai at the CutTheKnotMath facebook page.", "$GH$ is $\\sqrt{3}$. Think of the picture as one large equilateral triangle, $\\triangle{JKL}$ with the sides of $2\\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\\triangle{JKL}$ $\\dfrac{\\sqrt{3}}{4}(2\\sqrt{3}+1)^2=\\dfrac{12+13\\sqrt{3}}{4}$. $[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,W); label(\"E\",E,W); label(\"F\",F,E); label(\"G\",G,E); label(\"H\",H,SE); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,N); label(\"L\",L,SE); [/asy]$ Triangles $\\triangle{DIJ}$, $\\triangle{EFK}$, and $\\triangle{GHL}$ have sides of $\\sqrt{3}$, so their total area is $3(\\dfrac{\\sqrt{3}}{4}(\\sqrt{3})^2)=\\dfrac{9\\sqrt{3}}{4}$. Now, you subtract their total area from the area of $\\triangle{JKL}$: $\\dfrac{12+13\\sqrt{3}}{4}-\\dfrac{9\\sqrt{3}}{4}=\\dfrac{12+13\\sqrt{3}-9\\sqrt{3}}{4}=\\dfrac{12+4\\sqrt{3}}{4}=3+\\sqrt{3}\\implies\\boxed{\\textbf{(C)}\\ 3+\\sqrt3}$ ## Solution 3 We will use, $\\frac{1}{2}ab\\sin x$ to find the area of the following triangles. Since $\\angle A=360$, $\\angle EAF=360-90-90-60=120$. The Area of $\\triangle AEF$ is $\\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin(120)$. Noting, $\\sin (2x) = 2\\sin (x)\\cos (x)$, Area of $\\triangle AEF = \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot 2 \\cdot \\sin(60) \\cdot \\cos(60) = \\dfrac{\\sqrt{3}}{4}$, Area of $\\triangle ABC = \\frac{1}{2} \\cdot 1 \\cdot 1", "EF), i.e., the first set of $m/q=3$ edges where each vertex appears exactly once. Then, we rotate the diagram anti clock-wise (can be done clock-wise as well) one step at a time, and each rotation will provide a 1-factor that is disjoint from the set of edges that appeared before. In this example, rotating the diagram one step provides the 1-factor (BC, DF, AE), rotating two steps provides the 1-factors (AD, BE, CF), and so on until we get back to the starter diagram. Combining all of these 1-factors, we can obtain the complete hypergraph factorization of $K_m^q$ as AB, CD, EF, BC, DF, AE, AD, BE, CF, BD, AF, CE, BF, DE, AC. We can then cycle through this list to assign the individuals in order to achieve maximal balance as described above.", "B, S); dot(\"C\", C, dir(90)); dot(\"D\", D, dir(90)); dot(\"E\", E, S); dot(\"F\", F, dir(0)); dot(\"G\", G, N); dot(\"H\", H, W); dot(\"I\", I, SW); dot(\"J\", J, SW); [/asy]$$\\textbf{(A) } \\frac{7}{3} \\qquad \\textbf{(B) } 8-4\\sqrt2 \\qquad \\textbf{(C) } 1+\\sqrt2 \\qquad \\textbf{(D) } \\frac{7}{4}\\sqrt2 \\qquad \\textbf{(E) } 2\\sqrt2$ ## Problem 22 What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$? $\\textbf{(A) } 100 \\qquad\\textbf{(B) } 101 \\qquad\\textbf{(C) } 200 \\qquad\\textbf{(D) } 201 \\qquad\\textbf{(E) } 202$ ## Problem 23 Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: $L,$ a rotation of $90^{\\circ}$ counterclockwise around the origin; $R,$ a rotation of $90^{\\circ}$ clockwise around the origin; $H,$ a reflection across the $x$-axis; and $V,$ a reflection across the $y$-axis. Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$transformations chosen from $\\{L, R, H, V\\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)", "### Home > INT2 > Chapter 1 > Lesson 1.2.4 > Problem1-62 1-62. Copy $ΔABC$ below on graph paper, then read the Math Notes box in this lesson before completing the parts below. 1. Rotate $ΔABC\\ 90^\\circ$ counterclockwise (↺) about the origin to create $ΔA'B'C'$. Name the coordinates of $C'$. 2. Reflect $ΔABC$ across the vertical line $x=1$ to create $ΔA''B''C''$. Name the coordinates of the vertices. 3. Translate $ΔABC$ so that $A'''$ is at $(4,\\ -5)$. Name the coordinates of $B'''$. Use the eTool below to view the rotations for parts (a), (b), and (c). Click the link at right for the full version of the eTool: Int2 1-62 HW eTool", "fl, df, dr, db, dl}$ the set of $12$ edges of $C$ $\\displaystyle C_3$ $=:$ $\\displaystyle \\set {F, B, U, D, L, R}$ the set of $6$ faces of $C$ Let $r$ be the rotation by $180 \\degrees$ about the axis defined by the line through the points $\\tuple {\\dfrac 1 2, \\dfrac 1 2, 0}$ and $\\tuple {\\dfrac 1 2, \\dfrac 1 2, 1}$. This is a symmetry of $C$ which induces $3$ permutations: $f_0: C_0 \\to C_0$ $f_1: C_1 \\to C_1$ $f_2: C_2 \\to C_2$ $f_0$, $f_1$ and $f_2$ can be specified explicitly as follows: $\\begin{array} {|r|r|} \\hline v & \\map {f_0} v \\\\ \\hline O = dbl & P_4 = dfr \\\\ P_1 = dfl & P_2 = dbr \\\\ P_2 = dbr & P_1 = dfl \\\\ P_3 = ubl & P_7 = ufr \\\\ P_4 = dfr & O = dbl \\\\ P_5 = ufl & P_6 = ubr \\\\ P_6 = ubr & P_5 = dfl \\\\ P_7 = ufr & P_3 = ubl \\\\ \\hline \\end{array} \\qquad \\begin{array} {|r|r|} \\hline e & \\map {f_1} e \\\\ \\hline uf & ub \\\\ ur & ul \\\\ ub & uf \\\\ ul & ur \\\\ fr & bl \\\\ br & fl \\\\ bl & fr \\\\ fl & br \\\\ df &", "Theorem (first of many proofs): the left diagram shows that $c^2 = 4 \\cdot \\frac{ab}2 + (b-a)^2 = a^2 + b^2$, and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to $a^2 + b^2$, but it is also the re-arranged version of the oblique square, which has area $c^2$).[3] $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } { /* first picture */ filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); fill(sq1, rgb(0.95,1,0.95)); fill(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw((0,0)--top--(c,0)--a*expi(rot1)--cycle, rgb(0.5,0.9,0.5)); draw(sq1 ^^ sq2); draw(a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c)); label(\"a\",a/2*expi(rot1),NW,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",(c/2,0),S,sm); } { /* second picture */ fill(shift(shiftR)*sq1, rgb(0.95,1,0.95)); fill(shift(shiftR)*sq2, rgb(0.95,1,0.95)); filldraw(shift(shiftR)*rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw(shift(shiftR+(0,-c))*((0,0)--top--(c,0)--a*expi(rot1)--cycle), rgb(0.5,0.9,0.5)); filldraw(shift(shiftR+(0,-c))*((0,0)--(c,0)--a*expi(rot1)--cycle), rgb(1,0.85,0.7)); draw(shift(shiftR)*((0,0)--(c,0) ^^ sq1 ^^ sq2 ^^ a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c))); label(\"a\",shiftR+a/2*expi(rot1),NW,sm); label(\"b\",shiftR+a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",shiftR+(c/2,0),S,sm); } [/asy]$ Another proof of the Pythagorean Theorem (animated version). $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare, tri = (0,0)--(0,a)--(b,0)--cycle; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "following commands define B_1 as the intersection of k and a line segment %perpendicular to $\\ell$ through C_1. A line segment is drawn between %B_1 and C_1, and a right-angle mark is placed at C_1. The labels \"$B_{1}$\" and \"$C_{1}$\" %are typeset either directly above or directly below this line segment. \\path[name path=perpendicular_line_segment_from_C_1] (C_1) -- ($(C_1)!4cm!90:(C')$); \\coordinate[name intersections={of=perpendicular_line_segment_from_C_1 and ray_k, by={B_1}}]; \\draw (C_1) -- (B_1) node [midway,right]{$y_{1}$}; \\coordinate (U_1) at ($(C_1)!3mm!-45:(A)$); \\draw (U_1) -- ($(C_1)!(U_1)!(A)$); \\draw (U_1) -- ($(C_1)!(U_1)!(B_1)$); %These commands put a brace below line segment AC_1 and label the length $x_{1}$. This brace %needs to be shifted downward from AC_1 to avoid interference with the labels A, C, and C_1. %First, coordinates for A_shift and C_1_shift are defined to be 5mm from AC_1. A brace is drawn %between A_shift and D_shift. \\coordinate (A_shift) at ($(A)!5.5mm!-90:(C_1)$); \\coordinate (C_1_shift) at ($(C_1)!5.5mm!90:(A)$); \\draw[decorate,decoration={brace,raise=5pt,amplitude=5pt,mirror}] (A_shift) -- node[below right=4mm and 0.0625mm, fill=white, inner sep=1pt]{$x_{1}$} (C_1_shift); %This command draws the two sides of the triangle with a common vertex A. \\draw (A) -- (B_1); \\draw (A) -- (C_1); %These commands draw an invisible line segment 2.5mm above AB_1 for labels B and B_1. \\coordinate (A_shift_for_labels_version_1) at ($(A)!2.5mm!90:(B_1)$); \\coordinate (B_1_shift_for_labels) at ($(B_1)!2.5mm!-90:(A)$); \\coordinate (B_1_shift_for_labels_extension) at ($(B_1_shift_for_labels)!-0.25cm!(A_shift_for_labels_version_1)$); \\path[name path=path_for_labels_B_and_B_1] (A_shift_for_labels_version_1) -- (B_1_shift_for_labels_extension); \\coordinate (B_above) at ($(B)!-0.75cm!(C)$); \\path[name path=path_from_C_to_B_above] (C) -- (B_above); \\coordinate (B_1_above) at", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "circle, and $(0,1)$ represents a reflection across a line through two opposite vertices. Note that: $(a,b) = (a,0) \\ast (0,b) = (a+(-1)^00, 0+b)$, so that $(a,b)$ represents rotating, then flipping, if $b = 1$, in which case the direction of rotation is reversed. It is easy to see that: $(1,0)$ and $(0,1)$ generate this group: $(a,b) = (a,0) \\ast (0,b) = (1,0)^a\\ast(0,1)^b$; that: $(1,0)^n = (n,0) = (0,0) = (0,2) = (0,1)^2$ and: $(0,1)\\ast(1,0) = (-1,1) = (1,0)^{-1}\\ast(0,1)$ -that is we have a group of order $2n$ which satisifes the same presentation as $D_n$ ($(1,0)$ plays the role of $r$ and $(0,1)$ plays the role of $s$). It is easy to check that: $(1,0)\\ast(a,0)\\ast(-1,0) = (a,0)$ and that: $(0,1)\\ast(a,0)\\ast(0,1) = (-a,0) = (a,0)^{-1}$, so any subgroup that contains only rotations is normal. If a subgroup contains a reflection $(a,1)$ for it to be normal we must have: $(1,0)\\ast(a,1)\\ast(-1,0) = (a+2,1)$ in it as well. If $n$ is odd, this implies the only normal subgroup containing a reflection is $D_n$ itself. However, if $n$ is even it only gives a necessary (but not sufficient) condition. It turns out for even $n$ we get exactly two conjugacy classes of reflections: $\\{(a,1): a\\text{ is even.}\\},\\ \\{(a,1): a\\text{ is odd.}\\}$ (any normal subgroup must contain all of any given conjugacy class, so", "and C $=(8,2)$. If we move this figure by 90 degrees anti-clockwise, then all the three points move 90 degrees anti-clockwise, and the new points after the rotation will be A $= (-6,8)$ B $= (-2,4)$ and C $= (-2,8)$, as shown in the figure below. Similarly, if we move the polygon at 180 degree rotation, then the new points will be A $= (-8,-6)$, B $= (-4,-2)$ and C $= (-8,-2)$ and finally if we rotate it at 270 degree at clock-wise direction then the points will be A $= (6,-8)$ B $= (2,-4)$ and C $= (2,-8)$. Now that you have an understanding of how rotation works, you will find it a lot easier to understand the concept of $-90^{o}$ rotation. Practice Questions: 1. Rotate the following points by $-90^{o}$. a) $(6,1)$ b) $(-7,-6)$ c $(-2,3)$ d) $(3,-8 )$ 2. You are given a Quadrilateral with vertices A $= (-1,9)$, B $= (-3,7)$ and C $= (-4,7)$ and D = $(-6,8)$. The quadrilateral is first rotated at 90^{o} clockwise, and then it is rotated at $90^{o}$ anti-clockwise. You are required to determine the value of coordinates after the final rotation. The new point after $-90^{o}$ rotation will be a) $(1,-6)$ b) $(-6, 7)$ c) $(3,2)$ d) $(-8,-3)$. The vertices of the quadrilateral are first rotated at", "Home > INT2 > Chapter 3 > Lesson 3.2.3 > Problem3-97 3-97. On graph paper, plot $ABCD$ if $A(0,3),B(2,5),C(6,3)$, and $D(4,1)$. 1. Rotate $ABCD\\ 90°$ clockwise ($↻$) about the origin to form $A'B'C'D'$. Name the coordinates of $B'$. $(5, −2)$ 2. Translate $A'B'C'D'$ up $8$ units and left $7$ units to form $A''B''C''D''$. Name the coordinates of $C''$. $(−4, 2)$ 3. After rotating $ABCD$ $180^\\circ$ to form $A'''B'''C'''D'''$, Arah noticed that the image lined up exactly with the original. In this case, about what point was $ABCD$ rotated? How did you locate it? Use the eTool below to solve each part. Click the link at right for the full version of the eTool. Int2 3-97 HW eTool", "# 1979 AHSME Problems/Problem 23 ## Problem 23 The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one. Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$. $[asy] size(150); import patterns; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; aux=midpoint(AA--BB); draw(BB--DD); P=midpoint(AA--aux); aux=midpoint(CC--DD); Q=midpoint(CC--aux); draw(AA--CC,dashed); dot(P); dot(Q); fill(DD--BB--CC--cycle,pattern(\"hatch\")); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); //Credit to TheMaskedMagician for the diagram[/asy]$ $\\textbf{(A) }\\frac{1}{2}\\qquad \\textbf{(B) }\\frac{3}{4}\\qquad \\textbf{(C) }\\frac{\\sqrt{2}}{2}\\qquad \\textbf{(D) }\\frac{\\sqrt{3}}{2}\\qquad \\textbf{(E) }\\frac{\\sqrt{3}}{3}$ ## Solution 1 Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find this distance, consider triangle $\\triangle PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $\\triangle ABC$, $CP=\\frac{\\sqrt{3}}{2}$. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); draw(P--Q,dashed); draw(P--CC,dashed); draw(AA--CC,dashed); dot(P); dot(Q); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ Next, we need to find $\\cos(\\angle PCQ)$ in order to find $PQ$ by the Law of Cosines. To do so, drop down $D$ onto $\\triangle ABC$", "is a complex number that has been rotated $90\\,^{\\circ}$ counter-clockwise about the origin. This is a straight-forward application of De Moivre's Theorem, which students should be familiar with from precalculus. So if we start at $b$ and add a $90\\,^{\\circ}$-rotated version of the displacement from $A$ to $B$, we arrive at the complex number of the vertex opposite $A$. This must then correspond to the complex number $b + i(b-a)$. Remember, we are after the complex number that corresponds to $R$ (let's call it $r$), and $R$ is the midpoint of this aforementioned diagonal. Consequently, we take the average of the complex numbers that correspond to the diagonal's endpoints, which yields $$r = {a+b+i(b-a) \\over 2}$$ Now we can appeal to symmetry to do the rest. If we want the complex number that corresponds to $T$ (let's call it $t$), which is the center of the square that has $\\overline{CD}$ as an edge. Here $C$ is playing the role of $A$ from the previous step, and $D$ is playing the role of $B$. As such, $$t = {c+d+i(d-c) \\over 2}$$ Note the difference $(t-r)$ gives the complex number (let's call it $\\alpha$) matching the displacement from $R$ to $T$, and $$\\begin{array}{rcl} \\alpha &=& \\frac{c+d+i(d-c)}{2} - \\frac{a + b + i(b-a)}{2}\\\\\\\\ &=& \\frac{(c+d-a-b) + i(d-c-b+a)}{2}\\\\\\\\ \\end{array}$$ To find the", "chosen). In any cross-section (say at $z = t$), there are 8 (potentially overlapping or degenerate) line segment, one for each corner's tetrahedron, specifically: \\begin{align} y_1 &= t &+ x \\\\ y_2 &= -t &+ x \\\\ y_3 &= 1-t &- x \\\\ y_4 &= 1+t &- x \\\\ y_5 &= t &- x \\\\ y_6 &= 2-t &- x \\\\ y_7 &= -1+t &+ x \\\\ y_8 &= 1-t &+ x \\\\ \\end{align} (Just try $t=0$ and $t=1$ to convince yourself these are right) With this in hand, we can turn to the computer to draw all of these for us for any $z$. Here's an implementation in R showing 25 cross-sections, proceeding right and down: xx = seq(0, 1, length.out = 100L) col = c('#e6194b', '#3cb44b', '#ffe119', '#0082c8', '#f58231', '#911eb4', '#46f0f0', '#f032e6') par(mfrow = c(5, 5), mar = c(0, 0, 0, 0), oma = c(0, 0, 0, 0)) t = seq(0, 1, length.out = prod(par('mfrow'))) coef = data.frame( t_a = c(0, 0, 1, 1, 0, 2, -1, 1), t_b = c(1, -1, -1, 1, 1, -1, 1, -1), b = c(1, 1, -1, -1, -1, -1, 1, 1) ) for (t_i in t) { plot(NA, xlim = c(0, 1), ylim = c(0, 1), asp = 1, xaxt = 'n', yaxt = 'n', ylab = '', xlab", "# 2020 AMC 10B Problems/Problem 23 The following problem is from both the 2020 AMC 10B #23 and 2020 AMC 12B #19, so both problems redirect to this page. ## Problem Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: $L,$ a rotation of $90^{\\circ}$ counterclockwise around the origin; $R,$ a rotation of $90^{\\circ}$ clockwise around the origin; $H,$ a reflection across the $x$-axis; and $V,$ a reflection across the $y$-axis. Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\\{L, R, H, V\\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.) $\\textbf{(A)}\\ 2^{37} \\qquad\\textbf{(B)}\\ 3\\cdot 2^{36} \\qquad\\textbf{(C)}\\ 2^{38} \\qquad\\textbf{(D)}\\ 3\\cdot 2^{37} \\qquad\\textbf{(E)}\\ 2^{39}$ ## Solution Let (+) denote counterclockwise/starting orientation and (-) denote clockwise orientation. Let 1,2,3, and 4 denote which quadrant A is in. Realize that from any odd quadrant and any orientation, the 4", "$BA$ and then rotate $60^{\\circ}$ counterclockwise around point $A$. e) Reflect over line $BA$ and then translate by directed line segment $BA$. ##### Problem 6 6) Draw the image of figure $ABC$ when translated by directed line segment $u$. Label the image of $A$ as $A'$, the image of $B$ as $B'$, and the image of $C$ as $C'$. 7) Explain why the line containing $AB$ is parallel to the line containing $A'B'$. ##### Problem 7 8) There is a sequence of rigid transformations that takes $A$ to $A'$, $B$ to $B'$, and $C$ to $C'$. The same sequence takes $D$ to $D'$. Draw and label $D'$:" ]

[ "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "$GH$ is $\\sqrt{3}$. Think of the picture as one large equilateral triangle, $\\triangle{JKL}$ with the sides of $2\\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\\triangle{JKL}$ $\\dfrac{\\sqrt{3}}{4}(2\\sqrt{3}+1)^2=\\dfrac{12+13\\sqrt{3}}{4}$. $[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,W); label(\"E\",E,W); label(\"F\",F,E); label(\"G\",G,E); label(\"H\",H,SE); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,N); label(\"L\",L,SE); [/asy]$ Triangles $\\triangle{DIJ}$, $\\triangle{EFK}$, and $\\triangle{GHL}$ have sides of $\\sqrt{3}$, so their total area is $3(\\dfrac{\\sqrt{3}}{4}(\\sqrt{3})^2)=\\dfrac{9\\sqrt{3}}{4}$. Now, you subtract their total area from the area of $\\triangle{JKL}$: $\\dfrac{12+13\\sqrt{3}}{4}-\\dfrac{9\\sqrt{3}}{4}=\\dfrac{12+13\\sqrt{3}-9\\sqrt{3}}{4}=\\dfrac{12+4\\sqrt{3}}{4}=3+\\sqrt{3}\\implies\\boxed{\\textbf{(C)}\\ 3+\\sqrt3}$ ## Solution 3 We will use, $\\frac{1}{2}ab\\sin x$ to find the area of the following triangles. Since $\\angle A=360$, $\\angle EAF=360-90-90-60=120$. The Area of $\\triangle AEF$ is $\\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin(120)$. Noting, $\\sin (2x) = 2\\sin (x)\\cos (x)$, Area of $\\triangle AEF = \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot 2 \\cdot \\sin(60) \\cdot \\cos(60) = \\dfrac{\\sqrt{3}}{4}$, Area of $\\triangle ABC = \\frac{1}{2} \\cdot 1 \\cdot 1", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "real b, real c){ real p=(a+b+c)/2; return sqrt(p*(p-a)*(p-b)*(p-c)); } currentprojection=orthographic((1,1.6,1),center=true,zoom=.95); // Step 1: construct the base A,B,C on the plane z=0 real a=6, b=5, c=4; triple B=(0,0,0), C=(a,0,0); // Abc is the projection of A on the segment BC triple Abc=barycentric(B,C,O,1/(a^2+c^2-b^2),1/(a^2+b^2-c^2),0); real bt=abs(Abc-B); real hbc=sqrt(c*c-bt*bt); triple A=Abc+hbc*dir(90,90); draw(A--Abc,red); draw(A--B--C--cycle); label(\"$A$\",A,plain.S); label(\"$B$\",B,plain.E); label(\"$C$\",C,plain.W); label(\"$A_{bc}$\",Abc,plain.N,red); // Step 2: get the top point D real at=6, bt=7, ct=4; // H is the projection of D on the base ABC real Sdab=Heron(at,bt,c); real Sdbc=Heron(bt,ct,a); real Sdca=Heron(ct,at,b); triple H=barycentric(A,B,C,1/(Sdab^2+Sdca^2-Sdbc^2),1/(Sdab^2+Sdbc^2-Sdca^2),1/(Sdca^2+Sdbc^2-Sdab^2)); real ha=abs(H-A); real hd=sqrt(at*at-ha*ha); triple D=H+hd*Z; draw(D--A^^D--B^^D--C); draw(D--H,blue); label(\"$D$\",D,plain.N); label(\"$H$\",H,plain.W,blue); dot(H); PS1: Why Asymptote and why not TikZ? That drawing way of using barycentric coordinate can be coded in several drawing languages. TikZ does has barycentric coordinate; but its computation is quite weak, Dimension too large error may happen, even in 2D when I had tried drawing the Euler line of a triangle)! Asymptote has better accuracy, and available for 3D. PS2: There is another way based on origami, described in a book of Polya. I will do it in free time later. • You wrote \"Full code (still some mistake ^^)\" and \" I will do it in free time later.\" When you post correct answer? Oct 6, 2021 at 0:38 • Peter: As I said: \"in free time later\". There is a", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). ## Solution 3 $[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot(\"C\", C, dir(0)); dot(\"A\", A, dir(90));dot(\"B\", B, dir(180)); dot(\"D\", D, NE); dot(\"E\", E, dir(90));dot(\"F\", F, dir(270)); dot(\"M\", M, NE); dot(\"N\", N, dir(270));dot(\"L\", L, NW); dot(\"X\", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]$ Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\\sin(60)$. Similarly, $AB=3\\sin(60)$, and $BC=5\\sin(60)$. So $\\triangle ABC$ is a 3-4-5 right triangle", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "# Carnot's Theorem \"\" Carnot's Theorem\"\" states that in a triangle $ABC$, the signed sum of perpendicular distances from the circumcenter $O$ to the sides (i.e., signed lengths of the pedal lines from $O$) is: $OO_A+OO_B+OO_C=R+r$ $[asy] pair a,b,c,O,i,d,f,g; a=(0,0); b=(4,0); c=(1,3); O=circumcenter(a,b,c); i=incenter(a,b,c); draw(a--b--c--cycle); draw(circumcircle(a,b,c)); draw(incircle(a,b,c)); dot(i); dot(O); label(\"A\",a,W); label(\"B\",b,E); label(\"C\",c,N); label(\"I\",i,N); label(\"O\",O,N); d=foot(O,b,c); dot(d); draw(O--d); label(\"O_A\",d,N); draw(rightanglemark(O,d,b)); f=foot(O,a,b); dot(f); draw(O--f); draw(rightanglemark(O,f,a)); label(\"O_C\",f,S); g=foot(O,c,a); dot(g); draw(O--g); draw(rightanglemark(O,g,a)); label(\"O_B\",g,W); [/asy]$ where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. Explicitly, $OO_A+OO_B+OO_C=\\frac{abc(|\\cos{A}|+|\\cos{B}|+|\\cos{C}|)}{4|\\Delta|}$ where $\\Delta$ is the area of triangle $\\Delta ABC$. Weisstein, Eric W. \"Carnot's Theorem.\" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html # Below is not Carnot's Theorem Carnot's Theorem states that in a triangle $ABC$ with $A_1\\in BC$, $B_1\\in AC$, and $C_1\\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$. #### Proof Only if: Assume that the given perpendiculars are concurrent at $M$. Then, from the Pythagorean Theorem, $A_1B^2=BM^2-MA_1^2$, $C_1A^2=AM^2-MC_1^2$, $B_1C^2=CM^2-MB_1^2$, $A_1C^2=MC^2-MA_1^2$, $C_1B^2=MB^2-MC_1^2$, and $B_1A^2=AM^2-MB_1^2$. Substituting each and every one of these in and simplifying gives the desired result. If: Consider the intersection of the perpendiculars from $A_1$ and $B_1$. Call this intersection point $N$,", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "$\\triangle ABC$ is$$s = \\frac{AB + BC + CA}2 = \\frac{15}2,$$and Heron's Formula gives the area of $\\triangle ABC$ as$$\\sqrt{s(s-AB)(s-BC)(s-CA)} = \\frac{15\\sqrt7}4.$$This area is also $rs$ implying that $r = \\frac{\\sqrt7}2$. Stewart's Theorem gives $AD =3\\sqrt2$. Because the ratio of the areas of $\\triangle IBC$ and $\\triangle ABC$ is $\\frac{ID}{AD},$ it follows that$$ID = AD\\cdot\\frac{\\frac{r\\cdot BC}2}{\\frac{15\\sqrt7}4} = \\sqrt2.$$Thus $IM = MD - ID = \\frac{\\sqrt2}2$. $[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); dot(\"I\", I, dir(120)); dot(\"M\", M, W); [/asy]$ Note that $\\angle EFI = 90^{\\circ} - \\angle FIA = 90^{\\circ} - \\angle CID = 90^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle C}{2} = \\frac{\\angle B}{2} = \\angle IBC$. Thus $\\triangle IBC \\sim \\triangle IFE$. The height of $\\triangle IBC$ to $I$ is $r=\\frac{\\sqrt7}2$, and the height of $\\triangle IFE$ to $I$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "D, E, F, G, H; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); G = rotate(90,(A + C)/2)*(E); H = rotate(90,(A + C)/2)*(F); draw(A--B--C--D--cycle); draw(E--G--F--H--cycle); dot(\"A\", A, N); dot(\"B\", B, dir(0)); dot(\"C\", C, S); dot(\"D\", D, W); dot(\"E\", E, N); dot(\"F\", F, S); dot(\"G\", G, W); dot(\"H\", H, dir(0));[/asy]$ $\\overline{EF}$ is the diagonal of the square, with length $17\\sqrt{2}$; the answer is $EF^2 = (17\\sqrt{2})^2 = 578$. Solution 3 A slightly more analytic/brute-force approach: Drop perpendiculars from $E$ and $F$ to $I$ and $J$, respectively; construct right triangle $EKF$ with right angle at K and $EK || BC$. Since $2[CDF]=DF*CF=CD*JF$, we have $JF=5\\times12/13 = \\frac{60}{13}$. Similarly, $EI=\\frac{60}{13}$. Since $\\triangle DJF \\sim \\triangle DFC$, we have $DJ=\\frac{5JF}{12}=\\frac{25}{13}$. Now, we see that $FK=DC-(DJ+IB)=DC-2DJ=13-\\frac{50}{13}=\\frac{119}{13}$. Also, $EK=BC+(JF+IE)=BC+2JF=13+\\frac{120}{13}=\\frac{289}{13}$. By the Pythagorean Theorem, we have $EF=\\sqrt{\\left(\\frac{289}{13}\\right)^2+\\left(\\frac{119}{13} \\right)^2}=\\frac{\\sqrt{(17^2)(17^2+7^2)}}{13}$$=\\frac{17\\sqrt{338}}{13}=\\frac{17(13\\sqrt{2})}{13}=17\\sqrt{2}$. Therefore, $EF^2=(17\\sqrt{2})^2=578$. Solution 4 Based on the symmetry, we know that $F$ is a reflection of $E$ across the center of the square, which we will denote as $O$. Since $\\angle BEA$ and $\\angle AOB$ are right, we can conclude that figure $AOBE$ is a cyclic quadrilateral. Pythagorean Theorem yields that $BO=AO=\\frac{13\\sqrt{2}}{2}$. Now, using Ptolemy's Theorem, we get that $$AO\\cdot BE + BO\\cdot AE = AB\\cdot AO$$ $$\\frac{13\\sqrt{2}}{2}\\cdot 5+\\frac{13\\sqrt{2}}{2}\\cdot 12 = 13\\cdot OE$$" ]

[ "How do you find the exact value of csc180? csc 180 = 1/ sin 180 = 1/0 = $\\infty$ ( or rather undefined) sine of ${0}^{0}$ and ${180}^{0}$ come on the X axis on the right and left of unit circle. Both give value zero.", "# How do you find the exact value of csc^-1(-1)? ${\\csc}^{- 1} \\left(- 1\\right) = - \\frac{\\pi}{2}$ ${\\csc}^{- 1} \\left(- 1\\right)$ is the angle whose cosecant ratio is $- 1$, i.e. whose sine ratio is $- 1$. As $\\sin \\left(- \\frac{\\pi}{2}\\right) = - 1$, hence ${\\csc}^{- 1} \\left(- 1\\right) = - \\frac{\\pi}{2}$", "# If sec(t) = 3, how do you find the exact value of csc(90 - t )? $\\sec \\left(t\\right) = 3$ $\\implies \\cos \\left(t\\right) = \\frac{1}{3}$ now, we know, $\\sin \\left(90 - x\\right) = \\cos x$ $\\sin \\left(90 - t\\right) = \\cos \\left(t\\right) = \\frac{1}{3}$ $\\csc \\left(90 - t\\right) = \\frac{1}{\\sin} \\left(90 - t\\right) = 3$", "# How do you find the value of csc pi? $\\csc \\left(\\pi\\right)$ is undefined $\\csc \\left(x\\right) = \\frac{1}{\\sin} \\left(x\\right)$ We have $\\sin \\left(\\pi\\right) = 0$, therefore $\\csc \\left(\\pi\\right) = \\frac{1}{0}$ which is undefined and has different limits when approached from above and below ($\\pm \\infty$).", "+0 0 145 1 +102 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "# How do you use a calculator to find csc^-1(2pi/9)? Find ${\\csc}^{-} 1 \\left(\\frac{2 \\pi}{9}\\right)$ Ans: ${44}^{\\circ} 25$ $\\csc \\left(0.70\\right) = \\frac{1}{\\sin \\left(0.70\\right)}$ Use calculator to find --> sin x = 0.70 --> arc $x = {44}^{\\circ} 25$", "$0$ Complete the square to find the minimum or maximum value. $x^2-8x+6$ 1. $-22$ 2. $11/2$ 3. $16/3$ 4. $-10$ $x^2-3x+2$ 1. $-1/4$ 2. $-5$ 3. $-7/5$ 4. $17/4$", "principal value of sec$$^{-1}$$ (1) be α then, sec$$^{-1}$$ (1) = θ ⇒ sec θ = 1 = sec 0 [Since, 0 ≤ θ ≤ π] Therefore, the principal value of sec$$^{-1}$$ (1) is 0. 6. Find the principal values of the inverse trig function csc$$^{-1}$$ (- 1). Solution: If the principal value of csc$$^{-1}$$ x is α then we know, - $$\\frac{π}{2}$$ ≤ θ ≤ $$\\frac{π}{2}$$ and θ ≠ 0. Therefore, if the principal value of csc$$^{-1}$$ (- 1) be θ then csc$$^{-1}$$ (- 1) = θ ⇒ csc θ = - 1 = csc (-$$\\frac{π}{2}$$) [Since, - $$\\frac{π}{2}$$ ≤ θ ≤ $$\\frac{π}{2}$$] Therefore, the principal value of csc$$^{-1}$$ (- 1) is (-$$\\frac{π}{2}$$). Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.", "# Thread: Find the minimum value 1. ## Find the minimum value Find the minimum value of (5 cosh x + 3 sinh x). How can I solve this? 2. Originally Posted by geton Find the minimum value of (5 cosh x + 3 sinh x). How can I solve this? Differentiate, put to zero, test nature .... the usual deal. Note that to solve when the derivative is equal to zero, you should: 1. Replace cosh and sinh with their exponential definitions. 2. Multiply through by 2 e^{-x}. 3. Solve the resulting quadratic equation in e^x ..... 3. $\\cosh x = \\frac{e^x + e^{-x}}{2}$ $\\sinh x = \\frac{e^x - e^{-x}}{2}$ so, $f(x) = 5\\cosh x + 3 \\sinh x$ $f'(x) = 5\\sinh x + 3\\cosh x = \\frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} = \\frac{4e^{2x} - 1}{e^x}$ we want to find $x$ such that $f'(x) = 0$ since $e^x$ is never zero, we will solve $4e^{2x} - 1 = 0$.. solving, we shall obtain $x = -\\ln 2$... substitute that $x$ to find the minimum value.. -------------------------- to check, $f''(x) = f(x) = 5\\cosh x + 3 \\sinh x = 4e^x + e^{-x} > 0$ for any $x$. thus, $x = -\\ln 2$ gives the minimum value for $f(x)$ 4. Yes I got x = -", "maximum or absolute minimum values. Justify your answers. Find the extreme values if you can. $\\frac{1}{x \\sin x} \\text { on }(0, \\pi)$ Determine whether the given function has absolute maximum or absolute minimum values. Justify your answers. Find the extreme values if you can. $\\frac{1}{x \\sin x} \\text { on }(0, \\pi)$...", "+ \\ \\frac{\\sec(\\pi-A)}{\\csc(2\\pi+A)} \\ = \\ 0$", "# Math Help - Finding the minimum value 1. ## Finding the minimum value Hey there I need some help with this problem Find the minimum value of absolute value of sinx + cosx + tanx + cotx + secx + cscx I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks. 2. ## Re: Finding the minimum value Originally Posted by gfbrd Hey there I need some help with this problem Find the minimum value of absolute value of sinx + cosx + tanx + cotx + secx + cscx I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks. The derivative of: $f(x) = \\sin(x)+ \\cos(x) + \\tan(x) + \\cot(x) + \\sec(x) + \\csc(x)$ is: $\\frac{df}{dx} = \\cos(x)-\\sin(x)+\\tan^{2}(x)-\\cot^{^2}(x)+\\sec(x) \\tan(x)-\\csc(x) \\cot(x)$ The $\\frac{df}{dx}$ above can be written as(for $y > 0$): \\begin{align*}=& \\cos(x) - \\sin(x) + \\frac{\\sin^{2}(x)}{\\cos^{2}(x)} -\\frac{\\cos^{2}(x)}{\\sin^{2}(x)} + \\left(\\frac{\\sin(x)}{\\cos(x)}\\right)\\left(\\frac{1} {\\cos(x)}\\right) - \\left( \\frac{\\cos(x)}{\\sin(x)}\\right)\\left(\\frac{1}{\\sin( x)}\\right) \\\\ \\\\ =& \\cos(x) - \\sin(x) + \\frac{\\sin^{2}(x)}{\\cos^{2}(x)} -\\frac{\\cos^{2}(x)}{\\sin^{2}(x)} + \\frac{\\sin(x)}{\\cos^{2}(x)} - \\frac{\\cos(x)}{\\sin^{2}(x)} \\\\", "+0 # help 0 51 3 Find the minimum value of $$\\sin x + \\csc x)^2 + (\\cos x + \\sec x)^2$$ for $$0 < x < \\frac{\\pi}{2}.$$ Aug 6, 2022 #2 0 minimum value is 9 Aug 6, 2022 #3 0 Hello Guest! Remember: $$sin^2 x +cos^2 x = 1 \\\\ csc (x) = \\frac{1}{sin (x)} \\\\ sec(x) = \\frac{1}{cos (x)}$$ So: $$(sin(x)+csc(x))^2+(cos(x)+sec(x))^2 \\\\ =sin^2(x)+\\frac{1}{sin^2(x)}+2sin(x)\\frac{1}{sin(x)}+cos^2(x)+\\frac{1}{sin^2(x)}+2cos(x)\\frac{1}{cos(x)} \\\\ =(sin^2(x)+cos^2(x))+2+2+(\\frac{1}{sin^2(x)}+\\frac{1}{cos^2(x)}) \\\\ =1+4+\\frac{sin^2(x)+cos^2(x)}{sin^2(x)cos^2(x)} =5+\\frac{1}{(sin(x)cos(x))^2}$$ Now, we want the minimum value, so we have to minimize: $$\\frac{1}{(sin(x)cos(x))^2}$$ or equivalently, maximize: $$(sin(x)cos(x))^2$$ Recall: $$sin(2x)=2sin(x)cos(x) \\implies sin(x)cos(x)=\\frac{1}{2}sin(2x)$$ Hence, we want to maximize $$\\frac{1}{4}sin^2(2x)$$ $$0 < sin^2(2x) \\le 1$$ So, the upper bound is 1, which is satisfied when $$x=\\frac{\\pi}{4}$$ So, the maximum value of $$\\frac{1}{4}sin^2(2x)$$ is: $$\\dfrac{1}{4}*(1)=\\dfrac{1}{4}$$ Therefore, the minimum value of $$\\frac{1}{(sin(x)cos(x))^2}$$ is $$\\frac{1}{\\frac{1}{4}}=4$$ Thus, the minimum value of the whole expression in the question is: $$5+4=9$$ , as guest has found below. I hope this helps! Aug 7, 2022", "# MINIMUM! :D Algebra Level 3 find the minimum value of ${ \\tan }^{ 2 }x + 2 \\cot x$ when $$x$$ is in the interval $$(0,\\frac { \\pi }{ 2 } )$$ dont forget to share and like. :D ×", "a mistake here, (lntan(x))' = csc x sec x, so that is equal to $cscx sec^2x$", "Home > PC > Chapter 5 > Lesson 5.1.1 > Problem5-10 5-10. Find all values of $x$ where $0 ≤ x ≤ 2\\pi$ for the following equations. 1. $\\csc x = −2$ $\\frac{1}{\\sin x}=-2$ 1. $\\operatorname { tan } x = - \\frac { \\sqrt { 3 } } { 3 }$ What special angle is this? In what quadrants is tangent negative?", "# more than one minimum value? Let $f(x)=(p\\sec(x))^2 + (q\\csc(x))^2$ Now,$f(x)=(p\\sec(x) - q\\csc(x))^2+2p\\sec(x)q\\csc(x)$ Now, $(p\\sec(x) - q\\csc(x))^2\\geq0$ and has minimum value = $0$ when $p\\sec(x) = q\\csc(x) \\implies \\tan(x)=\\frac{q}{p}$ The minimum value of $f(x)$ will be $2p\\sec(x)q\\csc(x)$ where $\\tan(x)=\\frac{q}{p}$ $=\\frac{4pq}{\\sin(2x)} =2(p^2+q^2)$ or directly putting the value $\\tan(x)$, $p^2(1+\\frac{q^2}{p^2})+ q^2(1+\\frac{p^2}{q^2})=2(p^2+q^2)$ Again, $f(x)=(p\\sec(x))^2 + (q\\csc(x))^2 =p^2+q^2+ (p\\tan(x))^2 + (q\\cot(x))^2$ $(p\\tan(x))^2 + (q\\cot(x))^2 = (p\\tan(x) - q\\cot(x))^2+2pq$ will have minimum value $2pq$ for $\\tan^2(x)=\\frac{q}{p}$ Clearly, here the minimum value of $f(x)=p^2+q^2+2pq = (p+q)^2$ When I used derivative test, I reached the last value. To rephrase the question, let h(x)=$(p\\tan(x))^2 + (q\\cot(x))^2$ =>f(x)=$p^2+q^2$+h(x) As calculated above, the minimum value of f(x)=$2(p^2+q^2)$ and the minimum value of h(x)=2pq The minimum value of f(x) should have been $p^2+q^2+2pq$. In general, $(p+q)^2\\neq 2(p^2+q^2)$ unless $p=q$. Where have I gone wrong? - When you decided that you had to minimize $(p\\sec(x)-q\\csc(x))^2$ to minimize $f(x)$. Analogous example: $f(x)=(x)^2+(2x)$ is not minimum when $(x)^2=0$. – Did Jul 14 '12 at 13:59 @did, if f(x)=g(x)+c where c is a constant, the extreme values of f(x) & g(x) will have one-one correspondence? – lab bhattacharjee Jul 14 '12 at 14:32 True and unrelated since, in your case, the second term is not a constant. (Did you check the Analogous example in my comment? If not, why?) – Did Jul 14", "# How to find the value of x where 0<=x<=360^@ for csc x= -1.3737? Jun 1, 2015 $\\csc x = \\frac{1}{\\sin} x = - 1.3737$ -> $\\sin x = \\frac{1}{-} 1.3737 = - 0.728$ On the trig unit circle: -0.728 -> x = -46.71 or x = 360 - 46.71 = 313.29 deg x = 180 + 46.71 = 226.71. Within period$\\left(0 , 2 \\pi\\right)$, there are 2 answers: x = 226.71 and x = 313. 29 deg", "# A Trigoquality Algebra Level 3 Find the minimum value of $$\\large \\frac{ 4+ 9x^2\\sin^2 x}{x\\sin x}$$ for $$0 < x <\\pi$$. ×" ]

[ "# Integrate $\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}dx$ Solve the indefinite integral $$I=\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}\\;dx$$ My Attempt: \\begin{align} I&=\\int\\frac{1}{\\sin x+\\cos x+\\frac{1}{\\sin x \\cos x}+\\frac{\\sin x +\\cos x}{\\sin x\\cos x}}\\;dx\\\\ \\\\ &=\\int\\frac{\\sin x\\cos x}{\\left(\\sin x+\\cos x\\right)\\left(\\sin x\\cos x \\right)+1+\\left(\\sin x+\\cos x\\right)}\\;dx \\end{align} How can I complete the solution from here? • Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14 • Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45 You want \\begin{align} \\int\\frac{dx}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x} &= \\int\\frac{dx}{\\sin x+\\cos x+\\frac{\\sin x}{\\cos x} +\\frac{\\cos x}{\\sin x}+\\frac1{\\sin x}+\\frac1{\\cos x}}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin^2 x \\cos x+\\cos^2 x \\sin x+\\sin^2 x +\\cos^2 x+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin x \\cos x(\\sin x+ \\cos x)+1+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1}\\\\ \\end{align}. Applying substitutions $\\sin x = \\frac{2t}{1+t^2}, \\cos x = \\frac{1-t^2}{1+t^2}, dx = \\frac{2 dt}{1+t^2}$, this becomes \\begin{align} \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1} &=\\int\\frac{\\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\\frac{2t}{1+t^2}+ \\frac{1-t^2}{1+t^2})+1}\\\\ &=\\int\\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\\\ &=\\int\\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \\quad \\text{ (according to Wolfram Alpha)}\\\\ &=\\int\\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\\\ \\end{align}. According to Wolfram Alpha, that quartic can be factored as", "## welshfella one year ago Need help with this one:- Find limit as x --> 0 of 2csc^2 x - (1/2) csc^2 (1/2) x. 1. welshfella both terms tend to infinity as x approaches zero so i converted them to the form f(x)/g(x) in order to apply l'hopitals but and got it to the form (0/0). Thats as far as i got. 2. ganeshie8 i think we can try to simplify the expression using trig identities 3. ganeshie8 \\begin{align} 2\\csc^2 x - \\frac{1}{2} \\csc^2 \\frac{x}{2}&=\\frac{1}{2}\\left(\\dfrac{4}{\\sin^2x}-\\frac{1}{\\sin^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{4}{4\\sin^2\\frac{x}{2}\\cos^2\\frac{x}{2}}-\\frac{1}{\\sin^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{1-\\cos^2\\frac{x}{2}}{\\sin^2\\frac{x}{2}\\cos^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{1}{\\cos^2\\frac{x}{2}}\\right)\\\\~\\\\ \\end{align} we can take the limit now 4. welshfella yea 5. welshfella 1/2 6. thomas5267 Not really awake. Use with caution. \\begin{align*} &\\phantom{=}2\\csc^2(x)-\\frac{1}{2}\\csc^2\\left(\\frac{1}{2}x\\right)\\\\ &=\\frac{2}{\\sin^2(x)}-\\frac{1}{2\\sin^2\\left(\\frac{1}{2}x\\right)}\\\\ &=\\frac{2}{1-\\cos^2(x)}-\\frac{1}{1-\\cos(x)}\\\\ &=\\frac{2}{1-\\cos^2(x)}-\\frac{1+\\cos(x)}{1-\\cos^2(x)}\\\\ &=\\frac{1-\\cos(x)}{1-\\cos^2(x)}\\\\ &=\\frac{1}{1+\\cos(x)} \\end{align*} 7. anonymous Another approach using the same identity: \\begin{align*} 2\\csc^2x-\\frac{1}{2}\\csc^2\\frac{x}{2}&=\\frac{4}{1-\\cos2x}-\\frac{1}{1-\\cos x}\\\\[1ex] &=\\frac{3-4\\cos x+\\cos2x}{(1-\\cos2x)(1-\\cos x)}\\\\[1ex] &=\\frac{2(1-\\cos x)^2}{(2-2\\cos^2x)(1-\\cos x)} \\end{align*} which gives the last line as in thomas's answer. 8. ganeshie8 Ahh that identity looks much better as it avoids the half angles/fractions 9. welshfella i didn't know that identity on the second line @ganeshie 10. welshfella that half angle one 11. welshfella I searched the web to find a half angle identity for sin^ x but couldnt find one 12. ganeshie8 ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them 13. thomas5267", "cos^2x(1 - cos^2x) + (1 - cos^2x)}{cos^2x} = sec^2x \\end{align} (19) \\begin{align} \\frac{cos^4x + cos^2x - cos^4x + 1 - cos^2x}{cos^2x} = sec^2x \\end{align} (20) \\begin{align} \\frac{1 }{cos^2x} = sec^2x \\end{align} (21) $$sec^2x = sec^2x$$ We will now prove the other Pythagorean identity as being equivalent: (22) $$1^2 + cot^2x = csc^2x$$ (23) \\begin{align} (cos^2x + sin^2x) + \\frac{cos^2x}{sin^2x} = csc^2x \\end{align} (24) \\begin{align} \\frac{sin^2x(cos^2x + sin^2x)}{sin^2x} + \\frac{cos^2x}{sin^2x} = csc^2x \\end{align} (25) \\begin{align} \\frac{sin^2x(cos^2x + sin^2x) + cos^2x}{sin^2x} = csc^2x \\end{align} (26) \\begin{align} \\frac{sin^2xcos^2x + sin^4x + cos^2x}{sin^2x} = csc^2x \\end{align} (27) \\begin{align} \\frac{sin^2x(1-sin^2x) + sin^4x + (1 - sin^2x)}{sin^2x} = csc^2x \\end{align} (28) \\begin{align} \\frac{sin^2x -sin^4x + sin^4x + 1 - sin^2x}{sin^2x} = csc^2x \\end{align} (29) \\begin{align} \\frac{1}{sin^2x} = csc^2x \\end{align} (30) $$csc^2x = csc^2x$$ # Odd and Even Trigonometric Identities. The odd and even trigonometric identities are more difficult to prove. They are: (31) \\begin{align} -sin(x) = sin(-x) \\quad cos(x) = cos(-x) \\quad -tan(x) = tan(-x) \\end{align} ## Odd Functions First let's acknowledge what an odd and an even function is. An odd function is a function that satisfies the condition such that: (32) $$f(-x) = -f(x)$$ Additionally, functions that are odd can be rotated 180 degrees about the origin to obtain the same graph (look at the graph of cosx for reference). So let's", "\\sin^2 B + \\sin^2 C) &=6 ,\\\\ \\sin^2 A + \\sin^2 B + \\sin^2 C &=2 ,\\\\ 4R^2\\sin^2 A + 4R^2\\sin^2 B + 4R^2\\sin^2 C &=2\\cdot4R^2 ,\\\\ a^2+b^2+c^2&=8R^2 ,\\\\ 2\\rho^2-8r\\,R-2r^2&=8R^2 ,\\\\ \\rho^2-(r+2R)^2&=0 ,\\\\ \\frac{\\rho^2-(r+2R)^2}{4R^2} =0 . \\end{align} And since \\begin{align} \\cos A \\cos B \\cos C &=\\frac{\\rho^2-(r+2R)^2}{4R^2} , \\end{align} one of $\\cos A,\\cos B,\\cos C$ must be 0, hence, one of the angles must be $\\tfrac\\pi2$. COMMENT.- First note that if $C=90^{\\circ}$ then you have $\\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\\circ}$ you can consider that $\\sin C=\\sin (180^{\\circ}-(A+B))=\\sin(A+B)$ and $\\cos C= \\cos (180^{\\circ}-(A+B))=-\\cos(A+B)$ so you get $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B)=2(\\cos^2A+\\cos ^2 B+\\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish. EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\\sin^2 A+\\sin^2 B+\\sin^2 (A+B) = 2\\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\\dfrac{\\pi}{2}$ (the \"if\" is obvious and we need the \"only if\"). So put $A + B = \\dfrac{\\pi}{2}\\pm h$ with $h\\ne 0$ and look at the functions $$\\sin^2(x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h-x)+\\sin^2(\\dfrac{\\pi}{2}\\pm h),\\quad0\\lt x\\lt\\dfrac{\\pi}{2}$$ which becomes for $h$ positive and $h$", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "problem, the coefficients of $x$ can be dealt with easily: $$\\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} = \\frac{ \\cos (3x) - 1 }{(3x)^2}\\frac{2x}{\\sin(2x)}\\frac{3x}{\\tan(3x)} \\frac{3^2}{2\\cdot 3}$$ where the last (numeric) fraction has been inserted to keep the expression the same. Now compute the limit of each fraction and you're done. \\begin{align} \\lim_{x\\to0} \\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} &= \\lim_{x\\to0} \\frac{ -2\\sin^2 (\\dfrac{3x}{2}) \\cos (3x) }{ \\sin (2x) \\sin (3x)} \\\\ &= -2\\lim_{x\\to0} \\dfrac{\\sin^2 (\\dfrac{3x}{2}) }{(\\dfrac{3x}{2})^2} \\dfrac{(2x)}{\\sin (2x)} \\frac{ (3x)}{ ( \\sin (3x)}.\\dfrac{(\\dfrac{3x}{2})^2}{(2x)(3x)}\\cos (3x) \\\\ &=\\color{blue}{-\\dfrac{3}{4}} \\end{align} $$\\lim_{x\\rightarrow0}\\frac{\\cos 3x - 1}{ \\sin 2x \\tan 3x }=-\\lim_{x\\rightarrow0}\\frac{2\\sin^2\\frac{3x}{2}}{\\sin2x\\sin3x}=$$ $$=-\\frac{3}{4}\\lim_{x\\rightarrow0}\\frac{\\frac{\\sin^2\\frac{3x}{2}}{\\frac{9x^2}{4}}}{\\frac{\\sin2x}{2x}\\cdot\\frac{\\sin3x}{3x}}=-\\frac{3}{4}$$ Just another way using the standard Taylor expansions. $$\\cos(3x)=1-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)=2 x-\\frac{4 x^3}{3}+O\\left(x^5\\right)$$ $$\\tan(3x)=3 x+9 x^3+O\\left(x^5\\right)$$ makes $$\\cos(3x)-1=-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)\\tan(3x)=6 x^2+14 x^4+O\\left(x^6\\right)$$ Now, long division leads to $$\\frac{\\cos (3x) - 1}{ \\sin (2x) \\tan (3x) }=-\\frac{3}{4}+\\frac{37 x^2}{16}+O\\left(x^3\\right)$$ showing the limit and how it is appraoched. Just for illustration, use $x=\\frac \\pi {12}$ (quite far away from $0$); the exact value of the expression is $\\sqrt{2}-2\\approx -0.585786$ while the above approximation would give $\\frac{37 \\pi ^2}{2304}-\\frac{3}{4}\\approx -0.591504$. The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $\\frac{\\cos(3x)+1}{\\cos(3x)+1}$, we get $$f(x) = \\frac{\\cos^{2}(3x) - 1}{\\sin(2x) \\tan(3x)(\\cos(3x)+1)} =\\frac{- \\sin(3x) \\cos(3x)}{\\sin(2x) (\\cos(3x)+1)} = \\frac{- (2x)", "and the chain rule. 1 $f(x) = 2x sin(x)$ Solution $$f'(x) = 2 sin(x) + 2x cos(x)$$ 2 $f(x) = \\frac{3x^2}{cos(x)}$ Solution $$f'(x) = \\frac{6x cos(x) + 3x^2 sin(x)}{cos^2(x)}$$ 3 $f(x) = sin^2(x)$ Solution $$f'(x) = 2 sin(x) cos(x)$$ 4 $f(x) = tan^4(x)$ Solution Remember that $tan^4(x)$ means $[tan(x)]^4$. \\begin{align} f'(x) &= 4[tan^3(x)]^3 sec^2(x) \\\\ &= 4 tan^3(x) sec^2(x) \\end{align} 5 $f(x) = \\frac{cos^2(x)}{tan(x)}$ Solution $$f' = \\frac{-2tan(x)cos(x)sin(x) - cos^2(x)sec^2(x)}{tan^2(x)}$$ 6 $f(x) = sec(x)$ Solution \\begin{align} f'(x) &= \\frac{cos(x)(0) + sin(x)}{cos^2(x)} \\\\ \\\\ &= \\frac{1}{cos(x)}\\cdot\\frac{sin(x)}{cos(x)} \\\\ \\\\ &= sec(x) tan(x) \\end{align} 7 $f(x) = csc(x)$ Solution \\begin{align} f'(x) &= \\frac{sin(x)(0) + cos(x)}{sin^2(x)} \\\\ \\\\ &= -\\frac{1}{sin(x)}\\cdot\\frac{cos(x)}{sin(x)} \\\\ \\\\ &= -csc(x) cot(x) \\end{align} 8 $f(x) = \\sqrt{x} cos(x)$ Solution \\begin{align} f'(x) &= \\frac{1}{2} x^{-1/2} cos(x) - \\sqrt{x} sin(x)\\\\ \\\\ &= \\frac{cos(x)}{2 \\sqrt{x}} \\end{align} 9 $f(x) = sin(2x^2 - x^4)$ Solution \\begin{align} f'(x) &= cos(2x^2 - x^4)(4x - 4x^3)\\\\ &= 4x (1 - x^2) cos(2x^2 - x^4) \\end{align} 10 $f(x) = cos(sin(x))$ Solution $$f'(x) = -sin(sin(x))cos(x)$$ xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine,", "# How to integrate this : $\\int \\frac{x^2+\\cos^2 x}{1+x^2}\\csc^2xdx$ How to integrate : $\\int \\frac{x^2+\\cos^2x}{1+x^2}\\csc^2xdx$ I am not getting any clue how to move further, request you to please provide hint, will be of great help. Thanks.. • but that was $cosec^2x$ not cosecx dx\\$ – sultan Apr 17 '17 at 11:29 \\begin{align} \\int \\frac{x^2+\\cos^2x}{1+x^2}\\csc^2xdx&=\\int \\frac{x^2+(1-\\sin^2x)}{1+x^2}\\csc^2xdx\\\\ &=\\int \\frac{(x^2+1)-\\sin^2x)}{1+x^2}\\csc^2xdx\\\\ &=\\int\\bigg[1-\\frac{\\sin^2x}{1+x^2}\\bigg]\\csc^2xdx\\\\ &=\\int\\bigg[\\csc^2x-\\frac{1}{1+x^2}\\bigg]dx\\\\ &=-\\cot x-\\arctan x+C \\end{align} Let \\begin{align} f(x) &= \\frac{x^2+\\cos^2x}{1+x^2} \\csc^2x \\\\ &= \\frac{x^2+ 1- \\sin^2x}{1+x^2} \\cdot \\frac{1}{\\sin^2x}\\\\ &= \\frac{1}{\\sin^2x} - \\frac{1}{1+x^2}. \\end{align} These two terms can be integrated easily. Before integrating, let's do some changes $$\\frac{(x^2+\\cos^2 x)\\csc^2x}{1+x^2}=\\frac{x^2\\csc^2x+\\cos^2 x\\csc^2x}{1+x^2}=\\frac{x^2\\csc^2x+\\cot^2 x}{1+x^2}=\\frac{x^2\\csc^2x+\\csc^2 x-1}{1+x^2}=\\frac{-1+(1+x^2)\\csc^2x}{1+x^2}=-\\frac{1}{1+x^2}+\\csc^2x.$$ But $$\\int -\\frac{1}{1+x^2}dx=-\\arctan x+C$$ and $$\\int \\csc^2xdx=-\\cot x+C.$$ So $$\\int \\frac{(x^2+\\cos^2 x)\\csc^2x}{1+x^2}dx=-\\arctan x-\\cot x+C.$$", "part. 23. Mar 20, 2009 ### Cyosis Barring any mistakes from my side you can do it without using any trigonometric identities: \\begin{align*} \\int \\cos^2 x\\,dx &= \\int 1*\\cos^2 x\\,dx \\\\ &= x \\x cos^2 x + \\int 2x \\cos x \\sin x\\,dx \\\\ &= x \\cos^2 x+(x \\sin^2 x- \\int \\sin^2 x\\,dx) \\\\ &= x \\cos^2x + x \\sin^2 x - (-\\sin x \\cos x- \\int -\\cos^2 x\\,dx) \\\\ &= x (\\cos^2x + \\sin^2 x) +\\sin x \\cos x- \\int \\cos^2 x\\,dx \\end{align*} It follows that: $$2\\int \\cos^2 x\\,dx=\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x \\Rightarrow$$ \\begin{align*} \\int \\cos^2 x\\,dx &= \\frac{1}{2} (\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x) \\\\ &= \\frac{1}{2} (\\x x +\\frac{1}{2} \\sin {2x}) \\\\ &= \\frac{x}{2}+\\frac{1}{4} \\sin {2x} \\end{align} This is the same result you get by using the double angle identity from the start, as it should be. Last edited: Mar 20, 2009 24. Mar 20, 2009 ### itchy8me brilliant! using the coefficient 1. damn thats a sneaky one. thanks guys and girls ;) !! :-D 25. Jan 5, 2010 ### ntc2415 Just use substitution. This makes it all so much easier. First you set u to be cos x$$\\int(cos x)$$2 dx then substitute cos x with u x$$\\int(u)$$2 dx Then you take the derivative of both sides", "x}{\\sin x+\\cos x}dx=\\int\\frac{1}{u+1}du=\\ln(|u+1|)=\\ln(|\\tan x+1|).$$ Since $$\\tan x=\\frac{\\sin x}{\\cos x}$$, we find that this is simply equal to $$\\ln\\left(\\frac{\\sin x+\\cos x}{\\cos x}\\right)=\\ln\\left(|\\sin x+\\cos x|\\right)-\\ln\\left(|\\cos x|\\right).$$ • (+1) Nice! Your approach is better than mine. – José Carlos Santos Jul 7 at 19:28 • Is this missing some absolute value operators? $u+1 = \\tan x + 1$ can be negative. – aschepler Jul 8 at 4:19 • Thanks—I’ve fixed it! – boink Jul 8 at 13:48 I hope that you don't mind if I don't use a substitution. Note that\\begin{align}\\int\\frac{\\sec(x)}{\\sin(x)+\\cos(x)}\\,\\mathrm dx&=\\int\\frac1{\\sin(x)\\cos(x)+\\cos^2(x)}\\,\\mathrm dx\\\\&=\\int\\frac{\\cos(x)-\\sin(x)}{\\cos(x)+\\sin(x)}+\\frac{\\sin(x)}{\\cos(x)}\\,\\mathrm dx\\\\&=\\log\\bigl(\\lvert\\cos(x)+\\sin(x)\\rvert\\bigr)-\\log\\bigl(\\lvert\\cos(x)\\rvert\\bigr).\\end{align} You can observe that $$\\frac{\\sec x}{\\sin x+\\cos x}=\\frac{1}{\\cos x(\\sin x+\\cos x)}= \\frac{\\sin^2x+\\cos^2x}{\\cos x(\\sin x+\\cos x)}= \\frac{\\tan^2x+1}{\\tan x+1}$$ A rational function in the tangent can be integrated via $$u=\\tan x$$ so $$du=(1+\\tan^2x)\\,dx$$ or $$dx=\\frac{1}{u^2+1}\\,du$$ and the integral is now elementary.", "# Various ways to calculate $\\int \\sin(x) \\cos(x) \\, \\mathrm{d}x$ Consider the integral • Substitution with $$u = \\sin(x)$$ gives $$\\II = \\frac{\\sin^2(x)}{2} + C$$ • Substitution with $$u = \\cos(x)$$ gives $$\\II = - \\frac{\\cos^2(x)}{2} + C$$ • (Noted in comments by Koro) Make the substitution $$u = \\sec(x) \\implies \\d u = \\sec(x) \\tan(x) \\, \\d x= \\frac{\\sin(x)}{\\cos^2(x)} \\, \\d x$$ Then \\begin{align*} \\II &= \\int \\sin(x) \\cos(x) \\frac{\\cos^2(x)}{\\sin(x)} \\, \\d u\\\\ &= \\int u^{-3} \\, \\d u\\\\ &= - \\frac{1}{2\\sec^2(x)} + C\\\\ &= - \\frac{\\cos^2(x)}{2} + C \\end{align*} A similarly motivated substitution: $$u = \\csc(x) \\implies \\d u = -\\cot(x) \\csc(x) \\, \\d x = - \\frac{\\cos(x)}{\\sin^2(x)} \\, \\d x$$ yields \\begin{align*} \\II &= -\\int \\sin(x) \\cos(x) \\frac{\\sin^2(x)}{\\cos(x)} \\, \\d u\\\\ &= -\\int u^{-3} \\, \\d u\\\\ &= \\frac{1}{2\\csc^2(x)} + C\\\\ &= \\frac{\\sin^2(x)}{2} + C \\end{align*} • Using $$\\sin(2x) = 2 \\sin(x) \\cos(x)$$ readily leads to $$\\II = -\\frac{\\cos(2x)}{4} + C$$ • Integration by parts differentiating $$\\sin(x)$$ yields $$\\II = \\sin^2(x) - \\II$$ which will yield a previous solution on solving for $$\\II$$. • Integration by parts differentiating $$\\cos(x)$$ yields $$\\II = -\\cos^2(x)- \\II$$ which, similarly, yields a previous solution once we solve for $$\\II$$. • Using the Weierstrass substitution $$t = \\tan(x/2)$$ gives $$\\II = \\int \\frac{2t}{1+t^2} \\frac{1-t^2}{1+t^2} \\frac{2}{1+t^2} \\, \\mathrm{d}t = \\frac{2t^2}{(1+t^2)^2} +", "\\: dx \\\\ \\end{align} We can now evaluate this integral with rather tedious methods we've learned before. In fact, we're going to use the following reduction formula: (2) \\begin{align} \\quad \\int cos^n x \\: dx = \\frac{1}{n} \\cos ^{n - 1} x \\sin x + \\frac{n - 1}{n} \\int \\cos ^{n-2} x \\: dx \\end{align} We thus obtain: (3) \\begin{align} \\quad \\quad \\int \\sin ^4 x \\cos ^4x \\: dx = \\frac{1}{2} \\cos x \\sin x + \\frac{x}{2} - 2 [\\frac{1}{4} \\cos ^{3} x \\sin x + \\frac{3}{4} \\int \\cos ^{2} x \\: dx] + \\frac{1}{6} \\cos ^{5} x \\sin x + \\frac{5}{6} \\int \\cos ^{4} x \\: dx \\\\ \\quad \\quad \\int \\sin ^4 x \\cos ^4x \\: dx = \\frac{1}{2} \\cos x \\sin x + \\frac{x}{2} - \\frac{1}{2} \\cos ^{3} x \\sin x - \\frac{3}{2} \\int \\cos ^{2} x \\: dx + \\frac{1}{6} \\cos ^{5} x \\sin x + \\frac{5}{6} \\int \\cos ^{4} x \\: dx \\\\ \\quad \\quad \\quad \\int \\sin ^4 x \\cos ^4x \\: dx = \\frac{1}{2} \\cos x \\sin x + \\frac{x}{2} - \\frac{1}{2} \\cos ^{3} x \\sin x - \\frac{3}{2} [ \\frac{1}{2} \\cos x \\sin x + \\frac{x}{2} ] + \\frac{1}{6} \\cos ^{5} x \\sin x + \\frac{5}{6} [ \\frac{1}{4} \\cos ^{3} x \\sin x + \\frac{3}{4} \\int \\cos ^{2} x", "x)$$ \\begin{align}{5(\\cos^2x-\\sin^2x)\\over \\sin x+\\cos x} & = \\frac{5(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\sin x + \\cos x} \\\\ \\\\ & = \\require{cancel}\\frac{5(\\cos x - \\sin x)(\\cancel{\\cos x + \\sin x})}{\\cancel{\\cos x+ \\sin x}}\\\\ \\\\ & = 5(\\cos x-\\sin x)\\end{align} $a^2-b^2$= $(a+b)(a-b)$, hence $\\cos^{2}(x)-\\sin^{2}(x)$ = $(\\cos x+\\sin x)(\\cos x-\\sin x)$.", "x)$$ \\begin{align}{5(\\cos^2x-\\sin^2x)\\over \\sin x+\\cos x} & = \\frac{5(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\sin x + \\cos x} \\\\ \\\\ & = \\require{cancel}\\frac{5(\\cos x - \\sin x)(\\cancel{\\cos x + \\sin x})}{\\cancel{\\cos x+ \\sin x}}\\\\ \\\\ & = 5(\\cos x-\\sin x)\\end{align} $a^2-b^2$= $(a+b)(a-b)$, hence $\\cos^{2}(x)-\\sin^{2}(x)$ = $(\\cos x+\\sin x)(\\cos x-\\sin x)$.", "the Type $\\int\\sin^mx\\cos^nxdx$ Where $m,n$ Are Positive Integers Case 1. One of the integer powers, say $m$, is odd. $m=2k+1$ for some integer $k$. So, \\begin{align*} \\sin^mx&=\\sin^{2k+1}x\\\\ &=(\\sin^2x)^k\\sin x\\\\ &=(1-\\cos^2x)^k\\sin x. \\end{align*} Use the substitution $u=\\cos x$ in this case. Example. Evaluate $\\int\\sin^3x\\cos^2xdx$. Solution. \\begin{align*} \\int\\sin^3x\\cos^2xdx&=\\int \\sin^2x\\sin x\\cos^2xdx\\\\ &=\\int(1-\\cos^2x)\\cos^2x\\sin xdx\\\\ &=-\\int(1-u^2)u^2du\\ (\\mbox{substition}\\ u=\\cos x)\\\\ &=\\frac{u^5}{4}-\\frac{u^3}{3}+C\\\\ &=\\frac{\\cos^5x}{5}-\\frac{\\cos^3x}{3}+C, \\end{align*} where $C$ is a constant. Example. Evaluate $\\int\\cos^3xdx$. Solution. \\begin{align*} \\int\\cos^3xdx&=\\int\\cos^2x\\cos xdx\\\\ &=\\int(1-\\sin^2x)\\cos xdx\\\\ &=\\int(1-u^2)du\\ (\\mbox{substitution}\\ u=\\sin x)\\\\ &=u-\\frac{u^3}{3}+C\\\\ &=\\sin x-\\frac{\\sin^3x}{3}+C, \\end{align*} where $C$ is a constant. Case 2. If both $m$ and $n$ are even. In this case, use the trigonometric identities $$\\sin^2x=\\frac{1-\\cos 2x}{2},\\ \\cos^2x=\\frac{1+\\cos 2x}{2}.$$ Example. Evaluate $\\int\\sin^2x\\cos^4xdx$. $$\\frac{1}{16}\\left(x-\\frac{1}{4}\\sin 4x+\\frac{1}{3}\\sin^32x\\right)+C,$$ where $C$ is a constant. Integrals of Powers of $\\tan x$ and $\\sec x$ This type of integrals can be mostly done by using the trigonometric identity $$1+\\tan^2x=\\sec^2x.$$ Example. Evaluate $\\int\\tan^4xdx$. Solution. \\begin{align*} \\int\\tan^4xdx&=\\int\\tan^2x\\tan^2xdx\\\\ &=\\int\\tan^2x(\\sec^2x-1)dx\\\\ &=\\int\\tan^2x\\sec^2xdx-\\int\\tan^2xdx\\\\ &=\\int u^2du-\\int(\\sec^2x-1)dx\\ (\\mbox{substition}\\ u=\\tan x)\\\\ &=\\frac{\\tan^3x}{3}-\\tan x+x+C, \\end{align*} where $C$ is a constant. Example. Evaluate $\\int\\sec^3xdx$. Solution. \\begin{align*} \\int\\sec^3xdx&=\\int\\sec x\\sec^2xdx\\\\ &=\\sec x\\tan x-\\int\\tan^2x\\sec xdx\\ (\\mbox{integration by parts})\\\\ &=\\sec x\\tan x-\\int(\\sec^2x-1)\\sec xdx\\\\ &=\\sec x\\tan x-\\int\\sec^3xdx+\\int\\sec xdx+C’\\\\ &=\\sec x\\tan x-\\int\\sec^3xdx+\\ln|\\sec x+\\tan x|+C’, \\end{align*} where $C’$ is a constant. Hence, $$\\int\\sec^3xdx=\\frac{1}{2}\\sec x\\tan x+\\frac{1}{2}\\ln|\\sec x+\\tan x|+C,$$ where $C=\\frac{C’}{2}$. Products of Sines and Cosines This type of integrals include $\\int\\sin mx\\cos nxdx$, $\\int\\sin mx\\cos nxdx$, and $\\int\\cos mx\\cos", "\\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(2 \\pi (f_\\text{s}/2) t) + A_x \\sin(2 \\pi (f_\\text{s}/2) t) \\right) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} \\Re\\{X[k]\\} \\cos(2 \\pi (k/N) f_\\text{s} t) - \\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ \\end{align} where $A_x$ can be any number, because that term will be zero at the sampling instances, when $t = nT = n/f_\\text{s}$. looking at it from a magnitude/phase POV, for $N$ odd $$x(t) = \\frac{X[0]}{N} + \\frac2N \\sum\\limits_{k=1}^{\\frac{N-1}{2}} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\})$$ and for $N$ even \\begin{align} x(t) & = \\frac{X[0]}{N} + \\frac{X[N/2]}{N \\cos(\\theta)} \\cos(2 \\pi (f_\\text{s}/2) t + \\theta) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\}) \\end{align} where $$\\Re\\{X[k]\\} = |X[k]| \\cos(\\arg\\{X[k]\\})$$ $$\\Im\\{X[k]\\} = |X[k]| \\sin(\\arg\\{X[k]\\})$$ or $$|X[k]| = \\sqrt{\\Re\\{X[k]\\}^2 + \\Im\\{X[k]\\}^2}$$ $$\\arg\\{X[k]\\} = \\operatorname{atan2}(\\Im\\{X[k]\\}, \\Re\\{X[k]\\})$$ and $\\theta$ is the phase of the Nyquist component and can be any real value except $\\pm \\frac{\\pi}{2}$ or $\\frac{\\pi}{2}+ m \\pi$ for integer $m$ (because you cannot divide by zero). the samples $x[n] \\triangleq x(n/f_\\text{s})$ will come out the same regardless of the phase of the Nyquist component as long as the amplitude of the component is adjusted accordingly by dividing by $\\cos(\\theta)$. • For real-valued sequences with $N$ even, you know both", "## Precalculus (6th Edition) Blitzer Consider the identity: $\\csc x-\\cos x\\cot x=\\sin x$ Solve the left side of the above identity, \\begin{align} & \\csc x-\\cos x\\cot x=\\frac{1}{\\sin x}-\\cos x\\frac{\\cos x}{\\sin x} \\\\ & =\\frac{1}{\\sin x}-\\frac{{{\\cos }^{2}}x}{\\sin x} \\\\ & =\\frac{1-{{\\cos }^{2}}x}{\\sin x} \\end{align} Now, use the identity $1-{{\\cos }^{2}}x={{\\sin }^{2}}x$ and solve further, \\begin{align} & \\csc x-\\cos x\\cot x=\\frac{{{\\sin }^{2}}x}{\\sin x} \\\\ & =\\sin x \\end{align} Now, the right side of the identity $\\csc x-\\cos x\\cot x=\\sin x$ is also $\\sin x$ Thus, the left side and the right side of the identity $\\csc x-\\cos x\\cot x=\\sin x$ are equal. Hence, the identity $\\csc x-\\cos x\\cot x=\\sin x$ is verified.", "set $(2u)=s$, therefore: $\\frac{ds}{du}=2$ $$ds=2 \\ du \\Rightarrow du=\\frac{1}{2} \\ ds$$ therefore the underbraced portion can be written as: \\begin{align} \\int \\cos(2u) \\ du &=\\int \\cos(s) \\times \\frac{1}{2} \\ ds \\\\ &=\\frac{1}{2} \\int \\cos(s) \\ ds \\\\ &=\\frac{1}{2} \\sin(s)+c \\ \\ \\text{back-sub the substitution}\\\\ &=\\frac{1}{2} \\sin(2u)+c \\end{align} we can now do our integration by parts: $$=L \\int 1 \\ du + L \\int \\cos(2u) \\ du$$ $$L(u+c)\\Biggr|_{\\frac{\\pi}{2}}^{h'}+L\\left[\\frac{1}{2}\\sin(2u)+c\\right]_{\\frac{\\pi}{2}}^{h'}$$ *the $c$'s are the constants of integration and will cancel out when the integral is evaluated. $$=L\\left[u+\\frac{1}{2}\\sin(2u)\\right]_{\\frac{\\pi}{2}}^{h'}$$ *A double angle identity for $\\sin(2u)$: \\begin{align} \\sin(2u) &=2\\sin(u)\\cos(u)\\\\ &=2\\sin(u)(1-\\sin^2(u))^{\\frac{1}{2}} \\end{align} therefore $$=L\\left[u+\\frac{1}{2} \\times 2\\sin(u)(1-\\sin^2(u))^{\\frac{1}{2}}\\right]_{\\frac{\\pi}{2}}^{h'}$$ *Recall $y=\\sin(u) \\Rightarrow u=\\arcsin(y)$ \\begin{align} &=L\\left[\\arcsin(y)+y \\times (1-y^2)^{\\frac{1}{2}}\\right]_{-1}^{h} \\\\ \\\\ &=L\\left[\\left(\\arcsin(h)+h \\times (1-h^2)^{\\frac{1}{2}}\\right)-\\left(\\arcsin(-1)+(-1)(1-(-1)^2)^{\\frac{1}{2}}\\right)\\right] \\\\ \\\\ &=L\\left[\\frac{\\pi}{2}+\\left(\\arcsin(h)+h(1-h^2)^{\\frac{1}{2}}\\right)\\right] \\end{align} *For any tank with radius $R$ $$V=R^2 \\times L\\left[\\frac{\\pi}{2}+\\left(\\arcsin(h)+h\\sqrt{(1-h^2)}\\right)\\right]$$ Interestingly, I ran into this problem while out on a frac job and we had missplaced our dipstick to our acid transport. We needed to keep tabs on how much acid we were pumping into the well.", "O(x^6)-x^2}{\\left(\\arcsin \\:x\\right)^2-x^2} \\end{align} (The above is true as $\\ln(1+x) \\approx x-x^2/2 +O(x^3)$ when $x$ is small) \\begin{align} \\lim _{x\\to \\:0}\\frac{\\sin \\left(x^2\\right)-\\frac{\\sin^2(x^2)}{2} + O(x^6)-x^2}{\\left(\\arcsin \\:x\\right)^2-x^2} &= \\lim _{x\\to \\:0}\\frac{x^2 -x^4/2 + O(x^{6})-x^2}{\\left(x + \\frac{x^3}{6}+O(x^5)\\right)^2-x^2} \\\\ &=\\lim _{x\\to \\:0}\\frac{\\frac{-x^4}{2}}{\\frac{2x^4}{6}}\\\\ &= -\\frac{3}{2} \\end{align} (I used Taylor expansions) • Unfortunately the answer has errors. For example, $\\sin x^2\\ne \\sin^2 x$. $\\sin x^2=x^2-\\frac16x^6+O(x^{10})$. $\\sin ^2x=(x-\\frac16 x^3+O(x^5))^2=x^2-\\frac13x^4+O(x^6)$. And please explain the justification to arrive at the second to last equality. – Mark Viola Sep 23 '15 at 20:59 • the answer is $-\\dfrac{3}{2}$ so please would you provide more details – Educ Sep 23 '15 at 21:09 • Thanks. I made the changes and edited the answer. Hope it helps! – chandu1729 Sep 23 '15 at 21:15" ]

[ "+0 # help 0 51 3 Find the minimum value of $$\\sin x + \\csc x)^2 + (\\cos x + \\sec x)^2$$ for $$0 < x < \\frac{\\pi}{2}.$$ Aug 6, 2022 #2 0 minimum value is 9 Aug 6, 2022 #3 0 Hello Guest! Remember: $$sin^2 x +cos^2 x = 1 \\\\ csc (x) = \\frac{1}{sin (x)} \\\\ sec(x) = \\frac{1}{cos (x)}$$ So: $$(sin(x)+csc(x))^2+(cos(x)+sec(x))^2 \\\\ =sin^2(x)+\\frac{1}{sin^2(x)}+2sin(x)\\frac{1}{sin(x)}+cos^2(x)+\\frac{1}{sin^2(x)}+2cos(x)\\frac{1}{cos(x)} \\\\ =(sin^2(x)+cos^2(x))+2+2+(\\frac{1}{sin^2(x)}+\\frac{1}{cos^2(x)}) \\\\ =1+4+\\frac{sin^2(x)+cos^2(x)}{sin^2(x)cos^2(x)} =5+\\frac{1}{(sin(x)cos(x))^2}$$ Now, we want the minimum value, so we have to minimize: $$\\frac{1}{(sin(x)cos(x))^2}$$ or equivalently, maximize: $$(sin(x)cos(x))^2$$ Recall: $$sin(2x)=2sin(x)cos(x) \\implies sin(x)cos(x)=\\frac{1}{2}sin(2x)$$ Hence, we want to maximize $$\\frac{1}{4}sin^2(2x)$$ $$0 < sin^2(2x) \\le 1$$ So, the upper bound is 1, which is satisfied when $$x=\\frac{\\pi}{4}$$ So, the maximum value of $$\\frac{1}{4}sin^2(2x)$$ is: $$\\dfrac{1}{4}*(1)=\\dfrac{1}{4}$$ Therefore, the minimum value of $$\\frac{1}{(sin(x)cos(x))^2}$$ is $$\\frac{1}{\\frac{1}{4}}=4$$ Thus, the minimum value of the whole expression in the question is: $$5+4=9$$ , as guest has found below. I hope this helps! Aug 7, 2022", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and", "## welshfella one year ago Need help with this one:- Find limit as x --> 0 of 2csc^2 x - (1/2) csc^2 (1/2) x. 1. welshfella both terms tend to infinity as x approaches zero so i converted them to the form f(x)/g(x) in order to apply l'hopitals but and got it to the form (0/0). Thats as far as i got. 2. ganeshie8 i think we can try to simplify the expression using trig identities 3. ganeshie8 \\begin{align} 2\\csc^2 x - \\frac{1}{2} \\csc^2 \\frac{x}{2}&=\\frac{1}{2}\\left(\\dfrac{4}{\\sin^2x}-\\frac{1}{\\sin^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{4}{4\\sin^2\\frac{x}{2}\\cos^2\\frac{x}{2}}-\\frac{1}{\\sin^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{1-\\cos^2\\frac{x}{2}}{\\sin^2\\frac{x}{2}\\cos^2\\frac{x}{2}}\\right)\\\\~\\\\ &=\\frac{1}{2}\\left(\\dfrac{1}{\\cos^2\\frac{x}{2}}\\right)\\\\~\\\\ \\end{align} we can take the limit now 4. welshfella yea 5. welshfella 1/2 6. thomas5267 Not really awake. Use with caution. \\begin{align*} &\\phantom{=}2\\csc^2(x)-\\frac{1}{2}\\csc^2\\left(\\frac{1}{2}x\\right)\\\\ &=\\frac{2}{\\sin^2(x)}-\\frac{1}{2\\sin^2\\left(\\frac{1}{2}x\\right)}\\\\ &=\\frac{2}{1-\\cos^2(x)}-\\frac{1}{1-\\cos(x)}\\\\ &=\\frac{2}{1-\\cos^2(x)}-\\frac{1+\\cos(x)}{1-\\cos^2(x)}\\\\ &=\\frac{1-\\cos(x)}{1-\\cos^2(x)}\\\\ &=\\frac{1}{1+\\cos(x)} \\end{align*} 7. anonymous Another approach using the same identity: \\begin{align*} 2\\csc^2x-\\frac{1}{2}\\csc^2\\frac{x}{2}&=\\frac{4}{1-\\cos2x}-\\frac{1}{1-\\cos x}\\\\[1ex] &=\\frac{3-4\\cos x+\\cos2x}{(1-\\cos2x)(1-\\cos x)}\\\\[1ex] &=\\frac{2(1-\\cos x)^2}{(2-2\\cos^2x)(1-\\cos x)} \\end{align*} which gives the last line as in thomas's answer. 8. ganeshie8 Ahh that identity looks much better as it avoids the half angles/fractions 9. welshfella i didn't know that identity on the second line @ganeshie 10. welshfella that half angle one 11. welshfella I searched the web to find a half angle identity for sin^ x but couldnt find one 12. ganeshie8 ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them 13. thomas5267", "# Determine the Minimum Value #### anemone ##### MHB POTW Director Staff member Determine the minimum value of $$\\displaystyle \\frac{\\sec^4 a}{\\tan^2 b}+\\frac{\\sec^4 b}{\\tan^2 a}$$ over all $a,b \\ne \\frac{k \\pi}{2}$ where $k$ is in $Z$. #### Ackbach ##### Indicium Physicus Staff member Since $a$ and $b$ are symmetric in the expression, I will set $a=b$, and then define $$f(a)= \\frac{2 \\sec^{4}(a)}{ \\tan^{2}(a)}= \\frac{2}{ \\cos^{2}(a) \\sin^{2}(a)}.$$ Differentiating yields \\begin{align*} f'(a)&= \\frac{-2(-2 \\cos(a) \\sin^{3}(a)+2 \\sin(a) \\cos^{3}(a))}{ \\cos^{4}(a) \\sin^{4}(a)} \\\\ &= \\frac{-4 \\sin(a) \\cos(a) [ \\cos^{2}(a)- \\sin^{2}(a)]}{ \\sin^{4}(a) \\cos^{4}(a)} \\\\ &= \\frac{-4 \\cos(2a)}{ \\sin^{3}(a) \\cos^{3}(a)}. \\end{align*} Setting $f'(a)=0$ implies that $$2a= \\frac{(2j+1) \\pi}{2} \\implies a=\\frac{(2j+1) \\pi}{4}.$$ Just to make sure, let us look for values of $a\\in (0, \\pi) \\setminus \\{\\pi/2 \\}$. This turns out to require $j=0,1$, and therefore $a\\in \\{ \\pi/4, 3 \\pi/4 \\}$. Plugging either of these into $f$ yields the minimum value of $8$. To be complete, we should take the second derivative $f''(a)$, and show that it is positive at these values. I will leave that to the reader. #### Jester ##### Well-known member MHB Math Helper Here's my solution Using the usual identity we have $\\dfrac{(\\tan^2 a + 1)^2}{\\tan^2b}+\\dfrac{(\\tan^2 b + 1)^2}{\\tan^2a}.$ If we let $x = \\tan a$ and $y = \\tan b$ then we have $z = \\dfrac{(x^2 + 1)^2}{y^2}+\\dfrac{(y^2 + 1)^2}{x}.$ Using", "don't. A valid calculation of the limit would for example be: \\begin{align} &\\lim_{x\\to 0}{\\frac{\\tan{x}-\\sin x}{x^3}} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\frac{1-\\cos x}{x^2} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} \\\\=&1\\cdot \\frac 1 2 =\\frac 1 2. \\end{align} Here we can calculate each limit separately and multiply them afterwards, as both exist. (In the first step I just took the\\tan x$factor outside). Hint: Use the Taylor development:$sin(x)=x-x^3/6+O(x^3)$,$tan(x)=x+x^3/3+O(x^3)$. • the order of the truncation error is surely$O(x^5)\\$, isn't it? Mar 9 '18 at 18:47 we have $$\\frac{-\\sin(x){\\cos(x)}+\\sin(x)}{\\cos(x)x^3}=\\frac{\\sin(x)(1-\\cos^2(x))}{x^3\\cos(x)(1+\\cos(x))}=\\frac{\\sin^3(x)}{x^3}\\frac{1}{\\cos(x)(1+\\cos(x))}$$", "# Minimum value of Function I have a function, $f(x) = \\sec x + \\csc x$ on the interval $x \\in (0, \\pi/2)$. The derivative of it is, $f'(x) = \\csc^2(x) \\sec^2(x) \\left(\\sin^3 x + \\cos^3 x\\right)$ Of course, when I tried to solve for $f'(x) = 0$, I realize that $f'(x)=0$ does not exist. However, the question insists that the there is a minimum. Am I doing something wrong here? - I don't mean to give you too much of the answer, but Wolfram Alpha is a really powerful tool when you need more insight about these kinds of problems: wolframalpha.com/input/?i=f%28x%29+%3D+sec+x+%2B+csc+x It appears your derivative is incorrect. – Jonathan Rich Mar 14 '13 at 17:38 Hint: Derivative is: $tan(x) sec(x)-cot(x) csc(x)$, $x = \\pi/4$ – Amzoti Mar 14 '13 at 17:39 I expanded from tan(x)sec(x)−cot(x)csc(x) to (sin x)/(cos^2 x) + (cos x)/(sin^2 x). I am sorry for not formatting properly. – 0xFF Mar 14 '13 at 17:40 You lost a $-$ sign – Robert Israel Mar 14 '13 at 17:42 (sin x)/(cos^2 x) - (cos x)/(sin^2 x) => (sin^3x-cos^3x)/(cos^2 x)(sin^2 x) => (sec^2)(csc^2 x)(sin^3x-cos^3x) Am I wrong with my expansion? – 0xFF Mar 14 '13 at 17:45 We should have \\begin{align}f'(x) &= \\frac{\\sin x}{\\cos^2 x}-\\frac{\\cos x}{\\sin^2x}\\\\ &= \\frac1{\\cos^2x\\sin^2x}\\bigl(\\sin^3x-\\cos^3x\\bigr)\\\\ &= \\frac1{\\cos^2x\\sin^2x}(\\sin x-\\cos x)\\bigl(\\sin^2x+\\sin x\\cos x+\\cos^2x\\bigr)\\\\ &= \\frac1{\\cos^2x\\sin^2x}(\\sin x-\\cos", "+0 trig function 0 36 2 Let f(x) = 2 - cos(x) + sin(x)^2. Find the minimum and maximum values of f(x). Jan 8, 2021 #1 +10829 +1 Let f(x) = 2 - cos(x) + sin(x)^2. Find the minimum and maximum values of f(x). Hello Guest! $$f(x) = 2 - cos(x) + sin^2(x)\\\\ \\frac{df(x)}{dx}= sin(x)+2sin(x)cos(x) =0$$ $$sin(x)(1+2cos(x))=0$$ $$sin(x)=0\\\\ x=arcsin(0)\\\\ \\color{blue}x_1=0\\\\ \\color{blue}x_3=\\pi =180^\\circ\\\\ \\color{blue}x_5=2\\pi=360^\\circ$$ $$1+2cos(x)=0\\\\ cos(x)=-\\frac{1}{2}\\\\ x=arccos(-\\frac{1}{2})$$ $$x_2=\\frac{2}{3}\\pi=120^\\circ\\\\ x_4=\\frac{4}{3}\\pi=240^\\circ$$ ! Jan 8, 2021 edited by asinus Jan 8, 2021 edited by asinus Jan 8, 2021 #2 +172 0 $$\\displaystyle f(x)=2-\\cos(x)+\\sin^{2}(x) \\\\=2-\\cos(x)+1-\\cos^{2}(x) \\\\ = 3-\\{\\cos^{2}(x)+\\cos(x)+1/4-1/4\\} \\\\ = 3-\\{(\\cos(x)+1/2)^{12}-1/4\\} \\\\ =13/4-(\\cos(x)+1/2)^{2}.$$ So, max will be 13/4 when cos(x) = -1/2, and min will be 13/4 - 9/4 = 1, when cos(x) = 1. Jan 8, 2021", "deals with maximum too. – Archisman Panigrahi Sep 25 '15 at 13:17 Let $f(x)=2^{\\sin x}+2^{\\cos x},\\;$ so $f^{\\prime}(x)=(2^{\\sin x}\\cos x-2^{\\cos x}\\sin x)(\\ln 2)$. $\\textbf{A)}$ To find the maximum of $f(x)$, it is enough to consider $f$ on $[0,\\frac{\\pi}{2}]$. $\\;\\;\\;$Then $f(0)=3$ and $f(\\frac{\\pi}{2})=3$, and $\\;\\;\\;f^{\\prime}(x)=0\\iff 2^{\\sin x}\\cos x=2^{\\cos x}\\sin x\\iff\\frac{2^{\\sin x}}{\\sin x}=\\frac{2^{\\cos x}}{\\cos x}\\iff x=\\frac{\\pi}{4}$, $\\;\\;\\;$since if $\\displaystyle g(t)=\\frac{2^t}{t},\\;$ $\\displaystyle g^{\\prime}(t)=\\frac{2^t(t\\ln 2-1)}{t^2}<0$ for $0< t<1$ so $g$ is 1-1 on $(0,1]$. $\\;\\;\\;$Since $f(\\frac{\\pi}{4})=2\\cdot2^{\\sqrt{2}/2}=2^{1+\\frac{\\sqrt{2}}{2}}>3,\\;$ $f(\\frac{\\pi}{4})=2^{1+\\frac{\\sqrt{2}}{2}}$ is the maximum value. $\\textbf{B)}$ Similarly, to find the minimum of $f(x)$ it is enough to consider $f$ on $[\\pi, \\frac{3\\pi}{2}]$. $\\;\\;\\;$Then $f(\\pi)=\\frac{3}{2}$ and $f(\\frac{3\\pi}{2})=\\frac{3}{2}$, and $f^{\\prime}(x)=0\\iff\\frac{2^{\\sin x}}{\\sin x}=\\frac{2^{\\cos x}}{\\cos x}\\iff x=\\frac{5\\pi}{4}$ $\\;\\;\\;$ since $\\displaystyle g(t)=\\frac{2^t}{t}$ is 1-1 on $[-1,0)$ as above. $\\;\\;\\;$Since $f(\\frac{5\\pi}{4})=2\\cdot2^{-\\frac{\\sqrt{2}}{2}}=2^{1-\\frac{\\sqrt{2}}{2}}<\\frac{3}{2},\\;\\;$ $f(\\frac{5\\pi}{4})=2^{1-\\frac{\\sqrt{2}}{2}}$ is the minimum value. • Using the fact that $2^z/z$ is 1-to-1 for $0<|z|\\le 1 which is readily shown by differentiating. – DanielWainfleet Sep 25 '15 at 10:32 Using AM GM inequality $$2^{\\sin x}+2^{\\cos x}\\ge2\\cdot2^{\\frac{\\sin x+\\cos x}2}$$ Now the the minimum value i.e., equality occurs if$2^{\\sin x}=2^{\\cos x}\\iff \\sin x=\\cos x\\implies\\tan x=1\\implies\\sin x=\\cos x=\\pm\\dfrac1{\\sqrt2}$So, the minimum value $$=2^{1-\\frac{1+1}{\\sqrt2}}$$ Set$y=\\pi+x$to find the maximum value • But this should lead to minimum value, right? However this is the maximum value – Archisman Panigrahi Sep 24 '15 at 17:53 • To find the maximum value, you might want to study$y^2$(which attains max/min at", "&= \\int_0^1 ((x-c)^2+c^2)dx\\\\ &=\\left[ \\frac{(x-c)^3}{3} +c^2x \\right]_0^1\\\\ &=\\frac{(1-c)^3}{3} + c^2 + \\frac{c^3}{3}\\\\ &=2c^2-c+\\frac{1}{3}. \\end{align*} 1. What is the minimum value of $I(c)$ (as $c$ varies)? Using the expression for $I(c)$ we calculated above, and completing the square, we have \\begin{align*} I(c) &= 2c^2-c+\\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 - \\frac{1}{8} + \\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 + \\frac{5}{24}. \\end{align*} So the minimum value that $I(c)$ can take is $5/24$. (This occurs when $c=1/4$.) 1. What is the maximum value of $I(\\sin \\theta)$ as $\\theta$ varies? Note that from our work above, the MINIMUM value of $I(\\sin\\theta)$ happens when $\\theta = \\arcsin (0.25)$. Now $c=\\sin \\theta$ varies between between $-1$ and $1$, so we need to find the maximum value of $I$ over this range. We worked out earlier that the overall minimum of $I$ occurs at $c=1/4$, and since $I$ is quadratic, it is symmetric about this point, and increasing for larger and smaller values of $c$. So the maximum value of $I(c)$ over the range $-1 \\le c \\le 1$ occurs at $c=-1$, the furthest point from $x=1/4$ in the range. So the maximum value of $I(\\sin \\theta)$ is $I(-1)=10/3$.", "problem, the coefficients of $x$ can be dealt with easily: $$\\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} = \\frac{ \\cos (3x) - 1 }{(3x)^2}\\frac{2x}{\\sin(2x)}\\frac{3x}{\\tan(3x)} \\frac{3^2}{2\\cdot 3}$$ where the last (numeric) fraction has been inserted to keep the expression the same. Now compute the limit of each fraction and you're done. \\begin{align} \\lim_{x\\to0} \\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} &= \\lim_{x\\to0} \\frac{ -2\\sin^2 (\\dfrac{3x}{2}) \\cos (3x) }{ \\sin (2x) \\sin (3x)} \\\\ &= -2\\lim_{x\\to0} \\dfrac{\\sin^2 (\\dfrac{3x}{2}) }{(\\dfrac{3x}{2})^2} \\dfrac{(2x)}{\\sin (2x)} \\frac{ (3x)}{ ( \\sin (3x)}.\\dfrac{(\\dfrac{3x}{2})^2}{(2x)(3x)}\\cos (3x) \\\\ &=\\color{blue}{-\\dfrac{3}{4}} \\end{align} $$\\lim_{x\\rightarrow0}\\frac{\\cos 3x - 1}{ \\sin 2x \\tan 3x }=-\\lim_{x\\rightarrow0}\\frac{2\\sin^2\\frac{3x}{2}}{\\sin2x\\sin3x}=$$ $$=-\\frac{3}{4}\\lim_{x\\rightarrow0}\\frac{\\frac{\\sin^2\\frac{3x}{2}}{\\frac{9x^2}{4}}}{\\frac{\\sin2x}{2x}\\cdot\\frac{\\sin3x}{3x}}=-\\frac{3}{4}$$ Just another way using the standard Taylor expansions. $$\\cos(3x)=1-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)=2 x-\\frac{4 x^3}{3}+O\\left(x^5\\right)$$ $$\\tan(3x)=3 x+9 x^3+O\\left(x^5\\right)$$ makes $$\\cos(3x)-1=-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)\\tan(3x)=6 x^2+14 x^4+O\\left(x^6\\right)$$ Now, long division leads to $$\\frac{\\cos (3x) - 1}{ \\sin (2x) \\tan (3x) }=-\\frac{3}{4}+\\frac{37 x^2}{16}+O\\left(x^3\\right)$$ showing the limit and how it is appraoched. Just for illustration, use $x=\\frac \\pi {12}$ (quite far away from $0$); the exact value of the expression is $\\sqrt{2}-2\\approx -0.585786$ while the above approximation would give $\\frac{37 \\pi ^2}{2304}-\\frac{3}{4}\\approx -0.591504$. The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $\\frac{\\cos(3x)+1}{\\cos(3x)+1}$, we get $$f(x) = \\frac{\\cos^{2}(3x) - 1}{\\sin(2x) \\tan(3x)(\\cos(3x)+1)} =\\frac{- \\sin(3x) \\cos(3x)}{\\sin(2x) (\\cos(3x)+1)} = \\frac{- (2x)", "that all the terms (in your case $\\sec^6(x)$, $\\csc^6(x)$, $\\sec^6(x)\\csc^6(x)$) have the same value (this is not true for your function). Moreover, the lower bound you find might not take on its minimum value at that same value of $x$, so even then the bound may not be sharp. In general AM-GM will not give you a sharp lower bound on the value of a function. – kccu Apr 4 at 14:38 Others explain why $$48$$ is correct as a a lower bound but may not be the sharp lower bound. One way to get a lower bound of $$80$$ involves using the fact that each term is a cubed quantity. Start with the decomposition $$\\sec^6 x + \\csc^6 x + \\sec^6 x\\csc^6 x=A+B$$ $$A=\\sec^6 x + \\csc^6 x$$ $$B=\\sec^6 x\\csc^6 x$$ Factor $$A$$ as a sum of cubes: $$A=(\\sec^2 x + \\csc^2 x)(\\sec^4 x - \\sec^2 x\\csc^2 x + \\csc^4 x)$$ $$A=\\dfrac{(\\cos^2 x + \\sin^2 x)(\\cos^4 x - \\cos^2 x\\sin^2 x + \\sin^4 x)}{\\cos^6 x\\sin^6 x}$$ Plugging in $$\\cos^2 x +\\sin^2 x =1$$ and $$\\cos^4 x +2 \\cos^2 x\\sin^2 x + \\sin^4 x=(\\cos^2 x +\\sin^2 x)^2=1$$: $$A=\\dfrac{1 - 3\\cos^2 x\\sin^2 x}{\\cos^6 x\\sin^6 x}$$ We have $$(\\cos x-\\sin x)^2=1-2\\cos x\\sin x\\ge 0$$ forcing $$|\\cos x\\sin x|\\le 1/2$$. Thereby $$A\\ge (1 - 3/4)×(64)=16$$ For $$B$$, simply render $$B=\\dfrac{1}{\\cos^6 x\\sin^6", "\\sin x \\cos x+6 \\cos^2 x}.$$ $$f(x)=\\frac{10}{1-\\cos 2x-6 \\sin 2x +3(1+\\cos 2x)}=\\frac{10}{4+2\\cos 2x-6 \\sin 2x}$$ $$\\implies f(x)=\\frac{5}{2+\\cos 2x-3 \\sin 2x}=\\frac{5}{2+\\sqrt{10}(\\cos 2x-3\\sin 2x)/\\sqrt{10}}$$ $$\\implies f(x)=\\frac{5}{2+\\sqrt{10}\\cos(2x+a)}~~~~\\color{red}{(1)}$$ $$\\implies f_{\\min}=\\frac{5}{2+\\sqrt{10}},\\quad f_{\\max}=\\frac{5}{2-\\sqrt{10}}.$$ when $$\\cos (\\cdots)=\\mp 1$$. These are only local maximum.and minimum. But $$f(x)$$ $$\\color{red}{(1)}$$ can take values close to $$\\pm \\infty$$ real value so the Range it's range is $$(-\\infty,\\frac{5}{2-\\sqrt{10}}] \\cup [\\frac{5}{2+\\sqrt{10}}, \\infty )$$. • $$2\\cos2x-3\\sin2x\\le\\sqrt{2^2+3^2}\\cos(2x+\\arccos\\dfrac3{\\sqrt{13}})$$ right? – lab bhattacharjee Jul 13 '20 at 10:02 • What about the negative values ?? – Yves Daoust Jul 13 '20 at 10:07 • Even worse now ! – Yves Daoust Jul 13 '20 at 10:14 this $$f(x)=\\frac{5}{\\sin^2x-6\\sin x\\cos x+3\\cos^2x}$$ can be simplified more neatly, if you use: $$\\sin^2(\\theta)+ \\cos^2(\\theta)=1$$ $$\\cos(2\\theta)=2\\cos^2(\\theta)-1$$ $$\\sin(2\\theta)=2\\sin(\\theta)\\cos(\\theta)$$ And the simplified denominator would be in the form: $$a\\cos(2x)+b\\sin(2x)+c$$ Then you can use $$-\\sqrt{a^2+b^2} \\le a\\cos(x) \\pm b\\sin(x) \\le \\sqrt{a^2+b^2}$$", "and the chain rule. 1 $f(x) = 2x sin(x)$ Solution $$f'(x) = 2 sin(x) + 2x cos(x)$$ 2 $f(x) = \\frac{3x^2}{cos(x)}$ Solution $$f'(x) = \\frac{6x cos(x) + 3x^2 sin(x)}{cos^2(x)}$$ 3 $f(x) = sin^2(x)$ Solution $$f'(x) = 2 sin(x) cos(x)$$ 4 $f(x) = tan^4(x)$ Solution Remember that $tan^4(x)$ means $[tan(x)]^4$. \\begin{align} f'(x) &= 4[tan^3(x)]^3 sec^2(x) \\\\ &= 4 tan^3(x) sec^2(x) \\end{align} 5 $f(x) = \\frac{cos^2(x)}{tan(x)}$ Solution $$f' = \\frac{-2tan(x)cos(x)sin(x) - cos^2(x)sec^2(x)}{tan^2(x)}$$ 6 $f(x) = sec(x)$ Solution \\begin{align} f'(x) &= \\frac{cos(x)(0) + sin(x)}{cos^2(x)} \\\\ \\\\ &= \\frac{1}{cos(x)}\\cdot\\frac{sin(x)}{cos(x)} \\\\ \\\\ &= sec(x) tan(x) \\end{align} 7 $f(x) = csc(x)$ Solution \\begin{align} f'(x) &= \\frac{sin(x)(0) + cos(x)}{sin^2(x)} \\\\ \\\\ &= -\\frac{1}{sin(x)}\\cdot\\frac{cos(x)}{sin(x)} \\\\ \\\\ &= -csc(x) cot(x) \\end{align} 8 $f(x) = \\sqrt{x} cos(x)$ Solution \\begin{align} f'(x) &= \\frac{1}{2} x^{-1/2} cos(x) - \\sqrt{x} sin(x)\\\\ \\\\ &= \\frac{cos(x)}{2 \\sqrt{x}} \\end{align} 9 $f(x) = sin(2x^2 - x^4)$ Solution \\begin{align} f'(x) &= cos(2x^2 - x^4)(4x - 4x^3)\\\\ &= 4x (1 - x^2) cos(2x^2 - x^4) \\end{align} 10 $f(x) = cos(sin(x))$ Solution $$f'(x) = -sin(sin(x))cos(x)$$ xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine,", "## Precalculus (6th Edition) Blitzer Published by Pearson # Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1177: 84 #### Answer See the verification below. #### Work Step by Step Consider the identity: $\\csc x-\\cos x\\cot x=\\sin x$ Solve the left side of the above identity, \\begin{align} & \\csc x-\\cos x\\cot x=\\frac{1}{\\sin x}-\\cos x\\frac{\\cos x}{\\sin x} \\\\ & =\\frac{1}{\\sin x}-\\frac{{{\\cos }^{2}}x}{\\sin x} \\\\ & =\\frac{1-{{\\cos }^{2}}x}{\\sin x} \\end{align} Now, use the identity $1-{{\\cos }^{2}}x={{\\sin }^{2}}x$ and solve further, \\begin{align} & \\csc x-\\cos x\\cot x=\\frac{{{\\sin }^{2}}x}{\\sin x} \\\\ & =\\sin x \\end{align} Now, the right side of the identity $\\csc x-\\cos x\\cot x=\\sin x$ is also $\\sin x$ Thus, the left side and the right side of the identity $\\csc x-\\cos x\\cot x=\\sin x$ are equal. Hence, the identity $\\csc x-\\cos x\\cot x=\\sin x$ is verified. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "Find the maximum and minimum of $\\cos x\\sin y\\cos z$. Given $x\\geq y\\geq z\\geq\\pi/12$, $x+y+z=\\pi/2$, find the maximum and minimum of $\\cos x\\sin y\\cos z$. I tried using turn $\\sin y$ to $\\cos(x+z)$, and Jensen Inequality, but filed. Please help. Thank you. *p.s. I'm seeking for a solution without calculus. - You may rewrite the problem as finding the extrema of $f(x,z)=\\cos(x)\\cos(z)\\cos(x+z)$ with $x, z \\ge \\pi/12$ and $x+z \\le 5\\pi/12$. Note that $$\\cos(x)\\cos(z)\\cos(x+z)=\\frac12 [\\cos(x+z)+\\cos(x-z)] \\cos(x+z)$$ and so \\begin{align*} & [\\cos(5\\pi/12)+\\cos(3\\pi/12)] \\cos(5\\pi/12)\\\\ \\le& [\\cos(5\\pi/12)+\\cos(x-z)] \\cos(5\\pi/12)\\\\ \\le&[\\cos(x+z)+\\cos(x-z)] \\cos(x+z)\\\\ \\le& [\\cos(\\pi/6)+\\cos(x-z)] \\cos(\\pi/6)\\\\ \\le& [\\cos(\\pi/6)+\\cos(0)] \\cos(\\pi/6). \\end{align*} Hence the maximum of $f$ occurs at $x=z=\\pi/12$ and the minima occur at $(x,z)=(4\\pi/12,\\,\\pi/12)$ or $(\\pi/12,\\,4\\pi/12)$.", "# Math Help - Finding the minimum value 1. ## Finding the minimum value Hey there I need some help with this problem Find the minimum value of absolute value of sinx + cosx + tanx + cotx + secx + cscx I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks. 2. ## Re: Finding the minimum value Originally Posted by gfbrd Hey there I need some help with this problem Find the minimum value of absolute value of sinx + cosx + tanx + cotx + secx + cscx I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks. The derivative of: $f(x) = \\sin(x)+ \\cos(x) + \\tan(x) + \\cot(x) + \\sec(x) + \\csc(x)$ is: $\\frac{df}{dx} = \\cos(x)-\\sin(x)+\\tan^{2}(x)-\\cot^{^2}(x)+\\sec(x) \\tan(x)-\\csc(x) \\cot(x)$ The $\\frac{df}{dx}$ above can be written as(for $y > 0$): \\begin{align*}=& \\cos(x) - \\sin(x) + \\frac{\\sin^{2}(x)}{\\cos^{2}(x)} -\\frac{\\cos^{2}(x)}{\\sin^{2}(x)} + \\left(\\frac{\\sin(x)}{\\cos(x)}\\right)\\left(\\frac{1} {\\cos(x)}\\right) - \\left( \\frac{\\cos(x)}{\\sin(x)}\\right)\\left(\\frac{1}{\\sin( x)}\\right) \\\\ \\\\ =& \\cos(x) - \\sin(x) + \\frac{\\sin^{2}(x)}{\\cos^{2}(x)} -\\frac{\\cos^{2}(x)}{\\sin^{2}(x)} + \\frac{\\sin(x)}{\\cos^{2}(x)} - \\frac{\\cos(x)}{\\sin^{2}(x)} \\\\", "# Integrate $\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}dx$ Solve the indefinite integral $$I=\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}\\;dx$$ My Attempt: \\begin{align} I&=\\int\\frac{1}{\\sin x+\\cos x+\\frac{1}{\\sin x \\cos x}+\\frac{\\sin x +\\cos x}{\\sin x\\cos x}}\\;dx\\\\ \\\\ &=\\int\\frac{\\sin x\\cos x}{\\left(\\sin x+\\cos x\\right)\\left(\\sin x\\cos x \\right)+1+\\left(\\sin x+\\cos x\\right)}\\;dx \\end{align} How can I complete the solution from here? • Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14 • Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45 You want \\begin{align} \\int\\frac{dx}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x} &= \\int\\frac{dx}{\\sin x+\\cos x+\\frac{\\sin x}{\\cos x} +\\frac{\\cos x}{\\sin x}+\\frac1{\\sin x}+\\frac1{\\cos x}}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin^2 x \\cos x+\\cos^2 x \\sin x+\\sin^2 x +\\cos^2 x+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin x \\cos x(\\sin x+ \\cos x)+1+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1}\\\\ \\end{align}. Applying substitutions $\\sin x = \\frac{2t}{1+t^2}, \\cos x = \\frac{1-t^2}{1+t^2}, dx = \\frac{2 dt}{1+t^2}$, this becomes \\begin{align} \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1} &=\\int\\frac{\\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\\frac{2t}{1+t^2}+ \\frac{1-t^2}{1+t^2})+1}\\\\ &=\\int\\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\\\ &=\\int\\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \\quad \\text{ (according to Wolfram Alpha)}\\\\ &=\\int\\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\\\ \\end{align}. According to Wolfram Alpha, that quartic can be factored as", "\\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(2 \\pi (f_\\text{s}/2) t) + A_x \\sin(2 \\pi (f_\\text{s}/2) t) \\right) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} \\Re\\{X[k]\\} \\cos(2 \\pi (k/N) f_\\text{s} t) - \\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ \\end{align} where $A_x$ can be any number, because that term will be zero at the sampling instances, when $t = nT = n/f_\\text{s}$. looking at it from a magnitude/phase POV, for $N$ odd $$x(t) = \\frac{X[0]}{N} + \\frac2N \\sum\\limits_{k=1}^{\\frac{N-1}{2}} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\})$$ and for $N$ even \\begin{align} x(t) & = \\frac{X[0]}{N} + \\frac{X[N/2]}{N \\cos(\\theta)} \\cos(2 \\pi (f_\\text{s}/2) t + \\theta) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\}) \\end{align} where $$\\Re\\{X[k]\\} = |X[k]| \\cos(\\arg\\{X[k]\\})$$ $$\\Im\\{X[k]\\} = |X[k]| \\sin(\\arg\\{X[k]\\})$$ or $$|X[k]| = \\sqrt{\\Re\\{X[k]\\}^2 + \\Im\\{X[k]\\}^2}$$ $$\\arg\\{X[k]\\} = \\operatorname{atan2}(\\Im\\{X[k]\\}, \\Re\\{X[k]\\})$$ and $\\theta$ is the phase of the Nyquist component and can be any real value except $\\pm \\frac{\\pi}{2}$ or $\\frac{\\pi}{2}+ m \\pi$ for integer $m$ (because you cannot divide by zero). the samples $x[n] \\triangleq x(n/f_\\text{s})$ will come out the same regardless of the phase of the Nyquist component as long as the amplitude of the component is adjusted accordingly by dividing by $\\cos(\\theta)$. • For real-valued sequences with $N$ even, you know both", "49 views If $f(x) = \\dfrac{\\sqrt{3} \\sin x}{2+\\cos x}$, then the range of $f(x)$ is 1. the interval $[-1 , \\sqrt{3}{/2}]$ 2. the interval $[-\\sqrt{3}{/2}, 1]$ 3. the interval $[-1, 1]$ 4. none of these in Calculus recategorized | 49 views Answer: $\\mathbf C$ $f^2(x) =\\dfrac{3-3\\cos^2x}{\\cos^2x+4\\cos x+4}$ Now, Let t = $\\cos x$ then, $f(t) = \\dfrac{3-3t^2}{t^2+4t+4}$ $f_{max}.f'(t) = \\frac{-6(2t+1)}{t+2} = 0$, when$t = \\frac{-1}{2}$ $\\therefore f(-1) =0, f(\\frac{-1}{2}) = 1, \\;and\\; f(1) = 0$ So, $f^2\\text{max} = 1 \\implies f^2(x) \\leq 1 \\implies -1\\leq f(x) \\leq 1$ So, the range of $f$ is $[-1,1]$ $\\therefore \\mathbf C$ is the right option. by Boss (18.9k points) edited by To find the range of $f(x)$, actually we need to find the maximum and minimum value. Now \\begin{align}f(x)&=\\frac{\\sqrt{3}\\sin x}{2+\\cos x}\\\\ \\Rightarrow f'(x)&=\\sqrt{3}\\cos x \\cdot \\frac{1}{2+\\cos x}+\\sqrt{3}\\sin x \\cdot \\frac{-\\sin x}{(2+\\cos x)^2}\\\\ &=\\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} \\end{align} At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \\begin{align}\\therefore f'(x)&=0\\\\ \\Rightarrow \\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} &=0\\\\ \\Rightarrow 2\\cos x +1&=0;~[\\because \\sin^2 x+\\cos^2 x=1] \\\\ \\Rightarrow \\cos x &= -\\frac{1}{2}\\\\ \\Rightarrow x &= \\frac{2\\pi}{3}, \\frac{4\\pi}{3}\\end{align} Now putting those values of $x$ to $f(x)$, we obtain $f(\\frac{2\\pi}{3})=\\frac{\\sqrt{3}(\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=1$ and $f(\\frac{4\\pi}{3})=\\frac{\\sqrt{3}(-\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$. $$\\therefore f(x) \\in [-1,1]$$ So the correct" ]

[ "x}\\cos(2\\beta x - \\sin x \\cos x)dx = \\frac{\\pi}{\\sqrt{e}}\\frac 1 {\\beta!}\\frac 1 { 2^{\\beta -1}}$$ I'm not sure about the general formula you're proposing. Jul 19, 2021 at 6:44", "\\frac{a^4cos^2(\\beta)+b^4sin^2(\\beta)}{a^2cos^2(\\beta)+b^2sin^2(\\beta)}= \\frac{ (a^2cos(\\beta))^2+(b^2sin(\\beta))^2}{(a\\,cos(\\beta))^2+(b\\,sin(\\beta))^2}$ And, finally, get the formula from wikipedia $R =\\sqrt{\\frac{ (a^2cos(\\beta))^2+(b^2sin(\\beta))^2}{(a\\,cos(\\beta))^2+(b\\,sin(\\beta))^2}}$ The calculator above uses this formula.", "other two sides, we can conclude that $\\frac{a}{{\\sin \\alpha }} = \\frac{b}{{\\sin \\beta }} = \\frac{c}{{\\sin \\gamma }} = 2R.$ See solved problems on Page 2.", "DE:DB = 2.$$ Thus $$\\sin \\beta=\\frac{\\sqrt3}{4},\\cos \\beta=\\frac{\\sqrt{13}}4.\\tag{1}$$ On the other hand, ...", "\\beta^2} = \\frac{4}{5}$ for $\\beta$, $\\beta = \\frac{1}{2}$. [2 minutes total ]", "we can take $\\gamma= 0$ so $cos^2(\\alpha)+ cos^2(\\beta)= 1$ which means that $cos(\\beta)= sin(\\alpha)$.)", "# Let $\\cos (\\alpha + \\beta) = \\frac{4}{5}$ and let $\\sin (\\alpha - \\beta) = \\frac{5}{13}$, where $0 \\leq \\alpha, \\; \\beta \\leq \\frac{\\pi}{4}$, then $\\tan 2 \\alpha =$ ( A ) $\\frac{56}{33}$ ( B ) $\\frac{25}{16}$ ( C ) $\\frac{20}{7}$ ( D ) $\\frac{19}{12}$", "with $$\\sin(\\alpha)=\\frac{a}{2r},\\sin(\\beta)=\\frac{b}{2r}$$ we get $$A=\\frac{1}{2}4r^2\\sin(\\gamma)=2r^2\\sin(\\gamma)\\le 2r^2$$", "Browse Questions # If $\\alpha +\\beta+\\gamma= 2\\theta$, then $cos \\theta+ \\cos (\\theta- \\alpha)+ \\cos (\\theta- \\beta)+ \\cos (\\theta- \\gamma)=$ $\\begin {array} {1 1} (1)\\;4 \\sin \\frac{\\alpha}{2}. \\cos \\frac{\\beta}{2}. \\sin \\frac{\\gamma}{2} & \\quad (2)\\;4 \\cos \\frac{\\alpha}{2}. \\cos \\frac{\\beta}{2}. \\cos \\frac{\\gamma}{2} \\\\ (3)\\;4 \\sin \\frac{\\alpha}{2}. \\sin \\frac{\\beta}{2}. \\sin \\frac{\\gamma}{2} & \\quad (4)\\;4 \\sin \\alpha.\\sin \\beta . \\sin \\gamma \\end {array}$ $(2)\\;4 \\cos \\frac{\\alpha}{2}. \\cos \\frac{\\beta}{2}. \\cos \\frac{\\gamma}{2}$", "# If $\\cos\\;\\alpha + \\cos\\;\\beta + \\cos\\;\\gamma = 0 = \\sin \\;\\alpha + \\sin \\;\\beta + \\sin \\;\\gamma$, prove that $\\sin 2\\alpha + \\sin 2\\beta + \\sin 2\\gamma = 0$ (Hints: Take $a = \\textit{$\\cos$} \\;\\alpha, \\;b = \\textit{$\\cos$} \\;\\beta, c = \\textit{$\\cos$} \\; \\gamma$, $a+b+c = 0 => a^{3} + b^{3}+ c^{3} = 3abc$ $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} = 0 => a^{2}+b^{2}+c^{2}=0$ This is the fourth part of the multi-part Q3.", "\\omega^2 C_3\\cos{(\\alpha t)} + \\omega ^2 C_4\\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $(\\omega^2 C_3 - \\alpha ^2 C_3)\\cos{(\\alpha t)} + (\\omega ^2 C_4 - \\alpha ^2 C_4) \\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $C_3(\\omega^2 - \\alpha^2) \\cos{(\\alpha t)} + C_4(\\omega ^2 - \\alpha ^2)\\sin{(\\alpha t)} = \\frac{\\alpha A}{L}\\cos{(\\alpha t)} + 0\\sin{(\\alpha t)}$. By equating like coefficients, we find $C_3(\\omega^2 - \\alpha^2) = \\frac{\\alpha A }{L}$ and $C_4(\\omega^2 - \\alpha ^2) = 0$ We can therefore see that $C_4 = 0$ and $C_3 = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)}$. So now we have our particular solution: $I_P = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. Our general solution is the sum of the homogeneous solution and the particular solution. So $I = I_C + I_P$ $I = C_1\\cos{(\\omega t)} + C_2\\sin{(\\omega t)} + \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. It also seems that you need to revise solving second order linear ODEs.", "the wrong answer. 16. UsukiDoll |dw:1433044545197:dw| 17. UsukiDoll $\\sin(\\theta + \\beta) = \\frac{3}{5}\\frac{2\\sqrt{6}}{5}+\\frac{4}{5} (\\frac{1}{5})$ 18. UsukiDoll $\\frac{6\\sqrt{6}}{25}+\\frac{4}{25}$ 19. UsukiDoll I relooked at cos beta and realized that 5 was the longest side of the triangle and that 2 sqrt{6} was the adjcent part... needed to find the opposite part of the triangle so I could find sin beta. 20. UsukiDoll", "$$\\beta_1$$, we have that $$\\frac{\\partial \\sum_{t=2}^{n}(y_t-\\beta_1y_{t-1})^2}{\\partial\\beta_1} =-2\\sum_{t=2}^{n}(y_t-\\beta_1y_{t-1})y_{t-1}.$$ Then solving the following equation for $$\\beta_1$$ $$-2\\sum_{t=2}^{n}(y_t-\\beta_1y_{t-1})y_{t-1}=0,$$ we find that $$\\beta_1=\\frac{\\sum_{t=2}^{n}y_ty_{t-1}}{\\sum_{t=2}^{n}y_{t-1}^2}$$", "sin }^{ 2n } } \\theta$ and z = $\\sum _{ n=0 }^{ \\infty }{ { cos }^{ 2n }\\theta }$ sin2n$\\theta$, 0 < $\\theta$ < $\\frac { \\pi }{ 2 }$,then show that xyz = x+y+z Hint :1+x+x2+x3+.......= $\\frac { 1 }{ 1-x }$where $\\left| x\\right|$< 1]. 10. 3 x 3 = 9 11. If sin A = $\\frac{3}{5}$ and cos B = $\\frac{9}{41}$, 0 < A < $\\frac{\\pi}{2}$, 0 < B < $\\frac{\\pi}{2}$. Find the value of cos (A - B) 12. If $\\cos { \\left( \\alpha -\\beta \\right) } +\\cos { \\left( \\beta -\\gamma \\right) } +\\cos { \\left( \\gamma -\\alpha \\right) } =\\frac { -3 }{ 2 }$ then prove that $\\cos { \\alpha } +\\cos { \\beta } +\\cos { \\gamma } =\\sin { \\alpha } +\\sin { \\beta } +\\sin { \\gamma } =0$ 13. Prove that (1 + sec 2$\\theta$) (1 + sec 4$\\theta$) ... (1 + sec 2n$\\theta$) = tan 2n$\\theta$ . cot $\\theta$. 14. 2 x 5 = 10 15. if $\\frac { cos^{ 4 }\\alpha }{ { cos }^{ 2 }\\beta } +\\frac { { sin }^{ 4 }\\alpha }{ { sin }^{ 2 }\\beta } =1$ prove that ${ sin }^{ 4 }\\alpha +{ sin }^{ 4 }\\beta =2{ sin }^{ 2 }\\alpha { sin }^{ 2", "Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.", "help, but in the previous part of the question we've already proved that when: $ x=0, \\frac{dy}{dx}=-2 $ 2. Originally Posted by yianni Here it is: $ \\frac{d^2y}{dx^2}+6\\frac{dy}{dx}+25y=-292sin(4x) $ So the auxiliary/characteristic equation will look like: $ r^2+6r+25=0 $ Solving that gives: $ r^2+6r+25=0 $ $ r = 3\\pm4i $ And so we'll get the following general homogeneous solution: $ GS(H) = e^{3x}(Acos(4x)+Bsin(4x)) $ So now I try to find the particular non homogeneous solution, trying: $ y_P = \\alpha xcos(4x)+\\beta xsin(4x) $ Why is the trial particular solution not: $y_P = \\alpha \\cos(4x)+\\beta \\sin(4x)$ CB 3. Originally Posted by yianni Here it is: $ \\frac{d^2y}{dx^2}+6\\frac{dy}{dx}+25y=-292sin(4x) $ So the auxiliary/characteristic equation will look like: $ r^2+6r+25=0 $ Solving that gives: $ r^2+6r+25=0 $ $ r = 3\\pm4i $ And so we'll get the following general homogeneous solution: $ GS(H) = e^{3x}(Acos(4x)+Bsin(4x)) $ So now I try to find the particular non homogeneous solution, trying: $ y_P = \\alpha xcos(4x)+\\beta xsin(4x) $ I suspect you put that \"x\" multiplying the sine and cosine in because you saw \"(A cos(4x)+ Bsin(4x))\" in the homogeneous solution. However, you really have $e^{3x}cos(4x)$ and $e^{3x}sin(4x)$ which are NOT equivalent to \"cos(4x)\" and \"sin(4x)\". More simply, the solutions to you characteristic equation are 3+ 4i and 3- 4i while just sin(4x) and cos(4x) correspond", "\\\\ i \\frac{d C_2}{d\\bar x} & = -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta\\bar x) C_2 + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_1 e^{i\\bar x}.\\end{split}$ To solve them numerically, we could decompose it into four real equations. $\\begin{split}- \\frac{d C_{1,I}}{d \\bar x} &= \\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} \\cos(\\bar x) + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} \\sin(\\bar x) \\\\ \\frac{d C_{1,R}}{d\\bar x}& = \\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,I} - \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} \\sin(\\bar x) + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} \\cos(\\bar x) \\\\ -\\frac{d C_{2,I}}{d\\bar x}&= -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R}\\cos \\bar x - \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,I}\\sin \\bar x \\\\ \\frac{d C_{2,R}}{d\\bar x} & = -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R} \\sin \\bar x + \\frac{\\alpha \\sin 2\\theta_v}{2}\\cos (\\beta \\bar x) C_{1,I} \\cos \\bar x .\\end{split}$ Consequently, we have the matrix equation as the original equations, $\\begin{split}\\frac{d}{d\\bar x}\\begin{pmatrix} C_{1,R}\\\\ C_{1,I} \\\\ C_{2,R} \\\\ C_{2,I} \\end{pmatrix} = \\frac{1}{2}\\begin{pmatrix} 0 & \\alpha\\cos 2\\theta_v & - \\alpha \\sin 2\\theta_v\\sin \\bar x & \\alpha \\sin 2\\theta_v \\cos \\bar", "+ \\cos^2 x =1$$, $$\\frac{\\cos^2 \\beta+4\\sin^2 \\beta}{1-\\sin^2\\beta} = 7\\\\ \\frac{\\cos^2 \\beta+4\\sin^2 \\beta}{\\cos^2\\beta} = 7\\\\ 1+4\\tan^2\\beta = 7\\\\ 4\\tan^2\\beta = 6\\\\ \\tan^2\\beta = 1.5$$", "\\alpha (6+\\beta) = -6\\beta\\\\ \\alpha = \\frac {-6\\beta}{6+\\beta}$$ And now we can choose any integer value for $$\\beta$$, and find a corresponding $$\\alpha$$ $$b = \\frac{-\\beta^2}{6+\\beta}$$ And this should work for any integer $$\\beta$$", "# Prove the following Question: If $\\cos (\\alpha+\\beta)=\\frac{3}{5}, \\sin (\\alpha-\\beta)=\\frac{5}{13}$ and $0<\\alpha, \\beta<\\frac{\\pi}{4}$, then $\\tan (2 \\alpha)$ is equal to : 1. $\\frac{21}{16}$ 2. $\\frac{63}{52}$ 3. $\\frac{33}{52}$ 4. $\\frac{63}{16}$ Correct Option: , 4 Solution:" ]

[ "1 = 1 - 2sin2(u) tan(2u) = 2tan(u) 1- tan2(u) cos(2u) = 1 - tan2(u) 1 + tan2(u) sin(2u) = 2tan(u) 1 + tan2(u) 1 + cos(2u) = 2 cos2(u) 1 - cos(2u) = 2 sin2(u) sin3β = 3sinβ - 4 sin3β cos3β = 4cos3β - 4 cosβ $\\tan3\\beta=\\frac{3\\tan\\beta - \\tan^3\\beta}{1-3\\tan^2\\beta}$ $\\cot3\\beta=\\frac{\\cot^3\\beta-3\\cot\\beta}{3\\cot^2\\beta-1}$ #### Addition and multiplication of sin and cos $\\textrm{ sin } \\alpha + \\textrm{ sin }\\beta = 2 \\textrm{ sin }\\frac{\\alpha + \\beta}{2} \\textrm{ cos }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ sin } \\alpha - \\textrm{ sin }\\beta = 2 \\textrm{ sin }\\frac{\\alpha - \\beta}{2} \\textrm{ cos }\\frac{\\alpha + \\beta}{2}$ $\\textrm{ cos } \\alpha + \\textrm{ cos }\\beta = 2 \\textrm{ cos }\\frac{\\alpha + \\beta}{2} \\textrm{ cos }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ cos } \\alpha - \\textrm{ cos }\\beta = -2 \\textrm{ sin }\\frac{\\alpha + \\beta}{2} \\textrm{ sin }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ sin }\\alpha \\textrm{ sin }\\beta = \\frac{1}{2} (\\textrm{ cos }(\\alpha - \\beta) - \\textrm{ cos }(\\alpha + \\beta))$ $\\textrm{ cos }\\alpha \\textrm{ cos }\\beta = \\frac{1}{2} (\\textrm{ cos }(\\alpha - \\beta) + \\textrm{ cos }(\\alpha + \\beta))$ $\\textrm{ sin }\\alpha \\textrm{ cos }\\beta = \\frac{1}{2} (\\textrm{ sin }(\\alpha + \\beta) + \\textrm{ sin }(\\alpha - \\beta))$ Contact email:", "\\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$ ${\\displaystyle \\cot \\alpha }$ ${\\displaystyle \\sec \\alpha }$ ${\\displaystyle \\csc \\alpha }$ i grader", "=1\\,\\!}$ ${\\displaystyle \\sin(\\alpha +\\beta )=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ ### Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ ### Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ ### Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ ### Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! # Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$", "-c^{2}\\cos ^{2}\\beta }}{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}}\\cos \\alpha ,\\\\{\\frac {d\\omega }{ds}}&={\\frac {1}{\\sqrt {(a^{2}-b^{2})\\sin ^{2}\\omega +(b^{2}-c^{2})\\cos ^{2}\\beta }}}{\\frac {\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}}\\sin \\alpha ,\\\\{\\frac {d\\alpha }{ds}}&={\\frac {1}{((a^{2}-b^{2})\\sin ^{2}\\omega +(b^{2}-c^{2})\\cos ^{2}\\beta )^{3/2}}}\\times \\\\&\\quad {\\biggl (}{\\frac {(a^{2}-b^{2})\\cos \\omega \\sin \\omega {\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}}{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}}\\cos \\alpha \\\\&\\qquad +{\\frac {(b^{2}-c^{2})\\cos \\beta \\sin \\beta {\\sqrt {a^{2}-b^{2}\\sin ^{2}\\beta -c^{2}\\cos ^{2}\\beta }}}{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}}\\sin \\alpha {\\biggr )}.\\end{aligned}}} ### Jacobi's solution Jacobi showed that the geodesic equations, expressed in ellipsoidal coordinates, are separable. Here is how he recounted his discovery to his friend and neighbor Bessel (Jacobi 1839, Letter to Bessel), The day before yesterday, I reduced to quadrature the problem of geodesic lines on an ellipsoid with three unequal axes. They are the simplest formulas in the world, Abelian integrals, which become the well known elliptic integrals if 2 axes are set equal. Königsberg, 28th Dec. '38. The solution given by Jacobi (Jacobi 1839) (Jacobi 1866, §28) is {\\displaystyle {\\begin{aligned}\\delta &=\\int {\\frac {{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}\\,d\\beta }{{\\sqrt {a^{2}-b^{2}\\sin ^{2}\\beta -c^{2}\\cos ^{2}\\beta }}{\\sqrt {(b^{2}-c^{2})\\cos ^{2}\\beta -\\gamma }}}}\\\\&\\quad -\\int {\\frac {{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}\\,d\\omega }{{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}{\\sqrt {(a^{2}-b^{2})\\sin ^{2}\\omega +\\gamma }}}}.\\end{aligned}}} As Jacobi notes \"a function of the angle β equals a function", "$\\frac{ 1 - 2 \\sin^2 x \\cos^2 x + \\cos^2 x \\sin^2 x ( \\cos^2 x + \\sin^2 x)}{(1+ \\cos^2 x)(1+ \\sin^2 x) }$ $\\Big)$ $= \\Big($ $\\frac{ 1 - 2 \\sin^2 x \\cos^2 x + \\cos^2 x \\sin^2 x }{1 + \\cos^2 x + \\sin^x + \\cos^2 \\sin^2 x }$ $\\Big)$ $= \\Big($ $\\frac{ 1 - \\sin^2 x \\cos^2 x }{2 + \\cos^2 \\sin^2 x }$ $\\Big)$ $=$ RHS. Hence Proved. $\\\\$ Question 13: $(1 + \\tan \\alpha \\tan \\beta)^2 + (\\tan \\alpha - \\tan \\beta)^2 = \\sec^2 \\alpha \\sec^2 \\beta$ LHS $= (1 + \\tan \\alpha \\tan \\beta)^2 + (\\tan \\alpha - \\tan \\beta)^2$ $= 1 + \\tan^2 \\alpha \\tan^2 \\beta + 2\\tan \\alpha \\tan \\beta + \\tan^2 \\alpha + \\tan^2 \\beta - 2\\tan \\alpha \\tan \\beta$ $= 1 + \\tan^2 \\alpha \\tan^2 \\beta + \\tan^2 \\alpha + \\tan^2 \\beta$ $= 1 + \\tan^2 \\beta + \\tan^2 \\alpha ( 1 + \\tan^2 \\beta)$ $= ( 1 + \\tan^2 \\beta) ( 1 + \\tan^2 \\alpha)$ $= \\sec^2 \\alpha \\sec^2 \\beta$ $=$ RHS. Hence Proved. $\\\\$ Question 14: $\\frac{(1+\\cot x + \\tan x)(\\sin x - \\cos x)}{\\sec^3 x - \\mathrm{cosec}^3 x}$ $= \\sin^2 x \\cos^2 x$ LHS $=$ $\\frac{(1+\\cot x + \\tan x)(\\sin x - \\cos x)}{\\sec^3 x - \\mathrm{cosec}^3 x}$ $=$ $\\frac{(1+\\frac{\\cos x}{\\sin x} + \\frac{\\sin x}{\\cos", "# If 2 tan α=3 tan β, Question: If $2 \\tan \\alpha=3 \\tan \\beta$, then $\\tan (\\alpha-\\beta)=$ (a) $\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$ (b) $\\frac{\\cos 2 \\beta}{5-\\cos 2 \\beta}$ (c) $\\frac{\\sin 2 \\beta}{5+\\cos 2 \\beta}$ (d) none of these Solution: (a) $\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$ Given: $2 \\tan \\alpha=3 \\tan \\beta$ Now, $\\tan (\\alpha-\\beta)=\\frac{\\tan \\alpha-\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}$ $=\\frac{\\frac{3}{2} \\tan \\beta-\\tan \\beta}{1+\\left(\\frac{3}{2} \\tan \\beta\\right) \\tan \\beta}$ $=\\frac{3 \\tan \\beta-2 \\tan \\beta}{2+3 \\tan ^{2} \\beta}$ $=\\frac{\\tan \\beta}{2+3 \\tan ^{2} \\beta}$ $=\\frac{\\frac{\\sin \\beta}{\\cos \\beta}}{2+3 \\frac{\\sin ^{2} \\beta}{\\cos ^{2} \\beta}}$ $=\\frac{\\sin \\beta \\cos \\beta}{2 \\cos ^{2} \\beta+3 \\sin ^{2} \\beta}$ $=\\frac{\\sin \\beta \\cos \\beta}{2+\\sin ^{2} \\beta}$ $=\\frac{2 \\sin \\beta \\cos \\beta}{4+2 \\sin ^{2} \\beta}$ $=\\frac{\\sin 2 \\beta}{4+1-\\cos 2 \\beta}$ $\\therefore \\tan (\\alpha-\\beta)=\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$", "= \\tan \\quad \\quad \\tan(\\pi - \\theta) = -\\tan \\theta$$ $$\\sin(\\theta + \\frac{\\pi}{2}) = \\cos \\theta \\quad \\quad \\sin(\\frac{\\pi}{2} - \\theta) = \\cos \\theta$$ $$\\cos(\\theta + \\frac{\\pi}{2}) = -\\sin \\theta \\quad \\quad \\cos(\\frac{\\pi}{2} - \\theta) = \\sin \\theta$$ $$\\tan(\\theta + \\frac{\\pi}{2}) = -\\frac{1}{\\tan \\theta} \\quad \\quad \\tan(\\frac{\\pi}{2} - \\theta) = \\frac{1}{\\tan \\theta}$$ # 正弦定理 $$\\sin (\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$$ $$\\sin (\\alpha - \\beta) = \\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta$$ $$\\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta$$ $$\\cos (\\alpha - \\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$$ $$\\tan (\\alpha + \\beta) = \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}$$ $$\\tan (\\alpha - \\beta) = \\frac{\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta}$$ 2倍角の公式 $$\\sin2 \\alpha = 2\\sin \\alpha \\cos \\alpha$$ $$\\cos2 \\alpha = \\cos^2 \\alpha - \\sin^2 \\alpha = 1 - 2\\sin^2 \\alpha = 2\\cos^2 \\alpha - 1$$ $$\\tan2 \\alpha = \\frac{2\\tan \\alpha}{1 - tan^2 \\alpha}$$ $$\\sin^2\\frac{\\alpha}{2} = \\frac{1 - \\cos2\\alpha}{2}, \\quad \\cos^2\\frac{\\alpha}{2} = \\frac{1 + \\cos2\\alpha}{2}$$ 3倍角の公式 $$\\sin3\\alpha = 3\\sin\\alpha - 4\\sin^3\\alpha$$ $$\\cos3\\alpha = 4\\cos^3\\alpha - 3\\cos\\alpha$$ $$\\tan3\\alpha = \\frac{3\\tan\\alpha - \\tan^3\\alpha}{1 - 3tan^2\\alpha}$$ # 和と積の変換公式 $$\\sin\\alpha\\cos\\beta = \\frac{1}{2}\\{\\sin(\\alpha + \\beta) + \\sin(\\alpha - \\beta)\\}$$ $$\\cos\\alpha\\sin\\beta = \\frac{1}{2}\\{\\sin(\\alpha +", "+ \\cos \\alpha \\cos \\beta)$ $\\displaystyle \\Rightarrow a^2 + b^2= 2 + 2 \\cos ( \\alpha + \\beta )$ $\\displaystyle \\text{Similarly, } \\therefore b^2 - a^2 = (\\cos \\alpha + \\cos \\beta)^2 - (\\sin \\alpha + \\sin \\beta)^2$ $\\displaystyle \\Rightarrow b^2 - a^2 = \\cos^2 \\alpha + \\cos^2 \\beta + 2 \\cos \\alpha \\cos \\beta - \\sin^2 \\alpha - \\sin^2 \\beta - 2 \\sin \\alpha \\sin \\beta$ $\\displaystyle \\Rightarrow b^2 - a^2 = (\\cos^2 \\alpha - \\sin^2 \\beta) + (\\cos^2 \\beta - \\sin^2 \\alpha) + 2( cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta)$ $\\displaystyle \\Rightarrow b^2 - a^2 = \\cos (\\alpha + \\beta) \\cos(\\alpha - \\beta) + \\cos ( \\beta + \\alpha) \\cos ( \\beta - \\alpha) + 2 \\cos (\\alpha + \\beta)$ $\\displaystyle \\because \\cos (\\beta - \\alpha) = \\cos [ - ( \\alpha - \\beta ) ] = \\cos ( \\alpha - \\beta )$ $\\displaystyle \\Rightarrow b^2 - a^2 = \\cos (\\alpha + \\beta) [ 2 \\cos ( \\alpha - \\beta ) + 2]$ $\\displaystyle \\Rightarrow b^2 - a^2 = \\cos (\\alpha + \\beta) [ a^2 + b^2 ]$ $\\displaystyle \\Rightarrow \\cos ( \\alpha + \\beta ) = \\frac{b^2 - a^2}{b^2 + a^2}$ $\\displaystyle \\sin ( \\alpha + \\beta ) = \\sqrt{1 - [\\cos ( \\alpha + \\beta )]^2 }$ $\\displaystyle = \\sqrt{1 -", "agree that is is correct (I've used $$\\K\\$$ instead of $$\\I_0\\$$). Here is how I derived my formula: $$\\I_0\\sin(\\omega t + \\beta) \\cdot U_0 \\sin(\\omega t + \\alpha) \\$$ = $$\\\\dfrac{I_0 U_0}{2}(2\\sin(\\omega t + \\beta) \\sin(\\omega t + \\alpha)) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(2(-\\cos(\\omega t + \\alpha + \\omega t + \\beta) + \\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(\\omega t + \\alpha + \\omega t + \\beta) + 2\\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) -\\cos(\\omega t + \\alpha + \\omega t + \\beta) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) +\\sin(\\omega t + \\alpha) \\sin(\\omega t + \\beta) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha - (\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha - (\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\alpha - \\beta ) ) = (2.) \\$$ Regarding that only the difference in phase matters, I can't see how this can be. If we only imagine two sine curves being shifted back and forth along the time axis it seems obvious to me that the", "\\alpha \\sin \\alpha \\tan^2 \\beta }{ \\cos^2 \\alpha-\\sin^2 \\alpha \\tan ^2 \\beta}$ $=\\frac{ \\sin \\alpha \\cos \\alpha (1+\\tan^2 \\beta) -\\tan \\beta (\\sin^2 \\alpha + \\cos^2 \\alpha )}{ \\cos^2 \\alpha - \\sin^2 \\alpha \\tan ^2 \\beta}$$=\\frac{ \\sin \\alpha \\cos \\alpha(\\sec^2 \\beta)- \\tan \\beta (1) }{ \\cos^2 \\alpha- \\sin^2 \\alpha \\tan^2 \\beta}$ 33. prizzyjade $=\\frac{ \\frac{ \\sin \\alpha \\cos \\alpha }{ \\cos^2 \\beta }- \\tan \\beta }{ \\cos^2 \\alpha - \\sin^2 \\alpha \\tan^2 \\beta }$ $=\\frac{\\frac{ \\sin \\alpha \\cos \\alpha }{ \\cos^2 \\beta} -\\frac{ \\sin \\beta}{ \\cos \\beta }}{\\cos^2\\alpha- \\frac{ \\sin^ 2 \\alpha \\sin^2 \\beta }{\\cos^2 \\beta } }$$=\\frac{ \\frac{ \\sin \\alpha \\cos \\alpha- \\sin \\beta \\cos \\beta }{ \\cos^2 \\beta } }{ \\frac{ \\cos^2 \\alpha \\cos^2 \\beta - \\sin^2 \\alpha \\sin ^2 \\beta }{\\cos^2 \\beta } }$$=\\frac{ \\sin \\alpha \\cos \\alpha -\\sin \\beta \\cos \\beta }{ \\cos^2 \\alpha \\cos^2 \\beta - \\sin^2 \\alpha \\sin^2 \\beta }$$=\\frac{ \\frac{ \\tan \\alpha }{ 1+\\tan^2 \\alpha } - \\frac{ \\tan \\beta }{ 1+ \\tan^\\beta } }{ 2-\\sin^2 \\alpha - \\sin^ 2 \\beta + \\sin^2 \\alpha \\sin^2 \\beta-\\sin^2 \\alpha \\sin^2 \\beta}$ 34. anonymous work from left, sub in the addition formula for tan(A+B) #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "\\cos \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} = b$ … … … … … ii) Dividing i) by ii) we get $\\displaystyle \\tan \\frac{\\alpha + \\beta}{2} = \\frac{a}{b}$ $\\displaystyle \\text{We know } \\sin (\\alpha + \\beta) = \\frac{2 \\tan \\big( \\frac{\\alpha + \\beta}{2} \\big)}{1+ \\tan^2 \\big( \\frac{\\alpha + \\beta}{2} \\big)} = \\frac{2 \\frac{a}{b}}{1+\\frac{a^2}{b^2} } = \\frac{2ab}{a^2+b^2}$ $\\displaystyle ( \\sin \\alpha + \\sin \\beta)^2 + ( \\cos \\alpha + \\cos \\beta)^2 = a^2 +b^2$ $\\displaystyle \\Rightarrow \\sin^2 \\alpha + \\sin^2 \\beta + 2 \\sin \\alpha \\sin \\beta + \\cos^2 \\alpha + \\cos^2 \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\sin \\alpha \\sin \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 ( \\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\cos ( \\alpha - \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow \\cos ( \\alpha - \\beta) = \\frac{a^2 + b^2-2}{2}$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 39: } \\text{If } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} , \\text{ prove that } \\cos \\alpha = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$ $\\displaystyle \\text{Given } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} ,$ $\\displaystyle \\text{RHS } = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$", "cos2A) sin2B = cos2A - cos2A sin2B - sin2B + cos2A sin2B = cos2A - sin2B (M.H.S.) Again, cos2A - sin2B = (1 - sin2A) - (1 - cos2B) = 1 - sin2A - 1 + cos2B = cos2B - sin2A ∴ L.H.S. = M.H.S. = R.H.S. Proved Given, θ =α +β sinθ = sin(α + β)............................(1) And $\\frac {tanα}{tanβ}$ = $\\frac xy$ $\\frac {tanα - tanβ}{tanα + tanβ}$ = $\\frac {x - y}{x + y}$.............................(2) Now, R.H.S. = $\\frac {x - y}{x + y}$ sinθ =$\\frac {tanα - tanβ}{tanα + tanβ}$ sinθ =$\\frac {\\frac {sinα}{cosα} -\\frac {sinβ}{cosβ}}{\\frac {sinα}{cosα} + \\frac {sinβ}{cosβ}}$ sinθ = $\\frac{\\frac{sin\\alpha.cos\\beta - cos\\alpha.sin\\beta}{cos\\alpha.cos\\beta}}{\\frac{sin\\alpha.cos\\beta + cos\\alpha.sin\\beta}{cos\\alpha.cos\\beta}}$. sin$\\theta$ = $\\frac {sin\\alpha cos\\beta - cos\\alpha sin\\beta}{sin\\alpha cos\\beta + cos\\alpha sin\\beta}$ . sin$\\theta$ = $\\frac {sin (\\alpha - \\beta)}{sin(\\alpha + \\beta)}$ . sin$\\theta$ = $\\frac {sin(\\alpha - \\beta)}{sin \\theta}$. sin$\\theta$ = sin (α - β) Hence, L.H.S. = R.H.S. Proved L.H.S. = sin (A + B + C) = sin {A + (B + C)} = sinA . cos(B + C) + cos A . cos (B + C) = sinA {cosB cosC - sinB sinC} + cosA {sinB cosC + cosB sinC} = sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC = cosA cosB cosC[$\\frac {sinA cosB cosC - sinA", "$(iv)\\,\\,\\cos^3\\alpha+\\cos^3(120^{\\circ}+\\alpha)+\\cos^3(240^{\\circ}+\\alpha)=\\frac 34 \\cos 3\\alpha$ Sol. $\\,\\,\\cos^3\\alpha+\\cos^3(120^{\\circ}+\\alpha)+\\cos^3(240^{\\circ}+\\alpha)\\\\=\\frac 14 [\\cos 3 \\alpha+3\\cos\\alpha+\\cos(360^{\\circ}+3\\alpha)\\\\+3\\cos(120^{\\circ}+\\alpha)+\\cos(720^{\\circ}+3\\alpha)\\\\+3\\cos(240^{\\circ}+\\alpha)]\\\\=\\frac 14[3\\cos3\\alpha+3\\cos\\alpha+3\\{\\cos(120^{\\circ}+\\alpha)\\\\+\\cos(240^{\\circ}+\\alpha)\\}]\\\\=\\frac 14[3\\cos 3\\alpha+3\\cos \\alpha+3.2\\cos(180^{\\circ}+\\alpha) \\\\ \\times \\cos60^{\\circ}]\\\\=\\frac 34\\cos 3\\alpha$ Additional Notes : $\\,4\\cos^3(120^{\\circ}+\\alpha)=\\cos [3(120^{\\circ}+\\alpha)]\\\\+3 \\cos(120^{\\circ}+\\alpha),\\\\ 4\\cos^3(240^{\\circ}+\\alpha)=\\cos [3(240^{\\circ}+\\alpha)]\\\\+3 \\cos(240^{\\circ}+\\alpha), \\\\ \\cos 3 \\alpha=\\cos(360^{\\circ}+3\\alpha)=\\cos(720^{\\circ}+3\\alpha)$ $\\,(v)\\, \\tan 70^{\\circ}-\\tan 50^{\\circ}+\\tan 10^{\\circ}=\\sqrt3$ Sol. $\\,\\tan 70^{\\circ}-\\tan 50^{\\circ}+\\tan 10^{\\circ}\\\\=\\tan (60^{\\circ}+10^{\\circ})-\\tan(60^{\\circ}-10^{\\circ})+\\tan 10^{\\circ}\\\\=\\frac{\\tan 60^{\\circ}+\\tan 10^{\\circ}}{1-\\tan 60^{\\circ} \\tan 10^{\\circ}}-\\frac{\\tan 60^{\\circ}-\\tan 10^{\\circ}}{1+\\tan 60^{\\circ}\\tan 10^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{\\sqrt3+\\tan 10^{\\circ}}{1-\\sqrt3\\tan 10^{\\circ}}-\\frac{\\sqrt3-\\tan 10^{\\circ}}{1+\\sqrt3\\tan 10^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{\\sqrt3+3\\tan 10^{\\circ}+\\tan 10^{\\circ}+\\sqrt3 \\tan^210^{\\circ}-(\\sqrt3-3\\tan 10^{\\circ}\\\\-\\tan 10^{\\circ}+\\sqrt3 \\tan^210^{\\circ})}{1-3\\tan^210^{\\circ}}\\\\+\\tan 10^{\\circ}\\\\=\\frac{8\\tan 10^{\\circ}}{1-3\\tan^210^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{9 \\tan 10^{\\circ}-3\\tan^3 10^{\\circ}}{1-3\\tan^210^{\\circ}} \\\\=3\\left( \\frac{3\\tan 10^{\\circ}-3\\tan^310^{\\circ}}{1-3\\tan^210^{\\circ}}\\right)\\\\=3\\tan30^{\\circ}\\\\=\\sqrt3$ $\\,(vi)\\, \\tan \\theta \\tan(\\frac{\\pi}{3}+\\theta)\\tan(\\frac{\\pi}{3}-\\theta)=\\tan 3\\theta$ Sol. $\\,\\, \\tan \\theta \\tan(\\frac{\\pi}{3}+\\theta)\\tan(\\frac{\\pi}{3}-\\theta)\\\\=\\tan \\theta . \\frac{\\sqrt3+\\tan \\theta}{1-\\sqrt3 \\tan \\theta}. \\frac{\\sqrt3-\\tan \\theta}{1+\\sqrt3 \\tan \\theta}\\\\=\\frac{3\\tan \\theta-\\tan^3\\theta}{1-3\\tan^2\\theta}\\\\=\\tan \\theta$ $\\,(vii)\\, \\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+8\\cot 8\\theta=\\cot \\theta$ Sol. $\\, \\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+8\\cot 8\\theta\\\\=\\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+\\frac{8(1-\\tan^2 4\\theta)}{2\\tan 4\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta+\\frac{8\\tan^24\\theta+8-8\\tan^24\\theta}{2\\tan 2\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta+\\frac{4}{\\tan 2\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta +\\frac{4(1-\\tan^2 2\\theta)}{2\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{4 \\tan^2 2\\theta+4-4\\tan^2 2\\theta}{2\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{2}{\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{2(1-\\tan^2\\theta)}{2\\tan \\theta}\\\\=\\cot \\theta$ $\\,(viii)\\, \\cos \\frac{\\pi}{11}\\cos \\frac{2\\pi}{11}\\cos \\frac{3\\pi}{11}\\cos \\frac{4\\pi}{11}\\cos \\frac{5\\pi}{11} =\\frac{1}{32}$ Sol. Let $\\,\\,\\frac{\\pi}{11}=\\theta \\Rightarrow \\theta=11 \\pi \\\\ \\therefore P=\\cos \\theta \\cos 2\\theta \\cos 3\\theta \\cos 4\\theta \\cos 5\\theta \\,\\,,\\text{say}\\\\ \\Rightarrow 2P\\sin \\theta=(2 \\sin \\theta\\cos \\theta) \\cos 2 \\theta \\cos3 \\theta \\cos 4 \\theta\\cos 5 \\theta \\\\ \\Rightarrow 2P\\sin \\theta=\\sin 2 \\theta\\cos 2 \\theta\\cos3 \\theta\\cos4 \\theta\\cos5 \\theta \\\\ \\Rightarrow 2^2P\\sin \\theta=2\\sin 2 \\theta\\cos 2 \\theta\\cos3 \\theta\\cos4 \\theta\\cos5 \\theta \\\\ \\Rightarrow 2^2P \\sin \\theta=\\sin 4\\theta \\cos 4\\theta \\cos 3\\theta \\cos 5\\theta \\\\ \\Rightarrow", "+ \\ldots + \\sin (2n-1) x =$ $\\frac{\\sin^2 nx}{\\sin x}$ for all $n \\in N$ $\\\\$ Question 40: Prove that $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (n-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{n-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{n \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ Let $P(n) :$ $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (n-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{n-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{n \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ For $n = 1 , LHS = \\cos ( \\alpha + (1-1) \\beta) = \\cos \\alpha$ RHS $= \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{1-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{\\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ $= \\cos \\alpha$ Therefore LHS = RHS Hence $P(1)$ is true for $n = 1$ Let $P(n)$ is true for $n = k$ $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (k-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{k-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{k \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ … i) Now we have to show", "\\omega^2 C_3\\cos{(\\alpha t)} + \\omega ^2 C_4\\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $(\\omega^2 C_3 - \\alpha ^2 C_3)\\cos{(\\alpha t)} + (\\omega ^2 C_4 - \\alpha ^2 C_4) \\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $C_3(\\omega^2 - \\alpha^2) \\cos{(\\alpha t)} + C_4(\\omega ^2 - \\alpha ^2)\\sin{(\\alpha t)} = \\frac{\\alpha A}{L}\\cos{(\\alpha t)} + 0\\sin{(\\alpha t)}$. By equating like coefficients, we find $C_3(\\omega^2 - \\alpha^2) = \\frac{\\alpha A }{L}$ and $C_4(\\omega^2 - \\alpha ^2) = 0$ We can therefore see that $C_4 = 0$ and $C_3 = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)}$. So now we have our particular solution: $I_P = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. Our general solution is the sum of the homogeneous solution and the particular solution. So $I = I_C + I_P$ $I = C_1\\cos{(\\omega t)} + C_2\\sin{(\\omega t)} + \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. It also seems that you need to revise solving second order linear ODEs.", "\\beta } \\right)}^{2}}+{{\\left( {\\cos \\alpha +\\cos \\beta } \\right)}^{2}}\\\\\\\\=\\ \\ {{\\cos }^{2}}\\alpha +2\\cos \\alpha \\cos \\beta +{{\\cos }^{2}}\\beta \\\\\\ \\ \\ \\ +\\ {{\\sin }^{2}}\\alpha +2\\sin \\alpha \\sin \\beta +{{\\sin }^{2}}\\beta \\\\\\\\=\\ \\ {{\\sin }^{2}}\\alpha +{{\\cos }^{2}}\\alpha +{{\\sin }^{2}}\\beta +{{\\cos }^{2}}\\beta \\\\\\ \\ \\ \\ +2\\cos \\alpha \\cos \\beta +2\\sin \\alpha \\sin \\beta \\\\\\\\=\\ \\ 2+2\\left( {\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta } \\right)\\\\\\\\=\\ \\ 2\\left[ {1+\\cos \\left( {\\alpha -\\beta } \\right)} \\right]\\\\\\\\=\\ \\ 2\\left[ {1+\\cos 2\\left( {\\displaystyle \\frac{{\\alpha -\\beta }}{2}} \\right)} \\right]\\\\\\\\=\\ \\ 2\\left[ {1+\\left( {2{{{\\cos }}^{2}}\\left( {\\displaystyle \\frac{{\\alpha -\\beta }}{2}} \\right)-1} \\right)} \\right]\\\\\\\\=\\ \\ 2\\left[ {2{{{\\cos }}^{2}}\\left( {\\displaystyle \\frac{{\\alpha -\\beta }}{2}} \\right)} \\right]\\\\\\\\=4{{\\cos }^{2}}\\left( {\\displaystyle \\frac{{\\alpha -\\beta }}{2}} \\right)\\end{array}$ 8. Prove that $\\displaystyle \\frac{{\\sin \\theta +\\sin 2\\theta +\\sin 4\\theta +\\sin 5\\theta }}{{\\cos \\theta +\\cos 2\\theta +\\cos 4\\theta +\\cos 5\\theta }}=\\tan 3\\theta$ Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\displaystyle \\frac{{\\sin \\theta +\\sin 2\\theta +\\sin 4\\theta +\\sin 5\\theta }}{{\\cos \\theta +\\cos 2\\theta +\\cos 4\\theta +\\cos 5\\theta }}\\\\\\\\=\\ \\displaystyle \\frac{{\\sin 5\\theta +\\sin \\theta +\\sin 4\\theta +\\sin 2\\theta }}{{\\cos 5\\theta +\\cos \\theta +\\cos 4\\theta +\\cos 2\\theta }}\\\\\\\\=\\ \\displaystyle \\frac{{2\\sin \\displaystyle \\frac{{5\\theta +\\theta }}{2}\\cos \\displaystyle \\frac{{5\\theta -\\theta }}{2}+2\\sin \\displaystyle \\frac{{4\\theta +2\\theta }}{2}\\cos \\displaystyle \\frac{{4\\theta -2\\theta }}{2}}}{{2\\cos \\displaystyle \\frac{{5\\theta +\\theta }}{2}\\cos \\displaystyle \\frac{{5\\theta -\\theta }}{2}+2\\cos \\displaystyle \\frac{{4\\theta +2\\theta }}{2}\\cos \\displaystyle \\frac{{4\\theta -2\\theta }}{2}}}\\\\\\\\=\\ \\displaystyle \\frac{{2\\left( {\\sin 3\\theta \\cos 2\\theta +\\sin", "\\arctan {2\\frac{\\sqrt{h^{2}-ab}}{a+b}}$ or $\\tan{2\\alpha} = \\frac{2\\sqrt{h^{2}-ab}}{a+b}$ Now, $\\frac{2\\sqrt{h^{2}-ab}}{a+b}$ $= \\frac{2\\sqrt{\\tan^{2}{\\alpha}\\sin^{2}{\\theta}-(\\cos^{2}{\\theta}\\sin^{2}{\\alpha} + \\sin^{2}{\\theta}-\\tan^{2}{\\alpha})\\cos^{2}{\\theta}\\sin^{2}{\\alpha}}}{2\\cos^{2}{\\theta}\\sin^{2}{\\alpha}+\\sin^{2}{\\theta}-\\tan^{2}{\\alpha}}$ $= \\frac{2\\tan{\\alpha}\\sqrt{\\sin^{2}{\\theta}-\\cos^{4}{\\theta}\\sin^{2}{\\alpha}\\cos^{2}{\\alpha}-\\sin^{2}{\\theta}\\cos^{2}{\\theta}\\cos^{2}{\\alpha}+\\sin^{2}{\\alpha}\\cos^{2}{\\theta}}}{2\\cos^{2}{\\theta}\\sin^{2}{\\alpha}+1-\\cos^{2}{\\theta}-\\tan^{2}{\\alpha}}$ $= \\frac{2\\tan{\\alpha}\\sqrt{1-\\cos^{2}{\\theta}-\\cos^{4}{\\theta}{\\cos^{2}{\\alpha(1-\\cos^{2}{\\alpha})-(1-\\cos^{2}{\\theta})\\cos^{2}{\\theta}\\cos^{2}{\\alpha}+(1-\\cos^{2}{\\alpha})\\cos^{2}{\\theta}}}}}{1-\\tan^{2}{\\alpha}-\\cos^{2}{\\theta}(1-2\\sin^{2}{\\alpha})}$ $=\\frac{2\\tan{\\alpha}\\sqrt{1+\\cos^{4}{\\theta}\\cos^{4}{\\alpha}-2\\cos^{2}{\\theta}\\cos^{2}{\\alpha}}}{1-\\tan^{2}{\\alpha}-\\cos^{2}{\\theta}(\\cos^{2}{\\theta}-\\sin^{2}{\\alpha})}$ $= \\frac{2\\tan{\\alpha}(1-\\cos^{2}{\\theta}\\cos^{2}{\\alpha})}{(1-\\tan^{2}{\\alpha})(1-\\cos^{2}{\\theta}\\cos^{2}{\\alpha})}=\\tan{2\\alpha}$ Hence, the given lines include an angle $2\\alpha$. More later, Nalin Pithwa. ### Pair of straight lines — quick review for IITJEE Math Pair of Straight Lines: I: Equation of Family of Lines: A first degree equation $ax+by+c=0$ represents a straight line involving three constants a, b and c, which can be reduced to two by dividing both the sides of the equation by a non-zero constant. For instance, if $a \\neq 0$ we can write the equation as $x+\\frac{b}{a}y+\\frac{c}{a}=0$ or $x+By+C=0$ So, in order to determine an equation of a line, we need two conditions on the line to determine these constants. For instance, if we know two points on the line or a point on the line and its slope etc., we know the line. But, if we know just one condition, we have infinite number of lines satisfying the given condition. In this case, the equation of the line contains an arbitrary constant and for different values of the constant we have different lines satisfying the given condition and the constant is called a parameter. Some Equations of Family of Lines: 1. Family of lines with given slope. (Family of given lines) $y = mx +k$,", "– \\cos^2\\alpha – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma)}{\\sin^2\\beta}$$ $$=\\frac{a^2(1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma)}{\\sin^2\\beta}$$. Thus, $$h = \\frac{a}{\\sin\\beta}\\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$. Finally, $$V = \\frac{Ah}{3} = \\frac{b\\cdot c\\cdot \\sin\\beta \\cdot a}{3 \\cdot 2 \\cdot \\sin\\beta} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$ $$=\\frac{abc}{6} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$. So, given the three surface angles and three edge lengths from a given vertex of a tetrahedron, the volume of the tetrahedron is: $V=\\frac{abc}{6} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$ Clio Corvid 1 Comment 1. Carlos Eduardo Olivieri says: Very good! Congrats!", "\\cos^2 x + \\sin^2 x)}{\\cos^2 x - \\sin^2 x}$ $=$ $\\frac{2}{\\cos^2 x - \\sin^2 x}$ $=$ $\\frac{2}{\\cos 2x}$ $= 2 \\sec 2x =$ RHS. Hence proved. $\\\\$ Question 23: $\\cot^2 x - \\tan^2 x = 4 \\cot 2x \\mathrm{cosec} 2x$ LHS $= \\cot^2 x - \\tan^2 x$ $=$ $\\frac{\\cos^2 x}{\\sin^2 x}$ $-$ $\\frac{\\sin^2 x}{\\cos^2 x}$ $=$ $\\frac{\\cos^4 x - \\sin^4 x}{\\sin^2 x \\cos^2 x}$ $=$ $\\frac{(\\cos^2 x - \\sin^2 x)(\\cos^2 x + \\sin^2 x)}{\\sin^2 x \\cos^2 x}$ $=$ $\\frac{\\cos^2 x - \\sin^2 x}{\\sin^2 x \\cos^2 x}$ $=$ $\\frac{4 \\cos 2x}{(2\\sin x \\cos x)^2}$ $=$ $\\frac{4 \\cos 2x}{\\sin^2 2x}$ $= 4 \\cot 2x \\ \\mathrm{cosec} 2x =$ RHS. Hence proved. $\\\\$ Question 24: $\\cos 4x - \\cos 4 \\alpha = 8( \\cos x - \\cos \\alpha) ( \\cos x + \\cos \\alpha) (\\cos x - \\sin \\alpha)(\\cos x + \\sin \\alpha)$ LHS $= \\cos 4x - \\cos 4 \\alpha$ $= 2 \\cos^2 2x - 2 \\cos^2 2\\alpha$ $= 2 ( \\cos 2x + \\cos 2\\alpha)(\\cos 2x - \\cos 2 \\alpha)$ $= 2 ( 2 \\cos^2 x - 1 + 1 - 2 \\sin^2 \\alpha)( 2 \\cos^2 x - 1 - 2 \\cos^2 \\alpha + 1)$ $= 8 (\\cos^2 x - \\sin^2 \\alpha) ( \\cos^2 x - \\cos^2 \\alpha)$ $= 8 ( \\cos x - \\cos \\alpha)(\\cos x + \\cos", "& cos \\beta & cos\\gamma \\\\sin 2 \\alpha & sin 2 \\beta & sin 2 \\gamma \\end{vmatrix}$ In the same manner, for the cosine, Cos 2α = cos (α + α) cos 2α = (cos α cos α - sin α sin α) cos 2α = (cos2α – sin2α) Applying the property of Pythagorean, sin 2 α + cos 2α =1, we can obtain two additional cosine identities i.e. Cos 2α = cos2α – sin2α Cos 2α = [-1 sin2α] - sin2α Cos 2α = 1- 2 sin2α Cos 2α = cos2α – sin2α cos 2α = cos2α – [1 – sin2α] cos 2α = 2cos2α – 1 The half‐angle identities for the sine and cosine are derived from two of the cosine identities explained above. Cos 2α = 2cos2α – 1 Cos 2 (β/2) - 2cos2 (β/2) – 1 Cosβ = 2cos2 (β/2) – 1 2cos2 (β/2) = 1 ± cosβ Cos 2α = 1 - 2 sin2α Cos 2 (β/2) = 1 – 2 sin2 (β/2) Cosβ = 1 – 2 sin2 (β/2) 2 sin2 (β/2) = 1 Cosβ Cos2 (β/2) = 1 + cosβ/2 Cos (β/2) = ± √[1+ cos(β/2)] Similarly, sin2 (β/2) = 1 - cosβ/2 sin (β/2) = √[1 - cos(β/2)] The symbol of the two preceding functions is dependent on the quadrant" ]

[ "1 = 1 - 2sin2(u) tan(2u) = 2tan(u) 1- tan2(u) cos(2u) = 1 - tan2(u) 1 + tan2(u) sin(2u) = 2tan(u) 1 + tan2(u) 1 + cos(2u) = 2 cos2(u) 1 - cos(2u) = 2 sin2(u) sin3β = 3sinβ - 4 sin3β cos3β = 4cos3β - 4 cosβ $\\tan3\\beta=\\frac{3\\tan\\beta - \\tan^3\\beta}{1-3\\tan^2\\beta}$ $\\cot3\\beta=\\frac{\\cot^3\\beta-3\\cot\\beta}{3\\cot^2\\beta-1}$ #### Addition and multiplication of sin and cos $\\textrm{ sin } \\alpha + \\textrm{ sin }\\beta = 2 \\textrm{ sin }\\frac{\\alpha + \\beta}{2} \\textrm{ cos }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ sin } \\alpha - \\textrm{ sin }\\beta = 2 \\textrm{ sin }\\frac{\\alpha - \\beta}{2} \\textrm{ cos }\\frac{\\alpha + \\beta}{2}$ $\\textrm{ cos } \\alpha + \\textrm{ cos }\\beta = 2 \\textrm{ cos }\\frac{\\alpha + \\beta}{2} \\textrm{ cos }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ cos } \\alpha - \\textrm{ cos }\\beta = -2 \\textrm{ sin }\\frac{\\alpha + \\beta}{2} \\textrm{ sin }\\frac{\\alpha - \\beta}{2}$ $\\textrm{ sin }\\alpha \\textrm{ sin }\\beta = \\frac{1}{2} (\\textrm{ cos }(\\alpha - \\beta) - \\textrm{ cos }(\\alpha + \\beta))$ $\\textrm{ cos }\\alpha \\textrm{ cos }\\beta = \\frac{1}{2} (\\textrm{ cos }(\\alpha - \\beta) + \\textrm{ cos }(\\alpha + \\beta))$ $\\textrm{ sin }\\alpha \\textrm{ cos }\\beta = \\frac{1}{2} (\\textrm{ sin }(\\alpha + \\beta) + \\textrm{ sin }(\\alpha - \\beta))$ Contact email:", "# If 2 tan α=3 tan β, Question: If $2 \\tan \\alpha=3 \\tan \\beta$, then $\\tan (\\alpha-\\beta)=$ (a) $\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$ (b) $\\frac{\\cos 2 \\beta}{5-\\cos 2 \\beta}$ (c) $\\frac{\\sin 2 \\beta}{5+\\cos 2 \\beta}$ (d) none of these Solution: (a) $\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$ Given: $2 \\tan \\alpha=3 \\tan \\beta$ Now, $\\tan (\\alpha-\\beta)=\\frac{\\tan \\alpha-\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}$ $=\\frac{\\frac{3}{2} \\tan \\beta-\\tan \\beta}{1+\\left(\\frac{3}{2} \\tan \\beta\\right) \\tan \\beta}$ $=\\frac{3 \\tan \\beta-2 \\tan \\beta}{2+3 \\tan ^{2} \\beta}$ $=\\frac{\\tan \\beta}{2+3 \\tan ^{2} \\beta}$ $=\\frac{\\frac{\\sin \\beta}{\\cos \\beta}}{2+3 \\frac{\\sin ^{2} \\beta}{\\cos ^{2} \\beta}}$ $=\\frac{\\sin \\beta \\cos \\beta}{2 \\cos ^{2} \\beta+3 \\sin ^{2} \\beta}$ $=\\frac{\\sin \\beta \\cos \\beta}{2+\\sin ^{2} \\beta}$ $=\\frac{2 \\sin \\beta \\cos \\beta}{4+2 \\sin ^{2} \\beta}$ $=\\frac{\\sin 2 \\beta}{4+1-\\cos 2 \\beta}$ $\\therefore \\tan (\\alpha-\\beta)=\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}$", "=1\\,\\!}$ ${\\displaystyle \\sin(\\alpha +\\beta )=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ ### Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ ### Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ ### Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ ### Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! # Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$", "\\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$ ${\\displaystyle \\cot \\alpha }$ ${\\displaystyle \\sec \\alpha }$ ${\\displaystyle \\csc \\alpha }$ i grader", "-c^{2}\\cos ^{2}\\beta }}{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}}\\cos \\alpha ,\\\\{\\frac {d\\omega }{ds}}&={\\frac {1}{\\sqrt {(a^{2}-b^{2})\\sin ^{2}\\omega +(b^{2}-c^{2})\\cos ^{2}\\beta }}}{\\frac {\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}}\\sin \\alpha ,\\\\{\\frac {d\\alpha }{ds}}&={\\frac {1}{((a^{2}-b^{2})\\sin ^{2}\\omega +(b^{2}-c^{2})\\cos ^{2}\\beta )^{3/2}}}\\times \\\\&\\quad {\\biggl (}{\\frac {(a^{2}-b^{2})\\cos \\omega \\sin \\omega {\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}}{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}}\\cos \\alpha \\\\&\\qquad +{\\frac {(b^{2}-c^{2})\\cos \\beta \\sin \\beta {\\sqrt {a^{2}-b^{2}\\sin ^{2}\\beta -c^{2}\\cos ^{2}\\beta }}}{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}}\\sin \\alpha {\\biggr )}.\\end{aligned}}} ### Jacobi's solution Jacobi showed that the geodesic equations, expressed in ellipsoidal coordinates, are separable. Here is how he recounted his discovery to his friend and neighbor Bessel (Jacobi 1839, Letter to Bessel), The day before yesterday, I reduced to quadrature the problem of geodesic lines on an ellipsoid with three unequal axes. They are the simplest formulas in the world, Abelian integrals, which become the well known elliptic integrals if 2 axes are set equal. Königsberg, 28th Dec. '38. The solution given by Jacobi (Jacobi 1839) (Jacobi 1866, §28) is {\\displaystyle {\\begin{aligned}\\delta &=\\int {\\frac {{\\sqrt {b^{2}\\sin ^{2}\\beta +c^{2}\\cos ^{2}\\beta }}\\,d\\beta }{{\\sqrt {a^{2}-b^{2}\\sin ^{2}\\beta -c^{2}\\cos ^{2}\\beta }}{\\sqrt {(b^{2}-c^{2})\\cos ^{2}\\beta -\\gamma }}}}\\\\&\\quad -\\int {\\frac {{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega }}\\,d\\omega }{{\\sqrt {a^{2}\\sin ^{2}\\omega +b^{2}\\cos ^{2}\\omega -c^{2}}}{\\sqrt {(a^{2}-b^{2})\\sin ^{2}\\omega +\\gamma }}}}.\\end{aligned}}} As Jacobi notes \"a function of the angle β equals a function", "$\\frac{ 1 - 2 \\sin^2 x \\cos^2 x + \\cos^2 x \\sin^2 x ( \\cos^2 x + \\sin^2 x)}{(1+ \\cos^2 x)(1+ \\sin^2 x) }$ $\\Big)$ $= \\Big($ $\\frac{ 1 - 2 \\sin^2 x \\cos^2 x + \\cos^2 x \\sin^2 x }{1 + \\cos^2 x + \\sin^x + \\cos^2 \\sin^2 x }$ $\\Big)$ $= \\Big($ $\\frac{ 1 - \\sin^2 x \\cos^2 x }{2 + \\cos^2 \\sin^2 x }$ $\\Big)$ $=$ RHS. Hence Proved. $\\\\$ Question 13: $(1 + \\tan \\alpha \\tan \\beta)^2 + (\\tan \\alpha - \\tan \\beta)^2 = \\sec^2 \\alpha \\sec^2 \\beta$ LHS $= (1 + \\tan \\alpha \\tan \\beta)^2 + (\\tan \\alpha - \\tan \\beta)^2$ $= 1 + \\tan^2 \\alpha \\tan^2 \\beta + 2\\tan \\alpha \\tan \\beta + \\tan^2 \\alpha + \\tan^2 \\beta - 2\\tan \\alpha \\tan \\beta$ $= 1 + \\tan^2 \\alpha \\tan^2 \\beta + \\tan^2 \\alpha + \\tan^2 \\beta$ $= 1 + \\tan^2 \\beta + \\tan^2 \\alpha ( 1 + \\tan^2 \\beta)$ $= ( 1 + \\tan^2 \\beta) ( 1 + \\tan^2 \\alpha)$ $= \\sec^2 \\alpha \\sec^2 \\beta$ $=$ RHS. Hence Proved. $\\\\$ Question 14: $\\frac{(1+\\cot x + \\tan x)(\\sin x - \\cos x)}{\\sec^3 x - \\mathrm{cosec}^3 x}$ $= \\sin^2 x \\cos^2 x$ LHS $=$ $\\frac{(1+\\cot x + \\tan x)(\\sin x - \\cos x)}{\\sec^3 x - \\mathrm{cosec}^3 x}$ $=$ $\\frac{(1+\\frac{\\cos x}{\\sin x} + \\frac{\\sin x}{\\cos", "+ \\ldots + \\sin (2n-1) x =$ $\\frac{\\sin^2 nx}{\\sin x}$ for all $n \\in N$ $\\\\$ Question 40: Prove that $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (n-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{n-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{n \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ Let $P(n) :$ $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (n-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{n-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{n \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ For $n = 1 , LHS = \\cos ( \\alpha + (1-1) \\beta) = \\cos \\alpha$ RHS $= \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{1-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{\\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ $= \\cos \\alpha$ Therefore LHS = RHS Hence $P(1)$ is true for $n = 1$ Let $P(n)$ is true for $n = k$ $\\cos \\alpha + \\cos ( \\alpha + \\beta) + \\cos ( \\alpha + 2\\beta) + \\ldots + \\cos ( \\alpha + (k-1) \\beta) = \\frac{\\cos \\big\\{ \\alpha + \\big( \\frac{k-1}{2} \\big) \\beta \\big\\} \\sin \\big( \\frac{k \\beta}{2} \\big) }{\\sin \\big( \\frac{\\beta}{2} \\big) }$ … i) Now we have to show", "= \\tan \\quad \\quad \\tan(\\pi - \\theta) = -\\tan \\theta$$ $$\\sin(\\theta + \\frac{\\pi}{2}) = \\cos \\theta \\quad \\quad \\sin(\\frac{\\pi}{2} - \\theta) = \\cos \\theta$$ $$\\cos(\\theta + \\frac{\\pi}{2}) = -\\sin \\theta \\quad \\quad \\cos(\\frac{\\pi}{2} - \\theta) = \\sin \\theta$$ $$\\tan(\\theta + \\frac{\\pi}{2}) = -\\frac{1}{\\tan \\theta} \\quad \\quad \\tan(\\frac{\\pi}{2} - \\theta) = \\frac{1}{\\tan \\theta}$$ # 正弦定理 $$\\sin (\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$$ $$\\sin (\\alpha - \\beta) = \\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta$$ $$\\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta$$ $$\\cos (\\alpha - \\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$$ $$\\tan (\\alpha + \\beta) = \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}$$ $$\\tan (\\alpha - \\beta) = \\frac{\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta}$$ 2倍角の公式 $$\\sin2 \\alpha = 2\\sin \\alpha \\cos \\alpha$$ $$\\cos2 \\alpha = \\cos^2 \\alpha - \\sin^2 \\alpha = 1 - 2\\sin^2 \\alpha = 2\\cos^2 \\alpha - 1$$ $$\\tan2 \\alpha = \\frac{2\\tan \\alpha}{1 - tan^2 \\alpha}$$ $$\\sin^2\\frac{\\alpha}{2} = \\frac{1 - \\cos2\\alpha}{2}, \\quad \\cos^2\\frac{\\alpha}{2} = \\frac{1 + \\cos2\\alpha}{2}$$ 3倍角の公式 $$\\sin3\\alpha = 3\\sin\\alpha - 4\\sin^3\\alpha$$ $$\\cos3\\alpha = 4\\cos^3\\alpha - 3\\cos\\alpha$$ $$\\tan3\\alpha = \\frac{3\\tan\\alpha - \\tan^3\\alpha}{1 - 3tan^2\\alpha}$$ # 和と積の変換公式 $$\\sin\\alpha\\cos\\beta = \\frac{1}{2}\\{\\sin(\\alpha + \\beta) + \\sin(\\alpha - \\beta)\\}$$ $$\\cos\\alpha\\sin\\beta = \\frac{1}{2}\\{\\sin(\\alpha +", "\\cos \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} = b$ … … … … … ii) Dividing i) by ii) we get $\\displaystyle \\tan \\frac{\\alpha + \\beta}{2} = \\frac{a}{b}$ $\\displaystyle \\text{We know } \\sin (\\alpha + \\beta) = \\frac{2 \\tan \\big( \\frac{\\alpha + \\beta}{2} \\big)}{1+ \\tan^2 \\big( \\frac{\\alpha + \\beta}{2} \\big)} = \\frac{2 \\frac{a}{b}}{1+\\frac{a^2}{b^2} } = \\frac{2ab}{a^2+b^2}$ $\\displaystyle ( \\sin \\alpha + \\sin \\beta)^2 + ( \\cos \\alpha + \\cos \\beta)^2 = a^2 +b^2$ $\\displaystyle \\Rightarrow \\sin^2 \\alpha + \\sin^2 \\beta + 2 \\sin \\alpha \\sin \\beta + \\cos^2 \\alpha + \\cos^2 \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\sin \\alpha \\sin \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 ( \\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\cos ( \\alpha - \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow \\cos ( \\alpha - \\beta) = \\frac{a^2 + b^2-2}{2}$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 39: } \\text{If } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} , \\text{ prove that } \\cos \\alpha = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$ $\\displaystyle \\text{Given } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} ,$ $\\displaystyle \\text{RHS } = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$", "\\omega^2 C_3\\cos{(\\alpha t)} + \\omega ^2 C_4\\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $(\\omega^2 C_3 - \\alpha ^2 C_3)\\cos{(\\alpha t)} + (\\omega ^2 C_4 - \\alpha ^2 C_4) \\sin{(\\alpha t)} = \\frac{\\alpha A}{L} \\cos{(\\alpha t)}$ $C_3(\\omega^2 - \\alpha^2) \\cos{(\\alpha t)} + C_4(\\omega ^2 - \\alpha ^2)\\sin{(\\alpha t)} = \\frac{\\alpha A}{L}\\cos{(\\alpha t)} + 0\\sin{(\\alpha t)}$. By equating like coefficients, we find $C_3(\\omega^2 - \\alpha^2) = \\frac{\\alpha A }{L}$ and $C_4(\\omega^2 - \\alpha ^2) = 0$ We can therefore see that $C_4 = 0$ and $C_3 = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)}$. So now we have our particular solution: $I_P = \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. Our general solution is the sum of the homogeneous solution and the particular solution. So $I = I_C + I_P$ $I = C_1\\cos{(\\omega t)} + C_2\\sin{(\\omega t)} + \\frac{\\alpha A}{L(\\omega^2 - \\alpha^2)} \\cos{(\\alpha t)}$. It also seems that you need to revise solving second order linear ODEs.", "\\alpha \\sin \\alpha \\tan^2 \\beta }{ \\cos^2 \\alpha-\\sin^2 \\alpha \\tan ^2 \\beta}$ $=\\frac{ \\sin \\alpha \\cos \\alpha (1+\\tan^2 \\beta) -\\tan \\beta (\\sin^2 \\alpha + \\cos^2 \\alpha )}{ \\cos^2 \\alpha - \\sin^2 \\alpha \\tan ^2 \\beta}$$=\\frac{ \\sin \\alpha \\cos \\alpha(\\sec^2 \\beta)- \\tan \\beta (1) }{ \\cos^2 \\alpha- \\sin^2 \\alpha \\tan^2 \\beta}$ 33. prizzyjade $=\\frac{ \\frac{ \\sin \\alpha \\cos \\alpha }{ \\cos^2 \\beta }- \\tan \\beta }{ \\cos^2 \\alpha - \\sin^2 \\alpha \\tan^2 \\beta }$ $=\\frac{\\frac{ \\sin \\alpha \\cos \\alpha }{ \\cos^2 \\beta} -\\frac{ \\sin \\beta}{ \\cos \\beta }}{\\cos^2\\alpha- \\frac{ \\sin^ 2 \\alpha \\sin^2 \\beta }{\\cos^2 \\beta } }$$=\\frac{ \\frac{ \\sin \\alpha \\cos \\alpha- \\sin \\beta \\cos \\beta }{ \\cos^2 \\beta } }{ \\frac{ \\cos^2 \\alpha \\cos^2 \\beta - \\sin^2 \\alpha \\sin ^2 \\beta }{\\cos^2 \\beta } }$$=\\frac{ \\sin \\alpha \\cos \\alpha -\\sin \\beta \\cos \\beta }{ \\cos^2 \\alpha \\cos^2 \\beta - \\sin^2 \\alpha \\sin^2 \\beta }$$=\\frac{ \\frac{ \\tan \\alpha }{ 1+\\tan^2 \\alpha } - \\frac{ \\tan \\beta }{ 1+ \\tan^\\beta } }{ 2-\\sin^2 \\alpha - \\sin^ 2 \\beta + \\sin^2 \\alpha \\sin^2 \\beta-\\sin^2 \\alpha \\sin^2 \\beta}$ 34. anonymous work from left, sub in the addition formula for tan(A+B) #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "# Math Help - To Find the Value of tan 1. ## To Find the Value of tan If $\\cos (\\alpha + \\beta) = \\frac{4}{5}, \\sin(\\alpha - \\beta) = \\frac{5}{13}$ and $\\alpha, \\beta \\in (0, \\frac{\\pi}{4})$, then find the value of $\\tan (2 \\alpha)$. 2. Originally Posted by kjchauhan If $\\cos (\\alpha + \\beta) = \\frac{4}{5}, \\sin(\\alpha - \\beta) = \\frac{5}{13}$ and $\\alpha, \\beta \\in (0, \\frac{\\pi}{4})$, then find the value of $\\tan (2 \\alpha)$. if $\\cos(\\alpha + \\beta) = \\frac{4}{5}$ , then $\\sin(\\alpha + \\beta) = \\frac{3}{5}$ if $\\sin(\\alpha - \\beta) = \\frac{5}{13}$ , then $\\cos(\\alpha - \\beta) = \\frac{12}{13}$ $\\sin(2\\alpha) = \\sin[(\\alpha + \\beta) + (\\alpha - \\beta)] = \\sin(\\alpha + \\beta)\\cos(\\alpha - \\beta) + \\cos(\\alpha + \\beta)\\sin(\\alpha - \\beta) $ $\\cos(2\\alpha) = \\cos[(\\alpha + \\beta) + (\\alpha - \\beta)] = \\cos(\\alpha + \\beta)\\cos(\\alpha - \\beta) - \\sin(\\alpha + \\beta)\\sin(\\alpha - \\beta)$ finally, note that ... $\\displaystyle \\tan(2\\alpha) = \\frac{\\sin(2\\alpha)}{\\cos(2\\alpha)}$ 3. Thank you very much..", "agree that is is correct (I've used $$\\K\\$$ instead of $$\\I_0\\$$). Here is how I derived my formula: $$\\I_0\\sin(\\omega t + \\beta) \\cdot U_0 \\sin(\\omega t + \\alpha) \\$$ = $$\\\\dfrac{I_0 U_0}{2}(2\\sin(\\omega t + \\beta) \\sin(\\omega t + \\alpha)) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(2(-\\cos(\\omega t + \\alpha + \\omega t + \\beta) + \\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(\\omega t + \\alpha + \\omega t + \\beta) + 2\\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) -\\cos(\\omega t + \\alpha + \\omega t + \\beta) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha) \\cos(\\omega t + \\beta) +\\sin(\\omega t + \\alpha) \\sin(\\omega t + \\beta) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha - (\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\omega t + \\alpha - (\\omega t + \\beta) ) ) = \\$$ $$\\\\dfrac{I_0 U_0}{2}(-\\cos(2\\omega t + \\alpha + \\beta) + \\cos(\\alpha - \\beta ) ) = (2.) \\$$ Regarding that only the difference in phase matters, I can't see how this can be. If we only imagine two sine curves being shifted back and forth along the time axis it seems obvious to me that the", "$\\sin^2\\beta+\\cos^2\\beta\\equiv1$sin2β+cos2β1 to deduce $\\cos\\beta$cosβ. Hence, $\\left(\\frac{5}{9}\\right)^2+\\cos^2\\beta=1$(59)2+cos2β=1 and so, $\\cos\\beta=\\sqrt{1-\\left(\\frac{5}{9}\\right)^2}=\\frac{2\\sqrt{14}}{9}$cosβ=1(59)2=2149 Now, $\\csc\\beta=\\frac{1}{\\sin\\beta}=\\frac{9}{5},$cscβ=1sinβ=95, $\\sec\\beta=\\frac{1}{\\cos\\beta}=\\frac{9}{2\\sqrt{14}}=\\frac{9\\sqrt{14}}{28},$secβ=1cosβ=9214=91428, $\\tan\\beta=\\frac{\\sin\\beta}{\\cos\\beta}=\\frac{5}{2\\sqrt{14}}=\\frac{5\\sqrt{14}}{28},$tanβ=sinβcosβ=5214=51428, $\\cot\\beta=\\frac{1}{\\tan\\beta}=\\frac{2\\sqrt{14}}{5}$cotβ=1tanβ=2145 #### Worked Examples ##### QUESTION 1 Find the value of $\\cos\\theta$cosθ if $\\sec\\theta=\\frac{8}{3}$secθ=83. ##### QUESTION 2 Find the value of $\\cot\\theta$cotθ if $\\tan\\theta=0.1$tanθ=0.1 . ##### QUESTION 3 If $\\sin\\alpha=\\frac{7}{25}$sinα=725 and $\\cos\\alpha=\\frac{24}{25}$cosα=2425, find: 1. $\\tan\\alpha$tanα 2. $\\cot\\alpha$cotα 3. $\\sec\\alpha$secα 4. $\\csc\\alpha$cscα ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "\\\\ i \\frac{d C_2}{d\\bar x} & = -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta\\bar x) C_2 + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_1 e^{i\\bar x}.\\end{split}$ To solve them numerically, we could decompose it into four real equations. $\\begin{split}- \\frac{d C_{1,I}}{d \\bar x} &= \\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} \\cos(\\bar x) + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} \\sin(\\bar x) \\\\ \\frac{d C_{1,R}}{d\\bar x}& = \\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,I} - \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} \\sin(\\bar x) + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} \\cos(\\bar x) \\\\ -\\frac{d C_{2,I}}{d\\bar x}&= -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,R} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R}\\cos \\bar x - \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,I}\\sin \\bar x \\\\ \\frac{d C_{2,R}}{d\\bar x} & = -\\frac{\\alpha \\cos 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{2,I} + \\frac{\\alpha \\sin 2\\theta_v}{2} \\cos (\\beta \\bar x) C_{1,R} \\sin \\bar x + \\frac{\\alpha \\sin 2\\theta_v}{2}\\cos (\\beta \\bar x) C_{1,I} \\cos \\bar x .\\end{split}$ Consequently, we have the matrix equation as the original equations, $\\begin{split}\\frac{d}{d\\bar x}\\begin{pmatrix} C_{1,R}\\\\ C_{1,I} \\\\ C_{2,R} \\\\ C_{2,I} \\end{pmatrix} = \\frac{1}{2}\\begin{pmatrix} 0 & \\alpha\\cos 2\\theta_v & - \\alpha \\sin 2\\theta_v\\sin \\bar x & \\alpha \\sin 2\\theta_v \\cos \\bar", "cos2A) sin2B = cos2A - cos2A sin2B - sin2B + cos2A sin2B = cos2A - sin2B (M.H.S.) Again, cos2A - sin2B = (1 - sin2A) - (1 - cos2B) = 1 - sin2A - 1 + cos2B = cos2B - sin2A ∴ L.H.S. = M.H.S. = R.H.S. Proved Given, θ =α +β sinθ = sin(α + β)............................(1) And $\\frac {tanα}{tanβ}$ = $\\frac xy$ $\\frac {tanα - tanβ}{tanα + tanβ}$ = $\\frac {x - y}{x + y}$.............................(2) Now, R.H.S. = $\\frac {x - y}{x + y}$ sinθ =$\\frac {tanα - tanβ}{tanα + tanβ}$ sinθ =$\\frac {\\frac {sinα}{cosα} -\\frac {sinβ}{cosβ}}{\\frac {sinα}{cosα} + \\frac {sinβ}{cosβ}}$ sinθ = $\\frac{\\frac{sin\\alpha.cos\\beta - cos\\alpha.sin\\beta}{cos\\alpha.cos\\beta}}{\\frac{sin\\alpha.cos\\beta + cos\\alpha.sin\\beta}{cos\\alpha.cos\\beta}}$. sin$\\theta$ = $\\frac {sin\\alpha cos\\beta - cos\\alpha sin\\beta}{sin\\alpha cos\\beta + cos\\alpha sin\\beta}$ . sin$\\theta$ = $\\frac {sin (\\alpha - \\beta)}{sin(\\alpha + \\beta)}$ . sin$\\theta$ = $\\frac {sin(\\alpha - \\beta)}{sin \\theta}$. sin$\\theta$ = sin (α - β) Hence, L.H.S. = R.H.S. Proved L.H.S. = sin (A + B + C) = sin {A + (B + C)} = sinA . cos(B + C) + cos A . cos (B + C) = sinA {cosB cosC - sinB sinC} + cosA {sinB cosC + cosB sinC} = sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC = cosA cosB cosC[$\\frac {sinA cosB cosC - sinA", "\\alpha = 3 \\tan \\beta \\Rightarrow \\tan \\alpha = \\frac{3}{2} \\tan \\beta$ $\\displaystyle \\text{LHS } = \\tan (\\alpha - \\beta)$ $\\displaystyle = \\frac{\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta}$ $\\displaystyle = \\frac{\\frac{3}{2} \\tan \\beta - \\tan \\beta}{1 + \\frac{3}{2} \\tan \\beta \\tan \\beta}$ $\\displaystyle = \\frac{\\tan \\beta}{2 + 3 \\tan^2 \\beta}$ $\\displaystyle = \\frac{\\frac{\\sin \\beta}{\\cos \\beta}}{2 + 3 \\big( \\frac{\\sin \\beta}{\\cos \\beta} \\big)^2}$ $\\displaystyle = \\frac{\\sin \\beta \\cos \\beta}{2 \\cos^2 \\beta + 3 \\sin^2 \\beta}$ $\\displaystyle = \\frac{\\sin \\beta \\cos \\beta}{2 + \\sin^2 \\beta}$ $\\displaystyle = \\frac{2\\sin \\beta \\cos \\beta}{4 + 2 \\sin^2 \\beta}$ $\\displaystyle = \\frac{\\sin 2\\beta}{4 + 2 (1- \\cos^2 \\beta)}$ $\\displaystyle = \\frac{\\sin 2\\beta}{6 - 2 \\cos^2 \\beta} = \\text{ RHS. Hence proved. }$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 38: } \\text{If } \\sin \\alpha + \\sin \\beta = a \\text{ and } \\cos \\alpha + \\cos \\beta = b , \\text{ prove that: }$ $\\displaystyle \\text{i) } \\sin (\\alpha + \\beta) = \\frac{2ab}{a^2 + b^2}$ $\\displaystyle \\text{ii) } \\cos (\\alpha - \\beta) = \\frac{a^2 + b^2 - 2}{2}$ $\\displaystyle \\text{i) } \\text{Given } \\sin \\alpha + \\sin \\beta = a$ $\\displaystyle \\Rightarrow 2 \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} = a$ … … … … … i) $\\displaystyle \\text{Given } \\cos \\alpha + \\cos \\beta = b$ $\\displaystyle \\Rightarrow 2", "– \\cos^2\\alpha – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma)}{\\sin^2\\beta}$$ $$=\\frac{a^2(1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma)}{\\sin^2\\beta}$$. Thus, $$h = \\frac{a}{\\sin\\beta}\\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$. Finally, $$V = \\frac{Ah}{3} = \\frac{b\\cdot c\\cdot \\sin\\beta \\cdot a}{3 \\cdot 2 \\cdot \\sin\\beta} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$ $$=\\frac{abc}{6} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$$. So, given the three surface angles and three edge lengths from a given vertex of a tetrahedron, the volume of the tetrahedron is: $V=\\frac{abc}{6} \\sqrt{1-\\cos^2\\alpha – \\cos^2\\beta – \\cos^2\\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma}$ Clio Corvid 1 Comment 1. Carlos Eduardo Olivieri says: Very good! Congrats!", "# If tanα= Question: If $\\tan \\alpha=\\frac{1}{7}, \\tan \\beta=\\frac{1}{3}$, then $\\cos 2 \\alpha$ is equal to (a) $\\sin 2 \\beta$ (b) $\\sin 4 \\beta$ (c) $\\sin 3 \\beta$ (d) $\\cos 2 \\beta$ Solution: It is given that $\\tan \\alpha=\\frac{1}{7}$ and $\\tan \\beta=\\frac{1}{3}$. Now, $\\tan 2 \\beta=\\frac{2 \\tan \\beta}{1-\\tan ^{2} \\beta}$ $=\\frac{2 \\times \\frac{1}{3}}{1-\\frac{1}{9}}$ $=\\frac{\\frac{2}{3}}{\\frac{8}{9}}$ $=\\frac{3}{4}$ $\\therefore \\tan (\\alpha+2 \\beta)=\\frac{\\tan \\alpha+\\tan 2 \\beta}{1-\\tan \\alpha \\tan 2 \\beta}$ $=\\frac{\\frac{1}{7}+\\frac{3}{4}}{1-\\frac{1}{7} \\times \\frac{3}{4}}$ $=\\frac{\\frac{25}{28}}{\\frac{25}{28}}$ $=1$ $\\tan (\\alpha+2 \\beta)=1=\\tan \\frac{\\pi}{4}$ $\\Rightarrow \\alpha+2 \\beta=\\frac{\\pi}{4}$ $\\Rightarrow \\alpha=\\frac{\\pi}{4}-2 \\beta$ $\\Rightarrow 2 \\alpha=\\frac{\\pi}{2}-4 \\beta$ $\\Rightarrow \\cos 2 \\alpha=\\cos \\left(\\frac{\\pi}{2}-4 \\beta\\right)=\\sin 4 \\beta$ $\\therefore \\cos 2 \\alpha=\\sin 4 \\beta$ Hence, the correct answer is option $B$.", "$(iv)\\,\\,\\cos^3\\alpha+\\cos^3(120^{\\circ}+\\alpha)+\\cos^3(240^{\\circ}+\\alpha)=\\frac 34 \\cos 3\\alpha$ Sol. $\\,\\,\\cos^3\\alpha+\\cos^3(120^{\\circ}+\\alpha)+\\cos^3(240^{\\circ}+\\alpha)\\\\=\\frac 14 [\\cos 3 \\alpha+3\\cos\\alpha+\\cos(360^{\\circ}+3\\alpha)\\\\+3\\cos(120^{\\circ}+\\alpha)+\\cos(720^{\\circ}+3\\alpha)\\\\+3\\cos(240^{\\circ}+\\alpha)]\\\\=\\frac 14[3\\cos3\\alpha+3\\cos\\alpha+3\\{\\cos(120^{\\circ}+\\alpha)\\\\+\\cos(240^{\\circ}+\\alpha)\\}]\\\\=\\frac 14[3\\cos 3\\alpha+3\\cos \\alpha+3.2\\cos(180^{\\circ}+\\alpha) \\\\ \\times \\cos60^{\\circ}]\\\\=\\frac 34\\cos 3\\alpha$ Additional Notes : $\\,4\\cos^3(120^{\\circ}+\\alpha)=\\cos [3(120^{\\circ}+\\alpha)]\\\\+3 \\cos(120^{\\circ}+\\alpha),\\\\ 4\\cos^3(240^{\\circ}+\\alpha)=\\cos [3(240^{\\circ}+\\alpha)]\\\\+3 \\cos(240^{\\circ}+\\alpha), \\\\ \\cos 3 \\alpha=\\cos(360^{\\circ}+3\\alpha)=\\cos(720^{\\circ}+3\\alpha)$ $\\,(v)\\, \\tan 70^{\\circ}-\\tan 50^{\\circ}+\\tan 10^{\\circ}=\\sqrt3$ Sol. $\\,\\tan 70^{\\circ}-\\tan 50^{\\circ}+\\tan 10^{\\circ}\\\\=\\tan (60^{\\circ}+10^{\\circ})-\\tan(60^{\\circ}-10^{\\circ})+\\tan 10^{\\circ}\\\\=\\frac{\\tan 60^{\\circ}+\\tan 10^{\\circ}}{1-\\tan 60^{\\circ} \\tan 10^{\\circ}}-\\frac{\\tan 60^{\\circ}-\\tan 10^{\\circ}}{1+\\tan 60^{\\circ}\\tan 10^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{\\sqrt3+\\tan 10^{\\circ}}{1-\\sqrt3\\tan 10^{\\circ}}-\\frac{\\sqrt3-\\tan 10^{\\circ}}{1+\\sqrt3\\tan 10^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{\\sqrt3+3\\tan 10^{\\circ}+\\tan 10^{\\circ}+\\sqrt3 \\tan^210^{\\circ}-(\\sqrt3-3\\tan 10^{\\circ}\\\\-\\tan 10^{\\circ}+\\sqrt3 \\tan^210^{\\circ})}{1-3\\tan^210^{\\circ}}\\\\+\\tan 10^{\\circ}\\\\=\\frac{8\\tan 10^{\\circ}}{1-3\\tan^210^{\\circ}}+\\tan 10^{\\circ}\\\\=\\frac{9 \\tan 10^{\\circ}-3\\tan^3 10^{\\circ}}{1-3\\tan^210^{\\circ}} \\\\=3\\left( \\frac{3\\tan 10^{\\circ}-3\\tan^310^{\\circ}}{1-3\\tan^210^{\\circ}}\\right)\\\\=3\\tan30^{\\circ}\\\\=\\sqrt3$ $\\,(vi)\\, \\tan \\theta \\tan(\\frac{\\pi}{3}+\\theta)\\tan(\\frac{\\pi}{3}-\\theta)=\\tan 3\\theta$ Sol. $\\,\\, \\tan \\theta \\tan(\\frac{\\pi}{3}+\\theta)\\tan(\\frac{\\pi}{3}-\\theta)\\\\=\\tan \\theta . \\frac{\\sqrt3+\\tan \\theta}{1-\\sqrt3 \\tan \\theta}. \\frac{\\sqrt3-\\tan \\theta}{1+\\sqrt3 \\tan \\theta}\\\\=\\frac{3\\tan \\theta-\\tan^3\\theta}{1-3\\tan^2\\theta}\\\\=\\tan \\theta$ $\\,(vii)\\, \\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+8\\cot 8\\theta=\\cot \\theta$ Sol. $\\, \\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+8\\cot 8\\theta\\\\=\\tan \\theta+2\\tan 2\\theta+4 \\tan 4\\theta+\\frac{8(1-\\tan^2 4\\theta)}{2\\tan 4\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta+\\frac{8\\tan^24\\theta+8-8\\tan^24\\theta}{2\\tan 2\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta+\\frac{4}{\\tan 2\\theta}\\\\=\\tan \\theta+2\\tan 2\\theta +\\frac{4(1-\\tan^2 2\\theta)}{2\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{4 \\tan^2 2\\theta+4-4\\tan^2 2\\theta}{2\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{2}{\\tan 2\\theta}\\\\=\\tan \\theta+\\frac{2(1-\\tan^2\\theta)}{2\\tan \\theta}\\\\=\\cot \\theta$ $\\,(viii)\\, \\cos \\frac{\\pi}{11}\\cos \\frac{2\\pi}{11}\\cos \\frac{3\\pi}{11}\\cos \\frac{4\\pi}{11}\\cos \\frac{5\\pi}{11} =\\frac{1}{32}$ Sol. Let $\\,\\,\\frac{\\pi}{11}=\\theta \\Rightarrow \\theta=11 \\pi \\\\ \\therefore P=\\cos \\theta \\cos 2\\theta \\cos 3\\theta \\cos 4\\theta \\cos 5\\theta \\,\\,,\\text{say}\\\\ \\Rightarrow 2P\\sin \\theta=(2 \\sin \\theta\\cos \\theta) \\cos 2 \\theta \\cos3 \\theta \\cos 4 \\theta\\cos 5 \\theta \\\\ \\Rightarrow 2P\\sin \\theta=\\sin 2 \\theta\\cos 2 \\theta\\cos3 \\theta\\cos4 \\theta\\cos5 \\theta \\\\ \\Rightarrow 2^2P\\sin \\theta=2\\sin 2 \\theta\\cos 2 \\theta\\cos3 \\theta\\cos4 \\theta\\cos5 \\theta \\\\ \\Rightarrow 2^2P \\sin \\theta=\\sin 4\\theta \\cos 4\\theta \\cos 3\\theta \\cos 5\\theta \\\\ \\Rightarrow" ]

[ "+0 # Halp 0 94 1 The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations \\begin{align*} x &= at + b, \\\\ y &= ct + d, \\end{align*} where $0 \\le t \\le 1,$ and $t = 0$ corresponds to the point $(-2,7).$ Find $a^2 + b^2 + c^2 + d^2.$ Jan 6, 2022", "+0 # Halp 0 129 1 The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations \\begin{align*} x &= at + b, \\\\ y &= ct + d, \\end{align*} where $0 \\le t \\le 1,$ and $t = 0$ corresponds to the point $(-2,7).$ Find $a^2 + b^2 + c^2 + d^2.$ Jan 6, 2022 ### 1+0 Answers #1 0 The parametric equations are x = 3t - 2 and y = 5t + 7, so a^2 + b^2 + c^2 + d^2 = 3^2 +(-2)^2 + 5^2 + 7^2 = 87. Jan 6, 2022", "them each equal to zero: \\begin{align*} f_x(x,y) &=2x−2y−4 \\\\[4pt] f_y(x,y) &=−2x+8y−2. \\end{align*} Setting them equal to zero yields the system of equations \\begin{align*} 2x−2y−4 &=0\\\\[4pt] −2x+8y−2 &=0. \\end{align*} The solution to this system is $$x=3$$ and $$y=1$$. Therefore $$(3,1)$$ is a critical point of $$f$$. Calculating $$f(3,1)$$ gives $$f(3,1)=17.$$ The next step involves finding the extrema of $$f$$ on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph: $$L_1$$ is the line segment connecting $$(0,0)$$ and $$(4,0)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=0$$ for $$0≤t≤4$$. Define $$g(t)=f\\big(x(t),y(t)\\big)$$. This gives $$g(t)=t^2−4t+24$$. Differentiating $$g$$ leads to $$g′(t)=2t−4.$$ Therefore, $$g$$ has a critical value at $$t=2$$, which corresponds to the point $$(2,0)$$. Calculating $$f(2,0)$$ gives the $$z$$-value $$20$$. $$L_2$$ is the line segment connecting $$(4,0)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=4,y(t)=t$$ for $$0≤t≤2.$$ Again, define $$g(t)=f\\big(x(t),y(t)\\big).$$ This gives $$g(t)=4t^2−10t+24.$$ Then, $$g′(t)=8t−10$$. g has a critical value at $$t=\\frac{5}{4}$$, which corresponds to the point $$\\left(0,\\frac{5}{4}\\right).$$ Calculating $$f\\left(0,\\frac{5}{4}\\right)$$ gives the $$z$$-value $$27.75$$. $$L_3$$ is the line segment connecting $$(0,2)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=2$$ for $$0≤t≤4.$$ Again, define $$g(t)=f\\big(x(t),y(t)\\big).$$ This gives $$g(t)=t^2−8t+36.$$ The critical value corresponds to the point $$(4,2).$$ So, calculating $$f(4,2)$$ gives the $$z$$-value $$20$$.", "# Cubic Bezier curve and a straight line intersection Suppose that two ends of a cubic Bezier curve is connected by a straight line. Is there a simple way to find out whether this straight line intersects the Bezier curve (apart from the endpoints)? If it intersects then what will be the corresponding Bezier curve parameter's value? Given 2D cubic Bezier segment $\\mathcal{B}(t,A,B,C,D)=A\\,(1-t)^3+3B\\,(1-t)^2t+3C\\,(1-t)t^2+D\\,t^3$, $t\\in[0,1]$ and line segment $\\mathcal{L}(t,A,D)=A\\,(1-s)+D\\,s$, $s\\in[0,1]$, the value of $t:\\mathcal{B}(t,A,B,C,D)=\\mathcal{L}(s,A,D),\\ t\\ne0, t\\ne1$ is the same as the value of $t:\\mathcal{B}(t,0,B-A,C-A,D-A)=\\mathcal{L}(s,0,D-A)$. So, with a substitution $b=B-A,\\ c=C-A,\\ d=D-A$ we can solve a system of two equations with two unknowns $t,s$: \\begin{align} \\mathcal{B}(t,0,b_x,c_x,d_x) &= \\mathcal{L}(s,0,d_x) \\\\ \\mathcal{B}(t,0,b_y,c_y,d_y) &= \\mathcal{L}(s,0,d_y) \\end{align} which gives the value of parameter $t$ as \\begin{align} t &= \\frac{b_x\\,d_y-d_x\\,b_y}{b_x\\,d_y-c_x\\,d_y-d_x\\,b_y+d_x\\,c_y} \\end{align} If $0<t<1$ than the intersection point of $\\mathcal{B}$ and $\\mathcal{L}$ is $X=\\mathcal{B}(t,A,B,C,D)$. • It is very explanatory and convincing, thanks. For a perfection I would like to propose a little algebraic correction: The line segment should be L(t,A,D). Jun 24 '15 at 18:50 • @John Railman: Of course it is $AD$, great thanks for pointing this, fixed. Jun 24 '15 at 18:59 I just wanted to add to g.kov's good derivation, that the $t$ value is the parameter for the Bézier segment, not the line, so the range of $0 < t < 1$", "$(x_1, y_1)$ and $(x_2, y_2)$ \\begin{aligned} x(t) &= x_1 + t(x_2 - x_1)\\\\ y(t) &= y_1 + t(y_2 - y_1) \\end{aligned} This parameterization has the property that for values $0 \\leq t \\leq 1$, $(x(t), y(t))$ is on the line segment, between the two endpoints. On the other hand, if t is less than zero, or t is greater than one, $(x(t), y(t))$ is outside of the two endpoints. Below are points (from left to right) for t = -0.5 in red, t = 0, 0.25, 0.5, 0.75, and 1 in green, and for t = 1.5 in red again: At what point do two line segments intersect? Lines A and B intersect for parametric values s and t, such that \\begin{aligned} A_x(s) &= B_x(t)\\\\ A_y(s) &= B_y(t) \\end{aligned} I’ll use the parametric equation of lines A and B and solve for the values s and t for which the equality is satisfied. \\begin{aligned} A_{x1} + s(A_{x2} - A_{x1}) &= B_{x1} + t(B_{x2} - B_{x1})\\\\ A_{y1} + s(A_{y2} - A_{y1}) &= B_{y1} + t(B_{y2} - B_{y1}) \\end{aligned} then \\begin{aligned} s(A_{x2} - A_{x1}) - t(B_{x2} - B_{x1}) &= B_{x1} - A_{x1}\\\\ s(A_{y2} - A_{y1}) - t(B_{y2} - B_{y1}) &= B_{y1} - A_{y1} \\end{aligned} In matrix form $\\begin{bmatrix} (A_{x2} - A_{x1}) & -(B_{x2} - B_{x1}) \\\\ (A_{y2} - A_{y1}) &", "to zero: [\begin{align*} f_x(x,y) &=2x−2y−4 [4pt] f_y(x,y) &=−2x+8y−2. end{align*}] Setting them equal to zero yields the system of equations [\begin{align*} 2x−2y−4 &=0[4pt] −2x+8y−2 &=0. end{align*}] The solution to this system is (x=3) and (y=1). Therefore ((3,1)) is a critical point of (f). Calculating (f(3,1)) gives (f(3,1)=17.) The next step involves finding the extrema of (f) on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph: (L_1) is the line segment connecting ((0,0)) and ((4,0)), and it can be parameterized by the equations (x(t)=t,y(t)=0) for (0≤t≤4). Define (g(t)=f\big(x(t),y(t)\big)). This gives (g(t)=t^2−4t+24). Differentiating (g) leads to (g′(t)=2t−4.) Therefore, (g) has a critical value at (t=2), which corresponds to the point ((2,0)). Calculating (f(2,0)) gives the (z)-value (20). (L_2) is the line segment connecting ((4,0)) and ((4,2)), and it can be parameterized by the equations (x(t)=4,y(t)=t) for (0≤t≤2.) Again, define (g(t)=f\big(x(t),y(t)\big).) This gives (g(t)=4t^2−10t+24.) Then, (g′(t)=8t−10). g has a critical value at (t=frac{5}{4}), which corresponds to the point (left(0,frac{5}{4} ight).) Calculating (fleft(0,frac{5}{4} ight)) gives the (z)-value (27.75). (L_3) is the line segment connecting ((0,2)) and ((4,2)), and it can be parameterized by the equations (x(t)=t,y(t)=2) for (0≤t≤4.) Again, define (g(t)=f\big(x(t),y(t)\big).) This gives (g(t)=t^2−8t+36.) The critical value corresponds to the point ((4,2).) So, calculating (f(4,2)) gives the (z)-value (20). (L_4)", "# Lines and Planes A line starts at an initial point and proceeds straight in a constant direction. Thus, we can write the equation of a line as \\begin{align*} x(t)=mt+b \\end{align*} where $b = \\left< b_1, b_2, \\ldots, b_N \\right>$ is the initial point, $m = \\left< m_1, m_2, \\ldots, m_N \\right>$ is the constant direction in which the line travels, and $x(t) = \\left< x_1(t), x_2(t), \\ldots, x_N(t) \\right>$ is the point reached by traveling $t$ units away from $b$ in the direction of $m$. (Though $x(t)$ is actually a vector, we can also refer to it as the point where the vector lands when the vector is placed at the origin.) For example, to compute the line between the points $(1,2,3,4)$ and $(5,-2,3,7)$ in 4-dimensional space, we can start by computing the direction $m$ as the difference between the two points: \\begin{align*} m &= \\left< 5,-2,3,7 \\right> - \\left< 1,2,3,4 \\right> \\\\ &= \\left< 4,-4,0,3 \\right> \\end{align*} Taking $b=\\left< 1,2,3,4 \\right>$ as our initial point, then, we can express the line as \\begin{align*} x(t) = \\left< 4,-4,0,3 \\right> t + \\left< 1,2,3,4 \\right>. \\end{align*} If we wanted to find another point on the line, we could substitute another value for $t$, say, $t=-2$. \\begin{align*} x(2) &= \\left< 4,−4,0,3 \\right>(−2) + \\left< 1,2,3,4 \\right> \\\\ &= \\left<", "0$ from the point $(0, 0, 0)$ to $(3, 1, -2)$. Note that $x - y + z = 0$ and $x + y + 2z = 0$ are both planes and either intersect at a straight line, do not intersect, or coincide. It's not hard to see that these planes in fact intersect at a straight line. We can thus parameterize the line segment from $(0, 0, 0)$ to $(3, 1, -2)$ for $0 ≤ t ≤ 1$ as: (3) \\begin{align} \\quad \\vec{r}(t) = (1 - t)(0, 0, 0) + t(3, 1, -2) = (3t, t, -2t) \\end{align} Thus: (4) \\begin{align} \\quad \\int_C x^2 \\: ds = \\int_0^1 (3t)^2 \\sqrt{(3)^2 + (1)^2 + (-2)^2} \\: dt \\\\ \\quad \\int_C x^2 \\: ds = \\int_0^1 9 \\sqrt{14} t^2 \\: dt \\\\ \\quad \\int_C x^2 \\: ds = 9 \\sqrt{14} \\left [ \\frac{t^3}{3} \\right ]_{t=0}^{t=1} \\\\ \\quad \\int_C x^2 \\: ds = 3 \\sqrt{14} \\end{align}", "= 1 \\\\ \\quad z = 1 - t - t^2 \\end{align} Therefore a parameterization of the curve of intersection of these two surfaces is given by $\\vec{r}(t) = (t, 1 + t, 1 - t - t^2)$. ## Example 2 Parameterize the curve of intersection of the paraboloid $z = x^2 + y^2$ and the plane $2x - 4y - z = 1$. The second equation can be rewritten as $z = 2x - 4y - 1$. Substituting this into the first equation gives us: (4) \\begin{align} \\quad x^2 + y^2 = 2x - 4y -1 \\\\ \\quad x^2 - 2x + y^2 + 4y = -1 \\\\ \\quad (x - 1)^2 + (y + 2)^2 = 4 \\end{align} Let $x = 2\\cos t + 1$ and $y = 2 \\sin t - 2$. Then $z = 2x - 4y - 1$ implies that: (5) \\begin{align} \\quad z = 2(2 \\cos t + 1) - 4(2 \\sin t - 2) - 1 \\\\ \\quad z= 4\\cos t + 2 - 8 \\sin t + 8 - 1 \\\\ \\quad z = 4 \\cos t - 8 \\sin t + 9 \\end{align} Therefore the curve of intersection of the original two surfaces can be parameterized as $\\vec{r}(t) = (2\\cos t + 1, 2 \\sin t - 2, 4", "$0\\le t\\le1$ then there is an intersection along the line segment. Do this for all four line segments, and if any intersects the circle, then the rectangle intersects the circle. The above equation originated from the equation of the circle, with a generic point on the line (segment) substituted for $x$ and $y$. \\begin{align*} x &= (1-t)\\cdot x_1 + t\\cdot x_2 \\\\ y &= (1-t)\\cdot y_1 + t\\cdot y_2 \\\\ r^2 &= (x-x_C)^2 + (y-y_C)^2 \\end{align*} In contrast to this solution, I use $t=0$ to represent the preceived starting point $(x_1,y_1)$ of the line, and $t=1$ for the end point $(x_2,y_2)$. This choice has no impact on your problem, as the direction of your line segments is irrelevant. But it is more useful or common for other applications. If you have trouble understanding these parametric equations for $x$ and $y$, I suggest you simply choose end points with simple coordinates, and plug in some values of $t$, both from within and outside the specified range. You will see how all of these lie on the connecting line, and the range $0\\le t\\le 1$ corresponds to the line segment. Note that all of the above won't deal with the scenario where the circle is fully contained within the rectangle, or vice versa. So you might want to check whether", "+0 0 44 1 A segment with endpoints at $A(2, -2)$ and $B(14, 4)$ is extended through $B$ to point $C$. If $BC = \\frac{1}{3} \\cdot AB$, what are the coordinates for point $C$? Express your answer as an ordered pair. Mar 14, 2021", "$C$ is a. the straight-line segment $x = t , y = t / 2 ,$ from $( 0,0 )$ to $( 4,2 ) .$ b. the parabolic curve $x = t , y = t ^ { 2 } ,$ from $( 0,0 )$ to $( 2,4 )$ Regina H. ### Problem 20 Evaluate $\\int _ { C } \\sqrt { x + 2 y } d s ,$ where $C$ is a. the straight-line segment $x = t , y = 4 t ,$ from $( 0,0 )$ to $( 1,4 )$ . b. $C _ { 1 } \\cup C _ { 2 } ; C _ { 1 }$ is the line segment from $( 0,0 )$ to $( 1,0 )$ and $C _ { 2 }$ is the line segment from $( 1,0 )$ to $( 1,2 )$ . YZ Yiming Z. ### Problem 21 Find the line integral of $f ( x , y ) = y e ^ { x ^ { 2 } }$ along the curve $\\mathbf { r } ( t ) = 4 t \\mathbf { i } - 3 t \\mathbf { j } , - 1 \\leq t \\leq 2$ Regina H. ### Problem 22 Find the line integral of $f ( x , y", "# A line segment with endpoints at (5 , -9 ) and (2, -7 ) is rotated clockwise by pi/2. What are the new endpoints of the line segment? The endpoints of the new line segment are $\\left(9 , 5\\right)$ and $\\left(7 , 2\\right)$. Rotate each point $\\frac{\\pi}{2}$ radians, so I'll start with $\\left(2 , - 7\\right)$ because it's closer to the origin. $\\left(2 , - 7\\right)$ becomes $\\left(7 , 2\\right)$ $\\left(5 , - 9\\right)$ becomes $\\left(9 , 5\\right)$", "+0 # coordinates 0 118 1 A line segment of length 5 has one endpoint at (1,2) and the other endpoint at (6,b). Find all possible values of b, separated by commas. Jul 5, 2022 #1 +2602 0 Using the distance formula yields: $$5 = \\sqrt{({6 - 1})^2 + ({2 - b})^2}$$ Simplifying this equation gives us $$25 = 25 + (2-b)^2$$, meaning $$(2-b)^2 = 0$$ Can you take it from here? Jul 5, 2022", "segment EF is perpendicular to the line segment OA. 5. b is the distance from O to F. 6. c is the distance from F to A. 7. d is the distance from O to B. 8. $θθ$ is the measure of angle $∠COA.∠COA.$ The goal of this project is to parameterize the witch using $θθ$ as a parameter. To do this, write equations for x and y in terms of only $θ.θ.$ 2. Show that $d=2asinθ.d=2asinθ.$ 3. Note that $x=dcosθ.x=dcosθ.$ Show that $x=2acotθ.x=2acotθ.$ When you do this, you will have parameterized the x-coordinate of the curve with respect to $θ.θ.$ If you can get a similar equation for y, you will have parameterized the curve. 4. In terms of $θ,θ,$ what is the angle $∠EOA?∠EOA?$ 5. Show that $b+c=2acos(π2−θ).b+c=2acos(π2−θ).$ 6. Show that $y=2acos(π2−θ)sinθ.y=2acos(π2−θ)sinθ.$ 7. Show that $y=2asin2θ.y=2asin2θ.$ You have now parameterized the y-coordinate of the curve with respect to $θ.θ.$ 8. Conclude that a parameterization of the given witch curve is $x=2acotθ,y=2asin2θ,−∞<θ<∞.x=2acotθ,y=2asin2θ,−∞<θ<∞.$ 9. Use your parameterization to show that the given witch curve is the graph of the function $f(x)=8a3x2+4a2.f(x)=8a3x2+4a2.$ ### Student Project #### Travels with My Ant: The Curtate and Prolate Cycloids Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a", "line segment joining the points $(0,2)$ and $(3,0)$ The correct option is $D$", "we can use $t$ as a scale factor: $d(A, P) = t * d(A, B)$, where $d$ denotes the euclidean distance between two points. We will use that fact when determining the closest intersection point. Now we can easily see, that if we want the point to be contained on: • line: $t \\in \\mathbb{R}$, • ray: $t \\geq 0$, • line-segment: $0 \\leq t \\leq 1$. Having all of that sorted out, we can finally derive the parametric equation: \\begin{align*} \\overrightarrow{AP} &= t \\overrightarrow{AB} \\\\ P - A &= t (B - A) \\\\ P &= t(B - A) + A \\\\ \\end{align*} If you still do not know what happens here, I can recommend an awesome short lecture by Norm Prokup: Parametrizing a Line Segment - Concept. Calculating the intersection point Assume that point $P$ is the intersection point of line-segment defined by points $A$ and $B$, and ray defined by points $C$ and $D$. Point $P$ is then expressed by set of two equations: $\\begin{cases} P = s(B - A) + A\\text {, where } 0 \\leq s \\leq 1 \\\\ P = r(D - C) + C\\text {, where } r \\geq 0 \\end{cases}$ Solve for and Substitute into the second equation Substitute into the first equation Having $s$ and $r$ calculated, we can", "of two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ on each of these lines segments, we can use our usual slope formula in each case. \\begin{align}m=\\frac{y_2-y_1}{x_2-x_1}\\end{align} For the line segment $$AB$$ with endpoints $$(1,2)$$ and $$(3,5)$$, this formula gives \\begin{align}m&=\\frac{y_2-y_1}{x_2-x_1}\\\\&=\\frac{5-2}{3-1}\\\\&=\\frac{3}{2}\\\\&=1.5\\end{align} For the line segment $$BC$$ with endpoints $$(3,5)$$ and $$(5,4)$$, we obtain \\begin{align}m&=\\frac{y_2-y_1}{x_2-x_1}\\\\&=\\frac{4-5}{5-3}\\\\&=\\frac{-1}{2}\\\\&=-0.5\\end{align} For the line segment $$CD$$ with endpoints $$(5,4)$$ and $$(3,1)$$, we obtain \\begin{align}m&=\\frac{y_2-y_1}{x_2-x_1}\\\\&=\\frac{1-4}{3-5}\\\\&=\\frac{-3}{-2}\\\\&=1.5\\end{align} Finally, for the line segment $$DA$$ with endpoints $$(3,1)$$ and $$(1,2)$$, we obtain \\begin{align}m&=\\frac{y_2-y_1}{x_2-x_1}\\\\&=\\frac{2-1}{1-3}\\\\&=\\frac{1}{-2}\\\\&=-0.5\\end{align} (c) $$ABCD$$ is a parallelogram only if all pairs of opposite sides are parallel to each other, i.e. if all opposite sides have the same slope. Both $$AB$$ and $$CD$$ have the same slope of $$1.5$$. Both $$BC$$ and $$DA$$ have the same slope of $$-0.5$$. Therefore, as all opposite sides have the same slope, this quadrilateral is indeed a parallelogram. (d) $$ABCD$$ is also a rectangle if all adjacent sides are perpendicular to each other. To determine if this is the case, we can use the fact that two lines of slopes $$m_1$$ and $$m_2$$ respectively are perpendicular to each other only if they satisfy the following condition: \\begin{align}m_1\\times m_2=-1\\end{align} For example, the product of the slopes of the adjacent line segments $$AB$$ and $$BC$$ is \\begin{align}m_1\\times m_2 &=1.5\\times -0.5\\\\&=-0.75\\end{align} As $$m_1 \\times m_2 \\neq -1$$, we don’t have to", "+0 # Consider the line through points $A = (1, 1, 2)$ and $B = (2, 3, 4)$. This line can be parametrized as $(at +b, ct +d, et +f)$for some cho 0 155 1 Consider the line through points $$A = (1, 1, 2)$$ and .$$B = (2, 3, 4)$$ This line can be parametrized $$(at +b, ct +d, et +f)$$ as for some choices of constants $$a,b, c, d, e$$ and f. What is the ordered pair $$\\left(\\dfrac{c}{a}, \\dfrac{e}{a}\\right)$$? Mar 15, 2019", "\\cos t - 8 \\sin t + 9)$. ## Example 3 Parameterize the curve of intersection of the top unit hemisphere $z = \\sqrt{1 - x^2 - y^2}$ and the plane $x + y = 1$. Let $x = t$. Then the second equation has that $y = 1 - t$. Plugging these values of $x$ and $y$ into the second equation gives us: (6) \\begin{align} \\quad z = \\sqrt{1 - t^2 - (1 - t)^2} = \\sqrt{1 - t^2 - 1 + 2t - t^2} = \\sqrt{2t - 2t^2} = \\sqrt{2(t - t^2)} \\end{align} Therefore the curve of intersection of the original two surfaces can be parameterized as $\\vec{r}(t) = (t, 1 - t, \\sqrt{2(t - t^2)} )$. Note that $t - t^2 ≥ 0$, that is $t ≥ t^2$. This happens for $0 ≤ t ≤ 1$." ]

[ "+ y^2)$ $(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$ then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides. (Under the change of notation $a,b,c \\to -(x+y),x,y$) Let $$a,b,c$$ be roots of $$x^3+mx-n=0$$ From Vieta formula we get: $$0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$ so $$\\boxed{a^2+b^2+c^2 =-2m}$$ Further $$x^3 = n-mx\\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n$$ so $$\\boxed{a^3+b^3+c^3 =3n}$$ Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$ so $$\\boxed{a^5+b^5+c^5 =-5mn}$$ Finally $$x^7 = x(n-mx)^2 = n^2x-2mnx^2+m^2x^3$$ $$=n^2x-2mnx^2+m^2n-m^3x$$ so $$\\boxed{a^7+b^7+c^7 = 7m^2n}$$ and we are done.", "$b=5$ and $c=0$. (vi) Given, $3x+2=0$ This implies, $3x+(0)y+2=0$ Comparing $3x+(0)y+2=0$ with $ax+by+c=0$ We get, $a=3$, $b=0$ and $c=2$. (vii) Given, $y-2=0$ This implies, $(0)x+y+(-2)=0$ Comparing $(0)x+y+(-2)=0$ with $ax+by+c=0$ We get, $a=0$, $b=1$ and $c=-2$. (viii) Given, $5=2x$ We get, The linear equation in the form $ax+by+c=0$ as, $2x-5=0$ This implies, $2x+(0)y+(-5)=0$ Comparing $2x+(0)y+(-5)=0$ with $ax+by+c=0$ We get, $a=2$, $b=0$ and $c=-5$. Updated on 10-Oct-2022 13:40:03", "\\: \\text{and}\\:y_1-3=3$ $x_1=5 \\:\\text{and}\\:y_1=6$ $\\boxed{D(x_1, y_1) = D(5,6)}$", "doing so, we get $$(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-6rst+(r^3+s^3+t^3)$$ Recall that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Thus, we get $$(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-3rst+(r+s+t)(r^2+s^2+t^2-rs-st-tr)$$ Plugging in $(r+s+t)=0$ we get $$(r+s)^3+(s+t)^3+(t+r)^3=0-3rst+0=-3\\cdot -251=\\boxed{753}$$ ~firebolt360", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "got $t=0$, $s=1$ and $b=1$. • How did you find the $s(2, b+1, -1)$ ? – J.Se Jan 4 '18 at 17:16 • It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\\vec{AB}(2,b+1,-1)$. – Michael Rozenberg Jan 4 '18 at 17:17 To parametrise a line you need a point it passes through and its direction. You can get from the vector formed by the two points it passes through. Thus $L_2=(-1,-1,2)+t(2,b+1,-1).$L_1$is given by $$x=1-t$$$$y=1+t$$$$z=1-t$$ and$L_2$is given by $$x=-1+(1-(-1))s=-1+2s$$$$y=-1+(b-(-1))s=-1+(b+1)s$$$$z=2+(1-2)s=2-s$$ Now we solve $$1-t=-1+2s\\implies2-t=2s$$ $$1-t=2-s\\implies t=s-1$$ so using the equations for$x$and$z$, $$2-s+1=2s\\implies \\boxed{s=1}\\implies \\boxed{t=0}$$ Let's do the same for$y$. We solve $$1+t=-1+(b+1)s\\implies 1+0=-1+(b+1)(1)\\implies 1=-1+b+1=b$$ Hence intersection only occurs when $$\\boxed{b=1}$$ In other words, for an intersection,$L_2$must go through the points$(-1, -1, 2)$and$(1, 1, 1)\\$.", "y) = $$0 \\in V$$ either a = 0 , or ( x – y) = 0 if we take a = 0 and any x ≠ y for a.\\ Second part Here v denote the set of ordered pairs of real numbers. If $$(a_1, a_2) and\\ (b_1, b_2)$$ are element of v and $$c \\in R$$,define $$(a_1 , a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2)$$ and $$c(a_1, a_2) = (a_1, c.a_2)$$. $$V_{s1}: (a_1, a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2) = (b_1.a_1 , b_2 + a_2) = (b_1, b_2) + (a_1, a_2)$$ $$V_{s2}: \\in [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = (a_1.b_1, a_2 + b_2) + (c_1, c_2)$$\\ = $$(a_1.b_1.c_1 , a_2 + b_2 + c_2) = (a_1, a_2) + (b_1.c_1 , b_2 + c_2) = (a_1, a_2) + [(b_1, b_2) + (c_1 , c_2)]$$ $$V_{s3}: \\exists (d_1, d_2) \\in V$$ s.t $$(a_1, a_2) + (d_1, d_2) = (a_1, a_2)$$ Or,$$(a_1.b_1 , a_2 + b_2 ) = (a_1 , a_2) \\Rightarrow a_1.b_1 = a_1 \\Rightarrow b_1 = 1$$ Also,$$a_2 + b_2 = a_2 \\Rightarrow b_2 = 0$$ Hence $$(d_1, d_2) = (1, 0)$$ so, $$(a_1, a_2) + (d_1, d_2) ≠ (a_1 , a_2)$$ $$v_{s3}$$ does not satisfy. $$V_{s4}: \\forall (a_1 , a_2) \\exists (-a_1, -a_2) \\in V$$ s.t $$(a_1,", "= 3.d(x,x+1)/2 + a_1$$ At this point we have prooved that if r>=2 then y will be >=r. Lets calculate some trivial examples: $$f_0(y) = d(1,2) + a_0 = d(1,2) + 3.d(1,2)/2 + a_1 >> 5.d(1,2)/2 = 5*5/2 = 12.5$$ so $$y >> 11$$ $$f_1(y) = d(2,3) + a_0 = d(2,3) + 3.d(2,3)/2 + a_1 >> 5.d(2,3)/2 = 5*17/2 = 42.5$$ so $$y >> 3$$ and in all other cases y>x+1 from the comparison rule, so y>3. But how big must $$a_1$$ be for r=3 such as $$f_{x-1}^3(y) >> d^3(x,x+1)$$. $$2.2.f_{x-1}(y) = 4.(d(x,x+1) + a_0) = 4.(d(x,x+1) + 3.d(x,x+1)/2 + a_1) = 7.d(x,x+1) + 4.a_1) >> 3.3.d(x,x+1)$$ and $$7.d(x,x+1) + 4.a_1) >> 9.d(x,x+1)$$ so $$a_1 >> d(x,x+1)/2$$ lets assume $$a_1 = d(x,x+1)/2 + a_2$$ and $$a_0 = 2.d(x,x+1) + a_2)$$ and lets calculate: $$f_2(y) = d(3,4) + a_0 = d(3,4) + 2.d(3,4) + a_2 >> 3.d(3,4) = 3*5099 = 15297$$ so $$y >> 10$$ and in all other cases y>x+1 from the comparison rule, so y>6. Completion of Proof This proof by induction shows that each iteration of calculating $$a_n$$ for r>=4 will require the minimum bound for y to increase by >=1 every time. Which means the requirement that y>=r must change to y>r and: $$f_x(y+2) << d(x+1,y)$$ for all cases including any examples of $$f_{x-1}^r(y) >>", "\\\\ y=2$$ Substitute this in (2): $$x+2=11\\\\ x=9$$ Therefore, the system is: {eq}\\boxed{\\mathbf{-x+y=-7\\\\ x+y=11}} {/eq} and the required number is {eq}xy = \\boxed{\\mathbf{92}} {/eq}.", "+0 # hard 0 56 1 Suppose that 3^a = 2 and 3^b = 5. If 3^x = 150,then write an expression for x in terms of a and b Dec 4, 2021 By the law of exponents, if $$3^a=2,$$ and $$3^b=5,$$ then $$3^{a+2b}=2 \\cdot 5 \\cdot 5 = 50.$$ Now we know that $$50 \\cdot 3 = 150,$$ so $$3^{a+2b+1}=150,$$ so in terms of $$a$$ and $$b,$$ $$\\boxed{x=a+2b+1}$$", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "$GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\\boxed{\\mathrm{(B)}\\ 20}$", "+ x!2 dx=2ln x+5 !ln x!2 +C 9. 1 A B = + $1=A(t!1)+B(t+4) . (t+4)(t!1) t+4 t!1 1 1 t=1$ 1=5B$B= . t=!4$ 1=!5A$A=! . Thus, 5 5 1 !1/5 1/5 1 1 1 \" (t+4)(t!1) dt=\" t+4 + t!1 dt=! 5 ln t+4 + 5 ln t!1 +C or 5 ln 10. 11. 1 x !1 2 t!1 t+4 +C = 1 A B = + . Multiply both sides by (x+1)(x!1) to get 1=A(x!1)+B(x+1) . (x+1)(x!1) x+1 x!1 1 1 . Substituting !1 for x gives 1=!2A# A=! . Thus, 2 2 1 1 ln x+1 + ln x!1 2 2 3 2 Substituting 1 for x gives 1=2B# B= 3 \" 2 1 x !1 2 3 dx =\" 2 !1/2 1/2 + x+1 x!1 dx= ! = ! x!1 2 1 1 1 1 1 1 3 ln 4+ ln 2 ! ! ln 3+ ln 1 = (ln 2+ln 3!ln 4) or ln 2 2 2 2 2 2 2 12. = x +3x+2 Substituting !2 for x gives !3=!B # B=3 . Substituting !1 for x gives !2=A . Thus, A B + . Multiply both sides by (x+1)(x+2) to get x!1=A(x+2)+B(x+1) . x+1 x+2 2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial", "must hold for all $x,y$ we can put for example $x=0$ and $y=0$ (you culd put somthing else also). Then we get: $$f(0,0) = f(1,-1)-5 = f(-2,1)+8$$ thus we have $$c=a-b+c-5 = -2a+b+c+8$$ From $c=a-b+c-5$ we get $\\boxed{a-b=5}\\;\\;\\;\\;\\;\\; (1)$ and from $c=-2a+b+c +8$ we get $\\boxed{-2a+b=-8} \\;\\;\\;\\;\\;\\; (2)$ Now if you add (boxed) equations $(1)$ and $(2)$ you get $-a=-3$ thus $a=3$ and $b=-2$. Yet we don't have any other limitation for $c$, so $c$ can be any number. So solution of your problem is(are) function(s): $$f(x,y)=3x-2y+c,\\;\\;\\;\\;\\;\\; c\\in \\mathbb{R}$$ You can easly check that this function(s) satisfyes starting equations.", "y)^3 - 3(x^2y + xy^2)}$ Now, let $a = x+ y$ and $b = x^2y + xy^2$, then we have the system $3a - b = 5$ .....................(1) $a^3 - 3b = 5$ ....................(2) Solve this system for $a$ and you have your answer (since $a$ is the sum of the two numbers) But, of course, who said they had to be integers? 4. Originally Posted by Jhevon Alternatively, let the numbers be $x$ and $y$. we have $x^3 + y^3 = 5$ $x^2 + y^2 = 3$ So as Moo said, $5 = (x + y)(3 - xy)$ that is, $\\boxed{5 = 3(x + y) - (x^2y + xy^2)}$ Now, $(x + y)^3 = x^3 + y^3 + 3(x^2y + y^2x)$ which means $\\boxed{5 = (x + y)^3 - 3(x^2y + xy^2)}$ Now, let $a = x+ y$ and $b = x^2y + xy^2$, then we have the system $3a - b = 5$ .....................(1) $a^3 - 3b = 5$ ................(2) Solve this system for $a$ and you have your answer (since $a$ is the sum of the two numbers)", "5$t$ c. $t$ = 2$d$ d. $d$ = 2$t$", "$$x=10$$ and $$y=5$$ to get $$\\boxed{20}$$ (iii) Similarly $$x=5$$ and $$y=10$$ to get $$\\boxed{-5}$$ b) Substituting the variables in, we get $$\\boxed{-40}$$ . May 4, 2020 #9 0 https://vle.mathswatch.co.uk/images/questions/question17117.png question b do you do 1.5x3=4.5 Guest May 4, 2020 #8 +172 0 Any more work for me to do? May 4, 2020 #10 0 just nee help on the question above plz Guest May 4, 2020 #11 +172 0 For b), all you have to do is substitute the given values into the equation. $$u = 3 m/s \\\\ a = 1.5 m/s^2 \\\\ t = 7 s$$ Thus we get: $$v=3 m/s + 1.5m/s^2 \\cdot 7 s \\\\ v=\\boxed{13.5 m/s}$$ . May 4, 2020", "and $D_2=A=9 \\implies Y+C_1=6$ and $C_2=0$. Thus $S_3=12$ so $D_3=E=2$ which is already taken. If $P=3$, then $S_3=18+C_2$. If $C_2\\ge2$, then $C_3=2$ and $D=3$, which is already taken by $P$. If $C_2=1$ then $D_3=E=9$ which is already taken. If $C_2=0$, then $S_2=9+Y+C1 \\implies Y+C_1=0$. Thus, $Y=0$ and $S=D$, which is not allowed. If $P=4$, then $S_3=21+C_2 \\implies C_3=2 \\implies D=3$. $S_2=12+Y+C_1 \\implies Y+C_1=7$ so that $D_2=A=9$. But then $C_2=1$ and $D_3=2$, which is already taken by $H$. If $P=5$, then $S_3=24+C_2 \\implies C_3=2 \\implies D=3$. $S_2=15+Y+C_1 \\implies Y+C_1=4$ so that $D_2=A=9$. But then $C_2=1$ and $D_3=5$ which is already taken by $P$. ## Therefore $P=6$ From $S_2=18+Y+C_1$ and $D_2=A=9$, we know that $Y+C_1 \\in \\{1,11\\}$. If $Y+C_1=11$, then $Y=8$ and $C_1=3$. But then $S_1=24+S$. No value of $S$ can make $C_1=3$ and $D\\in \\{4,3\\}$. Therefore, $Y+C_1=1$, and $S_2=19$, so $C_2=1$. If $Y=0$, then $C_1=0$ as well, which is a contradiction, so $Y=1$. This means that $S_3=18+9+C_2=28$, so $D_3=E=8$ and $C_3=2$. Thus, $D=3$. Lastly, $S_1=3+S=D=3 \\implies S=0$. Full solution: $$H=2$$ $$A=9$$ $$P=6$$ $$Y=1$$ $$E=8$$ $$D=3$$ $$S=0$$ The corresponding numbers are: A = 9 D = 3 E = 8 H = 2 P = 6 S = 0 Y = 1 The summation then becomes: 29661 29661 29661 + 3910 ------ 92893", "= 4$ and $y = 8$ $x = 5$ and $y = 7$ $x = 6$ and $y = 6$ $x = 7$ and $y = 5$ $x = 8$ and $y = 4$ $x = 9$ and $y = 3$ $x = 10$ and $y = 2$ $x = 11$ and $y = 1$", "given sequence, $a_1=-1.2, a_2=-3.2, a_3=-5.2$ $a_2-a_1=-3.2-(-1.2)=-3.2+1.2=-2$ $a_3-a_2=-5.2-(-3.2)=-5.2+3.2=-2$ $a_2 - a_1 = a_3 - a_2$ $d=a_2 - a_1=-2$ $a_5=a_4+d=-7.2+(-2)=-7.2-2=-9.2$ $a_6=a_5+d=-9.2+(-2)=-9.2-2=-11.2$ $a_7=a_6+d=-11.2+(-2)=-11.2-2=-13.2$ (iv) Given sequence is $-10, -6, -2, 2, …..$ In the given sequence, $a_1=-10.2, a_2=-6, a_3=-2$ $a_2-a_1=-6-(-10)=-6+10=4$ $a_3-a_2=-2-(-6)=-2+6=4$ $a_2 - a_1 = a_3 - a_2$ $d=a_2 - a_1=4$ $a_5=a_4+d=2+4=6$ $a_6=a_5+d=6+4=10$ $a_7=a_6+d=10+4=14$ (v) Given sequence is $3, 3 + \\sqrt2, 3 + 2\\sqrt2, 3 + 3\\sqrt2, …..$ In the given sequence, $a_1=3, a_2=3 + \\sqrt2, a_3=3 + 2\\sqrt2$ $a_2-a_1=3 + \\sqrt2-3=\\sqrt2$ $a_3-a_2=3 + 2\\sqrt2-(3 + \\sqrt2)=\\sqrt2$ $a_2 - a_1 = a_3 - a_2$ $d=a_2 - a_1=\\sqrt2$ $a_5=a_4+d=3 + 3\\sqrt2+ \\sqrt2=3 + 4\\sqrt2$ $a_6=a_5+d=3 + 4\\sqrt2 + \\sqrt2=3 + 5\\sqrt2$ $a_7=a_6+d=3 + 5\\sqrt2+ \\sqrt2=3 + 6\\sqrt2$ (vi) Given sequence is $0.2, 0.22, 0.222, 0.2222, ……$ In the given sequence, $a_1=0.2, a_2=0.22, a_3=0.222$ $a_2-a_1=0.22-0.2=0.02$ $a_3-a_2=0.222-0.22=0.002$ $a_2 - a_1 ≠ a_3 - a_2$ Therefore, the given sequence is not an AP. (vii) Given sequence is $0, -4, -8, -12, …..$ In the given sequence, $a_1=0, a_2=-4, a_3=-8$ $a_2-a_1=-4-(0)=-4$ $a_3-a_2=-8-(-4)=-8+4=-4$ $a_2 - a_1 = a_3 - a_2$ $d=a_2 - a_1=-4$ $a_5=a_4+d=-12+(-4)=-16$ $a_6=a_5+d=-16+(-4)=-20$ $a_7=a_6+d=-20+(-4)=-24$ (viii) Given sequence is $-\\frac{1}{2}, -\\frac{1}{2}, -\\frac{1}{2}, -\\frac{1}{2}, …….$ In the given sequence, $a_1=-\\frac{1}{2}, a_2=-\\frac{1}{2}, a_3=-\\frac{1}{2}$ $a_2-a_1=-\\frac{1}{2}-(-\\frac{1}{2})=-\\frac{1}{2}+-\\frac{1}{2}=0$ $a_3-a_2=-\\frac{1}{2}-(-\\frac{1}{2})=-\\frac{1}{2}+-\\frac{1}{2}=0$ $a_2 - a_1 = a_3 - a_2$ $d=a_2 - a_1=-\\frac{1}{2}$ $a_5=a_4+d=-\\frac{1}{2}+0=-\\frac{1}{2}$ $a_6=a_5+d=-\\frac{1}{2}+0=-\\frac{1}{2}$ $a_7=a_6+d=-\\frac{1}{2}+0=-\\frac{1}{2}$ (ix) Given sequence is $1, 3, 9, 27, …….$" ]

[ "+0 # Halp 0 94 1 The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations \\begin{align*} x &= at + b, \\\\ y &= ct + d, \\end{align*} where $0 \\le t \\le 1,$ and $t = 0$ corresponds to the point $(-2,7).$ Find $a^2 + b^2 + c^2 + d^2.$ Jan 6, 2022", "+0 # Halp 0 129 1 The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations \\begin{align*} x &= at + b, \\\\ y &= ct + d, \\end{align*} where $0 \\le t \\le 1,$ and $t = 0$ corresponds to the point $(-2,7).$ Find $a^2 + b^2 + c^2 + d^2.$ Jan 6, 2022 ### 1+0 Answers #1 0 The parametric equations are x = 3t - 2 and y = 5t + 7, so a^2 + b^2 + c^2 + d^2 = 3^2 +(-2)^2 + 5^2 + 7^2 = 87. Jan 6, 2022", "# Cubic Bezier curve and a straight line intersection Suppose that two ends of a cubic Bezier curve is connected by a straight line. Is there a simple way to find out whether this straight line intersects the Bezier curve (apart from the endpoints)? If it intersects then what will be the corresponding Bezier curve parameter's value? Given 2D cubic Bezier segment $\\mathcal{B}(t,A,B,C,D)=A\\,(1-t)^3+3B\\,(1-t)^2t+3C\\,(1-t)t^2+D\\,t^3$, $t\\in[0,1]$ and line segment $\\mathcal{L}(t,A,D)=A\\,(1-s)+D\\,s$, $s\\in[0,1]$, the value of $t:\\mathcal{B}(t,A,B,C,D)=\\mathcal{L}(s,A,D),\\ t\\ne0, t\\ne1$ is the same as the value of $t:\\mathcal{B}(t,0,B-A,C-A,D-A)=\\mathcal{L}(s,0,D-A)$. So, with a substitution $b=B-A,\\ c=C-A,\\ d=D-A$ we can solve a system of two equations with two unknowns $t,s$: \\begin{align} \\mathcal{B}(t,0,b_x,c_x,d_x) &= \\mathcal{L}(s,0,d_x) \\\\ \\mathcal{B}(t,0,b_y,c_y,d_y) &= \\mathcal{L}(s,0,d_y) \\end{align} which gives the value of parameter $t$ as \\begin{align} t &= \\frac{b_x\\,d_y-d_x\\,b_y}{b_x\\,d_y-c_x\\,d_y-d_x\\,b_y+d_x\\,c_y} \\end{align} If $0<t<1$ than the intersection point of $\\mathcal{B}$ and $\\mathcal{L}$ is $X=\\mathcal{B}(t,A,B,C,D)$. • It is very explanatory and convincing, thanks. For a perfection I would like to propose a little algebraic correction: The line segment should be L(t,A,D). Jun 24 '15 at 18:50 • @John Railman: Of course it is $AD$, great thanks for pointing this, fixed. Jun 24 '15 at 18:59 I just wanted to add to g.kov's good derivation, that the $t$ value is the parameter for the Bézier segment, not the line, so the range of $0 < t < 1$", "them each equal to zero: \\begin{align*} f_x(x,y) &=2x−2y−4 \\\\[4pt] f_y(x,y) &=−2x+8y−2. \\end{align*} Setting them equal to zero yields the system of equations \\begin{align*} 2x−2y−4 &=0\\\\[4pt] −2x+8y−2 &=0. \\end{align*} The solution to this system is $$x=3$$ and $$y=1$$. Therefore $$(3,1)$$ is a critical point of $$f$$. Calculating $$f(3,1)$$ gives $$f(3,1)=17.$$ The next step involves finding the extrema of $$f$$ on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph: $$L_1$$ is the line segment connecting $$(0,0)$$ and $$(4,0)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=0$$ for $$0≤t≤4$$. Define $$g(t)=f\\big(x(t),y(t)\\big)$$. This gives $$g(t)=t^2−4t+24$$. Differentiating $$g$$ leads to $$g′(t)=2t−4.$$ Therefore, $$g$$ has a critical value at $$t=2$$, which corresponds to the point $$(2,0)$$. Calculating $$f(2,0)$$ gives the $$z$$-value $$20$$. $$L_2$$ is the line segment connecting $$(4,0)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=4,y(t)=t$$ for $$0≤t≤2.$$ Again, define $$g(t)=f\\big(x(t),y(t)\\big).$$ This gives $$g(t)=4t^2−10t+24.$$ Then, $$g′(t)=8t−10$$. g has a critical value at $$t=\\frac{5}{4}$$, which corresponds to the point $$\\left(0,\\frac{5}{4}\\right).$$ Calculating $$f\\left(0,\\frac{5}{4}\\right)$$ gives the $$z$$-value $$27.75$$. $$L_3$$ is the line segment connecting $$(0,2)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=2$$ for $$0≤t≤4.$$ Again, define $$g(t)=f\\big(x(t),y(t)\\big).$$ This gives $$g(t)=t^2−8t+36.$$ The critical value corresponds to the point $$(4,2).$$ So, calculating $$f(4,2)$$ gives the $$z$$-value $$20$$.", "got $t=0$, $s=1$ and $b=1$. • How did you find the $s(2, b+1, -1)$ ? – J.Se Jan 4 '18 at 17:16 • It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\\vec{AB}(2,b+1,-1)$. – Michael Rozenberg Jan 4 '18 at 17:17 To parametrise a line you need a point it passes through and its direction. You can get from the vector formed by the two points it passes through. Thus $L_2=(-1,-1,2)+t(2,b+1,-1).$L_1$is given by $$x=1-t$$$$y=1+t$$$$z=1-t$$ and$L_2$is given by $$x=-1+(1-(-1))s=-1+2s$$$$y=-1+(b-(-1))s=-1+(b+1)s$$$$z=2+(1-2)s=2-s$$ Now we solve $$1-t=-1+2s\\implies2-t=2s$$ $$1-t=2-s\\implies t=s-1$$ so using the equations for$x$and$z$, $$2-s+1=2s\\implies \\boxed{s=1}\\implies \\boxed{t=0}$$ Let's do the same for$y$. We solve $$1+t=-1+(b+1)s\\implies 1+0=-1+(b+1)(1)\\implies 1=-1+b+1=b$$ Hence intersection only occurs when $$\\boxed{b=1}$$ In other words, for an intersection,$L_2$must go through the points$(-1, -1, 2)$and$(1, 1, 1)\\$.", "a line that intersects; $L:x−1−1=y−1=z−1L:x−1−1=y−1=z−1$ 267. a. $3x−2y+4z=0;3x−2y+4z=0;$ b. $3x−2y+4z=03x−2y+4z=0$ 269. a. $(x−1)+2(y−2)+3(z−3)=0;(x−1)+2(y−2)+3(z−3)=0;$ b. $x+2y+3z−14=0x+2y+3z−14=0$ 271. a. $n=4i+5j+10k;n=4i+5j+10k;$ b. $(5,0,0),(5,0,0),$ $(0,4,0),(0,4,0),$ and $(0,0,2);(0,0,2);$ c. 273. a. $n=3i−2j+4k;n=3i−2j+4k;$ b. $(0,0,0);(0,0,0);$ c. 275. $(3,0,0)(3,0,0)$ 277. $x=−2+2t,y=1−3t,z=3+t,x=−2+2t,y=1−3t,z=3+t,$ $t∈ℝt∈ℝ$ 281. a. $−2y+3z−1=0;−2y+3z−1=0;$ b. $〈0,−2,3〉·〈x−1,y−1,z−1〉=0;〈0,−2,3〉·〈x−1,y−1,z−1〉=0;$ c. $x=0,y=−2t,z=3t,x=0,y=−2t,z=3t,$ $t∈ℝt∈ℝ$ 283. a. Answers may vary; b. $x−11=z−6−1,y=4x−11=z−6−1,y=4$ 285. $2x−5y−3z+15=02x−5y−3z+15=0$ 287. The line intersects the plane at point $P(−3,4,0).P(−3,4,0).$ 289. $16141614$ 291. a. The planes are neither parallel nor orthogonal; b. $62°62°$ 293. a. The planes are parallel. 295. $1616$ 297. a. $1829;1829;$ b. $P(−5129,13029,6229)P(−5129,13029,6229)$ 299. $4x−3y=04x−3y=0$ 301. a. $v(1)=〈cos1,−sin1,2〉;v(1)=〈cos1,−sin1,2〉;$ b. $(cos1)(x−sin1)−(sin1)(y−cos1)+2(z−2)=0;(cos1)(x−sin1)−(sin1)(y−cos1)+2(z−2)=0;$ c. ### Section 2.6 Exercises 303. The surface is a cylinder with the rulings parallel to the y-axis. 305. The surface is a cylinder with rulings parallel to the y-axis. 307. The surface is a cylinder with rulings parallel to the x-axis. 309. a. Cylinder; b. The x-axis 311. a. Hyperboloid of two sheets; b. The x-axis 313. b. 315. d. 317. a. 319. $−x29+y214+z214=1,−x29+y214+z214=1,$ hyperboloid of one sheet with the x-axis as its axis of symmetry 321. $−x2103+y22−z210=1,−x2103+y22−z210=1,$ hyperboloid of two sheets with the y-axis as its axis of symmetry 323. $y=−z25+x25,y=−z25+x25,$ hyperbolic paraboloid with the y-axis as its axis of symmetry 325. $x215+y23+z25=1,x215+y23+z25=1,$ ellipsoid 327. $x240+y28−z25=0,x240+y28−z25=0,$ elliptic cone with the z-axis as its axis of symmetry 329. $x=y22+z23,x=y22+z23,$ elliptic paraboloid with the", "$(x_1, y_1)$ and $(x_2, y_2)$ \\begin{aligned} x(t) &= x_1 + t(x_2 - x_1)\\\\ y(t) &= y_1 + t(y_2 - y_1) \\end{aligned} This parameterization has the property that for values $0 \\leq t \\leq 1$, $(x(t), y(t))$ is on the line segment, between the two endpoints. On the other hand, if t is less than zero, or t is greater than one, $(x(t), y(t))$ is outside of the two endpoints. Below are points (from left to right) for t = -0.5 in red, t = 0, 0.25, 0.5, 0.75, and 1 in green, and for t = 1.5 in red again: At what point do two line segments intersect? Lines A and B intersect for parametric values s and t, such that \\begin{aligned} A_x(s) &= B_x(t)\\\\ A_y(s) &= B_y(t) \\end{aligned} I’ll use the parametric equation of lines A and B and solve for the values s and t for which the equality is satisfied. \\begin{aligned} A_{x1} + s(A_{x2} - A_{x1}) &= B_{x1} + t(B_{x2} - B_{x1})\\\\ A_{y1} + s(A_{y2} - A_{y1}) &= B_{y1} + t(B_{y2} - B_{y1}) \\end{aligned} then \\begin{aligned} s(A_{x2} - A_{x1}) - t(B_{x2} - B_{x1}) &= B_{x1} - A_{x1}\\\\ s(A_{y2} - A_{y1}) - t(B_{y2} - B_{y1}) &= B_{y1} - A_{y1} \\end{aligned} In matrix form $\\begin{bmatrix} (A_{x2} - A_{x1}) & -(B_{x2} - B_{x1}) \\\\ (A_{y2} - A_{y1}) &", "+0 # A segment with endpoints at $A(2, -2)$ and $B(14, 4)$ is extended through $B$ to point $C$. If $BC = \\frac{1}{3} \\cdot AB$, what are the coo 0 57 1 A segment with endpoints at $A(2, -2)$ and $B(14, 4)$ is extended through $B$ to point $C$. If $BC = \\frac{1}{3} \\cdot AB$, what are the coordinates for point $C$? Express your answer as an ordered pair. Jan 18, 2021 #1 +116126 +1 A = (2, -2) B =(14,4) BC = (1/3) AB We can find the coordinates of C thusly : ( 14 + (1/3) ( 14 -2) , 4 + (1/3)(4 - -2) ) = (14+ 12/3 , 4 + 6/3) = (14 + 4 , 4 + 2) = (18, 6) Jan 18, 2021", "+0 # Consider the line through points $A = (1, 1, 2)$ and $B = (2, 3, 4)$. This line can be parametrized as $(at +b, ct +d, et +f)$for some cho 0 155 1 Consider the line through points $$A = (1, 1, 2)$$ and .$$B = (2, 3, 4)$$ This line can be parametrized $$(at +b, ct +d, et +f)$$ as for some choices of constants $$a,b, c, d, e$$ and f. What is the ordered pair $$\\left(\\dfrac{c}{a}, \\dfrac{e}{a}\\right)$$? Mar 15, 2019", "# Crossed Line Segments Geometry Level 3 Line segments $$AB$$ and $$CD$$ intersect (internally) at point $$E$$. We are given that $$AE=20$$, $$EB= 12$$, $$CE=10$$ and $$ED = 24$$. What are the last 3 digits of the expression $$AC \\times BD + BC \\times AD$$? Details and assumptions If you think that the last 3 digits are 012, then you can enter either 12 or 012 as your numerical answer. ×", "- 5x + 7y, \\end{align*} $$4y^2 - 100 = 0\\implies y = 5\\implies x = 11/3 \\\\ \\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \\implies \\boxed{045}.$$ ### Solution 3 Let the incircle of $ABM$ hit $AM$, $AB$, $BM$ at $X_{1},Y_{1},Z_{1}$, and let the incircle of $CBM$ hit $MC$, $BC$, $BM$ at $X_{2},Y_{2},Z_{2}$. Draw the incircle of $ABC$, and let it be tangent to $AC$ at $X$. Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$, and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$. By standard computations, $AX=\\dfrac{AB+AC-BC}{2}=7$ and $CX=\\dfrac{BC+AC-AB}{2}=8$. Now, let $AX_{1}=7x$ and $CX_{2}=8x$. We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$. Then, $BZ_{1}=12-7x$. We also have $CY_{2}=CX_{2}=8x$, so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$. Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$. Also,$X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$. Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range). Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$. To finish, we will compute area ratios. $\\dfrac{[ABM]}{[CBM]}=\\dfrac{AM}{MC}=\\dfrac{8-x}{7+x}$. Also, since their inradii are equal, we get $\\dfrac{[ABM]}{[CBM]}=\\dfrac{40-16x}{40-14x}$. Equating and cross multiplying yields the quadratic", "0$ from the point $(0, 0, 0)$ to $(3, 1, -2)$. Note that $x - y + z = 0$ and $x + y + 2z = 0$ are both planes and either intersect at a straight line, do not intersect, or coincide. It's not hard to see that these planes in fact intersect at a straight line. We can thus parameterize the line segment from $(0, 0, 0)$ to $(3, 1, -2)$ for $0 ≤ t ≤ 1$ as: (3) \\begin{align} \\quad \\vec{r}(t) = (1 - t)(0, 0, 0) + t(3, 1, -2) = (3t, t, -2t) \\end{align} Thus: (4) \\begin{align} \\quad \\int_C x^2 \\: ds = \\int_0^1 (3t)^2 \\sqrt{(3)^2 + (1)^2 + (-2)^2} \\: dt \\\\ \\quad \\int_C x^2 \\: ds = \\int_0^1 9 \\sqrt{14} t^2 \\: dt \\\\ \\quad \\int_C x^2 \\: ds = 9 \\sqrt{14} \\left [ \\frac{t^3}{3} \\right ]_{t=0}^{t=1} \\\\ \\quad \\int_C x^2 \\: ds = 3 \\sqrt{14} \\end{align}", "that the line is at $(3,0) (3,0)$ at $t=0, t=0,$ and at $(−2,−5) (−2,−5)$ at $t=1. t=1.$ 39. Parameterize the line from $(−1,0) (−1,0)$ to $(3,−2) (3,−2)$ so that the line is at $(−1,0) (−1,0)$ at $t=0, t=0,$ and at $(3,−2) (3,−2)$ at $t=1. t=1.$ 40. Parameterize the line from $(−1,5) (−1,5)$ to $(2,3) (2,3)$ so that the line is at $(−1,5) (−1,5)$ at $t=0, t=0,$ and at $(2,3) (2,3)$ at $t=1. t=1.$ 41. Parameterize the line from $(4,1) (4,1)$ to $(6,−2) (6,−2)$ so that the line is at $(4,1) (4,1)$ at $t=0, t=0,$ and at $(6,−2) (6,−2)$ at $t=1. t=1.$ #### Technology For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. 42. 43. For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 44. ${ x 1 ( t )=3 t 2 −3t+7 y 1 ( t )=2t+3 { x 1 ( t )=3 t 2 −3t+7 y 1 ( t )=2t+3$ $t t$ $x x$ $y y$ –1 0 1 45. ${ x 1 ( t )= t 2 −4 y 1 ( t )=2 t 2 −1 { x 1 ( t )= t 2 −4 y 1 ( t )=2 t 2 −1$ $t t$", "the line segment connecting the points (1,3) and $$(4,8) .$$ Use the fact that a point plus a vector yields another point. A vector between these points is $$<4-1,8-3>=<3,5>$$ Thus the point (1,3) plus $$t$$ times the vector $$<3,5>$$ will produce the point (4,8) when $$t=1$$ and the point (1,3) when $$t=0$$. $$(x, y)=(1,3)+t \\cdot<3,5>,$$ for $$0 \\leq t \\leq 1$$ You than then break up this vector equation into parametric form. $$x=1+3 t$$ $$y=3+5 t$$ $$0 \\leq t \\leq 1$$ ##### Review Eliminate the parameter in the following sets of parametric equations. 1. $$x=3 t-1 ; y=4 t^{2}-2 t$$ 2. $$x=3 t^{2}+6 t ; y=2 t-1$$ 3. $$x=t+2 ; y=t^{2}+4 t+4$$ 4. $$x=t-5 ; y=t^{3}+1$$ 5. $$x=t+4 ; y=t^{2}-5$$ For the parametric equation $$x=t, y=t^{2}+1,$$ graph over each interval of $$t$$. 6. $$-2 \\leq t \\leq-1$$ 7. $$-1 \\leq t \\leq 0$$ 8. $$-1 \\leq t \\leq 1$$ 9. $$-2 \\leq t \\leq 2$$ 10. $$-5 \\leq t \\leq 5$$ 11. Eliminate the parameter and graph the following parametric curve: $$x=\\sin t, y=-4+3 \\cos t$$ 12. Eliminate the parameter and graph the following parametric curve: $$x=1+2 \\cos t, y=1+2 \\sin t$$ 13. Using the previous problem as a model, find a parameterization for the circle with center (2, 4) and radius $$3 .$$ 14. Find the parameterization", "want to consider only $(\\delta x,\\delta y,\\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$. We want to find the point at which $\\delta f(x,y,z)=(0,1,1)\\cdot(\\delta x,\\delta y,\\delta z)=0$ for all $(\\delta x,\\delta y,\\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$. For that to be true, $(0,1,1)$ must be a linear combination of $(3,1,0)$ and $(x,y,z)$. This means that $(x,y,z)$ is also a linear combination of $(3,1,0)$ and $(0,1,1)$; that is, on the plane $x-3y+3z=0$. Thus, we are looking for a point on the unit sphere that is on the planes $x-3y+3z=0$ and $3x+y=3$. The intersection of these two planes is the line $(0,3,3)+(-3,9,10)t$. So we need to solve \\begin{align} 1&=(0-3t)^2+(3+9t)^2+(3+10t)^2\\\\ &=18+114t+190t^2 \\end{align} that is $t=\\dfrac{-57\\pm\\sqrt{19}}{190}$. Note that \\begin{align} f((0,3,3)+(-3,9,10)t) &=(0,1,1)\\cdot((0,3,3)+(-3,9,10)t)\\\\ &=6+19t \\end{align} Plugging in the values of $t$ give us the extreme values: $\\dfrac{3+\\sqrt{19}}{10}$ and $\\dfrac{3-\\sqrt{19}}{10}$ Another way. Given that you have $g = 3- 3x + z$ to find extrema, with the constraint $1 = x^2 + 9(1-x)^2 + z^2 = (10x^2 -18x + 9) + z^2 = 10(x-\\frac{9}{10})^2 + z^2 + \\frac{9}{10}$ Thus the constraint is $100(x-\\frac{9}{10})^2 + 10z^2 = 1$ Now we can let $x - \\frac{9}{10} = \\frac{\\sin(t)}{10}$ and $z = \\frac{\\cos(t)}{\\sqrt{10}}$ so that the constraint is satisfied. Then our objective becomes $g = 3 - 3(\\frac{9}{10}+\\frac{\\sin(t)}{10})+ \\frac{\\cos(t)}{\\sqrt{10}}$. $$\\implies 10g-3 =", "# Lines and Planes A line starts at an initial point and proceeds straight in a constant direction. Thus, we can write the equation of a line as \\begin{align*} x(t)=mt+b \\end{align*} where $b = \\left< b_1, b_2, \\ldots, b_N \\right>$ is the initial point, $m = \\left< m_1, m_2, \\ldots, m_N \\right>$ is the constant direction in which the line travels, and $x(t) = \\left< x_1(t), x_2(t), \\ldots, x_N(t) \\right>$ is the point reached by traveling $t$ units away from $b$ in the direction of $m$. (Though $x(t)$ is actually a vector, we can also refer to it as the point where the vector lands when the vector is placed at the origin.) For example, to compute the line between the points $(1,2,3,4)$ and $(5,-2,3,7)$ in 4-dimensional space, we can start by computing the direction $m$ as the difference between the two points: \\begin{align*} m &= \\left< 5,-2,3,7 \\right> - \\left< 1,2,3,4 \\right> \\\\ &= \\left< 4,-4,0,3 \\right> \\end{align*} Taking $b=\\left< 1,2,3,4 \\right>$ as our initial point, then, we can express the line as \\begin{align*} x(t) = \\left< 4,-4,0,3 \\right> t + \\left< 1,2,3,4 \\right>. \\end{align*} If we wanted to find another point on the line, we could substitute another value for $t$, say, $t=-2$. \\begin{align*} x(2) &= \\left< 4,−4,0,3 \\right>(−2) + \\left< 1,2,3,4 \\right> \\\\ &= \\left<", "such that $CD = 3.825 \\ in$ . 2. If $OP = 17$ and $QP = 6$ , what is $OQ$ ? 3. Make a sketch that matches the description: $S$ is between $T$ and $V$ . $R$ is between $S$ and $T$ . $TR = 6 \\ cm, \\ RV = 23 \\ cm,$ and $TR = SV$ . Then, find $SV, TS, RS$ and $TV$ . 4. For $\\overline{HK}$ , suppose that $J$ is between $H$ and $K$ . If $HJ = 2x + 4, \\ JK = 3x + 3,$ and $KH = 22$ , find the lengths of $HJ$ and $JK$ . 5. What is the distance between the two points shown below? 1.To draw a line segment, start at “0” and draw a segment to 3.825 in. Put points at each end and label. 2. Use the Segment Additional Postulate. $OQ + QP = OP$ , so $OQ + 6 = 17$ , or $OQ = 17 - 6 = 11$ . So, $OQ = 11$ . 3. Interpret the first sentence first: $S$ is between $T$ and $V$ . Then add in what we know about $R$ : It is between $S$ and $T$ . To find $SV$ , we know it is equal to $TR$ , so $SV = 6 \\", "13 If given a line equation $y = 2x+3$, then line’s image under the translation $T = (3, 2)$ is represented as $\\cdots \\cdot$ A. $y = 3x$ D. $y = 2x-4$ B. $y = 2x + 6$ E. $y = 2x-1$ C. $y = 2x-6$ Solution Suppose the line $y = 2x+3$ passes through the point $(x, y)$. After being translated by $T(3, 2)$, we found its image as shown in the diagram below. $(x, y) \\xrightarrow{T(3, 2)} (x+3, y+2)$ Hence, we have $x’ = x + 3$ and $y’ = y + 2$, or in the same manner, $\\begin{cases} x = x’-3 \\\\ y = y’-2 \\end{cases}$ Substitute $x$ and $y$ in terms of $x’$ and $y’$ to the initial line equation. \\begin{aligned} y & = 2x + 3 \\\\ y’-2 & = 2(x’-3) + 3 \\\\ y’ & = 2x’-6 + 3 + 2 \\\\ y’ & = 2x’-1 \\end{aligned} Thus, the image of the line $y = 2x+3$ after being translated by $T(3,2)$ is given by $\\boxed{y=2x-1}$ [collapse] Problem Number 14 The equation of the image of line $2x+y-1=0$ after being transformed by matrix $\\begin{pmatrix} 1 & 1 \\\\ 1 & 2 \\end{pmatrix}$ and then continued by the reflection over $X$-axis, is $\\cdots \\cdot$ A. $3x+y-1=0$ B. $5x-y+1=0$ C. $3x+y+1=0$ D. $5x+y-1=0$ E. $5x+y+1=0$", "b, c> &= \\textbf{v}\\\\&= <1, 2, -18> \\\\ (x_o, y_o, z_o) &= A \\\\&= (1, -2, 13)\\\\\\\\x&=x_o + at\\\\&= 1 +t\\\\y&=y_o + bt\\\\&= -2 + 2t\\\\z&=z_o + ct\\\\&= 13 – 18t\\end{aligned} Now that we have the parametric equations of the line, let’s use them to rewrite the equation of the plane. \\begin{aligned}3x + 2y – z + 10 &= 0\\\\3(1 +t) + 2(-2 + 2t) – (13 – 18t) + 10 &= 0\\\\3 + 3t – 4 +4t -13 + 18t + 10 &=0 \\\\25t&= 4\\\\t&= \\dfrac{4}{25}\\\\&= 0.16\\end{aligned} Find the coordinates of the point of intersection by substituting the parameter,t = 0.16, into the equation. \\begin{aligned}x&= 1 +t\\\\&= 1+ 0.16\\\\&=1.16\\\\y&= -2 + 2t\\\\&= -2 + 2(0.16)\\\\&= -1.68\\\\z&= 13 – 18t\\\\&= 13 – 18(0.16)\\\\&= 10.12 \\end{aligned} We can also double-check our answer by substituting the values into the equation of the plane. \\begin{aligned}3x + 2y – z + 10 &= 0\\\\ 3(1.16) + 2(-1.68) -10.12 + 10&= 0\\\\0 &\\overset{\\checkmark}{=}0\\end{aligned} This means that the line and the plane intersect at the point,(1.16, -1.68, 10.12)$. Example 4 Determine whether the line,$\\mathbf{r} = (1, -1, 2) + t(2, -4, -2)$, intersects the plane that contains the points,$(1, 2, -3)$,$(2, 3, 1)$, and$(0, -2, -1)$. If so, find their point of intersection. Solution Use the three points to find the normal vector of", "# Circle locus, how to satisfy the equation. $A(-3,1), B(0,-5), P(X,Y)$ If $|AP| = 2|BP|$ prove that $x$ and $y$ satisfy the equation: \\begin{aligned} \\ x^2+y^2-2x+14y+30 =0 \\end{aligned} I get as far as determining the co-ordinates like so \\begin{aligned} \\ \\sqrt{(x+3)^2+(y-1)^2}= 2\\sqrt{(x)^2+(y+5)^2} \\end{aligned} To \\begin{aligned} \\ x^2+y^2+2y+40-6x =0 \\end{aligned} Which gives me $(3, -1)$, this won't satisfy, is my method correct? - If we start from the equation $$\\sqrt{(x+3)^2+(y-1)^2}= 2\\sqrt{(x)^2+(y+5)^2},$$ and square both sides, we obtain the equivalent equation $$x^2+6x+9+y^2-2y+1=4x^2+4y^2+40y+100,$$ which simplifies to $$3x^2+3y^2-6x+42y+90=0,$$ which is equivalent to $$x^2+y^2-2x+14y+30=0.$$ Comment: Draw the line segment that joins $A$ to $B$. It is fairly easy to see that $(-1,-3)$ is the point on this line segment that is twice as far from $A$ as it is from $B$. So $(-1,-3)$ ought to be on our circle, and indeed it is. That provides a partial check on the correctness of our computations. No, when you square both sides you need to square the \"$2$\". If $a=2b$ then $a^2=(2b)^2=4b^2$. – André Nicolas Jan 22 '12 at 19:50" ]

[ "# Triggered Trigonometry #2 Algebra Level 2 Find the smallest positive integer value of $$\\theta$$ in degrees such that $81^{sin^2{\\theta}} + 81^{cos^2{\\theta}} = 30.$ ×", "# Triggered Trigonometry #2 Algebra Level 2 Find the smallest positive integer value of $\\theta$ in degrees such that $81^{sin^2{\\theta}} + 81^{cos^2{\\theta}} = 30.$ × Problem Loading... Note Loading... Set Loading...", "+0 # trig question 0 113 1 Find the measure, in degrees, of the smallest positive angle $\\theta$ for which $\\sin 3 \\theta = \\cos 17 \\theta$. Dec 15, 2021 #1 +117437 +1 $$sin\\;3\\theta = cos\\; 17\\theta\\\\ sin\\;3\\theta = sin\\; (90-17\\theta)\\\\$$ you can take it from there. Dec 15, 2021", "... theta is the angle between two sides. ...", "## Trigonometry (10th Edition) 1. Since the angle is larger than 360°, to find the smallest positive coterminal angle, we will subtract 360° until we will get positive angle smaller the 360 degrees 541° -360° = 181$^{\\circ}$", "## anonymous 4 years ago Find the smallest positive angle θ in the third quadrant for which tan(θ) = 1. in third quadrant so answer should lie between pi to 3pi we know tan pi/4= 1 and tan (pi+theta)= tan theta so tan(pi+pi/4)=1 so snawer pi+pi/4= 5pi/4 $5 \\pi/4$", "Trigonometry (11th Edition) Clone $\\theta=4000^{\\circ}$ $\\cos\\theta = 0.766$ $\\sin\\theta = 0.643$ As the $\\cos\\theta$ and $\\sin\\theta$ are both positive, the angle lies in the first quadrant.", "## Trigonometry (11th Edition) Clone $\\theta=1997^{\\circ}$ $\\cos\\theta = -0.956$ $\\sin\\theta = -0.292$ As the $\\cos\\theta$ and $\\sin\\theta$ are both negative, the angle lies in the third quadrant.", "positive measure that is coterminal with measure 820°. $$820° – 360° = 460°$$ $$460° – 360° = 100°$$ Take note that -520° is a negative coterminal angle. Since the given angle measure is negative or non-positive, add 360° repeatedly until one obtains the smallest positive measure of coterminal with the angle of measure -520°. $$-520° + 360° = -160°$$ $$-160° + 360° = 200°$$ As a result, the angles with measure 100° and 200° are the angles with the smallest positive measure that are coterminal with the angles of measure 820° and -520°, respectively. ### Example For Finding All Coterminal Angles With 120° For any integer k, $$120° + 360° k$$ will be coterminal with 120°. Shown below are some of the coterminal angles of 120°. At k = -2 $$\\angle \\theta = 120° + 360° (-2)$$ $$\\angle \\theta = 120° – 720°$$ $$\\angle \\theta = -600°$$ At k = -1 $$\\angle \\theta = 120° + 360° (-1)$$ $$\\angle \\theta = 120° – 360°$$ $$\\angle \\theta = -240°$$ At k = 1, $$\\angle \\theta = 120° + 360° (1)$$ $$\\angle \\theta = 480°$$ At k = 2. $$\\angle \\theta = 120° + 360° (2)$$ $$\\angle \\theta = 120° + 720°$$ $$\\angle \\theta = 840°$$ Therefore, the formula $$\\angle \\theta = 120° + 360° k$$ represents the coterminal angles", "## Trigonometry (10th Edition) 1. r is always positive 2. x is positive 3. Therefore, $\\cos\\theta=\\frac{x}{r}=\\frac{+}{+}=+$", "1. ## radian measure unit circle Refer to the figure, which indicates radian measure on the unit circle for angles in standard position. Determine whether the quantities are positive or negative or zero. cos 6 + cos 1 and cos 6 - cos 1 I really don't even know how to start on this one. Thanks guys 2. both values are positive. $\\theta = 6$ is an angle in quadrant IV ... its cosine value is close to 1 $\\theta = 1$ is an angle in quadrant I ... its cosine value is close to 1/2 ... just a matter of familiarity with the unit circle and radian values of angles.", "degrees.​ You could also subtract 360 degrees to get -325 degrees and subtract 720 degrees to get ​-685 degrees​. 2. Find the smallest positive angle, in degrees, coterminal with -15 radians. Add multiples of 2π until you get a positive angle. Since 2π = 6.28, we need to multiply by 3 to end up with a positive angle:", "$$\\displaystyle\\cos(x) \\approx 1$$ • I get it sir. Thanks – Aneek Aug 26 '15 at 6:25 If you use circular function approach to define $\\cos \\theta$ then for very small but positive $\\theta$, we have $\\cos \\theta = x$ where $x$ is the $x$ coordinate of the point on the unit circle which lies on the terminal side of the angle $\\theta$. From the unit circle, you can see that as $\\theta$ gets smaller and smaller, $x$ gets bigger and bigger and tends toward the point $(1,0)$ on the $x$ axis. This works like a proof of your claim. See Wikipedia Small-angle approximation — Geometric: in a right triangle the side opposite to the small angle is almost same length as the corresponding circular arc, which in turn is equal to the angle (see the definition of radian), hence sine approximately equal the angle. At the same time the long leg is almost equal to hypotenuse, making the cosine approximately equal $1$.", "+0 trig 0 39 1 Find the measure, in degrees, of the smallest positive angle theta for which sin (3 theta) = cos (6 theta). May 23, 2022 #1 +9457 0 We can make use of the identity $$\\cos \\theta = \\sin(90^\\circ - \\theta)$$. $$\\sin 3\\theta = \\cos 6\\theta\\\\ \\sin 3\\theta = \\sin(90^\\circ - 6\\theta)$$ To get the smallest positive angle theta, we remove the sin from both sides (note that it does not always work, but if we assume that 0 degrees < 3 theta < 90 degrees and 0 degrees < 90 degrees - 6 theta < 90 degrees, then it works.) $$3\\theta = 90^\\circ -6\\theta\\\\ 9\\theta = 90^\\circ\\\\ \\theta = \\boxed{10^\\circ}$$ May 23, 2022", "While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $$x$$ for which the sine of $$x$$ degrees is the same as the sine of $$x$$ radians are $$\\frac{m\\pi}{n-\\pi}$$ and $$\\frac{p\\pi}{q+\\pi}$$, where $$m$$, $$n$$, $$p$$, and $$q$$ are positive integers. Find $$m+n+p+q$$. (第二十届AIME2 2002 第10题)", "## Trigonometry (10th Edition) 1. Since this number is larger than 360°, to find its smallest positive coterminal positive angle, we will subtract 360 to get positive angle less than 360 $699^{\\circ}-360^{\\circ}= 339^{\\circ}$", "and the smallest angles (in degrees)is: A) 20∘ B) 25∘ C) 30∘ D) 35∘", "of $F$ and $Δx$ is the magnitude of $Δx$,and $\\theta$ is the angle between $F$ and $Δx$ Note, in particular that the magnitudes are positive by definition, so the $cos\\theta$ is negative if and only if $\\theta$ is between $90°$ and $180°$.When the angle has these ranges, the the force has a component perpendicular to the direction of motion, and a component opposite the direction of motion. The perpendicular component contributes nothing to the work, and the component opposite the motion contributes a negative amount to the work. Last edited: Feb 25, 2014 3. Feb 25, 2014 Thank you", "## Trigonometry (10th Edition) 1. r is always positive 2. y is negative 3. Therefore, $\\sin\\theta=\\frac{y}{r}=\\frac{-}{+}=-$", "and how come for 14 degrees, there isnt another angle of 180-14? 4. Hello tea123456 Originally Posted by tea123456 can i ask why there are 2 angles that results the same sine value? I don't know how far your study of trigonometry has gone, but if you're using the Sine and Cosine Rules, you need to know how to handle obtuse angles. The rules are: $\\sin(180^o - \\theta)=\\sin \\theta$ and $\\cos(180^o - \\theta)=-\\cos \\theta$ To understand the reasons for these rules, you'll need to look at the definitions of sine and cosine for angles of any size. These are based on the unit circle. You may find this page helpful. and how come for 14 degrees, there isnt another angle of 180-14? Because, in a triangle, the smallest angle is always less than $90^o$ (obviously!) and it is always opposite the shortest side (because of the Sine Rule, where each side is in proportion to the sine of the opposite angle). So the $14^o$ angle is the smallest angle here, because it's opposite the shortest side. And so it's $14^o$, not $166^o$." ]

[ "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$.", "# Difference between revisions of \"Dodecagon\" A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--G); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$.", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "n$ parallel to the bases of the half-circles are drawn such that the distances between $l_k$ and $l_0$ and $l_{-k}$ and $l_0$ are always the same for all $k=1\\to n$. The intersection points of $l_k$ with one of the half-circles are labeled $A_k, B_k$, and with the other half-circle at $C_k,D_k$, as shown in the diagram. Prove that $$\\prod_{k=-n}^n |A_kB_k|+|C_kD_k| \\ge \\prod_{k=-n}^n |A_kD_k|+|B_kC_k|$$ ## Picture 2 $[asy] for (int i=0;i<6;i=i+1){ draw(dir(60*i)--dir(60*i+60)); } draw(dir(120)--(dir(0)+dir(-60))/2); draw(dir(180)--(dir(60)+dir(0))/2); fill(dir(120)--dir(180)--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); fill((dir(0)+dir(-60))/2--dir(0)--(dir(60)+dir(0))/2--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); [/asy]$ $$\\text{Prove the shaded areas are equal.}$$ $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,black); } } for(int i = 1; i < 7; ++i){ filldraw((i,7-i)--(i+1,7-i)--(i+1,8-i)--(i,8-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((i,4-i)--(i+1,4-i)--(i+1,5-i)--(i,5-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((8-i,4+i)--(7-i,4+i)--(7-i,3+i)--(8-i,3+i)--cycle,white); } filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle,white); filldraw((0,1)--(0,2)--(1,2)--(1,1)--cycle,white); filldraw((7,8)--(6,8)--(6,7)--(7,7)--cycle,white); filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white); [/asy]$ ## physics problem $[asy]size(100,100); draw((0,0)--(0,10)); draw(Circle((1,1),1)); dot((0,0)); draw(9.5*dir(80)..9.5*dir(70)..9.5*dir(60),EndArrow()); label(\"A\",(-1,5),dir(180));[/asy]$ $[asy] size(85,85); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw(9.5*dir(50)..9.5*dir(40)..9.5*dir(30),EndArrow()); label(\"B\",(0,5),dir(0)); [/asy]$ $[asy] size(110,110); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); label(\"C\",(5,1.5),dir(90)); [/asy]$ ## Solution $[asy] import olympiad; size(350,350); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw((-1,0)--(7,0),grey); draw((0,0)--(sqrt(3),1),linetype(\"8 8\")); draw((0,0)--(0,5),grey); draw(sqrt(3)*dir(60)--(sqrt(3),1)--(sqrt(3),0),linetype(\"8 8\")); draw(anglemark((1,0),(0,0),dir(30))); label(\"\\varphi\",0.3*dir(15),dir(15)); draw(anglemark(dir(60),(0,0),(0,1))); label(\"\\theta\",0.3*dir(75),dir(75)); label(\"1\",(sqrt(3),0.5),dir(0)); label(\"\\frac{1}{\\tan\\varphi}\",(sqrt(3)/2,0),dir(-90)); [/asy]$ $[asy] import olympiad; size(300,300); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); draw((-1,0)--(12,0),grey); draw((0,0)--(0,3),grey); draw(anglemark(10*dir(11.42),(0,0),(0,1))); label(\"\\theta\",0.3*dir(50.71),dir(50.71)); draw((0,0)--(10,1),linetype(\"8 8\")); draw(10*dir(11.42)--(10,1)--(10,0),linetype(\"8 8\")); label(\"1\",(10,0.5),dir(0)); label(\"10\",(5,0),dir(-90)); label(\"\\varphi\",3.4*dir(2.855),dir(2.855)); markscalefactor=0.4; draw(anglemark((1,0),(0,0),dir(5.71))); [/asy]$ ## inscribed triangle $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(56)--dir(230),green); draw(dir(-23)--dir(-98),red);[/asy]$ $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(0)--dir(36),red); draw(dir(0)--dir(72),red); draw(dir(0)--dir(108),red); draw(dir(0)--dir(144),green);", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "10^{\\circ}(3-\\tan^2 10^{\\circ})}{1-3\\tan^2 10^{\\circ}}\\\\&=-\\tan (3\\times 10)^{\\circ}\\\\&=\\tan 150^{\\circ}\\end{align*} This gives $\\angle CDP=180^{\\circ}-150^{\\circ}=30^{\\circ}$. But $\\angle ACP=30^{\\circ}+40^{\\circ}=70^{\\circ}$, we get $\\angle ADC=70^{\\circ}-30^{\\circ}=40^{\\circ}$. ##### Well-known member Re: Find \\$\\angle ADC\\$ My solution (I just solved it): Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\\angle APB=90^{\\circ}$ and $BP=k$. \\begin{tikzpicture}[auto] \\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x}; \\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y}; \\coordinate[label=left:B] (B) at (0,0); \\coordinate[label=right] (P) at (8,0); \\coordinate[label=left: D] (D) at (5,0); \\coordinate[label=right:C] (C) at (8,-3); \\coordinate[label=right:A] (A) at (8,-8); \\draw (B) to (P); \\draw (B) to (A); \\draw (D) to (A); \\draw (B) to (C); \\draw (A) to (P); \\draw (C) to (D); \\draw [dashed] (D) -- (3.6, 1.4); \\node (1) at (1.5,-0.2) {$10^{\\circ}$}; \\node (2) at (1.5,-0.9) {$30^{\\circ}$}; \\node (3) at (7,-6.3) {$30^{\\circ}$}; \\node (4) at (7.7,-6) {$20^{\\circ}$}; \\draw (P) rectangle +(-0.5, -0.5); \\end{tikzpicture} We then have $C=\\left(k, -k\\tan 10^{\\circ}\\right)\\\\D=\\left(\\dfrac{k}{2\\sin 110^{\\circ}\\cos 40^{\\circ}}, 0\\right)=\\left(\\dfrac{k}{\\sin 150^{\\circ}+\\sin 70^{\\circ}}, 0\\right)=\\left(\\dfrac{k}{\\frac{1}{2}+\\cos 20^{\\circ}}, 0\\right)$ \\begin{align*}\\text{Gradient of } l_{CD}&=\\dfrac{-k\\tan 10^{\\circ}}{k\\left(1-\\dfrac{1}{\\frac{1}{2}+\\cos 20^{\\circ}}\\right)}\\\\&=\\dfrac{-\\tan 10^{\\circ}\\left(\\frac{1}{2}+\\cos 20^{\\circ}\\right)}{\\cos 20^{\\circ}-\\dfrac{1}{2}}\\\\&=\\dfrac{-\\tan 10^{\\circ}(4\\cos^2 10^{\\circ}-1)}{4\\cos^2 10^{\\circ}-3}\\\\&=\\dfrac{-\\tan 10^{\\circ}(4\\cos^2 10^{\\circ}-(\\sin^2 10^{\\circ} +\\cos^2 10^{\\circ}))}{4\\cos^2 10^{\\circ}-3(\\sin^2 10^{\\circ} +\\cos^2 10^{\\circ})}\\\\&=\\dfrac{-\\tan 10^{\\circ}(3\\cos^2 10^{\\circ}-\\sin^2 10^{\\circ})}{\\cos^2 10^{\\circ}-3\\sin^2 10^{\\circ}}\\\\&=\\dfrac{-\\tan 10^{\\circ}(3-\\tan^2 10^{\\circ})}{1-3\\tan^2 10^{\\circ}}\\\\&=-\\tan (3\\times 10)^{\\circ}\\\\&=\\tan 150^{\\circ}\\end{align*} This gives $\\angle CDP=180^{\\circ}-150^{\\circ}=30^{\\circ}$. But $\\angle ACP=30^{\\circ}+40^{\\circ}=70^{\\circ}$, we get $\\angle ADC=70^{\\circ}-30^{\\circ}=40^{\\circ}$. How do we know", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$,", "draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"A\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"B\",(168/5, 24/5),NE); dot((24,0)); label(\"C\",(24,0),NW); dot((40, 0)); label(\"D\",(40, 0),NE); dot((24, 12)); label(\"E\",(24, 12),NE); dot((24, -12)); label(\"F\",(24, -12),SE); dot((54/5, 72/5)); label(\"G\",(54/5, 72/5),NW); dot((0, 0)); label(\"O_1\",(0, 0),S); dot((30, 0)); label(\"O_2\",(30, 0),S); [/asy]$ $r_1 = O_1A = 24$$r_2 = O_2B = 6$$AG = BO_2 = r_2 = 6$$O_1G = r_1 - r_2 = 24 - 6 = 18$$O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$$\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$$\\frac{O_2D}{30} = \\frac{6}{18}$$O_2D = 10$ $CD = O_2D + r_1 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{\\textbf{192}}$ ~isabelchen ## Solution 2 Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$" ]

[ "or }x={{61.475}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\end{align*} 2. Solutions for the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{90}}^\\circ}} \\right]$: $\\scriptsize x ={{28.525}^\\circ}\\text{ or }x={{61.475}^\\circ}$ Back to Exercise 2.2 ### Unit 2: Assessment 1. . 1. . \\scriptsize \\begin{align*}2\\cos \\theta -1 & =0\\\\\\therefore \\cos \\theta & =\\displaystyle \\frac{1}{2}\\end{align*} Ref angle: $\\scriptsize \\theta ={{60}^\\circ}$ Cosine is positive in the first and fourth quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{60}^\\circ}={{300}^\\circ}$ General solution: $\\scriptsize \\theta ={{60}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{300}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ 2. . \\scriptsize \\begin{align*}2{{\\sin }^{2}}\\theta +\\sin \\theta -1 & =0\\\\\\therefore (2\\sin \\theta -1)(\\sin \\theta +1) & =0\\\\\\therefore \\sin \\theta =\\displaystyle \\frac{1}{2}\\text{ } & \\text{or }\\sin \\theta =-1\\end{align*} $\\scriptsize \\sin \\theta =\\displaystyle \\frac{1}{2}$: Ref angle: $\\scriptsize \\theta ={{30}^\\circ}$ Sine is positive in the first and second quadrants. $\\scriptsize \\theta ={{180}^\\circ}-{{30}^\\circ}={{150}^\\circ}$ General solution: $\\scriptsize \\theta ={{30}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{150}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\sin \\theta =1$: Ref angle: $\\scriptsize \\theta ={{90}^\\circ}$ General solution: $\\scriptsize \\theta ={{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ 3. . \\scriptsize \\displaystyle \\begin{align*}4\\sin x+3\\tan x & =\\displaystyle \\frac{3}{{\\cos x}}+4\\quad \\quad \\cos x\\ne 0\\therefore x\\ne {{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore 4\\sin x+3\\displaystyle \\frac{{\\sin x}}{{\\cos x}}-\\displaystyle \\frac{3}{{\\cos x}}-4 & =0\\\\\\therefore 4\\sin x\\cos x+3\\sin x-3-4\\cos x & =0\\\\\\therefore 4\\cos x(\\sin x-1)+3(\\sin x-1) & =0\\\\\\therefore (\\sin x-1)(4\\cos x+3) & =0\\\\\\therefore \\sin x=1\\text{ } & \\text{or }\\cos x=-\\displaystyle \\frac{3}{4}\\end{align*} $\\scriptsize \\sin x=1$: Ref angle: $\\scriptsize x={{90}^\\circ}$ General solution: $\\scriptsize x={{90}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos x=-\\displaystyle \\frac{3}{4}$: Ref angle: $\\scriptsize x={{138.59}^\\circ}$ Cosine is negative in the second", "-- cycle; \\draw[thick] (0,0) -- ({180-\\angle}:1) -- ({-cos(\\angle)}, 0); \\draw[->] ({\\angle/2}:.4) node {$\\theta$} (0:.3) arc (0:\\angle:.3); \\draw[->] ({(180+\\angle)/2}:-.14) node {$180^\\circ+\\theta$} (0:.1) arc (0:{180+\\angle}:.1) ; \\draw[->] ({(180-\\angle)/2}:.29) node {$180^\\circ-\\theta$} (0:.2) arc (0:{180-\\angle}:.2) ; \\end{tikzpicture} We have an equation of the form $\\cos\\theta = -\\cos\\phi$. Given a $\\theta$, for $\\cos\\theta$ to be equal to the opposite of another cosine, we can see that the other angle must either be $180^\\circ-\\theta$ or $180^\\circ+\\theta$. And we may have to add a multiple of $360^\\circ$, which is the period of the circle. So: \\begin{array}{lcl} \\cos(2x+20^\\circ)=-\\cos(x-11^\\circ) \\\\ 2x + 20^\\circ = 180^\\circ - (x-11^\\circ) + 360^\\circ k &\\lor& 2x + 20^\\circ = 180^\\circ + (x-11^\\circ) + 360^\\circ k \\\\ 3x = 171^\\circ + 360^\\circ k &\\lor& x = 149^\\circ + 360^\\circ k \\\\ x = \\frac 13\\cdot171^\\circ + 120^\\circ k &\\lor& x = 149^\\circ + 360^\\circ k \\\\ \\end{array} Thanks a lot Klaas van Aarsen, the alternative method is equally helpful and insightful. Much appreciated. #### Attachments • 28.4 KB Views: 28 #### skeeter ##### Well-known member MHB Math Helper small correction ... #### Attachments • 41.2 KB Views: 25", "solution is correct. ### Example 2.2 Determine the general solution for $\\scriptsize 7\\cos 2\\theta +4=0$. Solution \\scriptsize \\displaystyle \\begin{align*}7\\cos 2\\theta +4&=0\\\\\\therefore \\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\end{align*} Step 1: Use a calculator to determine the reference angle \\scriptsize \\displaystyle \\begin{align*}\\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\\\\\therefore 2\\theta &={{124.8}^\\circ}\\end{align*} Note: We keep working with the reference angle of $\\scriptsize 2\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\scriptsize \\displaystyle \\cos 2\\theta =-\\displaystyle \\frac{4}{7}$. $\\scriptsize \\cos 2\\theta \\lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant. Second quadrant: $\\scriptsize 2\\theta ={{124.8}^\\circ}$ Third quadrant: $\\scriptsize \\cos ({{360}^\\circ}-\\theta )=\\cos \\theta$ $\\scriptsize 2\\theta ={{360}^\\circ}-{{124.8}^\\circ}={{235.2}^\\circ}$. Step 3: Generate the general solution \\scriptsize \\begin{align*}2\\theta & ={{124.8}^\\circ}+k{{.360}^\\circ}\\text{ or 2}\\theta ={{235.2}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{62.4}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} Step 4: Check your general solution $\\scriptsize k=2$: $\\scriptsize \\therefore \\theta ={{62.4}^\\circ}+2\\times {{180}^\\circ}={{422.4}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+2\\times {{180}^\\circ}={{477.6}^\\circ}$ $\\scriptsize \\displaystyle \\cos (2\\times {{422.4}^\\circ})=-\\displaystyle \\frac{4}{7}$ $\\scriptsize \\displaystyle \\cos (2\\times {{477.6}^\\circ})=-\\displaystyle \\frac{4}{7}$ ### Take note! The general solutions for equations involving the three basic trigonometric ratios can be written as follows: If $\\scriptsize \\sin \\theta =x$ then: $\\scriptsize \\theta =\\left( {{{{\\sin }}^{{-1}}}x+k{{{.360}}^\\circ}} \\right)\\text{ or }\\theta =\\left( {\\left( {{{{180}}^\\circ}-{{{\\sin }}^{{-1}}}x} \\right)+k{{{.360}}^\\circ}} \\right),k\\in \\mathbb{Z}$ If $\\scriptsize \\cos \\theta =x$ then: $\\scriptsize \\theta", "# The angle $168^\\circ$ is constructible Prove that the angle $\\theta=168^\\circ$ is constructible using a straightedge and a compass. It is enough to show that the number $\\cos\\theta$ is constructible, and WolframAlpha gave $\\cos\\theta=\\frac18 (1-\\sqrt5)-\\frac14 \\sqrt{{\\frac32} (5+\\sqrt5)}$. I think after that it is immediate because we first add $\\sqrt5$ to $\\mathbb{Q}$ and then add $\\sqrt{{\\frac32} (5+\\sqrt5)}$, so both extensions are of degree $2$. But then how can we find that representation of $\\cos\\theta$? And is there a nicer way? EDIT I found online a cool criterion: can we just state that the order of $168^\\circ$ is $15$, since $168*15$ is divisible by 360, and then since $15=3\\times5$ is a product of Fermat primes we are done? And if so, Do we have to check all numbers $1,...,15$ to see $15$ is indeed the order? • You know(?) that you can construct a regular hexagon ($60°$), and a regular pentagon ($72°$). Then you can also construct $12° = 72° - 60°$. Constructing $12°$ is clearly equivalent to constructing $168° = 180° - 12°$. – Daniel Fischer Feb 9 '14 at 21:51 The canonical pentagon is indeed constructible, as mentioned by the OP, and so is its angle $$\\varphi=\\frac{3\\pi}{5}=\\frac{540^\\circ}{5}=108^\\circ$$ Also $60^\\circ$ is the angle of equilateral triangle, and so it is another constructible angle. Hence $\\vartheta=108^\\circ+60^\\circ=168^\\circ$ is also constructible. The", "$6\\times 4=24$ ones, is divisible by $13$. Also we can replace the rightmost $6\\times 4=24$ ones by zeros. So if the unknown digit is $d$ then $13\\mid 10^{24}(10+d)$. By Fermat's little theorem $10^{12}\\equiv 1\\pmod{13}$, so $13\\mid 10+d$. It follows that $d=3$. $\\square$ 4. Let $\\gcd(a,b)=g$, write $a=gm$ and $b=gn$ with $\\gcd(m,n)=1$. Now the equation rewrites to $(m+n)/mn = g/c$. Now $\\gcd(g,c)=\\gcd(a,b,c)=1$ and $\\gcd(m+n,mn)=\\gcd(m,n)=1$ so $m+n=g$. Substituting this value yields $a+b=gm+gn=g(m+n)=g^2$. $\\square$ 5. Notice that $\\angle AOB=2\\angle ADB=2\\angle ADE=2\\left(90^\\circ-\\angle DAE\\right)=180^\\circ-2\\angle DAC=180^\\circ-\\angle COD$. Now let $\\angle AOB=\\theta$. Now $\\left(EA^2+EB^2\\right)+\\left(EC^2+ED^2\\right)=AB^2+CD^2$ by Pythagorean theorem. Let $R$ be the radius. By law of cosine $OA^2+OB^2=AB^2+2R^2\\cos \\theta$ and $OC^2+OD^2=CD^2+2R^2\\cos\\left(180^\\circ-\\theta\\right)=CD^2-2R^2\\cos\\theta$. Adding these we have $OA^2+OB^2+OC^2+OD^2=AB^2+CD^2$ as well. $\\square$ 6. Clearly the circles touch at the midpoints of the sides. It is then a straight computation using cosine law to find that $r_1=a^2/8b+b/2$ and $r_2=b^2/8a+a/2$. Subbing and simplifying reduces the inequality to $a^2/b+b^2/a\\ge a+b$ which follows by Titu's lemma. $\\square$ - 4 years, 11 months ago I can't see why I mustn't call you god. - 4 years, 11 months ago I can't see why I mustn't call you god. - 4 years, 11 months ago I can't see why I mustn't call you god. - 4 years, 11 months ago I can't see why I mustn't call you god. - 4 years, 11 months ago", ". \\end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) Solution 6 (Law of Cosines) WLOG, let $P$ be on minor arc $\\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\\theta = \\angle AOP$. By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\\triangle AOP, \\, \\triangle COP, \\, \\triangle BOP, \\, \\triangle DOP$ to get \\begin{align*} PA^2 &= 2r^2(1 - \\cos \\theta), \\\\ PC^2 &= 2r^2(1 - \\cos (180 - \\theta)) = 2r^2(1 + \\cos \\theta), \\\\ PB^2 &= 2r^2(1 - \\cos (90 - \\theta)) = 2r^2(1 - \\sin \\theta), \\\\ PD^2 &= 2r^2(1 - \\cos (90 + \\theta)) = 2r^2(1 + \\sin \\theta). \\end{align*} Taking the products of the first two and last two equations, respectively, $$56^2 = (PA \\cdot PC)^2 = 4r^4(1 - \\cos \\theta)(1 + \\cos \\theta) = 4r^4(1 - \\cos^2 \\theta) = 4r^4 \\sin^2 \\theta,$$ and $$90^2 = (PB \\cdot PD)^2 = 4r^4(1 - \\sin \\theta)(1 + \\sin \\theta) = 4r^4(1 - \\sin^2 \\theta) = 4r^4 \\cos^2 \\theta.$$ Adding these equations, $$56^2 + 90^2 = 4r^4,$$ so $$2r^2 = \\sqrt{56^2+90^2} = 2\\sqrt{28^2+45^2} = 2\\sqrt{2809} = 2 \\cdot 53 = \\boxed{106}.$$ ~OrangeQuail9 Solution 7 (Subtended Chords) First draw a diagram. $[asy] pair A, B,", "and third quadrants. $\\scriptsize x={{360}^\\circ}-{{138.59}^\\circ}={{221.41}^\\circ}$ General solution: $\\scriptsize x={{138.59}^\\circ}+k{{.360}^\\circ}\\text{ or }x={{221.49}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ For the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{180}}^\\circ}} \\right]$: $\\scriptsize x={{90}^\\circ}\\text{ or }x={{138.59}^\\circ}$ 4. . \\scriptsize \\begin{align*}6\\cos \\theta -5 & =\\displaystyle \\frac{4}{{\\cos \\theta }}\\quad \\quad \\cos \\theta \\ne 0\\therefore \\theta \\ne {{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore 6{{\\cos }^{2}}\\theta -5\\cos \\theta -4 & =0\\\\\\therefore (3\\cos \\theta -4)(2\\cos \\theta +1) & =0\\\\\\therefore \\cos \\theta =\\displaystyle \\frac{4}{3}\\text{ } & \\text{or }\\cos \\theta =-\\displaystyle \\frac{1}{2}\\end{align*} $\\scriptsize \\cos \\theta =\\displaystyle \\frac{4}{3}$ – No solution $\\scriptsize \\cos \\theta =-\\displaystyle \\frac{1}{2}$: Ref angle: $\\scriptsize \\theta ={{120}^\\circ}$ Cosine is negative in the second and third quadrants. $\\scriptsize \\theta ={{360}^\\circ}-({{120}^\\circ})={{240}^\\circ}$ General solution: $\\scriptsize \\theta ={{120}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{240}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ 5. . \\scriptsize \\begin{align*}\\cos 2x-\\cos x+1 & =0\\\\\\therefore 2{{\\cos }^{2}}x-1-\\cos x+1 & =0\\\\\\therefore 2{{\\cos }^{2}}x-\\cos x & =0\\\\\\therefore \\cos x(2\\cos x-1) & =0\\\\\\therefore \\cos x=0\\text{ } & \\text{or }\\cos x=\\displaystyle \\frac{1}{2}\\end{align*} $\\scriptsize \\cos x=0$: Ref angle: $\\scriptsize x={{90}^\\circ}$ General solution: $\\scriptsize x={{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos x=\\displaystyle \\frac{1}{2}$ Ref angle: $\\scriptsize \\theta ={{60}^\\circ}$ Cosine is positive in the first and fourth quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{60}^\\circ}={{300}^\\circ}$ General solution: $\\scriptsize \\theta ={{60}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{300}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ For the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{360}}^\\circ}} \\right]$: $\\scriptsize x=\\text{6}{{\\text{0}}^\\circ}\\text{ or }x={{90}^\\circ}\\text{ or }x={{180}^\\circ}\\text{ or }x=\\text{30}{{\\text{0}}^\\circ}\\text{ or }x=\\text{36}{{\\text{0}}^\\circ}$ 2. $\\scriptsize \\cos 2\\theta =2{{\\cos }^{2}}\\theta -1$ 1. . \\scriptsize \\begin{align*}\\text{LHS}=\\cos 2\\theta +3\\cos \\theta -1\\\\=2{{\\cos }^{2}}\\theta -1+3\\cos \\theta -1\\\\=2{{\\cos }^{2}}\\theta +3\\cos", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "the sum of angles in a triangle is 180º. 180 – 48 – 90 = 180º so the third angle in the triangle is 42º. Since \\theta and 42^o are on a straight line, \\begin{aligned} \\theta&=180-42 \\\\\\\\ \\theta&=138^{\\circ} \\end{aligned} Common misconceptions • Mixing up angle facts There are a lot of angle facts and it is easy to mistake alternate angles with corresponding angles. To prevent this from occurring, think about co-interior angles being inside the C shape. • Using a protractor to measure an angle Using a protractor to measure an angle. Most diagrams are not to scale and so using a protractor will not result in a correct answer unless it is a coincidence. Practice co-interior angles questions \\theta =136^{\\circ} \\theta =44^{\\circ} \\theta =143^{\\circ} \\theta =37^{\\circ} \\theta and 44^{\\circ} are co-interior angles. Therefore, \\theta = 180-44=136^{\\circ} 2. Calculate the value of \\theta \\theta =125^{\\circ} \\theta =55^{\\circ} \\theta =70^{\\circ} \\theta =62.5^{\\circ} Using co-interior angles, we can calculate 180-125=55^{\\circ} Using co-interior angles again, we can see that \\theta=180-55=125^{\\circ} 3. Given the information in the diagram, calculate the size of angle \\theta \\theta =110^{\\circ} \\theta =70^{\\circ} \\theta =40^{\\circ} \\theta =140^{\\circ} Using co-interior angles, we can calculate 180-110=70^{\\circ} Since it is an isosceles triangle, the other angle at the bottom of the triangle is 70^{\\circ} as well. Then, using angles", "Exercise 2.1 ### Exercise 2.2 1. . 1. . \\scriptsize \\begin{align*}\\cos 2x & =\\sin {{32}^\\circ}\\\\\\therefore \\cos 2x & =0.530\\end{align*} Or \\scriptsize \\begin{align*}\\cos 2x & =\\sin {{32}^\\circ}\\\\\\therefore \\cos 2x & =\\cos ({{90}^\\circ}-{{32}^\\circ})\\\\&=\\cos {{58}^\\circ}\\end{align*} Ref angle: $\\scriptsize \\theta ={{58}^\\circ}$ Cosine is positive in the first and fourth quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{58}^\\circ}={{302}^\\circ}$ General solution: $\\scriptsize \\theta ={{58}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{302}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ So: \\scriptsize \\begin{align*}2x&={{58}^\\circ}+k{{.360}^\\circ}\\text{ or }2x={{302}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\\\x&={{29}^\\circ}+k{{.180}^\\circ}\\text{ or }x={{151}^\\circ}+k{{.180}^\\circ}\\text{, }k\\in \\mathbb{Z}\\end{align*} 2. . \\scriptsize \\begin{align*}\\cos 2x & =\\sin x\\\\\\therefore 1-2{{\\sin }^{2}}x & =\\sin x\\\\\\therefore 2{{\\sin }^{2}}x+\\sin x-1 & =0\\\\\\therefore (2\\sin x-1)(\\sin x+1) & =0\\\\\\therefore \\sin x=\\displaystyle \\frac{1}{2}\\text{ } & \\text{or }\\sin x=-1\\end{align*} $\\scriptsize \\sin x=\\displaystyle \\frac{1}{2}$: Ref angle: $\\scriptsize x={{30}^\\circ}$ Sine is positive in the first and second quadrants. $\\scriptsize x={{180}^\\circ}-{{30}^\\circ}={{150}^\\circ}$ General solution: $\\scriptsize x={{30}^\\circ}+k{{.360}^\\circ}\\text{ or }x={{150}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\sin x=-1$: Ref angle: $\\scriptsize x=-{{90}^\\circ}$ General solution: $\\scriptsize x={{270}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ 3. . \\scriptsize \\begin{align*}\\sin \\theta \\cdot \\sin 2\\theta +\\cos 2\\theta & =1\\\\\\therefore \\sin \\theta \\cdot 2\\sin \\theta \\cos \\theta +{{\\cos }^{2}}\\theta -{{\\sin }^{2}}\\theta -1 & =0\\\\\\therefore 2{{\\sin }^{2}}\\theta \\cos \\theta +{{\\cos }^{2}}\\theta -1-{{\\sin }^{2}}\\theta & =0\\\\\\therefore 2{{\\sin }^{2}}\\theta \\cos \\theta -{{\\sin }^{2}}\\theta -{{\\sin }^{2}}\\theta & =0\\\\\\therefore 2{{\\sin }^{2}}\\theta \\cos \\theta -2{{\\sin }^{2}}\\theta & =0\\\\\\therefore 2{{\\sin }^{2}}\\theta (\\cos \\theta -1) & =0\\\\\\therefore 2{{\\sin }^{2}}\\theta &=0\\text{ } & \\text{or cos}\\theta =1\\\\\\therefore \\sin \\theta &=0\\text{ } & \\text{or }\\cos \\theta =1\\end{align*} $\\scriptsize \\sin \\theta", "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", ")}$ and adjacent side of size ${\\displaystyle \\cos(\\theta )}$ in units of the hypotenuse. Neither sine nor cosine can ever exceed 1 and the closer one of them is to 1, the closer the other must be to 0. We can see this in two ways: • It follows immediately from the formula. As either sine squared or cosine squared gets closer to one the amount left for the other diminishes. • It can be seen from the geometry. The hypotenuse is one and is longer than either of the other sides. As one side gets closer to one, the other must get closer to 0. ## Using the Formula Cosine 30 degrees ${\\displaystyle \\sin(30^{\\circ })=0.5}$ . What is ${\\displaystyle \\cos(30^{\\circ })}$? Well, ${\\displaystyle \\sin ^{2}(30^{\\circ })+\\cos ^{2}(30^{\\circ })=1}$ ${\\displaystyle (0.5)^{2}+\\cos ^{2}(30^{\\circ })=1}$ ${\\displaystyle \\cos ^{2}(30^{\\circ })=1-(0.5)^{2}=0.75}$ ${\\displaystyle \\cos(30^{\\circ })={\\sqrt {0.75}}\\approx 0.866}$ In the very last step we had to decide between ${\\displaystyle 0.866}$ and ${\\displaystyle -0.866}$ . We chose the positive value because from looking at a triangle we know ${\\displaystyle \\cos(30^{\\circ })}$is positive. ## This formula and Circles In an earlier exercise we asked you to plot points ${\\displaystyle {\\big (}\\cos(t),\\sin(t){\\big )}}$ , and you got a circle - well, you should have done. For each point you plot, there is a right triangle with coordinates", "of the triangle: the horizontal side has length $x$ and the vertical side has length $y$. If we make the circle have radius $1$, then the hypotenuse of the triangle is $1$, and we have \\begin{align*} \\cos \\theta &= \\frac{x}{1} = x \\\\ \\sin \\theta &= \\frac{y}{1} = y \\end{align*} and our point $(x,y)$ can be written as $(\\cos \\theta, \\sin \\theta)$. Immediately, we notice two important things. First, using the Pythagorean theorem on the triangle, we see that \\begin{align*} \\sin^2 \\theta + \\cos^2 \\theta = 1. \\end{align*} This is a handy equation that can be useful in simplifying trigonometric expressions. For example, the expression $(\\sin \\theta + \\cos \\theta)^2 + (\\sin \\theta - \\cos \\theta)^2$ is actually just equivalent to $2$. \\begin{align*} &(\\sin \\theta + \\cos \\theta)^2 + (\\sin \\theta - \\cos \\theta)^2 \\\\ &= (\\sin^2 \\theta + 2\\sin \\theta \\cos \\theta + \\cos^2 \\theta) \\\\ &\\hspace{1cm} + (\\sin^2 \\theta - 2 \\sin \\theta \\cos \\theta + \\cos^2 \\theta) \\\\ &= 2 \\sin^2 \\theta + 2 \\cos^2 \\theta \\\\ &= 2 \\end{align*} Second, angles repeat every $360^\\circ$, since going $360^\\circ$ around the circle brings us back to the starting point of $0^\\circ$. That means, for example, that $\\sin 390^\\circ$ and $\\sin -330^\\circ$ are both equivalent to $\\sin 30^\\circ$. \\begin{align*} 390^\\circ &= 30^\\circ + 360^\\circ \\\\ -330^\\circ &=", "0 like 0 dislike 166 views Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. 1. $\\alpha=60^{\\circ}$ 2. $\\beta=-225^{\\circ}$ 3. $\\gamma=540^{\\circ}$ 4. $\\varphi=-750$ | 166 views 0 like 0 dislike 1. To graph $\\alpha=60^{\\circ}$, we draw an angle with its initial side on the positive $x$-axis and rotate counter-clockwise $\\frac{60^{\\circ}}{360^{\\circ}}=\\frac{1}{6}$ of a revolution. We see that $\\alpha$ is a Quadrant I angle. To find angles which are coterminal, we look for angles $\\theta$ of the form $\\theta=\\alpha+360^{\\circ} \\cdot k$ for some integer $k$. - When $k=1$, we get $\\theta=60^{\\circ}+360^{\\circ}=420^{\\circ}$. - Substituting $k=-1$ gives $\\theta=60^{\\circ}-360^{\\circ}=-300^{\\circ}$. - If we let $k=2$, we get $\\theta=60^{\\circ}+720^{\\circ}=780^{\\circ}$. 2. Since $\\beta=-225$ is negative, we start at the positive $x$-axis and rotate clockwise $\\frac{225}{360^{\\circ}}=\\frac{5}{8}$ of a revolution. We see that $\\beta$ is a Quadrant II angle. To find coterminal angles, we proceed as before and compute $\\theta=-225^{\\circ}+360^{\\circ} \\cdot k$ for integer values of $k$. Letting $k=1$ $k=-1$ and $k=2$, we find $135^{\\circ},-585^{\\circ}$ and $495^{\\circ}$ are all coterminal with $-225^{\\circ}$. 3. Since $\\gamma=540$ is positive, we rotate counter-clockwise from the positive $x$-axis. One full revolution accounts for $360^{\\circ}$, with $180^{\\circ}$, or half of a revolution, remaining. Since the", "of $\\theta$ will be $k$. Then according to the Pythagorean Theorem, the length of the hypotenuse will be $\\sqrt{13}k$. Then we have $\\sin(\\theta)=2\\sqrt{3}/\\sqrt{13}$ and $\\cos(\\theta)=1/\\sqrt{13}$. Now back to the original problem. Consider the right angled $\\triangle BEC$. We get $$\\angle EBC + \\angle BEC + \\angle BCE = 180^\\circ \\Rightarrow \\angle EBC = 30^\\circ$$ Now consider $\\triangle BFD$. We have $$\\angle BFD = 180^\\circ - (\\angle EBC + \\theta) = 180^\\circ - (30^\\circ + \\theta)$$ Apply sine rule to $\\triangle BFD$. \\begin{align} BD/\\sin(180^\\circ - (30^\\circ + \\theta)) &= BF/\\sin(\\theta)\\\\ BD/\\sin(30^\\circ + \\theta) &= BF/\\sin(\\theta)\\\\ BD/(\\sin(30^\\circ)\\cos(\\theta) + \\cos(30^\\circ)\\sin(\\theta)) &= BF/\\sin(\\theta)\\\\ BD/(1/2\\cdot1/\\sqrt{13} + \\sqrt{3}/2\\cdot2\\sqrt{3}/\\sqrt{13}) &= BF/(2\\sqrt{3}/\\sqrt{13})\\hspace{4cm}\\\\ BD/((1/2 + 3)/\\sqrt{13}) &= BF/(2\\sqrt{3}/\\sqrt{13})\\\\ BD/(7/2) &= BF/(2\\sqrt{3})\\\\ BF &= BD\\cdot2/7\\cdot2\\sqrt{3}\\\\ &= 60\\sqrt{3}/7\\text{ cm} \\tag{since BD=15\\text{ cm}} \\end{align} Now in the right angled $\\triangle BEC$, \\begin{align} \\sin(\\alpha) &= BE/BC\\\\ BE/BC &= \\sqrt{3}/2\\\\ BE &= 15\\sqrt{3}\\text{ cm}\\tag{since BC = 30\\text{ cm}} \\end{align} Now we wish to find $EF$, so $EF = BE-BF = (15-60/7)\\sqrt{3}\\text{ cm} = 45\\sqrt{3}/7\\text{ cm}$. This should be our final answer. But they say that my answer is wrong. Can anybody please point out the mistake in my arguments? - Just to clarify, in the sentence \"determine the distance from the foot of the perpendicular to this point\", by \"the foot of the perpendicular\" I mean the foot of the perpendicular from", "30^\\circ \\Omega \\: = \\: 57.735 \\: + \\: j33.333 \\Omega$ and $E_C = 1.5 \\angle -30^\\circ A \\: \\times \\: 159.155 \\angle -90^\\circ \\Omega \\: = \\: 238.732 \\angle -120^\\circ V \\: = \\: -119.366 \\: - \\: j206.748 V$ Now, for the tricky part: Since we know the angular relationship between $E_R$ and $E_C$, we know that $E_R$ is 90° ahead of $E_C$ which turns out to be (-120°) + 90° = (-30°) Also, we know that $E_L$ is 90° ahead of $E_R$, which is (-30°) + 90° = 60° Now, all we have to do is find the magnitude of those values: Since $E_L$ and $E_C$ are 180° apart, the sum of their angles must be either 60° or -120°, therefore it must be 60° so that, when added to [E_R] (which has an angle of -30°), we get a vector resulting with an angle of 0°. Also, the x-components of the sum of $E_L$, $E_C$, and $E_R$ must be 100 and the sum of their y-components must be zero. This gives us: $E_{L+C} \\: \\times \\: \\cos{60} \\: + \\: E_R \\: \\times \\: \\cos(-30) =100$ $\\frac{E_{L+C}}{2} \\: + \\: \\frac{E_R \\: \\times \\: \\sqrt{3}}{2} \\: = \\: 100$ $E_{L+C} \\: + \\: (E_R \\: \\times \\: \\sqrt{3}) \\: = \\: 200$ and $E_{L+C}", "Note: You can also express your reference angle as a positive angle – $\\scriptsize {{360}^\\circ}-{{68.2}^\\circ}={{291.8}^\\circ}$ General solution: $\\scriptsize \\theta =-{{68.2}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }$ 3. . \\scriptsize \\displaystyle \\begin{align*}8\\sin \\theta +3 & =0\\\\\\therefore \\sin \\theta & =-\\displaystyle \\frac{3}{8}\\end{align*} Ref angle: $\\scriptsize \\theta =-{{22.0}^\\circ}$ Note: You can also express your reference angle as a positive angle – $\\scriptsize {{360}^\\circ}-{{22.0}^\\circ}={{338}^\\circ}$ $\\scriptsize \\sin \\theta$ is negative in the third and fourth quadrants. $\\scriptsize \\theta ={{180}^\\circ}+{{22.0}^\\circ}={{202.0}^\\circ}$ General solution: $\\scriptsize \\theta =-{{22.0}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{202.0}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }$ 4. . \\scriptsize \\begin{align*}-6\\cos 2\\theta -3 & =0\\\\\\therefore \\cos 2\\theta & =-0.5\\end{align*} Ref angle: $\\scriptsize \\theta ={{120}^\\circ}$ $\\scriptsize \\cos \\theta$ is negative in the second and third quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{120.0}^\\circ}={{240}^\\circ}$ General solution: \\scriptsize \\begin{align*}2\\theta & ={{120}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{240}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{60}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{120}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\end{align*} 2. . 1. . \\scriptsize \\begin{align*}3\\cos \\theta -2 & =-3\\\\\\therefore \\cos \\theta & =-\\displaystyle \\frac{1}{3}\\end{align*} Ref angle: $\\scriptsize \\theta ={{109.5}^\\circ}$ $\\scriptsize \\cos \\theta$ is negative in the second and third quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{109.5}^\\circ}={{250.5}^\\circ}$ General solution: $\\scriptsize \\theta ={{109.5}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{250.5}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }$ Specific solution: $\\scriptsize \\theta ={{109.5}^\\circ}\\text{ or }\\theta ={{250.5}^\\circ}$ 2. $\\scriptsize \\tan (3\\theta -{{42}^\\circ})=3.4$ Ref angle: $\\scriptsize 3\\theta -{{42}^\\circ}={{73.6}^\\circ}$ General solution: \\scriptsize \\begin{align*}3\\theta -{{42}^\\circ} &={{73.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore 3\\theta &={{115.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore \\theta & ={{38.5}^\\circ}+k{{.60}^\\circ},k\\in \\mathbb{Z}\\end{align*} Specific solution: $\\scriptsize \\theta ={{38.5}^\\circ}\\text{ or }\\theta ={{98.5}^\\circ}\\text{ or }\\theta ={{158.5}^\\circ}$ Back to", "as in the figure below Keep in mind that in such problems, the given angles must be measured from a reference axis in a clockwise direction. Here, we choose the positive $x$ axis as the reference. Because it is easier to reach the $23^\\circ$ west of south from the $+x$ axis in a counterclockwise direction, so after adding $90^\\circ$ to $23^\\circ$, we must insert a negative in front of the result. Therefore, the angles of $23^\\circ$ west of south and $30^\\circ$ east of south are measured from $+x$ axis as $\\alpha=90^\\circ+23^\\circ=-113^\\circ$ and $180^\\circ+30^\\circ=210^\\circ$. For the first flight, the components of the displacement vector are obtained as below \\begin{align*} A_x&=|\\vec{A}|\\cos \\alpha\\\\&=250\\times \\cos (-113^\\circ) \\\\&=-97.68\\quad {\\rm m} \\\\\\\\ A_y&=|\\vec{A}|\\sin \\alpha\\\\&=250\\times \\sin (-113^\\circ) \\\\&=-230.12\\quad {\\rm m}\\end{align*} Similarly, for the second flight, the components for the displacement vector $\\vec{B}$ are \\begin{align*} B_x&=|\\vec{B}|\\cos \\theta\\\\&=200\\times \\cos (210^\\circ) \\\\&=-173.20\\quad {\\rm m} \\\\\\\\ B_y&=|\\vec{A}|\\sin \\theta\\\\&=200\\times \\sin (210^\\circ) \\\\&=-100\\quad {\\rm m}\\end{align*} Now, adding the corresponding components gives the components of the displacement vector $\\vec{D}$ \\begin{align*} D_x&=A_x+B_x\\\\&=-97.68-173.20\\\\&=-270.88\\quad {\\rm m}\\\\\\\\ D_y&=A_y+B_y\\\\&=-230.12-100\\\\&=-330.12\\quad {\\rm m}\\end{align*} Recall from physics that, the magnitude of the displacement vector is the direct distance between start and end points. Problem (11): A ball is fired with an initial speed of 15 m/s at an angle of $37^\\circ$ with the horizontal. Find the vertical and horizontal components", "part because the approach is so counter-intuitively simple, so much so that it took me days to convince other engineers it was right. A senior engineer, late to a meeting, even left me at his desk with a blank piece of paper and a pencil and the dismissive farewell of \"Try to prove to yourself why you're wrong before I come back\". In the engineering world, test cases can make a good case for the algorithm's veracity. Here is a batch of test cases. Here $\\delta$ is the \"original\" half angle of the wedge; it is not used except to help generate test cases. $\\mathrm{M}$ $\\delta$ $p$ $c$ $\\delta_1=(\\delta+p)-c$ $\\delta_2=(\\delta-p)+c$ $\\theta_1=\\theta(\\mathrm{M},\\delta+p)-c$ $\\theta_2=\\theta(\\mathrm{M},\\delta-p)+c$ $2$ $10^{\\circ}$ $0^{\\circ}$ $0^{\\circ}$ $10^{\\circ}$ $10^{\\circ}$ $\\theta_{10}-0^{\\circ}=39.31^{\\circ}$ $\\theta_{10}+0^{\\circ}=39.31^{\\circ}$ $2$ $10^{\\circ}$ $5^{\\circ}$ $5^{\\circ}$ $10^{\\circ}$ $10^{\\circ}$ $\\theta_{15}-5^{\\circ}=40.34^{\\circ}$ $\\theta_{5}+5^{\\circ}=39.30^{\\circ}$ $2$ $10^{\\circ}$ $5^{\\circ}$ $0^{\\circ}$ $15^{\\circ}$ $5^{\\circ}$ $\\theta_{15}-0^{\\circ}=45.34^{\\circ}$ $\\theta_{5}+0^{\\circ}=34.30^{\\circ}$ $2$ $10^{\\circ}$ $0^{\\circ}$ $5^{\\circ}$ $5^{\\circ}$ $15^{\\circ}$ $\\theta_{10}-5^{\\circ}=34.31^{\\circ}$ $\\theta_{10}+5^{\\circ}=44.31^{\\circ}$ $2$ $10^{\\circ}$ $5^{\\circ}$ $-5^{\\circ}$ $20^{\\circ}$ $0^{\\circ}$ $\\theta_{15}+5^{\\circ}=50.34^{\\circ}$ $\\theta_{5}-5^{\\circ}=29.30^{\\circ}$ $2$ $10^{\\circ}$ $-5^{\\circ}$ $5^{\\circ}$ $0^{\\circ}$ $20^{\\circ}$ $\\theta_{5}-5^{\\circ}=29.30^{\\circ}$ $\\theta_{15}+5^{\\circ}=50.34^{\\circ}$ $2$ $15^{\\circ}$ $-5^{\\circ}$ $5^{\\circ}$ $5^{\\circ}$ $25^{\\circ}$ $\\theta_{10}-5^{\\circ}=34.31^{\\circ}$ $\\theta_{20}+5^{\\circ}=58.42^{\\circ}$ Here, $\\theta(\\mathrm{M},\\delta)$ denotes a function to calculate shock angle from Mach number and wedge angle. This can be accomplished with equation 150a (the solution of which was the motivation for the previous post). The four test shock angle values are $\\theta_{5}=34.30^{\\circ}$, $\\theta_{10}=39.31^{\\circ}$, $\\theta_{15}=45.34^{\\circ}$, $\\theta_{20}=53.42^{\\circ}$. If you write your own tests,", ". The angle completes one revolution to give ${360}^{\\circ }$ and then continues to give the required angle. We get the following results: $\\begin{array}{ccc}\\hfill sin\\left({360}^{\\circ }+\\theta \\right)& =& sin\\theta \\hfill \\\\ \\hfill cos\\left({360}^{\\circ }+\\theta \\right)& =& cos\\theta \\hfill \\\\ \\hfill tan\\left({360}^{\\circ }+\\theta \\right)& =& tan\\theta \\hfill \\end{array}$ Note also, that if $k$ is any integer, then $\\begin{array}{ccc}\\hfill sin\\left(k{360}^{\\circ }+\\theta \\right)& =& sin\\theta \\hfill \\\\ \\hfill cos\\left(k{360}^{\\circ }+\\theta \\right)& =& cos\\theta \\hfill \\\\ \\hfill tan\\left(k{360}^{\\circ }+\\theta \\right)& =& tan\\theta \\hfill \\end{array}$ Write $sin{293}^{\\circ }$ as the function of an acute angle. 1. $\\begin{array}{ccc}\\hfill sin{293}^{\\circ }& =& sin\\left({360}^{\\circ }-{67}^{\\circ }\\right)\\hfill \\\\ & =& -sin{67}^{\\circ }\\hfill \\end{array}$ where we used the fact that $sin\\left({360}^{\\circ }-\\theta \\right)=-sin\\theta$ . Check, using your calculator, that these values are in fact equal: $\\begin{array}{ccc}\\hfill sin{293}^{\\circ }& =& -0,92\\cdots \\hfill \\\\ \\hfill -sin{67}^{\\circ }& =& -0,92\\cdots \\hfill \\end{array}$ Evaluate without using a calculator: ${tan}^{2}{210}^{\\circ }-\\left(1+cos{120}^{\\circ }\\right){sin}^{2}{225}^{\\circ }$ 1. $\\begin{array}{ccc}& & {tan}^{2}{210}^{\\circ }-\\left(1+cos{120}^{\\circ }\\right){sin}^{2}{225}^{\\circ }\\hfill \\\\ & =& {\\left[tan\\left({180}^{\\circ }+{30}^{\\circ }\\right)\\right]}^{2}-\\left[1+cos\\left({180}^{\\circ }-{60}^{\\circ }\\right)\\right]·{\\left[sin\\left({180}^{\\circ }+{45}^{\\circ }\\right)\\right]}^{2}\\hfill \\\\ & =& {\\left(tan{30}^{\\circ }\\right)}^{2}-\\left[1+\\left(-cos{60}^{\\circ }\\right)\\right]·{\\left(-sin{45}^{\\circ }\\right)}^{2}\\hfill \\\\ & =& {\\left(\\frac{1}{\\sqrt{3}}\\right)}^{2}-\\left(1,-,\\frac{1}{2}\\right)·{\\left(-,\\frac{1}{\\sqrt{2}}\\right)}^{2}\\hfill \\\\ & =& \\frac{1}{3}-\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\hfill \\\\ & =& \\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}\\hfill \\end{array}$ ## Reduction formulae 1. Write these equations as a function of $\\theta$ only: 1. $sin\\left({180}^{\\circ }-\\theta \\right)$ 2. $cos\\left({180}^{\\circ }-\\theta \\right)$ 3. $cos\\left({360}^{\\circ }-\\theta \\right)$ 4. $cos\\left({360}^{\\circ }+\\theta \\right)$ 5. $tan\\left({180}^{\\circ }-\\theta \\right)$" ]

[ "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "+0 # Help hard quick 0 68 0 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ Use the picture below to find vectors with integer components and magnitude no greater than 8 that satisfy $$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "+0 # math help 0 89 1 Let P be the matrix that projects onto j that is, we want P to satisfy $$\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ for all vectors v. Use the pictures below to calculate $$\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$$. Calculate the matrix P that projects onto j. Aug 15, 2019 $$\\text{v projected onto (0,1,0) is (v \\cdot (0,1,0))(0,1,0)}\\\\ v \\cdot (0,1,0) = v_y\\\\ v \\perp (0,1,0) = (0,v_y,0)\\\\ \\text{The matrix that will perform this is}\\\\ P=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "Vector to the Zero Vector Let $T : \\mathbb{R}^n \\to \\mathbb{R}^m$ be a linear transformation. Let $\\mathbf{0}_n$ and $\\mathbf{0}_m$ be zero vectors of $\\mathbb{R}^n$ and... Close", "# Is the zero Matrix a vector space? ## Homework Statement So I have these two Matrices: M = \\begin{pmatrix} a & -a-b \\\\ 0 & a \\\\ \\end{pmatrix} and N = \\begin{pmatrix} c & 0 \\\\ d & -c \\\\ \\end{pmatrix} Where a,b,c,d ∈ ℝ Find a base for M, N, M +N and M ∩ N. ## Homework Equations I know the 8 axioms about the vector spaces. ## The Attempt at a Solution I chose these fours matrices as a base for the first three vector spaces. \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix} I got that the only vector space satisfying M ∩ N is \\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} If M ∩ N constitutes a vector space, I can use the same base as the other three. But if it doesnt, I have to explain why. Not sure how to go about that...", "+0 # HELP ASAP PLEEASSE!!! 0 89 1 Let $$\\mathbf{A}$$ be a matrix such that $$\\mathbf{A}^{-1} = \\begin{pmatrix} 1 & 1 \\\\ 2 & x \\end{pmatrix}$$ for some value of $$x$$. What is the vector that $$\\mathbf{A}$$ maps to $$\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}?$$ Oct 13, 2019 The answer is $$\\begin{pmatrix} 3 \\\\ -7 \\end{pmatrix}$$.", "# Math Help - Projection Matrix 1. ## Projection Matrix The vectors a1= (1, ,1 0) and a2= (1, 1, 1) span a plan in R3. Find a projection matrix P onto the plane, and find a nonzero vector b that is projected to zero. I know a projection matrix is found with the formula A(A^tA)^-1A^t, but I don't really understand what the question is asking. I've come up with P=[ (.5, .5, 0), .5, .5, 0), (0, 0, 1] and b = [1, -1, 0]. Does this look OK?", "oldest newest most voted Q1. In his lecture, Prof. Strang begins with the simplest case: the orthogonal projection of a vector $\\mathbf{b}$ onto a one dimensional subspace $S=\\mathrm{span}\\langle\\mathbf{a}\\rangle$, $\\mathbf{a}$ being a non-null vector in $\\mathbb{R}^n$. The projection vector $\\mathbf{p}$ should be a multiple of $\\mathbf{a}$, so Prof. Strang writes it as $\\mathbf{p}=x\\mathbf{a}$. Here $x$ is a scalar number, not a matrix, and $\\mathbf{a}$ is a vector, identified with a $n\\times 1$ matrix. He wants to deduce that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The", "## A square root of the zero matrix Find a nonzero matrix $$\\mathbf A$$ such that $$\\mathbf A\\mathbf A=\\mathbf 0$$. For instance, $$\\mathbf A= \\begin{pmatrix} 1 & 1 \\\\ -1 & -1 \\\\ \\end{pmatrix}$$.", "### Solving Matrix Equations #### Definition Given matrices $\\mathbf{A}$ and $\\mathbf{B}$, we can solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$ for the unknown matrix $\\mathbf{X}$ when the matrix inverse of $\\mathbf{A}$ exists. We do this by multiplying both sides of the equation on the left by $\\mathbf{A}^{-1}$, providing it exists, to give \\begin{align} \\mathbf{A} \\mathbf{X} &= \\mathbf{B} \\\\ \\mathbf{A}^{-1} \\mathbf{A} \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\\\ \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\end{align} as $\\mathbf{A}^{-1} \\mathbf{A} = \\mathbf{I}$ and multiplying any matrix by the identity matrix leaves it unaltered. It is important that both sides of the equation are multiplied on the left by $\\mathbf{A}^{-1}$ because matrix multiplication does not commute - multiplying on the right could give a different result. #### Worked Example ###### Example 1 Let \\begin{align} \\mathbf{A} &= \\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}, \\\\ \\mathbf{B} &= \\begin{pmatrix}4\\\\1\\end{pmatrix}. \\end{align} Solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$. ###### Solution We have the equation $\\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}\\mathbf{X} = \\begin{pmatrix}4\\\\1\\end{pmatrix}$ To get $\\mathbf{X}$ on its own, multiply both sides by the inverse of $\\mathbf{A}$. Note that $\\lvert \\mathbf{A} \\rvert = -5-6 = -11$. Therefore, $\\mathbf{A}^{-1} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}$ So $\\mathbf{X} = \\mathbf{A}^{-1} \\mathbf{B} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}$ Now, use matrix multiplication to find $\\mathbf{X}$. \\begin{align} \\mathbf{X} &= -\\frac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}(-20)+(-2)\\\\(-12)+1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}-22\\\\-11\\end{pmatrix}\\\\ &= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\end{align} #### Simultaneous Equations We can apply this method to solving a system of", "Revision history [back] Q1. In his lecture, Prof. Strang begins with the simplest case: the orthogonal projection of a vector $\\mathbf{b}$ onto a one dimensional subspace $S=\\mathrm{span}\\langle\\mathbf{a}\\rangle$, $\\mathbf{a}$ being a non-null vector in $\\mathbb{R}^n$. The projection vector $\\mathbf{p}$ should be a multiple of $\\mathbf{a}$, so Prof. Strang writes it as $\\mathbf{p}=x\\mathbf{a}$. Here $x$ is a scalar number, not a matrix, and $\\mathbf{a}$ is a vector, identified with a $n\\times 1$ matrix. He wants to deduce that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage", "# Finding matrix linear transformation Question: Find the $3 \\times 3$ matrix $A$, associated with the linear transformation that projects vectors in $\\mathbb{R^3}$ (orthogonally) onto the plane $x+y+z=0$. I was given this question just as a review question for my class. I took linear algebra over 2 years ago, so my memory is really fuzzy. I was given the hints to find the matrix $A$ by thinking of it as a composition of a rotation, simpler projection, then another rotation. Thoughts: I know the normal vector of the plane is $<1,1,1>$. I think this may be a useful fact. Also after googling a bit I found that the matrix . Perhaps, I could work with this, although I honestly don't really know exactly how to proceed. I know there are many projections onto this plane, but I don't know exactly how to find an orthogonal one. Any help would be much appreciated. A projection matrix onto a plane in 3-D Euclidean space always has the form $$\\mathbf P = \\mathbf I - \\frac{\\mathbf n \\mathbf n^T}{\\mathbf n^T \\mathbf n},$$ where $\\mathbf n$ is the normal vector on the plane. Let's see why this is true. What do we expect if we apply $\\mathbf P$ to $\\mathbf n$? We should get zero right? Well, as you can verify we", "projection becomes a matrix that has no inverse and the programs stops. Can I assume this is the culmination of numerical error? I'm obviously doing this incorrectly so any advice would be much appreciated.", "vector projection onto a plane I'm learning Linear Algebra on my own time. Currently studying projections. I understand all of the derivation, but there is one piece of it that bothers me. Say I have a matrix $A$ that represents a plane in $3D$, (matrix has 2 columns and 3 rows), $b$ is a vector I'm trying to project onto a plane, $p$ is its projection and $e$ (error) is the vector normal to the plane, $e = b - p$ Given that $e$ is normal to the plane then it is normal to every vector in the plane, meaning the inner product with any vector in the plane is 0, i.e. $A^Te = 0$, substituting $e = b - p$ we get $$A^T(b - p)= 0$$ $$A^Tb - A^Tp = 0$$ $$A^Tb = A^Tp$$ $$b = p$$ My derivation clearly shows that projection equals to the vector being projected, but geometrically it is not so. What am I missing? You claim that $(A^Tb = A^Tp) \\Rightarrow (b=p)$. This isn't true. E.g. let $A = \\begin{pmatrix}1 & 0\\\\ 0 & 1\\\\ 0 & 0\\end{pmatrix}$. This has two columns and three rows as required. Let $b = \\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix}, c = \\begin{pmatrix}1\\\\ 1\\\\ 2\\end{pmatrix}$. Clearly $b \\neq c$. However, $A^Tb = \\begin{pmatrix}1 & 0 & 0\\\\ 0", "of the problem. Assume that two images are given, where $\\mx{P}_L$ is the camera matrix corresponding to the first image and $\\mx{P}_R$ is the camera matrix for the second image. Assume that one 3D point with homogeneous coordinate vector $\\vc{u}$ is seen in both images. In the first image, the homogeneous coordinate vector is $\\vc{v}_L$ and in the other image the vector is $\\vc{v}_R$. The camera matrix equations for these two images are $$\\lambda_L \\vc{v}_L = \\mx{P}_L \\vc{u}\\ \\ \\ \\ \\mathrm{and} \\ \\ \\ \\ \\lambda_R \\vc{v}_R = \\mx{P}_R \\vc{u}.$$ (7.92) These two matrix equations can be written as a joint matrix equation, i.e., $$\\begin{pmatrix} \\mx{P}_L & \\vc{v}_L & \\vc{0} \\\\ \\mx{P}_R & \\vc{0} & \\vc{v}_R \\end{pmatrix} \\begin{pmatrix} \\vc{u}\\\\ -\\lambda_L \\\\ -\\lambda_R \\end{pmatrix} = \\begin{pmatrix} \\vc{0}\\\\ \\vc{0} \\end{pmatrix} .$$ (7.93) Notice that the matrices and vectors are made out of blocks of other matrices or vectors. One often talks of block matrices. The matrix equation is of type $\\mx{A} \\vc{x} = \\vc{0}$. Here $\\mx{A}$ is a square matrix of size $6 \\times 6$. Notice that the matrix equation has a solution $\\begin{pmatrix}\\vc{u}\\\\-\\lambda_L \\\\-\\lambda_R \\end{pmatrix}$ that is not the zero vector. According to Theorem 7.10, this means that $\\det \\mx{A} = 0$. If we expand the determinant first along the fifth column and then along the sixth column", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "12, 2010 2. May 13, 2010 ### lanedance woops looked at it again - ok, so $$\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} 0 & 0 \\\\ 1 & 1 \\end{pmatrix} = \\begin{pmatrix} b & b\\\\ d & d \\end{pmatrix}$$ as you have sort of done, you can imagine the matricies as vectors in $$\\mathbb{R}^4$$ if should be clear that any matrix with b=d=0, will be mapped to the zero matrix, giving the null space if either b or d is non-zero, then the resultant transformation will be non-zero, giving the basis of vectors that will be mapped to non-zero vectors. input these individulaly to find a basis for the image space... (i think the one you presented is incorrect) Last edited: May 13, 2010" ]

[ "& \\sin\\theta \\\\ \\sin\\theta & 1 + \\cos\\theta \\end{pmatrix}$, $\\begin{pmatrix}a & b \\\\ c & 1 - a\\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\\beta)^\\textsf{T}(y - X\\beta)$, $\\hat\\beta = \\left(X^\\textsf{T}X\\right)^{-1}X^\\textsf{T}y$, $1 & -2 & -3 1 . 1 & 0 & 0 \\\\ {\\displaystyle A^{k}=A^{k-1}A=AA=A} a The matrix Z0Zis symmetric, and so therefore is (Z0Z) 1. 1 & -2 [math]P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. If a matrix $\\begin{pmatrix}a & b \\\\ c & d \\end{pmatrix}$ is idempotent, then $a = a^2 + bc,$ $b = ab + bd,$ implying $b(1 - a - d) = 0$ so $b = 0$ or $d = 1 - a,$ [1][2] That is, the matrix 2 Suppose that $A^{k-1} = A$. The projection transform establishes which part of the modeled scene will be visible, and what sort of projection will be applied. In linear algebra, M and P are both known as projection matrices, Ch 2 in DM, page 57, in particular gives the geometric interpretation. and Examples of $2 \\times 2$ idempotent matrices are: demonstrate on board. It is a bit more convoluted to prove that any idempotent matrix is the projection matrix for some subspace, but that’s also", "# Find a matrix for the linear transformation of reflection about a $\\theta$ line using the matrix for projection Projection matrix: $$\\begin{pmatrix} \\cos^2\\theta & \\cos\\theta \\sin\\theta\\\\ \\cos\\theta \\sin\\theta & \\sin^2\\theta\\\\ \\end{pmatrix}$$ Reflection matrix: $$\\begin{pmatrix} 2\\cos^2\\theta -1 & 2\\cos\\theta sin\\theta\\\\ 2\\cos\\theta \\sin\\theta & 2\\sin^2\\theta -1\\\\ \\end{pmatrix}$$ which is equivalent to $$\\begin{pmatrix} \\cos2\\theta & \\sin2\\theta\\\\ \\sin2\\theta & -\\cos2\\theta\\\\ \\end{pmatrix}$$ I can arrive at the second form geometrically, but the problem asks that I do it with the projection matrix and vector addition. Any hints/tips? It could also be that the problem is poorly worded, in which case a geometrical analysis is all that is necessary. \"Find a formula for R based on P. Explain your arguments by drawing a graph, using the rules of sums of vectors.\" I drew the graph and was able to successfully explain it in terms of sums of vectors. Hint: use this fact: if $X$ is a point, $X'$ its projection on the axis and $X''$ the reflection of $X$, than $X'$ is the middle point of $XX''$. so, if $P$ is the projection matrix and $R$ the reflection, we have: $$X'=PX \\qquad X''=RX$$ and, we write the fact that $X'$ is the midpoint as: $$X+X''=2X'$$ so: $$X+RX=2PX \\quad \\Rightarrow \\quad RX=2PX-X=(2P-I)X$$ • Yeah, it makes sense that the reflection is a projection onto the", "15.6: Orthogonal Matrices A nonsingular matrix is called orthogonal when its inverse is equal to its transpose: $\\mathbf{A}^T =\\mathbf{A}^{-1}\\rightarrow \\mathbf{A}^T\\mathbf{A} = \\mathbf{I}. \\nonumber$ For example, for the matrix $\\mathbf{A}=\\begin{pmatrix} \\cos \\theta&-\\sin \\theta \\\\ \\sin \\theta&\\cos \\theta \\end{pmatrix}, \\nonumber$ the inverse is $\\mathbf{A}^{-1} =\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix} =\\mathbf{A}^T \\nonumber$ We do not need to calculate the inverse to see if the matrix is orthogonal. We can transpose the matrix, multiply the result by the matrix, and see if we get the identity matrix as a result: $\\mathbf{A}^T=\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix} \\nonumber$ $\\mathbf{A}^T \\mathbf{A} =\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix}\\begin{pmatrix} \\cos \\theta&-\\sin \\theta \\\\ \\sin \\theta&\\cos \\theta \\end{pmatrix} =\\begin{pmatrix} (\\cos^2 \\theta+ \\sin^2\\theta)&0 \\\\ 0&(\\sin^2 \\theta+ \\cos^2\\theta) \\end{pmatrix} =\\begin{pmatrix} 1&0 \\\\ 0&1 \\end{pmatrix} \\nonumber$ The columns of orthogonal matrices form a system of orthonormal vectors (Section 14.2): $\\mathbf{M}=\\begin{pmatrix} a_1&b_1&c_1 \\\\ a_2&b_2&c_2 \\\\ a_3&b_3&c_3 \\end{pmatrix}\\rightarrow \\mathbf{a}=\\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix}; \\;\\mathbf{b}=\\begin{pmatrix}b_1\\\\b_2\\\\b_3\\end{pmatrix}; \\;\\mathbf{c}=\\begin{pmatrix}c_1\\\\c_2\\\\c_3\\end{pmatrix} \\nonumber$ $\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}=\\mathbf{c}\\cdot\\mathbf{b}=0 \\nonumber$ $|\\mathbf{a}|=|\\mathbf{b}|=|\\mathbf{c}|=1 \\nonumber$ Example $$\\PageIndex{1}$$ Prove that the matrix $$\\mathbf{M}$$ is an orthogonal matrix and show that its columns form a set of orthonormal vectors. $\\mathbf{M}=\\begin{pmatrix} 2/3&1/3&-2/3 \\\\ 2/3&-2/3&1/3 \\\\ 1/3&2/3&2/3 \\end{pmatrix} \\nonumber$ Note: This problem is also available in video format: http://tinyurl.com/k2tkny5 Solution We first need to prove that $$\\mathbf{M}^T\\mathbf{M}=1$$ $\\mathbf{M}=\\begin{pmatrix} {\\color{red}2/3}&{\\color{blue}1/3}&-2/3 \\\\ {\\color{red}2/3}&{\\color{blue}-2/3}&1/3 \\\\ {\\color{red}1/3}&{\\color{blue}2/3}&2/3 \\end{pmatrix}\\rightarrow \\mathbf{M}^T=\\begin{pmatrix}", "the diagonal with all of the other elements set to zero. \\phantom{-}\\cos \\Theta & \\sin \\Theta & 0 & 0 \\\\ \\begin{pmatrix} & z_n = 0 \\Rightarrow z_e = -f In math and physics, right-handed systems are the norm. \\begin{array}{lr} }\\right\\rbrack\\). \\end{pmatrix} Our vector is only valid when $$w=1$$. pt_{1} &= (\\matrix{0 & 0 & 0})\\\\ . Think about any first-person shooter game. pt_{3}S &= \\\\ 0 & -{\\text{focal length}} & 0 & 0\\\\\\ \\begin{pmatrix} y_c \\\\\\ 0 & 0 & 1 & 0 \\\\ \\right.\\\\\\ \\eqalign { -0.707 & 0.707 & 0 & 0 \\\\ When any valid input matrix is multiplied with the identity matrix, the result will be the unchanged input matrix. \\left\\lbrack \\matrix{ \\begin{array}{lr} \\matrix{\\phantom{-} \\cos \\Theta & \\sin \\Theta \\\\ -\\sin \\Theta & \\cos \\Theta} The positive $$z-axis$$ will point in the direction of your thumb, based upon the direction your coordinate space is oriented. T_c R_z T_o &= Perspective projection is shown below in figure 31. 0 & 0 & 1} Let’s start by finding the values of $$x$$ and $$y$$, the depth will come later. \\\\ \\begin{pmatrix} • Orthographic projection Oblique projection when the projection is when the projection is not perpendicular to the view perpendicular to the view plane plane 15 16. 0 & 0 & \\frac{n}{f-n} & \\frac{nf}{f-n}\\\\\\ Which", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "- i*\\sin(\\theta), 1, 1\\}$$ and $$\\operatorname{eig}(B) = \\{ -\\cos(\\theta) - i*\\sin(\\theta), -cos(\\theta) + i*\\sin(\\theta), 1, 1\\}$$ Both of the matrices have n distinct eigenvalues, so find nonsingular matrices that diagonalize them. $$S^{-1}AS = diag(-\\cos(\\theta) - i*\\sin(\\theta), -cos(\\theta) + i*\\sin(\\theta), 1, 1)$$ Then just multiply this matrix by $$I = \\begin{vmatrix} -1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{vmatrix}$$ Since similarity is transitive, you have that $A$ is similar to $B$. if A is diagonalizable, it's possible I think. You define the opposite of the projection on the subspace formed by your two eigenvalues with the cos and sin, and you change the base to be in the diagonalisation base: $E = P*D*P^{−1} , A = P*D1∗P^{−1} , A*E = P*D3∗P^{−1}$ , with the opposite eigenvalues ( the eigenvectors have been changed into their opposite thanks to the opposite of the projection)", "\\theta \\\\ \\sin \\theta & r \\cos \\theta \\end{pmatrix}?$ Edit: I think I see your problem. In your 2D example you seem to be treating $dxdy$ as the product of differentials. This is true for the right definition of product but it isn't the usual geometric product. In particular, $dx,dy$ are both linear functionals so it makes no sense to compose them. Your last line where your \"FOIL\" these is where this mistake becomes apparent. $dxdy$ is a bivector which is formally defined as the wedge product of $dx,dy$. However, it is often misleading to think of it as an object generated by a binary operation on $dx,dy$. It is better to think of it as just some linear functional which has 2 vectors as input (i.e. a linear functional on the wedge product of two vector spaces). See https://en.wikipedia.org/wiki/Bivector for a more thorough explanation. Last edited by SDK; October 23rd, 2018 at 06:14 AM. October 23rd, 2018, 06:31 AM #17 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Quote: Originally Posted by SDK . Just to be clear, are you asking me if I know how to compute the determinant of this matrix $\\begin{pmatrix} \\cos \\theta & -r \\sin \\theta \\\\ \\sin \\theta & r \\cos \\theta \\end{pmatrix}?$ See https://en.wikipedia.org/wiki/Bivector for a more", "# Projection matrix questions 1. Mar 29, 2007 ### EvLer I am studying for exam and something does not make sense anymore: looking at projection matrix, how come P=P2 where P2 = A(ATA)-1ATA(ATA)-1AT = A(ATA)-1AT = P but then they also say that cancelations (like distributing inverse operation and having AA-1 = I type things) are possible only if A is invertible, so does that mean that ATA is invertible? EDIT: might as well ask this: find matrix of projection p on plane x+y+2z = 0 my attempt: I know that the normal vector is (1, 1, 2) and then... not sure where to go with that.... Last edited: Mar 29, 2007 2. Mar 30, 2007 ### HallsofIvy Staff Emeritus Yes, although a projection map is not, in general, invertible, it's not too difficult to show that PTP always is. Yes, (1, 1, 2) is normal to the given plane and, since that has length $\\sqrt{6}$, $(1/\\sqrt{6},1/\\sqrt{6},2/\\sqrt{6})$ is a unit vector normal to the plane. The standard way to find the matrix for a linear transformation (in a given basis) is to see what it does to each of the basis vectors. Since the dot product of (1, 0, 0) with $(1/\\sqrt{6},1/\\sqrt{6},2/\\sqrt{6})$ is $1/\\sqrt{6}$, its projection onto that vector is (1/6, 1/6, 1/3) and so its projection onto", "# Thread: Linear Algebra projection 1. ## Linear Algebra projection Suppose E is a projection in M_2(C) with rank one. Show that there is a real number θ and a complex number β with |β| = 1 such that E = cos^2 βsinθcosθ Bsinθcosθ sin^2 θ M_2(C) represents an 2 x 2 matrix in the complex numbers. Β represents β conjugate. 2. ## Re: Linear Algebra projection It's impossible to be sure what you mean. Is it $\\begin{pmatrix}cos^2(\\theta) & \\beta sin(\\theta)cos(\\theta) \\\\ \\overline{\\beta}sin(\\theta)cos(\\theta) & sin^2(\\theta)\\end{pmatrix}$ ?", "• @l'ombradel'atzavara Thanks. The projection matrix is exactly defined like $\\mathbf{H}$, the hat matrix, even in books of linear algebra. I am using Strang for instance. You can check that it has all the properties of a projection matrix, namely symmetry and idempotence, while your definition of it does not. – JohnK Oct 26 '15 at 20:44 • You are right. $\\hat{\\mathbf{y}} = X \\hat{\\boldsymbol \\beta} = X \\left(X^\\top X \\right)^{-1} X^\\top \\mathbf{y}$. And the hat matrix is: $H = X \\left(X^\\top X \\right)^{-1} X^\\top$. – Antoni Parellada Oct 26 '15 at 21:02", "is still not a projection because it stretches vectors lying in the plane, e.g. it maps $\\begin{pmatrix}0 \\\\ 1 \\\\ 0\\end{pmatrix}$ to $\\begin{pmatrix}0 \\\\ 2 \\\\ 0\\end{pmatrix}$. So you also need a constraint ensuring that no such stretching (and also no rotating) occurs. This is neatly captured by the condition $T^2 = T$. Indeed, a matrix represents a projection if and only if it satisfies $T^2 = T$. So to summarize, if you want to show that a 3x3 matrix is a projection onto a particular plane, you need to verify all of the following: 1. $T^2 = T$, so $T$ is a projection 2. $\\dim(\\ker(T)) = 1$ or equivalently, $\\dim(\\text{im}(T)) = 2$, so $T$ projects onto a plane 3. $\\langle n, Tx\\rangle = 0$ where $n$ is normal to the plane and $x$ is arbitrary, so $T$ projects onto the specified plane You can find equivalent conditions which will allow you to do less work. [strike]For example, you can replace condition 3 with $Tn = 0$.[/strike] [correction: If condition 3 is replaced with $Tn=0$ then not only is $T$ a projection, it is in fact an orthogonal projection.] But the basic idea remains the same. Last edited: #### Seydlitz Thanks for the comprehensive information jbunniii, I really appreciate it. By the way I just realized the fact", "# Matrix Group induction proof and order of elements question $H$ is the set of the matrices $A$ of the form $$A= \\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}$$ where $0\\leq \\theta < 2\\pi$ is a group with respect to matrix multiplication. Question part (1) Prove by induction that for any $\\theta$ and any positive integer $n$, $$\\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}^{n} = \\begin{pmatrix} \\cos(n\\theta) & \\sin (n\\theta) \\\\ -\\sin (n\\theta) & \\cos (n\\theta)\\end{pmatrix}$$ When $n =1$ we have $$\\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}^{1} = \\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}$$ and clearly this holds for the base case. Now I'm going to assume true when $n=k$ and try and prove for $k+1$. So we have $$\\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}^{k}\\begin{pmatrix} \\cos(\\theta) & \\sin (\\theta) \\\\ -\\sin (\\theta) & \\cos (\\theta)\\end{pmatrix}^{1} = \\begin{pmatrix} \\cos(k\\theta + \\theta) & \\sin (k\\theta + \\theta) \\\\ -\\sin (k\\theta + \\theta) & \\cos (k\\theta + \\theta)\\end{pmatrix}$$ Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is $1^{k+1}$ and its the same on", "# How to find eigenvalues and eigenvectors of this matrix Can you help to find eigenvalues and eigenvectors of the following matrix? Here is the matrix: $$C = \\small \\begin{pmatrix} -\\sin(\\theta_{2} - \\theta_{M}) & \\sin(\\theta_{1} - \\theta_{M}) & 0 & \\ldots & 0 & \\sin(\\theta_{2} - \\theta_{1}) \\\\ \\sin(\\theta_{3} - \\theta_{2}) & -\\sin(\\theta_{3} - \\theta_{1}) & \\sin(\\theta_{2} - \\theta_{1}) & \\ldots & 0 & 0 \\\\ 0 & \\sin(\\theta_{4} - \\theta_{3}) & -\\sin(\\theta_{4} - \\theta_{2}) & \\ldots & 0 & 0 \\\\ 0 & \\ddots & \\ddots & \\ddots & \\ddots & \\vdots\\\\ \\sin(\\theta_{M} - \\theta_{M - 1}) & 0 & 0 & \\ldots & \\sin(\\theta_{1} - \\theta_{M}) & -\\sin(\\theta_{1} - \\theta_{M - 1}) \\\\ \\end{pmatrix}$$ where $0 = \\theta_{1} < \\ldots < \\theta_{M} < 2 \\pi$. I have found 2 vectors for eigenvalue 0: $$v_{1} = (1, \\cos(\\theta_{1}), \\ldots, \\cos(\\theta_{M - 1}))^{T}$$ $$v_{2} = (0, \\sin(\\theta_{1}), \\ldots, \\sin(\\theta_{M - 1}))^{T}$$ In case $\\theta_{j} = \\frac{2 \\pi (j - 1)}{M}, \\; j = 1, \\ldots, M$ the matrix becomes circulant matrix, and it's obvious to find all its eigenvectors and eigenvalues. If it's essential for somebody, I can explain the application where this matrix appears. In article by Lele, Kulkarni, Willsky - \"Convex-polygon estimation from support-line measurements\" this matrix represents the conditions of consistent support function measurements. Each", "$A = I + kJ$ where $I$ is the identity matrix , because the product of any other is not commutative : $AB \\neq BA$ unless one of them is $I$ ... 9. Originally Posted by simplependulum I have a quite stupid method : We know : $\\begin{pmatrix} \\cos{\\theta} & \\sin{\\theta} \\\\ -\\sin{\\theta} & \\cos{\\theta} \\end{pmatrix}^n = \\begin{pmatrix} \\cos(n\\theta) & \\sin(n\\theta) \\\\ -\\sin(n\\theta) & \\cos(n\\theta) \\end{pmatrix} $ The matrix $\\!\\left(\\!\\begin{array}{*{20}c}1&\\dfrac{x}{n}\\\\-\\dfrac{x}{n}&1\\\\\\end{array}\\!\\right)^n$ I first let $\\frac{x}{n} = \\tan{\\theta}$ I obtain $\\!\\left(\\!\\begin{array}{*{20}c}1&\\tan{\\theta}\\\\-\\tan{\\theta}&1\\\\\\end{array}\\!\\right)^n$ $= \\frac{1}{\\cos^n{\\theta}} \\!\\left(\\!\\begin{array}{*{20}c}\\cos{\\theta}&\\sin{\\ theta}\\\\-\\sin{\\theta}&\\cos{\\theta}\\\\\\end{array}\\!\\right)^n$ $= \\frac{1}{\\cos^n{\\theta}} \\!\\left(\\!\\begin{array}{*{20}c}\\cos(n\\theta)&\\sin( n\\theta)\\\\-\\sin(n\\theta)&\\cos(n\\theta)\\\\\\end{array}\\!\\right)$ As $n \\to \\infty$ $\\lim_{n\\to\\infty} \\cos^n{\\theta} = \\frac{1}{\\sqrt{1 + \\frac{x^2}{n^2} }^n} \\to e^{-\\frac{x^2}{2n} } = 1$ While $\\lim_{n\\to\\infty} n \\theta = \\lim_{n\\to\\infty} n \\tan^{-1}(\\frac{x}{n}) = x$ Therefore , the limit of your matrix is $1 \\!\\left(\\!\\begin{array}{*{20}c}\\cos{x}&\\sin{x}\\\\-\\sin{x}&\\cos{x}\\\\\\end{array}\\!\\right)$ $= \\!\\left(\\!\\begin{array}{*{20}c}\\cos{x}&\\sin{x}\\\\-\\sin{x}&\\cos{x}\\\\\\end{array}\\!\\right)$ This is very cute, but...would it be more or less plausible to arrive at this if we didn't know what the answer is beforehand? Tonio 10. Originally Posted by simplependulum It is really a great idea and an applicable method in solving other matrix problems , but i think we must put a matrix $A = I + kJ$ where $I$ is the identity matrix , because the product of any other is not commutative : $AB \\neq BA$ unless one of them is $I$ ... as long as the matrices $X$ and", "Commutativity of projection and rotation How to prove (if it is true … I suppose it is) that orthogonal projection from N-dimensional space onto 2-dimensional plane and rotation in this plane (I assume components of vectors out of the plane are not affected by the rotation) are commutative operations independently from dimension of N-dimensional vector space ? Solutions Collecting From Web of \"Commutativity of projection and rotation\" Let us work in $\\mathbb{R}^4$. Your question amounts to say that the following Projection and Rotation matrices $$P=\\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&0\\\\0&0&0&0\\end{bmatrix} \\ \\ \\text{and} \\ \\ R=\\begin{bmatrix}\\cos \\theta &-\\sin \\theta&0&0\\\\ \\sin \\theta& \\cos \\theta&0&0\\\\0&0&1&0\\\\0&0&0&1\\end{bmatrix}$$ commute. This is true because all the computation is in the product of the upper left $2 \\times 2$ blocks, which evidently commute. As you may know, matrices represent transformations with regard to a particular basis. In $\\mathbb{R}^n$ we usually work with the “standard” basis, which I am sure you know: ($e_1, e_2, \\dots, e_n$). Now, a projection matrix onto an arbitrary plane is pretty nasty looking. Depending on the plane it would be full of numbers, and we don’t want that. JeanMarie’s projection matrix looks like the identity matrix; nice and concise. Here we make use of a little theorem which says that you can construct an orthonormal basis from a set of $n$ linearly independent vectors in $\\mathbb{R}^n$.", "# Math Help - Matrix and Vector Magnitude 1. ## Matrix and Vector Magnitude Given matrix M and vector X where: $ $M = \\left( {\\begin{array}{cc} cos\\theta & sin\\theta \\\\ -sin\\theta & cos\\theta \\end{array} } \\right)$ \\textit{and} X = \\left( {\\begin{array}{cc} x \\\\ y \\end{array} } \\right) $ Compute: $ \\frac { \\left|{MX}\\right|}{\\left|{X}\\right|} $ So I found that the result is 1. The question is to explain the result and describe the action of M geometrically. I think the result means that $ $M = \\left( {\\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} } \\right)$ $ which, geometrically, doesn't change the length of vector X. Would that be correct? 2. $M=\\begin{bmatrix}{\\;\\;\\;\\cos \\theta}&{\\sin \\theta}\\\\{-\\sin \\theta}&{\\cos \\theta}\\end{bmatrix}=\\begin{bmatrix}{\\cos (-\\theta)}&{-\\sin (-\\theta)}\\\\{\\sin (-\\theta)}&{\\;\\;\\;\\cos (-\\theta)}\\end{bmatrix}$ represents a rotation around $(0,0)$ and angle $-\\theta$ so , $\\left\\|{MX}\\right\\|= \\left\\|{X}\\right\\|\\; \\forall{X}\\in{\\mathbb{R}^2}$ Fernando Revilla 3. Another way of seeing that is to note that the determinant is 1: $\\left|\\begin{array}{cc}cos(\\theta) & sin(\\theta) \\\\ -sin(\\theta) & cos(\\theta)\\end{array}\\right|= (cos(\\theta))(cos(\\theta))(- (sin(\\theta))(-sin(\\theta))= cos^2(\\theta)+ sin^2(\\theta)= 1$ and so the length of a vector is not changed. 4. But the length does not change by the fact of being $M$ orthogonal (in this case a rotation) and not by the fact of being $\\det M=1$ . For example: $M=\\begin{bmatrix}{1}&{1}\\\\{0}&{1}\\end{bmatrix}$ has determinant $1$ , however $\\left\\|{(0,1)^t}\\right\\|=1\\neq \\sqrt{2}=\\left\\|{M(0,1)^t}\\right\\|$ Fernando Revilla 5. Ouch! Good point!", "the determinant. This makes sense now. However, I cannot grasp just now what the visual connection is to the equality of transposed and inversed matrix. These are two different kinds of 'mapping reversion' or 'mapping change' and yet they are equal in the context of rectangular vectors (matrices). Why is that? – Taufi Apr 20 '16 at 18:42 • @Taufi In this case, you can think of the transpose of a transformation as its “mirror image.” For a rotation, this reverses the angle and thus undoes the rotation. For a reflection, it makes no difference (reflection matrices are symmetric), but reflections are their own inverses, anyway. – amd Apr 20 '16 at 19:12 Think of 2x2 orthogonal matrix. If it's determinant 1, then it will be a matrix of the form $$\\begin{pmatrix} cos(\\theta)&-sin(\\theta)\\\\ sin(\\theta)&cos(\\theta)\\\\ \\end{pmatrix}$$ So it transpose is $$\\begin{pmatrix} cos(\\theta)&sin(\\theta)\\\\ -sin(\\theta)&cos(\\theta)\\\\ \\end{pmatrix}$$ First matrix is rotation by $\\theta$ counter-clockwise and another is rotation by $\\theta$ clockwise, so it makes sense they are the inverse of each other.", "# Space Projection of singular matrix Describe the space that $$A$$ projects onto and the space that $$A$$ projects along$$A = \\begin{bmatrix} 0 & 1 \\\\0&1 \\end{bmatrix}.$$ A matrix, $$M$$ is a perpendicular projection operator on the column space of $$M \\iff MM = M$$ and $$M = M^T.$$ Clearly, $$A$$ does not satisfy this theorem. How do I describe the space that $$A$$ projects onto and the space that $$A$$ projects along? I need help. First you need to check that $$A$$ projects, that is $$A^2=A$$. That is true. Then the subspace it projects to is the range of $$A$$ (I hope you can find it) and the space it projects along is the nullspace of $$A$$ (you should be able to find it too). • The second answer there is correct. The first one is totally wrong provided $n\\ge 2$ and rank$(A)<n$ – JCAA Aug 30 at 20:20", "The additional condition of positive semi-definitivity means that all eigenvalues are non-negative. Since the trace is the sum of the eigenvalues, it follows that a trace of zero implies that all eigenvalues are 0. Still, this turns out to be irrelevant for your problem. #### caffeinemachine ##### Well-known member MHB Math Scholar Let $A = (a_{ij})$ be an $n\\times n$ symmetric matrix with trace zero. The first step is to find an orthogonal transformation that converts $A$ to a matrix with a zero on the $(1,1)$-position on the diagonal. If $a_{11}=0$ there is nothing to prove. Otherwise, choose $j>1$ such that $a_{jj}$ has the opposite sign to $a_{11}$. Such a $j$ must exist because the diagonal entries sum to zero. Now let $P_\\theta$ be the orthogonal matrix given by rotating the $1$ and $j$ coordinates through an angle $\\theta$ and leaving all the other coordinates alone. Specifically, the $2\\times2$ submatrix of $P_\\theta$ consisting of rows and columns $1$ and $j$ looks like $\\begin{bmatrix}\\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{bmatrix}$, $P_\\theta$ has $1$s in all the other diagonal places, and zeros everywhere else. You can check that the $(1,1)$-element of $P_\\theta AP_\\theta^{-1}$ is $a_{11}\\cos^2\\theta - 2a_{1j}\\cos\\theta\\sin\\theta + a_{jj}\\sin^2\\theta$. When $\\theta=0$ this is $a_{11}$. When $\\theta=\\pi/2$ it is $a_{jj}$ which has the opposite sign to $a_{11}$. By the intermediate", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019" ]

[ "15.6: Orthogonal Matrices A nonsingular matrix is called orthogonal when its inverse is equal to its transpose: $\\mathbf{A}^T =\\mathbf{A}^{-1}\\rightarrow \\mathbf{A}^T\\mathbf{A} = \\mathbf{I}. \\nonumber$ For example, for the matrix $\\mathbf{A}=\\begin{pmatrix} \\cos \\theta&-\\sin \\theta \\\\ \\sin \\theta&\\cos \\theta \\end{pmatrix}, \\nonumber$ the inverse is $\\mathbf{A}^{-1} =\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix} =\\mathbf{A}^T \\nonumber$ We do not need to calculate the inverse to see if the matrix is orthogonal. We can transpose the matrix, multiply the result by the matrix, and see if we get the identity matrix as a result: $\\mathbf{A}^T=\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix} \\nonumber$ $\\mathbf{A}^T \\mathbf{A} =\\begin{pmatrix} \\cos \\theta&\\sin \\theta \\\\ -\\sin \\theta&\\cos \\theta \\end{pmatrix}\\begin{pmatrix} \\cos \\theta&-\\sin \\theta \\\\ \\sin \\theta&\\cos \\theta \\end{pmatrix} =\\begin{pmatrix} (\\cos^2 \\theta+ \\sin^2\\theta)&0 \\\\ 0&(\\sin^2 \\theta+ \\cos^2\\theta) \\end{pmatrix} =\\begin{pmatrix} 1&0 \\\\ 0&1 \\end{pmatrix} \\nonumber$ The columns of orthogonal matrices form a system of orthonormal vectors (Section 14.2): $\\mathbf{M}=\\begin{pmatrix} a_1&b_1&c_1 \\\\ a_2&b_2&c_2 \\\\ a_3&b_3&c_3 \\end{pmatrix}\\rightarrow \\mathbf{a}=\\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix}; \\;\\mathbf{b}=\\begin{pmatrix}b_1\\\\b_2\\\\b_3\\end{pmatrix}; \\;\\mathbf{c}=\\begin{pmatrix}c_1\\\\c_2\\\\c_3\\end{pmatrix} \\nonumber$ $\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}=\\mathbf{c}\\cdot\\mathbf{b}=0 \\nonumber$ $|\\mathbf{a}|=|\\mathbf{b}|=|\\mathbf{c}|=1 \\nonumber$ Example $$\\PageIndex{1}$$ Prove that the matrix $$\\mathbf{M}$$ is an orthogonal matrix and show that its columns form a set of orthonormal vectors. $\\mathbf{M}=\\begin{pmatrix} 2/3&1/3&-2/3 \\\\ 2/3&-2/3&1/3 \\\\ 1/3&2/3&2/3 \\end{pmatrix} \\nonumber$ Note: This problem is also available in video format: http://tinyurl.com/k2tkny5 Solution We first need to prove that $$\\mathbf{M}^T\\mathbf{M}=1$$ $\\mathbf{M}=\\begin{pmatrix} {\\color{red}2/3}&{\\color{blue}1/3}&-2/3 \\\\ {\\color{red}2/3}&{\\color{blue}-2/3}&1/3 \\\\ {\\color{red}1/3}&{\\color{blue}2/3}&2/3 \\end{pmatrix}\\rightarrow \\mathbf{M}^T=\\begin{pmatrix}", "this means that the determinant is negative. • A matrix with determinant of 0 is called a singular matrix. • Singular matrices map the unit square to a line—or to (0,0) in the case of the zero matrix. In these cases, the transformed “parallelograms” therefore have area 0. • A matrix with a non-zero determinant is called non-singular. • # Part 6: Inverses of 2 × 2 matrices Given two matrices A and B, if AB = I, the identity matrix, then B is the inverse of A. We can denote the inverse of A as A-1 i.e. B = A-1. A square matrix M has an inverse, denoted M-1, if and only if its |M| ≠ 0. If the determinant of M is 0, then M has no inverse. Given a matrix M and its inverse, M-1, the following will be true: • MM-1 = I • M-1M = I The inverse of a 2 × 2 matrix $$\\textsf{M}$$ = $$\\begin{pmatrix} a & b\\\\c & d\\end{pmatrix}$$ can be found (where it exists) as follows: $$\\textsf{M}^{-1}$$ = $$\\dfrac{1}{\\begin{vmatrix} \\textsf{M} \\end{vmatrix}} \\begin{pmatrix} d & -b\\\\-c & a\\end{pmatrix}$$ = $$\\dfrac{1}{ad-bc} \\begin{pmatrix} d & -b\\\\-c & a\\end{pmatrix}$$. Extension 1: Verify that MM-1 = I. Extension 2: Verify that M-1M = I. Extension 3: Find the determinant of M-1. Questions 1. Can", "the diagonal with all of the other elements set to zero. \\phantom{-}\\cos \\Theta & \\sin \\Theta & 0 & 0 \\\\ \\begin{pmatrix} & z_n = 0 \\Rightarrow z_e = -f In math and physics, right-handed systems are the norm. \\begin{array}{lr} }\\right\\rbrack\\). \\end{pmatrix} Our vector is only valid when $$w=1$$. pt_{1} &= (\\matrix{0 & 0 & 0})\\\\ . Think about any first-person shooter game. pt_{3}S &= \\\\ 0 & -{\\text{focal length}} & 0 & 0\\\\\\ \\begin{pmatrix} y_c \\\\\\ 0 & 0 & 1 & 0 \\\\ \\right.\\\\\\ \\eqalign { -0.707 & 0.707 & 0 & 0 \\\\ When any valid input matrix is multiplied with the identity matrix, the result will be the unchanged input matrix. \\left\\lbrack \\matrix{ \\begin{array}{lr} \\matrix{\\phantom{-} \\cos \\Theta & \\sin \\Theta \\\\ -\\sin \\Theta & \\cos \\Theta} The positive $$z-axis$$ will point in the direction of your thumb, based upon the direction your coordinate space is oriented. T_c R_z T_o &= Perspective projection is shown below in figure 31. 0 & 0 & 1} Let’s start by finding the values of $$x$$ and $$y$$, the depth will come later. \\\\ \\begin{pmatrix} • Orthographic projection Oblique projection when the projection is when the projection is not perpendicular to the view perpendicular to the view plane plane 15 16. 0 & 0 & \\frac{n}{f-n} & \\frac{nf}{f-n}\\\\\\ Which", "& \\sin\\theta \\\\ \\sin\\theta & 1 + \\cos\\theta \\end{pmatrix}$, $\\begin{pmatrix}a & b \\\\ c & 1 - a\\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\\beta)^\\textsf{T}(y - X\\beta)$, $\\hat\\beta = \\left(X^\\textsf{T}X\\right)^{-1}X^\\textsf{T}y$, $1 & -2 & -3 1 . 1 & 0 & 0 \\\\ {\\displaystyle A^{k}=A^{k-1}A=AA=A} a The matrix Z0Zis symmetric, and so therefore is (Z0Z) 1. 1 & -2 [math]P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. If a matrix $\\begin{pmatrix}a & b \\\\ c & d \\end{pmatrix}$ is idempotent, then $a = a^2 + bc,$ $b = ab + bd,$ implying $b(1 - a - d) = 0$ so $b = 0$ or $d = 1 - a,$ [1][2] That is, the matrix 2 Suppose that $A^{k-1} = A$. The projection transform establishes which part of the modeled scene will be visible, and what sort of projection will be applied. In linear algebra, M and P are both known as projection matrices, Ch 2 in DM, page 57, in particular gives the geometric interpretation. and Examples of $2 \\times 2$ idempotent matrices are: demonstrate on board. It is a bit more convoluted to prove that any idempotent matrix is the projection matrix for some subspace, but that’s also", "y \\end{pmatrix}.$ Similarly, a shear parallel to the y axis has x' = x and y' = y + $\\lambda$x. In matrix form: $\\begin{pmatrix}x' \\\\ y' \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ \\lambda & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}.$ Clearly the determinant will always be 1, as no matter where the shear element is placed, it will be a member of a skew-diagonal that also contains zero elements (as all skew-diagonals have length at least two) hence its product will remain zero and won't contribute to the determinant. Thus every shear matrix has an inverse, and the inverse is simply a shear matrix with the shear element negated, representing a shear transformation in the opposite direction. In fact, this is part of an easily derived more general result: if S is a shear matrix with shear element $\\lambda$, then Sn is a shear matrix whose shear element is simply n$\\lambda$. Hence, raising a shear matrix to a power n multiplies its shear factor by n. Properties If S is an n×n shear matrix, then: • S has rank n and therefore is invertible • 1 is the only eigenvalue of S, so det S = 1 and trace S = n • the eigenspace of S has n-1 dimensions. • S is asymmetric • S", "whether $P_2$ lies on $S_1$ or vice-versa, which means that it is enough to check whether there exists a $t \\in \\mathbb{R}$ such that $P_2 = P_1 + t \\cdot \\vec{v}_1$, i.e. it is enough to check whether the directional vector between the two points $P_1$ and $P_2$ is colinear with the vector $\\vec{v}$. If the determinant of the matrix $\\begin{pmatrix}x_2-x_1&a\\\\y_2-y_1&b\\end{pmatrix}$ is not zero, then the vectors can't be colinear. If the determinant is zero, all that is left to do is to check whether there exists such a $t$ between $0$ and $1$. ### Linear Independence Now if the two directional vectors are linearly independent, the matrix $V$ is regular and the above matrix equation can easily be solved by computing the inverse $V^{-1}$ of the matrix $V$: $V\\vec{t}=\\vec{P} \\Leftrightarrow \\vec{t}=V^{-1} \\cdot \\vec{P}.$ ### Example Let $P_1:=(4,7)$, $P_2:=(1,1)$, $\\vec{v_1}:=\\begin{pmatrix}12 \\\\ -4\\end{pmatrix}$ and $\\vec{v_2}:=\\begin{pmatrix}16 \\\\ 4\\end{pmatrix}$, as depicted in the figure below: In this example, $V=\\begin{pmatrix}12&-16 \\\\ -4 & -4\\end{pmatrix}$ and $P=\\begin{pmatrix}-3 \\\\ -6\\end{pmatrix}$. The inverse of $V$ is $V^{-1}=\\dfrac{1}{28} \\begin{pmatrix}1&-4 \\\\ -1 & -3\\end{pmatrix},$thus\\begin{align*}\\vec{t}&=V^{-1} \\cdot \\vec{P} \\\\ &= \\dfrac{1}{28} \\begin{pmatrix}1&-4 \\\\ -1 & -3\\end{pmatrix} \\cdot \\begin{pmatrix}-3 \\\\ -6\\end{pmatrix} \\\\ &= \\dfrac{1}{4} \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}.\\end{align*}Plugging $t_1=\\frac{3}{4}$ into the equation of $S_1$, we get $I=\\begin{pmatrix}4 \\\\ 7\\end{pmatrix} + \\dfrac{3}{4} \\cdot \\begin{pmatrix}12 \\\\ -4\\end{pmatrix} = \\begin{pmatrix} 13 \\\\", "= \\mathbf{M} \\times \\mathbf{M}^{-1} = \\mathbf{I}$ For a two by two matrix, the inverse is given by $\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix}$ which we can easily confirm with \\begin{align*} \\frac{1}{ad-bc} \\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix} \\times \\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} &= \\frac{1}{ad-bc} \\begin{pmatrix} da - bc & db - bd\\\\ -ca + ac & -cb + ad \\end{pmatrix}\\\\ &= \\frac{1}{ad-bc} \\begin{pmatrix} ad - bc & 0\\\\ 0 & ad - bc \\end{pmatrix}\\\\ &= \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\end{align*} The denominator of the fraction, $$ad-bc$$, is known as the determinant of the matrix and must be non-zero for the matrix to have an inverse. Matrices with determinants of zero are known as singular matrices and project vectors onto a lower dimensional space. Note that matrices with very small determinants, known as near singular matrices, whilst mathematically having inverses will present significant numerical difficulties when trying to compute their inverses. This is because they squash vectors down to a volume that is very thin in some directions, losing many significant digits in those directions through cancellation error. Now, one useful application of inverses is for the solution of systems of simultaneous linear equations, which can be", "the determinant. This makes sense now. However, I cannot grasp just now what the visual connection is to the equality of transposed and inversed matrix. These are two different kinds of 'mapping reversion' or 'mapping change' and yet they are equal in the context of rectangular vectors (matrices). Why is that? – Taufi Apr 20 '16 at 18:42 • @Taufi In this case, you can think of the transpose of a transformation as its “mirror image.” For a rotation, this reverses the angle and thus undoes the rotation. For a reflection, it makes no difference (reflection matrices are symmetric), but reflections are their own inverses, anyway. – amd Apr 20 '16 at 19:12 Think of 2x2 orthogonal matrix. If it's determinant 1, then it will be a matrix of the form $$\\begin{pmatrix} cos(\\theta)&-sin(\\theta)\\\\ sin(\\theta)&cos(\\theta)\\\\ \\end{pmatrix}$$ So it transpose is $$\\begin{pmatrix} cos(\\theta)&sin(\\theta)\\\\ -sin(\\theta)&cos(\\theta)\\\\ \\end{pmatrix}$$ First matrix is rotation by $\\theta$ counter-clockwise and another is rotation by $\\theta$ clockwise, so it makes sense they are the inverse of each other.", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "# Find a matrix for the linear transformation of reflection about a $\\theta$ line using the matrix for projection Projection matrix: $$\\begin{pmatrix} \\cos^2\\theta & \\cos\\theta \\sin\\theta\\\\ \\cos\\theta \\sin\\theta & \\sin^2\\theta\\\\ \\end{pmatrix}$$ Reflection matrix: $$\\begin{pmatrix} 2\\cos^2\\theta -1 & 2\\cos\\theta sin\\theta\\\\ 2\\cos\\theta \\sin\\theta & 2\\sin^2\\theta -1\\\\ \\end{pmatrix}$$ which is equivalent to $$\\begin{pmatrix} \\cos2\\theta & \\sin2\\theta\\\\ \\sin2\\theta & -\\cos2\\theta\\\\ \\end{pmatrix}$$ I can arrive at the second form geometrically, but the problem asks that I do it with the projection matrix and vector addition. Any hints/tips? It could also be that the problem is poorly worded, in which case a geometrical analysis is all that is necessary. \"Find a formula for R based on P. Explain your arguments by drawing a graph, using the rules of sums of vectors.\" I drew the graph and was able to successfully explain it in terms of sums of vectors. Hint: use this fact: if $X$ is a point, $X'$ its projection on the axis and $X''$ the reflection of $X$, than $X'$ is the middle point of $XX''$. so, if $P$ is the projection matrix and $R$ the reflection, we have: $$X'=PX \\qquad X''=RX$$ and, we write the fact that $X'$ is the midpoint as: $$X+X''=2X'$$ so: $$X+RX=2PX \\quad \\Rightarrow \\quad RX=2PX-X=(2P-I)X$$ • Yeah, it makes sense that the reflection is a projection onto the", "\\mathbb {I} _{3}={\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{pmatrix}}}$, ... In general, the n subscript is included only if necessary; if the size of the identity matrix can be deduced from context, we omit the subscript. For example, we would most likely say: ${\\displaystyle \\scriptstyle A\\mathbb {I} =A=\\mathbb {I} A}$ since only one identity matrix is dimensionally compatible for multiplication. ### Zero matrix The additive identity under matrix addition is known as a zero matrix, and denoted ${\\displaystyle \\scriptstyle 0_{m,n}}$ for an m×n matrix. Its entries are all zeroes, so (for example) ${\\displaystyle 0_{3,2}={\\begin{pmatrix}0&0\\\\0&0\\\\0&0\\end{pmatrix}}}$ It is evident that for any m×n matrix A, ${\\displaystyle \\scriptstyle A+0_{m,n}=A}$ It is also clear that the product of any matrix with a zero matrix is another zero matrix, which may or may not have the same dimensions. As with the identity matrix, the subscript is omitted if the context admits only one zero matrix. In this example, any other zero matrix could not be added to A, so the subscript is redundant and we could say ${\\displaystyle A+0=A}$ ### Invertible matrix Some, but not all, matrices have a multiplicative inverse. That is, for a matrix A, there may exist a matrix ${\\displaystyle \\scriptstyle A^{-1}}$ such that ${\\displaystyle \\scriptstyle AA^{-1}=A^{-1}A=\\mathbb {I} }$ Only square matrices may be inverted. Furthermore, if the determinant of a square matrix is 0, it", "The additional condition of positive semi-definitivity means that all eigenvalues are non-negative. Since the trace is the sum of the eigenvalues, it follows that a trace of zero implies that all eigenvalues are 0. Still, this turns out to be irrelevant for your problem. #### caffeinemachine ##### Well-known member MHB Math Scholar Let $A = (a_{ij})$ be an $n\\times n$ symmetric matrix with trace zero. The first step is to find an orthogonal transformation that converts $A$ to a matrix with a zero on the $(1,1)$-position on the diagonal. If $a_{11}=0$ there is nothing to prove. Otherwise, choose $j>1$ such that $a_{jj}$ has the opposite sign to $a_{11}$. Such a $j$ must exist because the diagonal entries sum to zero. Now let $P_\\theta$ be the orthogonal matrix given by rotating the $1$ and $j$ coordinates through an angle $\\theta$ and leaving all the other coordinates alone. Specifically, the $2\\times2$ submatrix of $P_\\theta$ consisting of rows and columns $1$ and $j$ looks like $\\begin{bmatrix}\\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{bmatrix}$, $P_\\theta$ has $1$s in all the other diagonal places, and zeros everywhere else. You can check that the $(1,1)$-element of $P_\\theta AP_\\theta^{-1}$ is $a_{11}\\cos^2\\theta - 2a_{1j}\\cos\\theta\\sin\\theta + a_{jj}\\sin^2\\theta$. When $\\theta=0$ this is $a_{11}$. When $\\theta=\\pi/2$ it is $a_{jj}$ which has the opposite sign to $a_{11}$. By the intermediate", "# Math Help - Matrix and Vector Magnitude 1. ## Matrix and Vector Magnitude Given matrix M and vector X where: $ $M = \\left( {\\begin{array}{cc} cos\\theta & sin\\theta \\\\ -sin\\theta & cos\\theta \\end{array} } \\right)$ \\textit{and} X = \\left( {\\begin{array}{cc} x \\\\ y \\end{array} } \\right) $ Compute: $ \\frac { \\left|{MX}\\right|}{\\left|{X}\\right|} $ So I found that the result is 1. The question is to explain the result and describe the action of M geometrically. I think the result means that $ $M = \\left( {\\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} } \\right)$ $ which, geometrically, doesn't change the length of vector X. Would that be correct? 2. $M=\\begin{bmatrix}{\\;\\;\\;\\cos \\theta}&{\\sin \\theta}\\\\{-\\sin \\theta}&{\\cos \\theta}\\end{bmatrix}=\\begin{bmatrix}{\\cos (-\\theta)}&{-\\sin (-\\theta)}\\\\{\\sin (-\\theta)}&{\\;\\;\\;\\cos (-\\theta)}\\end{bmatrix}$ represents a rotation around $(0,0)$ and angle $-\\theta$ so , $\\left\\|{MX}\\right\\|= \\left\\|{X}\\right\\|\\; \\forall{X}\\in{\\mathbb{R}^2}$ Fernando Revilla 3. Another way of seeing that is to note that the determinant is 1: $\\left|\\begin{array}{cc}cos(\\theta) & sin(\\theta) \\\\ -sin(\\theta) & cos(\\theta)\\end{array}\\right|= (cos(\\theta))(cos(\\theta))(- (sin(\\theta))(-sin(\\theta))= cos^2(\\theta)+ sin^2(\\theta)= 1$ and so the length of a vector is not changed. 4. But the length does not change by the fact of being $M$ orthogonal (in this case a rotation) and not by the fact of being $\\det M=1$ . For example: $M=\\begin{bmatrix}{1}&{1}\\\\{0}&{1}\\end{bmatrix}$ has determinant $1$ , however $\\left\\|{(0,1)^t}\\right\\|=1\\neq \\sqrt{2}=\\left\\|{M(0,1)^t}\\right\\|$ Fernando Revilla 5. Ouch! Good point!", "+0 # Help hard quick 0 68 0 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ Use the picture below to find vectors with integer components and magnitude no greater than 8 that satisfy $$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$", "similarly for the nullspace of $1-P$. 1. since $P$ is a projection matrix $P^2=P$. Using this it is easy to show that (I-P)(I-P) = I-P. Hence I-P is also Projection (symmetry property also preserves). 2. Since the eigenvalues of a projection matrix are either ones or zeros you can use eigendecomposition: $P = U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T$ where $M$ is the dimension of the subspace to which the projectios is carried it's easy to show that $I - U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T = U \\begin{pmatrix} \\bf{0} & \\bf{0}\\\\ \\bf{0} & I_{N-M} \\end{pmatrix} U^T$. From here it is easy to extract the 4 sub-spaces of each projection operator. The above analysis is correct for orthogonal projectors. For oblique projectors a similar calculation can be done using singular value decomposition. • Oblique projectors don't diagonalise like this – Jazzmaniac Mar 27 '16 at 12:09 • Right. Obviously the problem consider orthogonal projectors. – Nir Regev Mar 27 '16 at 12:13 • I don't see any reference to orthogonal projection in the question – Jazzmaniac Mar 27 '16 at 12:14 • @Jazzmaniac you're right. I updated the answer – Nir Regev Mar 27 '16 at 12:29", "$$M=\\begin{pmatrix}\\cos\\theta & \\sin\\theta \\\\ -\\sin\\theta&\\cos\\theta\\end{pmatrix}$$ and $$X$$ is the vector $$X=\\begin{pmatrix}x\\\\y \\end{pmatrix}\\, .$$ Calculate all possible dot products between the vectors $$X$$ and $$MX$$. Compute the lengths of $$X$$ and $$MX$$. What is the angle between the vectors $$MX$$ and $$X$$. Draw a picture of these vectors in the plane. For what values of $$\\theta$$ do you expect equality in the triangle and Cauchy--Schwartz inequalities? 6. Let $$M$$ be the matrix $$\\begin{pmatrix} 1&0&0&1&0&0\\\\ 0&1&0&0&1&0\\\\ 0&0&1&0&0&1\\\\ 0&0&0&1&0&0\\\\ 0&0&0&0&1&0\\\\ 0&0&0&0&0&1 \\end{pmatrix}$$ Find a formula for $$M^{k}$$ for any positive integer power $$k$$. Try some simple examples like $$k=2,3$$ if confused. 7. $$\\textit{Determinants:}$$ The determinant $${\\rm det}\\, M$$ of a $$2\\times 2$$ matrix $$M=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}$$ is defined by $${\\rm det}\\, M =ad -bc\\, .$$ a) For which values of $${\\rm det}\\, M$$ does $$M$$ have an inverse? b) Write down all $$2\\times 2$$ bit matrices with determinant 1. (Remember bits are either 0 or 1 and $$1+1=0$$.) c) Write down all $$2\\times 2$$ bit matrices with determinant 0. d) Use one of the above examples to show why the following statement is FALSE. $$\\textit{Square matrices with the same determinant are always row equivalent.} 8. What does it mean for a function to be linear? Check that integration is a linear function from \\(V$$ to $$V$$, where $$V = \\{ f: \\mathbb{R} \\to", "of sets ${\\displaystyle \\left\\{\\omega _{1},-\\omega _{1},\\dots ,\\omega _{(n-1)/2},-\\omega _{n-1)/2}\\right\\}=\\left\\{1,\\dots ,(n-1)/2,(n+3)/2,\\dots ,n\\right\\},}$ and each diagonal block acts upon a pair of entries of v which are not both zero. So, for example, when n = 3, the matrix can be ${\\displaystyle {\\begin{pmatrix}0&-1&0\\\\1&0&0\\\\0&0&-1\\\\\\end{pmatrix}}.}$ For the vector (1, 0, 1/2), the resulting simplex has vertices ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1/\\surd 2\\end{pmatrix}},{\\begin{pmatrix}0\\\\1\\\\-1/\\surd 2\\end{pmatrix}},{\\begin{pmatrix}-1\\\\0\\\\1/\\surd 2\\end{pmatrix}},{\\begin{pmatrix}0\\\\-1\\\\-1/\\surd 2\\end{pmatrix}},}$ each of which has distance 2 from the others. ## Geometric properties ### Volume The volume of an n-simplex in n-dimensional space with vertices (v0, ..., vn) is ${\\displaystyle \\mathrm {Volume} ={\\frac {1}{n!}}\\left|\\det {\\begin{pmatrix}v_{1}-v_{0}&&v_{2}-v_{0}&&\\cdots &&v_{n}-v_{0}\\end{pmatrix}}\\right|}$ where each column of the n × n determinant is a vector that points from vertex v0 to another vertex vk. [10] This formula is particularly useful when ${\\displaystyle v_{0}}$ is the origin. The expression ${\\displaystyle \\mathrm {Volume} ={\\frac {1}{n!}}\\det \\left[{\\begin{pmatrix}v_{1}^{T}-v_{0}^{T}\\\\v_{2}^{T}-v_{0}^{T}\\\\\\vdots \\\\v_{n}^{T}-v_{0}^{T}\\end{pmatrix}}{\\begin{pmatrix}v_{1}-v_{0}&v_{2}-v_{0}&\\cdots &v_{n}-v_{0}\\end{pmatrix}}\\right]^{1/2}}$ employs a Gram determinant and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions, e.g., a triangle in ${\\displaystyle \\mathbb {R} ^{3}}$. A more symmetric way to compute the volume of an n-simplex in ${\\displaystyle \\mathbb {R} ^{n}}$ is ${\\displaystyle \\mathrm {Volume} ={1 \\over n!}\\left|\\det {\\begin{pmatrix}v_{0}&v_{1}&\\cdots &v_{n}\\\\1&1&\\cdots &1\\end{pmatrix}}\\right|.}$ Another common way of computing the volume of the simplex is via the Cayley–Menger determinant, which works even when the n-simplex's vertices are in a Euclidean space", "# Is the zero Matrix a vector space? ## Homework Statement So I have these two Matrices: M = \\begin{pmatrix} a & -a-b \\\\ 0 & a \\\\ \\end{pmatrix} and N = \\begin{pmatrix} c & 0 \\\\ d & -c \\\\ \\end{pmatrix} Where a,b,c,d ∈ ℝ Find a base for M, N, M +N and M ∩ N. ## Homework Equations I know the 8 axioms about the vector spaces. ## The Attempt at a Solution I chose these fours matrices as a base for the first three vector spaces. \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix} I got that the only vector space satisfying M ∩ N is \\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\\\ \\end{pmatrix} If M ∩ N constitutes a vector space, I can use the same base as the other three. But if it doesnt, I have to explain why. Not sure how to go about that...", "entire plane and giving back a 1 dimensional ‘space’ – a line through the origin. Another interesting thing to note is that both the points $\\begin{pmatrix}x\\\\ y\\end{pmatrix}$ and $\\begin{pmatrix}y\\\\ x\\end{pmatrix}$ both end up in the same point $\\begin{pmatrix}x+y\\\\ x+y\\end{pmatrix}$. This means given a point $\\begin{pmatrix}x+y\\\\ x+y\\end{pmatrix}$, we would not know which point on the plane it corresponds to, i.e, the inverse is not well-defined. (In fact, given any point $\\begin{pmatrix}p\\\\ p\\end{pmatrix}$ at the output there are an (uncountably) infinite number of points which could have produced that) When we say that a matrix is singular or non-invertible, this is what we mean. We normally check this by finding the determinant of the matrix (which is zero in the singular case), but this normally does not appeal much to intuition. To generalise this, any matrix that takes a higher dimensional space to a lower dimensional space is not invertible. The natural next question to ask is that if a matrix maintains the dimension of the output space equal to the input, is it invertible ? This can in fact be proved to be true, and this can be taken as a simple interpretation of the Rank-Nullity Theorem. Going back to our example, what if the point/vector at the input was already on the line $p-q=0$ ? For example, the", "entire plane and giving back a 1 dimensional ‘space’ – a line through the origin. Another interesting thing to note is that both the points $\\begin{pmatrix}x\\\\ y\\end{pmatrix}$ and $\\begin{pmatrix}y\\\\ x\\end{pmatrix}$ both end up in the same point $\\begin{pmatrix}x+y\\\\ x+y\\end{pmatrix}$. This means given a point $\\begin{pmatrix}x+y\\\\ x+y\\end{pmatrix}$, we would not know which point on the plane it corresponds to, i.e, the inverse is not well-defined. (In fact, given any point $\\begin{pmatrix}p\\\\ p\\end{pmatrix}$ at the output there are an (uncountably) infinite number of points which could have produced that) When we say that a matrix is singular or non-invertible, this is what we mean. We normally check this by finding the determinant of the matrix (which is zero in the singular case), but this normally does not appeal much to intuition. To generalise this, any matrix that takes a higher dimensional space to a lower dimensional space is not invertible. The natural next question to ask is that if a matrix maintains the dimension of the output space equal to the input, is it invertible ? This can in fact be proved to be true, and this can be taken as a simple interpretation of the Rank-Nullity Theorem. Going back to our example, what if the point/vector at the input was already on the line $p-q=0$ ? For example, the" ]

[ "+0 # In a polar coordinate system, the midpoint of the line segment whose endpoints are and is the point Enter 0 291 1 In a polar coordinate system, the midpoint of the line segment whose endpoints are $$\\left( 8, \\frac{5 \\pi}{12} \\right)$$ and $$\\left( 8, -\\frac{3 \\pi}{12} \\right)$$ is the point $$(r, \\theta)$$ Enter $$(r, \\theta)$$ where $$r>0$$ and $$0 \\le \\theta < 2 \\pi$$ Aug 28, 2020 edited by Guest Aug 28, 2020", "## Precalculus (6th Edition) Blitzer The polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$. The polar coordinates of the point are $\\left( r,\\theta \\right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\\tan \\theta =\\frac{y}{x}$ …… (2) Substituting the values of $x\\ \\text{ and }\\ y$ in (1) and (2), we get \\begin{align} & r=\\sqrt{{{x}^{2}}+{{y}^{2}}} \\\\ & =\\sqrt{{{\\left( 0 \\right)}^{2}}+{{\\left( -6 \\right)}^{2}}} \\\\ & =\\sqrt{36} \\\\ & r=6 \\end{align} And, \\begin{align} & \\tan \\theta =\\frac{y}{x} \\\\ & =\\frac{-6}{0} \\\\ & \\tan \\theta =\\infty \\left( \\text{undefined} \\right) \\end{align} Hence, $\\tan \\theta =\\infty$ And, $\\tan \\frac{\\pi }{2}=\\infty$ The $\\theta$ lies on the negative y-axis which means \\begin{align} & \\theta =\\pi +\\frac{\\pi }{2} \\\\ & =\\frac{3\\pi }{2} \\end{align} Therefore, the polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$.", "# Midpoint of two complex numbers in polar form Say we have two complex numbers: $re^{i\\theta}$ and $se^{i\\phi}$ Is there a straightforward way to find the polar form of the midpoint of these two complex numbers? I think I'm correct in saying that the argument of the midpoint will be: $\\frac{\\theta+\\phi}{2}$ I am asking this question as I had to find the midpoint between two \"consecutive\" (consecutive as in the sense of adjacent) roots of unity; in this case the moduli of the two complex numbers in question will be the same, but I was wondering if there was any method for the general case. I know we could do it by first converting to Cartesian, but this seems unsatisfactory. Thanks! • it is not true that argument of the midpoint of two number is the average of their arguments. take for example $1$ and $2i.$ – abel May 25 '15 at 20:02 • Where is the problem in trying to find a polar form of $$\\frac12 (re^{i\\theta} + se^{i\\phi})$$ – AlexR May 25 '15 at 20:03 • Not true take the two points $z_1=1$ and $z_2=2i$. The argument of the mid point is not $\\pi/4$ as claimed. There is no straightforward way to express the midpoint in polar form – marwalix May 25 '15 at 20:03 •", "# A line segment has endpoints at (2 ,1 ) and (3 ,5 ). If the line segment is rotated about the origin by ( pi)/2 , translated vertically by -8 , and reflected about the x-axis, what will the line segment's new endpoints be? Jan 24, 2016 $P \\left(1 , - 6\\right) \\mathmr{and} Q \\left(5 - 5\\right)$ #### Explanation: Let $P \\left(x , y\\right)$ then $P ' \\left(x ' , y '\\right) = R \\left(\\theta\\right) P \\left(x , y\\right)$ for $\\theta = \\frac{\\pi}{2}$ $P ' \\left(x ' , y '\\right) = R \\left(\\frac{\\pi}{2}\\right) P \\left(x , y\\right) = P ' \\left(- y , x\\right)$ this a special case of the generalized rotation operation. Now $P \\left(2 , 1\\right) \\to P ' \\left(- 1 , 2\\right)$ and $Q \\left(3 , 5\\right) \\to Q ' \\left(- 5 , 3\\right)$ To translate by -8 simply subtract 8 to y part $P ' \\left(- 1 , 2\\right) \\to P ' ' \\left(- 1 , - 6\\right)$ $Q ' \\left(- 5 , 3\\right) \\to Q ' ' \\left(- 5 , - 5\\right)$ reflection about x axis will change the sign of point x $P ' ' \\left(- 1 , - 6\\right) \\to P ' ' ' \\left(1 , - 6\\right)$ $Q ' ' \\left(- 5 , - 5\\right) \\to Q ' '", "to polar coordinates, we have \\begin{align*} \\iint_{R^2} e^{-10(x^2+y^2)}\\,dx \\, dy &= \\int_{\\theta=0}^{\\theta=2\\pi} \\int_{r=0}^{r=\\infty} e^{-10r^2}\\,r \\, dr \\, d\\theta = \\int_{\\theta=0}^{\\theta=2\\pi} \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) d\\theta \\\\ &=\\left(\\int_{\\theta=0}^{\\theta=2\\pi}\\right) d\\theta \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\lim_{a\\rightarrow\\infty}\\left(-\\frac{1}{20}\\right)\\left(\\left. e^{-10r^2}\\right|_0^a\\right) \\\\ &=2\\pi \\left(-\\frac{1}{20}\\right)\\lim_{a\\rightarrow\\infty}\\left(e^{-10a^2} - 1\\right) \\\\ &= \\frac{\\pi}{10}. \\end{align*} Exercise $$\\PageIndex{7}$$ Evaluate the integral $\\iint_{R^2} e^{-4(x^2+y^2)}dx \\, dy. \\nonumber$ Hint Convert to the polar coordinate system. $$\\frac{\\pi}{4}$$ ## Key Concepts • To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals. • The area $$dA$$ in polar coordinates becomes $$r \\, dr \\, d\\theta$$. • Use $$x = r \\, \\cos \\, \\theta, \\, y = r \\, \\sin \\, \\theta$$, and $$dA = r \\, dr \\, d\\theta$$ to convert an integral in rectangular coordinates to an integral in polar coordinates. • Use $$r^2 = x^2 + y^2$$ and $$\\theta = tan^{-1} \\left(\\frac{y}{x}\\right)$$ to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed. • To find the volume in polar coordinates bounded above", "The Cylindrical Coordinate System # The Cylindrical Coordinate System Recall that when we were dealing with polar coordinates in $\\mathbb{R}^2$ that we thought of every point $P$ on the plane as an ordered pair $(r, \\theta)$ where $r$ was the signed length of the ray extending from the origin to $P$, and $\\theta$ was the angle between the polar axis (or $x$-axis) and the ray. Furthermore, we realized that the polar coordinate system acted with parametric equations $x= r \\cos \\theta$ and $y = r \\sin \\theta$. We will not introduce a new coordinate system for $\\mathbb{R}^3$ that is analogous to the polar coordinate system in $\\mathbb{R}^2$. This new coordinate system is called the Cylindrical Coordinate System that represents every point $P$ in $\\mathbb{R}^3$ as an order triple $(r, \\theta, z)$ where $r$ represents the signed length of the projection of the ray from the origin to $P$ onto the $xy$-plane, $\\theta$ represents the angle from the positive polar axis to the projection of the ray onto the $xy$-plane, and $z$ represents the signed height of $P$ above/below the $z$-axis. This coordinate system can be represented with the following parametric equations: (1) \\begin{align} \\left\\{\\begin{matrix} x = r \\cos \\theta\\\\ y = r \\sin \\theta\\\\ z = z \\end{matrix}\\right. \\end{align} Notice that if $z = 0$ then we reduce", "# A line segment has endpoints at (4 ,0 ) and (2 ,9 ). If the line segment is rotated about the origin by ( pi)/2 , translated vertically by -8 , and reflected about the x-axis, what will the line segment's new endpoints be? Oct 11, 2016 The endpoints are $\\left(0 , 4\\right)$ and $\\left(- 9 , 6\\right)$ #### Explanation: Rotate $\\left(4 , 0\\right) b y \\frac{\\pi}{2}$: $| 4 , 0 | = 4$ $\\theta = {\\tan}^{-} 1 \\left(\\frac{0}{4}\\right) = 0$ The rotated angle ${\\theta}_{r} = \\frac{\\pi}{2}$ $\\left(4 \\cos \\left(\\frac{\\pi}{2}\\right) , 4 \\sin \\left(\\frac{\\pi}{2}\\right)\\right) = \\left(0 , 4\\right)$ Rotate $\\left(2 , 9\\right) b y \\frac{\\pi}{2}$: $| 2 , 9 | = \\sqrt{{2}^{2} + {9}^{2}}$ $| 2 , 9 | = \\sqrt{85}$ $\\theta = {\\tan}^{-} 1 \\left(\\frac{9}{2}\\right)$ The rotated angle ${\\theta}_{r} = {\\tan}^{-} 1 \\left(\\frac{9}{2}\\right) + \\frac{\\pi}{2}$ $\\left(\\sqrt{85} \\cos \\left({\\tan}^{-} 1 \\left(\\frac{9}{2}\\right) + \\frac{\\pi}{2}\\right) , \\sqrt{85} \\sin \\left({\\tan}^{-} 1 \\left(\\frac{9}{2}\\right) + \\frac{\\pi}{2}\\right)\\right) = \\left(- 9 , 2\\right)$ Translate both points vertically by -8: $\\left(0 , 4 - 8\\right) = \\left(0 , - 4\\right)$ $\\left(- 9 , 2 - 8\\right) = \\left(- 9 , - 6\\right)$ Reflect about the x axis: $\\left(0 , - 1 \\cdot - 4\\right) = \\left(0 , 4\\right)$ $\\left(- 9 , - 1 \\cdot - 6\\right) = \\left(- 9 , 6\\right)$", "## Precalculus (6th Edition) Blitzer The polar coordinates of $\\left( -\\sqrt{3},-1 \\right)$ are $\\left( 2,\\frac{7\\pi }{6} \\right)$. The polar coordinates of the point are $\\left( r,\\theta \\right)$. Now rewrite polar coordinates in terms of rectangular coordinates as below: $r=\\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\\tan \\theta =\\frac{y}{x}$ …… (2) Substituting the values of $x\\ \\text{ and }\\ y$ in (1) and (2), we get \\begin{align} & r=\\sqrt{{{x}^{2}}+{{y}^{2}}} \\\\ & =\\sqrt{{{\\left( -\\sqrt{3} \\right)}^{2}}+{{\\left( -1 \\right)}^{2}}} \\\\ & =\\sqrt{3+1}=\\sqrt{4} \\\\ & r=2 \\end{align} And, \\begin{align} & \\tan \\theta =\\frac{y}{x} \\\\ & =\\frac{-1}{-\\sqrt{3}} \\\\ & \\tan \\theta =\\frac{1}{\\sqrt{3}} \\end{align} Hence, $\\tan \\theta =\\frac{1}{\\sqrt{3}}$ And, $\\tan \\frac{\\pi }{6}=\\frac{1}{\\sqrt{3}}$ Also, $\\theta$ lies in the third quadrant which means \\begin{align} & \\theta =\\pi +\\frac{\\pi }{6} \\\\ & \\theta =\\frac{7\\pi }{6} \\end{align} Therefore, the polar coordinates of $\\left( -\\sqrt{3},-1 \\right)$ are $\\left( 2,\\frac{7\\pi }{6} \\right)$.", "(315-250)^\\circ} \\approx 8.16\\end{align*}. ## Points to Consider • How is the polar coordinate system similar/different from the rectangular coordinate system? • How do you plot a point on a polar coordinate grid? • How do you determine the coordinates of a point on a polar grid? • How do you calculate the distance between two points that have polar coordinates? ## Review Questions 1. Graph each point: 1. \\begin{align*}M (2.5, 210^\\circ)\\end{align*} 2. \\begin{align*}S \\left (-3.5, \\frac{5 \\pi}{6} \\right )\\end{align*} 3. \\begin{align*}A \\left (1, \\frac{3 \\pi}{4} \\right )\\end{align*} 4. \\begin{align*}Y \\left (5.25, - \\frac{\\pi}{3} \\right )\\end{align*} 2. For the given point \\begin{align*}A \\left (-4, \\frac{\\pi}{4} \\right )\\end{align*}, list three different pairs of polar coordinates that represent this point such that \\begin{align*}-2\\pi \\le \\theta \\le 2\\pi\\end{align*}. 3. For the given point \\begin{align*}B (2, 120^\\circ)\\end{align*}, list three different pairs of polar coordinates that represent this point such that \\begin{align*}-2\\pi < \\theta < 2 \\pi\\end{align*}. 4. Given \\begin{align*}P_1\\end{align*} and \\begin{align*}P_2\\end{align*}, calculate the distance between the points. 1. \\begin{align*}P_1 (1,30^\\circ)\\end{align*} and \\begin{align*}P_2 (6,135^\\circ)\\end{align*} 2. \\begin{align*}P_1 (2,-65^\\circ)\\end{align*} and \\begin{align*}P_2 (9,85^\\circ)\\end{align*} 3. \\begin{align*}P_1 (-3,142^\\circ)\\end{align*} and \\begin{align*}P_2 (10,-88^\\circ)\\end{align*} 4. \\begin{align*}P_1 (5,-160^\\circ)\\end{align*} and \\begin{align*}P_2 (16, -335^\\circ)\\end{align*} 1. \\begin{align*}& \\left (-4, \\frac{\\pi}{4} \\right ) && \\text{all positive} \\rightarrow && \\left (4, \\frac{5\\pi}{4} \\right ) \\\\ & && \\text{both negative} \\rightarrow && \\left (-4, \\frac{-7\\pi}{4} \\right ) \\\\ &", "Polar Coordinate Systems # Polar Coordinate Systems Thus far we have used what is known as a Cartesian coordinate system, or a two-dimensional plane to represent curves with coordinate pairs $(x, y)$. Now we are going to learn about a new coordinate system known as the Polar Coordinate System, which uses coordinate pairs $(r, \\theta)$, where $r$ represents the length of the line segment originating from the origin and terminating at the length of $r$, and where $\\theta$ represents the angle the line segment makes with what we call the polar axis (analogous to the $x$-axis). For example, let's look at the following polar coordinates where $r = 2$, and $\\theta = \\frac{\\pi}{3}$ or rather coordinates $P(2, \\frac{\\pi}{3} )$: The horizontal black line represents our polar axis, while the vertical black line is there just to define the angles of $\\frac{\\pi}{2}$ and $\\frac{3\\pi}{2}$. Also notice that the line segment is dotted. The line segment itself is not a part of the coordinate $(2, \\frac{\\pi}{3} )$, but it is instead just an indicator of how far away from the origin a point $P$ should be. The picture above illustrates the important of $r$, as it tells us the length away from the origin that a point $P$ is, while the angle $\\theta$ describes the inclination or declination of that", "\\right]\\end{align} Therefore, $$\\cos \\theta = 0$$ and $$\\sin \\theta = 1$$ Since, $$\\theta$$ lies in quadrant I, $$\\theta = \\frac{\\pi }{2}$$ Hence, \\begin{align}i&= r\\cos \\theta + ir\\sin \\theta \\\\&= \\cos \\frac{\\pi }{2} + i\\sin \\frac{\\pi }{2}\\end{align} Thus, this is the required polar form. Related Sections Related Sections", "To plot: Thepoints which has polar coordinate displaystyle{left({2},frac{{{7}pi}}{{4}}right)} also two alternaitve sets for the same. To plot: Thepoints which has polar coordinate $\\left(2,\\frac{7\\pi }{4}\\right)$ also two alternaitve sets for the same. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Dora Let the point $\\left(2,\\frac{7\\pi }{4}\\right)$ be denoted by P. Consider the point $\\left(2,\\frac{7\\pi }{4}\\right)$, it is 2 units from the Origin O and line OP makes an angle of $\\frac{7\\pi }{4}$ with the positive direction of the x-axis. Add pi to the angle and change cardinal coordinate to -2 to get one alternate solution. Therefore, the point P $\\left(2,\\frac{7\\pi }{4}\\right)$ can also be represented by $\\left(-2,\\frac{7\\pi }{4}+\\pi \\right)=\\left(-2,\\frac{11\\pi }{4}\\right).$ Change the radial coordinate to -2 and subtract $3\\pi$ from the angle to obtain the other alternate solution. Thus, the point $P\\left(2,\\frac{7\\pi }{4}\\right)$ can be given by $\\left(-2,\\frac{7\\pi }{4}-3\\pi \\right)$ $=\\left(-2,-\\frac{19\\pi }{4}\\right).$ Use online graphing calculator and plot the points of P as shown below in Figure 1. From the Figure 1 it can be noted that alternate representations are plotted on the same location.", "# A line segment has endpoints at (3 ,4 ) and (2 ,5 ). If the line segment is rotated about the origin by pi , translated horizontally by 2 , and reflected about the x-axis, what will the line segment's new endpoints be? Oct 20, 2016 $\\left(- 1 , 4\\right) \\mathmr{and} \\left(0 , 5\\right)$ #### Explanation: There is a special shortcut to rotate by $\\pi$ but I will do that rotation the long way, because it shows how do it for any angle. The first point rotated by $\\pi$: $r = \\sqrt{{3}^{2} + {4}^{2}}$ $r = 5$ $\\theta = {\\tan}^{-} 1 \\left(\\frac{4}{3}\\right) + \\pi$ $\\left(5 \\cos \\left({\\tan}^{-} 1 \\left(\\frac{4}{3}\\right) + \\pi\\right) , 5 \\sin \\left({\\tan}^{-} 1 \\left(\\frac{4}{3}\\right) + \\pi\\right)\\right) =$ $\\left(- 3 , - 4\\right)$ The second point rotated by $\\pi$: $r = \\sqrt{{2}^{2} + {5}^{2}}$ $r = \\sqrt{29}$ $\\theta = {\\tan}^{-} 1 \\left(\\frac{5}{2}\\right) + \\pi$ $\\left(\\sqrt{29} \\cos \\left({\\tan}^{-} 1 \\left(\\frac{5}{2}\\right) + \\pi\\right) , 5 \\sin \\left({\\tan}^{-} 1 \\left(\\frac{5}{2}\\right) + \\pi\\right)\\right) =$ $\\left(- 2 , - 5\\right)$ Translated horizontally by 2 means add two to the x coordinates: $\\left(- 3 , - 4\\right) \\to \\left(- 1 , - 4\\right)$ $\\left(- 2 , - 5\\right) \\to \\left(0 , - 5\\right)$ Reflected about the x axis means multiply the y coordinates by -1: $\\left(- 1 , - 4\\right) \\to", "# A line segment has endpoints at (6 ,5 ) and (2 ,7 ). If the line segment is rotated about the origin by pi , translated horizontally by 2 , and reflected about the x-axis, what will the line segment's new endpoints be? Jan 30, 2016 (-4, 7); (0, 7) #### Explanation: Use the Rotation, Translation and Reflection mattrix. But this a special case so you can do it in your head. $\\vec{V '} = {M}_{R \\left(\\theta\\right)} \\vec{V}$ $\\left(- {x}_{1} , - {y}_{1}\\right) = {M}_{R \\left(\\theta\\right)} \\left({x}_{1} , {y}_{1}\\right)$ (-6, -7); (-2,-7) $\\vec{V ' '} = {M}_{T \\left(2\\right)} \\vec{V '}$ $\\left(- {x}_{1} + 2 , - {y}_{1}\\right) = {M}_{T \\left(2\\right)} \\left(- {x}_{1} , - {y}_{1}\\right)$ (-4, -7); (0,-7) vec(V') = M_(RF(x) vec(V) $\\left(- {x}_{1} + 2 , {y}_{1}\\right) = {M}_{R F \\left(x\\right)} \\left(- {x}_{1} + 2 , - {y}_{1}\\right)$ (-4, 7); (0, 7)", "# 7: Polar and Spherical Coordinate Systems $$\\newcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$", "= r\\cos\\theta$ $\\sin\\theta= \\dfrac{O}{H} = \\dfrac{y}{r}$. Rearrange for $y$: $y = r \\sin \\theta$ #### Worked Examples ###### Example 1 Find the polar coordinates of the point with Cartesian coordinates $(2,2)$. ###### Solution \\begin{align} r &= \\sqrt{2^2+2^2}\\\\ &= \\sqrt{8}\\\\ &= 2\\sqrt{2} \\end{align} \\begin{align} \\theta &= \\tan^{-1}\\left(\\frac{2}{2}\\right)\\\\ &= \\tan^{-1}(1)\\\\ &= \\frac{\\pi}{4} \\end{align} So the polar coordinates are $\\left(2\\sqrt{2},\\dfrac{\\pi}{4}\\right)$. ###### Example 2 Find the Cartesian coordinates of the point with polar coordinates $\\left(1, \\dfrac{3\\pi}{4}\\right)$. ###### Solution \\begin{align} x &= 1 \\cos \\frac{3\\pi}{4}\\\\ &= \\cos \\frac{3\\pi}{4}\\\\ &= -\\frac{1}{\\sqrt{2} } \\end{align} \\begin{align} y &= 1 \\sin \\frac{3\\pi}{4}\\\\ &= \\sin \\frac{3\\pi}{4}\\\\ &= \\frac{1}{\\sqrt{2} } \\end{align} So the Cartesian coordinates are $\\left(-\\dfrac{1}{\\sqrt{2} },\\dfrac{1}{\\sqrt{2} }\\right)$. #### Video Example Hayley Bishop converts $(1,3)$ to polar coordinates, and $(2,\\frac{\\pi}{3})$ to Cartesian coordinates. #### Workbook This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples. #### More Support You can get one-to-one support from Maths-Aid.", "\\frac{5 \\pi}{6}\\right)$$ 2. $$\\left(2, \\frac{\\pi}{2}\\right)$$ 3. $$\\left(4,-\\frac{7 \\pi}{6}\\right)$$ 4. $$\\left(-2, \\frac{5 \\pi}{3}\\right)$$ Give two alternate sets of coordinates for each point. 5. $$\\left(2,60^{\\circ}\\right)$$ 6. $$\\left(5,330^{\\circ}\\right)$$ 7. $$\\left(2,210^{\\circ}\\right)$$ Graph each equation. 8. $$r=4$$ 9. $$\\theta=\\frac{\\pi}{4}$$ 10. $$r=2 \\theta$$ with $$\\theta: 0 \\leq \\theta \\leq 2 \\pi$$. Convert each point to rectangular form. 11. $$\\left(4, \\frac{2 \\pi}{3}\\right)$$ 12. $$\\left(3, \\frac{\\pi}{4}\\right)$$ 13. $$\\left(5, \\frac{\\pi}{3}\\right)$$ Convert each point to polar form using radians where $$0 \\leq \\theta<2 \\pi$$. 14. (1,3) 15. (1,-4) 16. (2,6) Convert each equation to polar form. 17. $$x=3$$ 18. \\(2 x+4 y=2\\ This page titled 10.1 Polar and Rectangular Coordinates is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.", "# What are the coordinates for the point p(45^circ) where p(theta)=(x,y) is the point where the terminal arm of an angle theta intersects the unit circl? Aug 3, 2018 $\\left(\\frac{1}{\\sqrt{2}} , \\frac{1}{\\sqrt{2}}\\right)$. See graphical depiction. #### Explanation: The polar equation of this unit circle is $r = 1$. A point of the radial line $\\vec{O P}$, in the direction $\\theta$ are $p \\left(\\theta\\right) = r \\left(\\cos \\theta , \\sin \\theta\\right)$. If P is the point of intersection with r = 1, $p \\left(\\theta\\right) = \\left(\\cos \\theta , \\sin \\theta\\right)$.. So, $p \\left(\\frac{\\pi}{4}\\right) = \\left(\\cos \\left(\\frac{\\pi}{4}\\right) , \\sin \\left(\\frac{\\pi}{4}\\right)\\right) = \\left(\\frac{1}{\\sqrt{2}} , \\frac{1}{\\sqrt{2}}\\right)$. See graphical depiction. graph{(x^2+y^2-1)(y-x)((x-1/sqrt2)^2+(y-1/sqrt2)^2-0.001)=0[0 2 0 1]}", "# What is the Cartesian form of ( 15 , ( 7pi)/3 ) ? Aug 14, 2017 $\\left(\\frac{15}{2} , 15 \\frac{\\sqrt{3}}{2}\\right)$ #### Explanation: In Polar coordinates a point is represented as $P \\left(r , \\theta\\right)$ In Cartesian form it is represented as $P \\left(x , y\\right)$ Where $x = r \\cos \\theta$ And $y = r \\sin \\theta$ $\\therefore$ Cartesian form of $P \\left(15 , \\frac{7 \\pi}{3}\\right)$ will be $x = 15 \\times \\cos \\left(\\frac{7 \\pi}{3}\\right) = \\frac{15}{2}$ $y = 15 \\times \\sin \\left(\\frac{7 \\pi}{3}\\right) = 15 \\frac{\\sqrt{3}}{2}$ $\\therefore$ The point in Cartesian coordinate will be $P \\left(\\frac{15}{2} , 15 \\frac{\\sqrt{3}}{2}\\right)$", "Center a coordinate system at the bottom left hand corner of the rectangle. Hence the coordinates of the (???) point are $(q,0)$ for some $q<w$. Note that the because theta is $45$ degrees, $y=w-q$ (Isosceles right triangle). Hence $q=w-y$, and our point is simply $(w-y, 0)\\dots$ - I edited the question to define (0,0). I don't know why you're assume that p2 always on the bottom line. It can be also on the left line. – stacker Jul 28 '10 at 2:53" ]

[ "= \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$. ## Solution 2 $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"O\",O,SW); pair M = (4,0);dot(M);label(\"M\",M,SE); pair N = (4,2);dot(N);label(\"N\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"B\",B,NE); pair C = (4,-3);dot(C);label(\"C\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"P\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"A\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"Q\",Q,S); pair R = (3,0); dot(R); label(\"R\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"D\",D,E); draw(A--D); [/asy]$ The above solution works, but is quite messy and somewhat difficult to follow. This solution provides", "as the center and draw segment $ON$ perpendicular to $AM$, $ON\\cap AM=N$, link $OM$. Then we have $OM\\parallel AD$. So $\\angle DAM=\\angle OMA$. Since $AD=2AM=2OM=1$, we have $\\cos\\angle DAM=\\cos\\angle OMP=\\frac{2}{\\sqrt{5}}$. As a result, $NM=OM\\cos\\angle OMP=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{5}}=\\frac{1}{\\sqrt{5}}.$ Thus $PM=2NM=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. Since $AM=\\frac{\\sqrt{5}}{2}$, we have $AP=AM-PM=\\frac{\\sqrt{5}}{10}$. Put $\\boxed{B}$. ~FANYUCHEN20020715 ## Solution 5 (Similar Triangles) Call the midpoint of $\\overline{AB}$ point $N$. Draw in $\\overline{NM}$ and $\\overline{NP}$. Note that $\\angle{NPM}=90^{\\circ}$ due to Thales's Theorem. $[asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw(circle((0.5,0.5),0.5)); draw((0,0)--(0,0.5)--(1,0.5)--cycle); label(\"A\",(0,0),SW); label(\"B\",(0,1),NW); label(\"C\",(1,1),NE); label(\"D\",(1,0),SE); label(\"M\",(1,0.5),E); label(\"P\",(0.2,0.1),S); label(\"N\",(0,0.5),W); draw((0,0.5)--(0.2,0.1)); markscalefactor=0.007; draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); [/asy]$ Using the Pythagorean theorem, $AM=\\frac{\\sqrt{5}}{2}$. Now we just need to find $AP$ using similar triangles. $$\\triangle APN\\sim\\triangle ANM\\Rightarrow\\frac{AP}{AN}=\\frac{AN}{AM}\\Rightarrow\\frac{AP}{\\frac{1}{2}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}}\\Rightarrow AP=\\boxed{\\textbf{(B) }\\frac{\\sqrt{5}}{10}}$$ ~QIDb602 ## Solution 6 (Intersecting Chords) Label the midpoint of $AD$ as $N$, and let $Q$ be the intersection of $ON$ and $AM$ Then $MQ=AQ=\\frac{1}{2}AM = \\frac{\\sqrt{5}}{4}$ and $NQ=\\frac{1}{4}$ Then $\\frac{1}{4}*\\frac{3}{4}=PQ*QM$ and $AP=AQ-PQ$ ~ERMSCoach ## Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem) Let $N$ be the midpoint of $\\overline{AB},$ from which $\\angle ANM=90^\\circ.$ Note that $\\angle NPM=90^\\circ$ by the Inscribed Angle Theorem. We have the following diagram: Since $AN=\\frac12$ and $NM=1,$ we get $AM=\\frac{\\sqrt5}{2}$ by the Pythagorean Theorem. Let $AP=x.$ It follows that $PM=\\frac{\\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\\triangle ANP$ gives $NP^2=\\left(\\frac12\\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\\triangle MNP$ gives $NP^2=1^2-\\left(\\frac{\\sqrt5}{2}-x\\right)^2.$ Equating the expressions for $NP^2$ produces \\begin{align*}", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(45), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, -3), SW); label(\"C_2'\", (9, -3), SE); label(\"C_3'\", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label(\"C_4'\", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$", "# Mock Geometry AIME 2011 Problems/Problem 11 ## Problem $C$ is on a semicircle with diameter $AB$ and center $O.$ Circle radius $r_1$ is tangent to $OA,OC,$ and arc $AC,$ and circle radius $r_2$ is tangent to $OB,OC,$ and arc $BC$. It is known that $\\tan AOC=\\frac{24}{7}$. The ratio $\\frac{r_2} {r_1}$ can be expressed $\\frac{m} {n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$ ## Solution $[asy] unitsize(4mm); draw(Arc((0,0),12,0,180)); draw((12,0)--(-12,0)); draw(circle((-6,9/2),9/2)); draw(circle((4,16/3),16/3)); draw((0,0)--(-84/25,288/25)); draw((0,0)--(-48/5,36/5)); draw((0,0)--(36/5,48/5)); draw((-6,9/2)--(-6,0)); draw((4,16/3)--(4,0)); label(\"C\",(-84/25,288/25),NNW); label(\"A\",(-12,0),WSW); label(\"B\",(12,0),ESE); label(\"O\",(0,0),S); label(\"D\",(-6,9/2),NNE); label(\"E\",(4,16/3),NNW); label(\"F\",(-6,0),S); label(\"G\",(4,0),S); [/asy]$ Let the circle with radius $r_1$ have center $D$ and the circle with radius $r_2$ have center $E$. Let the projections of $D$ and $E$ onto $AB$ be $F$ and $G$, respectively. $D$ is equidistant from $OA$ and $OC$, so it is on the angle bisector of $\\angle AOC$. We're given that $\\tan AOC=\\frac{24}{7}$, so $\\cos AOC=\\frac{7}{25}$. Now, we have $\\tan FOD=\\tan\\frac{AOC}{2}=\\sqrt{\\frac{1-\\cos AOC}{1+\\cos AOC}}=\\sqrt{\\frac{1-\\frac{7}{25}}{1+\\frac{7}{25}}}=\\frac{3}{4}$, which is positive because $\\angle FOD<\\frac{\\pi}{2}$. We therefore also have $\\sin FOD=\\frac{3}{5}$. Now, $DO=\\frac{FD}{\\sin FOD}=\\frac{r_1}{\\frac{3}{5}}=\\frac{5r_1}{3}$, and we see that the radius of the large semicircle is $\\frac{5r_1}{3}+r_1=\\frac{8r_1}{3}$. Similarly, $OE$ is the angle bisector of $\\angle BOC$. Now, $\\cos BOC=\\cos(\\pi-AOC)=-\\cos AOC=-\\frac{7}{25}$, and so $\\tan BOE=\\tan\\frac{BOC}{2}=\\sqrt{\\frac{1-\\cos BOC}{1+\\cos BOC}}=\\sqrt{\\frac{1+\\frac{7}{25}}{1-\\frac{7}{25}}}=\\frac{4}{3}$, again positive because $\\angle BOE<\\frac{\\pi}{2}$, and $\\sin BOE=\\frac{4}{5}$. Now, $OE=\\frac{EG}{\\sin BOE}=\\frac{r_2}{\\frac{4}{5}}=\\frac{5r_2}{4}$. The radius of the large semicircle is thus $\\frac{5r_2}{4}+r_2=\\frac{9r_2}{4}$. Since", "0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, 3), N); label(\"C_2'\", (9, 3), N); label(\"C_3'\", (15/2, 0), N); dot((15/2, 0)); [/asy]$ Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\\frac{3}{2}$. Now, we can determine the radius of $C_4$ using the formula $r = \\frac{k^2 \\cdot r'}{\\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \\frac{3}{2}$. $\\overline{OC}$ is just the distance from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7);", "(2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0, 3.6), S); [/asy]$ We know that $AU = UM$, $AV = VC$, so $UV = \\frac{1}{2} MC$. As $\\angle UPM = \\angle VPC = 90$, we can see that $\\triangle UPM \\cong \\triangle VPC$ and $\\triangle UVP \\sim \\triangle MPC$ with a side ratio of $1 : 2$. So $UP = VP = 4$, $MP = PC = 8$. With that, we can see that $S\\triangle UPM = 16$, and the area of trapezoid $MUVC$ is 72. As said in solution 1, $S\\triangle AMC = 72 / \\frac{3}{4} = \\boxed{\\textbf{(C) } 96}$. ## Solution 5 (Medians, Height, Pythagorean Theorem, Lots of symmetry, A little extra work) [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); draw((-2, 6), (2, 6)) label(\"K\", (0, 6), NE); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy] It is well known that medians divide each other into segments of $2:1$ ratio. From this, we have $PC=MP=8$ and $UP=UV=4$. From right triangle $\\triangle{MPC}$, $MC^2=MP^2+MC^2=8^2+8^2=128$, which implies $MC=\\sqrt{128}=8\\sqrt{2}$. Then the area of $\\triangle{AMC}$ is $\\dfrac{8\\sqrt{2} \\cdot AB}{2}$, so our goal is to find $AB$. Note that $AB=AP+BP$. Since $\\triangle{AMC}$ is isosceles, by symmetry $MB=MC$, because $AB$ is the altitude. Knowing this, $BP$", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, because $\\triangle RPM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED", "draw(A--D^^C--E^^D--O^^E--O^^B--O); label(\"A\", A, dir(point--A)); label(\"C\", C, dir(point--C)); label(\"O\", O, dir(point--O)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); label(\"B\", (1.33,1.04), SW); [/asy]$ Using the Base Altitude formula, where $DB$ and $BE$ are the bases and $OA$ and $CO$ are the altitudes, respectively, $[DBO] = [EBO] = \\frac{\\sqrt{3}-1}{2}$. The area of sector $DOE$ is $\\frac{1}{12}$ of circle $O$. The area of circle $O$ is $4\\pi$, and therefore we have the area of sector $DBE$ to be $\\frac{\\pi}{3} + 1 - \\sqrt{3} \\Longrightarrow \\boxed{A}$ Like the first solutions, you find that the area of sector $DOE$ is $\\frac{\\pi}{3}$. We also know that the triangles will not be in terms of ${\\pi}$. Looking at the answers, choices $\\text{(A)}$ and $\\text{(E)}$ both contain $\\frac{\\pi}{3}$. However, based on the diagram, we observe that the answer must be less than $\\frac {\\pi}{3}$. Only $\\boxed{A}$ consists of a value less than $\\frac{\\pi}{3}$.", "know that $\\overline{PC}\\parallel\\overline{OM}.$ We construct $D$ on $\\overline{PC}$ such that $\\overline{XD}\\perp\\overline{PC}.$ Clearly, $MXDC$ is a rectangle. Since $\\angle XPD=\\angle OXP$ by alternate interior angles, we have $\\triangle XPD\\sim\\triangle OXP$ by the AA Similarity, with ratio of similitude $\\frac{XP}{OX}=\\frac 35.$ Therefore, we get that $PD=\\frac 95$ and $PC=PD+DC=PD+MX=\\frac 95 + \\frac 32 = \\frac{33}{10}.$ The area of $\\triangle PQR$ is$$\\frac12(RQ)(PC)=\\frac12\\left(3\\sqrt3\\right)\\left(\\frac{33}{10}\\right)=\\frac{99\\sqrt3}{20},$$and the answer is $99+3+20=\\boxed{\\textbf{(D) } 122}.$ ~MRENTHUSIASM Solution 2 Diagram $[asy] draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label(\"C\", (7, 0), S); label(\"O\", (11, 3), E); label(\"P\", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label(\"Q\", Q, N); label(\"R\", R, E); [/asy]$ Solution Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\\overline{AB}$. Since $\\angle QPR = 60^{\\circ}$, we have $\\angle QOR = 2\\cdot 60^{\\circ}=120^{\\circ}$ and $\\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$. Since $QR=3\\sqrt{3},$ we know $OQ=OR=\\tfrac{3\\sqrt{3}}{\\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\\triangle", "7 \\right]$$ = $$\\scriptsize 6 \\left [ \\normalsize \\frac {11 \\: + \\: 21}{3}\\right]$$ = $$\\scriptsize 6 \\: \\times \\: \\left [ \\normalsize \\frac {32}{3}\\right] \\\\ \\scriptsize = 2 \\: \\times \\: 32 \\\\ \\scriptsize = 64 mm$$ Question 7 In the diagram above AB is a chord of a circle of radius 10cm, M is the mid-point of AB and If MN = 2cm, calculate (a) AB (b) The angle arc ANB subtends at the centre of the circle (c) The difference in length between AB and arc ANB to the nearest mm. Solution (a) Calculate AB Consider the sector ANBO AO, ON and OB are radii θ1 OB = OA = ON = 10cm OM = 10 – 2 = 8cm Using Pythagoras on right angled AMO AO2 = OM2 + AM2 AM2 = AO2 – OM2 AM = $$\\scriptsize \\sqrt{AO^2 \\: – \\: OM^2}$$ AM = $$\\scriptsize\\sqrt{10^2 \\: – \\: 8^2}$$ AM = $$\\scriptsize \\sqrt{36}$$ AM = $$\\scriptsize 6cm$$ AB = 2 x 6 = 12cm (b) Calculate The angle that are ANB subtends at the centre of the circle Consider Isosceles $$\\scriptsize \\Delta ABO \\; \\; \\theta_1 = \\theta_2$$ Cos θ1 = $$\\frac{adj}{hyp}\\\\ = \\frac {8}{10} \\\\ = \\frac {4}{5} \\\\ = \\scriptsize 0.8000$$ θ1 = cos – 1 (0.8000) θ1 = 36.9o Angle", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "\\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*} If the length of $AD$ is $200$, then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be $$\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}$$ Which is a contradiction. Therefore, our answer is $\\boxed{500}.$ ## Solution 3 (Trigonometry Bash) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,E); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$ Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on $\\Delta BOC$; let $\\theta = \\angle BOC$. We get the following equation: $$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta$$ Substituting the values in, we get $$(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta$$ Canceling out, we get $$\\cos\\theta=\\frac{3}{4}$$ Because $\\angle AOB$, $\\angle BOC$, and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$. To find", "0 \\Rightarrow x = \\frac{-\\sqrt{3}\\pm 3\\sqrt{7}}{2}$. Since $x$ must be positive, $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow (C)$. ## Solution 2 (without trigonometry) Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$. $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label(\"$$E$$\", EE,SE); [/asy]$ As proved in the first solution, $\\angle OCD = 150^\\circ$. That makes $\\Delta OCE$ a $30-60-90$ triangle, so $OE = \\frac{x}{2}$ and $CE= \\frac{x\\sqrt 3}{2}$ Since $\\Delta ODE$ is a right triangle, $\\left({\\frac{x}{2}}\\right)^2 + \\left({\\frac{x\\sqrt 3}{2} +1}\\right)^2 = 4^2 \\Rightarrow x^2+x\\sqrt3-15 = 0$ Solving for $x$ gives $x =\\frac{3\\sqrt{7}-\\sqrt{3}}{2}\\Rightarrow (C)$ ## Solution 3 (simply Pythagorean Theorem) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$E$$\",(0,4.17)); label(\"$$F$$\",(0.75,4.15),W); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$ By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\\angle OED = 90^\\circ$. Since $OD = 4$ and $DE = \\frac{1}{2}$, $OE = \\sqrt{16-\\frac{1}{4}} = \\frac{\\sqrt{63}}{2} = \\frac{3\\sqrt{7}}{2}$. Since $\\triangle CED$ is $30-60-90$, $CE = \\frac{\\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\\triangle CED$). Thus $OC =", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From similar triangles, we have that $PC=\\frac{x\\sqrt{x^2+18x}}{x+9}$. Similarly, $TP=QT=\\frac{9\\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem, $BQ=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem once again, $BP=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2+(\\frac{18\\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\\sqrt{405+\\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\\sqrt{432}$, so the answer is $(\\sqrt{432})^2=\\boxed{432}$. Solution 4 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\\frac{9x}{x+9}$. Now, apply law of cosines on $\\triangle PAB$. $$BP^2=\\left(\\frac{9x^2}{x+9}\\right)^2+18^2-2(18)\\left(\\frac{9x}{x+9}\\right)\\cos(\\angle PAB).$$ We can see that $\\cos(\\angle PAB)=\\cos(180^{\\circ}-\\angle", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$" ]

[ "as the center and draw segment $ON$ perpendicular to $AM$, $ON\\cap AM=N$, link $OM$. Then we have $OM\\parallel AD$. So $\\angle DAM=\\angle OMA$. Since $AD=2AM=2OM=1$, we have $\\cos\\angle DAM=\\cos\\angle OMP=\\frac{2}{\\sqrt{5}}$. As a result, $NM=OM\\cos\\angle OMP=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{5}}=\\frac{1}{\\sqrt{5}}.$ Thus $PM=2NM=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. Since $AM=\\frac{\\sqrt{5}}{2}$, we have $AP=AM-PM=\\frac{\\sqrt{5}}{10}$. Put $\\boxed{B}$. ~FANYUCHEN20020715 ## Solution 5 (Similar Triangles) Call the midpoint of $\\overline{AB}$ point $N$. Draw in $\\overline{NM}$ and $\\overline{NP}$. Note that $\\angle{NPM}=90^{\\circ}$ due to Thales's Theorem. $[asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw(circle((0.5,0.5),0.5)); draw((0,0)--(0,0.5)--(1,0.5)--cycle); label(\"A\",(0,0),SW); label(\"B\",(0,1),NW); label(\"C\",(1,1),NE); label(\"D\",(1,0),SE); label(\"M\",(1,0.5),E); label(\"P\",(0.2,0.1),S); label(\"N\",(0,0.5),W); draw((0,0.5)--(0.2,0.1)); markscalefactor=0.007; draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); [/asy]$ Using the Pythagorean theorem, $AM=\\frac{\\sqrt{5}}{2}$. Now we just need to find $AP$ using similar triangles. $$\\triangle APN\\sim\\triangle ANM\\Rightarrow\\frac{AP}{AN}=\\frac{AN}{AM}\\Rightarrow\\frac{AP}{\\frac{1}{2}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}}\\Rightarrow AP=\\boxed{\\textbf{(B) }\\frac{\\sqrt{5}}{10}}$$ ~QIDb602 ## Solution 6 (Intersecting Chords) Label the midpoint of $AD$ as $N$, and let $Q$ be the intersection of $ON$ and $AM$ Then $MQ=AQ=\\frac{1}{2}AM = \\frac{\\sqrt{5}}{4}$ and $NQ=\\frac{1}{4}$ Then $\\frac{1}{4}*\\frac{3}{4}=PQ*QM$ and $AP=AQ-PQ$ ~ERMSCoach ## Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem) Let $N$ be the midpoint of $\\overline{AB},$ from which $\\angle ANM=90^\\circ.$ Note that $\\angle NPM=90^\\circ$ by the Inscribed Angle Theorem. We have the following diagram: Since $AN=\\frac12$ and $NM=1,$ we get $AM=\\frac{\\sqrt5}{2}$ by the Pythagorean Theorem. Let $AP=x.$ It follows that $PM=\\frac{\\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\\triangle ANP$ gives $NP^2=\\left(\\frac12\\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\\triangle MNP$ gives $NP^2=1^2-\\left(\\frac{\\sqrt5}{2}-x\\right)^2.$ Equating the expressions for $NP^2$ produces \\begin{align*}", "7 \\right]$$ = $$\\scriptsize 6 \\left [ \\normalsize \\frac {11 \\: + \\: 21}{3}\\right]$$ = $$\\scriptsize 6 \\: \\times \\: \\left [ \\normalsize \\frac {32}{3}\\right] \\\\ \\scriptsize = 2 \\: \\times \\: 32 \\\\ \\scriptsize = 64 mm$$ Question 7 In the diagram above AB is a chord of a circle of radius 10cm, M is the mid-point of AB and If MN = 2cm, calculate (a) AB (b) The angle arc ANB subtends at the centre of the circle (c) The difference in length between AB and arc ANB to the nearest mm. Solution (a) Calculate AB Consider the sector ANBO AO, ON and OB are radii θ1 OB = OA = ON = 10cm OM = 10 – 2 = 8cm Using Pythagoras on right angled AMO AO2 = OM2 + AM2 AM2 = AO2 – OM2 AM = $$\\scriptsize \\sqrt{AO^2 \\: – \\: OM^2}$$ AM = $$\\scriptsize\\sqrt{10^2 \\: – \\: 8^2}$$ AM = $$\\scriptsize \\sqrt{36}$$ AM = $$\\scriptsize 6cm$$ AB = 2 x 6 = 12cm (b) Calculate The angle that are ANB subtends at the centre of the circle Consider Isosceles $$\\scriptsize \\Delta ABO \\; \\; \\theta_1 = \\theta_2$$ Cos θ1 = $$\\frac{adj}{hyp}\\\\ = \\frac {8}{10} \\\\ = \\frac {4}{5} \\\\ = \\scriptsize 0.8000$$ θ1 = cos – 1 (0.8000) θ1 = 36.9o Angle", "= \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$. ## Solution 2 $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"O\",O,SW); pair M = (4,0);dot(M);label(\"M\",M,SE); pair N = (4,2);dot(N);label(\"N\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"B\",B,NE); pair C = (4,-3);dot(C);label(\"C\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"P\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"A\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"Q\",Q,S); pair R = (3,0); dot(R); label(\"R\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"D\",D,E); draw(A--D); [/asy]$ The above solution works, but is quite messy and somewhat difficult to follow. This solution provides", "0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, 3), N); label(\"C_2'\", (9, 3), N); label(\"C_3'\", (15/2, 0), N); dot((15/2, 0)); [/asy]$ Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\\frac{3}{2}$. Now, we can determine the radius of $C_4$ using the formula $r = \\frac{k^2 \\cdot r'}{\\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \\frac{3}{2}$. $\\overline{OC}$ is just the distance from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7);", "from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(45), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, -3), SW); label(\"C_2'\", (9, -3), SE); label(\"C_3'\", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label(\"C_4'\", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$", "\\alpha\\log\\tan t/2$ we obtain \\displaystyle \\begin{aligned} \\gamma^\\phi(s) &= s \\\\ \\gamma^\\theta(s) &= 2\\arctan e^\\frac{s}{\\alpha} \\end{aligned} Let us draw such a curve. R.<t> = RealLine() ; print(R) Real number line R print(dim(R)) 1 c = S2.curve({polar: [2*atan(exp(-t/10)), t]}, (t, -oo, +oo), name='c') print(latex(c.display())) $\\displaystyle \\begin{array}{llcl} c:& \\mathbb{R} & \\longrightarrow & \\mathbb{S}^2 \\\\ & t & \\longmapsto & \\left({\\theta}, {\\phi}\\right) = \\left(2 \\, \\arctan\\left(e^{\\left(-\\frac{1}{10} \\, t\\right)}\\right), t\\right) \\\\ & t & \\longmapsto & \\left({\\xi}, {\\zeta}\\right) = \\left(-\\frac{1}{10} \\, t, t\\right) \\end{array}$ c.parent() $\\displaystyle \\mathrm{Hom}\\left(\\mathbb{R},\\mathbb{S}^2\\right)$ c.plot(chart=polar, aspect_ratio=0.1) It’s not totally clear this is curved so let’s try with another example. d = S2.curve({polar: [2*atan(exp(-t)), t]}, (t, -oo, +oo), name='d') print(latex(d.display())) $\\displaystyle \\begin{array}{llcl} d:& \\mathbb{R} & \\longrightarrow & \\mathbb{S}^2 \\\\ & t & \\longmapsto & \\left({\\theta}, {\\phi}\\right) = \\left(2 \\, \\arctan\\left(e^{\\left(-t\\right)}\\right), t\\right) \\\\ & t & \\longmapsto & \\left({\\xi}, {\\zeta}\\right) = \\left(-t, t\\right) \\end{array}$ d.plot(chart=polar, aspect_ratio=0.2) Now it’s clear that a straight line is curved in polar co-ordinates. But of course in Mercator co-ordinates, it is a straight line. This explains its popularity with mariners: if you draw a straight line on your chart and follow that bearing or rhumb line using a compass you will arrive at the end of the straight line. Of course, it is not the shortest path; great circles are but is much easier to navigate. c.plot(chart=mercator,", "\\} } (\\frac{2 h}{r} + o(h))$$ In which $\\frac{E(\\rho 1_{P1 \\in D, P2 \\in D_h \\backslash D})}{P\\{P1 \\in D, P2 \\in D_h \\backslash D \\} }$ is exact (with ac. $o(h)$) average distance between points on the disk of radius $r$ and on the edge. Which is easy to calculate in polar coordinates with the center on the edge of the disk. $$\\frac{1}{\\pi r^2} \\int^{\\pi/2}_{-\\pi/2} \\int^{2r cos(\\phi)}_{0} s^2 ds d\\phi = \\frac{32 r}{9 \\pi} + o(h)$$ So we have: $$f(r + h) = f(r)(1 - \\frac{4 h}{r} + o(h)) + 2 (\\frac{32 r}{9 \\pi} + o(h))(\\frac{2 h}{r} + o(h))$$ $$f(r+h) - f(r) = - \\frac{4 h}{r} f(r) + \\frac{128 h}{9 \\pi} + o(h)*...$$ $$f'(r) = - \\frac{4}{r} f(r) + \\frac{128}{9 \\pi} + o(1)$$ With $f(0) = 0$ we get: $$E \\rho = f(r) = \\frac{128 r}{45 \\pi}$$ Some notes about the integral: disk scheme in geogebra point F -- is the point in disk, point C -- is the point on the edge, distance between them is $s$, the angle between Ox and CF is $\\phi$. And since we transformed coordinates to polar, we should multiply by Jacobian matrix determinant, which is $s$. And finally $\\frac{1}{\\pi r^2}$ is density of point in the disk. Let $D$ be the disk, and let $$M = \\iint\\limits_{(x_0,y_0)\\in D} ~~\\iint\\limits_{(x,y) \\in D}dxdydx_0dy_0.$$", "for a few examples: For $P_1=(-3,2)$ and $P_2=(2,-3)$, we get the midpoint $M=\\left(\\frac{-3+2}2,\\frac{2+(-3)}2\\right)=\\left(\\frac{-1}2,\\frac{-1}2\\right)=(-0.5,-0.5)$. The points $P_1=(2,2)$ and $P_2=(2,4)$, we get the midpoint $M=\\left(\\frac{2+2}2,\\frac{2+4}2\\right)=\\left(\\frac{4}2,\\frac{6}2\\right)=(2,3)$. The midpoint of $P_1=(2,-4)$ and $P_2=(6,8)$ is given by $M=\\left(\\frac{2+6}2,\\frac{-4+8}2\\right)=\\left(\\frac{8}2,\\frac{4}2\\right)=(4,2)$. The midpoint of the last two points $P_1=(3,3)$ and $P_2=(-7,9)$ is given by $M=\\left(\\frac{3+(-7)}2,\\frac{3+9}2\\right)=\\left(\\frac{-4}2,\\frac{12}2\\right)=(-2,6)$. • #### Calculate the meeting point of Imke and her friend. ##### Tipps You have to add the coordinates and divide by $2$. The $x$-coordinate is a rational number while the $y$-coordinate is a natural number. Here you see the calculation of the midpoint of $(3,7)$ and $(7,8)$. ##### Lösung Here we have to determine the midpoint of $(2,7)$ and $(9,15)$. To do this, we use the midpoint formula. This means to get a coordinate of the midpoint we add the corresponding coordinates of those points and divide the resulting sum by $2$: $M=\\left(\\frac{2+9}2,\\frac{7+15}2\\right)=\\left(\\frac{11}2,\\frac{22}2\\right)=(5.5,11)$.", "is the point $(8,0)$, in polar coordinates, it can be any of the following: $8\\angle 0, 8\\angle 2\\pi, 8\\angle 4\\pi, \\cdots$ This is because we can go around the circle as many times as we want and return to the same point. Since we expect three roots, let's use the first three polar representations of 8: $$z^3 = 8 = \\left\\{ \\begin{array}{c} 8e^{0 i} \\\\ 8e^{2\\pi i} \\\\ 8e^{4\\pi i}\\\\ \\vdots \\end{array} \\right.$$ $$\\Rightarrow z = \\left\\{ \\begin{array}{lcl} \\left(8e^{0 i}\\right)^{\\frac{1}{3}} &=& 8^{\\frac{1}{3}} e^{\\frac{0}{3}} = 2\\\\ \\left(8e^{2\\pi i}\\right)^{\\frac{1}{3}} &=& 8^{\\frac{1}{3}} e^{\\frac{2\\pi}{3}i} = 2 \\left(\\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}\\right) = 2 \\left(-\\frac{1}{2} + i \\frac{\\sqrt{3}}{2}\\right) = -1 + i\\sqrt{3} \\\\ \\left(8e^{4\\pi i}\\right)^{\\frac{1}{3}} &=& 8^{\\frac{1}{3}} e^{\\frac{4\\pi}{3}i} = 2 \\left(\\cos \\frac{4\\pi}{3} + i \\sin \\frac{4\\pi}{3}\\right) = 2 \\left(-\\frac{1}{2} - i \\frac{\\sqrt{3}}{2}\\right) = -1 - i\\sqrt{3} \\\\ &\\vdots& \\end{array} \\right.$$ If we had used more than 3 numbers, the roots would have started repeating. If we had used less than 3, we would have missed some answers in the same way that there are two answers to $z^2 = 9$ (namely $z = \\pm 3$). ## Cartesian vs Polar Which is the best representation: Cartesian, $z = a + i b$, or Polar, $z = re^{i\\theta}$. As you might expect, it depends on what you're trying to do... For example, let's take:", "the $y$-coordinates of the two points divided by the absolute difference between the $x$-coordinates of the two points. This evaluates to $$\\frac{\\sin(36)+\\sin(56)}{\\cos(36)-\\cos(56)}$$ Now, using sum to product identities, we have this equal to $$\\frac{2\\sin(46)\\cos(10)}{-2\\sin(46)\\sin({-10})}=\\frac{\\sin(80)}{\\cos(80)}=\\tan(80)$$ so the answer is $\\boxed{\\textbf{(E) } 80}.$ ~lifeisgood03 Note: Though this solution is excellent, setting $M = (0,0)$ makes life a tad bit easier ~MathleteMA ## Solution 2 (Rotation, Isosceles Triangle, Parallel Lines) We will refer to the Diagram section. In this solution, all angle measures are in degrees. We rotate $\\triangle PUM$ by $180^\\circ$ about $M$ to obtain $\\triangle AU'M.$ Let $H$ be the intersection of $\\overline{PA}$ and $\\overline{GU'},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(375); pair P, A, T, U, G, M, N, U1, H; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); U1 = rotate(180,M)*U; H = intersectionpoint(P--A,G--U1); fill(U--P--M--cycle^^M--U1--A--cycle,yellow); dot(\"P\",P,1.5*SW,linewidth(4)); dot(\"A\",A,1.5*SE,linewidth(4)); dot(\"U\",U,1.5*(0,1),linewidth(4)); dot(\"G\",G,1.5*NE,linewidth(4)); dot(\"T\",T,1.5*(0,1),linewidth(4)); dot(\"M\",M,1.5*S,linewidth(4)); dot(\"N\",N,1.5*(0,1),linewidth(4)); dot(\"U'\",U1,1.5*S,linewidth(4)); dot(\"H\",H,1.5*NW,linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A); draw(G--U1,dashed); label(\"1\",midpoint(G--A),1.5*dir(30)); label(\"1\",midpoint(A--U1),1.5*dir(-30)); label(\"1\",midpoint(U--P),1.5*dir(150)); label(\"36^\\circ\",P,5*dir(18),fontsize(8)); label(\"56^\\circ\",A,2.5*dir(180-56/2),fontsize(8)); label(\"36^\\circ\",A,2.5*dir(180+25),fontsize(8)); Label L = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(U--M, 1, .5, 0, 8, red)); add(pathticks(M--U1, 1, .5, 0, 8, red)); [/asy] Note that $\\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\\angle AGU'=\\angle AU'G=\\frac{180-\\angle", "\\begin{align*}{r^2} & = {x^2} + {y^2}\\hspace{0.75in} r = \\sqrt {{x^2} + {y^2}} \\\\ \\theta & = {\\tan ^{ - 1}}\\left( {\\frac{y}{x}} \\right)\\end{align*} Let’s work a quick example. Example 1 Convert each of the following points into the given coordinate system. 1. Convert $$\\left( { - 4,\\frac{{2\\pi }}{3}} \\right)$$ into Cartesian coordinates. 2. Convert $$\\left( { - 1,-1} \\right)$$ into polar coordinates. Show All Solutions Hide All Solutions a Convert $$\\left( { - 4,\\frac{{2\\pi }}{3}} \\right)$$ into Cartesian coordinates. Show Solution This conversion is easy enough. All we need to do is plug the points into the formulas. \\begin{align*}x & = - 4\\cos \\left( {\\frac{{2\\pi }}{3}} \\right) = - 4\\left( { - \\frac{1}{2}} \\right) = 2\\\\ y & = - 4\\sin \\left( {\\frac{{2\\pi }}{3}} \\right) = - 4\\left( {\\frac{{\\sqrt 3 }}{2}} \\right) = - 2\\sqrt 3 \\end{align*} So, in Cartesian coordinates this point is $$\\left( {2, - 2\\sqrt 3 } \\right)$$. b Convert $$\\left( { - 1,-1} \\right)$$ into polar coordinates. Show Solution Let’s first get $$r$$. $r = \\sqrt {{{\\left( { - 1} \\right)}^2} + {{\\left( { - 1} \\right)}^2}} = \\sqrt 2$ Now, let’s get $$\\theta$$. $\\theta = {\\tan ^{ - 1}}\\left( {\\frac{{ - 1}}{{ - 1}}} \\right) = {\\tan ^{ - 1}}\\left( 1 \\right) = \\frac{\\pi }{4}$ This is not the correct angle however. This value", "# Chapter 6 - Section 6.3 - Polar Coordinates - Exercise Set - Page 742: 22 (a)$$(8, \\frac{13\\pi }{6})$$ (b)$$(-8, \\frac{7\\pi }{6})$$ (c)$$(8, -\\frac{11\\pi }{6})$$ #### Work Step by Step To plot the point $(r, \\theta )=(8, \\frac{\\pi }{6})$, begin with the $\\frac{\\pi }{6}$ angle. Because $\\frac{\\pi }{6}$ is a positive angle, draw $\\theta = \\frac{\\pi }{6}$ counterclockwise from the polar axis. Now consider $r=8$. Because $r \\gt 0$, plot the point by going out eight units on the terminal side of $\\theta$. Please note that if $n$ is any integer, the point $(r, \\theta )$ can be represented as$$(r, \\theta ) = (r, \\theta +2n\\pi ) \\\\ \\text{or} \\\\ (r, \\theta )= (-r, \\theta +\\pi + 2n\\pi ).$$So we have (a)$$(8, \\frac{\\pi }{6})= (8, \\frac{\\pi }{6}+2(1)\\pi)=(8, \\frac{13\\pi }{6}),$$(b)$$(8, \\frac{\\pi }{6})= (-8, \\frac{\\pi }{6}+\\pi +2(0)\\pi )=(-8, \\frac{7\\pi }{6}),$$(c)$$(8, \\frac{\\pi }{6})= (8, \\frac{\\pi }{6}+2(-1)\\pi)=(8, -\\frac{11\\pi }{6}).$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "SchachtN 2021-09-29 For the polar point $\\left(-4,-\\frac{3\\pi }{4}\\right)$ a) Give alternate coordinates with $r>0$ and $-2\\pi \\le \\theta <0$. b) Give alternate coordinates with $r<0$ and $0\\le \\theta <2\\pi$. c) Give alternate coordinates of your choice. Willie Step 1 Given: $P\\left(-4,-\\frac{3\\pi }{4}\\right)$ a) $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(4,-\\frac{3\\pi }{4}+\\pi \\right)$ $=\\left(4,\\frac{\\pi }{4}\\right)$ So, $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(4,\\frac{\\pi }{4}\\right)$ Step 2 b) $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(-4,-\\frac{3\\pi }{4}+2\\pi \\right)$ $=\\left(-4,-\\frac{5\\pi }{4}\\right)$ So, $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(-4,\\frac{5\\pi }{4}\\right)$ c) $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(4,-\\frac{3\\pi }{4}+3\\pi \\right)$ $=\\left(4,\\frac{9\\pi }{4}\\right)$ So, $\\left(-4,-\\frac{3\\pi }{4}\\right)=\\left(4,\\frac{9\\pi }{4}\\right)$ Jeffrey Jordon Answer is given below (on video) Do you have a similar question?", "\\\\\\\\ &=k\\int_{\\pi/6}^{5\\pi/6}\\cos \\phi \\left(\\frac12 (4\\sin^2\\phi -1)\\right)d\\phi\\\\\\\\ &=0 \\end{align} For the term $M_y$, we have that \\begin{align} M_y&=\\int_{\\pi/6}^{5\\pi/6} \\int_1^{2\\sin \\phi} \\frac{k\\rho \\sin \\phi}{\\rho}\\rho d\\rho d\\phi\\\\\\\\ &=k\\int_{\\pi/6}^{5\\pi/6} \\sin \\phi \\int_1^{2\\sin \\phi} \\rho d\\rho \\\\\\\\ &=k\\int_{\\pi/6}^{5\\pi/6}\\sin \\phi \\left(\\frac12 (4\\sin^2\\phi -1)\\right)d\\phi\\\\\\\\ &=\\sqrt{3}k \\end{align} You drew a good picture of the region of integration, though it might be helpful to you if you make your drawing larger. On a larger drawing it's easier to add details like these: The possible values of $\\theta$ for integration will range from the angle of the ray $\\overrightarrow{OA}$ to the angle of $\\overrightarrow{OB}$. We know that the length $OA = OB = 1$, and it should be clear by symmetry that the $y$-coordinate of $A$ is $\\frac12,$ and the same for $B$. The ray $\\overrightarrow{OA}$ therefore has $\\theta = \\arcsin \\frac12,$ and you should recognize that $\\arcsin \\frac12 = \\frac\\pi6.$ The ray $\\overrightarrow{OB}$ is in a symmetric position with $\\theta = \\pi - \\frac\\pi6.$ Within that range of $\\theta$ values, consider an arbitrary line $\\overleftrightarrow{OP}$. The radius of any point on that line within in the region of integration is at least $1$, but no greater than the distance $OP$. That distance is $r$ for a point with polar coordinates $(r,\\theta)$ on the circle $x^2 + y^2 = 2y.$ Subsituting $x^2 + y^2 = r^2$ and $y", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "# Convert the two rectangular (Cartesian) points to polar coordinates with $r > 0$and$0 ≤ θ < 2π.$ (a) Convert the two rectangular (Cartesian) points to polar coordinates with $r > 0$and$0 ≤ θ < 2π.$ $(4,-4)$ and $(-1,\\sqrt 3)$ (b) Convert the two polar points to rectangular (Cartesian) coordinates. $(-1,\\frac{\\pi}{2})$ and $(6,-\\frac{5\\pi}{4})$ (c) Sketch the set of points $\\left\\{ (r,\\theta ):1\\le r\\le 4,\\frac { \\pi }{ 4 } \\le \\theta \\le \\frac { 3\\pi }{ 4 } \\right\\}$ My attempt a) for $(4,-4)$ = $(r,\\theta)$= $(4\\sqrt 2,-\\frac{\\pi}{4})$ for $(-1,\\sqrt 3)$ = $(r,\\theta)$= $( 2,-\\frac{\\pi}{3})$ b)for $(-1,\\frac{\\pi}{2})$ = $(x,y)$ = $(0,1)$ for $(6,-\\frac{5\\pi}{4})$ = $(x,y)$ = $(-3\\sqrt 3,-3\\sqrt 3)$ Please verify my answers and also explain me how to do c) • I don't understand why it asks you sketch a set of infinitesimal points. Is it asking for a region shaded? – Andrew Li May 8 '18 at 13:05 • Your angle in the second point of (a) is wrong (you also want the first angle to be between the given values). I think your first one in part (b) is right, but it is not good practice to have a negative radius. Your second part of (b) is wrong. – Bill Wallis May 8 '18 at 13:07 For that b)for $(-1,\\frac{\\pi}{2})$ = $(x,y)$ = $(0,1)$ it", "the unit circle of the original map to the calculated ellipsoid. Using polar coordinates, we represent each point P of the standard reference map with phasor coordinates (gi, si) as $$P\\left( {g_i,s_i} \\right) = P\\left( {r_i \\ast {\\mathrm{cos}}\\theta _i,r_i \\ast {\\mathrm{sin}}\\theta _i} \\right)$$ (11) The ellipsoid has semi-major axes $$a = \\frac{\\omega }{2}$$ (12) and semi-minor axes $$b = \\frac{h}{2}$$ (13) Therefore, the ellipse equation becomes $$\\left( {\\frac{{g_i}}{{\\omega /2}}} \\right)^2 + \\left( {\\frac{{s_i}}{{h/2}}} \\right)^2 = rad^2$$ (14) Where rad is a ratio used to scale each radius ri in the reference map to a proportionally corresponding distance in the boundary box-adapted ellipse, which in polar coordinates becomes $${\\it{rad}}^2 = r_i^2 \\ast \\left( {\\left( {\\frac{{{\\mathrm{cos}}\\,\\theta _i}}{{\\frac{\\omega }{2}}}} \\right)^2 + \\left( {\\frac{{{\\mathrm{sin}}\\,\\theta _i}}{{\\frac{h}{2}}}} \\right)^2} \\right)$$ (15) Each point P(gi, si) of the standard reference map is geometrically scaled to a new coordinate (go, so) inside the ellipsoid using forward mapping, obtaining the equation $$\\left( {r_o,\\theta _o} \\right) = \\left( {\\sqrt {g_i^2 + s_i^2} /{\\it{rad}},{\\mathrm{tan}}^{ - 1}\\frac{{g_i}}{{s_i}}} \\right)$$ (16) This transform is applied to all standard reference maps to generate the respective scaled versions. ### Morph mode We linearly morph each point P(gi, si) to a new point P'(go, so) by utilizing a shifted-cone approach. Each standard map reference is first projected on to a 3D conical surface centered on the", "know: $A O = 6$, $O P = 3$, $A B \\bot O P$ and $A O \\bot B O$ Let's start: $\\because \\cos \\left(\\angle A O P\\right) = \\frac{P O}{A O} = \\frac{1}{2}$ I use degrees instead of radians here. :./_AOP=60° as $\\cos \\left(60\\right) = \\frac{1}{2}$ Using: $\\cos \\left(\\angle P O B\\right) = \\frac{O P}{O B}$ where $\\angle P O B = 90 - \\angle A O P = 30$ and $O P = 3$ $\\cos \\left(30\\right) = \\frac{3}{O B}$ $O B = \\frac{3}{\\cos \\left(30\\right)} = 2 \\sqrt{3}$ Since slopes are $\\pm \\frac{A O}{B O}$ where $A O = 6 \\mathmr{and} B O = 2 \\sqrt{3}$ $\\pm \\frac{A O}{B O} = \\pm \\frac{6}{2 \\sqrt{3}} = \\pm \\sqrt{3}$ Therefore the values of $m$ such that $y = m x + 6$ is tangent to ${x}^{2} + {y}^{2} = 9$ are $\\sqrt{3}$ and $- \\sqrt{3}$. But wait there's more Remember how I said there was an algebraic way to do it? And that it's overly complicated? Here it is (no diagrams): Any point on this circle can be expressed in polar coordinates: $\\left(r , \\theta\\right)$ where $r$ is the radius and $\\theta$ is the degrees offset from the $x$ axis. Therefore any point on the circle can be rewritten as: $\\left(3 , \\theta\\right)$. A tangent line on the circle", "- \\frac{\\rho}{1-\\rho}(O_2-O_1)$$, and power $$k = \\rho|IO_2^2 - r_2^2|$$. The second inversion has pole $$I = O_1 + \\frac{\\rho}{1+\\rho}(O_2-O_1)$$, and power $$k = \\rho|IO_2^2 - r_2^2|$$. O1 <- c(5,4); r1 <- 2 O2 <- c(8,5); r2 <- 3 rho <- r1/r2 I1 <- O1 - rho/(1-rho)*(O2-O1) k1 <- rho * abs(c(crossprod(I1-O2))-r2^2) I2 <- O1 + rho/(1+rho)*(O2-O1) k2 <- rho * abs(c(crossprod(I2-O2))-r2^2) opar <- par(mar = c(4,4,1,1)) plot(NULL, asp = 1, xlim = c(-3, 11), ylim = c(2,8), xlab = NA, ylab = NA) draw.circle(O1[1], O1[2], r1, lwd = 2) draw.circle(O2[1], O2[2], r2, lwd = 2) draw.circle(I1[1], I1[2], sqrt(k1), lwd = 3, border = \"red\") draw.circle(I2[1], I2[2], sqrt(k2), lwd = 3, border = \"red\") # check invertCircle(I1, k1, list(center=O1, radius=r1)) # should be C2 ## $center ## [1] 8 5 ## ##$radius ## [1] 3 invertCircle(I1, k1, list(center=O2, radius=r2)) # should be C1 ## $center ## [1] 5 4 ## ##$radius ## [1] 2 invertCircle(I2, k2, list(center=O1, radius=r1)) # should be C2 ## $center ## [1] 8 5 ## ##$radius ## [1] 3 invertCircle(I2, k2, list(center=O2, radius=r2)) # should be C1 ## $center ## [1] 5 4 ## ##$radius ## [1] 2 ## Tangent circles In the case when the two circles are tangent, there is only one inversion which swaps the two circles, and this is an inversion", "N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); [/asy]$ $C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius of $C_4$. We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\\Omega$, and all points on $\\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$, and since $\\Omega$ has radius of $6$, the resulting line must be $\\frac{36}{4} = 9$ units away from $O$. $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2," ]

[ "Inside Triangle) so P(Point Inside Triangle|Point Inside Quad) = P(Point Inside Triangle)/P(Point Inside Quad) which is why its a ratio of the areas axiomatically.", "× # Geometry In triangle ABC, Angle BAC = 60, AB=2AC.Point P is inside the triangle such that PA=$$\\sqrt3$$,PB=5 and PC=2.What is the area of triangle ABC? Note by Kishan K 3 years, 10 months ago Sort by:", "# Interior Point P! Find the Angle! Geometry Level 4 Suppose P is the point inside the triangle ABC so that $$\\angle PAB=10^o$$ , $$\\angle PBA=20^o$$, $$\\angle PCA=30^o$$, and $$\\angle PAC=40^o$$. Find the value of $$\\angle ABC$$ in degree ! ×", "# P and the Triangle Geometry Level 5 Let $$P$$ be a point in triangle $$ABC$$ with area $$1$$, and let $$f(P)=PA+PB+PC$$. For every triangle $$ABC$$, let $$h_{ABC}=\\min_P\\big[f(P)\\big]$$. Find $$\\displaystyle \\min_{ABC}\\left[h_{ABC}\\right]^4.$$ ×", "# find the point $P$ such that the expression has minimum value Let $ABC$ be a triangle with sides $$a,b,c.$$ Find a point $P$ inside the triangle such that $$a(PA)^2+b(PB)^2+c(PC)^2$$ is minimum • a simple calculation gives $P=(aA+bB+cC)/(a+b+c)$ - but it doesn't reveal the geometric meaning of $P$ :-) – user8268 May 9 '14 at 13:47 • It seems to me that P is the intersection of the bisectrices – user126154 May 9 '14 at 14:05 Let $I$ be the incenter of $ABC$. $I$ is the barycenter of the weighted points $(A;a)$,$(B;b)$ and $(C;c)$. This means that $$a\\overrightarrow{IA}+b\\overrightarrow{IB}+c\\overrightarrow{IC}=\\overrightarrow{0}\\tag{1}$$ Now \\eqalign{ a(PA)^2+b(PB)^2+c(PC)^2&=a(\\overrightarrow{IA}-\\overrightarrow{IP})^2 +a(\\overrightarrow{IB}-\\overrightarrow{IP})^2+ a(\\overrightarrow{IC}-\\overrightarrow{IP})^2\\cr &=a (IA)^2+b(IB)^2+c(IC)^2\\cr &\\phantom{=}-2\\underbrace{(a\\overrightarrow{IA}+b\\overrightarrow{IB}+c\\overrightarrow{IC})}_0 \\overrightarrow{IP}+(a+b+c)(IP)^2\\cr &=a (IA)^2+b(IB)^2+c(IC)^2+(a+b+c)(IP)^2 } Thus, the minimum is attained if and only if $IP=0$, that is, when $P$ is the incenter of the triangle. • Why is summation of a×IA term zero??? – avz2611 May 9 '14 at 14:23 • Because $I$ is the barycenter of $(A;a)$, $(B;b)$, $(C;c)$ see $(1)$. – Omran Kouba May 9 '14 at 14:43", "$P$ such that $PC=3,PA=4$ and $PB=5$. Find the perimeter of $\\triangle ABC$. The solution was simply amazing and gave me the incentives to solve this one বড় ভালবাসি তোমায়,মা", "A point P is chosen in the interior of triangle ABC so that when lines are drawn through P parallel to the sides of triangle ABC, the resulting smaller triangles have areas 4,9,49, respectively. Find the areas of triangle ABC. (第二届AIME1984 第三题)", "# Thread: Ratio of the area of triangle in terms of another triangle 1. I have trouble in determining the ratio of the area of $\\triangle PST$ in terms of $\\triangle PQR$ In the triangle PQR $QT=TR$, $PS=1 cm$ , $SQ=2 cm$ , How should I be writing the area of $\\triangle PST$ in terms of $\\triangle PQR$ What is known by me : Since $|\\overline{QT}|=|\\overline{TR}|$ it follows that $\\triangle PQT$ and $\\triangle PTR$ have equal areas, so $$[PQT]=\\frac12[PQR]$$ Has It got to do something with, Quote: If two polygons are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. So what must be done from here? 2. 3. Hi mathlearn, A key idea is that $\\triangle PST$ and $\\triangle PQT$ share the same altitude from vertex $T$. Using this fact, deduce that $\\operatorname{Area}(\\triangle PST) = \\frac{1}{3}\\operatorname{Area}(\\triangle PQT)$. Originally Posted by Euge Hi mathlearn, A key idea is that $\\triangle PST$ and $\\triangle PQT$ share the same altitude from vertex $T$. Using this fact, deduce that $\\operatorname{Area}(\\triangle PST) = \\frac{1}{3}\\operatorname{Area}(\\triangle PQT)$. Hi Euge , What I still don't understand is that , How did $\\operatorname{Area}(\\triangle PST) = \\frac{1}{3}\\operatorname{Area}(\\triangle PQT)$ get derived? From $|\\overline{PS}|=\\frac12|\\overline{SQ}|$ it follows that the area of $\\triangle PST$ is a half of the area of $\\triangle", "for a triangle as compared to the area formula for the circle? Can you write an area formula that is defined around the point P? Perhaps the areas of the three triangles PAB, PBC, PCA? Carl 5. Nov 5, 2005 ### klawesyn28 It's task from russian books with task from geometry. I don't see how your hints can help me.", "# Area without the required lengths. Geometry Level 3 Let $P$ be a point inside an equilateral triangle $ABC$ such that $PA = 3, PB =4, PC =5.$ Find the area of $\\triangle ABC$ to 2 decimal places. ×", "$P$ is a point inside a square $ABCD$ such that $PA= 1$, $PB = 2$ and $PC = 3$. How big is ${\\angle APB}$ ?", "Distance Tells Us The Area Geometry Level 3 Let $ABCD$ be a square and $P$ be a point inside the square such that $PB = 23$ and $PD = 29$ . Find the area of $\\triangle APC$ ×", "### Japan Finals 1991 - 2019 (JMO) 31p geometry problems from Japanese Mathematical Olympiads Finals (JMO Final) with aops links in the names 1991 - 2019 Let P,Q and R be points on the sides BC,CA and AB of a triangle ABC respectively, such that $\\overrightarrow{BP}:\\overrightarrow{PC}=\\overrightarrow{CQ}:\\overrightarrow{QA}=\\overrightarrow{AR}:\\overrightarrow{RB}=t:\\left( 1-t \\right)$ for some real number t. Prove that there is a triangle ∆ whose side lengths are AP,BQ,CR, and find the ratio of the area of triangle ABC to that of ∆ in terms of t. Let ABC be a given triangle with the area 1. Let D and E be points on sides AB and AC respectively. Lines BE and CD intersect at P. Find the maximum possible value of SPDE under the condition SBCED = 2SPBC. Let P0 be a point in the plane of triangle A0A1A2. Define Pi (i = 1,…,6) inductively as the point symmetric to Pi−1 with respect to Ak, where k is the remainder when i is divided by 3. (a) Prove that P6 ≡ P0. (b) Find the locus of points P0 for which PiPi+1 does not meet the interior of A0A1A2 for 0 ≤ I ≤ 5 . In a triangle ABC, M is the midpoint of BC. Given that ÐMAC = 15o, find the maximum value of ÐABC. Let ABCDE be a", "# In figure, P is a point in the interior of a parallelogram ABCD In figure, P is a point in the interior of a parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) = \\frac { 1 }{ 2 } ar (ABCD) (ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)", "for which the distances PA, PB, PC are in the ratio PA:PB:PC = 4:5:6, and then to scale the result up so that these distances become 4, 5 and 6. So let $A = \\bigl(\\frac12,\\frac{\\sqrt3}2\\bigl)$, $B = (0,0)$ and $C = (1,0)$. The set of points for which the ratio of distances PB:PC is 5:6 is a circle with equation $11(x^2+y^2) + 50x - 25 = 0$. The set of points for which the ratio of distances PB:PA is 5:4 is a circle with equation $9(x^2+y^2) - 25x - 25\\sqrt3y + 25 = 0$. Those circles intersect at the point $P = (x,y)$ given by $x = \\dfrac{-245 + 165\\sqrt{21}}{1204} \\approx 0.4245,\\quad y = \\dfrac{2835 - 435\\sqrt{21}}{1204\\sqrt3} \\approx 0.4036.$ The distance PB is then $d = \\sqrt{x^2+y^2}\\approx0.5857$. If we scale up the diagram to make PB = 5 then the side of the triangle becomes $5/d\\approx8.53635.$ If you do the calculation exactly, then the value for the side of the triangle comes out as $\\boxed{\\sqrt{\\dfrac{602}{77 - 15\\sqrt{21}}}}.$ Unless I have made arithmetical mistakes (something that unfortunately happens all too often), that expression cannot be simplified any further, which makes me think that this problem does not have a neat solution at all.", "Categories # Hidden triangular inequality (PRMO Problem 23, 2019) Problem Let ABCD be a convex cyclic quadrilateral . Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD? Topic Geometry Contest 2019 Mathematical circle . Sequential Hints The first step is very clear, you have to some how locate the point P . claim : The intersection of the diagonal is the required point . Proof : suppose that there is a point P’ other than P (the intersection of diagonal) . now try to apply triangular inequality in $\\triangle$PBD & $\\triangle$APC. (do the rest yourself , and complete the proof) so , the area of [ABCD]=$\\frac{1}{2}$$\\sin\\theta$(PA*PB+PB*PC+PC*PD+PD*PA) now, the area will maximum when $\\sin\\theta$=1 so , just put the value of PA, PB, PC,PD", "B}}$$ = $$\\vec{a}$$ $$\\overrightarrow{\\mathbf{A C}}$$ = $$\\vec{b}$$ Since ABCD is a parallelogram , we have Question 8. If $$\\overline{\\mathbf{P O}}+\\overline{\\mathbf{O Q}}$$ = $$\\overline{\\mathbf{Q O}}+\\overline{\\mathbf{O R}}$$, prove that the points P , Q , R are collinear. Let P, Q , R be the given points . Let O be the origin. Then the position vectors of P, Q, R are $$\\overrightarrow{\\mathrm{OP}}$$, $$\\overrightarrow{\\mathrm{OQ}}$$ and $$\\overrightarrow{\\mathrm{OR}}$$. ∴ P, Q, R lie on a straight line. Hence, P, Q, R are collinear. Question 9. If D is the midpoint of the aide BC of i triangle ABC , prove that $$\\overrightarrow{\\mathbf{A B}}$$ + $$\\overrightarrow{\\mathbf{A C}}$$ = 2 $$\\overrightarrow{\\mathbf{A D}}$$ Given D is the mid point of the side BC of a triangle ABC Question 10. If G is the centroid of a triangle ABC, prove that $$\\overrightarrow{\\mathbf{G A}}+\\overrightarrow{\\mathbf{G B}}+\\overrightarrow{\\mathbf{G C}}=\\overrightarrow{\\mathbf{0}}$$ Let ABC be the triangle with centroid G. Let $$\\vec{a}$$, $$\\vec{b}$$ and $$\\vec{c}$$ be the position vectors of the vertices A, B and C respectively with respect to the origin O. Then Question 11. Let A, B, and C be the vertices of a triangle. Let D, E, and F be the midpoints of the sides BC, CA, and AB respectively. Show that $$\\overrightarrow{\\mathbf{A D}}+\\overrightarrow{\\mathbf{B E}}+\\overrightarrow{\\mathbf{C F}}=\\overrightarrow{\\mathbf{0}}$$ Let ABC be the triangle. D, E, F are the mid points of the", "Let $$ABC$$ be equilateral, and $$D, E,$$ and $$F$$ be the midpoints of $$\\overline{BC}, \\overline{CA},$$ and $$\\overline{AB},$$ respectively. There exist points $$P, Q,$$ and $$R$$ on $$\\overline{DE}, \\overline{EF},$$ and $$\\overline{FD},$$ respectively, with the property that $$P$$ is on $$\\overline{CQ}, Q$$ is on $$\\overline{AR},$$ and $$R$$ is on $$\\overline{BP}.$$ The ratio of the area of triangle $$ABC$$ to the area of triangle $$PQR$$ is $$a + b\\sqrt {c},$$ where $$a, b$$ and $$c$$ are integers, and $$c$$ is not divisible by the square of any prime. What is $$a^{2} + b^{2} + c^{2}$$? (第十六届AIME1998 第12题)", "Regiomontanus In triangle ABC, suppose the ratio • Anonymous commented In triangle ABC, suppose the ratio • Anonymous commented It is not allowing me to post the whole question • Anonymous commented In triangle ABC suppose the ratio", "P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD? TopicGeometry..." ]

[ "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$", "B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); [/asy]$ Let $\\text{E}$ be the origin. Then, $\\text{D}=(1, 0)$ $\\text{A}=(0, 2)$ $\\text{B}=(1, 2)$ $\\text{F}=(2, 1)$ ${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\\frac{1}{2}x+2$ Subtracting the second equation from the first gives us $\\frac{5}{2}x-2=0$. Thus, $x=\\frac{4}{5}$. Plugging this into the first equation gives us $y=\\frac{8}{5}$. Since $\\text{C} (0.8, 1.6)$, $G$ is $(0.8, 2)$, ${AB}=1$ and ${CG}=0.4$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot 0.4=0.2=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\ \\frac15}$. ## Solution 3 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); label(\"H\",H,W); label(\"I\",I,E); [/asy]$ Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$ Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \\frac{3}{2} = 2: 3$. Based on similarity the length of the height of $GC$ is thus $\\frac{2}{5}\\cdot1 = \\frac{2}{5}$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{5}=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "label(\"C\",C,N); label(\"D\",DD,W); label(\"E\",EE,S); label(\"F\",FF,NE); label(\"1\",(A+EE)/2,S); label(\"2\",(EE+B)/2,S); [/asy]$ $\\textbf{(A) } \\frac{4}{9} \\qquad \\textbf{(B) } \\frac{1}{2} \\qquad \\textbf{(C) } \\frac{5}{9} \\qquad \\textbf{(D) } \\frac{3}{5} \\qquad \\textbf{(E) } \\frac{2}{3}$ ## Problem 21 How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11? $\\textbf{(A) }1\\qquad\\textbf{(B) }2\\qquad\\textbf{(C) }3\\qquad\\textbf{(D) }4\\qquad \\textbf{(E) }5$ ## Problem 22 Point $E$ is the midpoint of side $\\overline{CD}$ in square $ABCD,$ and $\\overline{BE}$ meets diagonal $\\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$ $[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(\"A\",(0,6),NW); label(\"B\",(6,6),NE); label(\"C\",(6,0),SE); label(\"D\",(0,0),SW); label(\"E\",(3,0),S); label(\"F\",(4,2),E); [/asy]$ $\\textbf{(A) } 100 \\qquad \\textbf{(B) } 108 \\qquad \\textbf{(C) } 120 \\qquad \\textbf{(D) } 135 \\qquad \\textbf{(E) } 144$ ## Problem 23 From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? $[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]$ $\\textbf{(A) } \\frac{2}{7} \\qquad \\textbf{(B) } \\frac{5}{42} \\qquad \\textbf{(C) } \\frac{11}{14} \\qquad \\textbf{(D) } \\frac{5}{7} \\qquad \\textbf{(E) } \\frac{6}{7}$ ## Problem", "(1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similary, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$.", "0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similarly, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$.", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "take $n-2$ more points, evenly spaced in [a,b] $u_i = k / {n-1} \\forall k \\in [1 .. n-1]$ $\\mathbf{B}_p(u (b - a) + a) = \\mathbf{B}_q(u_i)$ for each $u_i$ For a 4 control points bezier curve If we take $u$ in $(0, {1 \\over 3}, {2 \\over 3}, 1)$, hence $t_0$, $t_1$, $t_2$, $t_3$ respectively equal to $a$, $\\frac{b - 2 a}{3}$, $\\frac{2b - a}{3}$, $b$ $\\mathbf{Q}_0 = \\mathbf{B}_p(t_0)$ $\\mathbf{Q}_1 = {-15 \\mathbf{B}_p(t_0) + 54 \\mathbf{B}_p(t_1) -27 \\mathbf{B}_p(t_2) + 6 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_2 = {6 \\mathbf{B}_p(t_0) -27 \\mathbf{B}_p(t_1) + 54 \\mathbf{B}_p(t_2) -15 \\mathbf{B}_p(t_3)} / 18$ $\\mathbf{Q}_3 = \\mathbf{B}_p(b)$", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, because $\\triangle RPM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED", "# 2017 USAJMO Problems/Problem 3 ## Problem ($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$. $[asy] size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3)); draw(Circle(O, 2sqrt(3)), black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); [/asy]$ ## Solution (No Trig/Bash) Extend $DP$ to hit $EF$ at $K$. Then note that $[DEF]\\cdot\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=[ABC].$ Letting $BF=x$ and $PF=y$, we have that $\\frac{x+AB}y=\\frac{y+PC}x=\\frac{AC}{BP}.$ Solving and simplifying using LoC on $\\triangle BPC$ gives $\\frac{AB}{AF}=\\frac{PC}{PB+PC}.$ Similarly, $\\frac{AC}{AE}=\\frac{PB}{PB+PC}.$ Now we find $\\frac{AK}{DK}.$ Note that $\\frac{AD}{DP}=\\frac{AD}{BD}\\cdot\\frac{BD}{DP}=\\frac{AC}{PB}\\cdot\\frac{AB}{PC}=\\frac{AB^2}{PB\\cdot PC}.$ Now let $E'=DE\\cap AF$ and $F'=DF\\cap AE$. Then by an area/concurrence theorem, we have that $\\frac{DK}{AK}+\\frac{DE'}{EE'}+\\frac{DF'}{FF'}=1,$ or $\\frac{DK}{AK}+(1-\\frac{DP}{AP}-\\frac{DC}{BC})+(1-\\frac{DP}{AP}-\\frac{BD}{BC})=1.$ Thus we have that $\\frac{DK}{AK}=2\\cdot\\frac{DP}{AP}.$ Manipulating these gives $\\frac{AK}{DK}=\\frac{(PB+PC)^2}{2\\cdot PB\\cdot PC}.$ Thus $\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=\\frac12,$ and we are done. ~cocohearts ## Solution 1 WLOG, let $AB = 1$. Let $[ABD] = X, [ACD] =", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "Solving for $b$ in terms of $a$ yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from E to ML and label the point of intersection P: pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L) label(\"P\",P,E) label(\"$A$\",A,NW); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,NE); label(\"$E$\",E,dir(90)); label(\"$F$\",F,NE); label(\"$G$\",G,NE); label(\"$H$\",H,W); label(\"$J$\",J,S); label(\"$K$\",K,SE); label(\"$L$\",L,SE); label(\"$M$\",M,dir(90)); label(\"$N$\",N,dir(180)); (Error compiling LaTeX. label(\"P\",P,E) ^ error: could not load module '7a023ad7e27f5ed1c7813ddb2ed5dce918789de8.asy') This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "to $\\overrightarrow{AC} \\iff \\overrightarrow{AB} = k\\overrightarrow{AC}$ for some constant $k$. Now $\\overrightarrow{AB} = -3 \\mathbf{i} + 12 \\mathbf{j} + 6 \\mathbf{k},\\quad \\overrightarrow{AC} = (p-4) \\mathbf{i} + 8 \\mathbf{j} + 4 \\mathbf{k},$ and thus $2\\overrightarrow{AB} = -6 \\mathbf{i} + 24 \\mathbf{j} + 12 \\mathbf{k},\\quad 3\\overrightarrow{AC} = 3(p-4) \\mathbf{i} + 24 \\mathbf{j} + 12 \\mathbf{k}.$ Comparing coefficients, we can say that $A, B$ and $C$ are collinear if and only if $-6 = 3(p-4)$, which is true if and only if $p = 2$. Alternatively, we could you use the fact that $C$ lies on the straight line $AB$. We could find the equation of the line $AB$ and then substitute $C$ into this equation to find $p$. 1. If $p = -2$, find the position vector of $D$ so that $ABCD$ is a parallelogram. We have the following sketch. The position vector of $D$ is therefore \\begin{align*} \\mathbf{d} = \\mathbf{a} + \\overrightarrow{AD} = \\mathbf{a} + \\overrightarrow{BC}&= \\mathbf{a} + \\mathbf{c} - \\mathbf{b} \\\\ &= \\left( 4\\mathbf{i} - 9\\mathbf{j} - \\mathbf{k} \\right) + \\left( -2\\mathbf{i} - \\mathbf{j} + 3\\mathbf{k} \\right) - \\left( \\mathbf{i} + 3\\mathbf{j} + 5\\mathbf{k} \\right) \\\\ &= (4 - 2 - 1) \\mathbf{i} + (-9 - 1 - 3) \\mathbf{j} + (-1 + 3 - 5) \\mathbf{k} \\\\ &= \\mathbf{i} - 13 \\mathbf{j} - 3 \\mathbf{k}. \\end{align*} Check:", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = $1008-x$ , then you", "Find the components of Let $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components ofLet $\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components of (a) $\\mathbf{u}+\\mathbf{w}$ (b) $\\mathbf{v}-3 \\mathbf{u}$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector Show that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vectorShow that $\\mathbf{u}=\\frac{1}{\\|\\mathbf{v}\\|} \\mathbf{v}$ is a unit vector ... Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components ofLet $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of (a) $\\mathbf{v}-\\mathbf{w}$ (b) $6 \\mathrm{u}+2 \\ma ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$0 answers 58 views Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$Find the components of the vector$\\overrightarrow{P_{1} P_{2}}$when$P_1(-6,2)$and$P_2(-4,-1)$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$0 answers 88 views Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)$Find the norm of$v$, and a unit vector that is oppositely directed to$\\mathbf{v} = (1, -1, 2)\\$ ..." ]

[ "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "# 2017 USAJMO Problems/Problem 3 ## Problem ($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$. $[asy] size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3)); draw(Circle(O, 2sqrt(3)), black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); [/asy]$ ## Solution (No Trig/Bash) Extend $DP$ to hit $EF$ at $K$. Then note that $[DEF]\\cdot\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=[ABC].$ Letting $BF=x$ and $PF=y$, we have that $\\frac{x+AB}y=\\frac{y+PC}x=\\frac{AC}{BP}.$ Solving and simplifying using LoC on $\\triangle BPC$ gives $\\frac{AB}{AF}=\\frac{PC}{PB+PC}.$ Similarly, $\\frac{AC}{AE}=\\frac{PB}{PB+PC}.$ Now we find $\\frac{AK}{DK}.$ Note that $\\frac{AD}{DP}=\\frac{AD}{BD}\\cdot\\frac{BD}{DP}=\\frac{AC}{PB}\\cdot\\frac{AB}{PC}=\\frac{AB^2}{PB\\cdot PC}.$ Now let $E'=DE\\cap AF$ and $F'=DF\\cap AE$. Then by an area/concurrence theorem, we have that $\\frac{DK}{AK}+\\frac{DE'}{EE'}+\\frac{DF'}{FF'}=1,$ or $\\frac{DK}{AK}+(1-\\frac{DP}{AP}-\\frac{DC}{BC})+(1-\\frac{DP}{AP}-\\frac{BD}{BC})=1.$ Thus we have that $\\frac{DK}{AK}=2\\cdot\\frac{DP}{AP}.$ Manipulating these gives $\\frac{AK}{DK}=\\frac{(PB+PC)^2}{2\\cdot PB\\cdot PC}.$ Thus $\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=\\frac12,$ and we are done. ~cocohearts ## Solution 1 WLOG, let $AB = 1$. Let $[ABD] = X, [ACD] =", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "to the area of the triangle $O A C$ is $9: 16$ $(7)$ 9. (2018/june/paper01/q10) The points $A, B, C$ and $D$ are such that $$\\overrightarrow{A B}=5 \\mathbf{i}+5 \\mathbf{j} \\quad \\overrightarrow{A C}=-2 \\mathbf{i}+15 \\mathbf{j} \\quad \\overrightarrow{A D}=-7 \\mathbf{i} +10 \\mathbf{j}$$ (a) (i) Find $\\overrightarrow{D C}$ as a simplified expression in terms of $\\mathbf{i}$ and $\\mathbf{j}$. (ii) Hence show that $A B C D$ is a parallelogram. (4) (b) Find a unit vector parallel to $\\overrightarrow{B D}$ as a simplified expression in terms of $\\mathbf{i}$ and $\\mathbf{j}$. The point $E$ lics on $B D$ and $B E: E D-3: 10$ (c) Find $\\overrightarrow{A E}$ as a simplified expression in terms of $\\mathbf{i}$ and $\\mathbf{j}$. (2) The point $F$ is such that $D C F$ and $A E F$ are both straight lines. (d) Find $D C: C F$ 10. $(2019 /$ june/paper02/q1) Referred to a fixed origin $O$, the point $A$ has position vector $(4 \\mathbf{i}+3 \\mathbf{j})$ and the point $B$ has position vector $(\\mathbf{i}+7 \\mathbf{j})$ (a) Find $\\overrightarrow{A B}$ as a simplified expression in terms of $\\mathbf{i}$ and $\\mathbf{j}$ (b) Find a unit vector that is parallel to $\\overrightarrow{A B}$ 11. (2019/juneR/paper01/q7) $O, A, B$ and $C$ are fixed points such that $$\\overrightarrow{O A}=8 \\mathbf{i}-6 \\mathbf{j} \\quad \\overrightarrow{O B}=15 \\mathbf{i}-6 \\mathbf{j} \\quad \\overrightarrow{O C}=8 \\mathbf{i}+\\mathbf{j}$$ (a) Find $\\overrightarrow{B", "# 1989 AIME Problems/Problem 15 ## Problem Point $P$ is inside $\\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\\triangle ABC$. ## Solutions ### Solution 1 Let $[RST]$ be the area of polygon $RST$. We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$, and line $XP$ intersects line $YZ$ at point $L$, then $\\dfrac{XP}{PL} = \\frac{[XPY] + [ZPX]}{[YPZ]}.$ $[asy] size(170); pair X = (1,2), Y = (0,0), Z = (3,0); real x = 0.4, y = 0.2, z = 1-x-y; pair P = x*X + y*Y + z*Z; pair L = y/(y+z)*Y + z/(y+z)*Z; draw(X--Y--Z--cycle); draw(X--P); draw(P--L, dotted); draw(Y--P--Z); label(\"X\", X, N); label(\"Y\", Y, S); label(\"Z\", Z, S); label(\"P\", P, NE); label(\"L\", L, S);[/asy]$ This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ (as they share a common height from $Y$), and the same is true of triangles $ZPY$ and $LPZ$. We'll also use the related fact that $\\dfrac{[XPY]}{[ZPX]} = \\dfrac{YL}{LZ}$. This is slightly more well known, as it is used in the standard proof of Ceva's theorem. Now we'll apply these results to the", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\\cdot16=\\boxed{\\textbf{(C) }96}$ ~quacker88 ~IceMatrix", "and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\\cdot16=\\boxed{\\textbf{(C) }96}$ ~quacker88 ## Solution 4 (Triangles) $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\",", "ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{(B) }30}$. - fath2012 ## Solution 9 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$, we have $$\\dfrac{BF}{FC} \\cdot \\dfrac{CA}{DA} \\cdot \\dfrac{DE}{BE} = 3\\dfrac{BF}{FC} = 1 \\implies \\dfrac{BF}{FC} = \\dfrac13 \\implies \\dfrac{BF}{BC} = \\dfrac14.$$ Therefore, $$[EBF] = \\dfrac{BE}{BD}\\cdot\\dfrac{BF}{BC}\\cdot [BCD] = \\dfrac12 \\cdot \\dfrac 14 \\cdot \\left( \\dfrac23 \\cdot [ABC]\\right) = \\boxed{\\textbf{(B) }30}.$$ ## Solution 10 (Graph Paper) $[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F); draw(B--D);", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "(2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0, 3.6), S); [/asy]$ We know that $AU = UM$, $AV = VC$, so $UV = \\frac{1}{2} MC$. As $\\angle UPM = \\angle VPC = 90$, we can see that $\\triangle UPM \\cong \\triangle VPC$ and $\\triangle UVP \\sim \\triangle MPC$ with a side ratio of $1 : 2$. So $UP = VP = 4$, $MP = PC = 8$. With that, we can see that $S\\triangle UPM = 16$, and the area of trapezoid $MUVC$ is 72. As said in solution 1, $S\\triangle AMC = 72 / \\frac{3}{4} = \\boxed{\\textbf{(C) } 96}$. ## Solution 5 (Medians, Height, Pythagorean Theorem, Lots of symmetry, A little extra work) [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); draw((-2, 6), (2, 6)) label(\"K\", (0, 6), NE); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy] It is well known that medians divide each other into segments of $2:1$ ratio. From this, we have $PC=MP=8$ and $UP=UV=4$. From right triangle $\\triangle{MPC}$, $MC^2=MP^2+MC^2=8^2+8^2=128$, which implies $MC=\\sqrt{128}=8\\sqrt{2}$. Then the area of $\\triangle{AMC}$ is $\\dfrac{8\\sqrt{2} \\cdot AB}{2}$, so our goal is to find $AB$. Note that $AB=AP+BP$. Since $\\triangle{AMC}$ is isosceles, by symmetry $MB=MC$, because $AB$ is the altitude. Knowing this, $BP$", "(iii) Find the area of triangle $A B D$. [2] 19 (CIE $2014, \\mathrm{w}$, paper 23 , question 5) In the diagram $\\overrightarrow{O P}=\\mathbf{b}, \\overrightarrow{P Q}=\\mathbf{a}$ and $\\overrightarrow{O R}=3 \\mathbf{a}$. The lines $O Q$ and $P R$ intersect at $X$. (i) Given that $\\overline{O X}=\\mu \\bar{O} \\dot{Q}$, express $\\overrightarrow{O X}$ in terms of $\\mu, \\mathbf{a}$ and $\\mathbf{b}$.[1] (ii) Given that $\\overrightarrow{R X}=\\lambda \\overrightarrow{R P}$, express $\\overrightarrow{O X}$ in terms of $\\lambda_{\\text {, }} \\mathbf{a}$ and $\\mathbf{b}$. [2] (iii) Hence find the value of $u$ and of $\\lambda$ and state the value of the ratio $\\frac{R X}{X P}$. [3] 20 (CIE 2015 , s, paper 12, question 7) In the diagram $\\overrightarrow{A B}=4 \\mathrm{a}, \\overrightarrow{B C}=\\mathbf{b}$ and $\\overrightarrow{D C}=7 \\mathbf{a}$. The lines $A C$ and $D B$ intersect at the point $X$. Find, in terms of $\\mathbf{a}$ and $\\mathbf{b}$. (i) $\\overrightarrow{D A}$ (ii) $\\overrightarrow{D B}$ Given that $\\overrightarrow{A X}=\\lambda \\overrightarrow{A C}$, find, in terms of $\\mathbf{a}, \\mathbf{b}$ and $\\lambda$, (iii) $\\overrightarrow{A \\dot{X}}$ (iv) $\\overline{D X}$. Given that $\\overrightarrow{D X}=\\mu \\overrightarrow{D B}$ (v) find the value of $\\lambda$ and of $\\mu$. 21 (CIE 2015, s, paper 21 , question 7) (a) The four points $O, A, B$ and $C$ are such that $\\overrightarrow{O A}=5 \\mathbf{a}, \\quad \\overrightarrow{O B}=15 \\mathbf{b}, \\quad \\overrightarrow{O C}=24 \\mathbf{b}-3 \\mathbf{a}$. Show that $B$ lies on the", "\\frac{1}{2} (\\mathbf b + 2\\mathbf d).$$ Now is it evident that $\\overline{PA}$ is parallel to the vector $\\overline{XA}$? What does this mean? How does this fact help you? To make this into a more systematic proof, we will proceed roughly as J.M. suggests, but using the vector notation. Let $E$ be the point of intersection of $AP$ with $BD$. We want to find $E$. Since $E$ lies inside the segment $BD$, $E$ can be written as $\\alpha \\mathbf b + (1-\\alpha) \\mathbf d$ for some $\\alpha \\in [0,1]$ (the exact range in which $\\alpha$ lies is not that important). Then, $E$ will lie in $AP$ iff $\\overline{EA}$ is parallel to $\\overline{PA}$. But we can compute these vectors: $$\\overline {EA} = (\\mathbf b +\\mathbf d) - (\\alpha \\mathbf b + (1-\\alpha) \\mathbf d) = (1-\\alpha) \\mathbf b + \\alpha \\mathbf d,$$ and $$\\overline{PA} = (\\mathbf b +\\mathbf d) - \\frac{1}{2} \\mathbf b = \\frac{1}{2} \\mathbf b + \\mathbf d.$$ Now under what condition will these two vectors be parallel to each other? -", "the area of $\\triangle{PQR}$ is $\\boxed{\\textbf{(C) }2\\sqrt{2}}$. Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\\overrightarrow{PQ}=[-2,2,0]$ and $\\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\\text{area} = \\frac{1}{2}\\left|\\overrightarrow{PQ} \\times \\overrightarrow{PR}\\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\\sqrt{2}$, giving the area as $2\\sqrt{2}$ like before. ## Solution 2 As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\\triangle PQR$ is perpendicular to the plane determined by $ABCD$. To see this, consider the bird's-eye view, looking down upon $P$, $Q$, and $R$ projected onto $ABCD$: $[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label(\"A\", (0,0), SW); label(\"B\", (3,0), SE); label(\"C\", (3,3), NE); label(\"D\", (0,3), NW); label(\"P\", (2,0), S); label(\"Q\", (0,2), W); label(\"R\", (1,1), NE); dot((2,0)); dot((0,2)); dot((1,1)); draw((0,2)--(2,0)); [/asy]$ Additionally, we know that $PQ$ is parallel to the plane determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\\overline{PQ} = 2\\sqrt{2}$, so the area of $\\triangle PQR$ is $\\frac{1}{2}", "lies on $O A$ such that $\\overrightarrow{O F}=\\lambda \\overrightarrow{O A}$ and $F, E$ and $B$ are collinear. (b) Find the value of $\\lambda$. The area of triangle $O F B$ is 5 square units. (c) Find the area of triangle $O A D$. Give your answer in the form $\\frac{p}{q}$, where $p$ and $q$ are integers. (3) 18. (2017/june/paper01/q3) In Figure $1, \\overrightarrow{O A}=\\mathbf{a}$ and $\\overrightarrow{O B}=\\mathbf{b}$ The point $C$ is the midpoint of $O A$ and the point $D$ divides $O B$ in the ratio $2: 1$ (a) Find $\\overrightarrow{C D}$ in terms of $\\mathbf{a}$ and $\\mathbf{b}$ (2) The point $E$ lies on $A B$ produced such that $\\overrightarrow{O E}=2 \\mathbf{b}-\\mathbf{a}$ (b) Find $\\overrightarrow{C E}$ in terms of $\\mathbf{a}$ and $\\mathbf{b}$ (c) Hence show that $C, D$ and $E$ are collinear. 19. (2018/jan/paper02/q6) Figure 3 shows the triangle $O A B$ with $\\overrightarrow{O A}=a$ and $\\overrightarrow{O B}=\\mathbf{b}$. (a) Find $\\overrightarrow{A B}$ in terms of $\\mathbf{a}$ and $\\mathbf{b}$. The point $P$ is such that $\\overrightarrow{O P}=\\frac{3}{4} \\overrightarrow{O A}$, and the point $Q$ is the midpoint of $A B$ (b) Find $\\overrightarrow{P Q}$ as a simplified expression in terms of a and $\\mathbf{b}$. The point $R$ is such that $P Q R$ and $O B R$ are straight lines where $$\\overrightarrow{Q R}=\\mu \\overrightarrow{P Q} \\text { and } \\overrightarrow{B" ]

[ ", 6 )$ $u = AB = \\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 6 \\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1 } \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 5 & 6 \\\\ 2 & 4 \\end{bmatrix}= 20 \\: – \\: 12 = 8$ The area of the parallelogram is $8$. ## Numerical Result The area of the parallelogram is $8$. ## Example Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$ Solution A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram. $A = ( 0 , 0 )$ $B = ( 6 , 8 )$ $C = ( 5 , 4 )$ $D = ( 11 , 6 )$ $u = AB = \\begin{bmatrix} 6\\\\ 8 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 5\\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1", "+0 # help 0 46 1 What is the area of the parallelogram with vertices $$\\begin{pmatrix} 3 \\\\ 1\\\\ 2 \\end{pmatrix},$$ $$\\begin{pmatrix} 2 \\\\ 1\\\\4\\end{pmatrix}$$$$\\begin{pmatrix} 4\\\\2\\\\4 \\end{pmatrix},$$and $$\\begin{pmatrix} 5\\\\2\\\\2 \\end{pmatrix}?$$ Apr 29, 2020", "the area of the paralellogram and the original square. Since the original square has area $1$, the area of the parallelogram is the determinant of $\\bc{\\A}$. Working this out requires only a small amount of basic geometry. Here’s the simplest way to do it. We first name the four elements of our matrix: $\\bc{\\A} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ We can now name the four corners of our parallelogram in terms of these four scalars, by multiplying the corner points of the unit square by $\\bc{\\A}$: $\\begin{matrix} \\times & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}1\\\\ 0\\end{pmatrix} & \\begin{pmatrix}0\\\\ 1\\end{pmatrix} & \\begin{pmatrix}1\\\\ 1\\end{pmatrix} \\\\ \\bc{\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}} & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}a\\\\ c\\end{pmatrix} & \\begin{pmatrix}b\\\\ d\\end{pmatrix} & \\begin{pmatrix}a+b\\\\ c+d\\end{pmatrix} \\\\ \\end{matrix}$ This gives us the following picture. Here we can see the area of the paralellogram clearly: there is one large rectangle of area $\\oc{ad}$. To get from this area to the area of our parallelogram, we should subtract the area of the green triangle in the bottom, which is part of the rectangle but not the parallelogram. But then, there’s a green triangle at the top, with the same size, which is (mostly) part of the parallelogram but not of the rectangle, so these cancel each other out. We", "# Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cristiano Sears ###### Not exactly what you’re looking for? content_user Vertices are $$A(-3,0),\\ B(-1,6),\\ C(8,5)$$ and $$D(6,-1)$$ So $$\\vec{AB}=<-1-(-3),6-0>=<2,6>$$ $$\\vec{AD}=<6-(-3),-1-0>=<9,-1>$$ Hence area of parallelogram $$=|\\vec{AB}\\times\\vec{AD}|$$ $$\\vec{AB}\\times\\vec{AD}=\\begin{bmatrix}i&j&k\\\\2&6&0\\\\9&-1&0\\end{bmatrix}$$ $$=i(0-0)+j(0-0)+k(-2-54)$$ $$=-56k$$ Hence Area $$=|\\vec{AB}\\times\\vec{AD}|$$ $$=\\sqrt{(-56)^2}=56$$", "as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The formula for the area of a parallelogram is base x height. Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [closed], Soft question on the area of a parallelogram using vectors. Area of the Parallelogram: A parallelogram is a geometric structure having one side having opposite sides parallel to each other. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. #vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj# First of all, to simplify the task, let's move it to a position when its vertex A coincides with the origin of coordinates. So this's a B C on G E. Now the area of", "$$\\overline { OX }$$ at R and $$\\overline { OY }$$ at A 5. Join $$\\overline { IR }$$, $$\\overline { FR }$$, $$\\overline { FA }$$ and $$\\overline { AI }$$ 6. FAIR is the required parallelogram Calculation of area: Area of the Parallelogram FAIR = b x h sq. units = 5.4 x 4.3 = 23.22 sq. cm Question 7. GAIN, GA = 7.5 cm, GI = 9 cm and ∠GAI – 100° Solution: Given: In the parallelogram GAIN, GA = 7.5 cm GI = 9 cm and ∠GAI= 100° Construction: Steps: 1. Draw a line segment GA = 7.5 cm. 2. Make an angle GAI = 100° at A. 3. With G as centre, draw an arc of radius 9 cm cutting AX at I. Join GI. 4. Draw a line IY parallel to GA through I. 5. With I as centre, draw an arc of radius T.5 cm on IY cutting at N. Join GN 6. GAIN is the required parallelogram. Calculation of area: Area of Parallelogram GAIN = b x h sq.units = 7.5 x 3.9 = 29.25 sq. cm Question 8. HERB, HE = 6 cm, ∠EHB = 60° and EB = 7 cm. Solution: Given: In the parallelogram HERB, HE = 6 cm, ∠EHB = 60° and EB = 7 cm", "# Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4 Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4), (11,6) ## Want to know more about Matrix transformations? • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Aniqa O'Neill Parallelogram can be described with 4 vertices or 2 vectors. Since we are given 4 vertices(ABCD), let's find vectors u, v that describe parallelogram. Graph these points to see how vectors are described A=(0,0) B=(5,2) C=(6,4) D=(11,6) $$u=AB=\\begin{bmatrix}5\\\\2\\end{bmatrix}$$ $$u=AC=\\begin{bmatrix}6\\\\4\\end{bmatrix}$$ Area of parallelogram is absolute value of determinant $$\\begin{bmatrix}u_1 & v_1 \\\\u_2 &v_2 \\end{bmatrix}$$ $$\\begin{bmatrix}u_1 & v_1 \\\\u_2 &v_2 \\end{bmatrix}=det\\begin{bmatrix}5& 6 \\\\2 &4 \\end{bmatrix}=20-12=8$$ Result: 8", "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", "and $$d = 2$$: 1. $$a = b + c \\times d$$ 2. $$ab^2 = 2c - d \\div 2$$ 3. Solve the following equations: 1. $$-7x = 56$$ 2. $$4(x + 3) = 16$$ 4. Sipho, Fundiswa and Ntosh are brothers. Sipho earns Rx per month; Fundiswa earns R1 000 more than Sipho per month, and Ntosh earns double what Sipho earns. If you add their salaries together you get a total of R27 000. 1. Write this information in an equation using $$x$$. (2) 2. Solve the equation to find how much Fundiswa earns per month. (2) 5. Construct the following figure using only a pencil, ruler and compass. Do not erase any construction arcs. 1. An angle of 60° (2) 2. The perpendicular bisector of line VW, where VW = 10 cm 3. Triangle KLM, where KL = 8,3 cm; LM = 5,9 cm and KM = 7 cm 4. Parallelogram EFGH, where E = 60°, EF = 4,2 cm and EH = 8 cm 1. What is/are the property/properties that make a rhombus different to a parallelogram? (1) 2. True or false: a rectangle is a special type of parallelogram. (1) 6. Determine the size of x in each figure. Show all the necessary steps and give reasons. 7. Study the following diagram. Then", "# Math Help - geometry and indices/log problem 1. ## geometry and indices/log problem 1. A parallelogram PQRS with area ( $1+2\\sqrt{7}$) $units^2$ has length PQ ( $3+\\sqrt{7}$) units. Given that the distance between the two parallel sides, SR and PQ is ( $a+b\\sqrt{c}$) units where a, b and c are integers. Find, without using a calculator, the values of a, b and c. 2. Given that $y=18+ax^b$ , y=56 when x=4 and y=20 when x=2.5, find the value of a and b. 2. Originally Posted by wintersoltice 1. A parallelogram PQRS with area ( $1+2\\sqrt{7}$) $units^2$ has length PQ ( $3+\\sqrt{7}$) units. Given that the distance between the two parallel sides, SR and PQ is ( $a+b\\sqrt{c}$) units where a, b and c are integers. Find, without using a calculator, the values of a, b and c. 2. Given that $y=18+ax^b$ , y=56 when x=4 and y=20 when x=2.5, find the value of a and b. 1. $A_{\\mbox{parallelogram}} = Bh$ where $B= 3+\\sqrt{7}$, $h=a+b\\sqrt{c}$ and $A = 1+2\\sqrt{7}$ 2. just solve the system: 56 = 18 + a(4)^b 20 = 18 + a(2.5)^b", "# Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3). I can't figure out what I'm doing wrong.. I let: $a = |DC| = (-2,-1) - (-6,3) = (4,-4)$ $b = |AD| = (-1,1) - (-5,5) = (4,-4)$ so: $\\begin{bmatrix} 4&-4\\\\4&-4\\end{bmatrix} \\implies$ Area $= -16 + 16 = 0$ ? The determinant is the area of the parallelogram is it not? • Not sure how you're labelling vertices, but $a$ appears to be $|CD|$, while $b$ appears to be $|AB|$. You get a result of zero because those lines are parallel. – Joey Zou Jul 6 '16 at 22:08 • You are correct that was the problem. The answer is $|-12|$ – Yusha Jul 6 '16 at 22:11", "} \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 6 & 5 \\\\ 8 & 4 \\end{bmatrix}= 24 \\: – \\: 40 = 16$ The area of the parallelogram is $16$. 5/5 - (5 votes)", "Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11 # Prove that the Area of the Parallelogram Formed by the Lines A1x + B1y + C1 = 0, A1x + B1y+ D1 = 0, A2x + B2y + C2 = 0, A2x + B2y + D2 = 0 is - Mathematics Short Note Prove that the area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1yd1 = 0, a2x + b2y + c2 = 0, a2x + b2y + d2 = 0 is $\\left| \\frac{\\left( d_1 - c_1 \\right)\\left( d_2 - c_2 \\right)}{a_1 b_2 - a_2 b_1} \\right|$ sq. units. Deduce the condition for these lines to form a rhombus. #### Solution The given lines are a1x + b1y + c1 = 0 ... (1) a1x + b1y + d1 = 0 ... (2) a2x + b2y + c2 = 0 ... (3) a2x + b2y + d2 = 0 ... (4) The area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0 and a2x + b2y + d2 = 0 is given below: $Area = \\left| \\frac{\\left( c_1 - d_1 \\right)\\left( c_2 - d_2 \\right)}{\\begin{vmatrix}a_1 & a_2 \\\\ b_1 &", "### Find area of parallelogram using ratio Question Sample Titled 'Find area of parallelogram using ratio' In the figure, ${A}{B}{C}{D}$ is a parallelogram. ${E}$ and ${F}$ are points lying on ${A}{B}$ and ${C}{D}$ respectively. ${A}{D}$ is produced and ${E}{F}$ is produced to meet at ${G}$. It is given that ${D}{F}:{F}{C}={2}:{7}$ and ${A}{D}:{D}{G}={1}:{1}$. If the area of $\\triangle{D}{F}{G}$ is ${7}\\ \\text{ cm}^{{2}}$ , then the area of paralleogram ${A}{B}{C}{D}$ is ${A}$${B}$${C}$${D}$${E}$${F}$${G}$ A ${63}$ $\\text{cm}^{{2}}$ . B ${21}$ $\\text{cm}^{{2}}$ . C ${18}$ $\\text{cm}^{{2}}$ . D ${9}$ $\\text{cm}^{{2}}$ . ${A}$${B}$${C}$${D}$${E}$${F}$${G}$$\\theta$${x}$${x}$${7}{y}$${2}{y}$ Let ${A}{D}={x},{D}{F}={2}{y},\\angle{F}{D}{G}=\\theta.$ ∵ ${D}{F}:{F}{C}$ $={2}:{7},$ ∴ ${F}{C}$ $={7}{y}$ ∵ ${A}{D}:{D}{G}$ $={1}:{1},$ ∴ ${D}{G}$ $={x}$ Consider $\\angle{D}{F}{G},$ $\\dfrac{{1}}{{2}}{\\left({x}\\right)}{\\left({2}{y}\\right)}{\\sin{\\theta}}$ $={7}$ ${x}{y}{\\sin{\\theta}}$ $={7}$ $\\ldots{\\left(\\ast\\right)}$ The area of ${A}{B}{C}{D}$ $={2}\\times{\\left[\\dfrac{{1}}{{2}}\\times{C}{D}\\cdot{A}{D}\\cdot{\\sin{{\\left({180}^{\\circ}-\\theta\\right)}}}\\right]}$ $={\\left({D}{F}+{F}{C}\\right)}\\cdot{x}\\cdot{\\sin{\\theta}}$ ${\\sin{{\\left({180}^{\\circ}-\\theta\\right)}}}={\\sin{\\theta}}$ $={\\left({2}{y}+{7}{y}\\right)}\\cdot{x}\\cdot{\\sin{\\theta}}$ $={9}{x}{y}{\\sin{\\theta}}$ $={9}{\\left({7}\\right)}$ Substitute ${\\left(\\ast\\right)}$ $={63}$ $\\text{cm}^{{2}}$ # 專業備試計劃 Level 4+ 保證及 5** 獎賞 ePractice 會以電郵、Whatsapp 及電話提醒練習 ePractice 會定期提供溫習建議 Level 5** 獎勵：會員如在 DSE 取得數學 Level 5** ，將獲贈一套飛往英國、美國或者加拿大的來回機票，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 Level 4 以下賠償：會員如在 DSE 未能達到數學 Level 4 ，我們將會全額退回所有會費，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 # FAQ ePractice 是甚麼？ ePractice 是一個專為中四至中六而設的網站應用程式，旨為協助學生高效地預備 DSE 數學（必修部分）考試。由於 ePractice 是網站應用程式，因此無論使用任何裝置、平台，都可以在瀏覽器開啟使用。更多詳情請到簡介頁面。 ePractice 可以取代傳統補習嗎？ 1. 會員服務期少於兩個月；或 2. 交易額少於 HK\\$100。 Initiating... HKDSE 數學試題練習平台", "# Math Help - Parallelogram... 1. ## Parallelogram... Let f(t) be the area of the parallelogram with vertices v=(1,2t), w = (2t,1), (0,0), and v+w = (2t+1,2t+1). Find a formula for f(t). I'm assuming one can't simply 'stand up' the parallelogram so that it's a rectangle... 2. Let $O(0,0), \\ A(1,2t), \\ B(2t,1), C(1+2t, 1+2t)$ $Area[OACB]=2\\cdot Area[OAB]=2\\cdot\\frac{1}{2}|\\Delta|$ where $\\Delta=\\begin{vmatrix}0 & 0 & 1\\\\1 & 2t & 1\\\\2t & 1 & 1\\end{vmatrix}=1-4t^2$ Therefore $f(t)=|1-4t^2|$ 3. Another way to do it is using the cross product of v and w. That also works with 3D coordinates. 4. Hello, PTL! Yet another approach . . . Did you make a sketch? Let $f(t)$ be the area of the parallelogram with vertices: $\\vec v\\,=\\,(1,2t),\\;\\vec w\\,=\\,(2t,1),\\;(0,0)$, and $\\overrightarrow{v+w} \\:=\\:(2t+1,2t+1).$ Find a formula for $f(t).$ Code: | 2t+1 + - - - - - - - - - - o v+w | * *: | * * : | v * * : 2t + - - o * : | * * : 1+ * o w : | * * : : | * * : : |* * : : - - o - - + - - - - + - - * - - (0,0) 1 2t 2t+1 We have a rhombus. If $d_1,\\:d_2$ are diagonals of a rhombus,", "Construct a parallelogram ABCD in which AB = 4.3 cm, Question: Construct a parallelogram ABCD in which AB = 4.3 cm, AD = 4 cm and BD = 6.8 cm. Solution: Steps of construction: Step 1: Draw $A B=4.3 \\mathrm{~cm}$ Step 2: With $B$ as the centre, draw an arc of $6.8 \\mathrm{~cm}$. Step 3: With $A$ as the centre, draw another $\\operatorname{arc}$ of $4 \\mathrm{~cm}$, cutting the previous arc at $D$. Step 4: Join $B D$ and $A D$. Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with $D$ as the centre, draw an arc of $4.3 \\mathrm{~cm}$. Step 6: With $B$ as the centre, draw another $\\operatorname{arc} 4 \\mathrm{~cm}$, cutting the previous arc at $C$. Step 7: Join $C D$ and $B C$. ​then, ABCD is the required parallelogram.", "c & d \\end{pmatrix}$ (The rows here are the vectors defined by the edges of the parallelogram that meet the origin.) For example, the parallelogram with corners $$(0, 0), (1, 2), (0, 1)$$ and $$(1, 3)$$ can be represented by the array $\\begin{pmatrix} 1 & 2 \\\\ 0 & 1 \\end{pmatrix}$ And in three dimensions, we can describe a parallelepiped with one corner at the origin, $$(0, 0, 0)$$, by putting the coordinates of the three corners that share edges with the origin as the three rows of a $$3$$ by $$3$$ array. And in one dimension we can represent a line segment, which starts at the origin and goes to a point $$x$$, by the single entry, $$x$$. Thus, the three matrices $\\begin{pmatrix} 5 \\end{pmatrix}, \\begin{pmatrix} 1 & 2 \\\\ 0 & 1 \\end{pmatrix}, \\begin{pmatrix} 1 & 2 & 3 \\\\ 2 & 1 & 8 \\\\ 1 & 4 & 1 \\end{pmatrix}$ can represent a line segment, a parallelogram and a parallelepiped respectively. We give a uniform name to the (signed) length of the line segment, the area of the parallelogram and the volume of the parallelepiped, all with some appropriate sign. Each is called the determinant of the given array. And we can define the determinant just as well with similar meaning for larger square", "of the parallelogram with edges ${v=(3,2)}$ and ${w=(1,4)}$. • Find the area of the triangle with sides ${v,w,}$and ${v+w}$.Draw it. • Find the area of the triangle with sides ${v,w,}$and ${w-v}$.Draw it. Solution: • $\\displaystyle \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=10.$ • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture omitted. • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture Omitted. $\\Box$ Exercise 4.4.28 A box has edges from ${(0,0,0)}$ to ${(3,1,1),(1,3,1),}$and ${(1,1,3)}$.Find its volumn and also find the area of each parallelogram face. Solution: The volumn of the box is $\\displaystyle \\begin{vmatrix} 3&1&1\\\\ 1&3&1\\\\ 1&1&3 \\end{vmatrix}=20.$ The area of each parallelogram face is $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 3&1 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 1&3\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2}.$ In order to compute the area of each parallelogram face,we can also compute $\\displaystyle \\sqrt{\\det A^TA}=\\sqrt{\\det \\left(\\begin{bmatrix} 3&1&1\\\\ 1&3&1 \\end{bmatrix} \\begin{bmatrix} 3&1\\\\ 1&3\\\\ 1&1 \\end{bmatrix}\\right)}= 6 \\sqrt{2}$ etc.$\\Box$ Exercise 4.4.35 An ${n}$– dimensional cube has how many corners?How many edges?How many ${n-1}$-dimensional faces?The ${n}$– cube whose edges are the rows of ${2I}$ has volumn ${\\underline{~~~~~~~~~~}}$.A hypercube computer has parallel processors at the corners with connections along the edeges. Solution: An ${n}$-dimensional cube", "## Formula : $$c\\;=\\;a*b\\;=\\;|a|\\;*\\;|b|\\;*\\;sinθ * n$$ Simplify with example: $$\\text{C}\\;=\\;\\begin{vmatrix}i & j & k \\\\a_1 & a_2 & a_3 \\\\b_1 & b_2 & b_3 \\\\ \\end{vmatrix}$$ c=[(a2∗b3)−(a3∗b2)]i+[(a3∗b1)−(a1∗b3)]j+[(a1∗b2)−(a2∗b1)]k ## Example : Suppose that a parallelogram whose adjacent sides are defined by the two vectors a (6, 3, 1) and b (3, -1, 5).Calculate the area of the parallelogram. Solutions : such that a = 6i + 3j + 1k and b = 3i – 1j + 5k $$a*b\\;=\\;\\text{C}\\;=\\;\\begin{vmatrix}i & j & k \\\\6 & 3 & 1 \\\\3 & -1 & -5 \\\\ \\end{vmatrix}$$ a*b=i((3*5)–(1*(-1)))+j((1*3)–(6*5))+k((6*(-1))–(3*3)) a*b=i((15)–(-1)))+j((3)–(30))+k((-6))–(9)) a*b=i(15+1)+j(3-30)+k(-6-9) $$a*b=16i+j(-27)+k(-15)$$ $$a*b=16i-17j-15k$$", "+0 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) +1 136 1 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) part (a): Let $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$find a*b and the answer was 10 Apr 2, 2022 #1 +30 +2 $\\sin^2 θ = 1 - \\cos^2 θ = 1 - \\frac{4}{13} = \\frac{9}{13}$ every angle in a parallelogram is between 0 and π radians, so θ must be positive, so $\\sin θ = \\sqrt{\\frac{9}{13}} = \\frac{3}{\\sqrt{13}}$ area of parallelogram is $\\|{a}\\| \\|{b}\\| \\sin \\theta = \\sqrt{65} \\cdot \\sqrt{5} \\cdot \\frac{3}{\\sqrt{13}} = \\boxed{15}$ Jun 25, 2022" ]

[ "just that. His solution is shorter (so better) than mine: \\begin{align*}A(A+B)^{-1}B&=(A+B-B)(A+B)^{-1}(A+B-A)\\\\&=[I-B(A+B)^{-1}](A+B-A)\\\\&=A+B-A-B+B(A+B)^{-1}A\\\\&=B(A+B)^{-1}A\\end{align*} # Determinants as Area and Volume The Area of a Parallelogram Let $v=(v_1,v_2)$ and $w=(w_1,w_2)$ be two linearly independent vectors in $\\mathbb{R}^2$. Then they span a parallelogram as shown in Figure 1. Figure 1. Parallelogram spanned by two vector v and w. The area $A$ of the parallelogram is \\begin{align*} A&=||v||||w||\\sin\\theta\\\\ &=||v\\times w||\\\\ &=\\left|\\begin{array}{cc} v_1 & v_2\\\\ w_1 & w_2 \\end{array}\\right|. \\end{align*} In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the area. Exercise. Given two vectors $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ in $\\mathbb{R}^3$, show that $||v||||w||\\sin\\theta=||v\\times w||$ where $\\theta$ is the angle between $v$ and $w$. Hint. Note that $||v||^2||w||^2\\sin^2\\theta=||v||^2||w||^2-(v\\cdot w)^2$. The Volume of a Parallelepiped Let $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$, $w=(w_1,w_2,w_3)$ be three linearly independent vectors in $\\mathbb{R}^3$. Then they span a parallelepiped as shown in Figure 2. Figure 2. Parallelepiped spanned by vectors u, v and w. The volume $V$ of the parallelepiped is \\begin{align*} V&=||u||\\cos\\theta ||v\\times w||\\\\ &=u\\cdot (v\\times w)\\\\ &=\\left|\\begin{array}{ccc} u_1 & u_2 & u_3\\\\ v_1 & v_2 & v_3\\\\ w_1 & w_2 & w_3 \\end{array}\\right|. \\end{align*} In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for", "## Calculus (3rd Edition) $\\sqrt {35}$ The area of the parallelogram is calculated as: $||\\textbf{u}\\times\\textbf{v}||$ We take the cross product: $\\textbf{u}\\times\\textbf{v}=\\begin{vmatrix}\\textbf{i}&\\textbf{j}&\\textbf{k}\\\\1&0&3\\\\2&1&1\\end{vmatrix}$ $=\\textbf{i}(0\\times1-3\\times1)-\\textbf{j}(1\\times1-3\\times2)+\\textbf{k}(1\\times1-0\\times2)$ $=-3\\textbf{i}+5\\textbf{j}+\\textbf{k}$ $||\\textbf{u}\\times\\textbf{v}||=\\sqrt {(-3)^{2}+5^{2}+1^{2}}=\\sqrt {35}$", "## The Area of a Parallelogram Area of the parallelogram determined by two nonzero and nonparallel vectors $$a = [a_1, a_2]$$ and $$b = [b_1, b_2]$$ is: $\\text{Area} = \\left\\|{a}\\right\\| h = \\left\\|{a}\\right\\| \\left\\|{b}\\right\\|(sin\\theta) = \\left\\|{a}\\right\\| \\left\\|{b}\\right\\|\\sqrt{1-cos^2\\theta}$ Where $$\\theta$$ is the angle between two vectors. We obtain the last equation using trig identity $$sin^2x + cos^2x = 1$$. Now let’s square the area equation $\\text{Area}^2 = \\left\\|{a}\\right\\|^2 \\left\\|{b}\\right\\|^2 (1-cos^2\\theta)$ $= \\left\\|{a}\\right\\|^2 \\left\\|{b}\\right\\|^2 - \\left\\|{a}\\right\\|^2 \\left\\|{b}\\right\\|^2 cos^2\\theta$ $=\\left\\|{a}\\right\\|^2 \\left\\|{b}\\right\\|^2 - (a \\cdot b)^2$ This step is because recall $$a \\cdot b = \\left\\|{a}\\right\\| \\left\\|{b}\\right\\|(cos\\theta)$$ $=(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1b_1 + a_2b_2)^2$ And this is in fact the same as following $=(a_1b_2 - a_2b_1)^2$ We can now obtain the area equation for our parallelogram $\\text{Area} = |a_1b_2 - a_2b_1|$ This equation with absolute value bar is also known as the determinant if the matrix $A = \\left[\\begin{matrix} a_1 & a_2 \\cr b_1 & b_2 \\end{matrix}\\right]$ We denote it with $$det(A)$$ $det(A) = \\left|\\begin{matrix} a_1 & a_2 \\cr b_1 & b_2 \\end{matrix}\\right| = a_1b_2 - a_2b_1$ This is also known as second-order determinant. ## The Cross Product Second-order determinants can also be useful when trying to find perpendicular vectors in $$R^3$$. Let’s say we have two vectors in $$R^3$$, $$p$$ and $$q$$. To find a vector $$c$$ that is", "+0 # Area of Parallelogram with two adjacent 2d vectors. 0 67 1 Find the area of the parallelogram with u and v as adjacent edges. u = 5i -2j v = 6i -2j Jun 11, 2020 #1 +8340 0 The area is equal to $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right|$$. Notice that $$\\mathbf{u}\\times \\mathbf{v} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\6&-2&0\\end{bmatrix} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\1&0&0\\end{bmatrix} = 2\\mathbf{k}$$ Therefore $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right| = |2\\mathbf{k}| = 2$$ Therefore the area of the required parallelogram is 2 sq. units. Jun 11, 2020", "one) intersects the other vector. Then find the magnitude of a vector that would be the height of the parallelogram. To determine the area of our parallelogram, we need the length of its base #|AB|# and the altitude #|DH|# from vertex #D# to point #H# on side #AB# (that is, #DH_|_AB#). #S = |AB|*|DH|=|U_B*V_D-V_B*U_D|#, In terms of original coordinates, it looks like this: Protection against an aboleths enslave ability. #C[U_C=x_C-x_A,V_C=y_C-y_A]# #|AD|=sqrt(U_D^2+V_D^2)# A rectangle is 12cm longer than its wide. {\\displaystyle 15in^ {2}}, since. We note that the area of a triangle defined by two vectors $\\vec{u}, \\vec{v} \\in \\mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. 1. The parallelograms are differentiated by the size of the adjacent sides as well as by the angles, however, the opposite sides are equal. #sin^2(/_BAD)=1-cos^2(/_BAD)=# The length #AD# is the length of vector #q#: Is there a bias against mentioning your name on presentation slides? So in your case we have to write the points in $\\mathbb{R}^2$ as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ Thus we have the same", "back and think about what properties a mathematical object, describing area should have. Suppose we have two (possibly infinitesimal) vectors, forming a two-dimensional parallelogram and we need the mathematical object that tells us the area of this parallelogram: \\mathrm{area}(V,W) = \\text{ area of the parallelogram formed by V and W}. An obious property is that the number $\\mathrm{area}(V,W)$ ought to double if we double the length of one vector: \\mathrm{area}(2V,W) = 2 \\times \\text{ area of the parallelogram formed by V and W} \\mathrm{area}(V,W+Z) = \\mathrm{area}(V,W) + \\mathrm{area}(V,Z) A pictorial proof can be seen in the following figure: Together these properties are what we called linearity above, and we see that this fits perfectly with the defintion of a $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. Another important propery is that $\\mathrm{area}(V,W)$ must vanish if $V$ and $W$ are parallel, i.e. if $W=aV$ for some number $a$, we have \\mathrm{area}(V,W) =\\mathrm{area}(V,aV) \\stackrel{!}{=} 0. .Because we have $\\mathrm{area}(V,aV)=a \\times \\mathrm{area}(V,V) \\stackrel{!}{=} 0$, we start with $\\mathrm{area}(V,V) \\stackrel{!}{=} 0$ in the next step to see that $\\mathrm{area}(V,W)$ must be an antisymmetric $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. The proof is simple: From \\mathrm{area}(V,V) = 0, if we write $V= U+W$, it follows \\mathrm{area}(U+W,U+W) = 0. Using the linearity yields \\mathrm{area}(U,U) + \\mathrm{area}(U,W) + \\mathrm{area}(W,U) + \\mathrm{area}(W,W) = 0. Using now $\\mathrm{area}(V,V)", "+0 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) +1 136 1 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) part (a): Let $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$find a*b and the answer was 10 Apr 2, 2022 #1 +30 +2 $\\sin^2 θ = 1 - \\cos^2 θ = 1 - \\frac{4}{13} = \\frac{9}{13}$ every angle in a parallelogram is between 0 and π radians, so θ must be positive, so $\\sin θ = \\sqrt{\\frac{9}{13}} = \\frac{3}{\\sqrt{13}}$ area of parallelogram is $\\|{a}\\| \\|{b}\\| \\sin \\theta = \\sqrt{65} \\cdot \\sqrt{5} \\cdot \\frac{3}{\\sqrt{13}} = \\boxed{15}$ Jun 25, 2022", "\\right\\|=\\sqrt{{{a}^{2}}+{{b}^{2}}}$ The vector $\\mathbf{w}-\\mathbf{v}$ is $29.9\\mathbf{i}+77.2\\mathbf{j}$. Comparing it with the above equation gives \\begin{align} & a=29.9\\mathbf{i} \\\\ & b=77.2\\mathbf{j} \\end{align} Then magnitude of $\\mathbf{w}-\\mathbf{v}$ is given by $\\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|$. \\begin{align} & \\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|=\\sqrt{{{\\left( 29.9 \\right)}^{2}}+{{\\left( 77.2 \\right)}^{2}}} \\\\ & \\approx 82.8 \\\\ & \\approx 83 \\end{align} Then magnitude of $\\mathbf{w}-\\mathbf{v}$ is $83$. The speed of the plane is $83\\text{ mph}$. If the vector is $\\mathbf{F}=a\\mathbf{i}+b\\mathbf{j}$, then magnitude of $\\mathbf{F}$ is given by $\\left\\| \\mathbf{F} \\right\\|$. $\\left\\| \\mathbf{F} \\right\\|=\\sqrt{{{a}^{2}}+{{b}^{2}}}$ And the direction angle is given as $\\cos \\theta =\\frac{a}{\\left\\| \\mathbf{F} \\right\\|}$ The vector $\\mathbf{w}-\\mathbf{v}$ is $29.9\\mathbf{i}+77.2\\mathbf{j}$. Then direction angle of $\\mathbf{w}-\\mathbf{v}$ is given by \\begin{align} & \\cos \\theta =\\frac{a}{\\left\\| \\mathbf{w}-\\mathbf{v} \\right\\|} \\\\ & \\cos \\theta =\\frac{29.9}{82.8} \\\\ & \\theta ={{\\cos }^{-1}}\\left( \\frac{29.9}{82.8} \\right) \\\\ & \\theta =68.8{}^\\circ \\end{align}", "to {\\displaystyle {\\begin{aligned}\\mathbf {u\\times v} &=\\\\(u_{2}v_{3}\\mathbf {i} +u_{3}v_{1}\\mathbf {j} +u_{1}v_{2}\\mathbf {k} )-(u_{3}v_{2}\\mathbf {i} +u_{1}v_{3}\\mathbf {j} +u_{2}v_{1}\\mathbf {k} )\\\\&=\\\\(u_{2}v_{3}-u_{3}v_{2})\\mathbf {i} +(u_{3}v_{1}-u_{1}v_{3})\\mathbf {j} +(u_{1}v_{2}-u_{2}v_{1})\\mathbf {k} .\\end{aligned}}} Using cofactor expansion along the first row instead, it expands to[6] ${\\displaystyle \\mathbf {u\\times v} ={\\begin{vmatrix}u_{2}&u_{3}\\\\v_{2}&v_{3}\\end{vmatrix}}\\mathbf {i} -{\\begin{vmatrix}u_{1}&u_{3}\\\\v_{1}&v_{3}\\end{vmatrix}}\\mathbf {j} +{\\begin{vmatrix}u_{1}&u_{2}\\\\v_{1}&v_{2}\\end{vmatrix}}\\mathbf {k} }$ which gives the components of the resulting vector directly. ## Properties ### Geometric meaning Figure 1. The area of a parallelogram as the magnitude of a cross product Figure 2. Three vectors defining a parallelepiped The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides (see Figure 1): ${\\displaystyle \\left\\|\\mathbf {a} \\times \\mathbf {b} \\right\\|=\\left\\|\\mathbf {a} \\right\\|\\left\\|\\mathbf {b} \\right\\|\\sin \\theta .\\,\\!}$ Indeed, one can also compute the volume V of a parallelepiped having a, b and c as edges by using a combination of a cross product and a dot product, called scalar triple product (see Figure 2): ${\\displaystyle \\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=\\mathbf {b} \\cdot (\\mathbf {c} \\times \\mathbf {a} )=\\mathbf {c} \\cdot (\\mathbf {a} \\times \\mathbf {b} ).}$ Since the result of the scalar triple product may be negative, the volume of the parallelepiped is given by its absolute value. For instance, ${\\displaystyle V=|\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )|.}$ Because the", "the area of the blue region, which is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[5] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V", "$(\\pi)$ is not parallel to $AB_1B_1$ (as claimed in your text) because: $$\\vec{AB_1} \\times \\vec{AB_2}=\\begin{pmatrix}2\\\\4\\\\-4\\end{pmatrix} \\times \\begin{pmatrix}4\\\\-4\\\\-2\\end{pmatrix}=\\begin{pmatrix}8\\\\-12\\\\-24\\end{pmatrix}$$ is not proportional to the normal vector $\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}$ of plane $(\\pi)$.", "as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The formula for the area of a parallelogram is base x height. Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [closed], Soft question on the area of a parallelogram using vectors. Area of the Parallelogram: A parallelogram is a geometric structure having one side having opposite sides parallel to each other. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. #vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj# First of all, to simplify the task, let's move it to a position when its vertex A coincides with the origin of coordinates. So this's a B C on G E. Now the area of", "$\\\\ R \\times S = \\begin{vmatrix} 2 & 0 \\\\ 1 & 3 \\end{vmatrix} i - \\begin{vmatrix} -2 & 0 \\\\ -2 & 3\\end{vmatrix}j + \\begin{vmatrix} -2 & 2 \\\\ -2 & 1 \\end{vmatrix}k$ $\\\\ R \\times S = (2 * 3 - 0 * 1)i - (-2 * 3 - 0 * -2)j + (-2 * 1 - 2 * -2)k$ $\\\\ N = 6i + 6j + 2k$ $|N| = \\sqrt {{6^2} + {6^2} + {2^2}}$ $|N| = \\sqrt {{76}} = 8.7178$ $\\ {N_{unit}} = \\frac{1}{{|N|}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right]$ $\\ {N_{unit}} = \\frac{1}{{8.7178}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0.6882\\\\ 0.6882\\\\ 0.2294\\\\ \\end{array}} \\right]$ It is a common mistake to believe that if R and S are unit vectors, the cross product will also be a unit vector. The vector product equation shows that this is only true when the angle between the two vectors is 90 degrees and therefore the sinus of the angle theta is 1. $\\\\ |T| = |R||S|sin\\theta$ Areas The vector product might be used to determine the area of a parallelogram or a triangle (with the vertices at P1 - P3). Image 17 shows the two vectors helping to form a parallelogram and a triangle. Figure 17 - Deriving the Area of a Parallelogramm /", "+ 4-1k$ $\\\\ R \\times S = \\begin{vmatrix} 2 & 0 \\\\ 1 & 3 \\end{vmatrix} i - \\begin{vmatrix} -2 & 0 \\\\ -2 & 3\\end{vmatrix}j + \\begin{vmatrix} -2 & 2 \\\\ -2 & 1 \\end{vmatrix}k$ $\\\\ R \\times S = (2 * 3 - 0 * 1)i - (-2 * 3 - 0 * -2)j + (-2 * 1 - 2 * -2)k$ $\\\\ N = 6i + 6j + 2k$ $|N| = \\sqrt {{6^2} + {6^2} + {2^2}}$ $|N| = \\sqrt {{76}} = 8.7178$ $\\ {N_{unit}} = \\frac{1}{{|N|}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right]$ $\\ {N_{unit}} = \\frac{1}{{8.7178}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0.6882\\\\ 0.6882\\\\ 0.2294\\\\ \\end{array}} \\right]$ It is a common mistake to believe that if R and S are unit vectors, the cross product will also be a unit vector. The vector product equation shows that this is only true when the angle between the two vectors is 90 degrees and therefore the sinus of the angle theta is 1. $\\\\ |T| = |R||S|sin\\theta$ Areas The vector product might be used to determine the area of a parallelogram or a triangle (with the vertices at P1 - P3). Image 17 shows the two vectors helping to form a parallelogram and a triangle. Figure 17 - Deriving the Area of a", "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", "formula is $\\mathbf{a} \\cdot \\mathbf{b} = \\begin{vmatrix} {\\mathbf{a}} \\end{vmatrix}\\begin{vmatrix} {\\mathbf{b}} \\end{vmatrix} \\cos{\\theta}$ So what you really have is $\\cos{\\theta} = \\cfrac{-1}{\\sqrt{3}}$ Therefore $$\\sin{\\theta} = \\sqrt{1 - \\cos^2{\\theta}} = \\sqrt{1 - \\frac{1}{3}} = \\sqrt{\\frac{2}{3}} = \\frac{\\sqrt{2}}{\\sqrt{3}}$$ Finally, the area of the triangle is: $$Area = \\frac{1}{2} (5 \\sqrt{3}) (15) \\frac{\\sqrt{2}}{\\sqrt{3}} = \\frac{75 \\sqrt{2}}{2}$$ We can just multiply $\\frac{\\sqrt{2}}{\\sqrt{2}}$ to the area, and then we get the answer you posted: $$Area = \\frac{75 \\sqrt{2}}{2} \\left(\\frac{\\sqrt{2}}{\\sqrt{2}}\\right) = \\frac{75}{\\sqrt{2}}$$ Since your vectors are in $\\mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value: $Area=(1/2) | a \\times b |$. Alternative solution $$a = |OA| = \\sqrt{1^2 + 5^2 + 7^2} = \\sqrt{75} = 3 \\sqrt{5}$$ $$b = |AB| = \\sqrt{(10-1)^2 + (10--5)^2 + (5--7)^2} = \\sqrt{450} = 15\\sqrt{2}$$ $$c = |BO| = \\sqrt{10^2+10^2+5^2} = 15.$$ Now you can calculate the semiperimeter $$s$$ which is just $$\\frac{1}{2} \\left(3 \\sqrt{5} + 15 \\sqrt2 + 15 \\right)$$, and use Heron's formula to find the area: $$A = \\sqrt{s(s-a)(s-b)(s-c)}$$ • That's a nice contribution but the problem explicitly said vectors are to be used.. and the solutions to the OP were already presented Sep 1 '19 at 13:39 • If the solutions using vectors were", "Find the Area of the Parallelogram Whose Diagonals Are 3 ^ I + 4 ^ J and ^ I + ^ J + ^ K - Mathematics Sum Find the area of the parallelogram whose diagonals are $3 \\hat{ i } + 4 \\hat{ j } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Solution $\\text{ Let: }$ $\\vec{a} = 3 \\hat{ i } + 4 \\hat{ j } + 0 \\hat{ k }$ $\\vec{b} = \\hat{ i } + \\hat{ j } + \\hat{ k }$ $\\therefore \\vec{a} \\times \\vec{b} = \\begin{vmatrix}\\hat{ i } & \\hat{ j } & \\hat{ k } \\\\ 3 & 4 & 0 \\\\ 1 & 1 & 1\\end{vmatrix}$ $= \\left( 4 - 0 \\right) \\hat{ i } - \\left( 3 - 0 \\right) \\hat{ j } + \\left( 3 - 4 \\right) \\hat{ k}$ $= 4 \\hat{ i } - 3 \\hat{ j } - \\hat{ k }$ $\\Rightarrow \\left| \\vec{a} \\times \\vec{b} \\right| = \\sqrt{4^2 + \\left( - 3 \\right)^2 + \\left( - 1 \\right)^2}$ $= \\sqrt{26}$ $\\text{ Area of the parallelogram } =\\frac{1}{2}\\left| \\vec{a} \\times \\vec{b} \\right|$ $= \\frac{\\sqrt{26}}{2} \\text{ sq. units }$ Is there an error in this question or solution? APPEARS IN RD Sharma Class 12 Maths Chapter 25 Vector or", "is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[6] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V = \\begin{bmatrix} a_1 & a_2 & \\dots", "# Find the Area of the Parallelogram Whose Diagonals Are 2 ^ I + 3 ^ J + 6 ^ K and 3 ^ I − 6 ^ J + 2 ^ K - Mathematics Sum Find the area of the parallelogram whose diagonals are $2 \\hat{ i } + 3 \\hat{ j } + 6 \\hat{ k } \\text{ and } 3 \\hat{ i } - 6 \\hat{ j } + 2 \\hat{ k }$ #### Solution $\\text{ Let } :$ $\\vec{a} = 2 \\hat{ i } + 3 \\hat{ j } + 6 \\hat{ k }$ $\\vec{b} = 3 \\hat{ i } - 6 \\hat{ j } + 2 \\hat{ k }$ $\\therefore \\vec{a} \\times \\vec{b} = \\begin{vmatrix}\\hat{ i } & \\hat{ j } & \\hat{ k } \\\\ 2 & 3 & 6 \\\\ 3 & - 6 & 2\\end{vmatrix}$ $= \\left( 6 + 36 \\right) \\hat{ i } - \\left( 4 - 18 \\right) \\hat[{ j } + \\left( - 12 - 9 \\right) \\hat{ k }$ $= 42 \\hat{ i } + 14 \\hat{ j } - 21 \\hat{ k }$ $\\Rightarrow \\left| \\vec{a} \\times \\vec{b} \\right| = \\sqrt{{42}^2 + {14}^2 + \\left( - 21 \\right)^2}$ $= \\sqrt{2401}$ $= 49$ $\\text{ Area of the parallelogram } =\\frac{1}{2}\\left| \\vec{a} \\times \\vec{b} \\right|$ $= \\frac{49}{2}", "# Find the area of the parallelogram. Hit this in my book. Can someone help me out? Let \\begin{align} u &= i - j \\\\ v &= 2i - 3j + k. \\end{align} Find the area of the parallelogram defined by $u$ and $v$. Am I right to take a 3rd point $v \\times u$? - You are on the right track. The magnitude of the cross-product gives the area. Do you know how to compute $v \\times u$? – Sammy Black May 12 '13 at 8:05 It's best to think about $v \\times u$ as a vector (well, a pseudovector) rather than a point. – Sammy Black May 12 '13 at 8:07 The area of a parallelogram =$|\\vec u \\times\\vec v |$ where $\\vec u$ and $\\vec v$ are vectors of conjugate sides of parallelogram. $\\vec u=i-j$ and $\\vec v=2i-3j+k$ $$\\vec u \\times \\vec v= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\1 & -1 & 0 \\\\2 & -3 & 1 \\end{vmatrix}$$ $$\\vec u \\times \\vec v=\\mathbf{i}(-1-0)-\\mathbf{j}(1-0)+\\mathbf{k}(-3+2)$$ $$\\vec u \\times \\vec v=\\mathbf{-i}-\\mathbf{j}\\mathbf{-k}$$ $$|\\vec u \\times\\vec v |=\\sqrt3$$ area of parallelogram is $\\sqrt 3$ here $\\mathbf{i}$,$\\mathbf{j}$,$\\mathbf{k}$ are unit vectors." ]

[ "as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The formula for the area of a parallelogram is base x height. Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [closed], Soft question on the area of a parallelogram using vectors. Area of the Parallelogram: A parallelogram is a geometric structure having one side having opposite sides parallel to each other. To do this, take one of the top vertices, and find a vector that is orthogonal to it and find the point at which that vector (the orthogonal one) intersects the other vector. #vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj# First of all, to simplify the task, let's move it to a position when its vertex A coincides with the origin of coordinates. So this's a B C on G E. Now the area of", "+0 # Area of Parallelogram with two adjacent 2d vectors. 0 67 1 Find the area of the parallelogram with u and v as adjacent edges. u = 5i -2j v = 6i -2j Jun 11, 2020 #1 +8340 0 The area is equal to $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right|$$. Notice that $$\\mathbf{u}\\times \\mathbf{v} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\6&-2&0\\end{bmatrix} = \\det\\begin{bmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\5&-2&0\\\\1&0&0\\end{bmatrix} = 2\\mathbf{k}$$ Therefore $$\\left|\\mathbf{u} \\times \\mathbf{v}\\right| = |2\\mathbf{k}| = 2$$ Therefore the area of the required parallelogram is 2 sq. units. Jun 11, 2020", "one) intersects the other vector. Then find the magnitude of a vector that would be the height of the parallelogram. To determine the area of our parallelogram, we need the length of its base #|AB|# and the altitude #|DH|# from vertex #D# to point #H# on side #AB# (that is, #DH_|_AB#). #S = |AB|*|DH|=|U_B*V_D-V_B*U_D|#, In terms of original coordinates, it looks like this: Protection against an aboleths enslave ability. #C[U_C=x_C-x_A,V_C=y_C-y_A]# #|AD|=sqrt(U_D^2+V_D^2)# A rectangle is 12cm longer than its wide. {\\displaystyle 15in^ {2}}, since. We note that the area of a triangle defined by two vectors $\\vec{u}, \\vec{v} \\in \\mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. 1. The parallelograms are differentiated by the size of the adjacent sides as well as by the angles, however, the opposite sides are equal. #sin^2(/_BAD)=1-cos^2(/_BAD)=# The length #AD# is the length of vector #q#: Is there a bias against mentioning your name on presentation slides? So in your case we have to write the points in $\\mathbb{R}^2$ as vectors in $\\mathbb{R}^3$ and apply the formula: $\\vec{AB} = \\begin{pmatrix}8\\\\4\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix}$, $\\vec{AD} = \\begin{pmatrix}13\\\\8\\\\0\\end{pmatrix} -\\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} =\\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix}$, $A_\\text{parallelogram}= \\left|\\vec{AB}\\times\\vec{AD}\\right| = \\left| \\begin{pmatrix}4\\\\2\\\\0\\end{pmatrix} \\times \\begin{pmatrix}9\\\\6\\\\0\\end{pmatrix} \\right| = \\left|\\begin{pmatrix}0\\\\0\\\\6\\end{pmatrix} \\right| = 6$, You might have noticed that this simplifies to, $$A_\\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ Thus we have the same", "+0 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) +1 136 1 Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a) part (a): Let $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$find a*b and the answer was 10 Apr 2, 2022 #1 +30 +2 $\\sin^2 θ = 1 - \\cos^2 θ = 1 - \\frac{4}{13} = \\frac{9}{13}$ every angle in a parallelogram is between 0 and π radians, so θ must be positive, so $\\sin θ = \\sqrt{\\frac{9}{13}} = \\frac{3}{\\sqrt{13}}$ area of parallelogram is $\\|{a}\\| \\|{b}\\| \\sin \\theta = \\sqrt{65} \\cdot \\sqrt{5} \\cdot \\frac{3}{\\sqrt{13}} = \\boxed{15}$ Jun 25, 2022", "back and think about what properties a mathematical object, describing area should have. Suppose we have two (possibly infinitesimal) vectors, forming a two-dimensional parallelogram and we need the mathematical object that tells us the area of this parallelogram: \\mathrm{area}(V,W) = \\text{ area of the parallelogram formed by V and W}. An obious property is that the number $\\mathrm{area}(V,W)$ ought to double if we double the length of one vector: \\mathrm{area}(2V,W) = 2 \\times \\text{ area of the parallelogram formed by V and W} \\mathrm{area}(V,W+Z) = \\mathrm{area}(V,W) + \\mathrm{area}(V,Z) A pictorial proof can be seen in the following figure: Together these properties are what we called linearity above, and we see that this fits perfectly with the defintion of a $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. Another important propery is that $\\mathrm{area}(V,W)$ must vanish if $V$ and $W$ are parallel, i.e. if $W=aV$ for some number $a$, we have \\mathrm{area}(V,W) =\\mathrm{area}(V,aV) \\stackrel{!}{=} 0. .Because we have $\\mathrm{area}(V,aV)=a \\times \\mathrm{area}(V,V) \\stackrel{!}{=} 0$, we start with $\\mathrm{area}(V,V) \\stackrel{!}{=} 0$ in the next step to see that $\\mathrm{area}(V,W)$ must be an antisymmetric $\\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix}$ tensor. The proof is simple: From \\mathrm{area}(V,V) = 0, if we write $V= U+W$, it follows \\mathrm{area}(U+W,U+W) = 0. Using the linearity yields \\mathrm{area}(U,U) + \\mathrm{area}(U,W) + \\mathrm{area}(W,U) + \\mathrm{area}(W,W) = 0. Using now $\\mathrm{area}(V,V)", "+ 4-1k$ $\\\\ R \\times S = \\begin{vmatrix} 2 & 0 \\\\ 1 & 3 \\end{vmatrix} i - \\begin{vmatrix} -2 & 0 \\\\ -2 & 3\\end{vmatrix}j + \\begin{vmatrix} -2 & 2 \\\\ -2 & 1 \\end{vmatrix}k$ $\\\\ R \\times S = (2 * 3 - 0 * 1)i - (-2 * 3 - 0 * -2)j + (-2 * 1 - 2 * -2)k$ $\\\\ N = 6i + 6j + 2k$ $|N| = \\sqrt {{6^2} + {6^2} + {2^2}}$ $|N| = \\sqrt {{76}} = 8.7178$ $\\ {N_{unit}} = \\frac{1}{{|N|}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right]$ $\\ {N_{unit}} = \\frac{1}{{8.7178}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0.6882\\\\ 0.6882\\\\ 0.2294\\\\ \\end{array}} \\right]$ It is a common mistake to believe that if R and S are unit vectors, the cross product will also be a unit vector. The vector product equation shows that this is only true when the angle between the two vectors is 90 degrees and therefore the sinus of the angle theta is 1. $\\\\ |T| = |R||S|sin\\theta$ Areas The vector product might be used to determine the area of a parallelogram or a triangle (with the vertices at P1 - P3). Image 17 shows the two vectors helping to form a parallelogram and a triangle. Figure 17 - Deriving the Area of a", "+0 Area of a parallelogram with vector vertices 0 78 1 Find the area of the parallelogram whose vertices are $$0$$, $$a$$, $$b$$, and $$a + b$$, where $$a$$ and $$b$$ are the vectors $${a} = \\begin{pmatrix} \\phantom -2 \\\\ -6 \\\\ \\phantom -5 \\end{pmatrix} \\quad \\text{and} \\quad {b} = \\begin{pmatrix} -1 \\\\ -2 \\\\ \\phantom -0 \\end{pmatrix}.$$ Apr 9, 2021", ", 6 )$ $u = AB = \\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 6 \\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1 } \\\\ u _ { 2 } & v _ { 2 } \\end{bmatrix} = det \\begin{bmatrix} 5 & 6 \\\\ 2 & 4 \\end{bmatrix}= 20 \\: – \\: 12 = 8$ The area of the parallelogram is $8$. ## Numerical Result The area of the parallelogram is $8$. ## Example Find area of the parallelogram whose vertices are given. $( 0 , 0 )$, $( 5 , 2 )$, $( 6 , 4 )$ , $( 11 , 6 )$ Solution A parallelogram can be described by $4$ vertices or $2$ vectors. Since we have $4$ vertices $( ABCD )$, we find the vectors $u$, $v$ that describe the parallelogram. $A = ( 0 , 0 )$ $B = ( 6 , 8 )$ $C = ( 5 , 4 )$ $D = ( 11 , 6 )$ $u = AB = \\begin{bmatrix} 6\\\\ 8 \\end{bmatrix}$ $v = AC = \\begin{bmatrix} 5\\\\ 4 \\end{bmatrix}$ Area of parallelogram is the absolute value of the determinant. $\\begin{bmatrix} u _ { 1 } & v _ { 1", "$\\\\ R \\times S = \\begin{vmatrix} 2 & 0 \\\\ 1 & 3 \\end{vmatrix} i - \\begin{vmatrix} -2 & 0 \\\\ -2 & 3\\end{vmatrix}j + \\begin{vmatrix} -2 & 2 \\\\ -2 & 1 \\end{vmatrix}k$ $\\\\ R \\times S = (2 * 3 - 0 * 1)i - (-2 * 3 - 0 * -2)j + (-2 * 1 - 2 * -2)k$ $\\\\ N = 6i + 6j + 2k$ $|N| = \\sqrt {{6^2} + {6^2} + {2^2}}$ $|N| = \\sqrt {{76}} = 8.7178$ $\\ {N_{unit}} = \\frac{1}{{|N|}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right]$ $\\ {N_{unit}} = \\frac{1}{{8.7178}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0.6882\\\\ 0.6882\\\\ 0.2294\\\\ \\end{array}} \\right]$ It is a common mistake to believe that if R and S are unit vectors, the cross product will also be a unit vector. The vector product equation shows that this is only true when the angle between the two vectors is 90 degrees and therefore the sinus of the angle theta is 1. $\\\\ |T| = |R||S|sin\\theta$ Areas The vector product might be used to determine the area of a parallelogram or a triangle (with the vertices at P1 - P3). Image 17 shows the two vectors helping to form a parallelogram and a triangle. Figure 17 - Deriving the Area of a Parallelogramm /", "treat the top numbers and bottom numbers completely separately. \\begin{aligned} 2\\begin{pmatrix} 3\\\\\\text{-}1 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} &= \\begin{pmatrix} 6\\\\\\text{-}2 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} \\\\ &=\\begin{pmatrix} 6-7\\\\\\text{-}2-1 \\end{pmatrix} \\\\ &= \\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}} \\end{aligned} This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down: We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down. The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour. This is the vector: $\\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}}$ ### Question 4: The Area of a Triangle The area of a triangle can be found my multiplying the base by the height and dividing by 2. However, this requires us to know the perpendicular height of the triangle. If we don’t know this measurement, we can use a slightly different formula to find the area. Consider the triangle below: Any triangle can be labelled like this. The lowercase letters denote the sides and the uppercase letters denote the angles. Each angle is", "Find the Area of the Parallelogram Whose Diagonals Are 3 ^ I + 4 ^ J and ^ I + ^ J + ^ K - Mathematics Sum Find the area of the parallelogram whose diagonals are $3 \\hat{ i } + 4 \\hat{ j } \\text{ and } \\hat{ i } + \\hat{ j } + \\hat{ k }$ Solution $\\text{ Let: }$ $\\vec{a} = 3 \\hat{ i } + 4 \\hat{ j } + 0 \\hat{ k }$ $\\vec{b} = \\hat{ i } + \\hat{ j } + \\hat{ k }$ $\\therefore \\vec{a} \\times \\vec{b} = \\begin{vmatrix}\\hat{ i } & \\hat{ j } & \\hat{ k } \\\\ 3 & 4 & 0 \\\\ 1 & 1 & 1\\end{vmatrix}$ $= \\left( 4 - 0 \\right) \\hat{ i } - \\left( 3 - 0 \\right) \\hat{ j } + \\left( 3 - 4 \\right) \\hat{ k}$ $= 4 \\hat{ i } - 3 \\hat{ j } - \\hat{ k }$ $\\Rightarrow \\left| \\vec{a} \\times \\vec{b} \\right| = \\sqrt{4^2 + \\left( - 3 \\right)^2 + \\left( - 1 \\right)^2}$ $= \\sqrt{26}$ $\\text{ Area of the parallelogram } =\\frac{1}{2}\\left| \\vec{a} \\times \\vec{b} \\right|$ $= \\frac{\\sqrt{26}}{2} \\text{ sq. units }$ Is there an error in this question or solution? APPEARS IN RD Sharma Class 12 Maths Chapter 25 Vector or", "of the parallelogram with edges ${v=(3,2)}$ and ${w=(1,4)}$. • Find the area of the triangle with sides ${v,w,}$and ${v+w}$.Draw it. • Find the area of the triangle with sides ${v,w,}$and ${w-v}$.Draw it. Solution: • $\\displaystyle \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=10.$ • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture omitted. • ${\\frac{1}{2} \\begin{vmatrix} 3&2\\\\ 1&4 \\end{vmatrix}=5. }$Picture Omitted. $\\Box$ Exercise 4.4.28 A box has edges from ${(0,0,0)}$ to ${(3,1,1),(1,3,1),}$and ${(1,1,3)}$.Find its volumn and also find the area of each parallelogram face. Solution: The volumn of the box is $\\displaystyle \\begin{vmatrix} 3&1&1\\\\ 1&3&1\\\\ 1&1&3 \\end{vmatrix}=20.$ The area of each parallelogram face is $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 3&1 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 1&3\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2},$ $\\displaystyle \\sqrt{ \\begin{vmatrix} 3&1\\\\ 1&1 \\end{vmatrix}^2+ \\begin{vmatrix} 3&1\\\\ 1&3 \\end{vmatrix}^2+ \\begin{vmatrix} 1&1\\\\ 1&3 \\end{vmatrix}^2 }=6 \\sqrt{2}.$ In order to compute the area of each parallelogram face,we can also compute $\\displaystyle \\sqrt{\\det A^TA}=\\sqrt{\\det \\left(\\begin{bmatrix} 3&1&1\\\\ 1&3&1 \\end{bmatrix} \\begin{bmatrix} 3&1\\\\ 1&3\\\\ 1&1 \\end{bmatrix}\\right)}= 6 \\sqrt{2}$ etc.$\\Box$ Exercise 4.4.35 An ${n}$– dimensional cube has how many corners?How many edges?How many ${n-1}$-dimensional faces?The ${n}$– cube whose edges are the rows of ${2I}$ has volumn ${\\underline{~~~~~~~~~~}}$.A hypercube computer has parallel processors at the corners with connections along the edeges. Solution: An ${n}$-dimensional cube", "the area of the paralellogram and the original square. Since the original square has area $1$, the area of the parallelogram is the determinant of $\\bc{\\A}$. Working this out requires only a small amount of basic geometry. Here’s the simplest way to do it. We first name the four elements of our matrix: $\\bc{\\A} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ We can now name the four corners of our parallelogram in terms of these four scalars, by multiplying the corner points of the unit square by $\\bc{\\A}$: $\\begin{matrix} \\times & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}1\\\\ 0\\end{pmatrix} & \\begin{pmatrix}0\\\\ 1\\end{pmatrix} & \\begin{pmatrix}1\\\\ 1\\end{pmatrix} \\\\ \\bc{\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}} & \\begin{pmatrix}0\\\\ 0\\end{pmatrix} & \\begin{pmatrix}a\\\\ c\\end{pmatrix} & \\begin{pmatrix}b\\\\ d\\end{pmatrix} & \\begin{pmatrix}a+b\\\\ c+d\\end{pmatrix} \\\\ \\end{matrix}$ This gives us the following picture. Here we can see the area of the paralellogram clearly: there is one large rectangle of area $\\oc{ad}$. To get from this area to the area of our parallelogram, we should subtract the area of the green triangle in the bottom, which is part of the rectangle but not the parallelogram. But then, there’s a green triangle at the top, with the same size, which is (mostly) part of the parallelogram but not of the rectangle, so these cancel each other out. We", "the area of the blue region, which is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[5] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V", "## Calculus (3rd Edition) $\\sqrt {35}$ The area of the parallelogram is calculated as: $||\\textbf{u}\\times\\textbf{v}||$ We take the cross product: $\\textbf{u}\\times\\textbf{v}=\\begin{vmatrix}\\textbf{i}&\\textbf{j}&\\textbf{k}\\\\1&0&3\\\\2&1&1\\end{vmatrix}$ $=\\textbf{i}(0\\times1-3\\times1)-\\textbf{j}(1\\times1-3\\times2)+\\textbf{k}(1\\times1-0\\times2)$ $=-3\\textbf{i}+5\\textbf{j}+\\textbf{k}$ $||\\textbf{u}\\times\\textbf{v}||=\\sqrt {(-3)^{2}+5^{2}+1^{2}}=\\sqrt {35}$", "is the interior of the parallelogram • The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is $A_\\text{rect} = (B+A) \\times H\\,$ and the area of a single orange triangle is $A_\\text{tri} = \\frac{1}{2} A \\times H. \\,$ Therefore, the area of the parallelogram is \\begin{align} K &= A_\\text{rect} - 2 \\times A_\\text{tri} \\\\ &= \\left( (B+A) \\times H \\right) - \\left( A \\times H \\right) \\\\ &= B \\times H \\\\ \\end{align} • Another area formula, for two sides B and C and angle θ, is $K = B \\cdot C \\cdot \\sin \\theta.\\,$ • The area of a parallelogram with sides B and C (BC) and angle $\\gamma$ at the intersection of the diagonals is given by[6] $K = \\frac{|\\tan \\gamma|}{2} \\cdot \\left| B^2 - C^2 \\right|.$ ### The area on coordinate system Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^2$ and let $V = \\begin{bmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{bmatrix} \\in\\R^{2 \\times 2}$ denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to $|\\det(V)| = |a_1b_2 - a_2b_1|\\,$. Let vectors $\\mathbf{a},\\mathbf{b}\\in\\R^n$ and let $V = \\begin{bmatrix} a_1 & a_2 & \\dots", "# Find the Area of the Parallelogram Whose Diagonals Are 2 ^ I + 3 ^ J + 6 ^ K and 3 ^ I − 6 ^ J + 2 ^ K - Mathematics Sum Find the area of the parallelogram whose diagonals are $2 \\hat{ i } + 3 \\hat{ j } + 6 \\hat{ k } \\text{ and } 3 \\hat{ i } - 6 \\hat{ j } + 2 \\hat{ k }$ #### Solution $\\text{ Let } :$ $\\vec{a} = 2 \\hat{ i } + 3 \\hat{ j } + 6 \\hat{ k }$ $\\vec{b} = 3 \\hat{ i } - 6 \\hat{ j } + 2 \\hat{ k }$ $\\therefore \\vec{a} \\times \\vec{b} = \\begin{vmatrix}\\hat{ i } & \\hat{ j } & \\hat{ k } \\\\ 2 & 3 & 6 \\\\ 3 & - 6 & 2\\end{vmatrix}$ $= \\left( 6 + 36 \\right) \\hat{ i } - \\left( 4 - 18 \\right) \\hat[{ j } + \\left( - 12 - 9 \\right) \\hat{ k }$ $= 42 \\hat{ i } + 14 \\hat{ j } - 21 \\hat{ k }$ $\\Rightarrow \\left| \\vec{a} \\times \\vec{b} \\right| = \\sqrt{{42}^2 + {14}^2 + \\left( - 21 \\right)^2}$ $= \\sqrt{2401}$ $= 49$ $\\text{ Area of the parallelogram } =\\frac{1}{2}\\left| \\vec{a} \\times \\vec{b} \\right|$ $= \\frac{49}{2}", "(7), we obtain $$MN \\;|| \\;AD$$ and $$AM\\; ||\\; DN$$ Therefore, quadrilateral $$AMND$$ is a parallelogram. It can be observed that $$\\Delta APD$$ and parallelogram are lying on the same base and between the same parallel lines and . \\begin{align}\\therefore \\text{Area }\\!(\\Delta {APD})\\!=\\!\\frac{1}{2} \\text {Area} (AMND) \\ldots(8)\\end{align} Similarly, for $$\\Delta PCB$$ and parallelogram , \\begin{align}\\therefore \\rm{Area}\\,(\\Delta {PCB})\\!=\\!\\frac{1}{2}\\text { Area }({MNCB})...(9)\\end{align} Adding Equations () and (), we obtain \\begin{align}\\begin{bmatrix}\\text{Area }\\left(\\Delta {APD}\\right)+\\\\\\text { Area }(\\Delta {PCB})\\end{bmatrix}= \\frac{1}{2}\\begin{bmatrix} \\text { Area }({AMND})+\\\\\\text { Area }({MNCB})\\end{bmatrix}\\end{align} \\begin{align}\\begin{bmatrix}\\text{Area }\\left(\\Delta {APD}\\right)+\\\\\\text { Area }(\\Delta {PCB})\\end{bmatrix}=\\frac{1}{2} \\text { Area }({ABCD})...(10) \\end{align} On comparing Equations () and (), we obtain \\begin{align}\\begin{bmatrix}\\rm{Area}\\,\\,(\\Delta {APD})+\\\\\\text { Area }(\\Delta {PBC})\\end{bmatrix}=\\begin{bmatrix}\\text { Area }(\\Delta {APB})+\\\\\\text { Area }(\\Delta {PCD})\\end{bmatrix}\\end{align} ## Chapter 9 Ex.9.2 Question 5 In the given figure, $$PQRS$$ and $$ABRS$$ are parallelograms and $$X$$ is any point on side $$BR$$. Show that (i) \\begin{align} \\text{ar }\\left( \\text{PQRS} \\right)\\text{ = ar }\\left( \\text{ABRS} \\right)\\end{align} (ii)\\begin{align} ar \\left( \\Delta\\!\\!\\text{ }AXS \\right)= \\frac{\\text{1}}{2}ar \\left( PQRS \\right)\\end{align} ### Solution What is known? $$PQRS$$ and $$ABRS$$ are parallelograms and X is any point on side $$BR$$. What is unknown? How we can show that $$\\text{ar }\\left( \\text{PQRS} \\right)\\text{ = ar}\\left( \\text{ABRS} \\right)$$ and $$\\text{ar }\\left( \\text{ }\\!\\!\\Delta\\!\\!\\text{ }A\\text{XS} \\right)\\text{ = }\\frac{\\text{1}}{2}\\text{ ar}\\left( \\text{PQRS} \\right)$$ Reasoning: If a triangle and parallelogram are on same base", ".,\\, u,\\,.\\mid\\quad(A_2)\\quad+}$ $\\mathbf {\\qquad \\mid.,\\, u,\\, .,\\, t,\\, .\\mid\\quad(A_3)\\quad+}$ $\\mathbf {\\qquad \\mid.,\\, t,\\, .,\\, t,\\, .\\mid\\quad(A_4)}$ $\\mathbf {A_1}$ and$\\mathbf {A_4}$ are null because they have equal columns. Thus $\\mathbf {A_2 = -A_3}$, so that the determinant cancels out. The only difference between $\\mathbf {A_2}$ and $\\mathbf {A_3}$ is the swap of 2 columns. $\\bigstar \\bigstar \\bigstar$ We define above the fundamental determinant properties, based on the geometric paradigm. Now we are ready to prove the Laplace Expansion, again using the finite induction. Imagine a matrix $M$ with two dimensions $\\mathbf {[\\vec{v}, \\vec{w}]}$, Its determinant defines a parallelogram. The vector $\\mathbf {\\vec{v} = [v_1 v_2]}$ has a normal given by $\\mathbf {[-v_2, v_1]}$, which is the basis of the parallelogram formed by$\\mathbf {\\vec{v}}$ and $\\mathbf {\\vec{w}}$ in the direction of the parelogram. Thus the scalar product $\\mathbf {[-v_2, v_1]\\bullet [w_1 w_2] = v_1w_2 - v_2w_1}$, which corresponds to the projection of the lateral of the parelogram in the height multiplied by the base of the parallelogram, gives the parallelogram area. $\\mathbf {\\begin{bmatrix} v_1&w_1\\\\v_2&w_2 \\end{bmatrix} = \\begin{bmatrix} v_1&w_1\\\\0&w_2 \\end{bmatrix}+\\begin{bmatrix} 0&w_1\\\\v_2&w_2 \\end{bmatrix}}$ It is clear that the same can be found by the Laplace Expansion with $\\mathbf {v_1 w_2}$and $\\mathbf {v_2 w_1}$, except that in the second case there was one line change (2 with 1), according to Theorem 6, by", "S = \\begin{vmatrix} 2 & 0 \\\\ 1 & 3 \\end{vmatrix} i - \\begin{vmatrix} -2 & 0 \\\\ -2 & 3\\end{vmatrix}j + \\begin{vmatrix} -2 & 2 \\\\ -2 & 1 \\end{vmatrix}k$ $\\\\ R \\times S = (2 * 3 - 0 * 1)i - (-2 * 3 - 0 * -2)j + (-2 * 1 - 2 * -2)k$ $\\\\ N = 6i + 6j + 2k$ $|N| = \\sqrt {{6^2} + {6^2} + {2^2}}$ $|N| = \\sqrt {{76}} = 8.7178$ $\\ {N_{unit}} = \\frac{1}{{|N|}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right]$ $\\ {N_{unit}} = \\frac{1}{{8.7178}}\\left[ {\\begin{array}{*{20}{c}} 6\\\\ 6\\\\ 2\\\\ \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0.6882\\\\ 0.6882\\\\ 0.2294\\\\ \\end{array}} \\right]$ It is a common mistake to believe that if R and S are unit vectors, the cross product will also be a unit vector. The vector product equation shows that this is only true when the angle between the two vectors is 90 degrees and therefore the sinus of the angle theta is 1. $\\\\ |T| = |R||S|sin\\theta$ Areas The vector product might be used to determine the area of a parallelogram or a triangle (with the vertices at P1 – P3). Image 17 shows the two vectors helping to form a parallelogram and a triangle. Figure 17 – Deriving the Area of a Parallelogramm / Triangle with the" ]

[ "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "prove that $$\\det(B)\\ge\\det(A)$$ (so that $$\\det(B)>0$$). Let $$A=\\pmatrix{p&s_1a&s_2b\\\\ s_1a&q&s_3c\\\\ s_2b&s_3c&r} \\ \\text{ and }\\ B=\\pmatrix{p&a&b\\\\ a&q&c\\\\ b&c&r}$$ where $$p,q,r>0,\\,a,b,c\\ge0$$ and $$s_1,s_2,s_3\\in\\{1,-1\\}$$. Then \\begin{aligned} \\det(A) &=pqr+2s_1s_2s_3abc-(pc^2+qb^2+ra^2)\\\\ &\\le pqr+2abc-(pc^2+qb^2+ra^2)\\\\ &=\\det(B). \\end{aligned}", "# Does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? Let $k=\\overline{k}$. Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\\det(g),\\det(g)^{-1}]$. Now for $B=\\left\\{ \\left( \\begin{array}{cc} b_{11} & b_{12} \\\\ 0 & b_{22} \\\\ \\end{array}\\right) : b_{ii}\\not=0, b_{12}\\in k \\right\\}$, let $B'=\\{ b\\in B: \\det(B)=1 \\}$. Suppose $B$ acts on $B$ by left multiplication. Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? -", "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "all $\\mathbf {N}$ lines are multiplied by $\\mathbf {\\beta}$, we have : $\\mathbf {Det (A * B) = Det (A * \\beta I) = \\beta^N Det (A * I) = \\beta^ N Det (A)}$ $\\mathbf {Det (A) * Det (B) = Det (A)\\beta^N Det (I) = \\beta^N Det (A)}$ **c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\\mathbf {s}$ of small cube matrices $\\mathbf {B_k}$ ($\\mathbf {k}$ is an index from 1 to s) of dimension $\\mathbf {\\beta}$ (diagonal matrix with only $\\mathbf {\\beta}$value), where $\\mathbf {\\beta}$ is arbitrarily small and therefore $\\mathbf {s}$ arbitrarily large. Look at the volume below the colorful surface, above the plane $\\mathbf {z=0}$ (vertical axis), below the plane $\\mathbf {y=0}$ (negative y) to down and left, right to the place $\\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\\mathbf {s}$) of tiny cubes. $\\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \\beta^N \\qquad}$ (c1) $\\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)}$ By (b) above $\\mathbf {= \\beta^N Det(A) +", "# What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$? Let $A,B\\in M_2(\\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\\det A= \\det B$. Our goal is to compute $\\det(A^2 – B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true: 1. $\\det(A^2+B^2)+\\det(A^2-B^2)=2(\\det A^2+\\det B^2)$. (Is this well known?) 2. (1) $\\implies \\det(A^2-B^2)=0$. 3. If $A,B\\in M_2(\\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$. 4. $(A+B)^2=0 \\implies \\det(A^2-B^2)=0$. Can someone help me with justifying these statements? Edit: Doug M provided an explanation for (1) in the answers. Here is an explanation for (2): $A^2+B^2+2AB=O_2 \\implies A^2+B^2=-2AB$. So $\\det(A^2+B^2)=4\\det(AB)$. Now using (1), $\\det(A^2-B^2)= 2(\\det(A^2)-2\\det(AB)+\\det(B^2)) = 2((\\det(A)^2-\\det(B)^2) = 0$. #### Solutions Collecting From Web of \"What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$?\" Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\\det(A)=\\det(B)$. In fact, the interesting result is Proposition. Let $A,B\\in M_2(\\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$. Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\\square$ Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\\det(A+B)=0$, $\\det(A^2-B^2)=0$;", "the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . • Dec 18th 2011, 02:17 AM NonCommAlg Re: Find mod ( det A ) Quote: Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ • Dec 18th 2011, 05:28 AM FernandoRevilla Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "Math Help - Find determinant 1. Find determinant $det(2A^8 B^{-1})$ I tried to break it up but I'm not sure where the 2 goes. $det(2A^8) det(B^{-1})$ ? Any help is appreciated! 2. $ \\det(2A^{8}B^{-1})= 2^{3}\\det(A^{8})\\det(B^{-1})= 8\\frac{\\det(A)^{8}}{\\det(B)}= 8\\frac{ (-1)^{8}}{-2} = -4 $", "$1 = \\det(\\mathbf I_n) = \\det(\\mathbf{A} \\mathbf{A}^{-1}) = \\det(\\mathbf{A}) \\det(\\mathbf{A}^{-1}),$ and if det(A) is a unit then (*) above shows that $\\mathbf{A}^{-1} = \\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ ## Properties The adjugate has the properties $\\mathrm{adj}(\\mathbf{I}) = \\mathbf{I}\\,$ $\\mathrm{adj}(\\mathbf{AB}) = \\mathrm{adj}(\\mathbf{B})\\,\\mathrm{adj}(\\mathbf{A})\\,$ for all n×n matrices A and B. The adjugate preserves transposition: $\\mathrm{adj}(\\mathbf{A}^T) = \\mathrm{adj}(\\mathbf{A})^T\\,$. Furthermore $\\det\\big(\\mathrm{adj}(\\mathbf{A})\\big) = \\det(\\mathbf{A})^{n-1}.\\,$ If p(t) = det(A − tI) is the characteristic polynomial of A and we define the polynomial q(t) = (p(0) − p(t))/t, then $\\mathrm{adj}(\\mathbf{A}) = q(\\mathbf{A}) = -(p_1 \\mathbf{I} + p_2 \\mathbf{A} + p_3 \\mathbf{A}^2 + \\cdots + p_{n} \\mathbf{A}^{n-1}),$ where pj are the coefficients of p(t), $p(t) = p_0 + p_1 t + p_2 t^2 + \\cdots + p_{n} t^{n}.$ The adjugate also appears in the formula of the derivative of the determinant. ## References • Strang, Gilbert (1988). \"Section 4.4: Applications of determinants\". Linear Algebra and its Applications (3rd ed.). Harcourt Brace Jovanovich. pp. 231–232. ISBN 0-15-551005-3. Wikimedia Foundation. 2010. ### Look at other dictionaries: • Matrix - получить на Академике действующий промокод Pharmacosmetica или выгодно matrix купить со скидкой на распродаже в Pharmacosmetica • Matrix (mathematics) — Specific elements of a matrix are often denoted by a variable with two subscripts. For instance, a2,1 represents the element at the second row and first column of a matrix A. In", "# If A and B are 4 times 4 matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? det(A^T) = ? det(B^{-1}) = ? det(B^4) = ? If A and B are $4×4$ matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? $det\\left({A}^{T}\\right)=?$ $det\\left({B}^{-1}\\right)=?$ $det\\left({B}^{4}\\right)=?$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Corben Pittman Given that A and B are $4×4$ matrices: det(A)=1 : det(B)=4 Then (1) $det\\left(AB\\right)=det\\left(A\\right)\\cdot det\\left(B\\right)$ $=1\\cdot 4$ det(AB)=4 (2) $det\\left(3A\\right)=3\\cdot det\\left(A\\right)$ $=3\\cdot \\left(1\\right)$ det(3A)=3 (3) $det\\left({A}^{T}\\right)=det\\left(A\\right)$ $det\\left({A}^{T}\\right)=1$ (4)$det\\left({B}^{-1}\\right)=\\frac{1}{|B|}$ $det\\left({B}^{-1}\\right)=\\frac{1}{4}$ (5) $det\\left({B}^{4}\\right)=det\\left(B\\cdot B\\cdot B\\cdot B\\right)$ $=det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)$ $=4\\cdot 4\\cdot 4\\cdot 4$ =256 Jeffrey Jordon", "is enough to conclusively answer. For option $\\mathbf A$ we have $\\frac{1}{det(A)} = det(A^{-1}) {\\space\\overset{\\mathbf A}{=}\\space} det(\\frac{A}{25}) = \\frac{det(A)}{25^4}$. Similarly for the other answers. Here is a very simple although not as elegant answer: Whatever the inverse would be, multiplication must be $1$ for diagonal elements, in particular for the first element. We can call it $c_{11}$. This shields $\\frac{25}{x}$ where $x$ is the one between options that makes this element 1. So it's a or d. In any case, you need to calculate all elements for choosing between this two options.", "# Show that $\\det(A) = 0$ or $\\mathrm{tr}(A)=0$ Let $$A$$ and $$B$$ are two matrices in $$\\mathcal{M}_{2,2}(\\mathbb{R})$$ with $$B\\neq0$$ and $$AB=-BA$$. Show that either $$\\mathrm{tr}(A) = 0$$ or $$\\det(A)=0$$. I'm stuck since the order is 2. A few attempts are 1. From the hypothesis we imply that $$\\mathrm{tr}(AB) = 0$$, thus $$(AB)^2=(\\det AB)(AB)$$. Take $$\\det$$ of both sides, we get $$\\det AB = 0$$ or $$\\det AB = 1$$. which leads to no thing. 2. Suppose $$\\det A \\neq 0$$, then multiply both side by $$A^{-1}$$ to the left we have $$B=A^{-1}(-B)A$$. This implies that $$\\mathrm{tr}B= 0$$. But i'm still stuck afterward. FIRST PROOF : You have $$AB=-BA$$ so multiplying by $$A$$ on the left, you get $$A^2B = -ABA = BA^2$$. Now, by Cayley-Hamilton theorem, you know that $$A^2- \\mathrm{Tr}(A) A + \\mathrm{det}(A) I = 0 \\quad \\quad (1)$$ Multiplying by $$B$$ on the right, you get $$A^2B- \\mathrm{Tr}(A) AB + \\mathrm{det}(A) B = 0, \\quad \\text{ i.e.} \\quad BA^2+ \\mathrm{Tr}(A) BA + \\mathrm{det}(A) B = 0 \\quad \\quad (2)$$ And multiplying $$(1)$$ by $$B$$ on the left, you get $$BA^2- \\mathrm{Tr}(A) BA + \\mathrm{det}(A) B = 0 \\quad \\quad (3)$$ Now $$(2)-(3)$$ gives you $$\\mathrm{Tr}(A)BA=0$$ If $$\\mathrm{Tr}(A)=0$$, you are done. Otherwise, $$BA=0$$, so the equation $$(3)$$ gives you $$\\mathrm{det}(A)B=0$$, and because $$B \\neq 0$$, you deduce", "Yes. If you want to do it explicitly, just multiply both sides with the inverse metric to raise both indices. 0 Consider the following matrix $$\\mathbf{A} = \\begin{bmatrix} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33} \\end{bmatrix}$$ Then the $\\det(\\mathbf{A})$ can be written as \\begin{align} \\det(\\mathbf{A}) &= (-1)^0a_{11}(a_{22}a_{33}-a_{23}a_{32})+(-1)^1a_{12}(a_{21}a_{33}-a_{23}a_{31})\\\\ ... 0 Use $A=(A_{ij})$, for definition $$\\det(\\mathbf{A}) = \\varepsilon_{ijk} A_{1i}A_{2j}A_{3k}$$ Or $$3!\\det(\\mathbf{A}) = \\varepsilon_{lmn}\\varepsilon_{ijk} A_{il}A_{jm}A_{kn}$$ Top 50 recent answers are included", "b} \\cr } } \\right|$ $=$ (a + b + c) (a + b + c)2 $\\Rightarrow$ x $=$ $-$ 2(a + b + c) 4 ### JEE Main 2019 (Online) 11th January Evening Slot Let A and B be two invertible matrices of order 3 $\\times$ 3. If det(ABAT) = 8 and det(AB–1) = 8, then det (BA–1 BT) is equal to : A ${1 \\over 4}$ B 16 C ${1 \\over {16}}$ D 1 ## Explanation ${\\left| A \\right|^2}.\\left| B \\right| = 8$ and ${{\\left| A \\right|} \\over {\\left| B \\right|}} = 8 \\Rightarrow \\left| A \\right| = 4$ and $\\left| B \\right| = {1 \\over 2}$ $\\therefore$ det(BA$-$1. BT) $= {1 \\over 4} \\times {1 \\over 4} = {1 \\over {16}}$", "$\\forall s,t,\\ \\mathbf{Q}_1(s+t)=\\mathbf{Q}_1(s)\\mathbf{Q}_1(t)=\\mathbf{Q}_1(t)\\mathbf{Q_1}(s)$. (Note: I've already asked this question here but did not get any answer despite a bounty.) Edit Additional possible hint: I've found that for $n=2$ (maybe it's more general), the follow results stands, if $A=\\mathbf{Q}_2\\mathbf{Q}_1(t_1)-\\mathbf{Q}_1(t_2)^{-1}\\mathbf{Q}_2$ and $\\phi(x)=\\det(A-xI)$: $$\\phi'(0)^2=-\\phi(0)\\det(B)$$ where $B$ is the $A$ matrix if $\\mathbf{P}=\\mathbf{I}$... ! I have no clue where this comes from, maybe it's obvious for someone (maybe related to Lie algebras?). This would conclude the proof as $\\phi(0)=\\det(A(t_1,t_2))=0$ for the appropriate $t_1,t_2$, so that $0$ would be a double eigenvalue of $A$. • Sorry, this is sort of complicated so maybe I'm misreading it, but when $n=1$ it looks like $Q_2Q_1$ and $Q_1^{-1}Q_2$ both always have the same determinant but the kernel is almost never two dimensional. Can you tell me what I'm missing? Nov 2 '14 at 10:17 • @GabrielC.Drummond-Cole $\\forall n$, $\\det(Q1)=1$ and $\\det(Q2)=-1$ so $\\det(Q_2Q_1)=\\det(Q_1^{-1}Q2)=-1$ is always true. Nov 2 '14 at 21:18 • @GabrielC.Drummond-Cole [failed to edit the previous comment within 5min...] It's quite complicated that's true, thank you for reading! When $n=1$, $Q_1(t)=\\begin{bmatrix}\\cos(\\omega t)& \\sin(\\omega t)/\\omega \\\\ -\\omega \\sin(\\omega t) & \\cos(\\omega t)\\end{bmatrix}$ and $Q_2=\\operatorname{diag}(1,-1)$ and the computation yields, for this particular case, $Q_2\\,Q_1 - Q_1^{-1}\\,Q_2=0_{2}$, so the kernel is always of dimension 2. Nov 2 '14 at 21:24 • But in the setup you use", "$\\mod (\\det A)=n^{n/2}$ . 4. ## Re: Find mod ( det A ) Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ 5. ## Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "= -2ae - 3cd - 2bf + 2ce + 2bd + 3af = -3 det(V) = 3ae + 5cd + bf - ce - 3bd - 5af You can see that det(V) = -$\\frac{1}{2}$ det(P) - det(U). So we get det(V) = -5 + 3 = -2", "You can use $\\det(A \\cdot B)= \\det(A) \\cdot \\det(B)$. And consider that $$(\\mathbf v+9\\mathbf w, \\mathbf w+3\\mathbf u, \\mathbf u+3\\mathbf v) = (\\mathbf u, \\mathbf v, \\mathbf w) \\cdot \\begin{bmatrix}0&3&1\\\\ 1&0&3\\\\ 9&1&1\\end{bmatrix}$$ Ignore this you did edit ur post op so #### Petrus ##### Well-known member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ $= 81\\det(v, w, u) + 81\\det(v, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)+$ $81/det(w, w, u) + 81\\det(w, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)$ The next step im kinda unsure how to do it, I guess ima swap road or something and try get exemple $\\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column? #### Klaas van Aarsen ##### MHB Seeker Staff member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ You have a 9 instead of a 1. It should be: $\\det(1v+9w, w+3u, u+3v) =1\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ You have repeated the same mistake here. $= 81\\det(v, w, u) +", "+0 # Help with Matrices! 0 57 1 +115 Given that both $$\\det(\\mathbf{A})$$ and $$\\det(\\mathbf{A}^{-1})$$ are integers, find all possible values of $$\\det(\\mathbf{A})$$. Apr 4, 2020", "be of size $$1\\times1$$ : $$N = [0]$$ since $$N^0=[0]^0=1$$ and $$N^1=[0]$$. But from here I dont know how to continue. Thank you in advance. • One has $\\det (sE-A) = s^r \\det(A_{22})$ plus lower order terms. – daw Nov 19 '19 at 10:51 • Also $E$,A$have to be square. – daw Nov 19 '19 at 10:52 • Sorry, but could you clarify, which rule gets me$det(sE−A)=s^r det(A_{22}) + ...$I have trouble calculating it, since$ A_{22}$must not be the same size as the zero block in$E$. – bambuk Nov 20 '19 at 11:19 • The exercise only makes sense if the sizes of the blocks of$E$and$A\\$ are the same. – daw Nov 20 '19 at 12:34 • i dont see why, could you explain? – bambuk Nov 21 '19 at 10:31" ]

[ "= \\boxed{\\textbf{(B)}\\ 24}.$", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &\\ \\ \\ \\ + 5[6\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 2\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &= -6\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) - 2\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4) \\\\ &\\ \\ \\ \\ + 30\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 10\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)\\end{align} Antisymmetry The fact that $\\det$ is antisymmetric means that if you swap any two arguments of $\\det$, you'll negate the sign. Thus $$\\det(\\mathbf a_1, \\mathbf a_2) = -\\det(\\mathbf a_2, \\mathbf a_1)$$ Likewise \\begin{align}\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) &= -\\det(\\mathbf a_1, \\mathbf a_4, \\mathbf a_3, \\mathbf a_2) \\\\ &= \\det(\\mathbf a_4, \\mathbf a_1, \\mathbf a_3, \\mathbf a_2)\\end{align} So swapping an even number of columns doesn't change the sign, but swapping an odd number will produce a negative sign. One consequence of this antisymmetry is that if two columns are scalar multiples of each other, then the determinant must be zero. This is because after pulling the constant multiple out front the two columns will be the same. Then if you swap those two columns, the determinant will look the same -- but it'll have a minus sign out front. And thus the determinant will", "the sequence. $\\boxed{\\textbf{(B)}\\ 21}$.", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "C}}{3.21 \\mathrm{~\\mu F} }= \\boxed{28 \\,\\text{V}}$$", "B)}{P(B)} = \\frac{\\frac{1}{3}}{\\frac{19}{45}}=\\boxed{\\frac{15}{19} \\ \\mathbf{(D)}}$.", "x= \\boxed{\\mathbf{\\pm 4}}$$", "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "1\\} =$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} =$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} =$ #### Solution 1. $\\{0\\} \\cap \\varnothing = \\boxed{\\varnothing}$ 2. $\\{0, 1\\} \\cap \\{0, 1\\} = \\boxed{\\{0, 1\\}}$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} = \\boxed{\\{1\\}}$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} = \\boxed{\\{4, 6\\}}$", "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "+ b^d$$. Then $$S(b, d) = \\boxed{(2^{k-1}-1)(1 + b^d) + 1}$$. For the case where $$d = 3$$ and $$k = 2$$ such as when $$b = 21$$, the answer will be $$\\boxed{1002_b}$$ as above. Labels:", "\\boxed{x=1+18t, y=1-7t, z=1+t, t\\in\\mathbf{R}}. {/eq}", "\\dfrac{40}{42}$$ giving us the desired answer of $20+21=\\boxed{41}$.", "all $\\mathbf {N}$ lines are multiplied by $\\mathbf {\\beta}$, we have : $\\mathbf {Det (A * B) = Det (A * \\beta I) = \\beta^N Det (A * I) = \\beta^ N Det (A)}$ $\\mathbf {Det (A) * Det (B) = Det (A)\\beta^N Det (I) = \\beta^N Det (A)}$ **c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\\mathbf {s}$ of small cube matrices $\\mathbf {B_k}$ ($\\mathbf {k}$ is an index from 1 to s) of dimension $\\mathbf {\\beta}$ (diagonal matrix with only $\\mathbf {\\beta}$value), where $\\mathbf {\\beta}$ is arbitrarily small and therefore $\\mathbf {s}$ arbitrarily large. Look at the volume below the colorful surface, above the plane $\\mathbf {z=0}$ (vertical axis), below the plane $\\mathbf {y=0}$ (negative y) to down and left, right to the place $\\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\\mathbf {s}$) of tiny cubes. $\\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \\beta^N \\qquad}$ (c1) $\\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)}$ By (b) above $\\mathbf {= \\beta^N Det(A) +", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$", "# Does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? Let $k=\\overline{k}$. Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\\det(g),\\det(g)^{-1}]$. Now for $B=\\left\\{ \\left( \\begin{array}{cc} b_{11} & b_{12} \\\\ 0 & b_{22} \\\\ \\end{array}\\right) : b_{ii}\\not=0, b_{12}\\in k \\right\\}$, let $B'=\\{ b\\in B: \\det(B)=1 \\}$. Suppose $B$ acts on $B$ by left multiplication. Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? -", "= \\boxed{\\textbf{(C) }\\frac{85}{8}\\pi}$.", "$\\boxed{\\textbf{(D) }2\\le S<6}$. ~Baolan ~ pi_is_3.14" ]

[ "a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &\\ \\ \\ \\ + 5[6\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 2\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &= -6\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) - 2\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4) \\\\ &\\ \\ \\ \\ + 30\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 10\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)\\end{align} Antisymmetry The fact that $\\det$ is antisymmetric means that if you swap any two arguments of $\\det$, you'll negate the sign. Thus $$\\det(\\mathbf a_1, \\mathbf a_2) = -\\det(\\mathbf a_2, \\mathbf a_1)$$ Likewise \\begin{align}\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) &= -\\det(\\mathbf a_1, \\mathbf a_4, \\mathbf a_3, \\mathbf a_2) \\\\ &= \\det(\\mathbf a_4, \\mathbf a_1, \\mathbf a_3, \\mathbf a_2)\\end{align} So swapping an even number of columns doesn't change the sign, but swapping an odd number will produce a negative sign. One consequence of this antisymmetry is that if two columns are scalar multiples of each other, then the determinant must be zero. This is because after pulling the constant multiple out front the two columns will be the same. Then if you swap those two columns, the determinant will look the same -- but it'll have a minus sign out front. And thus the determinant will", "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "# Does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? Let $k=\\overline{k}$. Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\\det(g),\\det(g)^{-1}]$. Now for $B=\\left\\{ \\left( \\begin{array}{cc} b_{11} & b_{12} \\\\ 0 & b_{22} \\\\ \\end{array}\\right) : b_{ii}\\not=0, b_{12}\\in k \\right\\}$, let $B'=\\{ b\\in B: \\det(B)=1 \\}$. Suppose $B$ acts on $B$ by left multiplication. Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? -", "= \\boxed{\\textbf{(B)}\\ 24}.$", "all $\\mathbf {N}$ lines are multiplied by $\\mathbf {\\beta}$, we have : $\\mathbf {Det (A * B) = Det (A * \\beta I) = \\beta^N Det (A * I) = \\beta^ N Det (A)}$ $\\mathbf {Det (A) * Det (B) = Det (A)\\beta^N Det (I) = \\beta^N Det (A)}$ **c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\\mathbf {s}$ of small cube matrices $\\mathbf {B_k}$ ($\\mathbf {k}$ is an index from 1 to s) of dimension $\\mathbf {\\beta}$ (diagonal matrix with only $\\mathbf {\\beta}$value), where $\\mathbf {\\beta}$ is arbitrarily small and therefore $\\mathbf {s}$ arbitrarily large. Look at the volume below the colorful surface, above the plane $\\mathbf {z=0}$ (vertical axis), below the plane $\\mathbf {y=0}$ (negative y) to down and left, right to the place $\\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\\mathbf {s}$) of tiny cubes. $\\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \\beta^N \\qquad}$ (c1) $\\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)}$ By (b) above $\\mathbf {= \\beta^N Det(A) +", "prove that $$\\det(B)\\ge\\det(A)$$ (so that $$\\det(B)>0$$). Let $$A=\\pmatrix{p&s_1a&s_2b\\\\ s_1a&q&s_3c\\\\ s_2b&s_3c&r} \\ \\text{ and }\\ B=\\pmatrix{p&a&b\\\\ a&q&c\\\\ b&c&r}$$ where $$p,q,r>0,\\,a,b,c\\ge0$$ and $$s_1,s_2,s_3\\in\\{1,-1\\}$$. Then \\begin{aligned} \\det(A) &=pqr+2s_1s_2s_3abc-(pc^2+qb^2+ra^2)\\\\ &\\le pqr+2abc-(pc^2+qb^2+ra^2)\\\\ &=\\det(B). \\end{aligned}", "# What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$? Let $A,B\\in M_2(\\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\\det A= \\det B$. Our goal is to compute $\\det(A^2 – B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true: 1. $\\det(A^2+B^2)+\\det(A^2-B^2)=2(\\det A^2+\\det B^2)$. (Is this well known?) 2. (1) $\\implies \\det(A^2-B^2)=0$. 3. If $A,B\\in M_2(\\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$. 4. $(A+B)^2=0 \\implies \\det(A^2-B^2)=0$. Can someone help me with justifying these statements? Edit: Doug M provided an explanation for (1) in the answers. Here is an explanation for (2): $A^2+B^2+2AB=O_2 \\implies A^2+B^2=-2AB$. So $\\det(A^2+B^2)=4\\det(AB)$. Now using (1), $\\det(A^2-B^2)= 2(\\det(A^2)-2\\det(AB)+\\det(B^2)) = 2((\\det(A)^2-\\det(B)^2) = 0$. #### Solutions Collecting From Web of \"What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$?\" Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\\det(A)=\\det(B)$. In fact, the interesting result is Proposition. Let $A,B\\in M_2(\\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$. Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\\square$ Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\\det(A+B)=0$, $\\det(A^2-B^2)=0$;", "# If A and B are 4 times 4 matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? det(A^T) = ? det(B^{-1}) = ? det(B^4) = ? If A and B are $4×4$ matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? $det\\left({A}^{T}\\right)=?$ $det\\left({B}^{-1}\\right)=?$ $det\\left({B}^{4}\\right)=?$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Corben Pittman Given that A and B are $4×4$ matrices: det(A)=1 : det(B)=4 Then (1) $det\\left(AB\\right)=det\\left(A\\right)\\cdot det\\left(B\\right)$ $=1\\cdot 4$ det(AB)=4 (2) $det\\left(3A\\right)=3\\cdot det\\left(A\\right)$ $=3\\cdot \\left(1\\right)$ det(3A)=3 (3) $det\\left({A}^{T}\\right)=det\\left(A\\right)$ $det\\left({A}^{T}\\right)=1$ (4)$det\\left({B}^{-1}\\right)=\\frac{1}{|B|}$ $det\\left({B}^{-1}\\right)=\\frac{1}{4}$ (5) $det\\left({B}^{4}\\right)=det\\left(B\\cdot B\\cdot B\\cdot B\\right)$ $=det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)$ $=4\\cdot 4\\cdot 4\\cdot 4$ =256 Jeffrey Jordon", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "can establish the recurrence $$p_k=p_{k-1}\\cos\\frac{\\pi}{n}-\\frac{\\cos^2 \\theta_{k-1}}{4}p_{k-2}$$ and then $\\det\\mathbf T=p_n$. There is also an explicit formula for the first and last columns of the inverse of a tridiagonal matrix (the result of $\\mathbf T^{-1}\\mathbf U$) which I will have to look up (unless somebody beats me to it), and proving the nonnegativity of $\\det\\mathbf A$ boils down to proving that either $p_n$ and $\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$ have the same sign, or that one of the two is zero. - I guess your $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf A^{-1}\\mathbf U)$ should be $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. the Sherman-Morrison-Woodbury formula for determinants is usually for $U,V$ of rank one, nd.edu/~jeff/Teaching/CBE60478/ShermanMorrison.pdf . Can you show me the reference for your formula $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. – Sunni May 18 '11 at 0:18 Yes @Sunni; I'll correct that typo. Faddev/Faddeva or Aitken's old books should have it, but if you're knowledgable with Schur complements, it shouldn't be too hard to reckon it yourself. – Ｊ. Ｍ. Jun 2 '11 at 4:01 This is not an aswer, but was too long for a comment. I suspect that for $N>3$ the matrix is (semi)positive definite, a stronger property than what is asked. As I pointed in a comment, to prove this is equivalent to prove the", "You can use $\\det(A \\cdot B)= \\det(A) \\cdot \\det(B)$. And consider that $$(\\mathbf v+9\\mathbf w, \\mathbf w+3\\mathbf u, \\mathbf u+3\\mathbf v) = (\\mathbf u, \\mathbf v, \\mathbf w) \\cdot \\begin{bmatrix}0&3&1\\\\ 1&0&3\\\\ 9&1&1\\end{bmatrix}$$ Ignore this you did edit ur post op so #### Petrus ##### Well-known member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ $= 81\\det(v, w, u) + 81\\det(v, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)+$ $81/det(w, w, u) + 81\\det(w, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)$ The next step im kinda unsure how to do it, I guess ima swap road or something and try get exemple $\\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column? #### Klaas van Aarsen ##### MHB Seeker Staff member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ You have a 9 instead of a 1. It should be: $\\det(1v+9w, w+3u, u+3v) =1\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ You have repeated the same mistake here. $= 81\\det(v, w, u) +", "1\\} =$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} =$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} =$ #### Solution 1. $\\{0\\} \\cap \\varnothing = \\boxed{\\varnothing}$ 2. $\\{0, 1\\} \\cap \\{0, 1\\} = \\boxed{\\{0, 1\\}}$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} = \\boxed{\\{1\\}}$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} = \\boxed{\\{4, 6\\}}$", "the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . • Dec 18th 2011, 02:17 AM NonCommAlg Re: Find mod ( det A ) Quote: Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ • Dec 18th 2011, 05:28 AM FernandoRevilla Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "$\\frac{13 - b}{7 - b} = 2$ Now we simplify the fraction : $13 - b = 2(7 - b)$ $13 - b = 14 - 2b$ $- b + 2b = 14 - 13$ $\\boxed{b = 1}$ 2. Find $a$ Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ : $13 = 4a + b$ Since $b = 1$, we get : $13 = 4a + 1$ $12 = 4a$ $3 = a$ $\\boxed{a = 3}$ 3. Conclusion You are done : you have found the linear relationship between $x$ and $f(x)$ : $\\boxed{f(x) = 3x + 1}$ Let's check those results : $f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ --> $f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ --> $f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ --> $f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ --> Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ : $f(1) = 1 \\times 3", "x= \\boxed{\\mathbf{\\pm 4}}$$", "$\\forall s,t,\\ \\mathbf{Q}_1(s+t)=\\mathbf{Q}_1(s)\\mathbf{Q}_1(t)=\\mathbf{Q}_1(t)\\mathbf{Q_1}(s)$. (Note: I've already asked this question here but did not get any answer despite a bounty.) Edit Additional possible hint: I've found that for $n=2$ (maybe it's more general), the follow results stands, if $A=\\mathbf{Q}_2\\mathbf{Q}_1(t_1)-\\mathbf{Q}_1(t_2)^{-1}\\mathbf{Q}_2$ and $\\phi(x)=\\det(A-xI)$: $$\\phi'(0)^2=-\\phi(0)\\det(B)$$ where $B$ is the $A$ matrix if $\\mathbf{P}=\\mathbf{I}$... ! I have no clue where this comes from, maybe it's obvious for someone (maybe related to Lie algebras?). This would conclude the proof as $\\phi(0)=\\det(A(t_1,t_2))=0$ for the appropriate $t_1,t_2$, so that $0$ would be a double eigenvalue of $A$. • Sorry, this is sort of complicated so maybe I'm misreading it, but when $n=1$ it looks like $Q_2Q_1$ and $Q_1^{-1}Q_2$ both always have the same determinant but the kernel is almost never two dimensional. Can you tell me what I'm missing? Nov 2 '14 at 10:17 • @GabrielC.Drummond-Cole $\\forall n$, $\\det(Q1)=1$ and $\\det(Q2)=-1$ so $\\det(Q_2Q_1)=\\det(Q_1^{-1}Q2)=-1$ is always true. Nov 2 '14 at 21:18 • @GabrielC.Drummond-Cole [failed to edit the previous comment within 5min...] It's quite complicated that's true, thank you for reading! When $n=1$, $Q_1(t)=\\begin{bmatrix}\\cos(\\omega t)& \\sin(\\omega t)/\\omega \\\\ -\\omega \\sin(\\omega t) & \\cos(\\omega t)\\end{bmatrix}$ and $Q_2=\\operatorname{diag}(1,-1)$ and the computation yields, for this particular case, $Q_2\\,Q_1 - Q_1^{-1}\\,Q_2=0_{2}$, so the kernel is always of dimension 2. Nov 2 '14 at 21:24 • But in the setup you use", "which is precisely the given answer. Alternatively, use the properties: \\begin{align}\\text{adj}(cA)&=c^{n-1}\\text{adj}(A) \\quad (1)\\\\ \\det(cA)&=c^n\\det(A) \\quad (2)\\\\ \\det(\\text{adj}(A))&=(\\det(A))^{n-1} \\quad (3)\\\\ \\det(A^{-1})&=(\\det(A))^{-1} \\quad (4) \\end{align} to get: $$\\det(3\\,\\text{adj}(2A^{-1}))\\stackrel{(1)}{=}\\\\ \\det(3\\cdot 2^2\\,\\text{adj}(A^{-1}))\\stackrel{(2)}{=}\\\\ 12^3\\det(\\text{adj}(A^{-1}))\\stackrel{(3)}{=}\\\\ 12^3(\\det(A^{-1}))^2\\stackrel{(4)}{=}\\\\ 12^3((\\det(A))^{-1})^2=\\\\ 12^3\\cdot 4^{-2}=\\\\ 108.$$", "is enough to conclusively answer. For option $\\mathbf A$ we have $\\frac{1}{det(A)} = det(A^{-1}) {\\space\\overset{\\mathbf A}{=}\\space} det(\\frac{A}{25}) = \\frac{det(A)}{25^4}$. Similarly for the other answers. Here is a very simple although not as elegant answer: Whatever the inverse would be, multiplication must be $1$ for diagonal elements, in particular for the first element. We can call it $c_{11}$. This shields $\\frac{25}{x}$ where $x$ is the one between options that makes this element 1. So it's a or d. In any case, you need to calculate all elements for choosing between this two options." ]

[ "+0 # Convert $2 e^{\\pi i/6}$ to rectangular form. 0 132 1 +474 Convert $$2 e^{\\pi i/6}$$ to rectangular form. Jan 4, 2019 #1 +99333 +2 This is probably the easiest question you have ever asked. But why would I answer it? https://web2.0calc.com/questions/let-a-and-b-be-real-numbers-the-complex-number-4 Jan 4, 2019", "rectangular form back them up with references or personal experience MATH 1113 at of.", "length $1$ and argument $x$. Converting $e^{ix}$ to rectangular coordinates, we get $$e^{ix}=\\cos(x)+i\\sin(x)\\tag{7}$$ Comparing the real and imaginary parts of $(2)$ and $(7)$, we get the series for $\\sin(x)$ and $\\cos(x)$. -", "of the required form. $$\\square$$", "# How do you convert e^3e^{-4i} into cartesian form? ${e}^{3} {e}^{- 4 i} = {e}^{3} \\cos \\left(- 4\\right) + {e}^{3} \\sin \\left(- 4\\right) i$", "2.4.4.11. $\\square$", "e^{-i \\pi /2}$ Since |z_4-z_1| = |z_2-z_1| (side of square are equal ) Now you will get z_4 from above Use the same technique and find z_3", "Form Math Calculus Precalculus 11/12/15 #### Convert from Rectangular to Polar Form Convert the following from Rectangular to Polar Form: 2x-9y+3=0 Rectangular Form 05/10/15 #### approximate [100, angle 2.95 rad] Approximate in rectangular form to the nearest 0.01 Rectangular Form Parametric Equations 12/16/13 #### How do I convert from parametric equations to rectangular form? Write the rectangular form for the parametric equation x=cosθ ; y=4sinθ ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "1. ## the right question write in rectangular form (square root 2-i) 4th power 2. $\\left ( \\sqrt{2-i} \\right )^4$ = $(2-i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$ -Dan 3. Hello, Shorty! Is this what you meant? Write in rectangular form: $(\\sqrt{2} - i)^4$ Just expand it . . . what's stopping you? $(\\sqrt{2} - i)^4 \\;=\\;(\\sqrt{2})^4 - 4(\\sqrt{2})^3(i) + 6(\\sqrt{2})^2(i^2) - 4(\\sqrt{2})(i^3) + (i^4)$ . . . . . . . $= \\;16 - 4(2\\sqrt{2})(i) + 6(2)(-1) - 4(\\sqrt{2})(-i) + 1$ . . . . . . . $= \\;16 - 8\\sqrt{2}i - 12 + 4\\sqrt{2}i + 1$ . . . . . . . $= \\;5 - 4\\sqrt{2}i$", "We need to sketch r=sin theta-1. Using the same process for the earlier examples, we obtain: For the curve above, when θ = 0, r = −1, so the curve starts on the left side of the origin. ### Conversion to Rectangular Form To convert to rectangular form, we use r2 = x2 + y2 and sin^2theta=(y^2)/(r^2)=(y^2)/(x^2+y^2) In rectangular form, r = sin θ − 1 is: sqrt(x^2+y^2)=y/sqrt(x^2+y^2)-1 x^2+y^2=y-sqrt(x^2+y^2) x^2+y^2-y=-sqrt(x^2+y^2) (x^2+y^2-y)^2=x^2+y^2 x^4+2x^2y^2+y^4-2y(x^2+y^2)+y^2 =x^2+y^2 x^4+2x^2y^2+y^4-2y(x^2+y^2)-x^2 =0", "5 e^{\\frac{i\\pi}{2} } \\left( {e^{ - i\\phi } } \\right) = \\sqrt 5 e^{i\\left( {\\frac{\\pi}{2} - \\phi } \\right)}$. This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates. 8. Nevermind! I figured it out! Thanks!", "Convert the rectangular form to polar form and then use this realtionship $\\sqrt{(X\\angle \\theta)(Y\\angle \\phi )}=\\sqrt{XY}\\angle( \\frac{\\theta + \\phi}{2})$", "\\pi i \\mathrm{e}^{-1}$$", "2^t}=4^t=(2^2)^t=2^{2t}$ LHS = RHS $\\square$ -", "the form $1+2\\pi i n /\\ln 2, n\\in\\mathbb{Z}-\\{0\\}$. – anon Aug 11 '11 at 4:01", "be written in the form of $$e^{i \\pi}+1=0$$.", "$2{\\pi}$ when squared has an angle between $2{\\pi}$ and $4{\\pi}$.", "e n}{11 - 2 \\sqrt{30}}$", "and $12$ is a square in $\\mathbb{Q}(\\sqrt{3})$.", "the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i)." ]

[ "$$P(\\pi/4) = \\frac{16}{2\\sqrt2\\pi^2}\\\\ \\boxed{P(\\pi/4) = \\frac{4\\sqrt2}{\\pi^2}}$$", "+0 # what is 2 to the 4/12 0 89 1 what is 2 to the 4/12? May 13, 2021 $2^{\\frac{4}{12}}=2^{\\frac{1}{3}}=\\sqrt[3]{2^1}=\\boxed{\\sqrt[3]{2}}$", "however that $$7 \\pm \\sqrt{12} = (2 \\pm \\sqrt{3})^2$$ and so we get $$2 + 2\\sqrt{12n^2 + 1} = (2 + \\sqrt{3} )^{2N} + (2 - \\sqrt{3} )^{2N}) + 2\\cdot (2 + \\sqrt{3} )^N \\cdot (2 - \\sqrt{3} )^N = \\\\ = (\\,( 2 + \\sqrt{3})^N + (2-\\sqrt{3})^N)^2$$ and inside the bracket we have a natural number.", "You say that you have $z= \\pm\\sqrt{i}$. Surely you know that the simplest way to find square roots (or other roots) or complex numbers is to put them in polar form- if $z= re^{i\\theta}$, then $\\sqrt{z}= \\sqrt{r}e^{i\\theta/2}$? \"i\" has absolute value 1 and argument $\\pi/2$ so that $i=.(1)e^{i\\pi/2}$. So the square root of i is $\\sqrt{1}e^{i\\pi/4}= e^{i\\pi/4}= cos(\\pi/4)+ i sin(\\pi/4)= \\frac{\\sqrt{2}}{2}(1+ i)$. Of course, we can always add $2\\pi$ to the argument without changing the number, $e^{i\\pi/2}= e^{i\\pi/2+ 2i\\pi}= e^{3i\\pi/2}$, and so another square root is $e^{i3\\pi/4}= cos(3\\pi/4)+ i sin(3\\pi/4)= -\\frac{\\sqrt{2}}{2}(1+ i)$ Last edited by a moderator: May 21, 2015", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "i} = \\frac{\\pi}{4}$ So the given series $11 - \\frac{11}{3} + \\frac{11}{5} - \\frac{11}{7} + \\ldots = \\frac{11}{4} \\pi$ (b) sum_(n=2)^oo ((-1)^n(8^n))/(n!) = sum_(n=0)^oo (-8)^n/(n!)-(-8)^0/(0!)-(-8)^1/(1!) $q \\quad = {e}^{-} 8 - 1 + 8 = 7 + {e}^{-} 8$", "## Intermediate Algebra (12th Edition) We are given that $a^{\\frac{m}{n}}=\\sqrt[n] a^{m}=(\\sqrt[n] a)^{m}$, if all indicated roots are real numbers. Therefore, $-121^{\\frac{1}{2}}=-(121^{\\frac{1}{2}})=-\\sqrt[2] 121^{1}=-(\\sqrt 121)=-(11)=-11$. We know that $\\sqrt 121=11$, because $11^{2}=121$.", "\\sum_{n\\ge 0} \\frac{x^{4n+1}}{(4n+1)!} \\end{eqnarray*} so \\begin{eqnarray*} \\sum_{n\\ge 0} \\frac{a^n}{(4n+1)!}&=&\\frac{1}{4\\sqrt[4]{a}} (e^{\\sqrt[4]{a}}-ie^{i\\sqrt[4]{a}} -e^{-\\sqrt[4]{a}}+ie^{-i\\sqrt[4]{a}})\\\\ &=& \\frac{1}{2\\sqrt[4]{a}} (\\sinh \\sqrt[4]{a} + \\sin \\sqrt[4]{a}). \\end{eqnarray*}", "+ \\sqrt{6}} = 11.56011345$ $\\therefore \\frac{- 6 \\sqrt{3} - 9 \\sqrt{2}}{-} 2 = 11.56011345$ $\\therefore 3 \\sqrt{3} + \\frac{9 \\sqrt{2}}{2} = 11.56011345$", "a_{i-1} $a_{i-1}$ \\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$", "a_{i-1} $a_{i-1}$ \\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "\\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$", "\\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$", "\\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$" ]

[ "## Trigonometry (11th Edition) Clone The rectangular form is $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$ $\\sqrt{3}$ $cis$ $315^\\circ$ = $\\sqrt{3}$ $(cos 315^\\circ$ + $isin 315^\\circ)$ = $\\sqrt{3} \\cdot \\frac{\\sqrt{2}}{2}$ + $\\sqrt{3} \\cdot i \\cdot (- \\frac{\\sqrt{2}}{2})$ = $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$ The rectangular form is $\\frac{\\sqrt{6}}{2}$ $- \\frac{\\sqrt{6}}{2} i$", "and trials, now in $$sin x = \\sqrt{1-e^{i\\pi/3}} \\\\$$ express ##1-e^{i\\pi/3}## in ##x+iy## form. Not that complicated if you have put some more creativity and trials, now in $$sin x = \\sqrt{1-e^{i\\pi/3}} \\\\$$ express ##1-e^{i\\pi/3}## in ##x+iy## form. Hmmm... I'm new to complex number... and, I don't know how to convert ##1-e^{i\\pi/3}## to rectangular form since there is a constant (in this case, 1)... blue_leaf77 Homework Helper Hmmm... I'm new to complex number... and, I don't know how to convert ##1-e^{i\\pi/3}## to rectangular form since there is a constant (in this case, 1)... In rectangular form of a complex number ##x+iy##, both ##x## and ##y## are real and every complex number can be transformed to this form. To do what you need to do, transform ##e^{i\\pi/3}## into rectangular form using Euler formula. In rectangular form of a complex number ##x+iy##, both ##x## and ##y## are real and every complex number can be transformed to this form. To do what you need to do, transform ##e^{i\\pi/3}## into rectangular form using Euler formula. Alright, so it becomes like this ##1-e^{i\\pi/3} = 1 - cos \\frac{\\pi}{3} + i\\ sin \\frac{\\pi}{3} = \\frac{1}{2} + \\frac{1}{2}\\sqrt{3}i = \\frac{1}{2} (1+\\sqrt{3}i) \\\\ sin\\ x = \\sqrt{\\frac{1}{2} (1+\\sqrt{3}i)} ## and I convert cos x to rectangular form, ## cos\\ x = \\frac{1}{2}\\sqrt{3}+\\frac{1}{2}i## ##sin\\ 2x = 2\\", "# How do you find the corresponding rectangular coordinates for the point (-1, (-3pi)/4 )? Jan 23, 2018 The rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$. #### Explanation: We can use $x = r \\cos \\left(\\theta\\right)$ and $y = r \\sin \\left(\\theta\\right)$ to convert: $x = - 1 \\cos \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ $y = - 1 \\sin \\left(- \\frac{3 \\pi}{4}\\right) = - 1 \\left(- \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\sqrt{2}}{2}$ So the rectangular form is $\\left(\\frac{\\sqrt{2}}{2} , \\frac{\\sqrt{2}}{2}\\right)$.", "# How do you convert (4, 4pi/3) into rectangular forms? Dec 6, 2015 $x = - 2$ $y = - 2 \\sqrt{3}$ #### Explanation: If you know the poalr coordinates $\\left(\\setminus \\rho , \\setminus \\theta\\right)$, then the transformation to have the rectangular ones is $x = r \\cos \\left(\\theta\\right)$ $y = r \\sin \\left(\\theta\\right)$ $x = 4 \\cos \\left(\\frac{4 \\pi}{3}\\right)$ $y = 4 \\sin \\left(\\frac{4 \\pi}{3}\\right)$ since $\\cos \\left(\\frac{4 \\pi}{3}\\right) = - \\frac{1}{2}$, and $\\sin \\left(\\frac{4 \\pi}{3}\\right) = - \\frac{\\sqrt{3}}{2}$, we have $x = 4 \\cdot \\left(- \\frac{1}{2}\\right) = - 2$ $y = 4 \\cdot \\left(- \\frac{\\sqrt{3}}{2}\\right) = - 2 \\sqrt{3}$", "Home > PC > Chapter 11 > Lesson 11.2.1 > Problem11-71 11-71. Find the rectangular form of each complex number. Use exact values. 1. $z = 6(\\cos(\\frac { 5 \\pi } { 6 })+ i \\sin (\\frac { 5 \\pi } { 6 }))$ Substitute the exact values for sine and cosine of $5π/6$. $z=6\\left( -\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i \\right)$ Distribute. 1. $z = 7(\\cos(\\frac { 3 \\pi } { 2 })+ i \\sin(\\frac { 3 \\pi } { 2 }))$", "square is $r\\sqrt{2}$ and $p=\\frac{S_{square}}{S_{circle}}=\\frac{2 r^2}{\\pi r^2}=\\frac{2}{\\pi}$", "$\\frac{\\sqrt 3} 2 + \\frac 1 2 i, -1 + 0i$, and $\\frac{\\sqrt 3}2 - i\\frac 1 2$. In polar form, these are $e^{i\\pi/3}, e^{i\\pi}$, and $e^{i 5\\pi/3}$.", "+0 # Convert $2 e^{\\pi i/6}$ to rectangular form. 0 132 1 +474 Convert $$2 e^{\\pi i/6}$$ to rectangular form. Jan 4, 2019 #1 +99333 +2 This is probably the easiest question you have ever asked. But why would I answer it? https://web2.0calc.com/questions/let-a-and-b-be-real-numbers-the-complex-number-4 Jan 4, 2019", "You say that you have $z= \\pm\\sqrt{i}$. Surely you know that the simplest way to find square roots (or other roots) or complex numbers is to put them in polar form- if $z= re^{i\\theta}$, then $\\sqrt{z}= \\sqrt{r}e^{i\\theta/2}$? \"i\" has absolute value 1 and argument $\\pi/2$ so that $i=.(1)e^{i\\pi/2}$. So the square root of i is $\\sqrt{1}e^{i\\pi/4}= e^{i\\pi/4}= cos(\\pi/4)+ i sin(\\pi/4)= \\frac{\\sqrt{2}}{2}(1+ i)$. Of course, we can always add $2\\pi$ to the argument without changing the number, $e^{i\\pi/2}= e^{i\\pi/2+ 2i\\pi}= e^{3i\\pi/2}$, and so another square root is $e^{i3\\pi/4}= cos(3\\pi/4)+ i sin(3\\pi/4)= -\\frac{\\sqrt{2}}{2}(1+ i)$ Last edited by a moderator: May 21, 2015", "1. ## the right question write in rectangular form (square root 2-i) 4th power 2. $\\left ( \\sqrt{2-i} \\right )^4$ = $(2-i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$ -Dan 3. Hello, Shorty! Is this what you meant? Write in rectangular form: $(\\sqrt{2} - i)^4$ Just expand it . . . what's stopping you? $(\\sqrt{2} - i)^4 \\;=\\;(\\sqrt{2})^4 - 4(\\sqrt{2})^3(i) + 6(\\sqrt{2})^2(i^2) - 4(\\sqrt{2})(i^3) + (i^4)$ . . . . . . . $= \\;16 - 4(2\\sqrt{2})(i) + 6(2)(-1) - 4(\\sqrt{2})(-i) + 1$ . . . . . . . $= \\;16 - 8\\sqrt{2}i - 12 + 4\\sqrt{2}i + 1$ . . . . . . . $= \\;5 - 4\\sqrt{2}i$", "## Trigonometry (11th Edition) Clone $\\frac{3\\sqrt 3}{2}-\\frac{3}{2}i$ We know that $\\cos30^{\\circ}=\\frac{\\sqrt 3}{2}$ and $\\sin30^{\\circ}=\\frac{1}{2}$ Substituting these values in the expression and solving: $3(\\cos30^{\\circ}+i\\sin30^{\\circ})=3(\\frac{\\sqrt 3}{2}-\\frac{1}{2}i)=\\frac{3\\sqrt 3}{2}-\\frac{3}{2}i$ Therefore, the rectangular form is $\\frac{3\\sqrt 3}{2}-\\frac{3}{2}i$.", "### Home > PC3 > Chapter 11 > Lesson 11.2.3 > Problem11-115 11-115. Write the rectangular form of each complex number. Use exact values. 1. $z = 6(\\cos(\\frac { 5 \\pi } { 6 })+ i \\sin (\\frac { 5 \\pi } { 6 }))$ Substitute the exact values for sine and cosine of $5π/6$. $z=6\\left( -\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i \\right)$ Distribute. 1. $z = 7(\\cos(\\frac { 3 \\pi } { 2 })+ i \\sin(\\frac { 3 \\pi } { 2 }))$", "the square: $7t - 11 - 2t^2 = -2(t-\\frac{7}4)^2-\\ldots$ -", "# Thread: Converting to Trigonometric form 1. ## Converting to Trigonometric form I couldn't succeed converting the following complex number to it's trigonometric form, could someone help? √(2+√3) + i*√(2-√3) 2. Originally Posted by Fujiwara I couldn't succeed converting the following complex number to it's trigonometric form, could someone help? √(2+√3) + i*√(2-√3) Put $z=\\sqrt{2+\\sqrt{3}}+i\\sqrt{2-\\sqrt{3}}\\Longrightarrow |z|^2=2+\\sqrt{3}+2-\\sqrt{3}=4$, $\\arg(z)=\\arctan\\frac{\\sqrt{2-\\sqrt{3}}}{\\sqrt{2+\\sqrt{3}}}=\\arctan\\sqrt{7-4\\sqrt{3}}=\\frac{1}{12\\pi}$ , so $z=2cis(\\frac{1}{12\\pi})=2e^{i\\frac{1}{12\\pi}$ Tonio", "\\:=\\:x^2 \\quad\\Rightarrow\\quad x^2 - 4x + 2 \\:=\\:0$ Quadratic Formula: . $x \\;=\\;\\frac{\\text{-}(\\text{-}4) \\pm \\sqrt{(\\text{-}4)^2 - 4(1)(2)}}{2(1)} \\;=\\;\\frac{4 \\pm\\sqrt{8}}{2} \\:=\\:\\frac{4 \\pm2\\sqrt{2}}{2} \\:=\\:2 \\pm \\sqrt{2}$ The side of the triangle is: . $2 + \\sqrt{2}$ cm. The side of the square is: . $1 + \\sqrt{2}$ cm. The area of the square is: . $(1 + \\sqrt{2})^2 \\;=\\;1 + 2\\sqrt{2} + 2 \\;=\\;\\boxed{3 + 2\\sqrt{2}\\,\\text{ cm}^2}$ 4. Ok, thanks guys. I understand better now", "My Math Forum Rectangular Form Complex Analysis Complex Analysis Math Forum September 24th, 2013, 04:41 AM #1 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Rectangular Form Hi, does anyone know how to write $e^{\\frac{2-i \\pi}{4}}$ in rectagular form? September 24th, 2013, 04:53 AM #2 Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Rectangular Form You need to use Euler's formula: $e^{ix}=cos(x)+isin(x)$ September 24th, 2013, 05:08 AM #3 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Rectangular Form so if I write it $e^{\\frac{2-i\\pi}{4}}= e^{\\frac{2}{4}-{\\frac{i\\pi}{4}}} = e^{\\frac{2}{4}} e^{{\\frac{-i\\pi}{4}}$ I have $e^{\\frac{1}{2}} cos(\\frac{\\pi}{4}) - i e^{\\frac{1}{2}} sin(\\frac{\\pi}{4})$ but then I am lost there is that already rectangular form? I think it is bc the real part (cosine) is not in terms of z. September 24th, 2013, 07:23 AM #4 Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Rectangular Form Do you not know what \"rectangular form\" means? You have a+ bi with $a= e^{1/2}cos(\\pi/4)$ and $b= -e^{1/2}sin(\\pi/4)$. I don't know what you mean by \"the real part (cosine) is not in terms of z\". There is no \"z\" in anything you write. September 24th, 2013, 07:40 AM #5 Senior Member Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0", "$$P(\\pi/4) = \\frac{16}{2\\sqrt2\\pi^2}\\\\ \\boxed{P(\\pi/4) = \\frac{4\\sqrt2}{\\pi^2}}$$", "# How do you convert e^3e^{-4i} into cartesian form? ${e}^{3} {e}^{- 4 i} = {e}^{3} \\cos \\left(- 4\\right) + {e}^{3} \\sin \\left(- 4\\right) i$", "written in rectangular form? 6. Which of the following is equivalent to $-5\\sqrt{2} \\text{cis } \\dfrac{5\\pi}{4}$ when written in rectangular form? ### Open Problem Graph the following complex numbers on the complex plane and include their corresponding absolute value number. a. $-5 – 5i$ b. $-12i$ c. $2\\sqrt{3} – 2i$ ### Open Problem Solution Images/mathematical drawings are created with GeoGebra.", "\\sqrt{2}\\right) - \\left(\\sqrt{6} + \\sqrt{2}\\right) i\\right)$ $= \\frac{11 + 2 \\sqrt{3}}{4} + \\frac{11 \\sqrt{3} - 2}{4} i$" ]